ch-3 properties of steam

INDIABOILER DOT COMTUTORIAL FOR SECOND CLASS BOILER ENGINEER’S PROFICIENCY EXAMINATION

DLP/BOE-II/ 1- 01092001

CHAPTER – 3Properties of Steam

1.0 Introduction:

Water vapour is called “Steam”. It is the most commonly used working substance in the operation of steam engine and steam turbines. Steam which has just been formed at the saturation temperature is a vapour which is a partially evaporated liquid carrying in it particles of liquid and under these circumstances, it can be liquefied back into water by minor changes in temperature or pressure. Steam as a vapour would not obey the laws of perfect gases unless it is in a highly dried condition. Steam in such a dried state is known as Superheated Steam. Superheated Steam is assumed to behave like a perfect gas when highly superheated. It possesses properties like those of gases – relationship between pressure, volume, temperature, internal energy, enthalpy and entropy. But the pressure, volume and temperature of steam as a vapour is not connected by any simple relationship such as are expressed by the characteristic equation for a perfect gas.

Pressure, temperature and volume can be given their actual absolute value; where as enthalpy and entropy are purely relative quantities. They are measured relatively from convenient datum condition and calculated per kg of steam. For steam, datum point is arbitrarily fixed as the condition of the water at 0 0C. Thus, the enthalpy, the internal energy and the entropy of water at 00C are measured to be zero. All their values measured above this temperature are considered positive and those measured below are taken as negative. The properties of steam and the changes in the properties have been determined by laboratory experiments by competent bodies such as ASME and others and have been published in form of Steam Tables or Graphically in the form of Steam Charts.

1.1 Formation of steam from water at constant pressure:

We know that there are three phases of substance namely solid, liquid and gaseous. Water, is in a liquid phase, Ice is its solid phase and Steam is the gaseous phase. The changes of phases take place with change in its temperature and pressure on it.

Fig. 1 shows different phases of substance “water”. By adding heat AB, the temperature of solid substance (ice) increased from –ve 0C to 00C at constant atmospheric pressure (1.0332 kg/cm2 or 1.01325 bars). At point b solid phase starts changing from solid to liquid on further addition of heat. At point c it completely changes into liquid phase.

The addition of heat BC at constant temperature (00C) is called Latent Heat of Liquefaction of Ice. Its value is 80 kcal (334.96 kJ) per kg of water. From point c, further heat addition causes rise in temperature of water. At point ‘d’, on further adding of heat phase change starts from liquid to vapour at constant temperature 1000 C at constant pressure 1.01325 bar or 1.0332 kg/cm2. At point ‘e’, liquid

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Fig. 1

phase completely changes to vapour or gaseous phase, steam. Heat CD is called Sensible Heat. Heat DE is called latent heat of evaporation or latent heat. The condition of water at point ‘d’ is called Saturated Water. At this condition vapour formation starts on minor addition of heat and on removal of heat, formation of water particles starts. At point E there are no water particles in the steam and the steam is said to be dry and is known as Dry Saturated Steam. As heating continuous further, the temperature of steam begins to rise and steam is now known as Superheated Steam and behaves more or less as a perfect gas. Heat EF is known as Superheat.

Fig. 2 Shows graphically that what happens when heat is added to one kilogram of water initially at 0 0C

The temperature corresponding to point B is called saturated temperature. Saturation temperature increases with increase of pressure and decreases with decrease of pressure. The addition of heat “ab” is known as sensible heat and is denoted by the letter “h”. As specific heat of water is 1 kcal/ kg 0C or 4.187 kJ / kg 0C, heat required for raising the temperature of 1 kg of water through 10C is 4.187 kJ. On further addition of heat, no temperature rise takes place up to point ‘C’. At point B steam begins to be formed and at point C all water is converted into steam. All added heat bc is used to increase the kinetic and potential energy of water molecules. This heat is known as latent heat of evaporation or latent heat and is denoted by the letter “L”.

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Tsup

T

A B EC D F

Tem

per

atu

re

edts

b c00 c

f

H


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Fig. 2

Latent heat decreases with increase of pressure. The value of latent heat becomes zero at pressure 225.4 kg/cm2 abs. or 220.9 bar abs. This State is known as Critical State. This pressure is called Critical Pressure and corresponding saturated temperature, 374.140C, is called Critical Temperature of steam. Between point B and C content of water will decrease as we go toward point C. At point C, all the water, including those particles of water, held in suspension will be evaporated, the steam is said to be dry and is known as Dry Saturated Steam.

The steam which contains water particles in suspension is known as Wet Steam. On heating further from point C, the temperature of steam begins to rise again and steam is now known as superheated steam. The temperature corresponding to point D is known as Superheat Temperature (tsup). The temperature difference between temperature of Superheated Steam and Saturation Temperature (tsat) i.e., (tsup – tsat) is known as a degree of superheat. Greater the degree of superheat, the more will the steam acquire the properties of a perfect gas. Superheating is assumed to take place at constant pressure. The value of specific heat of superheated steam (Cp) varies with the pressure and degree of superheat. It increases with the increase of pressure and decreases with the increase of degree of superheat.

