FIRST LAW OF THERMODYNAMICS (Part-1)FIRST LAW OF THERMODYNAMICS (Part-1)

INTRODUCTIONINTRODUCTION

The First Law of Thermodynamics is a particular statement based on the principles of law of conservation of energy according to which the total amount of energy in any thermodynamic system remains constant i.e. “Energy cannot be created or destroyed”. None of the energy is gained or expended in the sense it is converted from one form to another. Initially the statement of conservation of energy was quantitatively analyzed for thermodynamic systems by J.P. Joule during the period of eighteen forties which has led to the statement of First law of thermodynamics.

Statement of First law of thermodynamicsStatement of First law of thermodynamicsThe transfer of heat and the performance of work may both cause the same effect in a system.

Energy which enters a system as heat may leave the system as work, or energy which enters the system as work may leave as heat. Hence, by the law of conservation of energy, the net work done by the system is equal to the net heat supplied to the system. The first law of thermodynamics can therefore be stated as follows:

“When a system undergoes a thermodynamic cyclic process, then the net heat supplied to the system from the surroundings is equal to the net work done by the system on its surrounding”.

Mathematicallyi.e., Q = W

where represents the sum for a complete cycle.

The first law of thermodynamics can not be proved analytically, but experimental evidence has repeatedly confirms its validity and since no phenomenon has been shown to contradict it, therefore the first law is accepted as a ‘law of nature’.

Joule’s Experiment

The English scientist J.P.Joule conducted a series of experiments in 1840’s, and these experiments led to the first law of thermodynamics. The objective of joules experiment was to establish a relation between amount of work spent to bring about the liberation of heat and the amount of heat liberated.

Fig-1: Joule’s Experiment

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The experimental set up is shown in Fig-1(a). Work is done on the fluid(system) kept in an insulated vessel by stirring of the paddle wheel. This work input to the fluid causes a rise in the temperature of the fluid. The amount of work on the fluid is calculated by the product of the weight and the vertical height(z) through which the weight descends. The system is then immersed in a water bath, as shown in Fig-1(b), after removing the insulation. The heat is transferred from the fluid to the water bath till the original temperature is reached, which will be indicated by the thermometer. Thus the system undergoes a complete cycle with two energy interactions i.e. definite amount of adiabatic work transfer(W1-2) to the system followed by an amount of heat transfer(Q1-2) from the system, shown in fig-1(c). The amount of heat rejected by the fluid is equal to the increase of energy of the water bath.

Fig-1(c): Cycle completed by a system with two energy interactions i.e., work transfer followed by heat transfer

Joule carried out many such experiments with different type of work interactions in a variety of systems, he found that the net work input to the fluid system was always proportional to the net heat transferred from the system regardless of work interaction. Based on this experimental evidence Joule stated that,

“When a system (closed system) is undergoing a cyclic process, the net heat transfer to the system is directly proportional to the net work done by the system”. This statement is referred to as the first law for a closed system undergoing a cyclic process.

i.e., Q W

J Q = W Where J = Joule’s equivalent or Mechanical equivalent of heat = 1Nm/J

If both heat transfer and work transfer are expressed in same units as in the S.I. units then the constant of proportionality in the above equation will be unity and hence the mathematical form of first law for a system undergoing a cyclic process can be written as

i.e., Q = W

If the cycle involves many more heat and work quantities as shown in fig-1(d), the same result will be found.

For this cyclic process the statement of first law can be written as

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Fig-1(d): Cyclic Process on a Property Diagram

The cyclic integral in the above equation can be split into a series of non cyclic integral as

or 1Q2 + 2Q3 + 3Q4 + 4Q1 = 1W2 + 2W3 + 3W4 + 4W1

i.e., Q = W

or (∑Q)cycle = (∑W)cycle

This is the first law for a closed system undergoing a cyclic process. i.e., it is stated as “When a closed system is undergoing a cyclic process the algebraic sum of heat transfers is equal to the algebraic sum of the work transfers”.

Equivalence of Heat and Work

Based on the results of a series of thoroughly conducted experiments J.P.Joule discovered a direct proportionality between the spent work ‘W’ and the quantity of heat obtained ‘Q’.

i.e., Q W

or Q = AW where ‘A’ is a Proportionality factor

Joule found that the proportionality factor ‘A’ remains same irrespective of the method of heat production, type of work, temperature of the body involved etc.

In other words, Joule established that when one and same amount of work is spent, one and same amount of heat is liberated. Thus the amount of liberated heat was shown to be equivalent to the amount equivalent of work spent; it is clear that this relation ship is true for the case when work is accomplished with the expense of heat.

