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SUPPLEMENTS

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HlllPS RESEARCH lABDRATORIE>.~' Illps Res. Repts Suppl.,nte In lbe Nclherlands•f •

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1961 No.1

All rights reserved byN.V. Philips' Gloeilampenfabrieken, Eindhoven, Netherlands.

Articles or illustrations reproduced, in whole or in part, must beaccompanied by full acknowledgement of the source:pmLIPS RESEARCH REPORTS SUPPLEMENTS

THE PHILIPS.STIRLING THERMAL ENGINE*)

ANALYSIS OF THE RHOMBIC DRIVE MECHANISMAND

EFFICIENCY MEASUREMENTS

BY

R. J. MEIJER

, *) Thesis, Technical University, Delft, November 1960.

Summary

Chapter I contains a shört history of the hot-gas engine and a discussion ofthe principle of the Stirling' system, followed by a description of a new drivemechanism for the displacer-piston engine which offers great advantages inconnection with the balancing of the engine and the reduction of the gas forces.In chapter 11 are given various calculations concerning this rhombic drive

mechanism: .a. the determination of the balancing conditions;b. determination of various quantities needed for further calculations from the

dimensions of the drive mechanism;c. calculation of the. variations in the pressure of the common buffer space of.

a multi-cylinder engine with arbitrarily chosen crank angles;d, calculation of the torque due to gas forces and inertia forces;e. calculation of the forces in the drive mechanism;f. estimate of the friction energy and the torque due to friction;g. some applications of the equations involving the torque.

This chapter ends with a sample calculation for a single-cylinder hot-gasengine which has been built and tested in the Research Laboratory of N.V.Philips' Gloeilampenfabrieken, Eindhoven.

Chapter III describes efficiencymeasurements which have been made on thisexperimental engine: after a short description of the construction of the engine,the measuring equipment is described; the results of the measurements aresummarized in tables and graphs.In the conclusions, some properties of the hot- gas engine are compared with

those of the internal-combustion piston engine. Appendix I gives a list of theequations which have been derived; the other three give tables and graphscontaining the coefficients of the series expansions used in the calculations ofchapter 11.

---~---~- - - -- ~- ~

CONTENTS

I. General introduetion1.1. A short history of the hot-gas engine .1.2. The principle of the Stirling engine. .1.3. Drive mechanisms used in displacer-piston enginesReferences. . . . . . . . . . . . . . . . . . . . .

3

371216

II. Analysis of the rhombic drive mechanism 17

n.l. Symbols. . . . . . . . . . . . . 17IJ.l.I. Symbols for the general form 17IJ.1.2. Symbols for the special form. 18

11.2. Balancing . . . . . . . . . . . . 26n.2.I. Introduction. . . . . . . . . . . . . . . 26IL2.2. The balancing conditions for the general form 27IJ.2.3. The balancing conditions fox: the special form . . . . . . . . . . 32

II.3. The determination of various quantities from the dimensions of the drivemechanism. . . . . . . . . . . . . . . . ~IJ.3.1. Introduction. . . . . . . . . . . . .IL3.2. S, 'Pdr and Xmax. • • • • • • • • • . •IJ.3.3. The volume of the expansion space, VE .IJ.3.4. The volume of the compression space, VcIJ.3.5. Vo, w, 'Pvo and ~ . . '. . . . . . . . .IJ.3.6. Summary of the equations derived above .

H.4. The pressure variation in circuit and buffer space .II.4.I. The pressure variation in the circuit. . . . . . . . . . . . . ..11.4.2. The pressure variation in the buffer space of a multi-cylinder engine

11.5. The torque on the crankshafts due to the gas forces and inertia forces. . .IL5.I. Introduction. . . . . . . . . . . . . . . . . . . . . . . . .11.5.2. The energy balance. . . . . . . . . . . . . . . . . . . . . .11.5.3. The torque on one crankshaft due to the gas forces of the circuit and

crankcase on the piston and the displacer . . . . . . . . . . . . 4811.5.4. The torque on one crankshaft due to the gas forces of the buffer space

acting on the first piston of a multi-cylinder engine . . . . . . . . 5011.5.5. The torque on one crankshaft due to the inertia forces . . . . . . 50

IL6. The forces on various parts of the drive mechanism due to gas forces andinertia forces. . . . . . . . . . . . . . . . . .11.6.1. Introduction. . . . . . . . . . . . . . .11.6.2. The forces on the piston drive mechanism . .11.6.3. The forces on the displacer drive mechanism .11.6.4. The forces on the crankshaft. . . . . . . .11.6.5. Simplified equations for the forces . . . . • . . . .

11.7. The power and the torque due to the friction on one crankshaftn.8. Some applications of the torques calculated above , .

II.8.I. Introduction. . . . . . . . . . . . .n.8.2. Motion of the crankshafts. . . . . . . .11.8.3: The torque on the gearwheels . . . . . .11.8.4. Determination of the size of the flywheels .II.8.5. Motion of the driven machine . . . . . . . . .IJ.8.6. The reaction force and reaction torque on the base

353536383942 ,42444445464647

5454545757585962626266676869

-1-

I1.9. Specimen calculation for ~ 'single-cylinder hot-gas engine. . . . . .II.9.1. Nominal data and the determination of numerical quantities.11.9.2. The torque (T, + Tr) for w = constant • . . . . .11.9.3. The effective moment of inertia Je . . . . . . . . .11.9.4. Calculation of the forces occurring in the system . . .11.9.5. An estimate of the torque Tr and the friction power PrI1.9.6. Motion of the crankshaft . . . . . .I1.9.7. Motion of the driven machine (brake) .11.9.8. The torque on the gearwheels .II.9.9. The reaction torque on the baseReferences . . . . .'. . . . . . .

7070727576778385868687

Ill. Efficiency measurements on a single-cylinder hot-gas engine withrhombic drive mechanism 88

. HI.1. Principal engine data . 88III.2. Constructional details . 88111.3. The measuring set-up 931II.4. Results ... 95

Conclusions 101

References 103

Appendix I. List of equations 104

Appendix 11. The coefficients Bn 114

Appendix Ill. The coefficients Gn 118

Appendix IV. The coefficients Rn 122

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I. GENERAL INTRODUCTION

1.1. A short history of the hot-gas engine

Hot-air engines were used in considerable numbers during the 19th century,and it was thought for a time that they might rival the steam engine as asource of power 1,2).

Slaby 3,5) has divided the systems appearing in this period into the followingtypes:a. Open systems in which a fresh charge of air is taken in for each cycle and

heated directly in a furnace situated outside the cylinder of the engine.b. Open systems in which a fresh charge of air is taken in for each cycle, but

is heated indirectly (i.e. by external combustion).c. Closed systems in which the same amount of air is used in successive cycles.The hot-air engine built by Sir George Cayley 1,3) in 1807, which wasprobably

the first one to work properly, belongs to group a.

Fig. 1. Hot-air engine (about 1880), built according to the design of Sir George Cayley,18076). A is the check valve through which the fresh charge of air is admitted; the outletvalve B and the inlet valve C are operated by the engine itself. As the piston P goes from rightto left, A and B are open, so new air is sucked into the cylinder via A while the air to the leftof the piston is forced out through B. As-the.piston goes the other way, A and B are closedand C is open. The fresh charge of air is thus blown through the :fire D and the hot flue gasesenter the cylinder to the left of P via C. The increased pressure due to this heating is the sameon both sides of the piston, but the area difference due to the presence of the piston rod

causes the piston to move to the right.

Figure 1 shows an example of this group, which was used for pumping waterand was still built in 1880.The four hot-air engines built by the Swede Ericsson in 1853 for the 2200-ton

ship which bore his name (see fig. 2) are typical examples of group b. Theseare probably the biggest hot-air engines ever made: the power pistons had adiameter of 4.2 m and a stroke of 1.8 m. These were meant to give an I.H:P.of 600, but were found on testing only to deliver about half this, However, thefuel consumption (about 1 kg coal per LH.P.) was considerably lower than thatofthe marine steam engines used at that time (about 1.4kg coal per LH.P.) 1,4,6).The ingenious invention of the Scottish minister Robert Stirling belongs to

-3-

Fig. 2. Ericsson's hot-air engine, 1833 4).The power piston A and the compressor piston B are rigidly connected by means of rods.The compressor sucks in air during its downward stroke and forces it through the regeneratorR during its upward stroke. This preheated air flows past the heated wall C to the powercylinder H, where it is further heated through the wall D. About 2/3 of the way through thestroke the valve E closes the openings F and G so that the air then expands in the hot cylinderH. E is pushed rapidly downwards at the end of the stroke, so the air can pass via F and theregenerator RI to the atmosphere during the downward stroke. The valves K and Ki arechanged after about 50 revolutions, so that the regenerator which gave up heat now accumulates

it and vice versa.

Fig. 3. The first Stirling engine according to the patent of 1816 6).A long vertical cylinder A is heated on top by the flue gases from the furnace B; the bottomis cooled either with water or by convection with the atmosphere. The cylinder contains thedisplacer C (of rather smaller diameter than A and centred by small rollers) and the powerpiston D. As the displacer moves to and fro the air in the cylinder A passes alternately fromthe cold space E (space between the bottom of the displacer and the top of the piston) andthe hot space F (above the displacer) via a regenerator and back. The regenerator (not shownhere) is in the annular space between the displacer and the cylinder, and probably consistedof thin wire wound around the displacer. The air is thus alternately in the cold and the hotspace, and undergoes temperature and pressure variations which cause it to perform work

on the piston D.

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·group c. He introduced his hot-gas engine, which contained a power piston,a displacer piston and a regenerator, in 1816. This is shown in fig. 3. The firstmodel was already in use in 1818, for pumping water out of a quarry. Togetherwith his brother James he introduced improvements into later models, suchas a compressor to increase the specificpower and a double-action power pistonto reduce leakage. The latter was of doubtful value, however, since it entailedthe provision of two displacer pistons in separate cylinders, thus increasing thedead space and decreasing the specific power.

Fig. 4. Example of a Stirling engine from about 18804).The displacer A and the piston B are both in the same cylinder. The displacer fits closely intothe cylinder at C, so that the air is forced through the cooler D and along the heater wall E.There is no separate regenerator. The hot flue gases from the furnace F flow round the heaterwall E and pass out of the chimney H via the damper G. If the engine is used for pumping

water, a pump rod can be fitted directlyon to the projecting end L of the rocker K.

The first, and in principle better, model of] 816 was forgotten until 1868,when W. Lehmann realized the advantage of a small dead space, and againplaced the displacer piston and the power piston in the same cylinder. Afterall sorts of mechanical improvements, this led to an engine as shown in fig. 4(usually used without the regenerator), which was sold in large numbers in thesecond half ofthe 19th century. According to the price-list of a German factoryfrom about 1885, the largest Stirling engine which they made gave 2 H.P.,weighed 4100 kg (with packing and masonry) and had a volume of 21 ma.

-5-

It is no wonder that the internal combustion engine with its much greaterspecific power and better efficiency displaced this hot-air engine, despite acertain charm which the latter possessed.

After the internal combustion engine had been brought to perfection, so thatall its advantages and disadvantages could be seen, intensive research on theStirling hot-air engine was carried out in the Research Laboratories of N.V.Philips' Gloeilampenfabrieken, Eindhoven 7-11). This was prompted by thefact that before the second world war Philips felt the need of a heat-drivensource of power for radios and similar equipment for use in parts of the worldwhere the fuel needed for such a device was easier to obtain than batteries.The only power sources which came into consideration, owing to the specialdemands made for this purpose, were the thermoelement, the closed-circuitsteam engine and the hot-air engine. The choice fell on the latter after preliminarystudies showed that the hot-air engines on the market at the time (see fig. 5)had been completely neglected by the development of modern materials andnew knowledge about fluid flow and heat transfer, and that in fact Stirling'ssystem was no longer fully used, since the regenerator has gradually fallen outof use.

Fig. 5. Hot-air engine with generator for a radio (1938).

The improvements in radio tubes and particularly the introduetion oftransistors, which reduced the powerconsumption ofbattery models enormously,made the hot-air engine less attractive for the purpose for which it had originallybeen intended, so the research then turned to higher-power models. Some smallgenerators working on this principle (see fig. 6) were however built on a specialoccasion in 1953, in memory of the original intention. The hot-air engines ofthese generators were the first to befitted with the rhombic drive mechanism 12,13).

The analysis of this drive mechanism and the test results of an experimentalhigh-power engine are the subjects of this thesis.

-6-

Fig. 6. Generator (with radio) built in 1953 and incorporating the first rhornbic drive for theStirling system.

1.2. The principle of the Stirling engine

An internal-combustion engine provides a surplus of work in virtue of thecompression at low temperature of a certain quantity of air, to which atomizedfuel is added eitber before or after the compression, the subsequent heating ofthe mixture by rapid combustion, and its expansion at high temperature.

The hot-gas engine is based on the same principle, i.e. the compression atlow temperature and expansion at high temperature of a given quantity of gas.The heating takes place, however, in an entirely different manner, the beat beingsupplied to the gas from outside, through a wall. For this reason the description"external-combustion engine" is appropriate. Owing to the high heat capacityof the wall, it is not of course possible to heat and cool the gas simply by rapidheating and cooling of the wall. We have already seen that Stirling made thegas temperature change periodically by causing a "dis placer piston" (hereaftersimply called "displacer") to transfer the gas back and forth between twospaces, one at a fixed high temperature and the other at a fixed low temperature- see fig. 7. If we raise the displacer in fig. 7, the gas will flow from the hot spacevia the heater and cooler ducts into the cold space. If now the displacer ismoved downwards the gas will return to the hot space along the same path.During the first transfer stroke the gas has to yield up a large quantity of heatto the cooler; an equal quantity of heat has to be taken up during the secondstroke from the heater. The regenerator shown in fig. 7 is inserted betweenthe heater duct and cooler duct in order to prevent unnecessary wastage of thisheat. It is a space filled with porous material to which the hot gas yields heatbefore entering the cooler; when the gas streams back, it takes up the storedheat again prior to its entry into the heater.

-7-

.2689

Fig. 7. Principle of the displacer system.As the displacer moves up and down, the gas flows from the hot space to the cold space via

the heater, regenerator and cooler, and back.

The displacer system, which serves to heat and cool the gas periodically, iscombined with a power piston (hereafter simply called the '·'piston") whichcompresses the gas while it is in the cold space and allows it to expand whilein the hot space (all dead spaces in cooler, heater etc. being disregarded). Sincecompression takes place at a lower temperature than expansion, a surplus ofwork results. Fig. 8 shows four stages of the cycle through which the whole

]/[ III 2690

Fig. 8. Principle of the hot-gas process.We will assume for the sake of simplicity that the piston and the displacer move discontinuously.The cycle can then be divided into four stages:I. Piston in its lowest position, displacer in its highest. All the gas is in the cold space.11. The displacer is still in its highest position; the piston has compressed the gas at low

temperature.Ill. The piston is still in its highest position; the displacer has transferred the gas from the

cold .space to the hot space.IV. The hot gas has expanded, and both the piston and the displacer are in their lowest

positions. The displacer will now return the gas to the cold space while the piston remainswhere it is, to give stage 1 again ..

-8-

-:---t 2691

Fig. 9. The discontinuous displacements of the piston CP) and the displacer (D) plotted asa function of time. Band E represents the volume variations of the hot space (VE), band Cthose of the cold space (Vc). These variations are plotted separately at the bottom of the

figure.

p

IIII

I

-v 2692

Fig. ID. The p,V diagram of the hot-gas cycle represented in fig. 9.

system passes if a discontinuous movement ofpiston and displacer ispresupposed.The displacements they are assumed to undergo are plotted as functions oftime in fig. 9; the ordinates in band E represent the variation in the volume ofthe hot space, and those in band C the variation in the volume of the cold space.The volume variations are plotted separately in the lower part of the diagram. 'Fig. 10 is the _p,Vdiagram of the cycle (V is the total volume of the gas).

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In a practical version of the engine the movements of piston and displacermust of course be continuous, not discontinuous, as they have been assumedto be in these figures; the continuous movements will be obtained with the aidof some kind of crank and connecting rod mechanism. It will not then bepossible to distinguish any sharp transitions between the four stages but thiswill not alter the principle ofthe cycle (or detract from its efficiency- see below).The movements of piston and displacer might now be as indicated in fig. 11,

Fig. 11. As fig. 9, except that the motion of the piston and displacer is now continuous, andthe displacements are plotted as a function of the crank angle IX.

The various stages of the cycle can no longer be clearly distinguished.

in which the volume variations of the cold and hot spaces have again beenplotted separately. The only essential condition for obtaining a surplus ofworkis that the volume variation of the hot space should have a phase lead withrespect to that of the cold space; this is equivalent to requiring that the ap-propriate p,V diagram, shown in fig. 12, should be traced out in the clockwisedirection. The variation of the pressure and the power may readily be calculatedif certain simplifying assumptions are made. We will assume that VE, the volumeof the hot (expansion) space, and Vc, the volume of the cold (compression)space, vary with the crank angle (X in a purely sinusoidal fashion:

VE = iVo (1 + cos IX)

VC =iwVo[l +COS(IX-<p)]

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The crank angle ex. (for constant angular velocity, proportional to the time:ex.= wt) is measured from the position at which VEhas its maximum value Ve ;cp is the phase angle between the volume variations of the hot and cold spaces,

p

I

_V 2694

Fig. 12. The p-V diagram for the cycle represented in fig. 11.

and w is the ratio between their maximum volumes. Another important quantityintroduced is the gas temperature ratio,

Tc7:=-_.

TE ' (7:< I)

The condition that the mass of the working fluid remains constant throughoutthe cycle now leads to the formula for the pressure p as a function of thecrank angle:

1-15p = pmax I+ IJ cos(ex.-O)

in which pmax is the maximum pressure occurring during the cycle and

IJ = V7:2 + w2 + 2 7:W cos cp7:+w+2s

tan 0 = w sin cp7: + w cos cp

(s = .~VVs:TTc, the relative dead spacereduced to temperature Ta, where Vsando· s

TBare the volume and absolute temperature of the dead spaces).

-11-• (,I

From the above we obtain the mean pressure,

1/ 1 - t5iJ= Pmax V, 1 + t5

and the power output

P = t wVoiJ(1-.) / sin e. 1+ (1-152)"

We are concerned here with a reversible cyclic process in which, in accordancewith the "idealized" conditions assumed (isothermal behaviour in cold andhot spaces and 100% efficient regenerator action), the supply of heat takesplace at only one temperature TEand the removal ofheat at only one temperatureTc. There is a theorem in thermodynamics which states that under theseconditions the efficiency with which heat is converted into work (the thermalefficiency) is that of the Carnot cycle:

TE-Tc1]= =1-.

TE

From this we may obtain the quantity of heat supplied per second,

qE = !_ = ~.wVo iJ {J sin ()1] 1+ V(1- (J2)

The equation for the power given above shows that the power is proportionalto the mean pressure in the engine. In order to obtain a large specificpower(power output per unit volume swept out by the piston), a high mean pressuremust therefore be used, among other things.

1.3. Drive mechanisms used in displacer-piston engines

There are various ways of making the piston and displacer perform thedesired movements. Fig. 3 shows the earliest method used, and fig. 4 showsone of the mechanisms in use at the end of the last century. These engines hadthe working gas at a low pressure (minimum pressure generally 1 atm. abs.).Stirling introduced a double-action piston with two separate displacers in 1827,in order to reduce the gas leaks at higher pressures and also to balance theforces exerted on the piston to a certain extent; this however also gave a largedead space.This disadvantage has been removed in the smaller hot-air engines and the

gas refrigeration machines made by Philips by placing the drive mechanism ina gastight crankcase and filling this with the working.gas at a pressure (the

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"Z6Q5

Fig. 13. The drive mechanism used in small hot-air engines.D = forked piston connecting rod, K = crankshaft, T = rocker, connected by hinged rods

S1 and S2 to the displacer rod V and to a point on D.

v

26Q6

Fig. 14. TIle drive mechanism used in the gas refrigerating machine.Two parallel connecting rods Dl and D2 on cranks Ki and K2 are connected to the pistonand one connecting rod V' on a crank K' between the other two is connected to the displacer

rod V.

buffer pressure) equal to the minimum or mean circuit pressure (figs. 13 and14). It is clear, however, that such a pressurized crankcase leads to rather

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heavy constructions, especially with large engines; and ifit is desired to operateat very high pressures in order to increase the specific power, this becomes agreat disadvantage. .

We shall describe below the principle of operation of a drive mechanismfor a displacer-piston engine which allows the buffer pressure to be applied,while keeping the crankcase at atmospheric pres~ure if desired. This drivemechanism also has the advantage of allowing even a single-cylinder engine tobe completely balanced. It consists of twin cranks and connecting-rod mecha-nisms offset from the central axis of the engine; the cranks rotate in oppositesenses and are coupled by two gearwheels.

Fig. IS. The rhombic drive mechanism.1 = piston. 6 = displacer. 5,5' = cranks on two shafts which rotate in opposite directionsand are coupled by the gear wheels 10 and 10'. 4,4' = connecting rods pivoted on the endsof a yoke 3 fixed to the hollow piston rod 2. 9,9' = connecting rods pivoted on the ends ofa yoke 8 fixed to the displacer rod 7, which passes through the hollow piston rod. 11, 12 =

stuffing boxes. 13 = buffer space filled with gas under high pressure.

Fig. 15 is a schematic diagram of the system. Fixed to the piston 1 by wayof piston rod 2 is a yoke 3. One end of the yoke is linked by connecting rod4 to crank 5, the other end by connecting rod 4' to crank 5'. The displacer isactuated by a precisely similar arrangement: the displacer rod 7; which passesthrough the hollow rod 2, has fixed to it a yoke 8, which is linked to cranks 5and 5' by connecting rods 9 and 9' respectively. If 9 and 9' are given the samelength as 4 and 4', the two pairs of connecting rods will form a rhombus, ofwhich only the angles vary when the system is in motion; it is for that reasonthat we have adopted the name "rhombic drive". Gearwheels 10-10' ensure

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_cl2698

exact symmetry of the system at all times. The two crankshafts being gearedtogether, the entire shaft output can be taken off either,The symmetry of the system and the coaxial arrangement of piston and

displacer rods make it an easy matter to avoid putting the crankcase under highpressure. The stuffing-box 11 for the displacer rod is inside the hollow pistonrod. One more seal, namely the stuffing-box round the piston rod (12in fig. IS),is all that is necessary to form a comparatively small cylindrical chamber 13under the piston, separate from the crankcase; this "buffer space" can be filledwith gas at the desired buffer pressure. The minimum permissible volume ofthebuffer space is determined only by the range within which it is desired to limitthe pressure variations inside the chamber. In a multi-cylinder engine the bufferchambers can be interconnected; this allows the volume of the individualspaces to be made even smaller.

Fig. 16. The volume variations VE and Vc of the hot and cold spaces respectively of a hot-gasengine with rhombic drive mechanism, plotted as in fig. 11.

The piston and displacer movements resulting from the new drive are dis-played graphically in fig. 16, in the same way as in figs. 9 and 11. Itwill be seenthat if the direction of rotation is as indicated in fig. IS, the volume variationof the hot space will have a phase lead with respect to that of the cold space,as is required. The piston and displacer do not by any means move in simple \\harmonic motion, but it is found that a very good version of the Stirling cycleis obtainable. This might seem surprising, for on the face of it the use of therhombic drive may appear to have restricted the freedom of choice with regardto parameters wand cp (the amplitude ratio and relative phase of the variationsin volumes VE and Vc - see above), which play a large part in the design of ahot-gas engine. However, we will see in 11.3that it is in fact possible, by altering

. the offset of the crankshafts, the proportions of cranks and connecting rods andthe ratio of piston and displacer diameters, to vary these parameters over quitea wide range.

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The balancing conditions for this rhombic drive mechanism will be deter-mined in section 11.2, from those for the general form (fig. 18), while aftercalculating the pressure variation in the buffer space in IIA the torque due togas forces and 'inertia forces will be treated in II.S. The forces occurring in thedrive mechanism will be calculated in 11.6, and with the aid of the equationsthus derived the friction losses will be estimated in 1I.7. Section 1I.8 gives someapplications of the torques calculated in 11.5 and 1I.7. The calculations in 1I.8are concluded with a specimen calculation for an engine which was constructedunder my direction in Philips' Research Laboratories, Eindhoven, whilechapter III will give the results of efficiencymeasurements carried out on thisengine.

