CHAPTER 2 ENERGY INTERACTION (HEAT AND WORK)

Energy can cross the boundary of a closed system in two ways: Heat and Work. WORK The work is done by a force as it acts upon a body moving in direction of force. The magnitude of force and the distance moved parallel to force is known as the magnitude of mechanical work. In thermodynamics the work transfer is considered to be occurring between the system and surroundings. Work is said to be done by system if the entire effect on the things external to the system can be equated to raising of weight (weights may not be actually raised but net effect external to the system would be raising of a weight). Let us consider a battery and the motor as shown in Figure 1 as a system. The motor is driving a fan. When a fan is replaced by a pulley and weight, the weight is raised by pulley driven by motor. In thermodynamics it is said that system is doing work as the sole effect external to the system can be equated to raising of a weight.

Battery

Motor

BOUNDARY

Fan

+ -

Battery

Motor

+ -

W

Figure 1

Suppose the fan driven by the motor battery arrangement becomes the part of the system itself as shown in Figure 2, no energy transfer takes place across the boundary of system and surrounding, therefore, no work is done by the system.

Battery

Motor

BOUNDARY

Fan

+ -

Figure 2 No work transfer takes place

Sign convention

Work done by system is arbitrarily taken to be positive, and when work is done on the system, it is taken to be negative.

P

ds

dA = PdV

2

1

dV

P

P

V

System System

W

Surroundings Surroundings

(a) W is positive (b) W is negative

W

Figure 3 Types of work

(1) Closed system work (non flow work)

(2) Open system work (flow work)

In isolated systems no energy is transferred. Therefore, no work is obtained. Closed system work Consider gas enclosed in a piston cylinder device as shown in Figure 4. The initial pressure of gas is ‘P’, the total volume is ‘V’, and the cross sectional area of piston is A. If a piston is allowed to move a distance ‘ds’ in a quasi static manner, the differential work done during this process is

dW Fds PAds PdV

Figure 4

Let a quasi static process occurs from 1 2 as shown in Figure 5

Figure 5 On P-V diagram, differential area dA is equal to PdV which is the differential work.

Work of process 1 – 2 = Area under curve 1 – 2 when projected on volume axis. 2 2

1-21 1

W dW= PdV

Closed system work = Area under P-V curve projected on volume axis

P

V

1

2

a

b

P1

P2

V1 V2

Conditions for applying W PdV

(1) The system must be closed system (2) The work should cross the boundary (3) Process must be reversible process (quasi static process)

PATH FUNCTION AND POINT FUNCTION Consider two different process (different paths) 1 – a – 2 and 1 – b – 2 with same end points as shown in Figure 6.

Figure 6

1-a-2W = Area under 1 – a – 2 on volume axis

1-b-2W = Area under 1 – b – 2 on volume axis

Since the area under each curve represents the work for each process, and is dependent on the path of system, the amount of work done is not a function of end states of the process and is depends on the path of system going from state 1 to state 2. For this reason work is

called path function and dW is an inexact differential. 2

1-2 2 11

dW=W W -W

Work is path function and thus inexact differential.

For a given state there is a definite value of property. The change in thermodynamic property of a system in any process is independent of the path of process during change of state and depends upon the initial and final state of the system. For example change in volume in process 1-a-2 and 1-b-2 is given by

12

V

VVVdV

2

1

Properties are state function or point function and does not depend upon path traced or in other words we can say that properties are exact differential.

To distinguish an inexact differential dW from an exact differential dV or dP, the differential line is being cut by a line at its top. dW PdV

1dV dW

P

1/P is called integrating factor. Inexact differential dW when multiplied by integrating factor 1/P becomes a exact differential dV.

For a cyclic process, the initial and final states of the system are the same and hence change in property is always zero.

dV 0, dP 0, dT 0

WORK DONE IN VARIOUS PROCESSES

Isochoric or Constant Volume or Isomeric Process

W PdV

Since change in volume is zero for constant volume process W 0

mRTPV TP Figure 7

Therefore for a process 1-2 as shown in figure 7

1

2

1

2

T

T

P

P

Isobaric or Constant Pressure or Isopiestic process PV = mRT Since pressure is constant V T For a process 1-2 as shown in figure 8

1

2

1

2

T

T

V

V

Figure 8

2

1

2

1

V

V

V

V

2 1

W PdV

W P dV

W P(V V )

Constant Temperature or Isothermal process PV = mRT Since T is constant and mass does not change in closed system PV Constant

Therefore for a process 1-2 as shown in figure 1 1 2 2PV =P V

V

CP

2

1

V

VW PdV

2

1

V

V

CW dV

V

2 1W C ln(V V ) Figure 9

2 11 1 1 1

1 2

V PW PV ln PV ln

V P

P 1

2

V

P

V

1 2

P

V

2

1

2 12 2 2 2

1 2

V PW P V ln P V ln

V P

2

1

VW mRTln

V

Expansion Compression

P

V

2

1

Figure 10

2

1

VW mRTln

V

2 1Since V V

W Positive Therefore, work is said to be done by

system in expansion process.