Using the results of these measurement. Joule calculated the magnitude of ‘A’ which is known as thermal equivalent of work and of ‘J’ referred to as the mechanical equivalent of heat.

A = 0.002342 kcal/kgm

J = 426.99 kgm/kcal

It is obvious that J = 1/A

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Note J = 426.99 kgm/kcal = 4.184 Nm/cal = 1 Nm/J

First law for a closed system undergoing a non-cyclic process (i.e., for a change of state)

The expression (ΣW)cycle = (ΣQ)cycle applies only to the system undergoing cycles, and the algebraic summation of all energy transfer across system boundaries is zero.

But if the system undergoes a change of states during which both heat and work transfer are involved, the net energy transfer will be stored or accumulated within the system. If ‘Q’ is the amount of heat transferred to the system and ‘W’ is the amount of work transferred from the system during the process shown in the fig-2 below,

The net energy transfer (Q-W) will be stored in the system. Energy in storage is neither heat nor work and is given the name internal energy or simply, the energy of the system.

Q-W = ∆E

where ∆E is the increase in the energy of the system

or Q = ∆E + W

Here Q, W and ∆E are all expressed in joules.

If there are more energy transfer involved in the process, as shown in the fig-3 below

Then the first law gives

(Q2 + Q3 – Q1) = ∆E + (W2 + W3 – W1 – W4)

i.e., energy is thus conserved in the operation. Therefore the first law is a particular formulation of the principle of the conservation of energy. It can be shown that the energy has a definite value at every state of a system and is therefore, a property of a system.

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Fig-2

Fig-3:

Energy- A Property of the system

Consider a system which changes its state from state from state-1 to state-2 by following the path L, and returns from state-2 to state-1 by following the path M (fig-4). So the system undergoes a cycle. Writing the first law for path L

QL = ΔEL + WL ----- (1)

And for path M

QM = ΔEM + WM ----- (2)

Fig-4: Energy-a property of system

The process L and M together constitute a cycle, for which

Q = W

WL + WM = QL + QM

or QL – WL = WM – QM ----- (3)

from the equation (1), (2), and (3), it yields

ΔEL = –ΔEM ----- (4)

Similarly, had the system returned from the state-2 to sate-1 by following the path N instead of path M

ΔEL = –ΔEN ----- (5)

For the equation (4) and (5)

ΔEM = ΔEN

Thus, it is seen that the change in energy between two states of a system is the same whatever the path the system may fallow in undergoing that change of state. If some arbitrary value of energy is assigned to the state-2, the value of energy at sate-1 is fixed independent of the path the system fallows.

Therefore, energy has definite value for every state of the system. Hence, it is a point function and a property of the system.

Modes of Energy

Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential, electric, magnetic, chemical and nuclear and their sum constitutes the total energy E of a system.

i.e., E = kinetic energy (KE) + Potential Energy (PE) + remaining forms of energy.

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The energy E is an extensive property and the specific energy e = E/m (J/kg) is an intensive property.

In thermodynamic analysis, it is often helpful to consider the various form of energy that make up the total energy of a system in two groups

Macroscopic energy Microscopic energy

The Macroscopic forms of energy are those a system possesses as a whole with respect to sum outside reference frames which are related to motion and the influence of some external effects such as gravity, magnetism, electricity and surface tension. Eg Kinetic energy and Potential energy

The Microscopic forms of energy are those related to the molecular structure of a system and the degree of molecular activity, and they are independent of outside reference frames. The sum of all the microscopic form of energy is called the internal energy (U).

Kinetic EnergyThe energy that a system possesses as a result of its motion relative to reference frames is called kinetic energy. When all parts of a system move with the same velocity, the kinetic energy(KE) is expressed as

KE = m V2/2 (kJ)

or, on a unit mass basis

ke = V2/2 (kJ/kg)

where V denotes the velocity of the system relative to some fixed reference frame

Potential energy

The energy that a system possesses as a result of its elevation in a gravitational field is called potential energy(PE) and is expressed as

PE = mgz (kJ)

or, on a unit mass basis

pe = gz (kJ/kg)

where g is the gravitational acceleration and z is elevation of the center of gravity of a system to some arbitrarily reference level

The magnetic, electric and surface tension effects are significant in some specialized cases only and usually ignored. In the absences of such effects, the total energy of a system is expressed as

E = KE + PE + U =mV2/2 + mgz + U

or, on unit mass basis

e = ke + pe + u = V2/2 + gz + u

Since the terms comprising E are point functions, we can write

ΔE or dE = d(KE) + d(PE) + dU

The first law of thermodynamics for a change of state of a system may therefore be written as