REFERENCES

1) W. J. M. RANKINE,Mechanics Magazine, London, 1854.. 2) J. N. O. B. KITCHlNG,Ericsson's Caloric engine, New York, 1860.3) J. O. KNOKE,Die Kraftmaschinen des Kleingewerbes, Berlin, 1887.4) Sm J. ALFREDEWINGK. C. B., The Steam-Engine and other Heat-Engines, Cambridge,

1926.5) T. FINKELSTEIN,The Engineer, 492-497, 522-527, 568-571,720-723, 1959.6) CH. DELAUNAY,Mechanica (translated by F. A. T. DELPRET),A. W. SYTHOFF,Leyden,

916-922,.1854.7) H. RINIA and F. K. DUPRÉ, Air engines, Philips Tech. Rev., 8, 129-136, 1946.8) H. DEBREYet al., Fundamentals for the development of the Philips' air engine, Philips

Tech. Rev., 9,97-104, 1947.9) F. L. VANWEENEN,The construction of the Philips' air engine, Philips Tech. Rev.,9,

125-134, 1947.10) J. W. L. KÖHLER and C. O. JONKERS,Fundamentals of the gas refrigerating machine,

Philips Tech. Rev., 16, 69-78, 1954.11) idem., Construction of a gas refrigerating machine, Philips Tech. Rev., 16,105-115,1954.12) R. J. MEUER,The Philips hot-gas engine with rhombic drive mechanism, Philips Tech. Rev.,

20, 245-262, 1958.13) Dutch Patent No., 84-033, R. J. MEuER.

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= cránk radius= length of piston connecting rod= length of displacer connecting rod= distance from crank pin to swivel point= eccentricity of piston drive mechanism= eccentricity of displacer drive mechanism= distance from centre of gravity of crankshaft to centre of

crankshaft= distance from centre of gravity of counterweight to centre

of crankshaft= crank angle (1p = 0 when the crank is at OT')= (constant) angle STP .= angle which the piston connecting rod makes with the

x-axis= angle which the displacer connecting rod makes with the

x-axis=,angle between radius of centre of gravity of counterweight

and crank radius

= angular velocity of crankshaft

rr, ANALYSIS OF THE RHOMBIC DRIVE MECHANISM

n.l. Symbols .

11.1.1. Symbols for the general form (see figs. 18 and 19)

rLcLSTee + LIerer

reo

Yo

d1p(J) = (it

. d21pw = dt21 L-Y=rc cL-Y=-r-

ST'j/=--

r

= angular acceleration of crankshaft

= relative length of piston connecting rod

= relative length of displacer connecting rod

= relative distance from crank pin to swivel point

ee = - = relative eccentricity of piston drive mechanismr

A e +LIe I' .. f d' Id' h'e + zrs = --- = re ative eccentricity 0 ISpacer nve mee amsmr

rereer =-- r

reoeeo =--

r

= relative distance from centre of gravity of crankshaft to'centre of crankshaft

= relative distance from centre of gravity of counterweightto centre of crankshaft

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ma

= mass concentrated in the point P: half of the mass of thepiston, piston rod and piston yoke and part of the massof the piston connecting rod

= mass concentrated in the point T: the rest of the mass ofthe piston connecting rod and part of the mass of thearm ST .

= mass concentrated in the point S: the rest of the mass ofthe arm ST and part of the mass of the displacer connect-,ing rod

= mass concentrated in the point Q: the rest of the mass ofthe displacer connecting rod and half of the mass of thedisplacer, displacer rod and displacer yoke

= mass of the crankshaft concentrated in Cer

= mass of the counterweight concentrated in Ceo

= abscissa of mi= abscissa of m2= abscissa of ma= abscissa of m,= abscissa of ms= abscissa of ms

Angles

/1.1.2. Symbolsfor the specialform (see figs. 20 and 21)

"P"Pm=wtLl1p"PI0(,

xXmax

qJvo

7'0

qJdr

= crank angle ("P = 0 when the crank is at OB')= mean angular velocity times time= "P-"Pm= crank angle for the first cylinder of a multi-cylinder engine= crank angle correspon ding to the first harmonic of the

volume of the expansion space (0(, = 0 when this volumeis maximum)

= angle which the connecting rods make with the x-axis= maximum value of X= phase angle by which the first harmonic of the volume of

the expansion space leads the first harmonic of the volumeof the compression space

= phase angle by which the cranks of the ith piston leadthose of the first piston

= angle between radius of centre of gravity of counterweightand crank radius (balancing conditions lead to ï'o = n)

= phase angle by which the first harmonic of the motion ofthe displacer leads that of the motion of the piston

-18-

Lengths, distances, reduced lengths

ep .;epm(XeLIep

;. Points <,

Ape

AdeBpe

Bdeepeede

rLdeLpeLpe = Lde = LeSp = Sd = Srer

reo

apeade1 LT=-;

eê=-

rrer

eer =--r

reoeeD =--

r

Xp

Xd

Xe, Ye

DiametersDp

Dd

= shaft angle of the driven machine= rot -(Xe= angle of twist= ep-epm

= middle of the small end of the piston connecting rod= middle of the small end of the displacer connecting rod= middle of the big end of the piston connecting rod= middle of the big end of the displacer connecting rod. centre of gravity of piston connecting rod= centre of gravity of displacer connecting rod

= crank radius= length of displacer connecting rod= length of piston connecting rod(balancing conditions)= eccentricity of drive mechanism= stroke of piston = stroke of displacer •= distance from centre of gravity of crankshaft to centre of

crankshaft= distance from centre of gravity of counterweight to centre

of crankshaft= distance from epe to Bpe= distance from ede to Bde

= relative length of connecting rod

= relative eccentricity of drive mechanism

= relative distance from centre of gravity of crankshaft tocentre of crankshaft

= relative distance from centre of gravity of counterweightto centre of crankshaft

= abscissa of piston= abscissa of displacer= co-ordinates of crank pin

= diameter of piston= diameter of displacer

-19-

dpdddppddPdepdes

del

Masses

mpmpem'pe

m"pe

mdmdemde

m"de

meomerme

mo

Radii of gyration

ipe

ipe.ide."lpe."ldeie

= diameter of piston rod= diameter of displacer rod= diameter of piston yoke pin= diameter of di:splacer yoke pin= diameter of crank pin= diameter of crank-pin bearing shell around which the

connecting rod of the displacer or the piston moves= diameter of crankshaft journal

= mass of piston, piston rod and piston yoke= mass of piston connecting rod= mass dynamically equivalent to part of mpe, concentrated

in Ape= mass dynamically equivalent to part of mpe; concentrated

in Bpe= mass of displacer, displacer rod and displacer yoke= mass of displacer connecting rod= mass dynamically equivalent to part of mde, concentrated

in Ade

= mass dynamically equivalent to part of mde, concentratedin Bde

= mass of counterweight= mass of crankshaft= mass of crankshaft and counterweight

2ape 2ade= mp + -L-- mpe = mu + --L- mde

= radius of gyration of the piston connecting rod about itscentre of gravity, Cpe

= radius of gyration of the displacer connecting rod aboutits centre of gravity, Cde

= radius of gyration of the reduced mass m'peabout Ape= radius of gyration of the reduced mass mde about Ade= radius of gyration of the reduced mass m;;e about Bpe= radius of gyration of the reduced mass mde about Bde= radius of gyration of tbe crankshaft with counterweight

about its centre

-20-

= cross-sectional area of displacer

Areas

Ad = : (D~-d~) = cross-sectional ar~a of displacer minus that of displacer rod

'JT, 2Ap =-Dp4 = cross-sectional area of piston

Ap = : (D~-d~)= cross-sectional area of piston minus that of displacer rod

A'; = : (D~-d~)= cross-sectional area of piston minus that of piston rod

f Ad=--

A'p

f' Ad=--

Ad

A"f" = __.R:._Kp

Volumes

VEVcVowVoVbVboVb!VbiVbcVbt

= volume of expansion space= volume of compression space= maximum value of first harmonic of VE

= maximum value of first harmonic of Vc-: varying volume of buffer space= maximum value of first harmonic of Vb= first and second harmonics of Vb (piston of first cylinder)= fust and second harmonics of Vb (piston of ith cylinder)= constant volume of buffer space= total buffer volume of multi-cylinder engine (constant

part + fust and second harmonics of varying part)= fictive dead space at the expansion end= fictive dead space at the compression end

Vbc= VbO

VSEVsc

Pressures

= pressure in expansion space= mean pressure in expansion space= pressure in compression space

-21-

Pc,PbPbpee

PEg=-PCForces

F"PFtpFtd

= mean pressure in compression space= pressure in buffer space= mean pressure in buffer space= pressure in crankcase

g' = Pb ,pc '

" peeg =-_-PC

= gas force on displacer= gas force on piston= gas force on piston due to circuit pressure + crankcase

pressure= gas force on piston due to buffer pressure= friction force on piston and piston rod= friction force of the displacer rod on the stuffing box in

the piston= vertical component of force' on piston yoke pin'= horizontal component of force on piston yoke pin= component ofthe force on the massless rod in the direction

of the piston connecting rod= component of the force on the massless rod perpendicular

to the piston connecting rod= component of the force of the big end Bpe of the piston

connecting rod on the crank pin (or crank-pin bearingshell) in the direction of the crank

= component of the force of the big end Bpe of the pistonconnecting rod on the crank pin (or crank-pin bearingshell) perpendicular to the crank

= vertical component of the force on the displacer yoke pin= horizontal component of the force on the displacer yoke

pin= component ofthe force on the massless rod in the direction

of the displacer connecting rod= component of the force on the massless rod perpendicular

to the displacer connecting rod= component of the force of the big end Bee of the displacer

connecting rod on the crank pin (or crank-pin bearingshell) in the direction of the crank

= component of the force of the big end Bde of the displacerconnecting rod on the crank pin (or crank-pin bearingshell) perpendicular to the crank

-22-

Fs

= component of the force of the main bearing on thecrankshaft journal in the direction of the crank

= component of the force of the main bearing on thecrankshaft journal perpendicular to the crank .

Torques on one crankshaft

TfdlTfd2Tfd3TCd3

= torque due to gas forces from circuit and crankcase= torque due to gas forces from buffer space= Tgc+ Tgb= torque due to inertia forces

= Tm-Jew= torque due to inertia forces at constant angular velocity= torque due to friction forces= mean value of Tf=Tg-Tr-Tmw= resultant torque

= torque due to Prei= torque due to PtP2= torque due to Prp3= torque due to Prp3

= torque due to Prei= torque due to Prd2= torque due to Piea= torque due to Pfd3

TgcTgbTgTm,

T~TmwTtTtTifrmTt

TtPlTrp2Trp3Tfp3

Tf4Tt5

= torque due to Pt4= torque due to Pf5

Other torques and momentsT = torque transmitted by the gearwheels of the drive mecha-

nism= reaction torque on the base= torque of the driven machine= mom~nt of the massless rod acting on m'pe= moment of the massless rod acting on mp'e= moment of the massless rod acting on mde= moment of the massless rod acting on m'de

Moments of inertia

Jpc = sum of the moments of inertia of m'peabout Ape and m/;eabout Bpc

-23-

Jde

Energies

I:EI:UEeEp

Ede

Powers

Prpl

Prp3

Prdl

Prd2

= sum of'the moments of inertia of mde about Ade and mdeabout Bde .

= effectivemoment of inertia of half of the drive mechanismabout the corresponding crankshaft

= moment of inertia of the flywheel which is not connectedwith the coupling .

= moment of inertia of the flywheel which is connected withthe coupling

= mean value of Je= je + JF!=je+1F2= moment of inertia of the driven machine= Jo +h+ J2

= total kinetic energy of half of the drive mechanism= total potential energy of half of the drive mechanism= kinetic energy of crankshaft and counterweight= half of the kinetic energy of the piston, piston rod and

piston yoke= kinetic energy of the piston connecting rod= half of the kinetic energy of the displacer, displacer rod

and displacer yoke= kinetic energy of displacer connecting rod

= total power consumed by friction of half of the drivemechanism

= half of the power consumed by friction of the piston andpiston rod

= power consumed by friction of the piston yoke pin= power consumed by friction of the crank-pin bearing shell

on the crank pin (crank-pin bearing shell being fixed topiston connecting rod)

= power consumed by friction of the big end of the pistonconnecting rod on the crank-pin bearing shell (the latterbeing fixed to the displacer connecting rod)

= half of the power consumed by the friction of the displacerrod on the stuffing box in the piston

= the power consumed by the friction of the displacer yokepin

-24-

Pfd3

Pfd3

Pr4

Pr5

= power consumed by the friction of the crank-pin bearingI shell on the crank pin (the crank-pin bearing shell beingfixed to the displacer connecting rod)

= power consumed by friction of the big end of the displacerconnecting rod on the crank-pin bearing shell (the latterbeing fixed to the piston connecting rod)

= power consumed by the friction ofthe crankshaftjournals ~in the main bearings

= power consumed in gearwheel transmission

Coefficients of friction

Further symbols

dV'W =w; =Tt. d2'1J'w = dt2U

NXpXdXcYeBnen, n,Gn

RnswgJ

k

= coefficient of friction corresponding to Pms= coefficient of friction corresponding to Pfp3= coefficient of friction corresponding to Pfp3 or Pfd3= coefficient of friction corresponding to Pres= coefficient of friction corresponding to Prd3= coefficient of friction corresponding to Pr4

= angular velocity of crankshaft

= angular acceleration of crankshaft= ratio of principal specific heats of working gas= number of cylinders= piston velocity= displacer velocity= acceleration of crank pin in the x direction= acceleration of crank pin in the y direction= coefficients (appendix II)= coefficients (appendix l)= coefficients (appendix Ill)= coefficients (appendix IV)= coefficient of non-uniformity= angular velocity of the driven machine= torsional stiffness of the coupling

-25-

= V k(Jo +h+ J2)JoCh + J2)

= fraction of power lost in gearwheel transmission, percrankshaft

.II.2. Balancing

IL2.l. Introduetion

As has already been mentioned in chapter J, each cylinder of a hot-gas enginecontains two reciprocating bodies with a phase difference between them, thepiston and the displacer. Since these two bodies have different functions, theyalso have different shapes.The displacer consists of a displacer body, which forms a moving seal between

the hot and cold gas, and a thermally insulating displacer dome. The pressuredifferences in the gas over the displacer are caused by the flow losses in theengine, which are usually small, so it is not generally necessaryto use pistonrings for this seal as long as- the displacer body fits closely into the cylinder.The displacer therefore has the form shown in fig. 17a.The piston has a more usual function, which may be compared with that of

the piston of a compressor, and it forms the moving seal between the cold gasof varying pressure in the circuit and the gas in the buffer space, which has a

a

Fig. l7a. The form of displacer used.Fig. 17b. The form of piston used.

Wo

more or less constant pressure. The piston is shown in fig. 17b.The masses of the displacer and the piston would in general be different if

they were both designed exclusively for the above-mentioned purposes; butthey can also be used for another purpose, i.e. balancing the motion of the

-26-

moving parts of the engine, which does not interfere with their basic functionssince they are differently shaped in any case. The ratio of the masses of thedisplacer and the piston must therefore satisfy certain conditions which arederived below.

11.2.2. The balancing conditions for the general form

The new drive mechanism described in section 1.3 is only one form of a moregeneral system. The general form of thedrive mechanism is shown in fig. 18.

Fig. 18. The general form of the rhombic drive mechanism.

This differs from the special form described above in that one of the connect-ing rods (considering one of the two symmetrical halves of the drive mechanism)is not connected to the crank pin, but to a point on the other connecting rod.Figure 18 shows the piston connecting rod equipped with such a swivel point S.The eccentricity of the piston yoke pin P and of the displacer yoke pin Q alsodiffer, being denoted by e and e + Lie respectively. The swivel point S on thepiston connecting rod is at a distance ST from the crank pin, and makes anangle cpo with the main direction TP of the connecting rod. The centre ofgravity ofthe crankshaft eer is at a distance rer from the centre 0 ofthe crank-shaft in the direction of the crank pin, while the position of the centre of

-27-

-28-

gravity of the counterweight with respect to 0 is given by the angle ?Jo and tbedistance reo: .We will now determine the conditions which must be imposed on the various

quantities in order to give complete balancing.This calculation is based on fig. 19, which only gives one half of the drive

mechanism, since the other half is its mirror-image. The lengths in this figureare expressed in terms of the crank radius.

Fig. 19. The general form of the rhombic drive mechanism, with dimensions expressedrelative to the crank radius.

It follows directly from the symmetry of the system that the sum of the inertiaforces in the y direction is zero: ~ Y = 0, and the sum of the torques due tothe inertia forces about the z-axis is also zero: ~ Tz = o.

In order to obtain balancing, therefore, it is only necessary that the sum ofthe forces in the x direction, ~ X, should be zero.

Since we are only concerned with forces, we may replace the mass of eachmoving part by two equivalent masses, which are chosen so that:1. the sum of the two masses is equal to the mass which they replace,2. the centre of gravity of the two masses coincides with the centre of gravity

of the part which they replace.

The drive mechanism may thus be represented by point masses mi, m2, ms,ma, !P5 and me with abscissae XI-X6 respectively.It may be seen from fig. 19 that:

. 1Xl = sm 'Ijl - - cos X, AX2 = sin 'Ijl

X3 = sin 'Ijl - V cos ex + f{Jo), cX4 = X3 +T cos ()

X5 = eer sin 'Ijl .

X6 = eeo sin ('Ijl + yo)and

sin X = A(e-cos 'Ijl)

sin (:)= _!_ [sin X - VA sin ex + cpo) + ALlelc~ cos X and ~ cos (:)can be expanded in series in terms of 'Ijl:

1 co-;;-cos X = ~ Bn cos tnp/L n=oC coJ: cos e = ~o An cos mp .

(1)

(2)

(3)

(4)

where Bn is a function of A and e, and An is a function of A, c, e, zls, V and cpo.If we substitute these expressions for sin X, cos X and cos (:)into the equations

for xi, X3 and X4, we obtain:co

xi = sin 'Ijl- ~ Bn cos n'ljln=o

X2 = sin 'Ijlco

X3 = vAe sin cpo-vA sin cpocos 'Ijl+ sin 'Ijl-VA cos cpo ~ Be cos n'ljln=oco co

X4 = vAe sin cpo- VA sin cpocos 'Ijl+ sin 'Ijl- VAcos f{Jo ~ Bncos n'ljl+ ~ An cos n'ljln=o n=oX5 = eer sin 'Ijl

X6 = eeo sin ('Ijl +yo)Differentiating twice with respect to time and multiplying by the correspond-

ing masses, we obtain the forces in the X direction.The sum of these forces must be zero:

6~x = ~Kmmm= 0

m=I

ximi =ml[w2{-sin 'Ijl+ ~n2Bncos n'ljl} +có{COS'ljl + ~nBnsin n'ljl}1K2m2 =m2[-w2sin 'Ijl+cócos'ljll

(5)

-29-

For n ~ 2, therefore, the following condition must hold:

Bn m4An = mi + 'IIÀ cos !po(m3 + ma) (7)

X3m3=ma[W2{'IIÀ sin !pocosV!-:-sin1p+ 'IIÀcos!po ~n2BncosnV!} ++w{'IIÀ sin<posin V!+cosV! + 'JIÀcos <po~nBnsinnV!}]

X4m4=ffi4[w2{'IIÀ sin !po cosV!-sinV! + 'IIÀcos!po ~n2BncosnV!-~n2AncosnV!} ++w{'IIÀ sin<posinV!+ cosV!+ 'IIÀ cos <po~nBnsin nV!- ~nAnsinnV!}]

. x5m5=ecrm5[-w2sinV!+wcosV!] .xGm6= ecom6[-w2sin (V!+Yo) + W cos ('IjJ+Yo)]

ex>

where ~ means ~ ..n=l

The counterweight ms can only give a harmonically varying force, whosemagnitude and phase can be chosen so as to make the first harmonic of theforces zero. The other parameters of the system must now be chosen so thatthe higher harmonics add up to zero at any time:

ex> 00

w2 [{mI + 'IIÀ cos !po(m3 + m4)} '~ n2Bncos n'IjJ- m, ~ n2Ancos n'IjJ]+n=2 n-2 (6)

ex> • 00

+ tb [{mI + 'IIÀ cos !PO(m3 + m4)} ~ nBn sin n'IjJ- ma 2: nAn sin n'IjJ]= 0n=2 n=2

It may be seen from equation (7) that Bn/An must have a value which isindependent ofn. We will now discover what this means in terms ofthe geome-try of the system.Put Bn/An = D.If this is substituted into equations (3) and (4), we have:

c ID T cos ()= T cos X + Clcos 'IjJ+ Co (8)

where

andCl= -BI + DAl

Co=-Bo + DAo

Equation (8) may be rewritten as follows:

cos ()= E cos(x- (3) + c~ (9)where

c~ = c~ (scj + co);1E=~=--~cD cos {3 (10)

tan {3 =-CI 1cos {3 = --;:::====

Vei + 1

. {3 ClSIn = - --;:::====-Vc~+ 1

We have already shown (2) that:

sin ()= _!_ {sin X - 'IIÀ sin(x + <po) + À4e}c

-30-

Tills may be written as:

.0 p.(') AAsm t =csm X-lX +cLlS (11) .

where p = yl- 2vA cos ({Jo+ v2A2 . and

VA sin ({Jotan « = . ;

1-VA cos ({Jo1

cos cc = -(1- vA cos ({Jo);p

. I, .sm cc = - Vil. Slo ({Jo

P

We now have the following equations for 0:

cos 0 = E cos(x- (3)+ c~

sin 0 = ~ sio(X-lX) + : zls

(9)

(11)

The balancing conditions can now be found by combining these two equations,making use of the relationship:

cos20 + sin20 = 1

Setting the sum of the coefficients of the first harmonic and of the secondharmonic of X equal to 0, and the sum of the constant terms equal to I, weobtain the following conditions:

E=R.=I·c '

o: = (3; Co = 0; ALls = 0

It now follows, from (10), that:D = 1/(cEcos(3),andsubstitutingtheabovevalues, wehaveD = 1/(I-vAcos({Jo).If this value of Bn/An is substituted in equation (7), it follows that:

ml- m4 + VA cos ({JO(m3+ 2fll4) = 0

We thus have the following conditions which must be imposed on the drivemechanism in order that the sum of the higher harmonics of the forces shouldbe zero (assuming that A =1= 0): .

zls = 0 (12)

(13)

(14)

c = yl- 2vA cos ({Jo+ v2A2

mi = ma - VA cos ({Jo(ms + 2fll4)

We still have the terms for the first harmonics to balance (see (6»:

w2[{ml + vÀ cos ({JO(m3+ fll4)} BI-1D4AI] cos 1p ++ dJ[{mi + VA cos ({Jo(ms + ill4)} BI -1D4AI] sin 1p

Now

[{mI + VA cos ({JO(m3+ m4)}B] - ill4AI] =m,( i -Al) = m4vA sin ({Jo

-31-

.Lle = 0 (12)

(13)

(14)

so the remaining first harmonics are:

Equation (5) thus becomes:

-w2[(ml + ms+ ma+ m, + [lerm5)sin 1p+ [leom6sin (1p+1'0) +-{vA.(ma+2m4) sin <po} cos 1p]+w[(ml +m2+ma+ffi4+ [lerm5)cos 1p++ [leom6cos("P+1'o) + {vA.(ma+2m4)sin <po} sin "p] =0

So1'0 =n-{};

MmG=--

[leo

where {}= tan-l vA.(ma+ 2m4) sin <pomi + m2 + ma + ma + [lerm5

and M = V(ml + m2 + ma + ma + [lerm5)2+ {vA.(ma+ 2m4)sin<po}2

In brief, then, the drive mechanism canbecompletelybalancedifthefollowingconditions are satisfied 1):

c = Vl- 2vA. cos <po + V2A.2

mi - ma + vA. (ma + 21114)cos <po = 0

while the position and mass of the counterweight are given by:

-1 vA.(ma+ 2m4) sin <po (15)1'0 =n-tanmi + m2 + ma + ma + [lerm5

ma = _1_ V(ml+ m2 + ma + ma + [lerm5)2+ {vA.(ma+ 2m4) sin <Pop (16)(!eo

The following remarks may be made about these conditions:a. The condition c = VI - 2vA. cos <p + V2A.2 implies that (if LIe= 0) PS

must be equal to SQ (see fig. 19).

PS = SQ = V ;2 - 2 ~ cos <po + v2,

b. If <po = 0° or 180°, S lies on the piston connecting rod or an extensionthereof. Then 1'0 = st, i.e. the centre of gravity. of the counterweight now liesdiametrically opposite the crankpin.

11.2.3. The balancing conditions for the special form

The rhombic drive mechanism described in section 1.3 (fig. 15) is a specialform of the general form discussed above. It differs from the general form in

-32-

that ST = 0 (see fig. 19). All the calculations given below refer to this form,which will be called simply the rhombic drive from now on (fig. 21)..

Since ST = Q (i.e. ')I = 0), equation (13) becomes c = 1, which means thatthe piston connecting rod is the same length as thé displacer connecting rod.The piston connecting rod of length L and mass mpe can be replaced by the

two masses:, apempe =--mpe

L

" (1 ape)mpe = -T mpe

while the corresponding masses for the displacer connecting rod are:, ade

mde =Tmde

" (1 ade )mue= . -T mde

(see fig. 20).

a bFig. 20a. The piston connecting rod with the equivalent mass system.Fig. 20b. The displacer connecting rod with its equivalent mass system.

Equation (14) then becomes:

apempe- ademde= t L (ma - mp)

(17)

(18)

(19)

(20)

(21)

where ma and mp are the masses of the moving parts of the displacer andaccessories and the piston and accessories respectively.The position of the counterweight is found by substituting ')I = 0 in equation

(15):Î'o = :n;

-33-

(22)

reomeo = r{mpe + mde + temp + md)} + rermer

or if the crankshaft is considered together with the counterweight:

reme = r{mpe + mde + temp + md)}

(23)

i.e. the centre of gravity of the counterweight must be diametrically oppositethe crank pin.

