Thermodynamics 1

ENGINEERING THERMODYNAMICS FOR YEAR I ELECTRICAL

1
INTRODUCTION AND THE FIRST LAW OF THERMODYNAMICS

All living things depend on energy for survival, and modern civilizations will continue to thrive only if existing sources of energy can be developed to meet the growing demands.

Energy exists in many forms, from the energy locked in the atoms of matter itself to the intense radiant energy emitted by the sun.

Many sources of energy exist: many are known, some perhaps unknown; but when an energy source exists, means must first be found to transform the energy into a form convenient to our purpose.

The chemical energy from the combustion of fossil fuels (oil. coal, gas) and waste
(agricultural, industrial, domestic) is used to produce heat which is converted to mechanical energy in turbines or reciprocating engines.

Uranium atoms are bombarded asunder and the nuclear energy released is used as heat.

The potential energy of large masses of water is converted into electrical energy as it passes through water turbines on its way from the mountains to the sea.

The kinetic energy of the wind is harnessed by windmills to produce electricity.

The energy of the waves of the sea is converted into electrical power by floating Turbines.

The tides produced by the rotation of the moon produce electrical energy by flowing through turbines in large river estuaries.

Hot rocks and trapped liquids in the depths of the earth are made to release their energy to be converted to electricity.

The immense radiant energy of the sun is tapped to heat water or by suitable device is converted directly into electricity.

Figure 1.1shows the various energy sources and the possible conversion paths with the more important transfers shown as bold.

Macroscopic versus microscopic viewpoint
A microscopic viewpoint is used to attempt to understand a process or system by considering the particle nature of matter.
This viewpoint might focus on molecules, atoms or even an electron and nucleus.
A complete description would require an enormous effort with suitable approximations.
A macroscopic consideration addresses the appropriate observable averages of the microscopic phenomena.
For example, the microscopic momentum transfer between gas molecules and a surface is observed on the macroscopic level as gas pressure on the surface.
Clearly, the macroscopic viewpoint is of direct consequence to the engineer.
Classical thermodynamics is a macroscopic science.
The fundamental statements or laws concern the macroscopic properties of matter.
Any atomic or microscopic concept must be exhibited in the macroscopic behavior of the system.
This does not imply that a microscopic viewpoint is inappropriate for thermodynamics.
A clear understanding of macroscopic phenomena is often possible through microscopic concepts.
Yet the overriding goal of engineering thermodynamics is to address macroscopic properties.
This course stresses the fundamental concepts from a macroscopic viewpoint.
References are made to microscopic behavior where it is helpful to clearly present the material.
Applied thermodynamics is the science on the relationship between heat, work, and the properties of systems.

It is concerned with the means necessary to convert heat energy from available sources such as fossil fuels into mechanical work.

A heat engine is the name given to a system which by operating in a cyclic manner produces net work from a supply of heat.

The laws of thermodynamics are natural hypotheses based on observations of the world in which we live.

It is observed that heat and work are two mutually convertible forms of energy and this is the basis of the First Law of Thermodynamics.

It is also observed that heat never flows unaided from an object at a low temperature to one at a high temperature, in the same way that a river never flows unaided uphill.

This observation is the basis of the Second Law of Thermodynamics which can be used to show that a heat engine cannot convert all the heat supplied to it into mechanical work but must always reject some heat at a lower temperature.
These ideas will be discussed and developed in due course but first some fundamental definitions must be made
1.1 Heat, Work and the System

In order to deal with the subject of applied thermodynamics rigorously, it is necessary to define the concepts used.

Heat is a form of energy which is transferred from one body to another body at a lower temperature, by virtue of the temperature difference between the bodies.

For example, when a body A at a certain temperature, say 20 °C, is brought into contact with a body B at a higher temperature, say 21 °C, then there will be a transfer of heat from B to A until the temperatures of A and B are equal
(Fig.1.2).

When the temperature of A is the same as the temperature of B, no heat transfer takes place between the bodies, and they are said to be in thermal equilibrium.

Heat is apparent during the process only and is therefore transitory energy.

Since heat energy flows from B to A there is a reduction in the intrinsic energy possessed by B and an increase in the intrinsic energy possessed by A.