1.2 Enthalpy:

The total heat content of the substance comprises three components namely, heat of water or Sensible Heat, heat of evaporation or Latent Heat and superheat – the additional heat imparted to the dry saturated steam to make it superheated called Super Heat. It is also know as enthalpy of water, enthalpy of super heated steam.

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Tsup

Liquid Vapour Gas

Evaporation

SensibleHeat

Latent Heat Super Heat

TsB

C

D

a b c α

Heat Added

Tem

pera

ture


DLP/BOE-II/ 1- 01092001

1.2.1 Enthalpy of Water:

The amount of heat absorbed by one kilogram of water being heated from the freezing point (00C) to the boiling point is known as the Enthalpy of the Saturated Water (sensible heat of water). It is denoted by the symbol ‘h’.

Sensible heat of 1 kg of water at 00C is zero and sensible heat of 1 kg of water at 1000C is 4.187 x 100 = 418.7 kJ/kg.

1.2.2 Enthalpy of Evaporation:

The enthalpy of evaporation (or Latent heat) is defined as the amount of heat required to convert one kilogram of water at the saturation temperature tsat

corresponding to its pressure into steam at the same temperature and pressure. Latent heat varies with the pressure. Its value decrease with increase of pressure. Its value at 1 bar absolute pressure is 2258 kJ/kg. Value of Latent heat corresponding to any pressure can be directly obtained from the steam tables.

1.2.3 Enthalpy of Dry Saturated Steam:

It is the sum of Enthalpy of Water and Enthalpy of Evaporation (i.e., Latent Heat). It is defined as the quantity of heat required to raise the temperature of one kilogram of water from freezing point to the saturation temperature and then convert it into dry saturated steam at that temperature and pressure. It is denoted by the symbol ‘Hs’.

Thus, Hs = h + L

This value ‘Hs’ can be directly obtained from the Steam Tables corresponding to given value of pressure or temperature of Steam.

1.2.4 Enthalpy of wet steam:

It is the amount of heat required to raise the temperature of one kilogram of water from freezing point (00C) to the boiling point ‘tsat’ and then to convert the boiling water into wet steam. Obviously, such wet steam shall not have absorbed full quantity of Latent Heat, the content of which will be proportional to the fraction of Dry Saturated Steam in the Wet Steam. If ‘x’ is the per unit content of Dry Saturated Steam in Wet Steam then, Hwet = h + x.L kJ/kg. The term ‘x’ is called Dryness Fraction and its value lies between 0 and 1. This is further explained below:

Wet steam is a mixture of water particle in suspension and dry steam. Dryness fraction of wet steam is a ratio of the mass of actual dry steam to mass of wet steam containing it.

If, ms= Mass of dry steam contained in the steam considered and m = mass of water in suspension in the steam considered;then, dryness fraction , x = ms/(ms + m)

1.2.5 Enthalpy of Superheated Steam:

It is defined as the amount of heat required to convert one kg of water at freezing temperature (00C) into superheated steam at a given pressure and temperature above saturation temperature. It is denoted by symbol Hsup.

Hsup = h + L + Cp (tsup – tsat) kJ/kg

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orHsup = Hs + Cp (tsup – tsat) KJ/kg

1.3 Specific Volume of Steam :

The volume of dry saturated steam in cubic metre per kg (m3/kg) is known as Specific Volume of dry Saturated Steam. Its symbol is Vs.

1.3.1 Specific Volume of Wet Steam:One kg of wet steam will consist of ‘x’ kg of dry steam and (1-x) kg of water in suspension where ‘x’ is a dryness fraction of wet steam.Specific volume of wet stream, having a dryness fraction of ‘x’

= Volume of dry steam + volume of water particles = x.Vs + (1-x).Vw

Where Vs and Vw denote the specific volume of steam and water respectively. As the volume of water at low pressure is very small compared with the steam, the term “(1-x).Vw” will become still smaller and can be neglected.

Hence, the specific volume of wet stream, neglecting volume of water particles = x.Vs m3/kg

and

density of wet steam = 1/(x.Vs) kg/m3

1.3.2 Specific Volume Of Super Heated Steam (Vsup):

Vsup = Vs . (Tsup/ Ts) m3/kg

Tsup = absolute temp of superheated steam

= absolute temp of saturated steam at same pressure.

Vs = Specific volume of dry saturated steam.

1.4 Steam Tables:

Steam Tables contain the properties of Steam in a tabular form. The properties of 1 kg of water, dry saturated steam and Specific Heat of Steam at Constant Pressure (Cp) corresponding to various temperatures as also specific volume are given in Steam Tables. The information given in these tables has been obtained experimentally by various competent bodies of international repute such as ASME. In the steam tables, the initial temperature of water is taken as 00C. The information contained in steam tables is as under:

Pressure of steam ( absolute) in bar p Saturation temperature of steam ts

Specific volume of dry saturated steam in m3/kg vs

Enthalpy of saturated water in KJ / kg h Enthalpy of dry saturated steam I KJ / kg Hs Enthalpy of evaporation of dry steam in KJ / kg L Entropy of water in KJ / kg K fw

Entropy of dry saturated steam in KJ / kg K fs

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1.5 Entropy:

1.5.1 Entropy of Water:

In general, addition of heat dQ to one kg of water will cause its temperature to rise by dT.