The mass of the counterweight follows from equation (16):

(24)where

me = meo + merreomeo- rermer

re =meo + mer

reo, rer and re are the distances from the centre ofthe crankshaft to the centresof gravity of the counterweight, crankshaft and counterweight + crankshaft;the corresponding masses are denoted by meo, mer and me.

---'--displac:~ryokc 2703

Fig. 21. The special form of the rhombic drive, which is obtained from the general formshown in fig. 18 by making point S coincide with point T.

-34-

11.3.1. IntroduetionThe design of the heat exchangers (heater, regenerator and cooler) of a

hot-gas engine depends to a very large extent on the purpose for which theengine is intended. In one case, for example, the stress may be laid on a largespecificpower or lowweight, while the efficiency is ofrelatively little importance;while in another case it is just the specific fuel consumption 'which is important,while the weight does not matter so much.Often other factors also play a part, e.g. the price, so a compromise must

be made between maximum specific power and minimum specific fuel con-sumption to give the best engine for the purpose in question.The dimensions of the heat exchangers which are determined by the above-

mentioned factors can be calculated to a good approximation from the firstharmonic of the volume variations and the corresponding phase angle.

11.3. The determination of various quantities from the dimensions of the drivemechanism

p

v+x 2704

Fig. 22. Part of the rhombic drive mechanism shown in several positions; the dimensions areexpressed relative to the crank radius.

-35-

We have seen in section L2 that the volumes of the hot and cold spaces canbe written as:

VE = tVo(l + cos IX)Vc = twVo {I + cos (IX - !pvo)}

respectively.The calculation of Vo, w, IX and !pvofrom the dimensions of the drive mecha-

nism is based on fig. 22, where the dimensions are given with respect to thecrankshaft radius r. Since the piston yoke is fixed rigidly on to the piston viathe piston rod, the piston yoke describes the same motions as the piston itself.The same is true of the displacer yoke and the displacer. We will thereforeconsider the motion of the corresponding yokes when it is wished to determinethat of the piston or the displacer.

11.3.2. S, !pdr and Xmax

a. The stroke of the piston and the displacerThe stroke of the piston is equal to that of the displacer, because of the

symmetry of the rhombic drive about the y-axis (see fig. 22). The displacer isin its lowest position when the displacer connecting rod forms a continuationof the crank (position VO), and in its top position when the displacer connectingrod covers the crank (position TO). .

SVT=-

rIt follows directly that:

(25)

(see fig. 23).b. The phase angle !pdr: the phase angle by which the first harmonic of thedisplacer motion leads the first harmonic of the piston motion.

If we call the abscissa of the displacer (yoke) Xd and that of the piston (yoke)Xp, then, with the sign convention for x shown in fig. 22, we have:

Xd. 1r = sm 7jJ+T cos X (27a)

xp. 1- = sm 7jJ - - cos Xr /l. (28a)

Now ~ cos X = V/l.; -(8- cos 7jJ)2

We will expand this function as a series in cos tnp:

s , 1 eoT cos X = ~ Ba cos 0'Ij!A )1-0

(26)

-36-

Fig. 23. The relative piston stroke (= relative displacer stroke) as a function of the relativelength of the connecting rods J. and the relatieve eccentricity e.

The coefficients Bn as functions of A and e àre given in Appendix 1I.It follows that:

Xd co- = sin 1P + 1: Bn cos ll'1pr n=o

(27)

Xp. * B- = sm '1{J- "" nCOS n'1{Jr n=o

(28)

orco

Bo + D cos('1{J - YE) + 1: Bn COS n'1{Jn=2

X co2 =-Bo - D COs('1{J+ YE) - 1: Bn COS n'1{Jr . n=2

Whence:

f{ldr = n-2YE

f{ldr =~tan-I. BI (29)

D = VB~+ 1

1YE = tan-I-

BI

(30)

(31) I

-37-

c. The angle Xmax: the maximum angle which the connecting rods make withthe x-axis.

The value of this angle follows directly from fig. 22:

Xmax = sirl Ä (s+ 1).

Lines of constant Xmax are drawn in fig. 25. It may be proved that at the pointsat which these curves cut the Bj-axis:

tan Xmax = BIÀ~O

We have seen under b thattan t f{Jdr= BI

sopdr = 2Xmax

at Ä = 0 (see fig. 25).

11.3.3. The volume of the expansion space VE

When the end of the displacer connecting rod is at T, the displacer is in itshighest position. At an arbitrary crank angle '1{Jthe volume VE of the expansionspace thus satisfies the equation:

VE-=SU-STrA ct

where Ad is the cross-sectional area of the displacer.

SU= sin '1{J+ ~ cos X

ST= y' (~ _ 1)2 _ S2

so VE=rAd{sin'1{J+ ~ COSX-y'(~ _1)2_S2}or, substituting

1 00 .

, cos X = 2: Bil cos n'1{J/L 11=0

(26)

into the above equation, we obtain the expression for VE as a function of '1{J:

00

VE = rAd {C + D cOS('1{J-YE) + 2: Bil cos n'1{J}n=2

(32)

where1 .

C= Bo-;: V(1-Ä)2_À2S2

D = VB~+ 11

YE = tan-l-BI

(30)

(31)

-38-

1[,3.4. The volume ofthe compression space Vc

This space is cont'ained between. the bottom of the displacer and the top ofthe piston. We may distinguish two cases:1. The diameter of the displacer is less than the diameter of the piston; or,

calling the effective area of the bottom of the displacer Ad and the effectivearea of the top of the piston A~, then Ad < A~.

This volume may be divided into two parts:a. VCl: this is the volume between the displacer and the piston, with cross-

sectional area A~.b. VC2: this is the volume between a fixed wall and the piston, with an annular

cross-sectional of area A~ - Ad.

a. The circular motion of the crank pin may be resolved into a vertical anda horizontal component. The vertical motion moves the whole drive mechanismup and down, and has no effect on the relative positions of the piston and the

+x2706

Fig. 24. Diagram drawn as an aid in determining the distance between the piston and the .displacer.

displacer. The variation of the volume between the piston and the displacer isdetermined by the horizontal component of the motion of the crank pin only(see fig. 24). Kl and K2 are the positions of the crank pin at which the volumeis minimum and maximum respectively. Because of the symmetry about the

-39-

y-axis the contribution of the piston is equal to that of the displacer, so thisvolume is:

VC! =2rA'd{YÄ,12-(e-I)2- ! cos X}

b. VC2is completely determined, by the motion of the piston. When thepiston connecting rod is a continuation of the crank, the piston is in its topmostposition, and will then be touching the fixed wall. At an arbitrary value of theangle the volume between the fixed wall and the annular surface A~ - Ad ofthe piston is:

VC2= PQr(A'p-Act)

VC2=r(A'p-A~I){Y(1 + !f-e2+sin1p-~ cos X}

or with

VC2 = r (1 - f)A'p{Y (1 + ~ f - e2 + sin 1p- ~ cos X}

If we t.hen substitute the s~ries expansion for ~ ~os X (26), we find for the

total volume:Vc = VCI+ VC2

co

Vc = rAp {E' - F cos(1p + yc)- (1 + f) ~ Bncos n1p}n=2

where

E' = -¥-Vl- Ä,2(e-I)2 + (1 ~f) V(1 + Ä,)2- Ä,2e2-(1 + f)Bo

F = V(1_f)2 + (1 +flB~. l-fyc = tarr+ (1 + f)BJ

2. The diameter of the displacer is greater than that of the piston, or, inthe notation of the previous part of this section,

Ad> A'pThis volume can also be divided into two parts:a. VCI: this is the volume between displacer and piston, with cross-sectional

~a~. .

-40-

b. V~2: this is the volume between a fixed wall and the displacer, of annularcross-sectional area A'd - A~.a. This volume is given by the same equation as in the previous case, .if Ad·

is replaced by A~.

Vb = 2rAp {Y ~ -(8-1)2_ ~ cos X}

b. This volume is completely determined by the motion of the displacer.When the displacer connecting rod is a continuation of the crank (VO in fig. 22),the displacer is in its lowest position, and will then be touching the :fixedwall.The volume VC2at a crank angle '1/)is then:

Vca = UT·r(Ad-A;)

vC2=rA'p(f-1){Y(1 + ~r-82-Sin'1/)- ~ cos X}

The total volume is thus found by combination of this equation with equation(26):

Vc =VÓ! + Vca00

Vc = rA'p{E" - F cos('1/)+ ye)- (1 + f) ~ Bn cos mp}n=2

2 ~~.~~~where E"=TVl-il.2(8-l)2-(1-f)V(I +1.)2-1.282-(1 +f)Bo

F = V(I-fi + (1 + f)2B~

l-fye = tan-I (1 + f) BI

We can thus in general write for Ve:00

Ve = rAp {E- F cos('1/)+ ye)- (1 + f) ~ Be cos mp} (33)n=2

2f (I-f)where for f ~ 1 :E = T Vl- ~2(8- 1)2+ -1.- V(il.+ 1)2- 1.282- (1 + f) Bo

forf ~ 1 :E = ~ V~- 1.2(8- 1)2- (1 I. f) V(il.+ 1)2- 1.282.- (1 + f)a,

F = V(1-f)2 + (1 + f)2B~

l....:...fye = tan-I (1 + f)BI

-41-

(38)

II.3.5. Vo, w, cpvo and cx

We can now easily determine Vo, w, cpvo and cc of the first harmonic of thevolume variation from (32) and (33), and also the fictive dead spaces VIlEandVse which arise from taking the first harmonic of the volume, V0, instead ofthe real volume. VE and Ve may be written as follows:

IX>

VE = rAdD {I + cos (1p-YE)} + rAd(C- D) + rAd ~ Bn cos n1pn=2

IX>

Ve = rA'pF {I + cos(1p + ye + n)} + rA'p(E- F)-(l +f)rA~ ~ Bs cosne. n=2

whenceVo = 2rAdD. 1 Fw =_._

f' Dwhere f' = Ad

Adcpvo = n-YE~yeo: = 1p-YE

Fictive dead space on the expansion side = VîlE= rAd (C - D).

Fictive dead space on the compression side = Vse = rA~(E- F).

1[,3.6. Summary of the equations derived above

(34)

(35)

Vo = 2rAdVB~+ 1

w = _!_ V (1- f)2 + (1 + f)2B~If' Bf + 1

-2Blcpvo = tan-1-----=----(1 + f)B~- (1~ f)

IYE= tan-1_Bl

oe = 1p- 'YE

(36)

(37)

. VEfirstlmrm.= 1VO {I + cos'«}

Vefirstharm.= ~.wVo{I + cos(cx - cp)}

(39)

(40)

(41)

(42)

, {V 2 1 V 2 22}VSE=rAd Bo- B1+1-T (I-À) -Àe (43)

-42-,-

.j>.w

-.a

Fig. 25a. The phase angle pdr between the piston and the displacer, the maximum angle whichthe connecting rods make with the vertical, Xmax, and BI (see equation 26) as funtions of À and 8.

Fig. 25b. The phase angle between the volumes of the expansion and compression spaces 'Pvo, andtheratio ofthe maximum values of these volumes, w, as functions of Bi and the relative areasfandf'.

to0IIIesQ.. -< -<to rn'rn' Cl,_.

ClIII"'1

I "Cl) I "'1

<19 . ....-.. "'1 ....-.. >...... ~ ...... oe'-e ----Cl) + +es >->IN >->I~..... ....., -::Jes '-' ~>~ to '"'-...... 0

:g I I I ......Cl) ~ >->"" '"'- Ies ....-.. >->Q.. ...... '(;;' ...... t<>:;<. I I I '(;;'.... ....., -::J It""''-' ......

"" <;» "" ......+ t<> + '-'

I ""....-.. ....-.. +......

>->II

......

+ +>->11_,:, _,:,

t<> t<>

to '"'- to~ ....-.. i-'''' '"'-...... --- ....-..

+ ,_.

~ +Ö' re Ö' ~"'1 I "'1 '"...., >-> ...., I\V ",'" //\ >->

2707 '" "':...... + ......+

....-.. ,--.,.j>. .j>..j>. .j>.CT' III.._, .._,

,n.4. The pressure yariation in circuit and buffer space

1l.4.1. The pressure variation in the circuit

We have seen in section 1.2 that the pressure in the circuit ~an be written as1- c5

p = Pmax1 + (j cos(a _ ())

if certain strongly simplifying assumptions are made.For the calculation of the forces operating in the engine, the real pressure

variation must be known, taking into account all the departures from idealityin the engine, such as flow losses, regenerator effects, finite heat transfer in thecylinders, non-ideal gases, higher harmonics of the motions, etc.After years of research in Philips' Research Laboratories in Eindhoven, it is

now possible to express all these effects formally, so that it is possible tocalculate the pressure variation in the compression and expansion spaces veryaccurately when designing an engine of completely new dimensions. Thesecalculations fall outside the scope of this thesis, so we will content ourselveswith giving the equations for the pressure in these two spaces:a. in the compression space:

00 00

pc = Pc {I + ~ pen cos n1p+ ~ qen sinne}n=l n=l

(45)b. in the expansion space:

00 00

PE = PE{I + ~PEn COS n1p+ ~ qEn sin n1p}n=l n=l

(46)

In order to give an impression of the way in' which these pressures actuallyvary, fig. 26 shows the Farnboro indicator diagrams of the pressure in the

r pressure p

2708

Fig.26.

Fig. 26. Farnboro indicator diagrams of a Stirling engine. Pc = pressure in compressionspace, PE = pressure in expansion space.

-44-

compression space and in the expansion space. These two curves were recordedsimultaneously.

1I.4.2. The pressure variation in the buffer space of a multi-cylinder engine

The spaces under the pistons in a multi-cylinder engine may be interconnected,so as .to give a common buffer space. The pressure in this buffer space willdepend on the positions of the various cranks, which can be chosen to give asuniform a torque as possible. In calculating the pressure variations in the bufferspace, we will assume that they are adiabatic, and that the flow losses, e.g. inthe connecting pipes, are negligible. '

We will consider an N-cylinder engine, and will give the first cylinder theindex 1.It may be seen from fig. 22 that the varying volume Ve of the buffer space

can be expressed as:

v, 1 . 1 V_- = -cos X-sm 'Ijl--(l- Ä)2-Ä2e2rA" ..1 ..1p .

or, making use of equation (26):00

Vb = rA'~ {C + D cos ('Ijl+YE)+ 2: Bn COSmp}n=2

where A~ is the effective area of the bottom of the piston (i.e. the total arealess the cross-sectional area of the piston rod) and C, D, YE and Bn have thesame significanee as in equation (32).

We will only consider the first and second harmonics of the volume variationsdue to the piston of the first cylinder:

Vbl = tVbO{bo + cos ('Ijll+ YE)+ b2 cos 21jJl}

VbO= 2rX;D = 2rA;{ VB~+ 11

C Bo- T V(l- ..1)2- Ä2e2bo = D = ----:-;:====~--

VB~+ 1

The phase lead of the piston of the ith cylinder with respect to that of thefirst cylinder is denoted by {Jr. It then follows for the ith cylinder that:

Vbi = !VbO{bo + COS(1jJI+ (h + YE)+ ba cos 2(1jJl+ (Ji)}

If the constant part of the volume of the buffer space is Vbe and ab = VbeIVbo,

-45-

00 00

Pb = Pb {I + ~ pbn cos n7p1+ ~ qbn sin n7p1}n=1 n=1

(47)

the total volume of the buffer space of an N-cylinder engine may be written as:N N

Vbt = ~-VbO{2ab + boN + ~COS(7p1+ (31 + YE) + b2 2: cos 2("1'1 + (31)}i=l 1=1

or

Vbt = tVbo {2ab + boN + Ri cos(7pJ+yI) + b2R2 cos 2(7p1+ Y2)}where

l/{N ·2 {N }2Rl = Y ~ cos {3iJ + ~ sin {31i=l 1-1

R2 = V{ ~ cos 2{31}2+ {~ sin 2{3;}21=1 i=l

N~ sin ({31+ YE)

YI = tan-1 :,,1=.;;.1 _N .~ COS({3i+ YE)i=lN~ sin 2{31

Y2 = tarr-! -,i;",..l _

~ cos 2{3i1-1

The adiabatic pressure variation in the buffer space is therefore:

Pb = constant X {I + Cl COS(7p1+ YI) + C2cos 2(7p1+ Y2)}-"where

cp" =-Cv

RlC1=----

2ab + boNb2R2

C2=----2ab + boN

This expression for the pressure may be expanded into a series, giving theequation:

where Pb is the mean pressure in the buffer space. Expressions for the coefficientsPbn and qbn are given in Appendix I.

11.5. The torque on the crankshafts due to the gas forces and inertia forces

11.5.1. Introduetion

The calculation of the torque is based on the energy balance of the engine.This neces~itates a knowledge of the coefficients of friction of piston and

-46-

bearings, which is largely lacking since these variables depend on a largenumber of factors which are as yet unknown, especially where plain bearingsare concerned. Moreover, the coefficient of friction is assumed to be constantin these calculations, which is certainly not the case for plain bearings, so inthis case the value taken for the coefficient of friction must be seen as anexperimental mean value. Since however the total friction power of a well-builtengine is only about 10% of the power delivered by the engine, the influenceof these simplifications on the final result will be slight.

In ball bearings, roller bearings and needle bearings the coefficient of frictionis more or less constant 3).

The gas force on the piston is the resultant of the forces due to the circuitpressure and the buffer-space pressure acting on the piston itself and the crank-case pressure acting on the piston rod.

The gas force on the displacer is due to the pressure difference between theexpansion space and compression space, which is caused by the flow resistancein the heat exchangers, and by the force on the displacer rod caused by, thepressure in the circuit and in the crankcase *).

The torques caused by the above-mentioned gas forces of the circuit and thecrankcase can be considered together, since the variation of these torques isthe same for each cylinder of a multi-cylinder engine, apart from the phasedifference. This is not so for the torque due to the pressure in the commonbuffer space of a multi-cylinder engine, however, since here the phase of thepressure variation is determined by the relative arrangement of the cranks ofthe various cylinders. The expression for this torque will therefore be derivedseparately.

IJ.5.2. The energy balance

If we consider half of the drive mechanism, with the directions of Fd, Fp andTt (the forces on the displacer and piston and the resultant torque respectively)given in fig. 21, the energy balance is given by the equation:

tFpdxp + ~Fddxd-Ttd1J!-Ptdt = d ~ U+ d ~ Eor

where Pr is the friction power and ~ U and ~ E are the potential and kineticenergies respectively of the drive mechanism.

We assume that the drive mechanism is completely balanced, i.e.

*) The pressure in the crankcase is usually about atmospheric.

-47-

If we write the force on the piston as the sum of two terms:

where F~ is the force due to the circuit pressure and the crankcase pressure,and F~ is the force due to the pressure in the buffer space, we may write theresultant torque as:

T, = ~-F'p xp + lFp"xp + .;tFaxa _ d ~ E _ Pr

~ to ~ W 2 ~ d'ljJ W

orTt + Tr = Tge + Tgb-Tm (48)

where

T J. F' Xp 1F xa (torque due to gas forces of circuit andge = 2- P - + 2- a-w w crankcase)

T iF" xpgb = 2 p~

(49)

(torque due to gas forces of buffer space) (50)

(torque due to inertia forces) (51)

, PrTr=-

t»(torque due to friction) (52)

1[,5.3. -The torque on the crankshaft due to the gas forces of the circuit andcrankcase on the piston and displacer

We see from equation (49) that the torque Tge may be written as:

a. Determination of t F~xp/wThe pressure in the crankcase, Pee, is assumed to be constant: Pee = g"pc.

If A~ is the cross-sectional area of the piston less the cross-sectional area ofthe displacer rod, and A~ is the cross-sectional area of the piston less that ofthe piston rod, then:

Fp = PcAp - g"Pc(Ap - A'~)

or writing Pc as in equation (45):

OCJ OCJ

Fp = pcAr {l- g"'+ g"f"+ ~ PCncos n'ljJ+ ~ qCn sin mp} (53). n=L . n=l

where f" = A;;A'p

-48-

Differentiating equation (28) with respect to the time, we obtain:

Xp= r {cos '1jJ+ ~ nBn sin n'1jJ}W n=l'

While combining (53) and (54), we have: .

t Fp Xp= t rpcA~ {l- g" + g"f" + ~ PCn cos n'1jJ+ ~ qCn sin n'1jJ}XW n~· n~

coX {CoS'1jJ+ L nBn sin n'1jJ}

n=l

Multiplying out, we obtain the expression:

-}F~ Xp = l rpcA'p { ~ C~ cos n'1jJ+ £ D~ sin n'1jJ} (55)W n=o n=l.

b. Determination of t FdXd/WUsing the symbols of II.3: Ad = cross-sectional area of displacer

Ad = Ad less cross-sectional area of displacer rod

Ad

A'p

and with

f' = AdAd

f

g

we obtain:

or writing Pc and PE as in equations (45) and (46):

Cl()

Fd = fpcA~ {gf' - 1 + g" - g"f' + L (gf'PE - pc)'; cos tnp +n=L

Cl()

+ L (gf''qn - qc)n sin n'1jJ}n=l

Differentiation of equation (27) with respect to time gives:

Xd Cl().

- = r{cos '1jJ- L nBn sin n'1jJ}W n=L

Combining (56) and (57) and multiplying out:,• Cl() Cl()

t Fd Xd= -lrpcApf{ L C~cos n'1jJ+ L D~sin n'1jJ} (58)W n=o n=L

-49-

(54)

(56)

(57)

,,~Addition of equations (55) and (58) gives the expression for the torque due tothe gas forces of the circuit and the crankcase:

Tge = .~.rpcA~ { ~ CIl COS n'lfl + ~Dil sin n'lfl}11=0 11=1

(59)

Expressions for the coefficients Cn and Dn are given in Appendix I.

11.5.4. The torque on one crankshaft due to thegas forces of the buffer spaceacting on thefirst piston of a multi-cylinder engine

According to equation (50),the torque T gb is:

T - rF;' Xpgb - -2' P W

The force on the piston due to the pressure in the buffer space is:

Expressing Pb as in equation (47), and writing:

, Pbg==-Pc

we obtain:00 00

Fi{= -g'f"pcA~ {I + ~Pbn cos n'lfl1 + ~ qbll sin n'lfl1} (60)11=1 11=1

Writing xp/was in equation (54), we have:

{

OO 00Tgb = ----{-rj)cA pg'f" ~ Cbll cos n'lfl1 + ~Dbll sin n'lfl1}

11=0 11=1(61)

Expressions for the coefficients Cbll and Dbl1 are given in Appendix I.Note. In order to calculate the torque for a given cylinder due to the pressure in the commonbuffer space of an N-cylinder engine, the crank angle belonging to this cylinder must be takenas the first angle. This means that Pb, pbn and qbn must be calculated afresh for eachcylinder for which the torque has to be determined.

11.5.5. The torque on one crankshaft due to the inertiaforces

For this calculation we need the total kinetic energy of half of the drivemechanism (see fig. 21). We already know the expressions for the purelyreciprocating parts, i.e.:

00

Xd = r {sin 'lfl + ~Bn cos tnp}11=0

00

Xp= r {sin 'lfl- ~ Bn cos n'lfl}11=0

(27)

(28)

We still need the kinetic energy of the connecting rods. The two-masssubstitution for a connecting rod used in 11.2.3' is only valid for forces. For the

-50-

calculation of moments, the masses must also be given the same total momentof inertia about an axis through the centre of gravity as the connecting roditself.

The masses replacing the piston connecting rod must in the first place satisfythe conditions:

, apempe = --mpe

L (17)

" (1 ape)mpe = - ---y:- mpe (18)

If we denote the moments of inertia of the masses m~e and m~e by i~~m~eand i~; m~e, where i~e and i~e are the radii of gyration of these masses, then theequivalent system of masses must also satisfy the condition:

.2 .,2, + .,,2 " + (L )2, + 2 "lpempe = lpempe lpempe - ape mpe apempe

Combining this with equations (17) and (18), we obtain:

.,2, ."2,, {.2 (L )} Jlpempe + lpempe = lpe - ape - ape mpe = pe (62)

Similarly, the following conditions must be satisfied by the displacer connectingrod:

, ademde = ---y:- mde

" (1 ade.)mde = - ---y:- mde

(19)

(20)

.,2, ."2.,, {.2 (L )} JIdemde + ldemde = ldc - ade - ade mde = de (63)

where i~e and i~e are the radii of gyration ofthe masses m~e and m~'erespectively.The connecting rods are thus now replaced by a set of masses and moments

of inertia, whose equations of motion can be expressed in a relatively simpleform. We may therefore write the kinetic energy ofthe crankshaft and counter-weight (Ee), piston connecting rod (Epe), displacer connecting rod (Ede), halfof the kinetic energy of the piston + piston rod + piston yoke (Ep) and halfof the kinetic energy of the displacer + displacer rod + displacer yoke (Ed) as:

El.2 2e = :rlemeCU

E J.. [{ 2(1 ape) } 2 ape .2 J .2]pc ='2 r -T mpe cu +TmpeXp + peXpe

Ep = impx;

El [{ 2(1 ade) } 2 +. ade .2 J.2 ]de = -2- r - T mde CU T mdeXde + deXde

El .2d = 4mdXd

-51-

or

• À sin 1pX = w cos X

• 2( À: ) cos- X = sin2 'W

(66)

(67)

.',\

where ipeand ide are the angular velocities of the piston connecting rod anddisplacer connecting rod respectively.

