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Engineering Thermodynamics Dr. Arif Al-Qassar 2016-2017

Definitions

Adiabatic process is a process during which there is no heat transfer. The word adiabatic

comes from the Greek word adiabatos, which means not to be passed.

Bar is the unit of pressure equal to 105 pascal.

Barometer is a device that measures the atmospheric pressure; thus, the atmospheric

pressure is often referred to as the barometric pressure.

Bernoulli equation for frictionless flow, it states that the sum of the pressure, velocity,

and potential energy heads is constant. It is also a form of the conservation of momentum

principle for steady-flow control volumes.

Boiler is basically a large heat exchanger where the heat originating from combustion

gases, nuclear reactors, or other sources is transferred to the water essentially at constant

pressure.

Boundary work (PdV work) is the work associated with the expansion or compression

of a gas in a piston-cylinder device. Boundary work is the area under the process curve

on a P-V diagram equal, in magnitude, to the work done during a quasi-equilibrium

expansion or compression process of a closed system.

Bourdon tube, named after the French inventor Eugene Bourdon, is a type of commonly

used mechanical pressure measurement device which consists of a hollow metal tube bent

like a hook whose end is closed and connected to a dial indicator needle.

British thermal unit BTU is the energy unit in the English system, representing the

energy needed to raise the temperature of 1 lbm of water at 68F by 1F.

Calorie (cal) is the amount of energy in the metric system needed to raise the

temperature of 1 g of water at 15 C by 1C.

Classical thermodynamics is the macroscopic approach to the study of thermodynamics

that does not require knowledge of the behavior of individual particles.

Clausius statement of the second law is expressed as follows: It is impossible to

construct a device that operates in a cycle and produces no effect other than the transfer

of heat from a lower-temperature body to a higher-temperature body.

Closed system consists of a fixed amount of mass (control mass), and no mass can cross

its boundary. But energy, in the form of heat or work, can cross the boundary.

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Compression ratio (r ) of an engine is the ratio of the maximum volume formed in the

cylinder to the minimum volume. Notice that the compression ratio is a volume ratio and

should not be confused with the pressure ratio.

Compressor is a device that increases the pressure of a gas to very high pressures

(typical pressure ratios are greater than 3).

Conduction is the transfer of energy from the more energetic particles of a substance to

the adjacent less energetic ones as a result of interaction between particles.

Conservation of energy principle states that, energy can change from one form to

another but the total amount of energy remains constant. That is, energy cannot be

created or destroyed (see first law of thermodynamics).

Convection is the mode of energy transfer between a solid surface and the adjacent fluid

that is in motion, and it involves the combined effects of conduction and fluid motion.

Cycle is a process, or series of processes, that allows a system to undergo state changes

and returns the system to the initial state at the end of the process. That is, for a cycle the

initial and final states are identical.

Density is defined as mass per unit volume.

Enthalpy H (from the Greek word enthalpien, which means to heat) is a property and is

defined as the sum of the internal energy U and the PV product.

Entropy can be viewed as a measure of molecular disorder, or molecular randomness. is

a property designated S and is defined as dS =(Q/T)int rev.

Equation of state is an equation that relates the pressure, temperature, and volume of a

substance, (PV=mRT).

Equilibrium implies a state of balance. In an equilibrium state there are no unbalanced

potentials (or driving forces) within the system. A system in equilibrium experiences no

changes when it is isolated from its surroundings.

First law of thermodynamics is simply a statement of the conservation of energy

principle; it may be expressed as follows: Energy can be neither created nor destroyed; it

can only change forms.

Gage pressure is the difference between the absolute pressure and the local atmospheric

pressure.

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Heat transfer (heat) is defined as the form of energy that is transferred between two

systems (or a system and its surroundings) by virtue of a temperature difference.

Ideal gas is a gas that obeys the ideal-gas equation of state.

Ideal gas specific heat relation is R = Cp - Cv

Internal energy (U) of a system is the sum of all the microscopic forms of energy.

Irreversible processes are processes which, once having taken place in a system, cannot

spontaneously reverse themselves and restore the system to its initial state.

Irreversibilities are the factors that cause a process to be irreversible. They include

friction, unrestrained expansion, mixing of two gases, heat transfer across a finite

temperature difference, electric resistance, inelastic deformation of solids, and chemical

reactions.

Isentropic process is an internally reversible and adiabatic process. In such a process

the entropy remains constant.

Iso prefix is often used to designate a process for which a particular property remains

constant.

Isobaric process is a process during which the pressure P remains constant.

Isochoric process (isometric process) is a process during which the specific volume v

remains constant.

Isolated system is a closed system in which energy is not allowed to cross the boundary.

Isothermal process is a process in which the temperature is maintained constant.

Joule (J) is a unit of energy and has the unit (N·m).

kelvin is the temperature unit of the Kelvin scale in the SI, (T=t+273).

Kinetic energy (KE) is energy that a system possesses as a result of its motion relative to

some reference frame. When all parts of a system move with the same velocity, the

kinetic energy is expressed as KE = m V2/2.

Mach number, named after the Austrian physicist Ernst Mach (1838–1916), is the ratio

of the actual velocity of the fluid (or an object in still air) to the speed of sound in the

same fluid at the same state.

Manometer is a device based on the principle that an elevation change of z of a fluid

corresponds to a pressure change of P/ g, which suggests that a fluid column can be

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used to measure pressure differences. The manometer is commonly used to measure

small and moderate pressure differences.

Mass flow rate (m.) is the amount of mass flowing through a cross section per unit time.

Open system is any arbitrary region in space through which mass and energy can pass

across the boundary.

Polytropic process is a process in which pressure and volume are often related by

PVn= C, where n and C are constants, during expansion and compression processes of

real gases.

Potential energy (PE) is the energy that a system possesses as a result of its elevation in

a gravitational field and is expressed as PE = mgz.

Power is the work done per unit time is called and has the unit kJ/s, or kW.

Pressure is defined as the force exerted by a fluid per unit area.

Pressure ratio is the ratio of final to initial pressures during a compression process.

Process is any change that a system undergoes from one equilibrium state to another. To

describe a process completely, one should specify the initial and final states of the

process, as well as the path it follows.

Pump is a steady flow device used to increase the pressure of a liquid while compressors

increase the pressure of gases.

Pure substance is a substance that has a fixed chemical composition throughout.

Radiation is the transfer of energy due to the emission of electromagnetic waves (or

photons).

Refrigerator is a cyclic device which causes the transfer of heat from a low-temperature

region to a high-temperature region. The objective of a refrigerator is to maintain the

refrigerated space at a low temperature by removing heat from it.

Specific heat is defined as the energy required to raise the temperature of a unit mass of a

substance by one degree. In general, this energy will depend on how the process is

executed.

Specific heat at constant pressure (Cp)is the energy required to raise the temperature of

the unit mass of a substance by one degree as the pressure is maintained constant.

Specific heat at constant volume (Cv)is the energy required to raise the temperature of

the unit mass of a substance by one degree as the volume is maintained constant.

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Specific heat ratio (γ or k) is defined as the ratio Cp/Cv.

Steady implies no change with time. The opposite of steady is unsteady, or transient.

Surroundings are everything outside the system boundaries.

Thermal efficiency th is the ratio of the net work produced by a heat engine to the total

heat input, th = Wnet/Qin.

Thermodynamics can be defined as the science of energy. Energy can be viewed as the

ability to cause changes. The name thermodynamics stems from the Greek words therme

(heat) and dynamis (power), which is most descriptive of the early efforts to convert heat

into power. Today the same name is broadly interpreted to include all aspects of energy

and energy transformations, including power production, refrigeration, and relationships

among the properties of matter. (i.e. thermodynamics: the science that deals with heat and

work and those properties of matter that relate to heat and work).

Work is the energy transfer associated with a force acting through a distance.

Working fluid is the fluid to and from which heat and work is transferred while

undergoing a cycle in heat engines and other cyclic devices.

Zeroth law of thermodynamics states that if two bodies are in thermal equilibrium with

a third body, they are also in thermal equilibrium with each other.