Then dQ = Cp. dT where Cp= specific heat of water .\ df= Cp (dT / T)

Integrating between temperature limits of T1 and T2

f2 - f1 = Cp T1òT2 dT/T = K {loge(T)}T1T2 = Cp loge(T2 /T1)

Entropy of water at 00C (freezing temperature T0 = 273 K) is taken as zero.Entropy of water at boiling temperature Ts is written as fw

fw = Cp loge(Ts /T0) = Cp loge(Ts /273) This value can be obtained directly from the steam tables.

1.5.2 Entropy of Evaporation (fe):

Heat addition during evaporation takes place at constant temperature if pressure is maintained constant. If the steam formed is wet, having dryness fraction ‘x’,

Entropy of evaporation, fe = xL / TsIf the steam formed is dry saturated, x=1,

then the entropy of evaporation is fe = L / TsThe value of fe = fs - fw may also be obtained from steam table.

1.5.3 Entropy of Dry Saturated Steam:

The entropy fs of dry saturated steam, reckoned from the freezing point of water (00C) is equal to the sum of water entropy (fw) and evaporation entropy (fe) fs = fw + fe = Cp loge(Ts /273) + L / Ts

The value of fs may also be obtained directly from the Steam Tables.

1.5.4 Entropy of Superheated Steam:

Entropy of Superheated Seam is calculated as follows:fsup = The total entropy of 1 kg of superheated steam reckoned above the

freezing temperature of water.

fsup = Cp loge Ts/ 273 + L / Ts + Cp loge Tsup / TS

= fw + fe + Cp loge Tsup / TS = fs + Cp loge Tsup / Ts

Cp is the specific heat of superheated steam.

1.6 External Work Done During Evaporation:

The enthalpy of evaporation (latent heat) absorbed by the steam during evaporation is utilized in two ways:

Firstly, in overcoming internal molecular resistance of water in changing its state from water to steam (vapour) and secondly, in overcoming the external resistance of the piston to its increasing volume during evaporation.

The first of this effect of enthalpy of evaporation (latent heat) is called Internal Work because changes have been brought within the body itself, and the second is

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called external work because work has been done on the body externally. The first represents the energy stored in the steam and is known as internal latent heat and the second represents the energy which has passed out of the steam having been utilized in doing work on the piston and is known as external work of evaporation.

The value of enthalpy of evaporation (latent heat) [= internal latent heat + external latent heat], can be directly obtained from the steam tables. In evaporating water to steam the volume increases from Vw to Vs under a constant pressure P; external work is thus done and the energy for performing this work is obtained during the absorption of enthalpy of evaporation. This work is known as external work of evaporation.

Work done per kg of dry saturated steam = 105xPx(Vs-Vw)/103 kJ/kg

where P = absolute pressure of steam in bar ,

Vs = volume of 1 kg of dry saturated steam in m3 at pressure p, and Vw= volume of 1kg of water in m3

At low pressure the volume of 1 kg of water is very small compared with the volume of the steam it forms. Hence, it is neglected from the above equation.And the equation may be written asWork of evaporation of dry steam = 105 pVs/103 kJ/kg

If steam is wet having a dryness fraction of x , the final volume of wet steam is xVs , then,

External work done per kg of wet steam = 105 P(xVs)/103 KJ / kg

External work done per kg of superheated steam = 105 PVsup/103 KJ/ kg

The above expression for external work of evaporation gives the portion of latent heat which has been transferred into mechanical work or represents the heat energy which has passed out of the steam. The remainder of the heat energy supplied remains as internal energy or energy stored in the steam and is known as internal latent heat. It is found by subtracting the external work of evaporation from the full latent heat.

i.e. Internal Latent heat = L - 105 PVsup/103 kJ / kg (for dry saturated steam)For wet steam having dryness fraction x, Internal latent heat = xL - P(xVs)/ 103 kJ/kg

1.7 Internal Energy of Steam:

The internal energy of steam is the actual heat energy above the freezing point of water stored in the steam and is the sum of internal latent heat and sensible heat reckoned from 00C.

The total heat of steam consists of the sensible heat plus the internal latent heat and the external work of evaporation. But the external work of evaporation is utilized in doing external work and represents the heat energy which has passed out of the system. Hence, the internal energy of steam denoted by symbol ‘u’ consists of two terms only namely, the internal latent heat and the sensible heat.

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Thus, internal energy is found by subtracting external work of evaporation from the enthalpy of steam.

If steam is dry saturated, enthalpy HS = ( h + L ) KJ/kg and Internal energy

If steam is wet having dryness fraction x,

enthalpy, Hwet = h + xL KJ/kg and Internal energy

If the steam is superheated to temperature tsup and volume Vsup m3 per kg.

Enthalpy of superheated steam Hsup = Hs+ Kp( tsup - ts)And Internal energy,

1.8 Vapour Formation Process:

The process of vapour (steam) formation from liquid state of water has the following important characteristics:

If the process of vaporisation is carried out at constant pressure, the temperature will remain constant until vaporization is complete and the liquid is completely evaporated.

During the process of superheating the vapour at constant pressure the volume will increase approximately in proportion to the absolute temperature, which indicates that the vapour is approaching the state of a perfect gas.