This intrinsic energy of a body is a function of temperature and must not be confused with heat.

Heat can never be contained in a body or possessed by a body.

A system may be defined as a collection of matter within prescribed and identifiable boundaries (Fig. 1.3).

The boundaries are not necessarily inflexible; for instance, the fluid in the cylinder of a reciprocating engine during the expansion stroke may be defined as a system whose boundaries are the cylinder walls and the piston crown.

As the piston moves so do the boundaries move (Fig. 1.4). This type of system is known as a closed system.

An open system is one in which there is a transfer of mass across the Boundaries.

Example: The fluid in a turbine at any instant may be defined as an open system whose boundaries are as shown in Fig. l.5

The pressure of a system is the force exerted by the system per unit area of its boundaries.

Units of pressure are for example, Pascal, Pa (where 1 Pa = 1N/m2) or bar. The symbol p will be used for pressure.

Pressure as defined here is called absolute pressure. A gauge for measuring pressure (e.g. as shown in Fig. 1.6(a) and 1.6(b)) records the pressure above atmospheric. This is called gauge pressure.

Absolute pressure equals gauge pressure plus atmospheric pressure.

The gauge shown in Fig. 1.6(b) is called a Bourdon gauge.

The absolute pressure of the system in a closed elliptical tube section forces the tube out of position against the pressure of the atmosphere.

The tube’s displacement is recorded by a pointer on a circular scale, which can be calibrated directly in bar.

When the pressure of a system is below atmospheric, it is called vacuum pressure (Fig. 1.7(a)).

When one side of a U-tube is completely evacuated and then sealed, the gauge will act as a barometer and the atmospheric pressure can be measured (Fig. 1.7(b)).

The gauges shown in Figs 1.6(a) and 1.7(a) measure gauge pressure in mm of a liquid of known relative density and are called manometers.

For example, when water is the liquid, then 1 mm or water = 1/103 x 9806.65 N/m2 = 9.81 N/m2 = 9.81 Pa where 1 m3 of water weighs 9810 N, say.

Mercury (Hg) is very often used in gauges. Taking the relative density of mercury as 13.6, then 1 mmHg = 1/103 x 13.6 x 9810 N/m2 = 133.4 N/m2 = 133.4 Pa

The specific volume of a system is the volume occupied by a unit mass of the system.

The symbol used is v and the units are in m3/ kg. The symbol V will be used for volume. (Note that the specific volume is the reciprocal of density.)

Work is defined as the product of a force and the distance moved in the direction of the force.

When a boundary of a closed system moves in the direction of the force acting on it, then the surroundings do work on the system.

When the boundary is moved outwards the work is done by the system on its surroundings.

The units of work are, for example, N m. If work is done on a unit mass of fluid, then the work done per kilogram of fluid has units of N m/kg.

Work is observed to be energy in transition. It is never contained in a body or possessed by a body.

Heat and work are both transitory energies and must not be confused with the intrinsic energy possessed by a system.

For example, when a gas contained in a well-lagged cylinder (Fig. 1.8(a)) is compressed by moving the piston to the left, the pressure and temperature of the gas are observed to increase, and hence the intrinsic energy of the gas increases.

Since the cylinder is well lagged, no heat can flow into or out of the gas.

The increase in intrinsic energy of the gas has therefore been caused by the work done by the piston on the gas.

As another example, consider a gas contained in a rigid container and heated (Fig. 1.8(b)).

Since the boundaries of the system are rigidly fixed then no work is done on or by the system. The pressure and temperature of the gas are observed to increase and hence the intrinsic energy of the gas will increase.

The increase in intrinsic energy has been caused by the heat flow to the system.

In the example of Fig. 1.8(a) the work done on the system is energy which is apparent only during the actual process of compression.

There is an intrinsic energy of the system initially and an intrinsic energy finally, but the work done appears only in transition from the initial to the final condition.

Similarly, in the example of Fig. 1.8(b), the heat supplied appears only in transition from one state of the gas to another.

Another way in which work may be transferred to a system is illustrated in Fig.1.9.

The paddle wheel imparts a change of momentum to the fluid and a work input is required to turn the shaft.

The kinetic energy attained by the fluid is dissipated by internal fluid friction and friction between the fluid and the container.