P

V

2

1

Figure 11

1

2

VW mRTln

V

2 1Since V V

W Negative

Therefore, work is said to be done on the system in compression process.

Adiabatic process A process is said to be an adiabatic process if there is no heat transfer from the system or to the system. For reversible adiabatic process

PV C

Where = adiabatic index ( always greater than 1)

= 1.4 for air

P V P V1 1 2 2W

1 2 1

Figure 10

Polytropic process

nPV C > n > 1

Where n is called polytropic index P V P V1 1 2 2W

1 2 n 1

General equation for all the processes is ( kPV C )

Constant pressure process kPV C

P

V

2

1

When k = 0 P = C

Constant volume process kPV C or k 1/ k 1/ k(PV ) C

1/ kP V C When k V = C

Isothermal process

kPV C When k = 1 PV = C Since PV = mRT

T C

Adiabatic process kPV C

When k =

PV C

Polytropic kPV C

When k = n nPV C

Figure 11 Slope of Isothermal process

The equation of isothermal process PV C represents the equation of a rectangular hyperbola. Differentiating the above equation

0VdPPdV VdPPdV

dP P

dV V

Where dP

dV

represents slope of isothermal process on P-V diagram.

Slope of Adiabatic process

PV C

dP P

dV V

Therefore, Slope of adiabatic curve = (Slope of Isothermal curve)

K = 0K = 1K = n

K =

K = 1

K = nK =

V

P

K = 8

Ideal gas equation for various processes (PV = mRT) Constant volume

1

2

1

2

T

T

P

P

Constant Pressure

1

2

1

2

T

T

V

V

Constant Temperature

2211 VPVP

Adiabatic

P V P V C1 1 2 2

or

PV C

mRTV C

V

1TV C

For a adiabatic process 1-2

1 1T V T V1 1 2 2

1T V

2 1T V1 2

1

T P2 2

T P1 1

1

V P γ1 2=V P

2 1

Polytropic process

n nP V P V1 1 2 2

n 1T V

2 1T V1 2

n 1

T P n2 2T P1 1

OPEN SYSTEM WORK Open system work is equal to area under P-V curve on pressure axis.

2

P

V

1

dA = VdPdP

V Figure 11

W VdP

(1) Isochoric process

2 1W V(P P )

1 2W (P P )V

Figure 12

(2) Constant pressure process

W VdP

Since dP = 0 W 0

Figure 13

3) Isothermal process PV=C Differentiating the above equation

0VdPPdV VdPPdV

VdPPdV

Figure 14

P

V

1

2

P

V

1 2

P

V

2

1

For Isothermal process open system work is equal to closed system work because rectangular hyperbola when projected on pressure axis and volume axis will give same area.

V V

2 2W mRTln C lnV V

1 1

V P V P

2 1 2 1W = P V ln = P V ln = P V ln = P V ln1 1 1 1 2 2 2 2V P V P

1 2 1 2

(4) Adiabatic

(P V P V )

1 1 2 2W1

(5) Polytropic

n(P V -P V )

1 1 2 2W = n-1

HEAT

Heat is defined as the form of energy transfer between two systems or a system and surrounding by virtue of to temperature difference. Heat is energy in transition (like work) and is recognized only as it crosses the boundary of system. Thus in thermodynamics the term heat simply means heat transfer.

Q m c T

Where Q is the amount of heat transfer to a body of mass m having specific heat c due to temperature difference T .

Specific heat (kJ/kg K) – Amount of heat required to raise the temperature of unit mass of a substance through unit degree temperature difference. Constant pressure process

When heat is added to the system as shown in Figure 15, the gas expands raising the piston and maintaining constant pressure inside the system.

p p f iQ mc T mc (T T )

where cp is the specific heat at constant pressure.

Figure 15

mti tf

HEAT

Constant volume process

Consider a process of heat transfer as shown in Figure 16. Stoppers do not allow the piston to move upwards (Expansion is restricted). Thus process can be called as constant volume heat addition.

v f iQ = m c T T where cv is the specific heat at constant volume.