If the pressure on the surface of the liquid is increased, the temperature at which evaporation is taking place also increases and vice versa.

If the pressure on the surface of the liquid is increased, latent enthalpy (latent heat) decreases.

The specific volume decreases with the increase of pressure and temperature.

At critical temperature, the specific volume of dry saturated vapour becomes equal to the specific volume of the liquid from which it is being formed.

1.9 Constant Volume Process:

If one kg of wet steam at initial pressureP1 and dryness fraction X1 is heated at constant volume to final pressure P2, its dryness fraction will change. Let X2 be the final dryness fraction of the steam. Then, since volume remains constant,

Volume before heating = volume after heating.

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u = ( Hs -105 PVs

103 ) KJ/kg

u ={Hwet - 105 P(xVs) 103 } KJ/kg

u ={Hsup - 105 PVsup

103 } KJ/kg


DLP/BOE-II/ 1- 01092001

i.e. X1 Vs1 = X2 VS2

X2 = (X1VS1) / VS2 -------------------------(1)

The values of the specific volume (VS1and VS2) for the pressure P1and P2 can be obtained from the steam tables and final state of the steam. i.e. X2 can be determined from the equation (1) for the given value of X1

Internal energy = [Enthalpy of 1 kg of steam] - [External work of evaporation]

and,

where P1 and P2 are in bar.

If P1 and P2 are in Kpa then U1 = H1 – P1 ( X1 VS1) kJ andU2 = H2 – P2 ( X2 VS2) kJ

Hence change in internal energy U2 - U1 may be calculated.

Since in this process there is no change of volume, no external work is performed. i.e. w = 0

Applying law of conservation of energy Heat supplied Q = ( U2-U1) + w = U2 – U1 + 0 = (U2-U1) KJ

1.9 Constant Pressure Process:

If heat is supplied to steam at constant pressure its dryness fraction and internal energy change and the work is done against constant external resistance (pressure). This represents the evaporation (or drying) of wet steam.

Let the dryness fraction of one kg of steam changes from X1 to X2

If the specific volume of steam at pressure Properties is Vs m3 /kg.

External work done,

Heat supplied Q = ( U2 – U1 ) + w

= H2 – H1 kJ

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U2 ={H2 - 105 P2(X2Vs2) 103 } KJ/kg

U1 ={H1 - 105 P1(X1Vs1) 103 } KJ/kg

w = 105 P (X2Vs2) 103

105 P (X1Vs1) 103-

105 PVs (X2 - X1) 103 kJ/kg=

═ {H2 - 105 P2(X2Vs2) 103 } ─ {H1 - 105 P1(X1Vs1)

103 } 105 PVs (X2 - X1) 103+


DLP/BOE-II/ 1- 01092001

This means that the heat supplied to steam at constant pressure is equal to the change in enthalpy of steam.

As long as steam is wet its temperature has a definite value at a given pressure and as such the temperature remains constant in a constant pressure process. From this fact it follows that for wet steam, a constant pressure process is also a constant temperature process or isothermal process.

If the wet steam is further heated after the dry saturation condition is reached, it will be superheated.

Let the absolute temperature of superheated steam be Tsup then,

External work done, W = (105 PVs) / 103 [(Tsup/Ts) – x1]

Heat supplied Q = Hsup – H1 KJWhere Hsup = enthalpy of superheated steam and H1 = enthalpy of wet steam

It may be noted that once the steam is superheated the expansion will no longer be isothermal.

1.10 Constant Temperature (Isothermal) Process:

As explained in the constant pressure process for wet steam, a constant temperature process is also a constant pressure process. As soon as the steam becomes superheated, it will approximately behave like a perfect gas and will follow hyperbolic law at constant temperature. The Isothermal process will therefore be hyperbolic process for superheated steam i.e., product of pressure and volume will remain constant.

1.11 Hyperbolic process:

For wet steam the hyperbolic process , i.e. PV = constant , is not an isothermal process as in the case of perfect gases. In case of superheated steam, hyperbolic process may be regarded as an isothermal process since superheated steam behaves like a perfect gas.

Let 1 kg of wet steam at initial pressure P1 and dryness fraction X1 be expanded in cylinder, according to the law PV = constant to the final pressure P2 and unknown dryness fraction X2 Then,

Initial volume of steam V1 = x1VS1 and

Final volume of steam V2 = x2 VS2

The values of specific volume (VS1 and VS2) for pressure P1 and P2 can be obtained from steam tables.

Then, as PV = constant, P1V1 =P2V2 i.e. P1( x1VS1) = P2 ( x2VS2)

From which the final state of steam, i.e. x2 can be calculated.

If x2 is found to be greater than unity, it follows that the steam is superheated then assuming the superheated steam to behave as a perfect gas,

final volume, V2 = VS2 x Tsup / Ts of 22


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Where Tsup = absolute temperature of superheated steam (unknown) and Ts = absolute temperature of formation of steam at pressure

P2 ( from steam tables)

Thus hyperbolic law is P1 ( x1VS1) = P2 VS2* Tsup

Ts --- (Eq. A)

From above Eq. A, the value of Tsup can be calculated. But, as P1 ( x1 VS1) = P2 ( x2 VS2) or

P2VS2* Tsup

Ts i.e. P1V1

= P2V2

It follows that the external work of evaporation is constant for a hyperbolic expansion.