When the container is well lagged, all the work input goes to increasing the intrinsic energy of the system.

Sign Convention

The sign convention used assumes that all external inputs to a system are positive.

That is Heat supplied to a system, Q is positive.

Work input to a system, W is positive.

When a system boundary is drawn to define the system then it follows that heat supplied, Q and work input, W will always be shown by arrows pointing into the system.

In algebraic equations it will be quite clear when numbers are substituted whether the value of Q and/or W is positive or negative.

A negative value for Q will indicate that heat is rejected from the system.

A negative value for W will indicate that work is done by the system on its surroundings.

In many cases it would cause unnecessary confusion by referring throughout to negative quantities.

For example, it is clear that for a device designed to produce power, such as an internal combustion engine or turbine, the work input to the system is always negative.

Although the above sign convention will be used for all algebraic equations it will be made clear in the wording that the system is producing a work output.

For example:
Work done by the system = - W

Similarly, for the case of a system designed specifically to cool a fluid, such as a condenser, it is clear that the heat supplied to the system is always negative.

Hence we can write, Heat rejected by the system = - Q

1.2 Units

Throughout this course SI units will be used although reference may be made to the old imperial units as necessary.

The International System of Units (Systeme International d'Unites, abbreviation SI) was adopted by the General Conference of Weights and Measures in 1960 and subsequently endorsed by the International Organization for Standardization.

It is a coherent system. In a coherent system all derived unit quantities are formed by the product or quotient of other unit quantities.

In SI units six physical quantities are arbitrarily assigned unit value and hence all other physical quantities are derived from these.

The six quantities chosen and their units are as follows: length (metre, m); mass (kilogram, kg); time (second, s); electric current (ampere, A); thermodynamic temperature (degree kelvin, K); luminous intensity (candela, cd).

Thus, for example, velocity = length/time has units of m/s; acceleration = velocity/time has units of m/s2; volume = length x length x length has units of m3; specific volume = volume/mass has units of m3/kg.

Force, energy and power

Newton's second law may be written as force α mass x acceleration for a body of constant mass, i.e. F = kma (1.1)

where m is the mass of a body accelerated with an acceleration, a by a force F; k is a constant.

In a coherent system of units such as SI, k = 1, hence F = ma

The SI unit of force is therefore kg m/s2. This composite unit is called the newton, N, i.e. 1 N is the force required to give a mass of 1 kg an acceleration of 1m/s2.

It follows that the SI unit of work (= force x distance) is the newton metre,
Nm.

As stated earlier heat and work are both forms of energy and hence both can have the units of kg m2 or N m.

A general unit for energy is introduced by giving the newton metre the name joule, J,

i.e. 1 joule, J = 1 newton x 1 metre or 1 J = 1 N m

The use of additional names for composite units is extended further by introducing the watt, W, as the unit of power, i.e. 1watt, W = l J/s = 1N m/s

Pressure

The unit of pressure (force per unit area) is N/m2 and this unit is sometimes called the pascal. Pa.

For most cases occurring in thermodynamics the pressure expressed in pascals would be a very small number.

A new unit is defined as follows: l bar = l05 N/m2 = l05 Pa

The advantage of using a unit such as the bar is that it is approximately equal to atmospheric pressure.

In fact the standard atmospheric pressure is exactly 1.01325 bar.

As indicated in section 1.1, it is often convenient to express a pressure as a head of a liquid.

We have:
Standard atmospheric pressure = 1.01325 bar = 0.76 m Hg

Temperature

The variation of an easily measurable property of a substance with temperature can be used to provide a temperature-measuring instrument.

For example, the length of a column of mercury will vary with temperature due to the expansion and contraction of the mercury.

The instrument can be calibrated by marking the length of the column when it is brought into thermal equilibrium with the vapour of boiling water at atmospheric pressure and again when it is in thermal equilibrium with ice at atmospheric pressure.

On the Celsius (or Centigrade) scale 100 divisions are made between the two fixed points and the zero is taken at the ice point.

The change in volume at constant pressure or the change in pressure at constant volume of a fixed mass of gas which is not easily liquefied (e.g. oxygen, nitrogen, helium, etc.) can be used as a measure of temperature.