We will assume that the drive mechanism is balanced (see section II.2.3), i.e.ipe = -ide. The total kinetic energy is thus:

,~ ] [.2 2{ ( ape' ade)}] 2,t., E = lî Ieme + r mpe + mde- --y:- mpe + L mde W +I 2(xp)2 ( ape 1_) 2 1 2(Xd)2( ade .1) 2+ li r rw --y:- ~pe + 2" mp w + 2" r rw L mde + 2 md w +

+ ~.iw2 (À~ r (Jpe + Jde)

Now itfollows from the balancing conditions that:

reme = r {mpe+ mde + t(mp + md)}

(64)

(24)and

(21)

Substituting these two expressions into the first term of equation (64), we have:

I {.2 + 2} 22" Ieme rreme- r me wwhere

2ape ,2ademo = mp +---mpe =md + ---mdeL L

The second and third terms of (64) may be taken together:

(65)

Differentiation with respect to time of the two equations:Xd. 1r = sin "p +T cos X (27a)

andxp. 1r = sin "p - T cos X (28a)

and substitution into (65) gives:

!r2mo{cos- v +(Lr(1- cos- X)}w2

It may be seen from fig. (21) that

sin X = À(e- cos 1p)

whence

,-52-

So the second and third terms of (64) can be written as:• 2

t r2mo {COS2 11' + ( L)}w2

and the total kinetic energy can be expressed as:• 2

~ E =l [(i~+ rre)me + r2mo(cos 21p-1) + (A~ ) {r2mo + i(Jpe + Jde) }]éi• 2 .

Let us now expand ( A: ) in a series.

Itfollows from equation (67) that ilAW = sin'1p/cos X, or expressed as a series:

_}.x = ~Gn sin tnp.W n=L

(68)

The values of Ge are given in Appendix Ill.Squaring this series, we obtain:

. 2 co

(-f-) = ~Hn cos mI'AW n=o

(69)

The values of Hn are given in Appendix IV.Substituting (69) into the expression for the total kinetic energy, we obtain .

the following expression in terms of the crank angle 1p:

~ E = lw2[(i~ + rre)me + ?mo {(I + z)Ho-l} +

2 1 co+(1 + z)r mo {Hl cos 1p+ (1 + z + H2)COS 21p + n~3Hn ~os n1p}] (70)

where

The effective moment of inertia of half ofthe drive mechanism about the crank-shaft is:

or Je = (i~+ rre) me + r2mo {(I + z) Ho - I} +2 1 co

+(1 + z)r mO{HI cos 1p+ (-1 + + H2)COS 21p + ~ Hn cos n1p} (71)z n=3

with the-mean value: Je = (i~ + rre)me + r2mo {(I + z)Ho-l}

The torque on one crankshaft due to the inertia forces follows directly fromthe differentiation of equation (70) with respect to 1p:

dJe .Tm =l--w2 +Jew

d1p

-53-

(72)

Fpi + Fp4 sin x- Fp:!cos X = m'pcxp

The equilibrium of the massless rod gives:

MIP + Mzp = Fp4L

(73)

or with only the varying part of Je:

, 1 dJe Z -.Tm =·9°--W +(Jc-Je)w- d'lP

and with a constant angular velocity w = ai:

Tm(.j=~} rZmoroZ(l + Z){HI sin 'lP+ 2 ( 1 : z + Rz) sin i 'lP+

(72a)

00

+ ~ nRn sin mp} (72b)11=3

II.6. The forces on various parts of the drive mechanism due to gas forces andinertia forces

1[,6.1. Introduetion

These calculations do not take the forces due to the friction and weight ofthe parts of the drive mechanism into account.

The forces and moments on the various parts of the drive mechanism areshown in fig. 27, with the various parts drawn separately for the sake of clarity.

The direction of the forces is mostly arbitrarily chosen, except for the forceson the piston and the displacer, Fp and Fa, which have the same directions asin section II.S, and the forces on the displacer connecting rod, which are themirror image of those on the piston connecting rod with respect to the y-axis.The expressions for the forces on the displacer connecting rod may thus beobtained directly from those for the forces on the piston connecting rod byreplacing 'Ijl, Xpand Fp by-'Ijl,-xd and=-Fa respectively. Theconnectingrodsare replaced by weightless rods with certain masses at the ends having certainmoments of inertia, as for the calculation of the torque due to the inertiaforces in II.S.S. This does not give the actual force distribution along theconnecting rods but a mean value, which is however quite sufficient for thestrength calculations 4).

1[,6.2. Theforces on the piston drive mechanism

These forces follow from Newton's laws together with the equations ofmotion of the various parts of the drive mechanism. The vertical componentof the force on the piston yoke pin, Fp!, is:

The equilibrium condition for the forces at Apc in the vertical direction gives:

-54-

M ".,,2 ..2p = -mpc1pcX

Fig. 27. Various forces and moments occurring in the rhombic drive mechanism.

where MIP and M2p are the moments ofthe massless rod acting on the niomentof inertia of the equivalent masses:

'M ' .,2 ••lp = -mpc1pcX

It follows that:

F ( , .,2 + ".,,2) Xp4= - mpc1pc mpc1pc L

or with the nomenclature used above:

F Jpc ..p4=-TX

Substitution of the expressions found for FpI and Fp4 in (73) gives:

F 1 (F '.. 2Jpc...)p3= --- p-moxp----X sm X. 2 cos X L

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Resolution of the forces at Apc in the horizontal direction gives:

Fp2 = Fp4 cos X + Fp3 sin X

and substituting the expressions for Fp4 and Fp3,

F 1 (F .. ) t Jpc Xp2= ·2· p-moXp an X--- ---L cos X

The equation of motion of the equivalent mass at Bpc in the horizontal directiongives:

-Fp3 sin X - Fp4 cos X + Fp5 cos VJ+ Fp6 sin VJ= mócyc

and in the vertical direction:

Fp3 cos x- Fp4 sin Z + Fp5 sin VJ- Fp6 cos VJ= m';cxc

where y o and Xcare the accelerations of the crank pin in the y and x directions.

Since

andit follows that

Yc= -rw2 cos VJ

Xc= -rw2 sin VJ

Fp5 = Fp3sin(X-VJ) + Fp4COS(X-VJ)-rw2mi{c

Fp6 = Fp3 cos(X- VJ)- Fp4 sin(x- VJ)

We can now substitute the series expansions found above, or expressionsderived from these, in the expressions for the forces.

Fp = Fp + Fi{ (53) and (60)or

00

Fp = pcAp {l- gft + g"f" - g'f" + ~ (Pc - g'f"Pb)n cos nVJ+n=I

00+ ~ (qC- g'f"qb)n sin nVJ}n=L

(74)

Moreover,00

Xp= rw (cos 1p+ ~nBn sin n1p)n=L

(54)

so00

Xp = rw2(- sin 1p+ ~n2Bn cos n1p)n=L

(75)

Finally,co

X = Ä.w }."; Gn sin nVJn=L

(68)

soco

X = Ä.w2 }."; nGn cos nVJn-I

(76)

-56-

00+ ~ (gfqs - qC)nsin n1p}11=1

11.6.3. The forces on the displacer drivé mechanism

The forces on thè various parts of the displacer drive mechanism can befound in the same way as those for the piston drive mechanism; but they mayalso be derived directly from the expressions for the forces on the piston drivemechanism by consideration of the symmetry of the system, as mentioned in11.6.1. This gives:

Fer = ----{·(Fd- mdxd)

Fd2 =-l(Fd-mOxd)tanx- JdC _i__2 L cos X

F 1 (F .. + 2 Jdc ..' )d3 =---- d-moXd ---x smX2 cos X . L

F Jdc ..d4=-TX

Fd5 = Fd3sin(X + 'lp) + Fcl4COS(x + 1p)-rw2mdc

FII6 = Fd3 cos (X + 1p)- Fd4 sin (X + 1p)

where Fd has the value found above (equation 56):00

Fd = fpcAp {gf" - 1+ g" - g"f' + ~ (gf'PE - PC)11cos n1p+11=1

Xd is found by differentiating (57)with respect to time:00

Xci= -rw2 (sin 1p+ ~n2Bn cos n1p)n=l

(77)

11.6.4. The forces on the crankshaft

It follows from the equilibrium of the crankshaft that:

F7 - Fp5 - Fd5 = m~rcw2

Fs + Fd6 - Fp6 = 0

Substituting the expressions for Fp5 and Fd5,

F7 = Fp3 sin(x-1p) + Fp4 cos(X-1p) + Fd3 sin(x + 1p)++ Fd4 cos (X + 1p)- w2 {r(m;;c + mdc) - mcrc}

The conditions for the mass distribution and balancing (18), (20), (21) and(24) give:

{ (" " ) } ( 2apc )r mpc + mdc - mcrc = -r -L--. + mp

which may be rewritten:

{r(m;;c + md'c)- mcrc} = -rm~

-57-

so: ''F7 = Fp3 sin (x-7p) +Fp4 cos (x-7p) +Fd3 sin (x+7p)+Fd4COS(x+7p)t morw2

Similarly:Fs = Fp3 cos (x - 7p).:_ Fp4 sin (x - 7p)- Fd3 cos (x + 7p)+ Fd4 sin (X + 7p)

11.6.5. Simplified equations for the forces

The contribution of the moment of inertia of the connecting rod to the forcesis always small, and may be neglected.

If we also replace the equivalent masses m~e and m~e by the expressionsgiven in (18) and (20), we obtain the following equations for the forces:

Fpl = HFp-mpxp)

Fp2 = ~(Fp - moxp) tan XFp3 =~(Fp - moxp) sec X

Fp4 =0Fp5 = Fp3 sin (X - 7p)- mpe (I' - a~e) rw2

FpG= Fp3 cos(X-7p)

Fui = --!(Fd - mdxd)

Fd2 = -HFd - mOxd) tan XFd3 = ~~(Fd - moxd) sec X

Fd4 = 0Fd5 = Fd3 sin (X + 7p)- mde (1- a{e) rw2

FdG = FrI3COS(X + 7p)

where

F7 = Fp3 sin(x-7p) + Fd3 sin(x + 7p)+ morw2

Fs = Fp3 cos(X-7p) - Fd1 cos (X + 7p)

00

Fp = peAp {1-g" + g"f" - g'f" + ~ (Pc - g'f"Pb)n cos n7p+n=l

00

+ ~ (qe- g'f"qb)n sin n7p} .n=l

(74)

00

Xp = rw2(- sin 7p+ L n2Bn cos n7p)n=L

(75)

00

Fd = fpeA'p {gf" - 1 + gft - g"f' + ~ (gf'PE - pe)n cos n7p+n=L

00

+ ~ (gf'qE- qe)n sin n7p}n=L

(56)

00

Xd= -rw2 (sin 7p+ ~n2Bn cos mI'), n=l

(77)

-58-

n.7. The power and the torque due to the friction on one crankshaft

As has been mentioned, it is usually only possible to calculate an approxi-mate value of the friction power.It has been known for a long time that the crank-connecting rod system is

very efficient 2). We have however given these calculations in connection withthe characteristic load of the Stirling system in order to give a qualitativepicture ofthe magnitude ofthe friction in the various bearings.Also, when carrying out optimum calculations concerning the overall

efficiency of the engine, the use of the most likely value of the coefficient offriction will give'a result which is nearer to the truth than if the friction poweris taken to be a constant percentage of the indicated power, or to be constant.

For the determination ofthe friction ofthe piston, see the relevant literature 4).The friction of the displacer without piston-rings will be neglected.We assume that the coefficient of friction is constant with respect to time.There are various possible ways of fixing the connecting rods on to the crank

pins. We will consider two cases: either the crank-pin bearing shell is fixed inthe big end of the piston connecting rod with the big end of the displacerconnecting rod swinging around it, or vice versa, see fig. 28.

a bFig. 28. Two ways of fitting the connecting rods round the crank pin:

a. With the crank-pin bearing shell :fixed in the big end of the piston connecting rod.b. With the crank-pin bearing shell :fixed in the big end of the displacer connecting rod.

The friction torque on the crankshaft from the various points of friction in•the drive mechanism is obtained by dividing the friction power by the angularvelocity w of the crankshaft (see equation 52).With the aid of the forces calculated above (1I.5) and the absolute value of

the first harmonic of the velocities, we may now calculate the followingquantities:

-59-

a. The friction power PtPI and the torque TrPI due to the friction of thepiston in the cylinder and the piston rod in the stuffing box (friction force Ftp):

PtPI = tFtP IXplsee eq~ation (54) for Xp.

PtPI = tFrp rwD [sin ('Ij) + YE)I'Trui = tFrp rD [sin ('Ij) + YE)I

where

and1

yE = tan-l-BI

See equations (30) and (31), and Appendix Il for BI.b. The friction power Pms and the torque Tres due to the friction of the

piston yoke pin of diameter dpp in the small end of the piston connecting rod(coefficient of friction fip2):

Prp2 = -~.fip2dppVF~1+ F~2 lxi for X see equation (68)

Ptp2 = -~-fip2dppÀwGIVF~1+ F;2lsin 'lj)1

Trp2 = -~fip2dppÀGIVF~1+ F;2 [sin 'Ij)I

for GI see Appendix Ill.c. The friction power P~P3and the torque T;P3due to the friction of the

crank-pin bearing shell on the crank pin of diameter dcp and coefficient offriction fi3 (crank-pin bearing shell fixed in big end of piston connecting rod).

d-.The friction power P~~3and the torque T;~3 due to the friction of thebig end of the piston connecting rod on the crank-pin bearing shell of dia-meter dcs and coefficient of friction fi~3 (which is fixed in the displacer con-necting rod): ._---

Prp3 = {..uP'3dcsVF~5 + F~612il

Prp3 = .uP'3dcsÀwG; VF;5 + F;6lsin 'lj)1

Trp3= .u~3dcsÀGIVF;5 + F;6lsin 'lj)1

-60-

e. The friction power Pnn and the torque' Trei due to the friction of thedisplacer rod on the stuffing box in the piston (friction force Fre)

Prei = iFfd Ixp - xdlfor Xd see (57)

Pfdl = Frd rwBI [sin 'IJl I

Tnn = Fre rBj lsin 'IJIIfor BI see Appendix 11.f, The friction power Pres and the torque Tres due to the friction of the

displacer yoke pin of diameter ddp in the small end of the displacer connectingrod (coefficient offriction ,ud2):

g. The friction power Pfd3 and the torque Tfd3 due to the friction of thecrank-pin bearing shell on the crank pin of diameter dep and coefficient offriction ,u3 (crank-pin bearing shell fixed in the big end of the displacer con-necting rod)

Prd3 = Prp3

Trd3 = Tep3h. The friction power P~d3 and the torque Tf~3 due to the friction of the big

end of the displacer connecting rod on the crank-pin bearing shell of diameter desand coefficient of friction ,urt3' which is fixed in the piston connecting rod:

Prd3 = -},udsdes VF~5+ F~612il

~fd3 = ,udsdcs },G1VF~5+ F~6lsin 'IJl I

i. The friction power Pf4 and the torque Tf4 due to the friction of thecrankshaft journals (diameter del) in the main bearings (coefficient of fric-tion ,u4)

Pf4 =! ,u4dcJw VF~ + F~

. Tf4 =.~,u4dcJVF~ + F~

-61-

j. The friction power Pr5 and the torque Tf5 due to the friction of the gear-wheels. It is assumed that the friction power Pf5 is proportional to the trans-mitted useful power; or, to a good approximation, that it is proportional tothe indicated power.

Pf5 = C·t rpcA'pCo co

Tr5= C·~·rpcA'pCo

The sum of these torques gives the total torque due to friction forces.

Tf = TfPI+ TfP2+ TfP3+ Tfdl + Tfd2+Tfd3+Tr4+ Tf5 (78)

IT.S. Some applications of the torques calculated above

11.8.1. Introduetion

If a highly variable torque acts on the crankshafts, the moment of inertiaof the drive mechanism is not enough to ensure the uniform rotation of thecrankshafts. It is therefore necessary to increase the moment of inertia bymeans of a flywheel. If all the torques acting on the flywheel of the engine areknown, then the maximum and minimum values of the angular velocity, orthe size of the flywheel if the angular velocity must lie within certain limits,can be calculated from the largest and smallest values of the integral of thetorque with respect to the crank angle.We are not only interested in the coefficient of nonuniformity of rotation

of the engine, but also in that of the driven machine, and in the variation of thetorque transmitted by the gear wheels of the drive mechanism; it is thereforenecessary to consider the two crankshafts and accessories, with a flywheel oneach shaft and the machine coupled to one of the shafts, in the calculation.It is important that the torque transmitted from one crankshaft to the othervia the gear wheels never changes sign, since if it does the effect of the backlashmakes the engine run noisily. It will be assumed in these calculations thatthe torsional stiffness of the drive mechanism is infinitely large compared tothat of the coupling. The damping of the various parts of the engine will notbe taken into account.

11.8.2. Motion of the crankshaftsThe calculation of the motion of the crankshafts is based on the system

shown in fig. 29, where the crankshaft A is drawn in the same line as crankshaftB for the sake of clarity. The moment of inertia of the crankshaft A consistsof the sum of the effective moment of inertia of half the drive mechanism, Je(equation 71) and the moment of inertia of the flywheel, hl. The moment ofinertia of the crankshaft B is equal to the sum of Je and the moment of inertiaof the flywheel of B, JF2. The shaft B is connected to the machine to be driven,

-62-

D, via a coupling C of torsional stiffness k. The moment of inertia of the ma-chine is Jo, and the torque acting on it is TI/"which is expressed as a functionof the shaft angle cp of the driven machine, of period 2n.

Fig. 29. Sketch of the coupling of the driven machine to the rhombic drive mechanismwhen all the power is taken from one crankshaft. JFl = moment of inertia of the flywheel onthe symmetrical half A of the drive mechanism. Je = effective moment of inertia of onesymmetrical half of the drive mechanism. JF2 = moment of inertia of the flywheel on thesymmetrical half B of the drive mechanism. C = coupling. Jo = moment of inertia of thedriven machine. T'P = half of the torque due to the gas forces, inertia forces and frictionforces of the engine. T = torque transmitted by the gearwheels. TIP = the torque of thedriven machine D. 1JI = crank angle of the engine. rp = shaft angle of the driven machine.

The torques acting on A are Tg, Tt and T. The torque due to the gas forces,Tg, is made up of the torque due to the gas forces of circuit and crankcaseT ge (equation 59) and the torque due to the gas forces of the buffer space Tgb(equation 61):

Tg = Tge + Tgb (79)

We will replace the friction torque Tr (equation 78) by its mean value Ttfor the purposes of calculation, since the variation of Tt is very small comparedto that of Tg, and is moreover not accurately known. The expression of theeffective moment of inertia Je as a function of 'Ijl is already known (see equation71 and fig. 34). The torque on the gear wheels is T, so we can write the equationof motion of A as follows:

Tg- Tt- T = i_ [l(Je + hl) ¥,2]d1p

T fT' T 1 dJe ·2 + J .. + J ..g- 1(- = '2' d1p 'Ijl e'ljl Fl1p

or

Je consists of a, constant part and a variable part, so the mean value of Je inthe term Je ip can be split off and combined with the term J Fl ij;.

Tg- Tr- T- ,}~~,¥,2_ (Je- Je) ij; = (hl + Je) ip (80)

We can then neglect the term (Je - Je) ij; in comparison with the muchlarger term (J Fl + Je) ip, which contains the moment of inertia of the flywheel.In the steady state the crank angle 1p may be written:

'Ijl = Ll1p + 'Ijlm (81),

where 'Ijlm = rot (ro being the mean angular velocity and t the time), and Ll'ljlis a periodic function of the time, of period 2n/w. The coefficient of non-

-63-

COmax-COmin .uniformity of rotation tJ (defined as tJ = _ ) is assumed to be

cosmall, so Ll1p is also small.

The influence of LI'1fJon the left-hand side of (80) can therefore be neglected,and this expression written as a function of 1jJm*).

The term HdJE/d1jJ) 'ljJ2is known, being the torque due to the inertia forces atconstant angular velocity, Tm'W(equatioi172b) **). The torque(Tg- 'fr- Tm6i)will be indicated by T 'I'm; this is already known. If we now write

hl + Je = J1 (82)

then equation (80) becomes:

T",m- T = hLliji

We can also write ep in a similar form to that for 1fJ:

(83)

(84)

where çs, = wt-£xc(£xc being the angle of twist). The torque Ttp may be thoughtof as given as a function of epm.

The equation of motion of B can be written in a similar way to that of A:

where

Ty•m + T - k(Ll1jJ- Liep +£xc)= JzLliji

J2 = Jra + Je

(85)

(86)

and the equation of motion of D is:

(87)

Since we have the equations of motion of the three bodies A, Band D, wecan solve for Ll1fJ.Addition of (83) and (85) gives:

2T.pm - k(Ll1fJ- Liep + £xc)= (h + J2) LIijl

while addition of (87) to this gives:

2 Ty•m - T'l'm = (h + J2) LIijl + JoLig;

It follows from (88) that:2 (Ji +J2) "

Lip = :11fJ- k T.",m+ k Ll1fJ+ £Xc

: (88)

(89)

(90)

*) The value of w~ given in equation (106) is usually accurate enough, but where the degreeof nonuniformity of rotation is large, the effect of the variation of the angular velocity onthe torque due to the inertia forces has to be taken into account by making use of a successiveapproximation. The value of w'" is first calculated from equation (106), making use o! theapproximations mentioned above, and isused to calculatetheterms jfdîsjde) ip2 and (Jo-Je) IJl.The torque Tlpmis now equal to Tg - Tr - T:n (see eqn. 72a), and this value of the torque isnow used to calculate w'I'. again, and so on. .

•• ) For a detailed treatment of this, see ref. 4)

-64-

Lt·· - Lt·· . 2 ï '+ ClJ + J2) iJiP'P - 'ljJ-i( >pm k

Substitution of (91) in (89) gives:

.. 2Jo·· Jo ....2T'/'m - T",m= (Jo +Jl + J2) Lt'ljJ-T T>pm+ k (Jl + J~) Lt'ljJ

Then:

T./,mis known in the form of a series which can be written as:<Xl <Xl

Tv/m= ~ an cos n'ljJm+ ~ bn sin n'ljJmn=ö n=l

Differentiating twice, we obtain

[

<Xl coïv/m= - w2 ~ n2an COSn'ljJm+ ~ n2bn sin n'ljJm]

n=l n=lThe torque T ",mof D is given by:

<Xl <Xl

T",m= ~ c~ cos n'Pm + ~ d~ sin n'Pmn=o n=l

or, putting 'Pm = 'ljJm-/Xc<Xl <Xl

T",m= ~ Cncos n'ljJm+ ~ dn sin n'ljJmn=o n=L

where cn = C~lcos niXc- d~ sin niXc

and dn = c~ sin niXc+ d~ cos niXc

Since in the stationary state kiXcis equal to the mean value2T>pm,it follows that: .

Co 2ao/Xc=Ï(=T

All the known terms in equation (92) are now written as series.Let us now write Lt'ljJas:

<Xl <Xl

LI'ljJ= ~ Xncos n'ljJin+ ~ Ynsin n'ljJmn=l n=l

(91)

(92)

(93)

(94)

(95)

(96)

(97)

(98)

of T9'm or of

(99)

(100)

where the coefficients Xn and Yn are as yet unknown. Differentiating twicegives:

Ltip =- w2 [ ~ n2xn cos n'ljJm + ~ n2 Ynsin n1pm]n=l n=l

and differentiating four times:

iJiP = w4 [ ~ n4xn cos n'ljJm+ ~ n4 Ynsin n'ljJm]n=l n=l

-65-

(lOl)

(l02)

Substituting the series (93), (94), (95), (101) and (102) into equation (92),we can solve directly for the coefficients Xnand Yn:

cn + 2an {(~o,)2 -1 )o

Xn= - ---------J (wn)2 {(~:)2-1 )

dn+2bn {(:o,) 2- 1 }o

Yn= - --------J (wn)2 {(~:r-1 )

(104)

where

(wo is the natural angular frequency of the engine + machine). Lbp is thusdefined as a function of t. By definition,

(81)differentiating once gives

w'" = Ll'IjJ + wwhere w'" is the fluctuating angular velocity of the crankshafts.Expressing this in series form,

(105)

w,,, = w [ 1 - ~ nx-, sin n1pm+ ~nYn cos n1pm] (106)n=l n=l

where the values of Xnand Ynare given by equations (103) and (l04).