Now, H1=h1+X1L1 and H2 = h2+X2L2

If the final condition of steam is wet or H2 =h2 +L2+KP(tsup-ts), If steam is superheated.

\ H2 – H1 = [ u2 + ] ─ [ u1 + ] ═ (u2 ─ u1) KJ

Thus .for hyperbolic expansion change of internal energy is equal to change in internal enthalpy.

Work done on piston, W ═ ═

Where r = V2 / V1 = P1 / P2

Applying law of conversation of energy , heat transferred during the process.

Q = w + ( U2 – U1 ) = [ + (U2 – U1) ] KJ

1.12 Polytropic process:

This process is covered by the general law PVn = constant during this process there is a transfer of heat between steam and the cylinder walls and external work is done on the piston. The law is applicable to saturated steam and superheated steam.

Let 1 kg of wet steam be expanded in a cylinder so that the expansion follows the law PVn = constant, the value of n being known. Let the initial condition of the steam be pressure P1 and dryness fraction X1 and let the steam be expanded to a pressure P2and unknown dryness fraction X2.

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105 P2V2

103

105 P1V1

103

105 P1( x1Vs1 ) log e (r) 103

105 P1V1 log e (r) 103

105 P1( x1Vs2 ) 103


DLP/BOE-II/ 1- 01092001

Now, P1V1n = P2V2

n i.e.

P1 ( x1VS1)n = P2 ( x2VS2)n

\x2 = ( P1/P2) 1/n * (x1VS1 / VS2 )

From which the value of X2 may be calculated. If X2 is found to be greater than unity, the steam will be superheated after the expansion. In that case,

P1( x1VS1)n = P2 ( VS2 * Tsup/ Ts)

\ Tsup = (P1 / P2) 1/n * X1 VS1 / VS2 * Ts

H1 = h1 + x1L1

H2 = h2 + x2L2 if wet after expansion.

H2 = HS2 + Cp ( tsup+ts) if superheated after expansion.

U1 = H1 - and U2 = H2 - (If wet after expansion) Or U2 = H2 - (If superheated after expansion )

WORK DONE ON PISTON DURING EXPANSION

105 P1V1 – P2V2 105 P1(x1VS1) – P2(x2VS2)W = 103 n - 1 = 103 n - 1

(If wet after expansion)

W = 105 / 103 P1 (x1 VS1) – P2 (VS2 * TSUP / Ts)

(If steam is superheated after expansion)

Applying the law of conservation of energy

Heat transferred, Q = W + ( U2 – U1) KJ

This will give the heat absorbed from or rejected to the cylinder walls by the steam during expansion. Positive sign will indicate that heat is received by the steam from the cylinder walls and negative sign will indicate that heat lost or rejected by the steam to the cylinder walls.

1.13 Isentropic Process:

Since during this process there is no transfer of heat between the medium (steam)and the cylinder walls, external work will be done on the piston during the process of expansion at the cost of internal energy , if friction is assumed to be

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105 P1( x1Vs1 ) 103

105 P2( x2Vs2 ) 103

105 P2( Vs2 - Tsup / Ts ) 103


DLP/BOE-II/ 1- 01092001

absent. The exponent of law, PVn = constant, no longer has a constant value as it has for a perfect gas, However, the external work can be calculated directly. Let 1 kg of steam at pressure P1 and dryness fraction X1 be expanded Isentropically internal a cylinder to a final pressure P2 and dryness fraction X2 at the end of expansion (internal which friction is totally absent) can be determined by applying the rule that entropy of the steam before expansion is equal to the entropy of the steam after expansion. i.e.fW1 + x1 fe1 = fW2 + x2 fe2

OR

fW1 + x1( fS1-fW1) = fW2 + x2( fS2-fW2) ------------(2)

If the steam was initially superheated to an absolute temperature Tsup , equ. (2) would become

fW1 + fe1 + Cp log e ( Tsup / Tsat ) = fW2 + x2 fe2

OR

fs1 + Cp log e ( Tsup / Ts ) = fW2 + x2( fs2 - f w2 )

Isentropic ( ideal adiabatic) expansion is represented by a vertical line on the T - f and H - f chart. Having determined the final dryness fraction ( X2) of the steam, the change in internal energy may now be calculated as under.

and

Heat transferred during isentropic ( frictionless adiabatic) process. Q = 0. Applying law of conservation of energy Heat transferred, Q = W+(U2 – U1 ) i.e. 0 = W+(U2 – U1 )Hence external work done, W = (U1 – U2 )

This means that “ during an isentropic process, the external work done is equal to the change in internal energy. For approximate result, it is often assumed that the law for isentropic process for steam is PV n = constant.

Where n = 1.13 for wet steam, and n = 1.3 for superheated steam.

Dr. Zeuner has suggested the well known equation n = 1.035 + 0.1 X where X is the initial dryness fraction of the steam.