Ministry of Higher Education and Scientific Research

University of Technology

Control and Systems Engineering Department

Third Year/ Control Engineering Branch

Engineering Thermodynamics

Asst. prof. Dr. Arif A. Al-Qassar

2016-2017

References:

Rayners Joel "Basic Engineering Thermodynamics in SI Units"

Yunus Çengel and Michael Boles "Thermodynamics: An Engineering Approach"

د. رحيم جوي "اساسيات ديناميك الحرارة "

الحرارية" واآلالتالثيرموديناميك د. مرتضى الكواكبي "

Chapter One

General Introduction

1.1. Introduction

Thermodynamics is a branch of physical science that treads with various phenomena

of energy and related properties of maters, especially the low of transformation of heat into

other forms of energy.

The name thermodynamics stems from the Greek words therme (heat) and dynamics

(power), which is most descriptive of the early efforts to convert heat into power. Today the

same name is broadly interpreted to include all aspects of energy and energy transformations,

including power production, refrigeration, and relationships among the properties of matter.

1.2. Working Substance

All thermodynamic systems require some working substance in order to perform

various operations. The working substances are, in general, fluids which are capable of

deformation in that they can readily be expanded and compressed. The working substance

also takes part in energy transfer. For example it can receive or reject heat energy or it

can be the means by which work i s d on e . Common examples of working substances

used in thermodynamic systems are air and steam.

1.3. Properties and State

Any characteristic of the substance which can be observed or measured is called a

property of the substance. Examples of properties are pressure, volume and temperature. This

type of property which is dependent upon the physical and chemical structure of the

substance is called an internal or thermostatic property.

Properties which are independent of mass, such as temperature and pressure, are said to be

intensive properties.

Properties which are dependent upon mass, such as volume and energy in its various forms,

are called extensive properties.

Knowledge of the various thermostatic properties of a substance defines the state of the

substance. If a property, or properties, are changed, then the state is changed.

1.4. Phase

When a substance is of the same nature throughout its mass it is said to be in a phase.

Matter can exist in three phases, solid, liquid and vapour or gas. If the matter exists in only

one of these forms then it is in a single phase. If two phases exist together then the substance

is in the form of a two-phase mixture. Examples of this are when a solid is being melted into

a liquid or when a liquid is being transformed into a vapour. In a single phase the substance is

said to be homogenous.

1.5. Process

When the state of a substance is changed by means of an operation or operations

having been carried out on the substance, then the substance is said to have undergone a

process. Typical processes are the expansion and compression of a gas or the conversion of

water into steam.

1.6. Cycle

If processes are carried out on a substance such that, at the end, the substance is

returned to its original state, then the substance is said to have been taken through a cycle.

This is commonly required in many engines. A sequence of events takes place which must be

repeated and repeated. In this way the engine continues to operate. Each repeated sequence of

events is called a cycle.

1.7. The System

All physical things in nature have some form of boundary whose shape in general

identifies it. Inside its boundary there are certain things with particular functions to carry

out. This inside arrangement is called a system. Outside the boundary of the object are the

surroundings.

If the mass of a system remains constant then the system is said to be a closed system,

for example a piston of a cylinder.

If the mass of a system changes, or is continuously changing, then the system is said to

be an open system. For example, an air compressor is an open system since air is

continuously streaming into and out of the machine, in other words, air mass is crossing its

boundary. This is called a two-flow boundary system. Another example is air leaving a

compressed air tank. This would be a one-flow boundary system since air is only leaving

the tank and none is entering. If the mass and energy cannot be transferred to or from the

surrounding, the system is said to be an insulated system. Closed and open systems are

illustrated in Fig. 1.1.

Fig. 1.1. Closed and Open Systems

1.8. Energy

Energy is defined as that capacity a body or working substance to do work in a

possess.

Here, work is defined, as in mechanics, as the result of moving a force through a distance.

The presence of energy can only be observed by its effects and these can appear in many

different forms.

An example where some of the forms in which energy can appear is in the motor car.

The petrol put into the petrol tank must contain a potential chemical form of energy

because by burning it in the engine, the motor car, through various mechanisms, is propelled

along the road. Thus, work, by definition. is being done because a force is being moved

through a distance.

As a result of burning the petrol in the engine, the general temperatures of the working

substances in the engine, and the engine, will be increased and this increase in

temperature must initially have been responsible for propelling the motor car.

Due to the increase in temperature of the working substances then. since the motor car is

moved and work is done, the working substance at the increased temperature must have

contained a form of energy resultant from this increased temperature. This energy content

resultant from the consideration of the temperature of a substance is called internal energy.

Some of this internal energy in the working substances of the engine will transfer to the

cooling system of the engine because the cooling water becomes hot. A transfer of energy in

this way, because of temperature differences, is called heat-transfer.

The motor car engine will probably have an electric generator which rotated by the engine and

is used to charge the battery. The battery, by its construction and chemical nature, stores

energy which can appear at the battery terminals as electricity. The electricity from the battery

can be used to rotate the engine starter which, in turn, rotates and starts the engine. By rotating

the engine to start it, the electric motor must be doing work and thus, electricity must have the

capacity for doing work, and hence is a form of energy.

To stop the motor car the brakes are applied. After the motor car has stopped the brake

drums are hot and thus, as discussed above, the internal energy of the brake drum

materials must have been increased. This internal energy increase resulted from the stopping

of the motor car and hence there must have been a type of energy which the motor car

possessed while it was in motion. This energy of motion is called kinetic energy.

From this discussion it will be seen that energy can appear in many forms and further, it

appears that energy, through the action of various devices, can be converted from one form

into another.

1.9. Temperature

Temperature is a measurement of the average kinetic energy of the molecules in an

object or system and can be measured with a thermometer or a calorimeter. It is a means of

determining the internal energy contained within the system.

The Celsius scale was the most commonly used scale in the countries using a metric

system of units. The thermodynamic temperature scale in the SI is the Kelvin scale. The

temperature unit on this scale is the kelvin, which is designated by K (not K).

The temperature scale in the America system is the Fahrenheit scale.

The Kelvin scale is related to the Celsius scale by

T(K) = T(C) + 273.15 --- (1.1)

T(F) = 1.8 T(C) + 32 --- (1.2)

Note that the magnitudes of each division of 1 K and 1C are identical. Therefore, when

we are dealing with temperature differences T, the temperature interval on both scales is the

same. T(K) = T(C) --- (1.3)

1.10. Pressure

Pressure is the force exerted normally on the unit area. We speak of pressure only

when we deal with a gas or a liquid. The counterpart of pressure in solids is stress. For a

fluid at rest, the pressure at a given point is the same in all directions. The pressure in

a fluid increases with depth as a result of the weight of the fluid. Since pressure is denned

as force per unit area (P =F/A), it has the unit of newton’s per square meter (N/m2), which

is called a Pascal (Pa). That is,

1 Pa = 1 N/m2

The pressure unit Pascal is too small for pressures encountered in practice. Therefore, its

multiples kilopascal (1 kPa = 103

Pa) and megapascal (1 MPa = 106

Pa) are commonly

used. Two other common pressure units are the bar and standard atmosphere:

1 bar = 105

Pa = 0.1 MPa = 100 kPa

1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bars

In the English system, the pressure unit is pound-force per square inch (lbf/in2, or psi), and

1 atm = 14.696 psi.

The actual pressure at a given position is called the absolute pressure. Pressures below

atmospheric pressure are called vacuum.

Absolute, gage, and vacuum pressures are all positive quantities and are related to each

other by

This is illustrated in Fig. 1.2.

Fig. 1.2. Absolute, gage, and vacuum pressures.

Small and moderate pressure differences are often measured by using a manometer, which

mainly consists of a glass or plastic U-tube containing a fluid such as mercury, water,

alcohol, or oil. To keep the size of the manometer at a manageable level, heavy fluids such as

mercury are used if large pressure differences are anticipated.

1.11. Volume

Volume is a property which is associated with cubic measure. The unit of volume

is the cubic metre (m3) together with its multiples and submultiples. Sometimes the litre

may be used. 1 litre = 1 dm3

= 10-3

m3.

If the volume of a substance increases then the substance is said to have been expanded. If

the volume of a substance decreases then the substance is said to have been compressed.

Specific volume is given the symbol v. The volume of any mass, other than unity, is given

the symbol V.

Pgage = Pabs - Patm (for pressures above Patm ) --- (1.4) Pvac = Patm - Pabs (for pressures below Patm ) --- (1.5)

1.12. Internal Energy

If a hot body is placed in contact with a cold body then the temperature of the hot

body begins to fall while at the same time the temperature of the cold body begins to rise. To

account for this it is said that the hot body gives up heat and hence its temperature falls, while

the cold body receives this heat and its temperature rises.