Figure 16 In constant pressure heating, more heat is required as compared to constant volume process for equal rise in temperature of a given mass of gas. This is attributed to the fact that constant pressure heating involves expansion (cooling effect).

pc always greater than vc because pc includes internal energy and external work

(expansion) where as vc includes only internal energy.

p airc 1.005 kJ/kgK

v airc = 0 .717 kJ/kgK

(R)air = pc - vc = 0.287 kJ/kgK

mti tf

HEAT

QUESTION BANK

Q1 A mass of gas is compressed in a quasi-static process from 80 KPa, 0.1m3 to 0.4 MPa, 0.03 m3. Assuming that the pressure and volume are related by pvn = constant, find the work done by the gas system.

(a) – 11.76 kJ

(b) – 8.73 kJ

(c) – 14.69 kJ

(d) – 16.52 kJ Solution Let subscript 1 denotes the initial condition and subscript 2 denotes the final condition

n n1 1 2 2

e

1 1 2 2

21 2

1

1 2

2 1

P V = P V

Taking log on both sides

ln P + n ln V = ln P + n ln V

P n ln V ln V = ln

P

V P n ln = ln

V P

0.1 400 n ln = ln

0.03 80

n = 1.3367 1.34

1 1 2 2P V P V Work done W =

n 1

80 0.1 400 0.03W = = 11.764 kJ

1.34 1

Q2 Determine the total work done by a gas system following an expansion process A-

B-C as shown in Figure.

(a) 0.25 MJ

(b) 1.00 MJ

(c) 2.25 MJ

(d) 2.30 MJ

Pv =C1.3

50

0.2 0.4V(m3)

0.8

C

A B

P(bar)

Solution

5 6Area under AB = 0.4 0.2 50 10 J = 10 W = 1 MJ

1.3 1.3B B C C

C

B B C C

5 5

P V P V

P 20.13 bar

P V P VArea under BC =

n 1

50 10 0.4 20.31 10 0.8Area under BC = 1.25MJ

1.3 1

Total work = Total area under the curve ABC = 1+1.25 = 2.25 MJ

Common data for questions 3 and 4

A rigid tank of 0.568 m3 volume contains air at 6.895 bar and 21.1 0C . The tank is equipped with a relief valve that opens at a pressure of 8.618 bars and remains open until the pressure drops to 8.274 bars. Assume that the temperature of the air remains constant during discharge and air in the tank behaves as ideal gas. Assume gas constant = 294.63 Nm/kg K for air.

If a fire causes the value to operate once as described. Q3 What is the temperature just before the value opens?

(a) 328 K

(b) 348 K

(c) 368 K

(d) 388 K Q4 What is the mass of air lost due to fire?

(a) 0.08 kg

(b) 0.18 kg

(c) 0.28 kg

(d) 0.38 kg Solution 3

CASE 1 BEFORE FIRE 5 2

1

31

1

51 1

11

Pressure, P 6.895 bar 6.895 10 N/m

Volume, V 0.566m

Temperature, T 21.2 273 2941.1 K

P V 6.895 10 0.566mass, m 4.5 kg

RT 294.63 294.1

CASE 2 JUST BEFORE THE VALVE OPENS

5 22

32 1

2 1

2 2 2 2

52 2

22

Pressure, P 8.618 bar 8.618 10 N/m

V V 0.566 m

m m

P V m RT

P V 8.618 10 0.566T 367.9 K

m R 4.5 294.63

Solution 4 5 2

3Pressure, P 8.274 bar 8.274 10 N/m

33 2 1

3 2

Volume, V V V 0.566 m

Temperature, T T 367.9 K

53 3

33

1 3

p V 8.274 10 0.566mass, m 4.32 K

RT 294.63 367.9

Hence, mass of air lost due to fire m m 4.5 4.32 0.18 kg

Q5 At the beginning of the compression stroke of a two-cylinder internal combustion engine the air is at a pressure of 101.325 kPa. Compression reduces the volume to 1/5 of its original volume, and the law of compression is given by pv1.2 = Constant. If the bore and stroke of each cylinder is 0.15 m and 0.25 m, respectively determine the power absorbed in kW by compression strokes when the engine speed is such that each cylinder undergoes 500 compression strokes per minute.

Solution

22

3 31

0.15dInitial Volume V = L 0.25 m 0.00442 m

4 4

1Initial pressure P = 101.325 kPa.

