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*P K Nag Exercise problems - Solved
*

Thermodynamics

*Contents
**Chapter-1: Introduction
*

*Chapter-2: Temperature
*

*Chapter-3: Work and Heat Transfer
*

*Chapter-4: First Law of Thermodynamics
*

*Chapter-5: First Law Applied to Flow Process
*

*Chapter-6: Second Law of Thermodynamics
*

*Chapter-7: Entropy
*

*Chapter-8: Availability & Irreversibility
*

*Chapter-9: Properties of Pure Substances
*

*Chapter-10: Properties of Gases and Gas Mixture
*

*Chapter-11: Thermodynamic Relations
*

*Chapter-12: Vapour Power Cycles
*

*Chapter-13: Gas Power Cycles
*

*Chapter-14: Refrigeration Cycles
*

*Solved by
*

*Er. S K Mondal
**IES Officer (Railway), GATE topper, NTPC ET-2003 batch, 12 years teaching
experienced, Author of Hydro Power Familiarization (NTPC Ltd) *

*Benefits of solving Exercise (unsolved) problems of P K Nag
*

• * The best ways is to study thermodynamics is through problems, you must
know how to apply theoretical concepts through problems and to do so you
*

*must solve these problems*

• *It contains Expected Questions of IES, IAS, IFS and GATE examinations*

• *It will enable the candidates to understand thermodynamics properly*

• *It will clear all your doubts*

• *There will be no fear of thermodynamics after solving these problems*

• *Candidate will be in a comfortable position to appear for various competitive
examinations*

• *Thermodynamics- “the Backbone of Mechanical Engineering” therefore
Mastering Thermodynamics is most important many of the subjects which
*

*come in later like Heat and Mass Transfer, Refrigeration and Air
*

*Conditioning, Internal Combustion Engine will require fundamental
*

*knowledge of Thermodynamics*

*Every effort has been made to see that there are no errors (typographical or otherwise) in the
*

*material presented. However, it is still possible that there are a few errors (serious or
*

*otherwise). I would be thankful to the readers if they are brought to my attention at the
*

*following e-mail address: [email protected]
*

*S K Mondal
*

Page 2 of 265

** Introduction****By: S K Mondal****Chapter 1
**

** 1. ** Introduction
Some Important Notes

Microscopic thermodynamics or statistical thermodynamics Macroscopic thermodynamics or classical thermodynamics A quasi-static process is also called a reversible process

**Intensive and Extensive Properties
Intensive property : **Whose value is independent of the size or extent i.e. mass of the system.
e.g., pressure

*p*and temperature

*T.*

**Extensive property:**Whose value depends on the size or extent i.e. mass of the system (upper case letters as the symbols). e.g., Volume, Mass (V, M). If mass is increased, the value of extensive property also increases. e.g., volume

*V,*internal energy

*U,*enthalpy

*H,*entropy S, etc.

**Specific property:**It is a special case of an intensive property. It is the value of an extensive property per unit mass of system. (Lower case letters as symbols) e.g: specific volume, density (

*v*, ρ).

**Concept of Continuum**The concept of continuum is a kind of idealization of the continuous description of matter where the properties of the matter are considered as continuous functions of space variables. Although any matter is composed of several molecules, the concept of continuum assumes a continuous distribution of mass within the matter or system with no empty space, instead of the actual conglomeration of separate molecules. Describing a fluid flow quantitatively makes it necessary to assume that flow variables (pressure, velocity etc.) and fluid properties vary continuously from one point to another. Mathematical descriptions of flow on this basis have proved to be reliable and treatment of fluid medium as a continuum has firmly become established. For example density at a point is normally defined as

0
lim *m*ρ

∀→

⎛ ⎞= ⎜ ⎟∀⎝ ⎠

Here ∀ is the volume of the fluid element and m is the mass
If ∀ is very large ρ is affected by the in-homogeneities in the fluid medium. Considering
another extreme if ∀ is very small, random movement of atoms (or molecules) would change
their number at different times. In the continuum approximation point density is defined at the
smallest magnitude of ∀ , before statistical fluctuations become significant. This is called
continuum limit and is denoted by *C*∀ .