Q = ΔE + W

Q = [dU + d(KE) + d(PE)] + W

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In words this equation states that as a system undergoes a change of state, energy may cross the boundary as either heat or work, and each may be positive or negative. The net change in the energy of the system will be exactly equal to the net energy that crosses the boundary of the system. The energy of the system may change in any of three ways, namely, by a change in IE, KE or P.E

Substituting for KE and PE in the above equation

Q = dU + + d (mgz) + W

In the integral form this equation is, assuming ‘g’ is a constant

Q1-2 = U2 – U1 + + mg (z2 – z1) + W1-2

In most of the situations the changes in KE and PE are very small, when compared with the changes in internal energies. Thus KE and PE changes can be neglected.

Q = dU + W

or Q1-2 = U2 – U1 + W1-2

Corollaries of First Law

Corollary 1: System Executing a Process

For a system executing a process, change in stored energy of the system is given as ΔE = δQ – δW Where E = energy stored in the system

Corollary 2: Isolated systemFor an isolated system,

δQ = 0 and δW = 0Hence ΔE = 0 or E = constant

Thus, the energy of an isolated system remains unchanged.

Therefore, the first law of thermodynamics may also be stated as follows, “Heat and work are mutually convertible but since energy can neither be created nor destroyed, the total energy associated with an energy conversion remains constant”.

Corollary 3: Perpetual Motion Machine of First Kind (PMM-1)

Definition: - It is an imaginary device which produces a continuous supply of work without absorbing any energy from the surrounding or from the system. Such a machine in effect creates energy from nothing and violates the first law of thermodynamics.

It is impossible to construct a perpetual motion machine of first kind. The PMM-1 violates the first law.

As per the law of conservation of energy, no engine can produce mechanical work continuously without some other form of energy disappearing simultaneously.

The Pure Substance

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The system encountered in thermodynamics is often quite less complex and consists of fluids that don not change chemically, or exhibit significant electrical, magnetic or capillary effects. These relatively simple systems are given the generic name the Pure Substance.

DefinitionA system is set to be a pure substance if it is (i) homogeneous in chemical composition, (ii) homogeneous in chemical aggregation and (iii) invariable in chemical aggregation.

Homogeneous in chemical composition means that the composition of each part of the system is same as the composition of any other part. Homogeneous in chemical aggregation implies that the chemical elements must be chemically combined in the same way in all parts of the system. Invariable in chemical aggregation means that the chemical aggregation should not vary with respect to time.

Satisfies condition (i) Satisfies condition (i) Does not satisfies condition (i) Satisfies condition (ii) Does not satisfies condition (ii) Satisfies condition (iii)

Fig-5: Illustration of the definition of pure substance

In fig-5 three systems are shown. The system (a) shown in the figure is a mixture of steam and water. It is homogeneous in chemical composition because in every part of the system we have, for every atom of oxygen we have two atoms of hydrogen, whether the sample is taken from steam or water. The same is true for system (b) consisting of water and uncombined mixture of hydrogen and oxygen. System (c) however is not homogeneous in chemical composition because in the upper part of the system hydrogen and oxygen are present in the ratio 1:1 where as in the bottom portion they are present in the ratio 2:1.

System (a) also satisfies condition (ii), because both hydrogen and oxygen have combined chemically in every part of the system. System (b) on the other hand does not satisfies condition (ii) because the bottom part of the system has two elements namely hydrogen and oxygen have chemically combined where as in the upper part of the system the two elements appear as a mixture of two individual gases.

Invariable in chemical aggregation means that the state of chemical combination of the system should not change with time. Thus the mixture of hydrogen and oxygen, if it is changing into steam during the time the system was under consideration, then the system’s chemical aggregation is varying with time and hence this system is not a pure substance. Thus the system (a) is a pure substance whereas the systems (b) and (c) are not pure substances.

The Two Property Rule for a Pure Substance

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The two property rule states that,

The thermodynamics state of a pure substance of a given mass can be fixed by specifying two independent properties provided (i) the substance is in equilibrium and (ii) the effects of gravity, motion, capillarity, electricity and magnetism are negligible.

The above rule indicates that if the values of two properties of a pure substance are fixed then the values for all other properties are fixed. This means that there is a definite relation between the two independent properties and each of the other properties. Each of these relations is called “Equation of state” for a pure substance. The equation of state for a pure substance can be in any one of the following forms:

(i) Algebraic equation (example: perfect gas equation), (ii) Tables (example: steam tables) and (iii) Charts (example: Mollier chart for steam).