105( P1* x1Vs1 – P2* X2Vs2) Then, work done on piston = 103 (n-1)

1.14 Throttling process:

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U1 ═ {H1 - 105 P1(X1Vs1) 103 }

U2 ═ {H2 - 105 P2(X2Vs2) 103 }


DLP/BOE-II/ 1- 01092001

When steam is forced through a small orifice under pressure, the steam is said to be throttled. Throttling process occurs when steam passes through a partially open valve. At the orifice, eddies are formed which eventually reconvert kinetic energy into heat at the lower pressure, So that throttling is always a wasteful operation which lowers the quality of heat. There is no change of potential energy, no external work is done and if the pipe is well lagged there is no heat transferred. This process is also called constant enthalpy process. Since no work is done during the expansion, the enthalpy remains constant. During the process of throttling the temperature of the steam is slightly reduced but this is compensated by the heat produced due to the friction offered throttling the steam by the surface of the very small orifice through which the steam has to pass. This heat due to friction, dries the steam ( i.e. increases the dryness fraction of wet steam ) and once the steam becomes completely dry it will be superheated by further throttling.

1.15 Methods of determination of dryness fraction of steam:There are four methods of determining the dryness fraction of steam. The methods are as follows:(i) Bucket calorimeter(ii) Separating calorimeter(iii) Throttling calorimeter(iv) Combined separating and throttling calorimeter.

1.16 Bucket calorimeter:This method of determining the dryness fraction of steam is not very accurate. The value of dryness fraction obtained by this method is lower than the actual value. A known weight of steam is passed through the water where steam is completely condensed. The heat lost by steam is equated to the heat gained by the water.

1.17 Separating calorimeter:This calorimeter is generally used to measure the probable value of dryness fraction of steam when the steam is very wet.It consists of a device for trapping the water and draining it off. The steam which escapes the instrument is condensed in a condenser. While calculating the dryness fraction of the steam it is assumed that all the water particles in suspension are

separated out and the steam leaving the separator is dry. This assumption is not correct so the results obtained are approximate.

1.18 Throttling calorimeter:

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It is shown combined with separating calorimeter in the figure.

When steam is allowed to pass through a restricted opening such as a partially closed valve, the steam is said to be throttled. The pressure of steam decreases but the volume increases. If no heat is lost by leakage to atmosphere, the total energy of the steam is the same before throttling and after throttling. Since the total heat of steam remains the same after throttling, the steam becomes drier. If the dryness fraction of steam is high initially, then after throttling the steam gets superheated and the temperature of superheater heat can be measured by a thermometer. The pressure of the steam after throttling can be read by a manometer. As the pressure and the temperature of the steam after throttling are known, the enthalpy of steam after throttling can be calculated. Since the enthalpy before throttling = enthalpy after throttling, the unknown quantity, the dryness fraction of steam before throttling can be calculated.Enthalpy before throttling = hf1 + x1 x hfg1

Where p1 = absolute pressure of steam before throttling hf1 = sensible heat of steam corresponding to p1,hfg1 = latent heat of steam corresponding to p1,and x1 = dryness fraction of steam (at p1) entering the throttling unit.Enthalpy after throttling (steam being superheated)= hg2 + cp (tsup – tsat)where p2 = pressure of steam, after throttling,hg2 = enthalpy of 1 kg of dry saturated steam corresponding to p2

tsup = temperature of superheated steam after throttling.tsat = temperature of saturated steam at constant pressure.Enthalpy before throttling = Enthalpy after throttling

Therefore, hf1 + x1 x hfg1 = hg2 + cp (tsup – tsat) = hsup2

Therefore, x1 = {hg2 + cp (tsup – tsat) – hf1 / hfg1} = (hsup2 – hf1) / hfg1

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1. Sampling inlet2. Shut off valve3. Separating unit4. Sight valve5. Connection to pressure gauge6. Throttling nozzle7. Thermometer pocket8. Manometer9. Connetion for condenser 10. Throttling unit11. Drain valve for water


DLP/BOE-II/ 1- 01092001

Examples

Question: Differentiate between Enthalpy and Entropy.

Answer:

Enthalpy EntropyIt is the summation of internal energy and flow work

It is not energy but simply a property

Mathematically it is defined as h=u+pv

Mathematically it is defined asdØ=dQ/T

It is one form of stored energy, which can be converted to another form like work, kinetic energy etc.

It can not be converted into another form but it can be created.

It is not used for deciding the direction of the process.

It is useful for deciding the direction and effectiveness of the process and measures the molecular disorder.

For mixing it obeys algebraically additive law

It does not obey the algebraically additive law in case of mixing

It’s unit is kJ/kg It’s unit is kJ/kg K

Question:. Differentiate between the following;Dry steam and Superheated steam

Answer:

Dry Steam Superheated SteamTotal heat content per kg of steam is equal to h + L, and is lower than superheated steam

Total heat content per kg of steam is equal to h + L + Cp( tsup- tsat)

Temperature of steam is equal to saturated temp. of steam.

Temperature of steam is higher than saturated temperature.

Sp. Volume of dry steam is lower than superheated steam at same pressure.

Sp.volume of superheated steam is higher than dry steam at same press. of 1 kg sup. steam.

Temp. & total heat is directly available in the steam table.

Total heat is to be calculated and not directly available in the steam table.