The store of energy which results from the motion of the atoms and molecules of a body

referred to as internal energy (U). Specific internal energy is designated u, the internal

energy of unit mass.

1.13. Enthalpy

Is a property of a substance and is a form of energy. It has been shown that internal

energy, pressure and volume are properties. The combination H= U + PV is enthalpy.

Specific enthalpy is h = u + Pv. It should be noted that enthalpy was at one time referred to as

total heat.

1.14. Work

Is the energy transfer associated with a force acting through a distance, work is done

by or on the system. Work is therefore a transient quantity being descriptive of that

process by which a force is moved through a distance. Work, being a transient quantity, is

therefore not a property.

Work is given the symbol W. If it is required to indicate a rate at which work is being done

then a dot is placed over the symbol W. Thus W. = work done/unit time

1.15. Heat (Heat transfer)

Is defined as the form of energy that is transferred between two systems (or a system

and its surroundings) due to the temperature difference.

Note that heat is a transient quantity, it being simply descriptive of the energy transfer

process through a system boundary resulting from temperature difference. If there is no

temperature difference, then there is no heat transfer, thus heat is not a property.

Heat energy is given the symbol Q. The unit of heat is (Joul). To indicate a rate of heat

transfer, a dot is placed over the symbol Q, thus Q.

= heat transfer/unit time.

For one kilogram of a substance is the specific heat and denoted (q), q= Q/m [J/kg].

1.16. Relationship between heat and work

Fig. 1. 3 shows two containers each containing a mass of water = m and each having a

thermometer inserted such that temperature measurement can be made. In each case, the mass

of water is the system.

(a) (b)

Fig. 1.3. Example of energy transfer

At (a) it is arranged that an external heater can transfer heat energy Q through the system

boundary into the water. At (b) it is arranged that a paddle wheel is immersed in the water

such that external paddle or stirring work W is done when the wheel is rotated.

In each case it is assumed that there is no energy loss from the system. Consider the

arrangement at (a). It is common experience to heat water in some containing vessel by

means of some external heating device.

Let the initial temperature as recorded on the thermometer be t1, and after heating, in

which heat energy = Q is transferred into the water, let the final temperature be t2.

Consider, now, the arrangement at (b). The container once again contains a mass of water

m but in this case a paddle wheel is introduced into the water. It is common

experience that friction makes things warm. The simple experience of rubbing one's hands

together in a brisk manner will show this. In the case under consideration it is possible to

rotate the paddle wheel against the frictional resistance of the water. Assume that the initial

temperature of the water is t1 and, after doing an amount of work = W on the paddle wheel,

the final temperature is t2.

Now a similar effect has been produced in both cases (a) and (b) in that a mass of water

m starting at a temperature = t1 has experienced a rise in temperature = (t2 — t1)

On the one hand, however, it was heat which was transferred to produce the effect but on the

other, it was a work transfer which produced an exactly similar effect.

The conclusion from this must therefore be that there is a relationship between heat and

work. If the unit of energy for both work and heat is the same, then, since the same effect

was produced in each case, the relationship is of the for (Q=W). The unit of energy in the SI

system of units is the joule (J) 1J=Nm

1.17. Specific heat capacity

The specific heat capacity of a substance is defined as the amount of heat which

transfers into or out of unit mass of the substance while the temperature of the substance

changes by one degree. Thus, if:

c = specific heat capacity of substance

m = mass of substance

t1 = original temperature of substance

t2 = final temperature of substance

Q = heat transferred to produce temperature change then,

Q = m c (t2 - t1) --- (1.6)

Specific heat capacity may vary with temperature, it is common practice to take an

average value within the temperature range considered and then use this average value as

being constant. The basic unit for specific heat capacity is joules/ kilogram kelvin (J/kgK).

Such a multiple as kilojoules/kilogramme kelvin (kJ/kgK) may also be employed.

Example 1.1:

5 kg of steel, specific heat capacity 480 J/kgK, is heated from 15C to 100C. How

much heat is required?

Solution: Q = m c (t2 - t1)

= 5 480 (100 - 15)

= 204000 J = 204 kJ.

Example 1.2:

An iron casting of mass 10 kg has an original temperature of 200C. If the casting

loses heat to the value 715.5 kJ, what is the final temperature?

Solution: From table

Specific heat capacity of cast iron = 477 J/kgK

Heat transferred from casting = m c (t2 - t1)

Heat transferred = -715.5 kJ = -715500 J

(Note: the negative sign indicates a heat loss)

Therefore, -715500 = 10 x 477 x (t2 - 200) , then t2 = 50C.

1.18. Work and the P-V diagram

Consider Fig. 1.4. In the lower half of the diagram is shown a cylinder in which a

fluid at pressure P is trapped by means of a piston of area A.

Fig. 1.4. Example of P-V diagram

Force on piston = PA

Now let the piston move back a distance L along the cylinder while at the same time the

pressure of the fluid remains constant. The force on the piston will have remained constant.

Work done = Force x distance = P x A L

AL = volume swept out by the piston, called the 'swept' or 'stroke' volume = (V2-Vl)

Work done = P(V2 - V1) --- (1.7)

Above the diagram of the piston and cylinder is shown a graph of the operation

plotted with the axes of Pressure and Volume. Such a graph is called a P-V diagram or graph.

The graph appears as horizontal straight line ab whose height is at pressure P and whose

length is from original volume V1 to final volume V2.

Now consider the area abcd under this graph.

Area = P(V2 - V1) --- (1.8)

But this is the same as the work done. Hence it follows that the area under a P-V diagram

gives the work done.

1.19. The Conservation of Energy

We have already known that, by designing suitable devices, then one form of energy

can be transformed into another. In a power-station, the potential chemical energy in the fuel

produces a high-temperature furnace. From the furnace, heat energy is transferred into the

steam being formed, which is passed into a turbine where some of it is converted into work.

The work is put into an alternator where some is converted into electrical energy. The

electricity generated is then passed out of the station to the public, who use it in various

devices to produce heat, light and power. Actually, not all the energy which is put into the

furnaces of the power-station ultimately appears as electrical energy. There are many losses

through the plant, as indeed there are in any power plant.

However, it is found that, in any energy transformation system, if all the energy forms are

added up, and including any energy losses which may have occurred, then the sum is

always equal to the energy input. Naturally, all the various energies must appear with the

same units in order to do this.

The fact that the total energy in any one energy system remains constant is called the

principle of the Conservation of Energy. This states that energy can neither be created nor

destroyed but can only be changed in form.

1.20. The Steady-Flow Energy Equation

This equation is a mathematical statement of the principle of the Conservation of

Energy as applied to the flow of a fluid through a thermo-dynamic system. The various forms

of energy which the fluid can have are as follows.

(a) Potential energy

If the fluid is at some height Z above a given datum level, then, as a result of its mass

it possesses potential energy with respect to that datum. Thus, for unit mass of fluid, in the

close vicinity of the earth,

Potential energy = gZ

(b) Kinetic energy

If the fluid is in motion then it possesses kinetic energy. If the fluid is flowing with

velocity C then, for unit mass of fluid,

Kinetic energy = C2

2

(c) Internal energy

All fluids store energy. The store of energy within any fluid can be increased or

decreased as the result of various processes carried out on or by the fluid. The energy stored

within a fluid which results from the internal motion of its atoms and molecules is called its

internal energy. It is usually designated by the letter U.

If the internal energy of unit mass of fluid is being discussed then this is called the

specific internal energy and is designated by u.

(d) Flow or displacement energy

Any volume of fluid entering or leaving a system must displace an equal volume ahead

of itself in order to enter or leave the system, as the case may be. The displacing mass must

do work on the mass being displaced, since the movement of any mass can only be

achieved at the expense of work.

Thus, if the volume of unit mass of liquid (its specific volume) is v1, at entry and its

pressure is P1 then in order to enter a system it must displace an equal specific volume v1

inside the system. Thus work to the value P1v1 is done on the specific volume inside the

system by the specific volume outside the system. This work is called flow or displacement

work and at entry it is energy received by the system.