Specific Heat (c)

Specific heat of a substance may be defined as the amount of heat required to raise the temperature of a unit mass of the substance through one degree.

The SI unit of specific heat is

The solids and liquids have only one specific heat whereas gases have two specific heats, which are based on the type of the heating process. i.e., specific heat at constant volume (cv) and specific heat at constant pressure (cp)

If δQ is the amount of heat transferred to raise the temperature of 1 kg of substance by dT, then, specific heat c = δQ/dT

or, δQ = c dT

the amount of treansferred to raise temperature of mass ‘m’ of a substance by dT, is given as,

δQ = m c dT

As we know, the specific heat of gas depends not only on the temperature but also upon the type of the heating process. i.e., specific heat of a gas depends on whether the gas is heated under constant volume or under constant pressure process.

We have δQ = m cv dT for a rev. non-flow process at constant volumeand δQ = m cp dT for a rev. non-flow process at constant pressure

For a perfect gas, cp & cv are constant for any one gas at all pressure and temperatures. Hence, integrating above equations.

Flow of heat in a reversible constant pressure process Q1-2 = m cp (T2 – T1)Flow of heat in a reversible constant volume process Q1-2 = m cv (T2 – T1)

Joule’s Law

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Joule’s law states as follows:

“The internal energy of a perfect gas is a function of the absolute temperature only.” i.e., u = f (T)

To evaluate this function let 1kg of a perfect gas be heated at constant volume.

According to non flow energy equation,

δQ = du + δW

δW = 0, since the volume remains constant

δQ = du

At constant volume the heat transferred for a perfect gas of 1 kg is given as,

δQ = cv dT

δQ = du = cv dT

and integrating

u = cv T + K (K = constant) i.e., internal energy, u = cv dT for a perfect gas

or For mass m, of a perfect gas Internal energy, U = m cvT

For a perfect gas, in any process between state-1 and state-2, change in internal energy is given as

(U2 – U1) = m cv (T2 – T1)

Relation between Specific heats cp & cv :

Consider a perfect gas being heated at constant pressure from T1 to T2.

According to non flow equation

Q1-2 = (U2 – U1) + W1-2 ------ (1)

Also for perfect gas,

U2 – U1 = m cv (T2 – T1)

In constant pressure process, the work done by the fluid,

W 1-2 = p (V2 – V1)

= m R (T2 – T1) [ pV = mRT, and p1 = p2 = p in this case] Heat transferred in the constant pressure process is given as

Q1-2 = m cp (T2 – T1)

By equating the three expressions in (1)

m cp (T2 – T1) = m cv (T2 – T1) + m R (T2 – T1)

m cp (T2 – T1) = (T2 – T1) (m cv + m R)

m cp = m(cv + R)

cp = cv + R

or cp – cv = R

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Dividing both sides by cv, we get

( where )

Similarly, dividing both sides by cp, we get

Specific Heat of a Pure Substance

Specific heat at constant volume

Specific heat at constant volume, cv of a pure substance is defined as the ratio of change of the specific internal energy of the substance with temperature, when the specific volume is held constant.

Symbolically it is written as,

Specific heat at constant pressure

Specific heat at constant pressure, cp of a pure substance is defined as the ratio of change of the specific enthalpy of the substance with temperature, when the pressure of the substance is held constant.

Symbolically it is written as,

Enthalpy

One of the most important fundamental quantities which occur consistently in thermodynamics is the internal energy (U) and pressure volume product (pV). This sum is called Enthalpy (H).

i.e. Enthalpy, H = U + pV

When the above is divided by the mass m of the system we have

i.e. H/m = h = u + pv

where h = specific enthalpy of the substance, u = specific internal energy and v = specific volume of the substance.The enthalpy of a fluid is a property of the fluid, since it consists of sum of a property and the product of the two properties. Since enthalpy is property like internal energy, pressure, specific volume and temperature, it can be introduced into any problem whether the process is a flow or a non flow process.

For a perfect gas referring the above equation

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h = u + pv

h = cvT + RT [ from Joules law for a perfect gas inernal energy, u = cvT & pv = RT]

h = (cv + R) T

h = cpT

and H = mcpT

Application of First law of Thermodynamics to Non-flow processes or Closed systemApplication of First law of Thermodynamics to Non-flow processes or Closed system

According to first law of thermodynamics, for a closed system, we have

Q1-2 –W1-2 = (ΔE)system

Where (ΔE)system = U2 – U1 + + mg (z2 – z1)

ΔE or dE = dU+ d(KE) + d(PE)