IJ.8.3. The torque on the gearwheelsNow that we know Ll1p as a function of the time, the torque T transmitted

by the gearwheels can be obtained directly by solution of the equation ofmotion for A:

T = T'I'm-hLlip

Use of equations (93) and (lOl) gives:

(83)

co coT = I:Pn cos n1pm+ I:Qn sin n1pm

n=o n=l(107)

where P~ = an + h(wn)2 Xn .

Qn = bn +h (wn)2 Ynand

-66-

Hence P _ + h (co - 2ao)0- ao J (l08)

And according to (99),

so (l08) becomes:

(107) therefore becomes:

Co= 2ao,

Po = ao.

where

co coT = ao + ~ Pn cos n'ljlm + ~ Qn sin n'ljlm

n=1 n=1(109)

and

r, [cn ~ 2an {(:~r-I}]o

Pn = an - -----------

J {(~:r-I}Jl [dn + 2bn {(:nr -I)].

oQn = bn - ---------

J {(~:r-I}

(110)

(111)

Remark. It follows from equations (110) and (111) that when, the flywheelsare equal (i.e. Jl = J2) and the torsional stiffness ofthe coupling is low (w~ -+ 0)or the speed of the engine is high (w -+ 00), the torque transmitted by thegearwheels, and therefore the force on the teeth, becomes constant. This isalso the case ifthe moment ofinertia ofthe flywheels is very large (Ji = J2 -+ 00).

II.8.4. Determination of the size of theflywheels

The minimum size of the flywheels depends on how large the coefficient ofnonuniformity of rotation may be. It is also necessary to know how thetorque of the driven machine acts on the engine. Strictly speaking, we canonly talk of the coefficient of nonuniformity of rotation of the engine in con-junction with a given machine driven by the engine. If the engine is consideredas a separate entity, then the coefficient of nonuniformity of rotation mustbe understood as referring to the engine working against a given constanttorque.

In the present case, the above-mentioned'definitions mean that:"\,

So:

and -2bnYn = (wn)2 (h + J2)~,

-67-

Substituting into equation (100):

_ 2. [00 an 00 bn . ]Lhp = -2 (J + J) Z 2 cos n1pm+ ~ 2sm n1pm

W 1 2 n=Ln n=l n

Differentiating once:

A • 2 [ ~ an . ,,00 bn ]LJ1p= _ (J + J) £J - sm n1pm- £J - cos n1pm

W 1 2 n=L n n=L n(112)

or(113)

where ~ represents the series jn brackets in equation (112).The coefficient of nonuniformity of rotation of the engine is defined as:

(j = WIJl max - WIJl mlnjjj

It follows from equation (105) that this can be written as:

(j = Ll1jJmax - Ll1jJmlnjjj

(114)

~ follows from (82), (86), (113) and (114) that the moment of inertia of thetwot flywheels is:

J + J - 2 [( ~max- ~mln)Fl F2 - w2(j

(115)

where ~max and ~mln are the maximum and minimum values of ~.If we consider the nonuniformity of rotation of the engine together with

the driven machine, the value of (JFl + JF2) obtained from equation (115) canbe substituted into equation (106) in order to check the coefficient of non-uniformity of rotation from the variation of wIJl.

11.8.5. Motion ofthe driven machineIn fact, what we are really interested in is the coefficient of nonuniformity

of rotation of the driven machine, and the calculated coefficient of nonunifor-mity of rotation of the engine can be used to test whether the assumptionsmentioned in II.8.2 are justified. If the coupling is not solid (infinitely stiff),the coefficient of nonuniformity of rotation of the driven machine may incertain cases differ considerably from that of the engine. The angular velocity,of D (fig. 29) should therefore be determined as a function of time, in order tocheck the value of (j for D.

(90)

-68-

Filling in the known series, we obtain:

OCJ [{ (wn)2} 2an]LIep= n~1 1- w" Xn- T cos ll1Pm+• 0 .

OCJ [{ (Wn)2} 2bn].+n~1 1- w" Yn-T smn'ljlmo

I where

" 11kWo = Vh+J;

By definition, ep= LIep+ (wt-ac)and differentiating, we have:

wp = LIep + wwhere wp is the fluctuating angular velocity of the driven machine D.This can be written in series form:

[

OCJ -2 2wp = w 1~~1n [{I- (::) }Xn- ~n] sin n'ljlnr+

1I.8.6. The reactio~force and reaction torqu'f on the base

It is of interest to know the forces and torques which this drive mechanismexerts on the supports, which might give rise to undesirable vibrations.If we neglect the play in the bearings, then thanks to the complete balancing

there is no variable force present, but only a constant force due to the weightof the engine. In order to determine the reaction torque, we must rememberthat the crankshaft A in fact rotates in the opposite direction to B (fig. 29).The reaction torque is equal to minus the sum of the component torques.If we take the direction of rotation of B as positive, then using equations

(83) and (85), the reaction torque Tr is found to be:

Tr = (h-J2)Llip-k(LI'Ijl-Llep)-kac (119)or in series form:

Tr = - 2T = - 2 (ao + -£ Pn cos n'ljlm + £ Qn sin n 'Ijlm) (120)n=L 11=1

where Pn and Qn are given' in equations (110) and (111).

It obviously makes for simplicity of manufacture if both the flywheels arethe same (i.e. h = J2), and it may be seen from equation (119) that it is alsobest to make h equal to J2 on account of the vibrations on the base. Theremaining variable torque, - k(LI'Ijl - LIep), can be minimized by making ksmall, i.e, by using a coupling ofIow torsional stiffness.

-69-

(116)

(84)

(118)

11.9. Specimen calculation for a single-cylinder hot-gas engine

As an example of the method of calculation, we will consider the enginewhich is discussed further in section Ill. All quantities are expressed in C.g.S.units.

II.9.1. Nominal data and determination of numerical quantities

Dp - 8.8 cm dp = 2.8 cm

Dd 8.8 cm dd = 1.6 cmr 2.65 cm dpp = 2.6 cm

e 3.92 cm ddP = 1.7 cm

L 8.8 cm dep = 4.8 cm

Vbe 3700 cmê des = 5.6 cmmp 4068.5 g del = 5.0 cm

md 4074.5 g ape = 1.75 cmmpe 1219.0 g ade = 2.15 cm

mde 980.0 g ipe = 5.01 cm

me = 15150.0 g ide =4.93 cm're 1.097 cm ie = 4.54 cm

Working gas: hydrogenMean pressure: 104 X 106 dyne/cm-n = 1500 and 2500 r.p.m. *)I

From the data given above, we can directly calculate the following:À. = 0.301 Ad 60.82 cm2.e = 1.478 Ap 60.82 cm2

f = 1.0000 Ad 58.81 cm2

f' = 1.0340 A~ 58.81 cm''A~ 54.66 cm2

The values of Bn corresponding to the above values of À. and ê are foundfrom appendix 11.

Bo = 2.8634Bi = 0.5266B2 =-0.1126

B3 = 0.0105B4 =-0.0021B5 = 0.0005

We can now determine the following quantities (Il.3.6):. ;, Vo = 364.3 em3 YE = 62° 14'

w 0.901 V~iE= - 9.2 cm"rpvo = 117° 46' V~c = -;- 32.2 cmê

The heat exchangers can now be designed knowing the quantities givenabove, which are obtained by taking the first harmonics, and the various dead

*) The values for 2500 r.p.m. are given in brackets.

-70-

spaces which are included for constructional reasons. When the dimensionsof the heat exchangers are known, PE and pc can be determined by thermody-namic and aerodynamic calculations, if the mean pressure and the speed of theengine ~re known. These calculations are based on an "undisturbed" pressure,which is "disturbed" by the pressure differences caused by the flow resistance.The mean value of the undisturbed pressure is called simply the mean pressureof the engine.

Pressure in the expansion space:

PE = 104.14 X 106 dyne/cm-(1.04.27 X 106)

PEl = 0.2700; PE2 = -0.0441; PE3 -0.0048; PE4 = 0.0006(0.2499) (-0.0430) (-0.0033) (-0.0008)

PE5 = -0.0001(0.0008)

qEl = -0.1629; qE2 = -0.0372; qE3 = 0.0035; qE4 = 0.0009(-0.1563) (-0.0404) (0.0044) (0.0015)

qE5 = -0.0004(-0.0004)

Pressure in the compression space:

pc = 103.96 X 106 dyne/erns(103.84 X 106)

pC! = 0.2752; PC2 = -0.0442; PC3 = -0.0065; PC4 = 0.0014(0.2626) (-0.0432) (-0.0075) (0.0012)

PC5 ;= -0.0003(0.0002)

qCl = -0.1733; qC2 = -0.0329; qC3 = 0.0035; QC4 = 0.0003(-0.1817) (-0.0297) (0.0044) (-0.0003)

QC5 = -0.0002(0.0001)

Pressure in the buffer space (II.4.2):

VbO = 327.4 cm3

bo 0.9501ba = -0.0996

ab = 11.3fh = 0 (single-cylinder engine)N = 1 (single-cylinder engine)

" = 1.4 (hydrogen) RI = 1R2 = 1

Hence: Cl 0.0425C2 = -0.0042

d = 1.0015Yl = 62°14'Y2 = 0°

-71-

The coefficients of the series expansion of the pressure in the buffer space arethus:

. Pbl = -0.0276Pb2 = 0.0052Pb3 = -0.0001

qbl 0.0523qb2 = -0.0012qb3 0.0003

The mean pressure in the buffer space is still to be chosen. Two pressurelevels can easily be maintained:1. The buffer pressure approximately equal to the minimum pressure of

the circuit; this may be achieved by connecting the buffer space and thecircuit via a check valve.2. The mean buffer pressure is equal to the mean pressure in the compression

space; this can be achieved by connecting the buffer space and the circuit bya capillary tube.The engine considered here uses the latter method, so Pb = Pc.The crankcase is at atmospheric pressure: Pee = 106 dyne-erns,

Il.9.2. The torque T, + Tt for w = constant (Il.5)

a. The torque Tgc due to the gas forces of the circuit and crankcase (11.5.3)

The following ratios can be calculated from the data given above:g = 1.0017 g" = 0.0096

(1.0041)f = 1.0000 f' = 1.034

a' = 1.0348(1.0372)

f" = 0.9294a" = 0.9639

(0.9614)

.}rpeAp = 8101 X 106 dyne-cm(8092)

* 0 * 0Pn, Pn, Qn and Qn can now be calculated from the coefficients pEn, pen, qEnand qCn:

pt = 0.2797 ; P~ = 0.2707 ; Qt = -0.1688 ; Q~ =-0.1778(0.2595) (0.2657) (-0.1623) (-0.2009)

P; = -0.0457 ; P~ = -0.0427 ; Q; = -0.0385 ; Q~ = -0.0273(-0.0447) (-0.0417) (-0.0420) (-0.0174)

P; = -0.0050 ; pg = -0.0080 ; Q; = 0.0036 ; Qg = 0.0034(-0.0034) (-0.0116) (0.0046) (0.0042)

pt = 0.0006 ; pg = 0.0016 ; Qt = 0.0009 ; Q~ =-0.0003(~0.0008) (0.0032) (0.0016) (-o.oq22)

P; = -0.0001 ; pg = -0.0005 ; Q; = -0.0004 ; Q~ 0.0000(0.0008) (-0.0004) (-0.0004) (0.0006)

-72-

These values and those of Bn can then be used to calculate Ce and Dn:Co = 0.0963

(0.0790). Cl = 1.0252

(1.0333)C2 = 0.1823

(0.1789)Ca = -0.0345

(-0.0406)C4 = -0.0037

(-0.0016)C5 = 0.0005

CO.0002)

0.4675(0.4439)

D2 = -0.2218(-0.2182)

Da = -0.0316(-0.0326)

D4 = 0.0028(0.0029)

D5 = 0.0011(0.0014)

We thus obtain the following expression for Tgc [dyne-cm]:

:rgC X 10-6 =780 + 8305 cos 1p + 1477 cos 21p- 279 cos 31p- 30 cos 41p + 4 cos 51p •••(639) (8361) (1447) (- 329) (- 13) (+ 2)

3787 sin 1p- 1797 sin 21p - 256 sin 31p + 23 sin 41p + 9 sin 51p...(3592) (- 1766) (- 264) (23) (11)

.oef

h

\ IT~"} \ /

\ l1:-N \ .J- r-,D<,-<, V '\-ji'-"

"--

:0000

.IOOD

-40000 30 60 90 120 t50 1110 210 240 270 300 3JO 360_crri ....... ,dc'.

Fig. 30. The torque due to the gas forces(circuit and buffer space) and the torquedue to the inertia forces (constant angularvelocity) on one crankshaft at 1500 r.p.m.

oooo.d

OOD

0001\ J I\ t: I 1/

1000

\ -,f. \ j -,D

1\ x_ / ',,\1000

/.._.

. o 30 60 90 IlO t50 tlO 210 2.a :no 300 330 360_ttan. ontI''' .• t· 2712

Fig. 31. As fig. 30, at 2500 r.p.m.

b. The torque Tgb due tà the gasforces ofthe buffer space (Il.5.4)

g' = 1.0000-:-lrpcA~g'f" =- 7529 X 106 [dyne-cm] *).

Cbn and Dnn can be calculated from the known coefficients pbn, qbn and Be ;*) The change in Pc from 1500 r.p.m. to 2500 r.p.m. can be neglected. .1

-73-

CbO = 0CbI = 0.9964 Cb2 = -0.0276 Cb3 = 0.0085 CM = -0.0007 Cb5 = 0.0002Dei = 0.5277 Db2 = -0.2063 Db3 == 0.0340 Db4 = -0.0094 Db5 = 0.0027whence the following expression for Tgb [dyne-cm]:T~b X 10-6 =

, -7502 cos 'ljJ + 208cos 2'ljJ- 64 cos 3'ljJ + 5 cos 4'ljJ- 2 cos 5'ljJ •••

-3973 sin 'ljJ + 1553sin 2'ljJ - 256 sin 3'ljJ + 71 sin 4'ljJ - 20sin 5'ljJ •••

The torque Tg = Tgc + Tgb is given in figs. 30 and 31.

c. The torque Tm;;; due to the inertia forces (II.5.5)mo - 4553g z = 0.0719Jpc = 15554gcm'' w2 = 24674rad2fsec2Jdc = 9800gems (68539)

The values of Hn corresponding to the known values of À. and e are foundfrom Appendix IV:Hl =-0.1276 H2 =-0.6196 H3 = 0.1145 H4 =-0.0416 H5 = 0.0117So Tmw [dyne-cm] is:Tmw X 10-6 = 54sin 'ljJ -265 sin 2'ljJ -145 sin 3'ljJ + 70sin 4'ljJ -25 sin 5'ljJ •••

(150) (-736) (-403) (195) (-69)see figs. 30 and 31.

4ooo)l1::/'

30001\, I /',

?NV

\\ n." \0 ~ 1//n.os lo 1/ ,x.. \ ,

- I .~ = --\ PI~ I

I

\, 1/ \ W\

\ 1\ -,-1

'-I~ 1 \ I

\,--~

I I Io ~ W ~ ~ ~ ~ ~ ~ = ~ = ~--..crank eng!.:11' ,dtg. 2713

Fig. 32. The resultant torque due to gas forces and inertia forces acting on one crankshaft at1500 and 2500 r.p.m. (see also figs. 30 and 31).

d. Combining the results ofa, band c, we obtain the torque on one crank-shaft due to gas and inertia forces, (Tr + Tr) [dyne-cm]:

(T, + Tr) X 10-6 =780+ 803cos 'ljJ+ 1685cos 21/)-343cos 31p-25 cos 4'ljJ + 2 cos 5'ljJ •••

(639) (859) (1655) (-393) (-8) (0)-240 sin 'ljJ + 21 sin 2'ljJ -367 sin 3'tp+ 24 sin 4'ljJ + 14sin 5'ljJ •••

(-531) (523) (-117) (-101) (60)

-74-

-The variation of the torque T, + Tr with at 1500 and 2500 r.p.m. is shownin fig. 32.

Fig. 33 shows the influence of the size of the buffer space on the torqueT, + Tr at 1500 r.p.m. Curves are shown for ab = 11.3 (this example), ab = 2and as = 00.

90 120 lSO lBO 210 240 2_... crankonglc '" ,«kg.

300 3 ,•. 27,14

Fig, 33. The effect 'of the size of the buffer space on the resultant torque due to gas forces andinertia forces on one crankshaft at 1500 r.p.m.

11.9.3. The effective moment of inertia Je (11.5.5)

Je as a function of 1fJ is directly given by equation (71). From Appendix IV,Ho is found to be 0.6660; the other coefficients and quantities have alreadybeen found for the determination of the torque due to the inertia forces.r &-c~

0.4

o.

,te!

-Î"'-- .__ ..-/ <, --- ..-/

2

•t

0.3

o

°0 30 60 90 120 tSO 180 210 240 270 300 330 360_crQ"kang1c~.dcg. 2115

Fig. 34. The effective moment of inertia of half of the drive mechanism as a function of thecrank angle lp.

-75-

Je [g-ems] is thus:Je X 10-6 = 0.3472 - 0.0044 COS 1p + 0.0107 COS 21p + 0.0039 COS 31p +

- 0.0014 cos 41p + 0.0004 cos 51p ...Je is plotted as a function of 1p in fig. 34.

11.9.4. Calculation oftheforces occurring in the system (11.6.5) .

a. Forces on the piston drive mechanism [dynes]

Fp X 10-6 = 432 + 1895 cos 1p - 300 cos 21p - 39 cos' 3'1/)+ 9 cos 4'1/) ••.(1763) (-293) (--45) (7)

- 1357 sin 1p - 194 sin 2'1/) + 20 sin 3'1/)+ 2 sin 41p •••(-1408) (-175) (26) (2)

Xp =-65386 sin 1p + 34432 cos '1/)-29450 cos 2'1/)+ 6179cos 31p-2197 cos41p+ 817cos 5'1/).

(-181628) (95645) (-81815) (17164) (-6103) (2270)

Table I shows values of the forces calculated at various crank angles

TABLE IThe forces on the piston drive mechanism at various crank angles [dynes]

'Ijl 0' 30' 60' 90' 120' 1500 180' 210' 240' 270' 300' 330' I

Fpl X 10-6 951 557 156 -242 -574 -751 -681 -392 175 870 1271 1250(878) (584) (235) (-138) (-415) (-460) (-373) (-305) (15) (559) (963) (1046)

Fp2 x 10-6 138 105 50 -115 -416 -730 -743 -388 119 416 385 232(126) (112) (78) (-56) (-282) (-409) (-362) (-297) (-19) (246) (280) (190)

Fp3 x 10-6 959 571 171 -259 -699 -1034 -996 -550 199 935 1308 1260(880) (606) (266) (-126) (-473) (-579) (-485) (-421) (-32) (554) (949) (1031)

Fp5 x 10-6 74 -254 -180 168 630 934 679 77 15 773 1211 756(-51) (-378) (-358) (-290) (293) (382) (185) (-69) (-lOO) (319) (748) (494)

Fp6 x 10-6 949 539 125 -115 -80 269 663 532 -183 -416 291 956(871) (572) (195) (-56) (-54) (151) (323) (407) (29) (-246) (211) (783)

b. Forces on the displacer drive mechanism [dynes] .

Fd X 10-6 = 220 + 27 cos 'I/) - 9 cos 2'1/)+ 10 cos 31p - 5 cos 41p •••(235) (-19) (- 9) (25) (- 13)

+ 28 sin 'I/) - ~4 sin 2'1/) + 1 sin 31p + 4 sin 41p ...(119) (- 7~) (1) (11)

Xd=-65386sin '1/)-34432 cos 1p+ 29450cos 2'1/)-6179 cos 31p +2197 cos4'IfJ-817 cos 5'1/)~

(-181628) (-95645) (81805) (-17164) . (6103) (-2270)

-76-

, ,

The values of the forces calculated at various crank angles are given intable 11.

TABLE IIThe forces on the displacer drive mechanism at various crank angles [dynes]

'IJ! 0' 30' 60' 90' 120' ISO' 1800 210' 240' 270' 300' 330'

'dl X IQ-6 -141 -213 -257 -315 -264 -98 65 77 25 -21 -60 -92(-166) (-384) (-594) (-698) (-574) (-129) (314) (354) (249) (157) (76) (-12)

id2 x IQ-6 -21 -42 -93 -167 -206 -95 92 94 27 -6 -16 -17(-25) (-78) (-201) (-378) (-455) (-123) (405) (403) (212) (90) (30) (0)

ida X 10-6 -145 -228 -316 -376 -346 -135 124 133 46 -13 -55 -90(-174) (-423) (-680) (-849) (-765) (-174) (543) (571) (354) (203) (100) (-1)

id5 X 10-6 -69 -196 -356 -385 -186 -13 -141 -176 -94 -36 -11 -18(-158) (-408) (-796) (-893) (-437) (-88) (-538) (-684) (-485) (-315) (-201) (-133)

id6 X 10-6 -143 -173 -70 167 317 130 -83 -35 5 -6 -40 -85(-172) (-321) (-151) (378) (702) (168) (-362) (-149) (40) (90) (73) (1)

c. Forces on the crankshaft [dynes]

F7 and Fs can be calculated directly from the values of Fp3 and Fd3; seetable Ill.

TABLE IIIThe forces on the crankshaft at various crank angles [dynes]

'IJ! 0' 30' 60' 90' 1200 ISO' 1800 210' 240' 270' 300' 330'

:<7 X 10-6 415 -40 -126 193 854 1331 948 311 331 1147 1610 1148(928) (351) (-17) (180) (993) (1431) (784) (384) (462) (1141) (1684) (1498)

Ps X IQ-6 1092 712 195 -282 -397 139 746 567 -188 -410 331 1041(1043) (893) (346) (-434) (-756) (-17) (685) (556) (-11) (-336) (138) (782)

The resultant forces on the various pins, shells and bearings are shown inpolar diagrams (figs. 35-42) for .the parts of the drive mechanism sketched infig. 28. The fullline represents the forces at 1500 r.p.m., and the broken linethe forces at 2500 r.p.m.

11.9.5. An estimate ofthe torque Tt and thefrictionpower Pr (11.7)

As was mentioned in 11.7, there are various ways of connecting the pistonconnecting rod and the displacer connecting rod to the crank pin. The con-struction used here is with the crank-pin bearing shell :fixedin the big end ofthe piston connecting rod.The value of Gr is found from Appendix III to be 1.15. D = 1.13, and

yE = 62°14'.The ratio of the various friction torques to the corresponding friction force

-77-

b slxlX!o'Force scale @vnc:s]

Fig. 35. Polar diagram indicating the magnitudeand direction of the resultant force of the piston

connecting rod on the piston yoke pin.

500Xld'

Force scale ldync:~

Fig. 37. Polar diagram indicating the magnitudeanddirection of the resultant force of the displacerconnecting rod on the crankpin bearing shell of

the piston connecting rod.

-78-

2716

POLE

o 50ox,06Force seak @yne~

2717

Fig. 36. Polar diagram indicatingthe magnitude and direction of theresultant force of the displacer con-necting rod on the displaceryoke pin.

l..:t!, '.

.tI·l t"""'I I, ,

!I' I·30·r :, • II II II II

o 500 x ,rJForce scale ~M:~

2718 2119

Fig. 38. Polar diagram indicating themagnitude and ·direction of the resultantforce of the crank-pin bearing shell on

the displacer connecting rod.

,,r,,

/ O'----~--~~SOOXIc/'

" Force: sealt ~YM~

/',,.,,,,,·:,~~O·

-'--r..r--------'

o

Fig. 39. Polar diagram indicating the magnitude and direction of the resultant force of thepiston connecting rod on the crank pin.

Fig. 40. Polar diagram in-dicating the magnitude anddirection of the resultantforce of the crank pin on

"" the crank-pin bearing shellofthe piston connecting rod.

----',"" "~---,,,,

\\\.IIIIII,

I

,;1210-

,/

3Q'

,,~i!..."''''- ':"'\''Ï,ff

.. ,'" "" ,00/,:' " 3d'" 330~'I ,, ": '

IIII\\\\,,,,

" ,120~"~',--~':-:-

...................

,,,,\

'oi?7C·40· \ ,

--f+.HI+---t,~ ---._--,,'\ ISO-~').'.!< t! ..\

:lA,!G ............... _ ....:iI.. ':

'10

oForce sealt [dynn]

-79-

---"'",,,,,

\00"'i\IIIIIIIII

-f---I,,,,

I,J'(&tJa

2720

~~-boXIc/'Force: scale: @ync:~

2722

10·,sd';;,,I,,r!,O·,,,

o ~Xlab ~td

Pqrce seete &ync~ ~'I20.

2723 ·i'oo·

0'

90'

Fig. 41. Polar diagram indicating the mag-nitude and direction of the resultant forceof the piston yoke pin on the small end of

the piston connecting rod.

Fig. 42. Polar diagram indicating the mag-nitude and direction of the resultant forceof the displacer yoke pin on the small end

of the displacer connecting rod.

or to the coefficient of friction can now be calculated directly as a functionof the crank angle with the aid of the values of the forces calculated above.These ratios are given in table IV.