lim
*C
*

*m*ρ
∀→ ∀

⎛ ⎞= ⎜ ⎟∀⎝ ⎠

Page 3 of 265

** Introduction****By: S K Mondal****Chapter 1
**One of the factors considered important in determining the validity of continuum model is
molecular density. It is the distance between the molecules which is characterized by mean free
path (λ). It is calculated by finding statistical average distance the molecules travel between
two successive collisions. If the mean free path is very small as compared with some
characteristic length in the flow domain (i.e., the molecular density is very high) then the gas
can be treated as a continuous medium. If the mean free path is large in comparison to some
characteristic length, the gas cannot be considered continuous and it should be analyzed by the
molecular theory.
A dimensionless parameter known as Knudsen number, Kn = λ / L, where λ is the mean free
path and L is the characteristic length. It describes the degree of departure from continuum.
Usually when Kn> 0.01, the concept of continuum does not hold good.
In this, Kn is always less than 0.01 and it is usual to say that the fluid is a continuum.
Other factor which checks the validity of continuum is the elapsed time between collisions. The
time should be small enough so that the random statistical description of molecular activity
holds good.
In continuum approach, fluid properties such as density, viscosity, thermal conductivity,
temperature, etc. can be expressed as continuous functions of space and time.

**The Scale of Pressure **

Absolute Pressure

Gauge Pressure

Vacuum Pressure

Local atmospheric Pressure

Absolute Pressure

Absolute Zero (complete vacuum)

**At sea-level, the international standard atmosphere has been chosen as****Patm = 101.325 kN/m2
**

Page 4 of 265

** Introduction****By: S K Mondal****Chapter 1
Some special units for Thermodynamics
**

kPa m /kg 3

**Note:** Physicists use below units

Universal gas constant, Ru= 8.314 kJ/kmole − K

Characteristic gas constant, *c u
R
*

*R
M
*

=

For Air R = 8.314 29

= kJ/kmole-K kg/kmole

= 0.287 kJ/kg- K

For water R = 8.314 18

kJ/kmole-K kg/kmole

= 0.461 kJ/kg -K

Units of heat and work is kJ

Units of pressure is kPa

1 atm = 101.325 kPa

1 bar = 100 kPa

1 MPa =1000 kPa.

Page 5 of 265

** Introduction****By: S K Mondal****Chapter 1
**

Questions with Solution P. K. Nag
**Q1.1 A pump discharges a liquid into a drum at the rate of 0.032 m3/s. The
**

**drum, 1.50 m in diameter and 4.20 m in length, can hold 3000 kg of the
liquid. Find the density of the liquid and the mass flow rate of the liquid
handled by the pump.
**

(**Ans. **12.934 kg/s)

**Solution:
**2dVolume of drum = h

4 π

×

2 3

3

3 3

.50= m 4

7.422 m mass 3000 kg kgdensity 404.203m mVolume 7.422

mass flow rate Vloume flow rate density kg= 0.032 404.203 s

kg12.9345 s

π ×1 × 4.2

= = =

= ×

×

=

**=
**

**Q1.2 The acceleration of gravity is given as a function of elevation above sea
**

**level by
**6g = 980.6 – 3.086 × 10 H−

** Where g is in cm/s2 and H is in cm. If an aeroplane weighs 90,000 N at
sea level, what is the gravity force upon it at 10,000 m elevation? What is
the percentage difference from the sea-level weight?
**

(**Ans. **89,716.4 N, 0.315%)
**Solution:** 6g´ 980.6 3.086 10 10,000 100−= − × × ×
2 2cm m977.514 9.77514s s= =

sea 90,000W 90,000 N kgf 9.806

= =

9178.054 kgf= eteW 9178.054 9.77514 N= × 89716.765 N=

( )

90,000 89716.765% less 100% 90,000

0.3147% less

− = ×

=

**
Q1.3 Prove that the weight of a body at an elevation H above sea-level is given
**

**by
2
**

**0 2
mg dW
g d H
**

⎛ ⎞= ⎜ ⎟+⎝ ⎠

** Where d is the diameter of the earth.
Solution:** According to Newton’s law of gravity it we place a man of m at an height of H

then

Page 6 of 265

** Introduction****By: S K Mondal****Chapter 1
**

Force of attraction = ( )2

GMm d H2 +

…

(i) If we place it in a surface of earth

then

( )