If kinetic energy and potential energy are negligible, then

Q1-2 –W1-2 = U2 – U1

1. Constant Volume (Isometric or Isochoric Process)

In the Constant volume process the working substances is contained in the rigid vessel as shown in the fig-6 below, let Q be the amount of heat supplied to the system. This results in increase of pressure and temperature at constant volume as represented by the process 1-2 on the p-V diagram. Since boundaries of the system are immovable and no work can done or by the system, it will be assumed that constant volume implies zero work unless stated otherwise

Fig-6: Constant Volume process Applying 1st law of thermodynamics to the process,

Q1-2 = U2 – U1 + W1-2

= U2 – U1 + 0

i.e., Q1-2 = cv (T2 – T1)

For mass ‘m’ of a substance, Q = mcv (T2 – T1)

2. Constant Pressure Process (or Isobaric Process)

Note: The term pV is sometimes referred to as “flow energy or flow work” which will be evident when we discuss first law for an open system,

The term U +pV is also called Total heat content in the system

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Q

Consider the cylinder with a piston carrying perfect gas shown in fig-7a. The gas pressure in the cylinder is maintained constant by adding weight on the piston. When heat is supplied to the system, its temperature will rise and it will expand forcing the piston to move upward. Thus the displacement work is done by the system against a constant force. The above process is shown in fig-7b on the pV diagram.

Fig-7: Constant Pressure Process

Work done by the system, δW = pdV

or W1-2 = = p(V2–V1)

From the first law of thermodynamics, we have

Q1-2 = W1-2 + ΔU

Q1-2 = W1-2 + (U2 – U1)

= p(V2–V1) + (U2 – U1)

= (U2 + pV2) – (U1 + pV1)

Q1-2 = H2 – H1

= m cp (T2 – T1)

3. Constant Temperature Process (Isothermal Process)

A process at a constant temperature is called an isothermal process, or constant temperature process. In this process the gas or vapour is heated at constant temperature and there shall be no change in internal energy. The work done will be equal to the amount of heat supplied. For this case since the gas is assumed an ideal gas and it implies that the gas follows Boyle’s law i.e. pV = C (T = Constant)

i.e pV = p1V1 = p2V2 = C or p =

Work done by the system W1-2 = = = [since pV = C]

From the first law of thermodynamics, we have

Q1-2 = W1-2 + ΔU Q1-2 = W1-2 + (U2 – U1)

Q1-2 = W1-2 + m cv (T2 – T1)

Since T is constant, i.e. T1 = T2

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Q

Q1-2 = W1-2 + 0

or Q1-2 = W1-2 =

Q1-2 = W1-2 =

4. Adiabatic process:

An adiabatic process is the thermodynamic process in which there is no heat interaction during the process, i.e. during the process, Q = 0. In these processes the work interaction is there at the expense of internal energy. The adiabatic process follows the law pVγ = constant where γ is called adiabatic index and is given by the ratio of two specific heats (cp/cv). Thus, it is obvious that adiabatic expansion shall be accompanied by the fall in temperature while temperature will rise during adiabatic compression. The adiabatic expansion process is shown on fig-9 below.

Fig-9: Adiabatic Process

Work done by the system W1-2 =

But,

i.e

W1-2 =

= =

=

W1-2 =

or W1-2 = (since pV = mRT)

From the first law of thermodynamics, we have

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Fig-8: Constant Temperature Process

Q1-2 = W1-2 + ΔU

Q1-2 = W1-2 + (U2 – U1)

Since there is no heat interaction i.e. Q = 0

0 = W1-2 + (U2 – U1)

W1-2 = (U1 – U2)

W1-2 = m cv (T1 – T2)

i.e = m cv (T1 – T2)

In an adiabatic expansion, the work done W1-2 by the fluid is at the expense of a reduction in the internal energy of the fluid. Similarly in an adiabatic composition process, all the work done on the fluid goes to increase the internal energy of the fluid.

To derive pV = C: For a reversible adiabatic process

To obtain the law relating p & V for a reversible adiabatic process, let us consider the non flow energy equation in the differential from.

Q = dU + W

For the reversible process W = pdV

Q = dU + pdV

For an adiabatic process Q = 0

dU + pdV = 0

Also for a perfect gas

pV = mRT or p =

Also U = m cv T (from Joul’s law)

dU = m cv dT

Hence on substituting,

m cv dT + = 0

cv dT + = 0

Dividing both sides by T, we get

cv + = 0

Integrating

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cv loge T + R loge V = constant

Substituting T =

cv loge + R loge V = constant

Dividing throughout both sides by cv

cv loge + loge V = constant

Again or [from the relation between the two specific heats cp & cv]

Hence substituting

loge + (γ – 1) loge V = constant

loge + loge V (γ – 1) = constant

loge = constant

i.e. loge = constant

i.e.

or pVγ = constant

For unit mass, pvγ = constant where v = specific volume

Relation between T & V, and T & p :

Now or or

For a perfect gas

or

5. Polytropic Process

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Polytropic process is the most commonly used process in practice. In this case the thermodynamic process is said to be governed by the law pVn = constant where n is the index which can vary from –∞ to +∞. Fig 10 shows some typical cases in which the value of n is varied and the type of process indicated for different values of n. Thus the various thermodynamics processes discussed above are special cases of polytropic process.