Such an instrument is called a gas thermometer.

It is found for all gases used in such thermometers that if the graph of temperature against volume in the constant pressure gas thermometer is extrapolated beyond the ice point to the point at which the volume of the gas would become zero, then the temperature at this point is -273 oC approximately (Fig. 1.10).

Similarly if the graph of temperature against pressure in the constant volume gas thermometer is extrapolated to zero pressure, then the same zero of temperature is found.

Therefore, an absolute zero of temperature has therefore been fixed and an absolute scale of temperature can be defined.

Temperature on the absolute Celsius scale can be obtained by adding 273 to all temperatures on the Celsius scale; this scale is called the Kelvin scale.

The unit of temperature is the degree Kelvin and is given the symbol K.

Since the Celsius scale which is used in practice has a different zero the temperature in degrees Celsius is given the symbol C

(e.g.20°C = 293 K approximately; also, 30°C -20 C = 10 K).

In this course, capital T is used for absolute temperature and small t for other temperatures.

Later in Module 5, an absolute scale of temperature will be introduced as a direct consequence of the Second Law of Thermodynamics.

It is found that the gas thermometer absolute scales approach the ideal scale as a limit.

Multiples and sub-multiples

Multiples and sub-multiples of the basic units arc formed by means of prefixes and the ones most commonly used as follows:

For example, power can be expressed in either megawatts, MW or kilowatts, kW or watts, W.

In the measurement of length the millimetre, mm, the metre, m and the kilometre, km are usually adequate.

For areas, the difference in size between the square millimetre, mm2 and the square metre, m2 is large (a factor of 106) and an intermediate size is useful; the square centimetre, cm2 is recommended for limited use only.

For volumes, the difference between the cubic millimetre, mm3, and the cubic metre, m3 is much too great (a factor of 109) and the most commonly used intermediate unit is the cubic decimetre, dm3 which is equal to one-thousandth of a cubic metre (i.e. 1 dm3 = 10-3 m3 ).

The cubic decimetre can also be called the litre,l. i.e. 1 litre, l = 1 dm3 = 10-3 m3

(Note, for very precise measurements, 1 litre = 1.000028 dm3.)

Certain exceptions to the general rule of multiplying factors are inevitable.

The most obvious example is in the case of the unit of time.

Instead of the centisecond, kilosecond or megasecond for instance: the minute, hour, day, etc. are used.

Similarly, a mass flow rate may be expressed in kilograms per hour, kg/h if this gives a more convenient number than when expressed in kilograms per second, kg/s.

Also the speed of road vehicles is expressed in kilometres per hour, km/h since this is more convenient than the normal unit for velocity which is metres per second, m/s.

1.3 The state of the working fluid

In all problems in applied thermodynamics we are concerned with energy transfers to or from a system.

In practice the matter contained within the boundaries of the system can be liquid, vapour, or gas and is known as the working fluid.

At any instant the state of the working fluid may be defined by certain characteristics called its properties.

Many properties have no significance in thermodynamics (e.g. electrical resistance), and will not be considered.

The thermodynamic properties introduced in this module are as follows; pressure, temperature, specific volume, specific internal energy, specific enthalpy and specific entropy.

It has been found that, for any pure working fluid, only two independent properties are necessary to define completely the state of the fluid.

Since any two independent properties suffice to define the state of a system, it is possible to represent the state of a system by a point situated on a diagram of properties.

For example, a cylinder containing a certain fluid at pressure, p1 and specific volume, v1 is at state 1, defined by point 1 on a diagram of p against v as shown on (Fig. 1.11(a)).

Since the state is defined, then the temperature of the fluid, T is fixed and the state point can be located on a diagram of p against T and T against v (Figs 1.11(b) and 1.11(c)).

At any other instant the piston may be moved in the cylinder such that the pressure and specific volume are changed to p2 and v2. State 2 can then be marked on the diagrams.

Diagrams of properties are used continually in applied thermodynamics to plot state changes.

The most important are the pressure-volume and temperature-entropy diagrams, but enthalpy-entropy and pressure-enthalpy diagrams are also used frequently.

1.4 Reversibility

In section 1.3 it was shown that the state of a fluid can be represented by a point located on a diagram using two properties as coordinates.