Assuming the friction force to be constant, we obtain the following mean *)values for the piston and the displacer respectively [cm]

(TrPl)Frp av

0.94(0.94) , (

Trdl) 0.87Frd av = (0.87)

And assuming that the coefficients of friction are constant, we have thefollowing mean values for the bearings and the gearwheels [dyne-cm]:

*) These are averages of the values given in table IV, which in this case is a sufficiently goodapproximation to the mean with respect to time.,

-80-

TABLE IVThe ratio of the friction torque to the coefficient of friction [dyne-cm], and to the friction force [cm], at various crank angles

00....

lp O' 30' 60' 90' I 1200 I 1500 I 1800 I 2100 I 240' I 270' I 300' I 3300

TrPi 1.325 1.496 1.266 0.698 0.058 0.799 1.325 1.496 1.266 0.698 0.058 0.799Frp (1.325) (1.496) (1.266) (0.698) (0.058) (0.799) (1.325) (1.496) (1.266) (0.698) (0.058) (0.799)

Trp2 0 128 64 121 276 236 0 124 83 434 518 286-- (0) (134) (97) (67) (196) (139) (0) (96) (10) (275) (319) (239)Pp2

T(P3 2621 2371 1909 1150 1859 2621 2208 1143 343 1322 2092 2532-- (2554) (3351) (3759) (4071) (2402) (828) (1850) (1857) (1134) (527) (948) (1708)Pp3

Trdi 0 0.698 1.208 1.395 1.208 0.698 0 0.698 1.208 1.395 1.208 0.698Frd (0) (0.698) (1.208) (1.395) (1.208) (0.698) (0) (0.698) (1.208) (1.395) (1.208) (0.698)

Trd2 0 32 69 105 85 20 0 18 9 6 16 14-- (0) (58) (160) (234) (187) (26) (0) (79) (83) (53) (21) (2)Pd2,

T(;3 0 253 610 814 618 127 0 174 158 72 70 84'" (0) (503) (1360) (1881) (1389) (183) (0) (679) . (818) (636) (359) (129)Pd3

Tr4 2920 1785 580 855 , 2355 3345 3015 1618 950 3045 4108 3875-- (3490) (2398) (865) (1175) (3120) (3578) (2603) (1690) (1155) (2973) (4225) (4225)p4

Trs 780 (constant)-C- (639) ,

---------

(Trp2)' 189 x 106{lp2 av = (137 X 106); (

Tep3) 1847 X 106 (Trd2) 31 X 106---;;;- I).V = (2082 X 106); {ld2 av = (75 X 106)

(Tfd3) = 248 X 106 •

/J" (661 X 106) ,,-d3 av

(Tf4) 2371 X 106{l4 ILV = (2625 X 106);

Tr5 780 X 106Y = (639 X 106)

This shows that in order to reduce the friction torque great care shouldbe taken with the construction of the bearings with coefficients of friction {l3

and {l4, in order to minimize the friction at these points in particular.In order to get an idea of the order of magnitude of the mechanical losses,

we will suppose that the drive mechanism is fitted with roller bearings andball bearings, whose coefficients of friction are roughly constant and areknown 3). According to these authors we may take the following coefficientsof friction for well lubricated bearings:a. self-aligning ball bearings {l = 0.0015 (for the main bearings of the crank-

shaft)b. needle bearings {l = 0.007 (for all the other bearings of the drive mechanism).

According to Horgen 6) the coefficient of friction of the piston rings in ourcase (4 piston rings, axial height 3.5 mm, viscosity of lubricating oil 35 centi-poise and zlps-r = 4.7 X 106 dyne-cmêj is zz= 0.10 at l500r.p.m. and jz = 0.12at 2500 r.p.m.

This corresponds to a friction force of 16 X 106 and (20 X 106) dynes respec-tively.

We found by experiment that the friction force of the piston-rod seal was5 X 106 and (6 X 106) dynes, and for the displacer rod 3 X 106 and (3.5 X 106)dynes.

So Frp = 21 X 106 dynes(26 X 106)

Frd = 3 X 106 dynes(3.5 X 106)

and

If we also assume that the losses due to the coupling of the two crankshaftsby the gearwheels is 2% of the power to be transmitted when all the poweris taken from one crankshaft, then:

t; = 0.01 . Tl5 = 8 X 106 dyne-cm. (6 X 106)

The values of Tl as a function of"P calculated from these data are plottedin fig. 43, while the mean value of the friction torque is:

Tl = 51 X 106 dyne-cm(58 X 106)

. -82-

The mechanical efficiencyis thus:

'YJm = 1_: 'If = 1_ 51 (58) X 106

~·rpcA~Co 780 (639) X 106

'YJm ~ 0.93(0.91)

(trpc A~Co is the constant part of the torque due to the gas forces of thecircuit on one crankshaft).

0'1.106

0

-: r-, n.1SO rpm

1\ v-r-l'\ T, V\ / ...____

i\ lZ,; '-

0

60

so

40

20

10

°0 30 60 90 120 ISO 160 2'0 240 270 300 330 360_c:ronk anglelP,!kg. '2724

Fig. 43. The torque due to the friction on one crankshaft at 1500 r.p.m.

The mean values of the separate losses for the piston and the displacer(in Watts) are thus:

PfP1 = 317;(640)

Pfd1 = 41(80)

and for the bearings and gearwheels:PfP2 = 21 ; PfP3 = 203 ;

(25) (381)Pfd2 = 3

(14)Pf5 = 125

(157)Pees = 27;

(121)Pf4 = 56;

(103)

Il.9.6. Motion ofthe crankshaft (Il.8.2)

We will consider here the set-up used for the testing of the engine describedin section Ill. The engine drives an eddy-current brake via a coupling whichis provided with rubber balls; it is assumed that this gives a constant reactiontorque. The torsional stiffness of such a coupling depends on the torque to be

-83-

transmitted and the frequency 7). We will assume here that the torsionalstiffnessisk = 4.7 X 1010 dyne-cm/rad; moreover,

(3.3 X 1010)

}o = 4.5 X 106 g-cm>; JF! = 8.0 X 106 g-cm>; Jra = 8.0 X 106 g-cmê.

T",m = Te + Tr+- Tt; for T, + Tr see 11.9.2and for Tt see 11.9.5.T'I'm = Co = 1458 X 106dyne-cm; Je = 0.3 X 106g-ems (see 11.9.3)

(1162 X 106)

SoJl = 8.3 X 106g-cmê ; J2 = 8.3 X 106 g-ems

W~ = 102.2 rad/sec;(85.6)

Wo = 115.2 rad/sec(96.6)

0.

..0

6

/ ~'",

/~~~ 1500 .~ V2

'\ lt~ I>soo-, \V".Jn pm' '-- --- _/ - --0---

'\ "" -- --- --- V -\/

.". ;,/ / 1Jn- 500r Pm / <,2

1/ " V --I'-- ~ 1\• \

6 JV -,

•

0.

0.

-0.

-().

_0.o 30 60 90 120 150 180 2$0 240 270 300 330 360_crankongl~.,., ,cXg. 2725

Fig. 44. The deviation (Lhp) of the crank angle from its mean value and (dlp) of the shaftangle from its mean, as functions of the mean crank angle of the engine, !pm.

Hence the coefficients Xn and Yn [rad]

Xn X 102:Xl = -0.489;

(-Q.156)X4 = 0.000;

(0.000)

X3 = 0.018(0.007)

X2 = -0.212;(-0.073)

and Yn X 102:

Yl = 0.146;(0.096)

Y4 =-0.000;(0.001)

Y2 =~.002;(-0.023)

Y3 = 0.020(0.002)

-84-

This deviation from the mean angular velocity is shown in fig. 45.

The motion of the crankshaft may therefore be expressed as follows [rad]:Lhp X 102 = -0.49 cos 1jlm -0.21 cos 21jlm + 0.02 cos 31jlm •••

(-0.16) (-0.07) (0.01)

+ 0.15 sin 1jlm-O.OO sin 21jlm + 0.02 sin 31jlm •••

(0.10) (-0.02) (0.00)

The variation of .d1jl is plotted in fig. 44.

The expression for w'" derived from this is thus:

~ X 102= 100+ 0.49 sin 1jlm + 0.42 sin 21jlm -0.05 sin 31jlm •.•

(0.16) (0.1~) (- 0.02)

+ 0.15 cos 1jlm - 0.01 cos 21jlm + 0.06 cos 31jlm ..•

(0.10) (-0.05) (0.01)

I~1010, r-r-,---

V""nalSC Orpm

I~- r\ SOO'pm ,-" --,

...

I~ /, -- \ VJ ......--_,'

5

Î"--V0

lOOS

'.000

0.99

0.99 0 30 60 90 120 ISO 160 210 240 270 300 330 360_crank angle""m,<kg. 2721)

Fig. 45. The ratio of the angular velocity of the engine to the mean angular velocity, as a, function of the mean crank angle, !pm.

IJ.9.7. Motion ofthe driven machine (brake) (11.8.5)

With w;; = 53.2 rad/sec and the values calculated in II.9.6, we have:(44.6)

.drp [rad]

.drp X 102 = 0.36 cos 1jlm + 0.03 cos 21jlm .•.

(0.02) (0.00)The variation of .drp is given in fig. 44.

- 0.11 sin 1jlm •.•

(- 0.01)

The expression for wrp derived from this is:wrp. X 102 = 100-0.36 sin 1jlm - 0.05 sin 21jlm •••

jjj (0.02) (-0.00)- 0.11 cos 1jlm •••

(- 0.01)

-85-

The deviation from the mean angular velocity is to be seen in fig. 46.

1.00

~

5

.250( /V V r-.I rpm

-- -- VI'--. ~r--.. I-- ,,50( 'pm5

0.

tOl0

1000

0.99

0.99 0 30 60 90 120 tSO 180 210 240 270 300 330 360_ crank angle '+'. ,!kg. 2727

-Fig. 46. The ratio of the angular velocity of the driven machine (brake) to the mean angular

velocity, as a function of the mean crank angle 1JIm.

11.9.8. The torque on the gearwheels (II.8.3)

The torque T [dyne-cm] is:

T X 10-6 = 729 - 199 cos 'ljJm-56 cos 2'ljJm+ 5 cos 3'ljJm...(581) (-29) (-12) (1)

+ 60 sin 'ljJm-Î sin 2'ljJm+ 5 sin 3'ljJm..., (18) (-4) (0)

It may be seen from fig. 47 that even at 1500 r.p.m. the torque is alwayspositive. At 2500 r.p.m. the torque on the gearwheels is nearly constant.

'iM-tmo 30 60 90 120 tSO lBO 210 240 270 300 330 360_crankanglc '+'mo<kg, 2728

Fig. 47. The torque T on the gearwheels as a function of the mean crank angle 1JIm.

11.9.9. The reaction torque on the base. (11.8.6)

Tr [dyne-cm] is:

Tr X 10-6 = - 1458 + 399 cos 'ljJm+ 112 cos 2'ljJm- 10 cos 3'ljJm...(- 1162) (58) (25) (-2)

- 119 sin 'ljJm+ 2 sin 2'ljJm- 10 sin 3'ljJm...(- 36) (8) (-1)

..,...86-

The reaction torque at 1500 and 2500 r.p.m. is shown in fig. 48 as a functionof 1flm. Itmay be seen that the rather stiff coupling gives rise to an appreciablevariable torque on the base at. 1500 r.p.m., though not at 2500.

-.-.

ooc ,10'I---

k...""", --I----6p<>/ <, ,

"""/ -t::~~ -,

f, ..-'P<>f7 - - - - ~--- i""~

BP<>

400

-.

o 30 60 90 120 150 180 2tO 240 270 300 330 360

_cronk a~1c W. ,«g. 2729

Fig. 48. The reaction torque T r on the base, as a function of the mean crank angle 1JIm.

REFERENCES1) French Patent No. 1,219,491 R. J. MEUER2) M. TOLLE,Regelung der Kraftmachine 3. Auflage 12-14, 1921.3) M. C; SHAW and F. MACKS,Analysis and lubrication of bearings, McCraw-Hill, 10-18,

1949.4) C. B. BIEZENOand R. GRAMMEL,Technische Dynamik, Julius Springer, XII. 6, 1939.5) C. B. BIEZENOand R. GRAMMEL,Technische Dynamik, Julius Springer, XII.13, 1939.6) H. HORGEN, Versuche über Kolbenringreibung und Undichtigkeitsverluste. Thesis.

Zürich 1942.7) B. I. C. E. R. A. compiled by E. J. NESTORlDES,A handbook on torsional vibration,

Cambridge, 1958.

-87-

8.8 cm8.8 cm6.0 cm6.0 cm1500 r.p.m.300cm/min140 kg/ern»

Ill. EFFICIENCY MEASUREMENTS ON A SINGLE-CYLINDER

HOT-GAS ENGINE WITH RHOMBIC DRIVE MECHANISM.

III.1. Principal engine data

Diameter of the displacer DeDiameter of the piston DpStroke of the displacer SdStroke of the piston SpNominal speed nMean piston velocity v pPermissible gas pressureCompression ratio (ratio of the highest and lowest values of

the gas pressure in the circuit, Pmax/pmin)Nominal temperature of heater tubesNominal temperature of cooling waterNominal flow rate of cooling waterWorking gas

2.0700°C15°C

0.8 I/sec.hydrogen

The mean pressure of the buffer space is equal to the mean pressure of thecircuit.

>0

Fig. 49. The rhombic drive of the test engine in two positions.

I11.2. Constructional details

2760

As has already been mentioned in section 1I.2, an experimental engine

-88-

incorporating the rhombic drive mechanism has been built in the ResearchLaboratories of N.V. Philips, Eindhoven. This engine has been designed forhigh pressures (maximum 140 kg/ern"), and the working gas is hydrogen.Fig. 49 shows the construction of the drive mechanism, whose mode of opera-tion is illustrated by photographing it in two positions (the crankshafts andgearwheels have been omitted). The general appearance and layout of the

Fig. 50. The test engine on the test bench.

engine is shown in fig. 50, and a sectional view is given in fig. SI. The heaterwith pre-heater is drawn separately in fig. 52, for the sake of clarity. Thevarious parts of the drive mechanism are indicated with the same numbers asin fig. IS, with the addition of the counterweight, 14. The cooler, regeneratorand heater all have an annular configuration, and are mounted around theexpansion-compression cylinder. The regenerator is divided into twelve small

-89-

-90~

chambers ("cups"), in order to keep the thermal stresses in the regeneratorwalllow 1).The design of the cooler has been adapted to that of the regenerator. The

cooler housing contains a number of tubes bunched together in groups (cooler"units"), the end of each unit protruding into one of the regenerator cups. Thetubes of each cooler unit are mounted with a sliding fit into the cooler housing,in order to take up any differences in the axial thermal expansion of thecylinder 15 on the one hand and the tubes 20 and regenerator housing 18(fig. 53) on the other 2).The displacer consists of a displacer body 6a and a thermally insulating

.dome 6b. The displacer body fits into the. cylinder 16 like a normal piston,while there is a narrow slit between the dome 6b and the (cylinder) wall 15of the hot space, so that the dome never touches the cylinder wall, even whenthe displacer is hot. Under the piston we see the buffer space 13 3).The heater consists of two sets of tubes 20 and 21, arranged alternately

with a small clearance, between which the hot combustion gases from theburner are led. The tubes of the one set 20 terminate in the regenerator cups(see figs. 51 and 53); the gas coming from the regenerator flows up thesetubes and via the common annular channel 22 into the downwardly directedtubes ofthe other set 21, which are soldered into the top ofthe hot cylinder 15.A series of :fins are brazed on to the tubes of set 20 in order to improve thetransfer of heat from the combustion gases.The preheater (27 in :fig.52a) is situated around the heater. This is so con-

structed that the air necessary for combustion enters the lower end of channelsAl situated around the periphery of the pre-heater (figs. S2a and b). The airleaves these channels sideways, and passes via the spaces A between thespirally curved partitions to the channels A2 around the inside wall of thepreheater and thence to the burner. The exhaust gases, which leave the burnerwith a temperature slightly higher than that of the heater, enter the channels BIaround the inside wall and thence through the spirally curved spaces B whichare arranged alternately between the spaces A to the outlet channels B2 aroundthe periphery, i.e. in counter-current with tlie fresh charge of air. Fig. S2ashows the flow of the combustion air on the right-hand side, and that of theexhaust gases on the left-hand side. This flow is obtained by sealing the upperends of the channels Al and BI and the lower ends of the channels A2 and B2.This construction gives a very compact preheater (fig. 52c) which has a lowflow resistance, and which needs very little insulation 4). The burner (24 in:fig. 52a) is of the "swirl-chamber" type with an atomizer 25, which sends afine spray-of liquid fuel into the upper part of the burner with the aid of com-pressed air.

--17

2730

Fig. 51. Cross-section of the engine shown in fig. 50, omitting the burner and air preheater,which are shown separately in figs. 52 and 53. The spaces filled with hydrogen are shadedgrey. 1 = piston, 6a and 6b = parts of the displacer, 12 = piston-rod stuffing box, 13 = bufferspace, 14 = counterweight, 15, 16, 17 = walls of the cylinder in which the piston anddisplacer move, 18 = regenerator housing, J 9 = cooler housing, 20, 21, 22 = heater tubes,23 = fins, 26 = tube for temperature probe; for significanee of other numbers, see fig. J 5.

-91-

exhaust gases

26-- _

a

b cFig.52.

a. The upper part of the engine, with burner 24, atomizer 25 and air preheater 27. Signifi-cance of other numbers as in fig. SI.

b. Horizontal section through the air preheater with the thin spiral spaces A and B, throughwhich the air for the combustion and the exhaust combustion gases flow in counter-current.These gases are taken in via the channels Al and BI, and leave via the channels A2 and B2,in the axial direction.c. View of air preheater, which fits over the heater shown in fig. 53.

-92-

111.3. The measuring set-up

In order to reveal possible faults in the efficiency measurements or in theengine itself, and in order to get a picture of the behaviour of the engine undervarious loads, all the losses occurring in the system were determined as well as

2763

Fig. 53. Construction of the regenerator housing and heater. The regenerator is divided intoa number of small chambers ("cups") which are placed a small distance away from thecylinder (bottom of figure). Three tubes may be seen coming out of each regenerator cup:the continuation of these form the vertically ascending tubes of the heater. The commonannular channel at the top connects these tubes with the descending tubes, which are placedin between the ascending ones and which come out in the top of the hot space of the cylinder.The fins (23 in fig. SI) which are brazed on to the walls of the tubes to improve the heattransfer from the combustion gases to the tubes may be seen above the regenerator cups.

the fuel supply and the useful power, so that a heat balance could be set up.Some of the mechanical losses could not be measured, but these can be foundapproximately by difference from the heat balance.

The following quantities are involved in the balance:a. the energy supplied in the form of fuelb. the useful powerc. the energy carried off in the cooling water of the circuit coolerd. the energy carried off in the cooling water of the cylinder coolere. the energy carried off from the preheater by the exhaust gases, by radiation

and by convectionf the mechanical losses.

-93-

The power of the engine was determined by means of four measurements:1. the gas pressure in the engine Pc (see II. 4.1)2. the speed of the engine3. the temperature of the heater tubes4. the temperature of the cooler tubesThese measurements were carried out, and the load on the engine was set

to the desired value, by means of equipment whose functions are indicatedin fig. 54. The temperature of the cooler tubes was not measured, but was

Fig. 54. The measuring set-up for the test engine. 1 = scales with stopwatch, 2 = eddy-curren-brake, 3 = thermopile for circuit cooler, 4 = flowmeter for circuit cooling water, 5 = thertmocouple for inlet temperature of circuit cooler water, 6 = flowmeter for burner air, 7 =twelve thermocouples for heater temperature, 8 = thermocouple for surface temperature ofair preheater, 9 = thermopile for exhaust gases from air preheater, 10 = % C02 and %CO + H2 meter, 11 = revolution counter, 12 = therrnostat, 13 = thermopile for cylindercooler, 14 = flowmeter for cylinder cooling water, 15 = thermocouple for inlet temperatureof cylinder cooling water, 16 = pressure gauges for maximum, minimum and mean circuit

pressures, 17 = pressure gauge for mean pressure of buffer space.

deduced from the inlet temperature of the cooling water and the rate of flowof the water in the circuit cooler. The measuring equipment indicated in fig. 54is:

1. Scales and stopwatch for measuring the fuel consumption. The stopwatchis electrically triggered by the pointer of the scales, which cuts off a narrow beamof light directed on to a photocell as it passes a certain point on the scale.

2. Eddy-current brake, controlled by thyratrons: Heenan Dynamatic

-94-

Dynamometer, type GVAL Mark I; the speed of the motor is thus keptconstant at the desired level. .3. Thermopile for measuring the difference between the inle't and outlet

temperatures of the circuit cooling water. The thermopile consists of twoidentical elements, one of which is placed in the inlet pipe and the other in theoutlet pipe of the circuit cooler. Each element contains one of the junctions of9 copper-constantan thermocouples; the construction is based on a goodmixing of the water and the attainment of a constant temperature at thethermocouple junctions.4. Flowmeter for the cooling water of the circuit cooler.5. Thermocouple for measuring the inlet temperature of the cooling water of

the circuit cooler.6. Flowmeter for burner air.7. 12 thermocouples (chromel-alumel) distributed over the heater for

. measuring the mean heater tube temperature.8. Thermocouple for measuring the surface température of the preheater.9. Thermopile consisting of 4 thermocouples (chromel-alumel) attached to a

thick perforated copper plate.10. % C02 and % CO + H2 meter (Siemens).11. Revolution counter belonging to the brake 2.12. Thermostat for keeping the heater tube temperature constant; the tem-

perature probe is in the gas stream in one of the heater tubes and is led outthrough the tube 26 (fig. 52). The thermostat only controls the fuel supply,while the air supply is controlled manually.

13. Thermopile for measuring the temperature difference between the inletand outlet cooling water of the cylinder cooler.

14. Flowmeter for the cooling water of the cylinder cooler.15. Thermocouple for measuring the inlet temperature of the cooling water

of the cylinder cooler.16. Pressure gauges for measuring the maximum, minimum and mean

pressures of the circuit.The maximum and minimum pressure gauges are fitted with check valves.

The mean pressure is measured via a capillary.17. Pressure gauge for measuring the mean pressure of the buffer space.

The cold junctions of all the thermocouples are placed in melting ice.

I1I.4. ResultsThe measuring instruments were calibrated before each test, and the cali-

bration was checked at regular intervals during the test. The burner was fedwith diesel oil. The lower heat value of a sample of this fuel was determinedby the Heat Laboratory of the Centraal Technisch Instituut T.N.O., Delft, and

-95-

was found to be 10230± 30 kcal/kg. The cooling water and the burner airwere obtained from the laboratory mains; the fuel pump and the lubricatingoil pump were electrically driven. The power used for this purpose was nottaken into account in the calculation of the results.

TABLE VThe most important test results for the test engine at various pressures and engine speeds

Pressure Speed Dynamo- Circuit 1.lIl circuit Cyllnder LIt Burner AtomIzer Temp, of Ambient Prchcater Fuelpm~ r.p.m. meter cooling coottns cooltea cylinder etr nlr exhaust temp. temp. grams/sec.

kg/cm:l :1:2 kg-m water water water coolJng gramsl gramsi gases oe oe:I: 0.2 gramsfscc. oe grams/sec. water sec. sec. oeoe50 250 4.53 703.7 1.33 16.9 5.12 5.76 0.50 177 21 174 0.165

80 250 7.39 671.4 1.81 16.1 5.48 5.76 .50 179 21 176 .223

110 250 10.17 663.2 2.25 15.7 5.18 6.37 .50 177 21 172 .269

140 250 12.87 684.7 2.56 16.4 5.06 8.45 .SO 173 21 169 .320

, 50 500 4.87 703.7 1.86 16.5 4.30 7.06 .54 192 21 185 .251

80 500 8.13 703.7 2.56 16.4 4.66 7.94 .45 192 21 183 .347

110 500 11.23 667.8 3.44 15.2 5.39 12.34 .45 179 21 171 .445

140 500 14.18 728.0 3.75 17.3 5.45 12.01 .45 184 21 180 .522

50 1000 5.06 838.8 2.46 19.6 5.66 8.17 .60 163 19 164 .407

80 1000 8.42 860.2 3.48 20.8 6.34 11.84 .60 163 19 167 .600

110 1000 11.69 875.0 4.45 22.1 6.34 17.95 .60 169 22 162 .772

140 1000 14.82 860.2 5.43 21.7 7.13 20.36 .60 172 22 164 .956

50 1500 4.82 832.0 3.16 24.1 6.39 12.25 .60 159 20 162 .562

80 1500 8.10 860.2 4.90 20.9 7.85 15.67 .60 179 22 177 .827

110 1500 11.28 845.8 6.45 22.4 7.73 21.63 .60 155 19 152 1.103

140 1500 14.36 838.8 8.13 21.0 9.66 25.88 .60 175 21 167 1.368

50 2000 4.21 838.8 4.47 18.5 9.34 15.04 .62 164 19 164 0.698

80 2000 7.26 832.0 6.56 19.1 10.54 23.08 .62 175 19 174 1.046

110 2000 10.50 860.2 8.84 18.5 11.45 29.30 .62 169 19 167 1.432

140 2000 13.12 781.6 11.81 19.2 13.74 34.58 .45 194 21 189 1.765

50 2500 3.71 818.5 5.78 19.5 10.66 19.11 .57 171 20 187 0.847

80 2500 6.69 805.6 8.67 20.4 12.05 27.18 .57 171 19 187 1.303

110 2500 9.49 805.6 11.86 20.6 13.55 33.39 .57 197 20 207 1.763.