Fig-10: Polytropic process

Work interaction in case of polytropic process can be given as

W1-2 =

But, [ where ‘n’ is called index of compression or expansion ]

i.e

W1-2 =

= =

=

W1-2 =

or W1-2 = (since pV = mRT)

From the first law of thermodynamics, we have

Q1-2 = W1-2 + ΔU Q1-2 = W1-2 + (U2 – U1)

Q1-2 = + m cv (T2 – T1)

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= – m cv (T1 – T2)

Since [from the relation between the two specific heats cp cv]

Q1-2 = – (T1 – T2)

=

=

=

= .

Q1-2 = . W1-2

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Application of First law of Thermodynamics to Open system or Flow processes Application of First law of Thermodynamics to Open system or Flow processes

Flow Process

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In the case of closed system there is only energy transfer across the system boundary. In most of the engineering applications, we come across equipment through which there a continuous flow of material in and out along with energy transfer. Such processes are called flow processes.

For example consider a portion of power plant is in fig-11, high pressure steam enters the turbine at section-1, expands while doing work on the rotor, and then leaves at low pressure at section-2.

Fig-11 Turbine Plant: A Flow Process

The flow process are generally analyzed by the concept of control volume and a control surface, in this approach, we focus attention upon the fixed region in space through which fluid flow takes place. This fixed region is called the control volume and its surface is called the control surface. (Shown in fig-11)

The method of analysis is to select a control volume and then account for all quantities of energy entering and leaving this volume. Note that in addition to flow of matter across the control surface, heat and work transfer also take place.

Like a closed system the control volume is also defined by a boundary ( the control surface ). There is however differences are

The closed system boundary may change shape, position and orientation, the control volume boundary usually does not.

The matter flows across the control volume boundary but not across the closed system boundary. The control volume is also called an open system.

The advantage and simplification resulting from the control volume concept can be understood by the turbine plant shown above (in fig-10). The average change of state of steam from entrance to exit can be related to the shaft work delivered by the turbine and heat transfer from the turbine across the control volume boundary. Thus we need not to bother about the numerous changes of state and heat & work interactions experienced by the steam during its passage through different parts of the turbine.

Flow Work or Flow Energy

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Unlike closed systems, control volumes involve mass flow across their boundaries, and some work is required to push the mass into or out of the control volume. This work is known as the flow work, or flow energy, and is necessary for maintaining a continuous flow through a control volume.

Fig-12: Flow Work

To obtain a relation for flue work, consider a fluid element of volume ‘V, as shown in fig-12. The fluid immediately upstream will force this fluid element to enter the control plume; thus, it can be regarded as an imaginary piston. The fluid element can be chosen to be sufficiently small so that it has uniform properties throughout.

If the fluid pressure is ‘p’ and the cross-sectional area of the fluid element is ‘A’, the force applied on the fluid element by the imaginary

F = p .A

To push the entire fluid element into the control volume, this force must act through a distance ‘L’. Thus, the work done in pushing the fluid element across the boundary (i.e., the flow work) is

Wflow = F.L = p. A. L = p. V (kJ)

The total work per unit mass is obtained by dividing both sides of the above equation by the mass of the fluid element:

wflow = p.v (kJ/kg) (where v = specific volume)

The flow work relation is same whether the fluid is pushed into or out of control volume.

(The source of this flow energy is a pump located somewhere in the surroundings.)

Energy accompanying mass in a flow process

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In a flow process the energy contained within the control volume is influenced by the following energy transfers across the control surface:

a) Heat transferb) Work transferc) Energy flow accompanying mass flow at entrance and exit of the control volume

It is, therefore, necessary to determine the energy flow per unit mass flowing at various positions coordinates of the control volume. The energy accompanying unit mass consists of the following parts:

1) Internal energy: Each kg of matter has the internal energy ‘u’ and as the matter crosses the system boundary the energy of the system changes by ‘u’ for every kg mass of the matter that crosses the system boundary.