Similarly, at exit, flow work must be done by the fluid inside the system on the fluid

outside the system in order that it may leave. Thus, if at exit, the pressure is P2 and the

specific volume is v2 then

Flow or displacement work rejected = P2 v2

(e) Heat received or rejected

During its passage through the system the fluid can have a direct reception or

rejection of heat energy through the system boundary. This is designated by Q. This must be

taken in its algebraic sense. Thus if,

Heat is received then Q is positive. Heat is rejected then Q is negative. If heat is

neither received nor rejected then Q = 0.

(f) External work done

During its passage through the system the fluid can do external work or have external

work done on it. This is usually designated by W. This must be taken in its algebraic sense.

Thus if,

External work is done by the fluid then W is positive. External work is done on the fluid

then W is negative. If no external work is done on or by the fluid then W = 0.

Fig. 1.5. A thermodynamics system

Fig. 1.5 illustrates some thermodynamic system into which is flowing a fluid with

pressure P1 specific volume v1 specific internal energy u1 and velocity C1. The entry is at a

height Zl above some datum level.

In its passage through the system, external work W may be done on or by the fluid and

also heat energy Q may be received or rejected by the fluid from or to the surroundings.

The fluid then leaves the system with pressure P2, specific volume v2, specific internal energy

u2 and velocity C2. The exit is at height Z2 above the datum level. Applying the principle of

conservation of energy to the system, then,

Total energy entering the system = Total energy leaving the system or, for unit mass of

substance,

--- (1.9)

This is called the steady-flow energy equation.

From this then, it will be noted that the steady-flow energy equation can also be written,

--- (1.10)

since heat is lost from the steam to the s

Example. 1.3:

In a steady-flow system, a substance flows at the rate of 4 kg/s. It enters at a pressure

of 620 kN/m2, a velocity of 300 m/s, internal energy 2100 kJ/kg and specific volume

0.37 m3/kg. It leaves the system at a pressure of 130 kN/m

2, a velocity of 150 m/s, internal

energy 1500 kJ/kg and specific volume l.2 m3/kg. During its passage through the system the

substance has a loss by heat transfer of 30 kJ/kg to the surroundings. Determine the power of

the system in kilowatts, stating whether it is from or to the system. Neglect any change in

potential energy.

Solution: Neglecting change in potential energy, the steady-flow energy equation

becomes, per unit mass of substance,

Q is written negative since 30 kJ/kg are lost to the surroundings.

Specific W = (u1 u 2 ) (P1v1 P2 v 2 )

2 2

C1 C2 Q 2

(3002 150

2 )

= (2100 -1500) + (620 0.37 - 130 1.2) +

= 676.75 kJ/kg.

The substance flows at the rate of 4 kg/s.

2 103

- 30

Output (since W is positive) = 676.75 4 = 2707 kJ/s = 2707 kW.

Example 1.4:

Steam enters a turbine with a velocity of 16 m/s and specific enthalpy 2990 kJ/kg. The

steam leaves the turbine with a velocity of 37 m/s and specific enthalpy 2530 kJ/kg. The heat

lost to the surroundings as the steam passes through the turbine is 25 kJ/kg. The steam flow

rate is 324000 kg/h. Determine the work output from the turbine in kilowatts.

Solution: Neglecting changes in potential energy,

Q is negative surroundings

= (2990 - 2530) + 16

2 103

Steam flow rate = 324000 kg/h = 324000/3600 = 90 kg/s

W = 434.443 90 = 39099.87 kJ/s (or) 39099.87 kW

39100 kW 39.1 MW

1.25. The non-flow energy equation

It has been shown that the steady-flow energy equation connecting the energies before

and after the flow of unit mass of substance through a system is of the form,

In the case of a closed system, however, in which the fluid mass remains constant, no

substance passing through the system boundary, the flow terms in eqn.1.9 will not apply.

Thus the terms Pv and C2/2 are neglected. The system is then said to be non-flow.

During a non-flow process, there may be a change of mass centre, in which case there

is a change in potential energy. However, in thermodynamic systems, the density of the

substance used is quite small and any change in potential energy which may occur would be

very small. Thus the terms gZ can be neglected.

Thus, from eqn. 1. , the energy equation for the non-flow case becomes,

ul + Q = u2 + W --- (1.11)

from which,

Q = (u2 - u1) + W (or) Q = u + W --- (1.12)

This is for unit mass. If any mass is being considered, then eqn.(1.12) becomes,

Q = U + W --- (1.13)

In words, for a non-flow process, this means that,

Heat transferred through

the system boundary =

Change of internal

energy +

Work transferred through the

system boundary

Remember, however, that the terms in eqn.1.13 must be taken their algebraic sense.

Thus, if,

Heat is received: Q is positive

Heat is rejected: Q is negative

Heat neither received nor rejected (adiabatic): Q = 0

Internal Energy increases: U is positive

Internal Energy decreases: U is negative

Internal Energy neither increases nor decreases: U = 0

Work done by system: W is positive

Work done on system: W is negative

No work done on or by system: W = 0

A typical non flow process is the expansion or compression of a substance in a cylinder.

Example1.5

During the compression stroke of an engine, the work done on the working substance

in the engine cylinder is 75 kJ/kg and the heat rejected to the surroundings is 42 kJ/kg. Find

the change of internal energy, stating whether it is an increase or decrease.

Solution:

This is a non-flow process and, hence, from eqn.(1.12),

Q = u + W (or)

u = Q - W

= -42 - (-75) Note: Heat is lost. Work is negative.

u = 33 kJ/kg This is an increase in internal energy.

1.26. The first law of thermodynamics

In section 1.16 it was shown that there is a relationship between heat and work. This

relationship was shown to be of the form,

W = Q --- (1.14)

where, W = Work

Q = Heat

The fact that there is a relationship between heat and work, is a statement of the First

Law of Thermodynamics.

Eqn. 1.14 is not meant to imply that if a certain amount of work is done on a system,

then it is all converted into heat or, conversely, if a certain amount of heat is supplied to a

system, then it is all converted into work.

Eqn. 1.14 simply means that if some work is converted into heat or some heat is

converted into work then the relationship between the heat and work so converted will be of

the form W = Q.

Actually, it is possible to convert work completely into heat by friction, for example.

The reverse process of converting heat completely into work is impossible.

An extension of eqn. 1.13 may be found in the Non-Flow Energy Equation,

Q = U+W

This equation takes into account that not all the energy need be of h eat-work or work-

heat conversion form. Some energy may be concerned with an internal energy change. A

further extension of eqn. appears in work with thermoynamic cycles. To complete a

cycle a working substance is taken through a sequence of events and is returned to its original

state. If the working substance is returned to its original state then its final properties are

identical with its orginal properties before the cycle.

If work is transferred during the cycle then, since there is no final change in the

properties of the working substance, the energy to provide the work must have been

transferred as heat and must exactly equal the work. Now, during some processes in a cycle,

work will be done by the substance, while, during others, work is done on the substance.

Similarly, during some processes heat is transferred out of the substance, while, during

others, it is transferred into the working substance.

Thus, for a cycle, since there is no nett property change,

Nett heat = Nett work

or as it is written symbolically,

Q W

The symbol means the summations round cycle.

1.13

Exercise:

1.1. 4.5 kg of water are heated from 15°C to 100°C. How much heat transfer is required?

1.2. A vacuum gage connected to a tank reads 30 kPa at a location where the barometric

reading is 755 mmHg. Determine the absolute pressure in the tank.

Take Hg = 13590 kg/m3.

1.3. A pressure gage connected to a tank reads 500 kPa at a location where the

atmospheric pressure is 94 kPa. Determine the absolute pressure in the tank.

1.4. A manometer containing oil (= 850 kg/m3) is attached to a tank filled with air. If the

oil-level difference between the two columns is 45 cm and the atmospheric pressure is

98 kPa, determine the absolute pressure of the air in the tank.

1.5. In a non-flow process there is a heat loss of 1055 kJ and an internal energy increase of

210 kJ. How much work is done?

1.6. Air passes through a gas turbine system at the rate of 4.5 kg/s. It enters with a velocity

of 150 m/s and a specific enthalpy of 3000 kJ/kgK. At exit, the velocity is 120 m/s

and the specific enthalpy is 2300 kJ/kg. The air has a heat loss to the surroundings of

25 kJ/kg as it passes through the turbine. Determine the power developed by the

turbine.

Chapter Two

Gases and Single Phase Systems

2.1. Boyle's law

With any mass of gas it is possible to vary the pressure, volume and temperature. In

this experiment it is arranged that the temperature of a fixed mass of gas remains constant

while corresponding changes in pressure and volume are observed.