For producing dry steam no need of superheat, only steam seperator is required to install in the drum

For producing superheated steam, super heater is additionally required to be installed.

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Example:(1) Two boilers A & B are delivering steam in equal amount to a common mains steam pipe both at a pressure of 14 kg / cm2 ab. Boiler A has a superheater and boiler B is without superheater. The temperature of steam supplied by the boiler A is 300 Deg.C the temperature of the resulting mixture of steam internal the steam mains is 235

Deg.C Assuming Kp for specific heat of steam supplied by the boiler B without the superheater. (Aug.88)

Solution: Let us assume each boiler delivers 1 kg of steam .Then the mass of the mixture of steam = 1 + 1 = 2 kg. Since boiler ‘A’ is having superheater, its quality of steam is

Pressure - 14 kg / cm2

Temp. - 300 0C.Boiler B doesn’t have a superheater, so its quality of steam may be wet or dry saturated and temp. of the mixture of steam is 235 0C.Now from steam table, at 14 kg/cm2 ab.

ts = 194.14 0Ch = 197.1 Kcal / kgL = 468.9 Kcal / kgH = 666.0 Kcal / kg

As ts is 194.14 0C, the quality of the mixture of steam is superheated.(1) Total heat of 1 kg of steam of boiler A

= H + Kp (tsup - tsat)= 666 + 0.5 (300 – 197.1)= 717.45 Kcal / kg

(2) Total heat of 1 kg of steam of mixture of steam

= H + Kp (tsup - tsat)= 666 + 0.5 ( 235 – 197.10)= 684.95 Kcal / kg.

(3) Total heat of 2 Kg of mixture of steam = 2 * 684.95= 1369.9 --------------------

Now the quality of steam of boiler B may have some dryness fraction \ Total heat of 1kg steam of boiler B = h+XL

= (197.1 +X * 468.9) Kcal /kgNow Total heat of mixture of 2 kg steam

=Total heat of 1 kg steam of boiler A + Total heat of 1 kg steam of boiler B=(717.45 + 197.1 + X * 468.9) Kcal /kg

But we have calculated the total heat of 2 kg mixture of steam .Which is equal to 1369.90 Kcal.Equating both,

717.45 + 197.1+ X * 468.9 = 1369.90 X * 468.9 = 1369.90 – 717.45 – 197.1 X * 468.9 = 455.35

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X = 455.35/ 468.9 = 0.9711=97.11 %

Example :(2) 1 m3 of steam at a pressure of 8 ata and having dryness fraction of 0.9 losses 1000 Kcal of heat at constant pressure .Calculate the dryness fraction at the end of the heat loss. (Aug.89)

Solution: Volume of steam = 1.0 m3

Pressure of steam = 8 ataDryness fraction of steam before heat loss = 0.9

Heat loss = 1000 Kcal Dryness fraction of steam after heat loss = ?

From steam tables , at pressure 8 atats = 169.61 Deg Constanth = 171.3 Kcal /kgL = 489.5 Kcal /kgVs = 0.2448 m3 / kg

Mass of 1m3 steam = 4.08496Total heat of 1 kg steam before heat loss = h + XL

= 171.3 + 0.9 * 489.5 = 611.85 Kcal /kg

\ Total heat of 4.08496 kg of steam = 4.08496 * 611.85 = 2499.3827 Kcal. Total heat of 4.08496 kg of steam after heat loss

= 2499.3827 -1000

= 1499.3827 Kcal /kg -----------But total heat of 4.08496 kg of steam after heat loss

= 4.08496 (h + XL) = 4.08496 ( 171.3 + X * 489.5) ------------

Equating & 1499.3827 = 4.08496 (171.3 + X * 489.5)\171.3 + X * 489.5 = 1499.3827 / 4.08496 = 367.0495

\X = 367.0495 - (171.3 / 489.5) = 0.3998 = 0.40

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DLP/BOE-II/ 1- 01092001

Examples for practice

Question: Explain the following terms. (Aug. 96)1) Critical pressure and critical temperature at steam.2) Sensible and Latent heat of steam.3) Specific volume of steam.4) Wire drawing ( throttling )

Question: What do we understand by the term specific volume? How will you determine it for

(a) wet steam,(b) dry and saturated steam, and(c) superheated steam?

Question: Explain clearly the terms ‘ Dryness fraction’ and superheat minus the external wok done. (Aug., 00)

Example:1. Steam at a pressure of 7 at a is allowed to pass through a throttling calorimeter and temperature is 112c. Determine the dryness fraction of steam. Assume the specific heat of superheated steam as 0.5 (Aug-1994).