2) Kinetic energy: Since the matter that crosses the system boundary will have some velocity say each kg of matter carries a K.E. thus causing the energy of the system to change by this amount for every kg of matter entering the system boundary.

3) Potential energy: P.E. is measured with reference to some base. Thus ‘z’ is the elevation of the matter that is crossing the system boundary, then each kg of matter will possess a P.E. of gz.

4) Flow energy or Flow work: It is defined as the work required either to push a certain quantity of fluid into or out of the system. The flow work per unit mass of a fluid is given by the product of pressure and specific volume (i.e. p.v); Thus external to the system there must be some force which forces the matter across the system boundary and the energy associated with this is called flow energy.

Thus energy accompanying a unit mass flow is given by

eflow = u + + gz + pv [ where eflow is energy accompanying a unit mass flow ]

= ( u + pv ) + + gz

= h + + gz

The total energy accompanying the mass flow is

Eflow =

where is the mass flow rate in the system ( kg/sec)

Steady Flow Process

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The processes occurring in an open system which permits the transfer of mass as well as energy across its boundaries are known as flow processes. A process is said to be steady flow process if the conditions in the control volume remain unchanged with time, i.e. which the mass flow rate into the system is equal to mass flow rate out of the system. The common examples of steady flow processes are: nozzles, open system turbines, compressors, pumps

The following conditions exist in a steady flow process: (assumptions)

The mass flow rate remains constant within the system. The fluid is uniform in composition. The only interaction between the system and the surrounding are work and heat. The state of the fluid at any point remains constant with the time. In the analysis only potential, kinetic and flow energies are considered.

Consider a control volume of a thermodynamic open system as shown in a fig-13, through which a working fluid is passing at a steady rate.

Fig-13: Steady flow open system

Let= mass flow rate, kg/s

p = pressure, Pav = specific volume, m3/kgC = velocity of fluid, m/su = specific internal energy, J/kgz = height from datum, mh = specific enthalpy of fluid, J/kgQ = heat supplied, Jq = heat supplied per unit mass, J/kgW = work transferred, Jw = work transferred per unit mass, J/kgA = area of flow, m2

1, 2 = subscripts for inlet and outlet respectively.

a. Steady flow Mass Balance

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For the conservation of mass, if there is no accumulation of mass within the control volume, the mass flow rate for a single stream entering must equal the mass flow rate leaving or

or

This equation is called as the ‘equation of continuity’.When more than one stream of fluid enters the control volume, then

b. Steady Flow Energy Equation

The total energy entering the system,

The total energy leaving the system,

According to the first law of thermodynamics,

E1 = E2

=

Now h = u + pv

=

− =

The above equation is called the Steady Flow Energy Equation (SFEE),For a unit mass, we have

If we consider the flow of a fluid through a pipe of cross-sectional area ‘A’, having specific volume, ‘v’ and the fluid velocity ‘C’ then mass flow rate is given as follows,

mass flow rate, = Volume flow rate × density

= (Area × velocity) × density

= A × C × ρ

or, = ρAC = (kg/sec)

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In differential form, SFEE can be written as:

When more than one stream of fluid enters the control volume, then

Mechanical work in a Steady flow process

The steady flow process energy equation per unit mass is:

In differential form, it can be written as:

=

For a closed system, according to first law of thermodynamics, we have

Hence the substituting the above equation in the differential form of SFEE

If kinetic and potential energies are neglected, then

Or

Integrating, we get

The work done in a steady flow process is shown in fig-14, where the area under the curve (1-2) on the

ordinate represent the work –

Applications of Steady Flow Energy Equation

1. Nozzle & Diffuser

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Fig-14: Work done in a steady flow process

Fig-15

A nozzle (fig-15) is a passage of varying cross-section by means of which the pressure energy of the flowing fluid is converted into kinetic energy. The main use of the nozzle is to produce a jet of high velocity to derive turbine and to produce thrust. From the SFEE per unit mass, we have

Here w = 0 and q = 0. Also z1 = z2. Hence

i.e., the gain in KE during the process is equal to the decrease in enthalpy of the fluid

Diffuser: A diffuser is a passage of varying cross section by means of which the kinetic energy of the flowing fluid is changed into pressure energy. The energy equation for a steady flow can be applied and similar results as for the nozzle may be obtained.