Fig. 2.1. PV diagram of a gas

Taking any point on the curve of fig. 2.1, 1 say, the product of its corresponding

pressure and volume P and V will equal some number, C say. Investigation of other points

such as 2 and 3, shows that, within the limits of experimental error, the products of their

corresponding pressures and volumes also equal this same number, or,

P1V1 = P2V2 = P3V3 = C, a constant --- (2.1)

Further experiments at different fixed temperatures, with different fixed masses

and also with different gases yield the same result although the constant C will be different

with each quantity of gas taken, each temperature fixed and each type of gas used.

From the results then, a general statement may be made as follows: During a change

of state of any gas in which the mass and the temperature remain constant the volume varies

inversely as the pressure, or,

PV = C, a constant --- (2.2)

This is known as Boyle's Law.

19

Example. 2.1

During an experiment on Boyle's Law, the original volume of air trapped in the

apparatus, with the two mercury levels of the same, was 20000 mm3. The apparatus was then

modified such that the volume of air became 17000 mm3, while the temperature remained

constant. If the barometer reading was 765 mm Hg, what was the new pressure exerted on the

air in mm Hg? Also, what was the difference in the two mercury column levels?

Solution:

Since both levels of mercury are the same at the beginning, then

Pi = atmospheric pressure = 765 mm Hg

Now Boyle's Law states that PV = C, a constant

From this then,

P1V1 = P2V2 (or) 17000

20000765VVPP

2

112

×== = 900 mm Hg

Since the final pressure P2 = 900 mm Hg and the atmospheric pressure = 765 mm Hg,

then, Difference in height of the two mercury columns = 900 - 765 = 135 mm

Example. 2.2.

A gas whose original pressure and volume were 300 kN/m2 and 0.14 m3 is expanded

until its new pressure is 60 kN/m2 while its temperature remains constant. What is its new

volume?

Solution: Since the temperature remains constant then this is an expansion according to Boyle's Law.

P1V1 = P2V2 or 2

112 P

PVV = 7.0

6030014.0 =×= m3

2.2. Charle's Law

.

Fig. 2.2. V-T diagram when pressure remains constant

20

The relationship between volume and temperature of a fixed mass of gas when the

pressure remains constant is of the linear form:

V = Ct + V0 --- (2.3)

where, V = volume,

t = temperature,

C = slope,

V0 = intercept on V axis

An interesting point however is that if all the straight lines obtained are produced back

to cut the temperature axis, they all cut this axis at the same point. It would be better to use

this point as a new origin and the law of the graph would then become:

V = C T --- (2.4)

where, T is the temperature recorded from the new origin.

It follows that, TV = C, a constant --- (2.5)

In words this may be stated as follows: During the change of state of any gas in which

the mass and pressure remain constant, the volume varies in proportion with the absolute

temperature.

This is known as Charle's law.

Example. 2.3.

During an experiment on Charles's Law, the volume of gas trapped in the apparatus

was 10000 mm3, when the temperature was 18° C. The temperature of the gas was then

raised to 85° C. What was the new volume of gas trapped in the apparatus if the pressure

exerted on the gas remained constant?

Solution:

Now according to Charle's law,

TV = C, a constant

From this, then 2

2

1

1

TV

TV

= .

In order to use this equation the temperatures T1 and T2 must be absolute

temperatures.

T1 = 18 + 273 = 291 K and T1 = 85 + 273 = 358 K

1230229135810000

TT

VV1

212 =×== mm3.

21

Example. 2.4.

A quantity of gas whose original volume and temperature are 0.2 m3 and 303° C,

respectively, is cooled at constant pressure until its volume becomes 0.1 m3. What will be the

final temperature of the gas?

Solution: Again, this is a change according to Charles's Law.

2

2

1

1

TV

TV

= (or) 11

22 T

VV

T ×=

The temperatures must be absolute temperatures.

T1 = 303 + 273 = 576 K

2885762.01.0T2 =×= K

t2 = 288 - 273 = 15° C.

2.3. The characteristic equation of a perfect gas

Consider a gas whose original state is pressure P1, volume V1 and temperature T1 and

let this gas pass through a change of state such that its final state is P2, V2 and T2. Inspection

of Fig. 2.3 will show that there are an infinite number of possible paths connecting states 1

and 2 when the process is shown on a P-V diagram. The concern at the moment, however, is

not in how the state changed from 1 to 2, but in the fact that since states 1 and 2 can exist for

the same mass of gas, then is there any law connecting these two states? This being the case

then, a choice of path from 1 to 2 is quite arbitrary, and it is therefore reasonable to assume a

path about which something is already known.

Fig. 2.3. Possible paths of connecting states 1 and 2

Boyle's and Charles's laws supply the answer. Figure 2.4 shows that it is quite

possible to move from 1 to 2 by first carrying out a Boyle's Law change down to some

intermediate state A, say, and then carrying out a Charles's Law change to the final condition.

22

Consider the Boyle's Law change from 1 to A. In this case the temperature remains

constant at T1. Also

P1V1 = PAVA --- (2.6)

Fig. 2.4. Boyle's Law change and Charle's Law change in PV diagram

All the pressure change must take place during this process since there will be no

change in pressure during the Charle's Law process which follows. In this case PA = P2.

∴eqn. (2.6) becomes P1V1 = P2VA (or) 2

11A P

VPV = --- (2.7)

Consider now the Charle's Law change from A to 2. In this case the pressure remains

constant at P2. Also

2

2

A

A

TV

TV

= --- (2.8)

During the Boyle's Law change from 1 to A the temperature remained constant.

∴TA = T1

from which eqn. (2.8) becomes 2

2

1

A

TV

TV

= --- (2.9)

But 2

11A P

VPV = from eqn. (2.7) and substituting this in eqn. (2.9)

2

2

12

11

TV

TPVP

= from which

2

22

1

11

TVP

TVP

= -- (2.10)

Now any change of state from state 1 would produce a similar result, and hence

eqn.(2.10) could be extended to read

etc. ,TVP

TVP

TVP

TVP

4

44

3

33

2

22

1

11 L==== --- (2.11)

where 3 and 4 represent other new conditions of state of the same mass of gas.

23

From eqn.(2.11) then, it follows that for any fixed mass of gas, changes of state are

connected by the equation, constantT

PV= --- (2.12)

Now sooner or later it will be necessary to know the actual mass of gas used during

any particular process.

Let v = volume of 1 kg of gas, the specific volume. Then from eqn.(2.12),

constantTPv

= --- (2.13)

When 1 kg of gas is considered, this constant is written R and is called the

characteristic gas constant, sometimes, the specific gas constant.

∴for 1 kg of gas,

RTPv

= --- (2.14)

Now consider the case when there are m kg of gas. Multiply both sides of eqn.(2.14)

by m, then RmT

Pmv= (or) PV = mRT --- (2.15)

This is known as the characteristic equation of a perfect gas.

The units of R can be obtained from eqn.(2.14). If pressure is in N/m2, specific

volume is in m3/kg and temperature in K, then,

kgKJ

kgKm.N

K1

kgm

mNR

TPv 3

2 ==××==

For air, the value of R is usually of the order of 0.287 kJ/kg K.

Example. 2.5

A gas whose original pressure, volume and temperature were 140 kN/m2, 0.1 m3 and

25°C, respectively, is compressed such that its new pressure is 700 kN/m2 and its new

temperature is 60°C. Determine the new volume of the gas.

By the characteristic equation,

2

22

1

11

TVP

TVP

=

0223.01.0298333

700140V

TT

PPV 1

1

2

2

12 =××== m3

Example 2.6.

A quantity of gas has a pressure of 350 kN/m2 when its volume is 0.03 m3 and its

temperature is 35°C. If the value of R = 0.29 kJ/kg K, determine the mass of gas present. If

24

the pressure of this gas is now increased to 1.05 MN/m2 while the volume remains constant,

what will be the new temperature of the gas?

By the characteristic equation,

PV = mRT

m = 118.03081029.0

03.010350RTPV

3

3

=××××

= kg

For the second part of the problem

2

22

1

11

TVP

TVP

= and in this case, V1 = V2

2

2

1

1

TP

TP

= (or) 1

212 P

PTT = = 924

1035.01005.1308 6

6

=××

× K = 651°C.