140 2500 11.44 768.6 15.74 18.7 15.98 45.49 .45 208 21 210 2.197

-96-

Measurements were carried out at four pressure levels, pmax= 50, 80, 110and 140 kg/cm», and at six engine speeds for each pressure: 250, 500, 1000,1500, 2000, 2500 r.p.m. The nominal temperature of the heater tubes was700°C, with a flow rate of 0.8 litres per second for the cooling water of thecircuit cooler and an inlet temperature of 15°C for this water. The mainresults of these measurements are given in table V. The brake horsepower andoverall efficiency calculated from these data are plotted against the enginespeed in figs. 55 and 56. The measured torque is plotted as a function of the' ,engine speed for the various pressure levels in fig. 57. The data plotted infigs. 55 and 56 are used to construct the lines of constant specificfuel consump-tion in the b.m.e.p. - n diagram of fig. 58. The b.m.e.p. (brake mean effectivepressure) is defined in the same way as for an internal combustion engine,with respect to the swept. volume of the piston:

b - 45.104• b.h.p. _ C b.h·p·[k / 2].m.e.p. - - -- gcmn 2 n4(Dp-dd) Sp'n

where b.h.p. = brake horse power (Continental) and C = 1275.

5

° V35 /- I'maLQI4 !<I/cm'

/ L25 / /

1// 7i-

j 1/ /V15

I~V .>~I

/;fy V i-:5

~,/

°

30

20

o 500 '000 isoo 2000 2SOO 3000- cnglnupctd ",rpm 2734

Fig. SS. Brake horsepower of the test engineplotted as a function of the engine speed

n, for various values of pmsx.

°Pmax.=t4

~ ~5

~<,

~.' tlO/ _"..-- Î'-

o ~ r,/. r-.5

~I/

5

°

5

° 500 1

20

° 000 1500 2000 2500-_cngincspt:lEd n,rpm 2735

Fig. 56. Efficiency (i.e. brake horsepowerdivided by the heat obtained from the fuelin unit time) of the test engine plotted as afunction of the engine speed n for various

values of pmsx. .

-97-

Idynomomt:tt:rtorqu<

I5 _kg-rn

V I---- -~. "4O'g,l:.

r---..t--... r--=t-V,._ ~

0 r-- ~.- ~p:../80

5 -- ---<---

-1-/ r=::::::

5 /50-~-,_

--250 --

12.5

o 500- t:nglnc speld n, rpm

750 1000 1250 1500 1750 2000 2250 2500

2736

Fig. 57. Dynamometer torque in kg-m plotted as a function of the speed n for various valuesof pmnx. For Pmnx = 140 kg/cm read Pmnx= 140 kgrcrns,

TABLE VIThe energies calculated from the measured data for

a. constant speed and various pressures

Pressure Speed Supplied Brake In circuit In cvtlnder In exhaust Rest b.rn.e.p,Pmu r.p.m, ± 2 by fuel load cooling water coollng water gases kwatts kg/cm'

kg/cm' ± 0.2 k watts k waus kwatts k watts k watts

50 1500 23.69 7.43 11.00 0.65 3.13 1.48 8.59

80 1500 35.43 12.48 17.64 0.68 4.02 0.61 14.43

110 1500 47.26 17.38 22.83 0.73 4.55 1.77 20.09

140 1500 58.61 22.12 28.54 0.85 5.77 1.33 25.57

b. constant pressure and various speeds

Pressure Speed Supplied Brake In cIrcuit In cylinder In exhaust Restpm ..:c r.n.m. ± 2 by fuel load coolIng water cooling water gases k waus

kg/cm' ± 0.2 k wens k watts k watts k waus k watts

140 250 13.70 3.30 7.33 0.35 2.63 0.09

140 500 22.36 7.28 11.42 0.40 3.41 -0.15

140 1000 40.93 15.22 19.55 0.65 4.68 0.83

140 1500 58.61 22.12 28.54 0.85 5.77 1.33

140 2000 75.57 26.96 38.63 1.10 8.03 0.85

140 2500 94.09 29.38 50.63 1.25 10.90 1.93

-98-

The lines of constant power in fig. 58 are drawn according to the equationb.m.e.p. = C X b.h.p./n. The upper limit of the family of curves shown infig. 58 is set by the fact that the maximum permissible pressure is 140 kg/cm'',The lower limit is not a real-one for this engine, but is simply caused by thefact that the lowest value of pmax used in these measurements was 50 ~g/cli12.The continuation of the curves of constant fuel consumption beyond this limit(broken line) is an extrapolation.

Table VI shows the heat balance calculated from the measurements for twocases:a. constant engine speed (1500 r.p.m.) and various pressures (Pmax = 50, ·80,

110 and 140 kg/cm''),b. constant pressure (pmax = 140 kg/cm-) and various engine speeds (n = 250,

500, 1000. 1.500,2000 and 2500 r.p.m.).

bm~p

27.5I-'k"".km"'-·-r----r---r--.,----.--~

~r--+--+--~~-+--+-~

O~C~'9Ud~~ '500 2000 2500 30002737·

Fig. 58. Lines of constant specific fuel consumption (g/b.h.p.-hr) and constant brake horse-power as a function of the mean effective pressure and the engine speed n (fuel consumption

based on a lower heat value of 10,000 kcal/kg).

The heat balances are plotted in figs. 59and 60.The heat losses due to radiation and convection from the preheater were

measured separately, by removing the preheater from the engine, insulating it

-99-

underneath, and heating it electrically from inside so that the surface tem-perature is about the same as when it is in use. The electrical energy suppliedshould then be equal to the heat losses in practice.

This is not a very accurate way of measuring this quantity, since the condi-, tions are doubtless rather different when the preheater is actually in use; but

since the quantity itself is generally small compared with the other factors, theeffect of this error is negligible. The relative effect of these losses is greatestat low engine loads, since their absolute value is more or less independent of theload on the engine.

The mechanicallosses occur in the heat balance as the sum of the measured

TABLE VIISome test results for the test engine at various heater-tube temperatures and various inlet

temperatures of the circuit cooling water (1500 r.p.m., Pmax= 140 kg/cm2)

Fuel Dynamometer Temp. hen Ier Inlet temp, ofgrams/sec. kg-m tubes cooüns wateroe oe

0.8703 1.84 257 15

1.0427 5.72 360 IS

I.I594 8.40 452 15

1.2578 10.92 547 15

1.3689 14.36. 705 IS

1.4035 15.15 750 15

L3689 14.36 705 15

1.3717 13.58 705 39-

1.3404 12.59 705 57

-100-

1 puunloflotal Mat1'~--r-ïl~L:~·~~=!f='k=t~}n===F~1

..0---- uhoust Sosu. ra latlon80,l-_J--~~~~if-==jf-~-~-~

''''''Ioi°0 4 8 12

-bmrpkglcm&16 20 24 28

Fig. 59. Heat balance plotted as a functionof the mean effective pressure

(n = 1500 r.p.m.).

_ c:ngiM s~c:d n, rpm 2738

Fig. 60. Heat balance plotted as a functionof the engine speed n(pmax = 140 kg/cm2).

piston friction losses and the energy which is not otherwise accounted for; thevalue thus obtained is clearly not very accurate. The piston friction losses aretaken as being equal to the heat carried off by the cylinder cooler, but thisquantity doubtless contains contributions due to heat passing from the com-pression space and the buffer space to the piston and liner, and heat conductedfrom the crankcase.

5

9 l/V

5

l70

1/ v';;V

5

t/V

5jI

t

to

% 00 ~ ~ ~ ~ ~ ~ ~_ Molu temp. ~c]

Fig. 61. Brake horsepower and efficiencyplotted as functions of the heater temper-

ature (n = 1500 r.p.m. andpmnx = 140 kg/ern").

I~

30

Ol--

------ r----.~ r--..-I--5

40

"\

15 30

0

5

0 0

o 20

20

to

010 20 3040 so_Inkl temp.of cooIrngwater [Oe]

60 70273'

Fig. 62. Brake horsepower and efficiencyplotted as functions ofthe inlet temperature

of the circuit cooling water (n = 1500r.p.m. and pmnx = 140 kgfcm2).

Some measurements were also made of the power and efficiencyas functionsof the heater tube temperature and the inlet temperature of the circuit coolingwater. These measurements were only made at one speed (1500 r.p.m.) andone pressure (Pmax = 140 kg/cm''), The results are given in table VII and areplotted as functions of the relevant temperature in figs. 61 and 62.

CONCLUSIONS

The engine described here, which is one of the possible ways of realizing theStirling process (see references 5, 7, 8 and 9 of chapter I) has many advantagesfrom an aerodynamic and thermodynamic point of view as a result of itscoaxial construction. Moreover, designing the piston and the displacer oneabove the other allows the temperature and pressure variations of the workingmedium to be separated: apart from the pressure differences caused by flowlosses, the displacer separates the regions of high and Iow temperature butlias the same pressure on both sides, while the piston separates the regions ofdifferent pressure (circuit and buffer space), but has practically the same

-101-

temperature on both sides. This reduces the heat losses in the working mediumand simplifies the construction of the displacer and the piston.However, the placing of the displacer and the piston in the same cylinder

makes a special drive mechanism necessary, since the displacer must be drivenwith a definite phase difference with respect to the piston, by a rod whichpasses through the latter.The use of the rhombic drive allows the piston and the displacer not only

to carry out the above-mentioned tasks, but also to balance the higher harmo-nics of the inertia forces. The first harmonic is balanced by the counterweights.

This complete balancing per cylinder unit is important, for multi-cylinderengines as well as for single-cylinder ones. The hot-gas engine can be usedas a single-cylinder engine up to high powers, since apart from the completebalancing, it possesses the property of giving a very low variation of thetorque as a function of the crank angle (see fig. 32). A single-cylinder hot-gasengine with the mean buffer pressure equal to the mean circuit pressure givesa better performance in this respect than a high-speed four-cylinder four-stroke diesel 5).

200°1 I I'600 -r-'- ----'20J-t . _ - ,- -- ._--- -J

:1I1 n_±-~- --1-o 30 60 90 120 lSO 180 210 240 270 300 330 360_crcnkongle.V ,d(g 2740

Fig. 63. The resultant torque due to the gas forces and inertia forces acting on one crankshaft,calculated-for a four-cylinder hot-gas engine built up of four single-cylinder engines of thetype descftbed here at 1500 and 2500 r.p.m. Crank angles of successive cylinders differ by 90°.

I~ a multi-cylinder engine the complete balancing per cylinder allows therelative crank positions to be chosen so that the torque is as constant aspossible as a function of the crank angle. If we take a four-cylinder engine asan example, then inspection of the Fourier series for the torque (see II.9.2.d)shows that if the crank angles of successive cylinders differ by 90°, then onlythe small 4th harmonic is left. In the sample calculation given in this thesis,

-102-

the amplitude of the 4th harmonic is only 5% of the mean torque at 1500r.p.m. and 12% at 2500 r.p.m. Fig. 63 shows the variation ofthetorque ofsuchan engine.The reaction torque on the base, which might give rise to undesired vibra-

tions of the surroundings, can be made arbitrarily small by use of this drivemechanism together with a coupling of Iow torsional stiffness (see 11.8.6). Thisis true for a single-cylinder engine as well as for a multi-cylinder one.Although the rhombic drive is more complicated than the normal drive

mechanism of an internal combustion engine, the facts mentioned above showsthat it has several advantages to compensate for this.The measurements made on the test engine showed that its overall efficiency

when used at high speeds is about the same as that of a high-speed diesel(maximum efficiency = 35-40%, depending on the size; this is equivalent to180-158 gfb.h.p.-hr). The economic efficiency (the price per b.h.p.-hr) is betterthan that of a high-speed diesel, however, even when 3% has been subtractedfor the running of the auxiliary equipment, since the lubricating oil consump-tion is practically nil, while it is 2-3 gfb.h.p.-hr for a diesel. Lubricating oil costsabout ten times as much as diesel oil, so 20-30gfb.h.p.-hr must be added to thefuel consumption of the diesel for the purposes of comparison.

One disadvantage of the hot-gas engine is that the efficiency falls as thetemperature of the cooling water rises, see fig. 62.It may be seen from figs. 55-57 that in principle the hot-gas engine can be

used over a wide range of speeds with a high efficiency and with a nearlyconstant mean torque. The maximum and minimum efficiencies of the testengine in the speed range from 250 to 2500 r.p.m. were 38 and 25%, and themaximum and minimum torques calculated on the basis ofpmax = 140 kgfcm2

were 15 and 11.5 kg-m. .Since the hot-gas engine has no rapid changes of pressure in the working

medium, and no rapidly accelerating parts, it is a quiet engine. In order tomake it as quiet as possible, much care must be taken in the construction of thegearwheels and of the blower which delivers the burner air for the continuous .combustion of the fuel.

The external combustion used in this engine is a very important feature.This, and the noiselessness, distinguish the hot-gas engine completely fromthe internal combustion engine. The use of external combustion makes thehot-gas engine universal: it can use not only a wide range of conventionalfuels, but also other energy sources such as solar and nuclear energy.

REFERENCES1) French Patent No. 1,182,569 R. J. MEIJER.2) Belgian Patent No. 576,158 R. J. MEIJER.3) French Patent No. 1,184,034 R. J. MEIJER.4) Belgian Patent No. 573,547 R. J. MEIJERand M. L. HERMANS.5) C. B. BIEZENOand R. GRAMMEL,Technische Dynamik, Julius Springer, XII, 13, 1939.

-103-

, V 2 . IV ·2 22VSE = rAd [Be - Bl + 1 - ;; (1 - A.) -.Ä. e }

APPENDIX I

Summary of the equations

Balancing

Lpe = Lee = L

apempe-ademde = tL(md-mp)

reme = r{mpe + mde + !Cmp + md)}

me = mer +meo

re = reomeo- rermermeo + mer

Masses and moments of inertia

2ape 2ademo = mp +Tmpe = ma +Tmde

Jpe = {i~e- ape (L- ape)} mpe

Jde = {i~e- ade (L - ade)} mde

Volumes

Expansion space00

VE = rAd {C + D cos (1p- YE) + L Bn COS n1p}n=2

1C = Bo-;; V(1- .Ä.)2- A.2e2

D = VB~+ 11

YE= tan-1-Bl

The first harmonic

VE = tv0 {I + cos ~ }

v: = 2rAd VB~ + 1

Fictive adiabatic dead space

-104-

Compression space

f~ 1

00

Ve=rAp{E-Fcos('ljJ+ye)-(l +f) ~Bncosmp}n=2 '

2f l-f: E =}" VI-,1~ (e-l)2 + -,1- V(1 + ,1)2- ,12e2- (1 + f) Bo

2 1- f r.::--:--=,,---_=_-.,,-: E = '1 Vl- ,12(e-l)2 - -,1- V(1 + ,1)2- ,12e2- (1 + f) Bo

F = V(1- f)2 + (1 + f)2 B~

_ -1 l-f ,ye - tan (1 + f) BI

See Appendix II for BnThe first harmonic

f~ 1

Vc = twVo{l + cos (cx;-cp)}

w = _!_ 1 / ei - f)2 + (1 + f)2 BIf/f V B~ + 1

-2BIcpvo= tan-l ------

(1 + f) B~- (1- f)

see fig. 25

see fig. 25

Fictive adiabatic dead space

{2f l-f

f~ 1 : Vse=rA/p T Vl- ,12(e-1)2 + -,1- V(1+ ,1)2-,12e2

V 2 '2 2}- (1 + f) Bo - (1 - f) + (1 + f) BI

f~ 1 : Vso = rA'p {~VI-,12(e-l)2 - 1 À f V(1+ ,1)2-,12e2 +- (1 + f) Bo-V(1- f)2 + (1 + f)~ B~}

Buffer space

Vbt = tVbO{2ab + boN + Rl cos ('lJli + yI) + b2R2 cos 2 ('lfi + Y2) ... }

VbO= 2rA'~ VB~+ 1

1Bo-'1 V(1- ,1)2- ,12e2

bo = ----;=====:----VB~+ 1

B2b2= --VB~+ 1

-105-

Rl = Y{ ~ COS fh}2 + {~ sin ()j}21=1 1-1

R2 = Y{ ~ cos 2Pl}2 + {~ sin 2PI}2j-I 1~1 .

N

~ sin (PI + YE)1=1Yl = tan-l--'N~-----~ cos (PI + YE)1=1

N~ sin 2p!

1 11-1y2 = "2" tan- --'N~---

~ cos 2p!1=1

See Appendix 11 for values of Be.Area ratios:

f= A~l .,A'p

f' =Ad .,Ad

K'f" =2

A'p

see fig. 23

Stroke, phase angle of drive mechanism and maximum slant of connecting rods

Stroke of piston = stroke of displacer =

cpdr = 2 tarr+ BI .

Xmax = sin-1 A(e + 1) .

Pressures

Expansion space00 00

PE = PE {I + ~pEn COS mp + ~qEn sin mp}n=1 n=1

PE, pEn and qEn follow from the thermodynamic and aerodynamic calculationsofthe circuit.Compression space

00 00

Pc = Pc {I + ~Pen cos mp + ~qen sin mp}n=1 n=1

pc, Pen and qen follow from the thermodynamic and aerodynamic calculationsofthe circuit.

-106-

Buffer space .' f ~~lUps'~lUf/lA MPfNFmlfl£1IPb = Pb {I + ~ pbn COSmp + ~ qbn sin mp}

n=l n=I

Pb can be chosen arbitrarily.

1 3 2Pbl = - d [{"Cl + 1" e" + 1) e" + 2) GCI + ClC2)}cos Yl +

+ !" e" + 1) ClC2cos eYl- 2Y2) .. .]1 2

Pb2 = d [1" (x + 1) Cl cos 2Yl- "C2 cos 2Y2 +- 1"e" + 1) e" + 2) elc~ + C~C2)cos 2Y2 .. ·l

113Pb3 = d [-l" (x + 1) ClC2cos eYl+ 2Y2) -24" e" + 1) e" + 2) Cl cos 3Y1 +

- 1r" (" + 1) e" + 2) C1C2cos eYl- 4Y2) .. .]

qbl = ~ [{"Cl + i" (" + 1) e" + 2) e!c~ + ClC~} sin Yl ++ -}"e" + 1) C1C2sin (Yl- 2Y2) .. ·l

1 . Iqb2 =- d [i" e" + 1) c~ sin 2Yl- "C2 sin 2Y2 +'

- 1" (" + 1) (x + 2) (.~c~+ C~C2)sin 2Y2 .. ·lqb3 = - ~ H-" e" + 1)ClC2sin (Yl + 2Y2) - ~4" e" + 1)(" + 2) c~ sin 3Yl +

- 1r" e" + 1) (" + 2) C1C2sin (Yl- 4Y2) . ".ld = 1 + t" ("+ 1) ec~ + c~)- 1r" (" + 1) (x + 2) C~C2cos 2 (Yl- Y2) ...

See Volume of buffer space forbI, b2, Rl, R2 and N.

Crankcase

Pee is constant

Pressure ratios

g' =Pb .Pc'

" Peeg=-::-Pc

Torques

Torque due to the gas forces of the circuit and the crankcase

Tgc = {-rpcA'p L~oc, cos mp +n~pnsin n'lJ'}

, 1 * lQO ° ° 3QOCl = a + 2"P2 +"2 2Bl +QIB2 + Q3B2 + '2 2B3+ ...

-107-

1 * 1 * 1pO 1 0 " 3 0D2 = "2Q1 + "2Q3 +"2 1B1- "2P3B1 + 2a B2 + 2" P1Ba + ...1* 1* 1010 0 " 0D3 = "2Q2 + "2Q4 + "2 P2B1-"2 P4B1 + P1B2 + 3a B3 + 2P1B4 + ...

1* 1* 10 0 ao "D4 = "2Qa + "2 Q5 + "2 P3B1 + P2B2 + "2 P1Ba + 4a B4 + ...

Pri= (1- f) POn + ff'gPEn

p~ = (1 + f) POn- ff'gPEnQri = (1- f) qOn + ff'gqEn

Q~= (1 + f) qOn- ff'gqEn

a' = 1- f - g" + fg" + f"g" + ff'g - ff'g"

a" = 1+ f - g" - fg" + f"g" - ff'g + ff'g"

Torque due to the gas forces of the buffer spaceGO GO

Tgb = - !rpoA pg'f" ( ~. Cbn cos mp +~Due sin n'1jJ)• n-o n-1

1 1Cbo = "2 Pb1 + "2 qb1B1 + qb2B2 + ...

1 1 3CbI = 1 +"2 Pb2 + "2qb2B1 + qb1B2 + qbaB2 + 2" qb2Ba + ...

1 ill aCb2 ="2 Pb1 +"2 pba-"2 qb1B1 +"2 qbaBl +"2 qb1Ba + ...

1 1 1 1Cba = "2Pb2 + "2 Pb4 - "2 qb2B1 + "2 qb4B1- qb1B2 + 2qb1B4-+ ...

1 1 1 aCM = "2pba + "2Pb5-"2 qbaBl - qb2B2- "2qb1B3 + ...

1 1 aCb5 = "2Pb4- "2 qMBl - qbaB2- "2qb2Ba- 2qb1B4 + ...

- 1 1 3Dbl = "2qb2 + BI - 2" Pb2BI + Pb1B2- PbaB2 + 2" Pb2Ba + ...

III 1 3., Db2 = 2" qb1 + "2qba + "2PblBl - "2PbaBl + 2B2 + 2" PblBa + ...

III 1 -Dba = "2qb2 + "2 qM + "2 Pb2Bl - "2 Pb4BI + PblB2 + 3Ba + 2Pb1B4 + ...

-108-

Torque due to the inertia forces

Trnw =-ir2w2mo{l +z) {Hl sin "I' +2 (1 ~ z +H2) sin21p + n~3n Hs sin ne }

).2 (Jpe + Jde)Z = -~--___:_

. r2mo

See Appendix IV for the values of Hn.Effective moment of inertia of half the drive mechanism about the correspond-ing crankshaft:

Je = (ie + rre) me + rms {(I + z) Ho-I} +

+ r2rno (1 + z) {Hl cos "I' + (1 ~ z + H2) cos 2"1' + n~3Hn cos n1p}

).2 (Jpe + Jde)z=-~::'__:'--==r2rno

See Appendix IV for values of Hs.

Forces

Forces on the piston drive mechanism (Jpe = 0)

Fni = i (Fp-mpxp)

F p2= t (F p - moxp) tan X

Fp3 = i (Fp-moxp) sec X

Fp4 = 0

Fp5 = Fp3sin(X-1p)-rnpe (1- at)rw2

Fp6 = Fp3 cos (X-1p)r 00

Fp = pcA'p {1-g" + g"f" - g'f" + ~ (Pc- g'f"Pb)n cos n1p+n-l 00

+ ~(qe-g'f"gb)nsinn1p}n=l00

Xp = rw2 (- sin "I' + ~ n2Bn cos n1p)rr=L

Forces on the displacer drive mechanism (Jde = 0)

Fdl =- i (Fd - rndXd)Fd2 =- i (Fa - rnoXd) tan.gFd3 = - i (Fd - rnoXd) sec X

Fd4 = 0

-109-

coXd = - rw2 (sin e + ~n2Bn cos n'1/))

rt=L

co+ :E (gf'qE- qc)n sin n'1/)}

n=L

Fd5 = Fd3 sin (X + '1/))- mdc (1- a~!:) rw2

FdG = Fd3 cos (X + '1/))co

Fd = fpcA'p {gf' - 1 + g" - g"f' + :E (gf'PE - Pc)n cos n'1/)+n=L

Forces 011 the crankshaft (Jpc = 0; Jde = 0).F7 =Fp3sin(X-'1/)) +Fd3sin(x +'1/)) +morw2

Fs = Fp3 cos (X- '1/))- Fd3 cos (x + '1/))

X = sin-l Ä(8- cos 1Jl)

See Appendix II for values of Be.PE, pEn, qEn, pC, pCn and qCn follow from the thermodynamic and aerody-

namic calculations of the circuit.For pbn and qbn see pressure of buffer space.

Friction powers

Prm = ~-Fep rwD [sin ('1/)+ YE)!

Pfv3 = ~-,u3wdcp V(FP5 + Fd5)2 + (FpG- Fd6)2 (1 + ÄGI sin '1/))

Prp3 = ,uP'3ÄwdCSGlV Fp~ + F;6 [sin '1/)!

Prei = rwBlFrd [sin '1/)!

Prea = ~-,ud2ÄwddpGIVF~l + F~2 [sin '1/)!

Prd3 = ,ud3ÄwdcsGl VF~5 + F!6 [sin '1/)!