312

1.2 1.21 21 2

1.211

2 1.22

1 1 2 2

VFinal volume V = = 0.000884 m

5

P V = P V

P VOr P = 699.41 700 kPa

V

1.2Work done/unit stroke/unit cylinder (W) = P V P V

1.2 1

101.325 0.00442 700 0.000884W 1.

1.2 1

2 1.0256 kJ

ve work means work is done on the system

W 500 2Power = kW

60

17.09 kW

Q6 A closed system undergoes 3 process as shown in Figure

1 - 2 Isobaric 2 – 3 Polytropic( n = 1.4) 3 – 1 Isothermal process

Find

(a) 2V

(b) Net work Comment on the result.

Solution (a)

3 3 1 1

33

Process 3 1 Isothermal process

P V P V

4 1 V 4 m

1

Process 2 3 Polytropic process

P = P = 4 bar1 2

P = 13 bar

1 2

3

V 2 V 3V 1= 1 m3

n

1.4 1.42 32 3

1.4 1.42

32

PV = C

P V = P V

4 V 1 1

V = 1.485 m

(b)

12 23 31

1 2 2 1

2 2 3 323

131 3 3

3

Net work = W + W + W

W = P V V 400 1.486 1 194.4 kJ work done by system

P V P V 400 1.486 100 4 W 486 kJ work done by system

n 1 1.4 1

VW = P V ln 554.5 kJ work don

V

net 12 23 31

e on system

W = W + W + W 125.8 kJ

A clockwise cycle on P – V diagram is a work producing cycle and anticlockwise cycle is a work consuming cycle. Net work is the area of closed region for a cycle.

Q7 An ideal gas is heated at constant volume until its temperature is 3 times the original temperature, then it is expanded isothermally till it reaches its original pressure. The gas is then cooled at constant pressure till it restored to its original state. Determine the net work done per kg of gas if initial temperature is 350 K.

[IES 2003:10 marks]

Solution

1-2Constant volume heating

2-3 Isothermal expansion

3-1Constant pressure cooling

Given that 1T 350 K

Process 1-2

1

2

3

V

P

2 1

2 2

1 1

2

1 2

1

T 3T 3 350 =1050 K

For a constant volume process

P T3

P T

w Pdv 0 (dv = 0)

Process 2-3

2 2 3 3

32 11 3

3 2 2

32 3 2

2

P V P V

PV P 1 (P P )

V P P 3

Vw RT ln .287 1050 ln(3) 331.1 kJ/kg

V

Process 3-1

1

3 1 1 3 1 3

3

net 1 2 2 3 3 4

w Pdv P v v R T T 0.287 350 1050 200.9 kJ / kg

w w w w 130.2 kJ/kg

GATE QUESTIONS

Q1 A 2 kW, 40 litres water heater is switched on for 20 minutes. The specific heat cp

for water is 4.2 kJ/kgK. Assuming all the electrical energy has gone into heating

the water, increase of the water temperature in degree centigrade is

(a) 2.7 (b) 4.0 (c) 14.3 (d) 25.25

[2003: 1 Mark]

Solution (c)

Heat Supplied Power time 2 kW 20 60 sec 2400 KJ

p

0

Heat Supplied mc T

2400 40 4.2 T (1 liter 1kg)

T 14.3 C

Q3 A frictionless piston cylinder device contains a gas initially at 0.8 MPa and 0.015

m3. It expands quasi-statically at constant temperature to a final volume of 0.03 m3. The work output (in KJ) during this process will be (a) 8.32 (b) 12 (c) 554.67 (d) 8320

[2009: 1 marks] Solution (a)

1

31

32

P = 0.8 MPa

V = 0.015 m

V = 0.03 m

For isothermal process

21 1

1

VWork output, W = P V ln

V

3 2 3 0.03W 0.8 10 kN/m 0.015 m ln = 8.32 kN.m = 8.32 kJ

0.015

Q4 A compressor undergoes a reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2 respectively. Potential and kinetic energy changes are to be ignored. The following notations are used:

v = specific volume and P = pressure of the gas. The specific work required to be supplied to the compressor for this gas compression process is

V

P(P ,V )1 1

(P ,V )2 2

(a)

2

1

Pdv (b)

2

1

vdP (c) 1 2 1v (P P ) (d) 2 1 2P (v v )

[2009: 1 Mark] Solution (b)

c

Open system work vdp

Since work is done on compressor

w vdp

Q5 Heat and work are

(a) Intensive properties

(b) Extensive properties

(c) Point functions

(d) Path functions

[2011: 1 Mark] Solution (d)