2.4. The specific heat capacities of a gas

The specific heat capacity of a substance may be defined as the amount of heat

transfer required to raise unit mass of a substance through 1 degree difference in temperature.

(a) The specific heat capacity at constant volume

This is defined as the amount of heat which transfers to or from unit mass of gas

while the temperature changes by 1 degree and the volume remains constant. This is

written cv.

(b) The specific heat capacity at constant pressure

This is defined as the amount of heat which transfers to or from unit mass of gas

while the temperature changes by 1 degree and the pressure remains constant. This is

written cp.

Both the specific heat capacities at constant volume and at constant pressure rise in

value with temperature. For calculation purposes, it is usual to assume an average value of

specific heat capacity within the temperature range being considered. Average values

commonly assumed for air are of the order cp = 1.005 and cv = 0.718, Unit, kJ/kg K.

2.5. The constant volume heating of a gas.

Let a mass of gas m be heated at constant volume such that its temperature rises from

T1 to T2 and its pressure rises from P1 to P2. Then

Heat received by

the gas = Mass ×

Specific heat capacity at

constant volume × rise in temperature

= m cv (T2 - T1) --- (2.16)

25

Now constant volume heating is a particular case of a non-flow process carried out on

a gas. Consider, then, the non-flow energy equation applied to constant volume heating.

Q = ∆U + W --- (2.17)

Now no external work is done during constant volume heating. This can be seen by

inspecting fig. 2.5, in which pressure is plotted against volume. The process appears as a

vertical straight line. There is no area beneath this line and hence there is no external work

done.

Fig. 2.5. PV diagram of constant volume heating process

∴W = 0

From this then, eqn. 2.17 becomes

Q = ∆U (or) mcv(T2 - T1) = U2 - U1 --- (2.18)

It follows, therefore, that all the heat added during constant volume heating goes

completely into increasing the stock of internal energy of the gas. Conversely, if a gas is

cooled at constant volume, the heat rejected will be at the expense of the stock of internal

energy of the gas. If the new pressure is required then this may be found by the application of

the characteristic equation of a perfect gas from which,

2

22

1

11

TVP

TVP

= and for this case, V1 = V2.

∴2

2

1

1

TP

TP

= or 1

212 T

TPP =

Example. 2.7.

2 kg of gas, occupying 0.7 m3, had an original temperature of 15°C. It was then

heated at constant volume until its temperature became 135°C. How much heat was

transferred to the gas and what was its final pressure? Take, cv = 0.72 kJ/kgK and R = 0.29

kJ/kgK.

26

Solution:

Heat transferred at constant volume = mcv(T2 - T1)

= 2 × 0.72 × (135 - 15) = 172.8 kJ.

Now P1V1 = m R T1

6.2387.0

28829.02V

mRTP

1

11 =

××== kN/m2

Since the volume remains constant, then,

2

2

1

1

TP

TP

= or 1.3382884086.238

TT

PP1

212 =×== kN/m2

2.6. The constant pressure heating of a gas

Let a mass of gas m be heated at constant pressure such that its temperature rises from

T1 to T2 and its volume increases from V1 to V2. Then

Heat received by

the gas = Mass ×

Specific heat capacity at

constant pressure × rise in temperature

= mcp (T2 - T1)

Now constant pressure heating is a particular case of a non-flow process carried out

on a gas. Consider, then, the non-flow energy equation applied to constant pressure heating.

Q = ∆U + W --- (2.19)

Fig. 2.6. PV diagram of constant pressure heating process

In this case external work is done by the gas. Fig. 2.6 shows a graph of a constant

pressure process plotted on a PV diagram. This graph shows that there is a definite area

beneath the constant pressure line, which gives the work done = P (V2 - V1), where P = the

constant pressure = P1 = P2. In this constant pressure case then, eqn. 2.19 becomes,

mcp (T2 - T1) = (U2 - U1) + P (V2 - V1) --- (2.20)

= (U2 + PV2) - (U1 + PV1)

= H2 - H1 --- (2.21)

i.e. Heat transferred at constant pressure = Change of enthalpy

27

From eqn. (2.19)

U2 - U1 = mcp (T2 - T1) - P(V2 - V1) or

U2 - U1 = mcp (T2 - T1) - mR(T2 - T1) --- (2.22)

since PV = mRT

If the new volume is required after a constant pressure process then this again may be

obtained by the use of the characteristic equation of a perfect gas, from which,

2

22

1

11

TVP

TVP

= and for this case, P1 = P2

∴2

2

1

1

TV

TV

= or 1

212 T

TVV =

Example. 2.8.

A gas whose pressure, volume and temperature are 275 kN/m2, 0.09 m3 and 185°C,

respectively, has its state changed at constant pressure until its temperature becomes 15°C.

How much heat is transferred from the gas and how much work is done on the gas during the

process? Take R = 0.29 kJ/kgK, cp = 1.005 kJ/kgK.

Solution:

First determine the mass of gas use.

Now, P1V1 = m R T1

∴4581029.009.010275

RTVPm 3

3

1

11

××××

== = 0.186 kg.

Heat transferred = m cp (T2 - T1)

= 0.186 × 1.005 × (288 - 458)

= -31.78 kJ

Notice the negative sign indicating that the heat has been extracted from the gas.

Since the pressure remains constant, then,

2

2

1

1

TV

TV

= ∴45828809.0

TT

VV1

212 ×== = 0.0566 m3

Work done = P(V2 - V1)

= 275 × (0.0566 - 0.09)

= -9.19 kJ

2.7. The difference of the specific heat capacities of a gas

It has been shown above that if a mass of gas m has its temperature changed from T1

to T2 then the change of internal energy can be determined by the expressions,

28

U2 - U1 = mcv(T2 - T1) --- (2.23)

and U2 - U1 = mcp (T2 - T1) - mR(T2 - T1) --- (2.24)

If the temperature change is the same for both expressions then it follows that eqn.

2.23 = eqn. 2.24, for the change of internal energy, is a function of temperature only.

mcv(T2 - T1) = mcp (T2 - T1) - mR(T2 - T1)

from which, cv = cp - R

since m(T2 - T1) is common throughout.

∴ cp - cv = R --- (2.25)

2.8. The polytropic process and a gas

A gas is no exception to this concept and so, if a mass of gas is expanded or

compressed, the general law of expansion or compression is of the polytropic form,

PVn = C --- (2.26)

For two state points 1 and 2, then n22

n11 VPVP = --- (2.27)

Further, Work done = 1n

VPVP 2211

−− --- (2.28)

Now, by the characteristic equation, PV = mRT --- (2.29)

Substituting eqn. 2.29 in 2.28

Work done = ( )

1nTTmR 21

−− --- (2.30)

Applying the non-flow energy equation,

Q = ∆U + W

= (U2 - U1) + W

= mcv(T2 - T1) + 1n

VPVP 2211

−− --- (2.31)

= mcv(T2 - T1) + ( )

1nTTmR 21

−− --- (2.32)

2.9. The combination of the polytropic law PVn = C and the characteristic equation

of a perfect gas

The law PVn = C wil lenable calculations to be made of the changes in pressure and

volume which occur during a polytropic process. Combining this with the characteristic

equation of a perfect gas will enable variations in temperature to be determined.

29

Consider a polytropic process in which the state of a gas changes from P1, V1, T1 to

P2, V2, T2.

By the polytropic law, n22

n11 VPVP = --- (2.33)

By the characteristic equation, 2

22

1

11

TVP

TVP

= --- (2.34)

From eqn. 2.34, 22

11

2

1

VPVP

TT

= --- (2.35)

From eqn. 2.33, n

1

2

2

1

VV

PP

⎟⎟⎠

⎞⎜⎜⎝

⎛= --- (2.36)

Substituting eqn. 2.36 in eqn. 2.35. 1n

2

1

2

1

VV

TT

−

⎟⎟⎠

⎞⎜⎜⎝

⎛= --- (2.27)

From eqn. 2.36, n/1

2

1

1

2

PP

VV

⎟⎟⎠

⎞⎜⎜⎝

⎛= or

n/1

1

2

2

1

PP

VV

⎟⎟⎠

⎞⎜⎜⎝

⎛= --- (2.28)

Substituting eqn. 2.28 in eqn. 2.35, n

1n

2

1

2

1

PP

TT

−

⎟⎟⎠

⎞⎜⎜⎝

⎛= --- (2.29)

Combining eqn. 2.27 and 2.29 then,

1n

1

2n

1n

2

1

2

1

VV

PP

TT

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= --- (2.30)

Example. 2.9.