Pe4 = t ,u4wdcJ VF~+ F:

D = VB~+ 1

1yE = tan-l-

BI

See Appendix II for values of BI and Appendix III for values of Gi,

-110-

klXc = Co = 2ao

Torque due to friction

TrPI = !FrprD [sin (71' +YE)I

Trp2 =! ,up2AdppGI VF~2 + F~2 Isin 71'1

Tfp3 =~.,u3dcp V(F ps + F dS)2+ (F ce - F (6)2 (1 + AGI sin 71')

Trp3 = ,uP3AdcsG.lV F~s + F;6 [sin 71'1

Trui = rBIFrd [sin 71'1,.--.--

Trd2 = !,ud2AddpGl V F~l + F~2 [sin 71'1

Tfd3 = Trp3

Tfd3 = ,ud3AdcsGl V F~s + F~6 [sin 71'1

Tr4 =~.,u4dcJVF~ + F:

Tt = Trpi + Trp2 + Trp3 + Tral + Tra2 + Tra3

D = VB~+ 11

yE = tan-l-BI

See Appendix 11for BI and Appendix III for GI.

Total torque on the crankshaft (for OJ = constant)

Tgc + Tgb-Tr-Tmw = Trpm00 00

T ••m = ~ an COSn7pm+ ~bn cos n7pmn=ö n=l

Torque of the driven machine (shaft angle = rpm)00 00

T 'I'm= ~ Cn COSnrpm + ~dn sin nrpmn=o n=l

Torque transmitted by the gearwheels00 00

T = ao + ~ Pn cos n7pm+ ~ Qe sin n7pmn=l n=l

Ir [cn + 2an {( ~~ f - I}]Pn - an =--,,-_0 _

J {(~~r-I}o

-111-

Ir [dn+ ze, {e;r -I}]Qn = bn - ----------

J {(~:r-lrReaction torque on the base

Tr = (Ji -J2) Llip-k (.Lhp-Llep)- klXc

Te =- 2T (see line 2 from bottom, page 111).

Motions

CrankshaftsLI"P= "P- "Pm; "Pm= rut00 00

LI"P= ~Xn cos n"Pm + ~Yn sin n"Pmn=l n=l

00 00

WV1 = ru [1 - ~ nxs sin n"Pm + ~nYn cos n"Pm]n-l h=l

Driven machineLI<p= ep-epm; epm= äit-lXc

Liep=00 [{ (ruO)2} 2an] 00 [{ (run)2} 2bn].n~l 1- Wil Xn- k cos n"Pm + n~l 1- Wil Yn-"""IC sm 0"Pm

o 0

00 [{ (run)2} 2bn] ]+n~ln 1- w~ Yn-T cosn"Pm

. {(run)2 }Cn + 2an CO' -1o

Xn =. {(ruO)2}J (run)2 Wo --:- 1

dn + 2bn {(!~)2- I}, 0

Yn

cn = c~ cos nlXc- d~ sin nIXc'

dn = c~ sin nIXc+ d~cos nIXc

-112-

" 1~Wo = Vh+h

Moment of inertia of flywheels

[~max-~min

JFl + JF2 = 2 w2(j

00 an • * bn~= ~ - SIn ll1pm + ""- cos ll1pmn=l II n=l II

•

-113-

tB.

APPENDIX II

0.

-.. 1\

\,0c-- 1\'f--- -. .\. I

~~...J

· 1

· ~-- ~-

~- t--

''''--

~,_ i~~~~ J

'0.,:~~

0.'" .1 "

The coefficients Bs ofthe series expansion1 00- cos z = I:Be cos neA n-oas functions of A and s,

-Á

0.2 o-n 0.3 0.35 Q.4 OAS o.s2741

.' /'_ .0

./ V

· /......-V V 17 7'· [/7 V /" ........

· ~V/ :;.7 17 V --- ~

, ~ V.,....V V .-v~~ ~

r.-: .......-~ v- z=:,

~~V V

_

~ VV v-I----d0 17/ /' ./ - -· f/ / ../" , L---sf-- V/ ......-· / /' fOt [.0: ~"",

· /

170. 005 0.' O.IS .

-Árus OA OAS o.s

2743

.d---_A

0.2$ 0.3 0.3$ 0.04 OilS o.s2742

"--'1--- -- -/~r7F77'7·.d h~~~~~0 -~ ~ ~ -0

~s·a· .#

1/ --I50 0'" O. OIS c.a 0.2S Q,

PPEND1X II Ba .E 0 0.25 0.50. 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

~5 19.9875 19.9859 19.9812 19.9734 19.9624 19.9483 19.9311. 19.9106 19.8871 19.8603 19.8303

10 9.9750 9.9718 9.9624 9.9466 9.9244 9.8959 9.8609 9.8194 9.7713 9.7165 ·9.6548

15 6.6290 6.6242 6.6099 6.5859 6.5522 6.5087 6.4550 6.3909 6.3161 6.2302 6.1327

?O 4.9496 4.9432 4.9238 4.8912 4.8453 4.7855 4.7114 4.6220 4.5164, 4.3934 4.2509

'5 3.9367 3.9285 3.9038 3.8621 3.8029 3.7252 3.6276 3.5079 3.3634 3.1893 2.9778

0 3.2570 3.2469 3.2164 3.1648 3.0908 2.9922 2.8657 2.7053 2.5002 2.2188 -

135 2.7675 2.7554 2.7187 2.6559 2.5646 2.4399 2.2724 2.0368 - - -0 2.3968 2.3825 2.3387 2.2631 2.1502 1.9876 1.7156 - - - -5 2.1050 2.0882 2.0365 1.9451 1.8016 - - - - - -

0 1.8684 1.8488 1.7875 1.6751 1.4588 - - - - - -

BI .E 0 0.25 . 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

~5 0 0.0125 0.0250 0.0376 0.0501 0.0627 0.0753 0.0879 0.1006 0.1133 0.1261

Ö 0 .0251 .0503 .0755 .1009 .1265 .1523 .1785 .2050 .2319 .2593

5 0 .0378 .0759 .1142 .1531 .1926 .2331 .2747 .3178 .3625 .4092

0 0 .0508 .1021 .1542 .2075 .2627 .3204 .3812 .4461 .5164 .5937

5 0 .0642 .1292 .1959 .2654 .3390 .4182 .5054 .6040 .7194 .8618

0 0 .0779 .1574 .2402 .3284 .4249 .5342 .6639 .8300 1.0836 -5 0 .0923 .1873 .2880 .3990 .5271 .6862 .. 9175 - - -0 0 .1074 .2192 .3410 .4820 .6617 .9875 - - - -5 0 .1234 .2540 .4018 .5885 - - - - - -0 0 .1407 .2928 .4761 .7881 - - - - - -

-B2c 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

5 0.0125 0.0125 0.0125 0.0125 0.0126 0.0126 0.0126 0.0127 0.0127 0.0127 0.0128

0 .0251 .0251 .0252 .0253 .0254 .0257 .0259 .0263 .0267 .0271 .0276

5 .0377 .. 0378 .0380 .0385 .0391 .0398 .0408 .0421 .0436 .0454 .0476

0 .0505 .0507 .0513 .0523 .0538 .0559 .Q586 .0621 .0665 .0723 .0799

5 .0635 .0639 .0651 .0673 .0705 .0750 .0813 .0900 .1023 .1206 .1497

b .0768 .0775 .0797 .0837 .0899 .0992 .1132 .1353 .1742 .2648 -

15 .0903 .0915 .0953 .1023 .1137 .1326 .1661 .2412 - - -P .1043 .1062 .1123 .1241 .1453 .1869 .3341 - - - -I; .1188 .1217 .1313 .1511 .1930 - - - - -

.1339 .1382 .1531 .1877 .3181 - - - - - -

_.- ~- ~

_A 2746

'r:---r- I I / / 7·· -!-.:--. .: "I .01

· IU I Ir ....

· IJ IJ I ~ /

I ~ VIVIV/1/ p, I I1 17'· ,...- II/l / / -:· 1/ -~

'11 1 / ./ ./II '// / V V ./

· IliJ V~ / 1/ /v

~~ ~ V V ... - 0,-0

:, /oos 0' 0 .• 0. 0 .. 0 0 . .J lS 0 OS 0.5

2644

I

,U ~ /· ~· !51· // I .I1 Cl,II / / !I Vl· '// Ij / V

· f-!· I I /· -I-· 11 / I / /

.1--. LL 1/ / / 1/ V. V-If//J rij I/j/ /,

1'/// I / / --·· "".0: .... 0· I / /· II j, 1/// / V•0 VI~tv1/0 0.05 0,0 o.~ 0' 0", 03 0" O. 0" o.

._-- -

Ba• 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

05 0 0 0 0 0 0 0 0 0 0 0

10 '0 0 0.0001 0.0001 0.0001 0.0002 0.0002 0.0002 0.0003 0.0003 0.0004

15 0 0.0001 .0002 .0003 .0005 .0006 .0007 .0009 .0011 .0013 .0016t

20 0 .0003 .0005 .0008 .0012 .0015 .0020 .Q026 .0033 .0043 .0056

25 0 .0005 .0011 .0017 .0025 .0034 .0047 .0066 .0093 .0139 .0225

130 0 .0009 .0020 .0032 .0048 .0071 .0108 .0171 .0304 .0728 -135 0 .0015 .0033 .0056 .0090 .0147 .0262 .0602 - - -~O 0 .0024 .0053 .0095 .0168 .0330 .1242 - - - -5 0 .0037 .0084 .0162 .0336 - - - - - -0 0 .0054 .0129 .0282 .1106 - - - - - -

-B4B 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

5 0 0 0 0 0 0 0 0 0 0 0

0 0.00002 0.00002 0.00002 0.00002 0.00002 0.00002 0.00002 0.00002 0.00002 0.00002 0.00002

5 .00005 .00005 .00006 .00006 .. 00006 .00007 .00008 .00009 .00010 .00012 .00015

0 .00013 .00013 .00014 .00015 .00017 .00021 .00025 .00032 .00041 .00055 .00078

5 .00026 .00026 .00029 .00034 .00041 .00Q54 .00073 .00106 .00164 .00278 .00535

0 .00045 .00047 .00054 .00068 .00092 .00134 .00216 .00391 .00856 .02940 -5 .00074 .00079 .00095 .00130 .00199 .00348 .00736 .02329 - - -

0 .00114 .00124 .00161 .00246 .00448 .01055 .07144 - - - -5 .00168 .00189 .00267 .00474 .01139 - - - - - -

0 .00240 .00280 .00443 .00979 .06436 - - - - - -

B5c 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

15 0 0 0 0 0 0 0 0 0 0 0

b 0 0 0 0 0 0 0 0 0 0 0

b 0 0 0 0 0 0 0 0 0.00001 0.00001 0.00001

P 0 0 0 0.00001 0.00001 0.00001 0.00002 0.00003 .00004 .00007 .00011

0 0.00001 0.00001 .00002 .00004 .00006 .00010 .. 00016 .00029 .00059 .00140

). , 0 .00002 .00004 .00007 .00012 .00022 .00042 .00094 .00264 .01338 -. ,

0 .00004 .00009 .00018 .00036 .00080 .00219 .01003 - - -b 0 .00008 .00021 .00046 .00111 .00356 .04747 - - - -

0 .00016 .00044 .00116 .00394 - - - - - -h 0 .00031 .00095 .00317 .04255 ..!.. - - - - -

APPENDIX IJ[

The coefficients Gil ofthe series expansion7. 00u: = l: Ge sin I1Ip asÀw n=L

functions of Ä and E.

Id'L- __~ __ _LULLL-LLL __ _£ ~-L~~ __ _L ~ __ ~ ~

o 0.05-).

0.35 CA OAS 0.5

2746

10~L_--~0~0~5--~0~j----70.~'5~---).

0.4 OAS 0.52749

035

PENDIX 1II Glc 0 0.25 I 0.50 I 0.75 1.00 I 1.25 I 1.50 I 1.75 1 2.00 I 2.25 2.50

5 1.00031 1.00039 1.00063 1.00102 1.00157 1.00228 1.00315 1.00417 1.00537 1.00672 1.00824

0 1.00125 1.00157 1.00252 1.00411 1.00635 1.00925 1.01283 1.01711 1.02212 1.02789 1.03446

5 1..00284 1.00356 1.00573 1.00939 1.01459 1.02138 1.02989 1.04022 1.05255 1.06709 1.08409

0 1.00508 1.00639 1.01035 1.01708 1.02672 1.03956 1.05596 1.07644 1.10170 1.13276 1.17104

5 1.00800 1.01011 1.01652 1.02749 1.04351 1.06536 1.09420 1.13185 1.18120 1.24717 1.33921

0 1.01165 1.01479 1.02442 1.04115 1.06620 1.10163 1.15102 1.22106 1.32645 1.51625 -"

5 1.01606 1.02053 1.03434 1.05888 1.09697 1.15417 1.24276 1.40147 - - -

0 1.02131 1.02746 1.04672 1.08201 1.14004 1.23799 - - - - -5 1.02747 1.03574 1.06219 1.11294 1.20580 - - - - - -0 1.03463 1.04564 1.08181 1.15684 - - - - - --., -

-G2s 0 0.25 I 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

5 0 0.00031 0.00063 0.00094 0.00126 0.00157 0.00189 0.00222 0.00254 0.00287 0.00321

0 0 .00126 .00253 .00381 .00512 .00645 .00782 .00924 .01072 .01227 .01389

5 0 .00287 .00577 .00875 .01185 .01512 .01860 .02237 .02649 .03107 .03621

0 0 .00518 .01048 .01603 .02200 .02856 .03595 .04447 .05455 .06682 .08220

5 0 .00825 .01682 .02608 .03647 .04858 .0633.0 .08199 .10698 .14261 .-19836

0 0 .01218 .02509 .03958 .05682 .07866 .10832 .15238 .22735 .40110 -

5 0 .01710 .03568 .05762 .08593 .12634 .19276 .33763 - - -p 0 .02318 .04920 .08210 .12972 .21331 - - - - -~ 0 .03068 .06661 .11655 .20391 - - - - - _.

.~ 0 .03993 .08950 .16899 - - - - - - -

G3·1 0 0.25 0.50 0.75 1.00 I 1.25 1.50 1.75 2.00 2.25 2.50

:> 0.00031 0.00031 0.00031 I 0.00031 0.00032 0.00032 0.00032 0.00032 0.00033 0.00033 0.00034

0 .00126 .00126 .00127 .00129 .00132 .00135 .00139 .00144 .00151 .00158 .. 00167

5 .00285 .00287 .00292 .00302 .00316 .00335 .00359 .00390 .00430 .00480 .. 00542

P .00512 .00518 .00536 .00568 .00617 .00685 .00778 .00906 .01083 .01329 .01684

~ .00810 .00825 .00873 .00958 .01092 .01293 .01593 .02052 .02782 .04031 .06419

P .01185 .01219 .01325 .01522 .01851 .02392 .03312 .05019 .08735 .20909 -.01646 .01711 .01925 .02346 .03111 .04558 .07687 .17328 - - -

D .02200 .02321 .02728 .03587 .05373 .09749 - - - - -.02862 .03074 .03819 .05569 .10159 - - - - - -

P .03647 .04007 .05352 .09093 - - - - - - -

,6

. -G.

/ ,.7~rr-1 / .-1-1// / / 7 7iV/II/j I/j V v·

, / // 70/ I?I / / 7 7/I 1/ / 5)' V /

/ "! / 71// 1/ / / IqY /

1/ / / Ij / J /JII Ij / / / aY

~ Ij/;Ij /. 7. /,ö0, 0.05 0.1 0,15 Q2 0.25 0.3 0.35 0.4 0,45 0.5-~ 2750

I/I 1/ / 1/ / / /I / / / / /

o I / 1/ / / 17 IJ.~ !el nl~I"l~/~ ,S?! I / /11/ / / 7 ~ RI 7 7I II 1// / ei

'rI ./. .,1'/

.. //iij jll Ir//0;~~ ~ ~ ~ ~ ~ = ~ ~ ~ =-).. 2751

- --

', \

-G4

• 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

P5 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0.00001 0.00001 0.00002 0.00003 0.00003 0.00004 0.00005 0.00005 0.00006

5 0 0.00002 .00005 .00008 .00011 .00014 .00018 .00023 .00029 .00037 .00046-0 0 .00008 .00016 .00026 .00038 .00052 .00071 .00097 .00135 .00189 .00275

5 0 .00020 .00043 .00070 .00106 .00157 .00235 .00362 .00584 .01018 .02003

0 0 .00044 .00095 .00163 '.00266 .00435 .00750 .01420 .03205 .11318 -5 0 '.00086 .00192 .00352 .00632 .01209 .02694 .08864 - - -0 0 .00157 .00367 .00732 .01523 .03858 - - - - -5 0 .00272 .00677 .01522 .04094 - - - - - -0 0 .00456 .01232 .03357 - - - - - - -

G5

c 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

5 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

5 0.00001 0.00001 0.00001 0.00001 0.00002 0.00002 0.00002 0.00003 0.00003 0.00004 0.00005

0 .00004 .00004 .00004 .00005 .00006 .00008 '.00011 .00015 .00022 .00033 .00052

5 .00010 .00010 .00012 .00015 .00021 .00030 .00047 .00078 .00141 .00283 .00672

0 .00021 .00023 .00028 .00040 .00062 .00106 .00202 .00449 .01270 .06516 -5 .00040 .00045 .00061 .00097 .00178 .00384 .01048 .04862 - - -0 .00072 .00083 .00125 .00232 .00535 .01707 - - - -- -5 .00121- .00146 .00250 .00570 .01896 - - - - - -0 .Q0195 .00251 .00503 .01538 - - - - - - -

The coefficientsHn ofthe series expansion ,o'f--+--l--+r-/--7t--¥...,.",r--/4--:;4---!-7.4--j---j

APPENDIX IV

(:i. )2 00- = 1: He cos ntpitoo n-o

asfunctionsof itand E.

t- H,3~'--~---'~--'r---~

lÖ~~~O~.O~S-~~LW~~~~_~~~~-~~~+--~-~-A

t - (H,+ 0.5)•

10'

h.:Vvb-2

r/Vv'h vV'H-Vv 1\,r-, --'7) '7 7~/' l..---\

/ // /fY /,../

~-175 /7' 7 ,v -:/-/ / "'"

~ ~

/ V v-3 ---;-- 7; / - I-- -,I---

2 / --- 1--

///;~ V/ VV

<jfo.,10 o 0.05 eoo-À

0.15 0,20 0.25 0.30 0.35 0.40 0.45 0.50

2754

-

PPENDIX IV Ho ,

• 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

05 0.50031 0.50039 0.50063 0.50102 0.50157 0.50228 0.50315 0.50419 0.50538 0.50675 0.50828

10 .50126 .50157 .50253 .50413 .50638 .50932 .51295 .51730 .52243 .52836 .53515

15 .50284 .50357 .50577 .50948 .51477 .52173 .53051 .54129 .55429 .56983 .58829

20 .50510 .50644 .51048 .51737 .52734 .54078 .55820 .58039 .60842 .64390 .68919

25 .50807 .51023 .51683 .52825 .54519 .56876 .60077 .64412 .70374 .78876 .91870

30 .51179 .51505 .52512 .54290 .57019 .61019 .66887 .75848 .91001 1.26150 -..35 .51633 .52103 .53576 .56256 .60587 .67515 .79419 1.05974 - - -

40 .52178 .52837 .54940 .58941 .65983 .79474 - - - - -45 .52825 .53733 .56710 .62780 .75399 - - - - - -50 .53590 .54829 .59068 .68826 - - - - - - -

-Hl

• I 0 0.25 0.50 I 0.75 1.00 1.25 1.50 1.75 2.00 2.25 I 2.50

05 0 0.00031 0.00063 0.00094 0.00126 0.00158 0.00190 0.00223 0.00256 0.00289 0.00323

10 0 .00126 .00254 .00383 .00515 .00652 .00793 .00941 .01097 .01263 .01440

15 0 .00289 .00582 .00886 .01206 .01549 .01923 .02336 .02800 .03330 .03946

20 0 .00524 .01064 .01640 .02273 .02989 .03824 .04828 .06071 .07660 .09769

25 0 .00840 .01725 .02705 .03846 .05241 .07031 .09456 .12951 .18405 .27982

30 0 .01252 .02605 .04183 .06169 .08864 .12853 .19449 .32472 .72698 -

35 0 .01776 .03763 .06246 .09714 .15218 .25678 .55331 - - -40 0 .02440 .05295 .09206 .15576 .28945 - - - - -45 0 .03280 .07357 .137i5 .27173 - - - - - -

50 0 .04355 .10234 .21456 - - - - - - --H2

• I 0 I 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

05 0.50000 0.50008 0.50031 0.50070 0.50125 0.50196 0.50283 0.50386 0.50505 0.50641 0.50794

10 .50000 .50031 .50125 .50282 .50505 .50793 .51150 .51579 .52083 .52666 .53333

15 .49998 .50069 .50281 .50639 .51149 .51819 .52663 .53696 .54940 .56421 .58173

20 .49995 .50120 .50499 .51144 .52074 .53321 .54928 .56956 .59487 .62638 .66571

25 .49987 .50182 .50777 .51800 .53302 .55364 .58105 .61700 .66408 .72588 .80632

30 .49972 .50252 .51112 .52608 .54850 .58008 .62341 .68180 .75528 .76675 -

35 .49947 .50325 .51494 .53555 .56694 .61172 .67056 .69732 - - -40 .49905 .50393 .51904 .. 54587 .58629 .63536 - - - - -45 .49840 .50441 .52301 .55522 .59374 - - - - - -

,50 .49742 .50449 .52586 .55659 - - - - - - -

tÖ20~O~.()5_u"O.'!Ll,--Lfh,L~~:+"-~-~---"-L__,,..L0.~5-05, _}. 2155

/ / I I -; / // 1/ / / 1/ 1/ 1/

Iq,vl7//1/ /I I 1:Jf .if //11 1/ / I ., 7 /

/ W 1/ /I/I I / / I. 0'}' /

/ 1// / I / / ,,"/1I/jj 1// 1/ / V#jV;jr/:1/ V

5 35 0.. OAS 0.5

-A

,(/

O.~S 0.52756

0.15 . Q2 Q2 0.3 0.2757

H~·11 0 0.25 0.50 I 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50

05 0 0.00031 0.00063 0.00094 0.00126 0.00158 0.00190 0.00222 0.00255 0.00289 0.00323

10 0 .00126 .00253 .00381 .00513 .00648 .00789 ,.00936 .01091 .01255 .01431

15 0 .00285 .00575 .00876 .01192 .01530' .01897 .02303 .02758 .03276 .03875

20 0 .00513 .01042 .01604 .02220 .02915 .Ö3720 .04681 .05861 .07352 .09300

25 0 .00813 .01667 .02608 .03694 .05007 .06666 .08864 .11930 .16473 .2373~

30 0 .01192 .02472 .03949 .05771 .08177 .11581 .16796 .25564 .39835 -35 0 .01657 .03489 .05723 .08716 .13132 .20368 .32576 - - -4Q 0 .02219 .04759 .08070 .12972 .21175 - - - - -45 0 .02891 .06337 .11199 .19152 - - - - - -

50 O· .03686 .08295 .15331 - - - - - - --H4 \

·1 0 0.25 0.50 I 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50.05 0.00031 0.00031 0.00031 0.00032 0.00032 0.00032 0.00032 0.00033 0.00033 0.00034 0.00034

10 .00126 '.00126 .00128 .00.130 .00133 .00138 .00144 .00151 .00159 .00170 .00182

15 .00284 .00287 .00294 .00307 .00326 .00351 .00385 .00428 .00484 .00555 .00647

0 .00510 .00518 1.00543 .00586 .00651 .00744 .00875 .01058 .01318 .01692 .02249

"5 .00806 .00826 .00889 .01003 .01184 .01463 .01892 .02569 .03689 .05678 .09612

0 .01178 .01221 .01360 .01622 .02069 .02826 .04160 .06716 .12347 .27600 -5 .01631 .01715 .01992 .02547 .03582 .05599 .10014 .21859 - - -0 .02174 .02327 .02845 .03962 .06330 .12026 - - - " - -5 .02816 .Ö3079 .04008 .06215 .11810 - - - - - -0 .03571 .04005 .05629 .10019 - - - - - - -

H5• 0 0.25 I 0.50 I 0.75 I 1.00 1.25 1.50 1.75 2.00 2.25 2.50

5 0 0 0 0 0 0 0 0 0 0 0

0 0 0.00001 0.00001 0.00002 0.00003 0.00003 0.00004 0.00005 0.00006 0.00007 0.00009 -5 0 .00003 .00007 .00010 .00015 .00019 .00025 .00033 .00042 .00054 .00069

0 0 .00011 .00022 .00035 .00052 '.00073 .00102 .00143 .00204 .00297 .00449

5 0 .00027 .00057 .00095 .00148 .00225 .00349 .00559 .00947 .01741 .03629

0 0 .00058 .00127 .00224 .00376 .00641 .01159 .02316 .05497 .18343 -.5 0 .00113 .00258 .00487. .00910 .01829 .04279 .13690 - - -0 0 .00205 .00493 .01018 .02217 .05818 - - - - -5 0 .00354 .00906 .02118 .05840 - - - - - -0 0 .00589 .01637 .04567 - - - - - - -

.. . ."

.