Q6 A cylinder contains 5 m3 of an ideal gas at a pressure of 1 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 5 bar. The work in KJ required for this process is (a) 804.7 (b) 953.2 (c) 981.7 (d) 1012.2

[2012: 1 Mark] Solution (a)

31

1

2

2isothermal 1 1

1

V 5 m

P 1 bar 100 kPa

P 5 bar 500 kPa

P 5W P V ln 100 5 ln 804.7 kJ

P 1

Q7 A reversible thermodynamic cycle containing only three processes and producing

work is to be constructed. The constraints are: (1) There must be one isothermal process

(2) There must be one isentropic process

(3) The maximum and minimum cycle pressures and the clearance volume are

fixed

(4) Polytropic process are not allowed

Then the number of possible cycles are:

(a) 1

(b) 2

(c) 3

(d) 4

[2005: 2 Marks] Solution (c)

1-2 adiabatic

2-3 Isothermal

3-1 Constant volume/Pressure process

4 Such work producing cycles are possible with clockwise direction. Q8 Match items from groups 1, 2, 3, 4 and 5

Group 1 Group 2 Group 3 Group 4 Group 5

When added to

the system, is

Differential Function Phenomenon

E – Heat G Positive I Exact K Path M Transient

F Work H Negative J Inexact L Point N Boundary

(a) F – G – J – K – M F – H – I – K – N

(b) E – G – I – K – M E – G – I – K – M

(c) F – H – J – L – N

E – H – I – L – M

(d) E – G – J – K – N F – H – J – K – M

[2006: 2 Marks] Solution (d) Q9 In a steady state steady flow process taking place in a device with a single inlet and

a single outlet, the work done per unit mass flow rate is given by outlet

inlet

w vdp,

Where v is the specific volume and p is the pressure, The expression for w given above

V

P1

2

3

1

2

3

3

2

1 2

1

3

Pmax

Pmin

(a) Is valid only if the process is both reversible and adiabatic

(b) Is valid only if the process is both reversible and isothermal

(c) Is valid for any reversible process.

(d) Is incorrect; it must be

[2008: 2 Marks]

Solution (c)

outlet

inlet

The expression

w vdp is valid for any reversible process. It does not depend whether

the process is isothermal or adiabatic.

w pdv work is for a reversible process for a closed system.

Q10 A gas expands in a frictionless piston cylinder arrangement. The expansion process

is very slow, and is resisted by an ambient pressure of 100 KPa. During the expansion process, the pressure of the system (gas) remains constant at 300 KPa. The change in volume of gas is 0.01 m3. What is the maximum amount of work that could be utilized from the above process? (a) 1 kJ (b) 2kJ (c) 3kJ (d) 4kJ

[2008: 2 Marks] Solution (b)

A gas expands due to pressure difference. The net pressure difference responsible

for expansion is 300 100 = 200 kPa

For a constant pressure process W P V

W 200 0.01 = 2 kJ

Linked Answer Questions 11 and 12 A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 150C. When the game started next day, the air temperature at the stadium was 50C. Assume that the volume of the football remains constant at 2500 cm3.

Q11 The amount of heat lost by the air in the football and the gauge pressure of air in

the football at the stadium respectively equal (a) 30.6 J, 1.94 bar

outlet

inlet

w pdv

(b) 21.8 J, 0.93 bar (c) 61.1 J, 1.94 bar (d) 43.7 J, 0.93 bar

[2006: 2 Marks] Solution (d)

1 gauge

1 absolute 1 atmgauge

01

02

3 6 3

1 3absolute

1

Given

P 1 bar

(P ) P P 1 1.013 2.013 bar 201.3 kPa

T 15 C 288 K

T 5 C 278 K

V 2500 cm 2500 10 m

P V 201.3 2500Mass of air in football 6.045 10 kg

RT 0.287 288

Heat loss

3v 1 2to surrounding Q mc T T 6.045 10 0.718 10

Q 0.0473 kJ

2 2

1 1

2

gauge absolute atm

For a constant volume process

T P

T P

278P 2.013 1.93 bar (absolute)

288

P P P 0.93 bar

Q12 Gauge pressure of air to which the ball must have been originally inflated so that it

would be equal 1 bar gauge at the stadium is (a) 2.23 bar (b) 1.94 bar (c) 1.07 bar (d) 1.00 bar

[2006: 2 Marks] Solution (c)

2 gauge

2 absolute atm 2 gauge

1 absolute 1

2 2absolute

1 absolute

2 2 absolute atmgauge

P 1 bar

(P ) P P 2.013 bar

P T

P T

P 2.08 bar

P (P ) P 1.07 bar