When a system changes state in such a way that at any instant during the process the state point can be located on the diagram, then the process is said to be reversible.

The fluid undergoing the process passes through a continuous series of equilibrium states.

A reversible process between two states can therefore be drawn as a line on any diagram of properties (Fig. 1.12(a)).

In practice, the fluid undergoing a process cannot be kept in equilibrium in its intermediate states and a continuous path cannot be traced on a diagram of properties.

Such real processes are called irreversible processes.

An irreversible process is usually represented by a dotted line joining the end states to indicate that the intermediate states are indeterminate (Fig. 1.12(b)).

A more rigorous definition of reversibility is as follows:

When a fluid undergoes a reversible process, both the fluid and its surroundings can always be restored to their original state.

The criteria of reversibility are as follows:

(a) The process must be frictionless. The fluid itself must have no internal friction and there must be no mechanical friction (e.g. between cylinder and piston).

(b) The difference in pressure between the fluid and its surroundings during the process must be infinitely small.
This means that the process must take place infinitely slowly, since the force to accelerate the boundaries of the system is infinitely small.

(c) The difference in temperature between the fluid and its surroundings during the process must be infinitely small.
This means that the heat supplied or rejected to or from the fluid must be transferred infinitely slowly.

It is obvious from the above criteria that no process in practice is truly reversible.

However, in many practical processes, a very close approximation to an internal reversibility may be obtained.

In an internally reversible process, although the surroundings can never be restored to their original state, the fluid itself is at all times in an equilibrium state and the path of the process can be exactly retraced to the initial state.

In general, processes in cylinders with a reciprocating piston are assumed to be internally reversible as a reasonable approximation.

However, processes in rotary machinery (e.g. turbines) are known to be irreversible due to the high degree of turbulence and scrubbing of the fluid.

1.5 Reversible work

Consider an ideal frictionless fluid contained in a cylinder behind a piston.
Assume that the pressure and temperature of the fluid are uniform and that there is no friction between the piston and the cylinder walls.

Let the cross-sectional area of the piston be A, let the pressure of the fluid be p, let the pressure of the surroundings be (p + dp) (Fig. 1.13).

The force exerted by the piston on the fluid is pA. Let the piston move under the action of the force exerted a distance dl to the left.

Then work done on the fluid by the piston is given by force times the distance moved.

i.e. Work done, dW = - (pA) x dl= -pdV

where dV is a small increase in volume. The negative sign is necessary because the volume is decreasing.

Or for a mass, m: dW= -mp dv

where v is the specific volume. This is only true when criteria (a) and (b) hold as stated in section 1.4.

When a fluid undergoes a reversible process a series of state points can be joined up to form a line on a diagram of properties.

The work done on the fluid during any reversible process, W is therefore given by the area under the line of the process plotted on a p - v diagram (Fig. 1.14), i.e.

(1.2)

When p can be expressed in terms of v then the integral, m12pdv can be evaluated.

Example 1.1

Unit mass of a fluid at a pressure of 3 bar and with a specific volume of
0.18 m3/kg, contained in a cylinder behind a piston expands reversibly to a pressure of 0.6 bar according to a law p = c/v2, where c is a constant. Calculate the work done during the process.

Solution: Referring to Fig 1.15

When an expansion process takes place reversibly (see Fig. 1.16), the integral, 12pdv is positive, i.e.

W = -m 12p dv = -m (shaded area on Fig. 1.16)

A process from right to left on the p - v diagram is one in which there is a work input to the fluid (i.e. W is positive). A compression process in which the pressure increases and the volume decreases is an example of this process.

Conversely, a process from left to right is one in which there is a work output from the fluid (i.e. W is negative). An expansion process in which the pressure decreases and the volume increases is an example of this process.

When a fluid undergoes a series of processes and finally returns to its initial state, then it is said to have undergone a thermodynamic cycle.

A cycle which consists only of reversible processes is a reversible cycle.

A cycle plotted on a diagram of properties forms a closed figure and a reversible cycle plotted on a p - v diagram forms a closed figure, the area of which represents the net work of the cycle.

For example, a reversible cycle consisting of four reversible processes
1 to 2, 2 to 3, 3 to 4, and 4 to 1 is shown in Fig. 1.17. The net work input (+W) is equal to the shaded area.