Example:2. 20 m3 of steam is at a pressure of 10 ata having dryness fraction of 0.95.This steam losses 2000 kcal of heat at constant pressure. Calculate the dryness fraction of the steam after the completion of the process. (Feb-1995) Example: 3. Steam at 18bar pressure and dryness 0.9, if heated at constant pressure, until dry and saturated. Find the increase in Volume, heat supplied and work done/kg of steam. If the volume is not kept constant, find how much heat must be extracted to reduce the pressure to 14 bar. (Sept-1997)Example: 4. Determine total heat and volume of 1kg of steam at 8.7 kg cm 2 ab and 225 0c take specific heat at constant pressure for superheated steam as 0.5. (Feb. 91)Example: 5. Two boilers discharge equal amount of steam into the main steam pipe the steam from one is at 14 kg/cm2 and 290c and from the other at 144kg/cm2 and 92% dry what will be the resulting quality of steam heat of superheated steam as 0.52 (Aug 1994)Example: 6. A Copper vessel of mess 3kgf contains 6 kg of water, if the initial temperature of vessel plus water is 20 oc and the final temperature is 90 oc, how much heat is added to accomplish these changes? (Aug 1999)Example: 7. The feed water to boiler enters the feed heater at 21c where the exhaust steam of a Boiler feed water pump is blown into it. This exhaust steam is 0.9 day at 1 bar and its Mass is 3% of the steam output of a Boiler. Negating losses, Calculate the temperature of the feed water leaving the feed heater. Take the specified heat capacity of water as 4.187 KJ / kg K. (Aug. 2000)

Example: 8. Find the enthalpy of one m3 of steam at 12 bar absolute and dryness

fraction 0.8 find also the Internal energy per m3 steam. (Aug-2000) of 22


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Example: 9. Calculate the total heat of 1 kg. of steam at a pressure of 8kg/cm2 abs. having a dryness function of 0.8 (Feb-1995)

Example: 10. A closed vessel of 0.24 m3 capacity contains steam at pressure of 11 ata and temperature of 200’c Find out the quantity of steam in the vessel. At what pressure in the Vessel will the steam by dry and saturated, if the vessel is cooled? The cooling is further continued till temperature of steam in the vessel becomes 116.3 oc Find out the pressure and condition of steam at the end of operation. (Sept 1997)

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ASSESSMENT SHEET

Question: State whether the statements are True or False, If False correct and rewrite them.

1) Water at pressure of 4 kg / cm2 and 140 0C. temperature when exposed to atmosphere will cool down. (Aug.94)

2) The dry saturated steam at very high pressure ( 150-200 kg/cm2) when throttled to atmosphere will become wet. (Oct.95, Aug, 96, Sep.97)

3) If X is the weight of dry steam and Y is the weight of water internal suspension then dryness fraction is equal to X-Y / X (Oct.95 & Aug 94)

4) The presence of impurities in ice raises its melting point. (Aug.94, Oct, 95)5) Specific heat capacity may vary with volume. (Aug., 96)6) Total heat of dry saturated steam is higher than that of superheated steam at

the same pressure. (Feb.88)7) Sensible heat of water at pressure of 3 kg /cm2 ab. Is 120 0C (Feb.88)8) The maximum temperature to which water can be heated in an economiser is

100 0 C. below the steam formation temperature not corresponding to the working pressure of the boiler. (Sept., 97, Feb., 2000)

9) Specific heat of water is constant but it increases with the decrease in the saturation temperature.

10) The dryness fraction of steam is defined as the product of the volume of dry steam internal a certain quantity of steam to the volume of total steam. (June, 98)

11) Throttling is the process of reversible drop in the pressure of a gas or steam flowing through a contraction. (June. 98)

12) Temperature of 0.9 dry steam at 5.0 bar is 135 0 C. (Feb., 99)13) At critical conditions latent heat of steam is minimum. (Feb., 99)14) The internal energy is the total heat of the steam minus the external work done. (Aug., 00)15) As the steam pressure increases Latent heat also increases. (Aug., 00)

Question: Explain the following terms referred to steam.a. Enthalpy of water.b. Enthalpy of evaporationc. Enthalpy of dry and saturated steam. (Aug. 96)

Example: 1 Find total heat required to produce 5 kg of steam at a pressure of 10 kg/cm2

ab. and temperature of 250 0C. from water at 35 0C. Take the specific heat of steam as 0.55 Kcal / kg 0C. ( Feb. 1988)

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Example: 2. Determine total heat and volume of 1kg of steam at 8.7 kg /cm 2 ab and dryness fraction 0.9. (Feb. 1991)

Example: 3. Find heat required to produce 3kg of steam at pressure of 7kg /cm2ab and temp of 225 oc from water at 27 oc . Take specific heat at constant pressure of super heated steam as 0.55 (March 1992)

Example: 4. Find out the internal energy of 1kg of steam at a pressure of 10ata when the condition of steam is

(a) Wet at dryness fraction 0.85(b) Dry at saturated (c) Super heated the degree of super heat being soc (d) Super heated the degree of super heat being soc

The specific heat of superheated steam at constant pressure is 0.54 (Aug -94)

Example: 5. Calculate the temperature, the molecular mass, the specific volume and the molecular volume of air. If 1.8 kg of air occupies 0.12 m3 at Pressure 30 bar absolute Assuming R = 0.287 kg/kg (Aug 1999)

Example: 6. Two boilers delivers stream in a common main. The stream supplied by 1st Boiler is at a pressure by 2nd boiler is 0.95 dry and at pressure of 10.5 bar. What will be the temperature of stream in the common main if the stream generation rate of 1st Boiler is 3000 kg/kh and of 2nd Boiler is 2400 kg/m. Assume specific heat of super hearted stream to be 2.30 KJ /kg K ( Feb. 2000 )

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ch-3 properties of steam

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