2. Boiler

Fig-16: Boiler

A boiler is a device to generate steam from feed water by heating due to burning of fuel. The steam can be used to derive steam engine or a stem turbine. For the boiler shown in the fig-16, applying SFEE per unit mass, we have

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Since no work is done in a boiler, therefore, w = 0. also changes in potential and kinetic energies are zero. Thus

kJ/kg

or, kJ/s or kW

where = mass flow rate of steam, kg/s

3. Turbine and Compressors

Turbine:

A turbine converts the heat energy of steam or gas into useful work. In this a steam or gas is passed through the turbine and part of its energy is converted into work in the turbine. This output of the turbine runs a generator to produce electricity as shown in fig-17. The steam or gas leaves the turbine at low pressure and temperature and the changes in potential energy, kinetic energy & heat transfer are negligible. Applying SFEE per unit mass, we have

Fig-17: Steam or Gas Turbine

Since z1 = z2 and q = 0

kJ/kg

and kJ/s or kW

where = mass flow rate of steam or gas, kg/s

Compressor

The function of the compressor is to compress air or gas from low pressure to high pressure with the help of work input, which is accompanied by increase in temperature. There are two types of compressor based on the type of compression

a. Rotary or Centrifugal Compressor

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b. Reciprocating Compressor

a. Rotary or Centrifugal compressor

A rotary compressor can be regarded as a reversed turbine (fig-18). The function of rotary compressor is to compressor is to compress the air or gas with the help of a rotating element called rotor followed by a stationary element called stator. This increases the energy level of the working fluid.

Fig-18: Centrifugal compressor

Applying SFEE per unit mass, we have

Since z1 = z2

The ‘q’ is taken as negative since heat is lost from the system and ‘w’ is taken as negative since work is supplied to the system.

or, kJ/kg

or, kJ/s or kW

where = mass flow rate of air, kg/s

b. Reciprocating Compressor

The reciprocating compressor (fig-19) draws in air from atmosphere and supplies at considerable higher pressure in small quantities (compared with centrifugal compressor). The reciprocating compressor can be considered as steady flow system provided the control volume includes the receiver which reduces the fluctuations of the flow considerably.

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Fig-19: Reciprocating Compressor

Applying SFEE per unit mass, we have

ΔPE = 0 and ΔKE = 0 since these changes are negligible compared with other energies The ‘q’ is taken as negative since heat is lost from the system and ‘w’ is taken as negative since

work is supplied to the system.

kJ/kg

or, kJ/s or kW

where = mass flow rate of air, kg/s

4. Heat Exchanger:

A heat exchanger is a device in which heat is transferred from one fluid to another. It is used to add or reduced heat energy of the fluid flowing through the device. Radiator in an automobile, condenser in a steam power and refrigeration plants, evaporator in a refrigerator are examples of heat exchangers. There will be no work interaction during the flow of the fluid through any heat exchanger.

a. Steam Condenser

The condenser is used to condense the steam in case of steam power plant and condense the refrigerant vapour in the refrigeration system using water or air as cooling medium. In the condenser shown in the fig-20, there is negligible change in potential and kinetic energies. The ‘q’ is taken as negative since heat is lost from the steam (system) and the work done is zero. Therefore, z1 = z2, C1 = C2, and w = 0.

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Fig-20: Condenser

Hence, applying steady flow energy equation we have

or, kJ/kg

or kJ/s or kW

where = mass flow rate of steam or gas, kg/s h1 & h2 = enthalpy of steam at entry & exit

Assuming there is no other interaction except the heat transfer between steam and water, then Q = heat gained by water passing through condenser

=

Substituting the value of Q in the above equation, we have

=

Where = mass flow rate of the cooling water cw = specific heat of water hw1 & hw2 = enthalpy of water at entry & exit

= temperature of the water at entry & exit

b. Evaporator

An evaporator is a device used in refrigeration system to take away heat from the refrigerator to maintain low temperature. In the evaporator shown in fig-21, there is no change in potential and kinetic energies. The work done is also zero. Therefore, z1 = z2, C1 = C2, and w = 0.

Fig-20: Evaporator

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Hence, applying steady flow energy equation we have

or, kJ/kg

or, kJ/s or kW

where = mass flow rate of refrigerant, kg/s h1 & h2 = enthalpy of refrigerant at entry & exit

5. Throttling Process

Throttling process is an irreversible process in which a fluid flows across a restriction in such a manner that there is drop of pressure without any change in enthalpy of the fluid, kinetic energy, potential energy, and there is no work or heat transfer. Such a process occurs in flow through porous plug, a partially closed valve or a small orifice. (Fig-21).

F-g-21: Throttling Process

Applying SFEE per unit mass, we have

Since, q = 0; w = 0; z1 = z2, & C1 C2

0 – 0 = h2 – h1 + 0 + 0

i.e., h1 = h2

In a throttling process, the enthalpy remains constant.

This process has wide application in engineering practice, on typical application of the throttling process is when fluid pass in a pipe line through a partially opened valve.

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