If the cycle were described in the reverse direction (i.e. 1 to 4, 4 to 3, 3 to 2, and 2 to 1), then the shaded area would represent net work output (-W) from the system.

The rule is that the enclosed area of a reversible cycle represents net work input, (+W) (i.e. net work done on the system) when the cycle is described in an anticlockwise manner

The enclosed area represents work output, (-W) (i.e. work done by the system) when the cycle is described in a clockwise manner.

Example 1.2

Unit mass of a certain fluid is contained in a cylinder at an initial pressure of 20 bar. The fluid is allowed to expand reversibly behind a piston according to a law pV2 = constant until the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston regains its original position; heat is then supplied reversibly with the piston firmly locked in position until the pressure rises to the original value of 20 bar. Calculate the net work done by the fluid for an initial volume of 0.05 m3.

Solution: Referring to Fig 1.18

It has been stated above that work is given by - p dv for a reversible process only.

It can easily be shown that - pdv is not equal to the work done if a process is irreversible.

For example, consider a cylinder, divided into a number of compartments by sliding partitions (Fig. 1.19).

Initially, compartment A is filled with a mass of fluid at pressure p1.

When the sliding partition 1 is removed quickly, then the fluid expands to fill compartments A and B.

When the system settles down to a new equilibrium state, the pressure and volume are fixed and the state can be marked on the p- V diagram (Fig. 1.20).

Sliding partition 2 is now removed and the fluid expands to occupy compartments A, B. and C.

Again the equilibrium state can be marked on the diagram. The same procedure can be adopted with partitions 3 and 4 until finally the fluid is at p2 and occupies a volume V2 when filling compartments A, B. C, D and E.

The area under the curve 1-2 on Fig. 1.20 is given by 12p dv , but no work has been done (apart from the negligible work required to move partitions).

No piston has been moved, no turbine wheel has been rotated; in other words, no external force has been moved through a distance.

This is the extreme case of an irreversible process in which pdv has a value and yet the work done is zero.

When a fluid expands without a restraining force being exerted by the surroundings, as in the example above, the process is known as free expansion.

Free expansion is highly irreversible by criterion (b), section 1.4.

In many practical expansion processes, some work is done by the fluid which is less thanpdv .

In many practical compression processes work is done which is greater than pdv.

It is important to represent all irreversible processes by dotted lines on a p-v diagram as a reminder that the area under the dotted line does not represent work.

1.6 Conservation of energy and the First Law of Thermodynamics

The concept of energy and the hypothesis that it can neither be created nor destroyed were developed by scientists in the early part of the nineteenth century and became known as the Principle of the Conservation of Energy.

The First Law of Thermodynamics is merely one statement of this general principle with particular reference to thermal energy (i.e. heat) and mechanical energy (i.e. work).

When a system undergoes a complete thermodynamic cycle, the intrinsic energy of the system is the same at the beginning and end of the cycle.

During the various processes that make up the cycle, work is done on or by the fluid and heat is supplied or rejected.

The network input can be defined as W.

The net heat supplied as Q where the symbol represents the sum for a complete cycle.

Since the intrinsic energy of the system is unchanged the First Law of Thermodynamics slates that:

When a system undergoes a thermodynamic cycle, then the net heat supplied to the system from its surroundings plus the net work input to the system from its surroundings must equal zero.

That is Q + W = 0 (1.3)

Example 1.3

In a certain Steam Plant the turbine develops 1000 kW. The heat supplied to the steam in the boiler is 2800 kJ/kg, the heat rejected by the steam to the cooling water in the condenser is 2I00 kJ/kg and the feed-pump work required to pump the condensate back into the boiler is 5 kW. Calculate the steam mass flow rate.

Solution

The cycle is shown diagrammatically in Fig. 1.21. A boundary is shown which encompasses the entire plant. Strictly, this boundary should be thought of as encompassing the working fluid only. For unit mass flow rate

1.7 The non-flow equation

In section 1.6 it is stated that when a system possessing a certain intrinsic energy is made to undergo a cycle by heat and work transfer, then the net heat supplied plus the net work input is zero.