A gas whose original pressure and temperature were 300 kN/m2 and 25°C,

respectively, is compressed according to the law PV1.4 = C until its temperature becomes

180°C. Determine the new pressure of the gas.

Solution:

It has been shown that for a polytropic compression, the relatioship between pressure

and temperature is

n1n

2

1

2

1

PP

TT

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

From this, 1n

n

2

1

2

1

TT

PP −

⎟⎟⎠

⎞⎜⎜⎝

⎛=

30

∴ 1299298453300

TTPP

4.04.1

1nn

1

212 =⎟

⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

kN/m2 = 1.299 MN/m2

Example. 2.10.

A gas whose original volume and temperature were 0.015 m3 and 285°C, respectively,

is expanded according to the law PV1.35 = C until its volume is 0.09 m3. Determine the new

temperature of the gas.

Solution:

The relationship between volume and temperature during a polytropic expansion of a

gas is, 1n

1

2

2

1

VV

TT

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

4.29809.0015.0558

VVTT

135.11n

2

112 =⎟

⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

K

t2 = 298.4 - 273 = 25.4°C.

Example. 2.11.

0.675 kg of gas at 1.4 MN/m2 and 280°C is expanded to four times the original

volume according to the law PV1.3 = C. Determine:

(a) the original and final volume of the gas,

(b) the final pressure of the gas,

(c) the final temperature of the gas.

Take R = 0.287 kJ/kgK.

Solution:

(a) Now P1V1 = m R T1

∴ 0765.0104.1

55310287.0675.0P

mRTV 6

3

1

11 =

××××

== m3 = original volume.

Since the gas is expanded to four times its original volume, then,

V2 = 4V1 = 4 × 0.0765 = 0.306 m3 = final volume

(b) n22

n11 VPVP =

231.0414.1

VVPP

3.1n

2

112 =⎟

⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛= MN/m2 = 231 kN/m2 = final pressure

(c) 2

22

1

11

TVP

TVP

=

31

∴ 36544.1

231.0553VPVP

TT11

2212 =××== K

t2 = 365 - 273 = 92°C = final temperature.

2.10. The adiabatic process and a gas

When dealing with the general case of a polytropic expansion or compression it was

stated that this process followed a law of the form PVn = C. Now the adiabatic process can be

a particular case of the polytropic process in which no heat is allowed to enter or leave during

the progress of the process. From this, then, it appears that there should be a particular value

of the index n which will satisfy this condition. An investigation is therefore necessary to see

if this is the case.

Consider an adiabatic expansion or compression in which a change of state occurs

from P1, V1, T1 to P2, V2, T2. Then

Change of internal energy = m cv (T2 - T1) --- (2.31)

Also Work done during the process = 1

VPVP 2211

−γ−

= ( )

)1(TTmR 21

−γ−

--- (2.32)

where γ (gamma) is the particular index which will satisfy the case of an adiabatic process.

From the polytropic law then, if γ is the adiabatic index γγ = 2211 VPVP --- (2.33)

Also, from the polytropic law,

1

1

2

1

2

1

2

1

VV

PP

TT

−γγ−γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= --- (2.34)

and by the characteristic equation 2

22

1

11

TVP

TVP

= --- (2.35)

Applying the non-flow energy equation Q = ∆U + W

For an adiabatic process Q = 0. ∴0 = ∆U + W

or W = -∆U --- (2.36)

i.e. Work is done at the expense of internal energy during an adiabatic expansion. Internal

energy increases at the expense of work during an adiabatic compression.

Substituting eqn. 2.31 and 2.32 in 2.36

( )

)1(TTmR 21

−γ−

= - m cv (T2 - T1)

32

∴( )

)1(TTmR 21

−γ−

= - m cv (T1 - T2)

from which vc)1(

R=

−γ (since m(T1 - T2) is a common term to both sides.)

From this, )1(cR

v

−γ= (or) v

v

v ccR

1cR +

=+=γ --- (2.37)

Now R = cp - cv and hence, substituting in eqn. 2.37

v

p

v

vvp

cc

cccc

=+−

=γ --- (2.38)

From this, then, the law for an adiabatic expansion or compression of a gas is PVγ = C,

where γ = v

p

cc

, the ratio of the specific heat capacities at constant pressure and constant

volume. The theoretical adiabatic is sometimes said to be a frictionless adiabatic. The average

value of γ, the adiabatic index, for air is of the order of 1.4.

Example. 2.12

A gas expands adiabatically from a pressure and volume 700 kN/m2 and 0.015 m3,

respectively, to a pressure of 140 kN/m2. Determine the final volume and the work done by

the gas. What is the change of internal energy in this case? Take cp = 1.046 kJ/kgK, cv =

0.752 kJ/kgK.

Solution:

Adiabatic index = γ = cp/cv = 1.046/0.752 = 1.39

For an adiabatic expansion,

γγ = 2211 VPVP

from which γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

/1

2

112 P

PVV = 048.0

140700015.0

39.1/1

=⎟⎠⎞

⎜⎝⎛× m3 = final volume

Workdone = 1

VPVP 2211

−γ−

= 69.9139.1

048.0140015.0700=

−×−× kJ

For an adiabatic expansion,

W = -∆U (or) ∆U = -W

∴Change of internal energy = -9.69 kJ

This is a loss of internal energy from the gas.

33

2.11. The isothermal process and a gas

An isothermal expansion or compression is defined as a process carried out such that

the temperature remains constant throughout. This is evidently the same as a process carried

out according to Boyle's Law. The law for an isothermal expansion or compression of a gas is

therefore,

PV = C, a constant --- (2.39)

Thus, for a change of state from 1 to 2,

P1V1 = P2V2 --- (2.40)

T1 = T2 = T = constant temperature --- (2.41)

Now the law PV = C is that of a rectangular hyperbola.

Work done = PV ln 1

2

VV --- (2.42)

This, therefore, is the expression which will give the work done during an isothermal process

on a gas.

From the characteristic equation,

PV = mRT --- (2.43)

Substituting eqn. 2.43 in 2.42

Work done = mRT 1

2

VV --- (2.44)

Applying the non-flow energy equation, Q = ∆U + W.

Since for an isothermal process T = constant, and by Joule's Law, the internal energy of a gas

is a function of temperature only then, if T = constant, then there is no change of internal

energy. Hence, for an isothermal process, ∆U = 0

∴The energy equation becomes, Q = W = PV ln 1

2

VV = mRT

1

2

VV --- (2.45)

From this, then, it follows that during an isothermal expansion all the heat received is

converted into external work. Conversely, during an isothermal compression, all the work

done on the gas is rejected by the gas as heat.

Example. 2.13.

A quantity of gas occupies a volume of 0.3 m3 at a pressure of 10 kN/m2 and a

temperature of 20°C. The gas is compressed isothermally to a pressure of 500 kN/m2 and

then expanded adiabaticallyto its initial volume. Determine, for this quantity of gas:

(a) the heat received or rejected (state which) during the compression,

34

(b) the change of internal energy during the expansion

(c) the mass of gas.

Assume for this gas that γ = 1.4 and cp = 1.0 kJ/kgK.

Solution:

(a) For the isothermal compression,

P1V1 = P2V2

V2 = V1 × P1/P2 = 0.3 × 100/500 = 0.06 m3

Now Q = ∆U + W and for an isothermal process on a gas ∆U = 0

∴Q = W

= PV ln r = PV ln 2

1

PP = 100 × 0.3 × ln(100/500) = -48.3 kJ.

This is heat rejected.

(b) For the adiabatic expansion,

γγ = 3322 VPVP

6.523.0

06.0500VV

PP4.1

3

223 =⎟

⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ

kN/m2

Now Q = ∆U + W and for an adiabatic process Q = 0

∴0 = ∆U + W or ∆U = -W = ( )

1VPVP 3322

−γ−−

= 5.3514.1

)3.06.5206.0500(−=

−×−×− kJ.

This is a loss of internal energy.

(c) cp - cv = R and cp/cv = γ

from which, cv = cp/γ

∴(cp - cp/γ) = R = cp (1 - 1/γ ) = 1.0(1 - 1/1.4) = 0.286 kJ/kgK

P1V1 = m R T1

358.029310286.03.010100

RTVPm 3

3

1

11 =××××

== kg.