This is true for a complete cycle when the final intrinsic energy of the system is equal to its initial value.

Consider now a process in which the intrinsic energy of the system is finally greater than the initial intrinsic energy.

The sum of the net heat supplied and the net work input has increased the intrinsic energy of the system,

i.e.

Gain in intrinsic energy = Net heat supplied + net work input

When the net effect is to transfer energy from the system, then there will be a loss in the intrinsic energy of the system.

When a fluid is not in motion, then its intrinsic energy per unit mass is known as the specific internal energy of the fluid and is given the symbol u.

The specific internal energy of a fluid depends on its pressure and temperature, and is itself a property.

The internal energy of mass, m of a fluid is written as U, i.e. mu = U.

The units of internal energy, U are usually written as kJ.

Since internal energy is a property, then gain in internal energy in changing from state 1 to state 2 can be written U2 – U1

Also, gain in internal energy = net heat supplied + net work input, i.e.

This equation is true for a process or series of processes between state 1 and state 2 provided there is no flow of fluid into or out of the system.

In any one non-flow process, there will be either heat supplied or heat rejected, but not both.

Similarly there will be either work input or work output, but not both.

Hence, U2 – U1 = Q + W for a non-flow process

Or for unit mass, Q = W = u2 – u1 (1.4)

This equation is known as the non-flow energy equation. Equation (1.4) is very often written in differential form.

For a small amount of heat supplied dQ, a small amount of work done on the fluid dW and a small gain in specific internal energy du, then

dQ + dW = du (1.5)
Example 1.4

In the compression stroke of an internal-combustion engine the heat rejected to the cooling water is 45 kJ/kg and the work input is 90 kJ/kg. Calculate the change in specific internal energy of the working fluid stating whether it is a gain or a loss.

Solution Q = -45 kJ/kg (- ve sign since heat is rejected).

W = 90 kJ/kg

Using equation (1.4), Q + W = u2 – u1

- 45 + 90 = u2 – u1

Therefore, u2 – u1 = 45 kJ/kg

i.e. Gain in internal energy = 45 kJ/kg

Example 1.5

In the cylinder of an air motor the compressed air has a specific internal energy of 420 kJ/kg at the beginning of the expansion and a specific internal energy of 200 kJ/kg after expansion. Calculate the heat flow to or from the cylinder when the work done by the air during the expansion is 100 kJ/kg.

Solution

From equation (1.4), Q + W = u2 – u1

i.e. Q - 100 = 200 - 420

therefore, Q= -120 kJ/kg

i.e. Heat rejected by the air = +120 kJ/kg

It is important to note that equations (1.3), (1.4), and (1.5) are true whether or not the process is reversible. These are energy equations.

For reversible non-flow processes we have, from equation (1.2)
W = -m 12p dv

or in differential form, dW = - m p dv

Hence for any reversible non-flow process for unit mass, substituting in equation (1.5), dQ = du + pdv (1.6)

or substituting in equation (1.4), Q = (u2 – u1 ) +12p dv (1.7)

Equations (1.6) and (l.7) can only be used for ideal reversible non-flow processes.

1.8 The steady-flow equation

In section 1.7, the specific internal energy of a fluid was said to be the intrinsic energy of the fluid due to its thermodynamic properties.

When unit mass of a fluid with specific internal energy, u, is moving with velocity C and is a height Z above a datum level, then it possesses a total energy of u + (C2/2) + Zg.

where C2/2 is the kinetic energy of unit mass of the fluid and Zg is the potential energy of unit mass of the fluid.

In most practical problems the rate at which the fluid flows through a machine or piece of apparatus is constant. This type of flow is called steady flow.

Consider a fluid flowing in steady flow with a mass flow rate, m through a piece of apparatus (Fig. 1.22).

This constitutes an open system as defined in section 1.2.

The boundary is shown cutting the inlet pipe at section 1 and the outlet pipe at section 2. This boundary is sometimes called a control surface and the system encompassed a control volume.

Let it be assumed that a steady rate of flow of heat Q units is supplied and that W is the rate of work input on the fluid as it passes through the apparatus.

Now in order to introduce the fluid across the boundary at the inlet an expenditure of energy is required.

Similarly in order to push the fluid across the boundary at exit, an expenditure of energy is required. The inlet section is shown enlarged in Fig. 1.23.

Fig. 1.23 Section at the inlet to the system in Fig.1.22

Consider an element of fluid of length, l and let the cross-sectional area of the inlet pipe be A1.

Then we have

Energy required to push element across boundary = (p1A1) x l = p1 x (volume of fluid element) = p1V1

Therefore
Energy required for unit mass flow rate of fluid = p1V1/m = p1 v1

where v1 is the specific volume of the fluid at section 1.

Similarly it can be shown that

Energy required at exit to push unit mass flow rate of fluid across the boundary = p2v2

Consider now the energy entering and leaving the system.

The energy entering the system consists of the energy of the flowing fluid at inlet:

and the energy term m p1v1, the heat supplied Q and the rate of work input, W.

The energy leaving the system consists of the energy of the flowing fluid at the outlet section:

and the energy term m p2v2.

Since there is steady flow of fluid into and out of the system and there are steady flows of heat and work, then the energy entering must exactly equal the energy leaving.

(1.8)

In nearly all problems in applied thermodynamics, changes in height are negligible and therefore the potential energy terms can be omitted from the equation.

The terms in u and pv occur on both sides of the equation and always will do so in a flow process since a fluid always possesses a certain internal energy and the term pv always occurs at inlet and outlet as seen in the above proof.

The sum of specific internal energy and the pv term is given the symbol h, and is called specific enthalpy:

i.e. Specific enthalpy, h = u + pv (1.9)

The specific enthalpy of a fluid is a property of the fluid since it consists of the sum of a property and the product of two properties.

Since specific enthalpy is a property like specific internal energy, pressure, specific volume and temperature.

It can be introduced into any problem whether the process is a flow process or a non-flow process. The enthalpy of mass, m of a fluid can be written as H
(i.e. mh = H).

The units of h are the same as those of internal energy.

Substituting equation (1.9) in equation (1.8) (1.10)

Equation (1.10) is known as the steady-flow energy equation. In steady flow the rate of mass flow of fluid at any section is the same as at any other section.

Consider any section of cross-sectional area A, where the fluid velocity is C, then the rate of volume flow past the section is CA. Also, since mass flow is volume flow divided by specific volume

Mass flow rate, m = CA/v = ρCA (1.11)

where v is the specific volume at the section and ρ the density at the section.

This equation is known as the continuity of mass equation.
With reference to Fig. 1.22

Example 1.6

In the turbine of a gas turbine unit, the gases flow through the turbine at
17 kg/s and the power developed by the turbine is 14000 kW. The specific enthalpies of the gases at inlet and outlet are 1200kJ/kg and 360kJ/kg respectively and the velocities of the gases at inlet and outlet are 60 m/s and 150 m/s respectively. Calculate the rate at which heat is rejected from the turbine. Find also the area of the inlet pipe given that the specific volume of the gases at inlet is 0.5 m3/kg.

Solution:

A diagrammatic representation of the turbine is shown in Fig. 1.24. From equation (1.10), neglecting changes in height.

Example 1.7

Air flows steadily at the rate of 0.4 kg/s through an air compressor, entering at 6 m/s with a pressure of 1 bar and a specific volume of 0.85 m3/kg and leaving at 4.5 m/s with a pressure of 6.9 bar and a specific volume of 0.16 m3/kg. The specific internal energy of the air leaving is 88 kJ/kg greater than that of the air entering Cooling water in a jacket surrounding the cylinder absorbs heat from the air at the rate of 59 kW. Calculate the power required to drive the compressor and the inlet and outlet pipe cross-sectional areas.

Solution

In this problem it is more convenient to write the flow equation as in equation 1.8, omitting the Z terms,

A diagrammatic representation of the compressor is shown in Fig. 1.25. Note that the heat rejected across the boundary is equivalent to the heat removed by the cooling water from the compressor. For unit mass flow rate:

C12/2 = (6x6)/2= 18 J/kg

In Example 1.7, the steady-flow energy equation has been used despite the fact that the compression consists of suction of air, compression in a closed cylinder and discharge of air.

The steady-flow equation can be used because the cycle of processes takes place many times in a minute and therefore the average effect is a steady flow of air through the machine.