2.12. The non-flow energy equation and the polytropic law PVn = C

Consider the expansion or compression of a gas according to the law PVn = C in

which the state changes from P1, V1, T1 to P2, V2, T2.

It has been shown that the change of internal energy

= ∆U = mcv (T2 - T1) --- (2.46)

35

Also, the work done during the change

= W = ( )

1nTTmR 21

−− --- (2.47)

Substituting eqn. 2.46 and 2.47 in the non-flow energy equation, Q = ∆U + W, then,

Q = mcv (T2 - T1) + ( )

1nTTmR 21

−− --- (2.48)

Now cp - cv = R and cp/cv = γ then cv = )1(

R−γ

--- (2.49)

Substituting (2.49) into eqn. (2.48)

)TT()1n(

Rm)TT()1(

RmQ 2112 −−

+−−γ

=

)TT()1(

Rm)TT()1(

Rm 1212 −−γ

−−−γ

=

( ) ( )⎭⎬⎫

⎩⎨⎧

−γ−

−−=

11

1n1)TT(mR 21

( )( )

( )( )⎭

⎬⎫

⎩⎨⎧

−γ−

−−−γ

−=11n

1n1)TT(mR 21

( )( )⎭⎬⎫

⎩⎨⎧

−γ−−γ

−=11n

)n()TT(mR 21

( ) ( )1n)TT(mR

1)n( 21

−−

−γ−γ

= --- (2.50)

or Q ( ) ×−γ−γ

=1

)n( Polytropic work

Now from this equation it is possible to examine what happens to the hat received or

rejected during an expansion or compression of a gas if the value of the index n is varied. For

a compression the work done is negative. For an expansion the work done is positive.

If n = γ, then ( )1)n(

−γ−γ = 0 and hence Q = 0, i.e. this is the adiabatic case.

If n = 1, then ( )1)n(

−γ−γ =1 and hence Q = Workdone, i.e. this is the isothermal case.

Now, substituting eqn. 2.49 in eqn. 2.50.

( )( )( )1n

)TT(1mc1

)n( 21v

−−−γ

−γ−γ

=

36

( ) )TT(1n)n(mc 21v −

−−γ

= --- (2.51)

)TT(mc 21n −= --- (2.52)

where cn = ( )1n)n(cv −

−γ , is called the polytropic specific heat capacity.

Example 2.14.

A gas expands according to the law PV1.3 = C, from a pressure of 1 MN/m2 and a

volume 0.003 m3 to a pressure of 0.1 MN/m2. How much heat was received or rejected by the

gas during this process? Determine, also, the polytropic specific heat. Take, γ = 1.4, cv =

0.718 kJ/kgK.

Solution:

Now, n22

n11 VPVP =

n/1

2

112 P

PVV ⎟⎟⎠

⎞⎜⎜⎝

⎛= = 0.003× (1/0.1)1/1.3 = 0.0176 m3

Heat received or rejected

Q ( ) ×−γ−γ

=1

)n( Work done ( ) ×−γ−γ

=1

)n( 1n

VPVP 2211

−−

( ) ×−−

=14.1

)3.14.1( 13.1

0176.01.0003.01−×−× = 0.00103 MJ = 1.03 kJ.

This is positive and hence heat is received by the gas.

cn = ( ) 239.013.13.14.1718.0

1n)n(cv =

−−

×=−−γ kJ/kgK.

Exercises:

2.1. A quantity of gas has an initial pressure of 140 kN/m2 and volume 0.14 m3. It is then

compressed to a pressure of 700 kN/m2 while the temperature remains constant.

Determine the final volume of the gas.

2.2. A quantity of gas has an initial volume of 0.06 m3 and a temperature of 15°C. It is

expanded to a volume of 0.12 m3 while the pressure remains constant. Determine the

final temperature of the gas.

2.3. A mass of gas has an initial pressure of 1 bar and a temperature of 20°C. The

temperature of the gas is now increased to 550°C while the volume remains constant.

Determine the final pressure of the gas.

37

2.4. A mass of air has an initial pressure of 1.3 MN/m2, volume 0.014 m3 and temperature

135°C. It is expanded until its final pressure is 275 kN/m2 and its volume becomes

0.056 m3. Determine:

(a) the mass of air;

(b) the final tempetature of the air.

Take R = 0.287 kJ/kgK.

2.5. A quantity of gas has an initial pressure and volume of 0.1 MN/m2 and 0.1 m3,

respectively. It is compressed to a final pressure of 1.4 MN/m2 according to the law

PV1.26 = constant. Determine the final volume of the gas.

2.6. A mass of gas has an initial pressure and temperature of 0.11 MN/m2 and 15°C,

respectively. It is compressed according to the law PV1.3 = constant until the

temperature becomes 90°C. Determine the final pressure of the gas.

2.7. 0.23 kg of air has an initial pressure of 1.7 MN/m2and a temperature of 200°C. It is

expanded to a pressure of 0.34 MN/m2 according to the law PV1.35 = constant.

Determine the work done during the expansion. Take R = 0.29 kJ.kgK.

2.8. A certain mass of air, initially at a pressure of 480 kN/m2 is expanded adiabatically to

a pressure of 94 kN/m2. It is then heated at constant volume until it attains its initial

temperature, when the pressure is found to be 150 kN/m2. State the type of

compression necessary to bring the air back to its original pressure and volume. Using

the infomation calculate the value of γ. If the initial temperature of the air is 190°C,

determine the work done/kg of air during the adiabatic expansion. Take R for air =

0.29 kJ/kgK.

2.9. One kilogramme of a certain gas is at 0.11 MN/m2 and 15°C. It is compressed until its

volume is 0.1 m3. Calculate the final pressure and temperature if the compression is

(a) isothermal, (b) adiabatic. Calculate, also, the work done, change of internal energy

and heat transfer in each case. Distinguish between positive and negative quantities.

Take, cp = 0.92 kJ/kgK; cv = 0.66 kJ/kgK/

2.10. A certain gas has a density of 0.09 kg/m3 at 0°C and 1.013 bar. Calculate the

characteristic gas constant and hence find the volume of 1 kg of this gas at 70°C and

2.07 bar. If a volume of 5.6 m3 of this gas at 1.02 bar and 0°C is heated at constant

pressure to 50°C, calculate:

(a) the heat transfer; (b) the change of internal energy; (c) the work done.

Take cv = 10.08 kJ/kgK.

38

2.11. At the beginning of compression in an oil engine cylinder, the pressure is 95 kN/m2,

the volume 14 litre and the temperature 100°C. The index for the compression curve

is 1.3 and the volume ratio of the compression is 14 to 1. Calculate the work done,

change of internal energy and the heat transfer during compression. Take R = 0.28

kJ/kgK; cv = 0.72 kJ/kgK.

2.12. Prove the relationship between the two principal specific heat capacities and the

characteristic gas constant for a perfect gas.

2.13. 0.45 kg of a certain gas expands adiabatically in a cylinder fitted with a piston. The

initial pressure and temperature are 690 kN/m2 and 185°C, respectively. The final

pressure is 138 kN/m2 and the temperature falls by165°C during the expansion. If the

gas does 53 kJ of work on the piston, calculate the two principal specific heat

capacities of the gas.

2.14. An engine cylinder of 75 mm diameter and 150mm stroke contains 1 g of air at a

temperature of 15°C. Calculate the work required to compress the air isothermally to

a pressure of 700 kN/m2. Neglect the clearance volume.

Take cp = 1.005 kJ/kgK; cv = 0.717 kJ/kgK.

2.15. A quantity of air occupies a volume of 30 litre at a temperature of 38°C and a pressure

of 104 kN/m2. The temperature of the air can be raised,

(a) by heating at constant volume until the pressure is 208 kN/m2 ; or,

(b) by adiabatic compression until the volume is 6 litre.

Find, for each case, the final temperature, the external work done, the change of

internal energy, and the heat transferred.

2.16. 0.25 kg of air at a pressure of 140 kN/m2 occupies 0.15 m3 and from this condition it

is compressed to 1.4 MN/m2 according to the law PV1.25 = C. Determine:

(a) the change of internal energy of the air,

(b) the work done on or by the air,

(c) the heat received or rejected by the air.

Take cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK.

1.13