( )

o2

o 2

GMmForce of attraction mg d

2

GMor g d

2

= =

=

**d
**

**H
**

**1.3
**

m

( ) ( )

( ) ( )

( )

2

2

o

2

2

o

GMmWeight W d H2

dmg 2 from equation... i d H2

dmg Pr oved. d 2H

∴ = +

= +

⎛ ⎞= ⎜ ⎟+⎝ ⎠

**Q1.4 The first artificial earth satellite is reported to have encircled the earth
**

**at a speed of 28,840 km/h and its maximum height above the earth’s
surface was stated to be 916 km. Taking the mean diameter of the earth
to be 12,680 km, and assuming the orbit to be circular, evaluate the value
of the gravitational acceleration at this height.
**

** The mass of the satellite is reported to have been 86 kg at sea-level.
Estimate the gravitational force acting on the satellite at the operational
altitude.
**

(**Ans. **8.9 m/s2; 765 N)
**Solution: **Their force of attraction = centrifugal force

2mvCentirfugal force

r =

2

3 3

28840 100086 60 60 N

12680 10 916 10 2

760.65 N (Weight)

×⎛ ⎞× ⎜ ⎟×⎝ ⎠= ⎛ ⎞×

+ ×⎜ ⎟ ⎝ ⎠

=

**
Q1.5 Convert the following readings of pressure to kPa, assuming that the
**

**barometer reads 760 mmHg:
**(a) 90 cmHg gauge (b) 40 cmHg vacuum
(c) 1.2 m H2O gauge (d) 3.1 bar
**
Solution: **760 mm Hg = 0.760 × 13600 × 9.81 Pa
= 10139.16 Pa
101.4 kPa Page 7 of 265

** Introduction****By: S K Mondal****Chapter 1
**
(a) 90 cm Hg gauge
= 0.90 × 13600 × 9.81 × 10-3 + 101.4 kPa
= 221.4744 kPa
(b) 40 cm Hg vacuum
= (76 – 40) cm (absolute)
= 0.36 × 43.600 × 9.81 kPa
= 48.03 kPa
(c) 1.2 m H2O gauge
= 1.2 × 1000 × 9.81 × 10-3 + 101.4 kPa
= 113.172 kPa
(d) 3.1 bar = 3.1 × 100 kPa = 310 kPa
**
Q1.6 A 30 m high vertical column of a fluid of density 1878 kg/m3 exists in a
**

**place where g = 9.65 m/s2. What is the pressure at the base of the column.
**(

**Ans.**544 kPa)

**Solution: ** p = z ρg
= 30 × 1878 × 9.65 Pa
= 543.681 kPa
**
Q1.7 Assume that the pressure p and the specific volume v of the atmosphere
**

**are related according to the equation 1.4 52.3 10 pv **= ×

**, where**

*p*is in N/m2 abs and*v*is in m3/kg. The acceleration due to gravity is constant at 9.81 m/s2. What is the depth of atmosphere necessary to produce a pressure of l.0132 bar at the earth’s surface? Consider the atmosphere as a fluid column.(**Ans. **64.8 km)

Page 8 of 265

** Introduction****By: S K Mondal****Chapter 1
Solution:** dp dh g= ρ

1.4 3

1 n 1.4

n

n

1or dp dh g v

gdhor v dp

pv 2.3 10 2300

2300 2300 1or v where n p p 1.4

g dh 2300or dp p

2300or g dh dp p

= × ×

=

= × =

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

⎛ ⎞= ⎜ ⎟ ⎝ ⎠

⎛ ⎞= ⎜ ⎟ ⎝ ⎠

**h
**

**dh
p
**

**p + dp
**

**Zero line
**

**p = h g**ρ

**h
**

**dh
**

**p = h g**ρ

**H -hO
**

( ) ( ) ( )

H 101320n

n 0 0

n 1 n

2300 dpor dh g p

2300or h 101320 0 2420m 2.42 km g 1 n

−

=

⎡ ⎤= − = =⎣ ⎦−

∫ ∫

**Q1.8 The pressure of steam flowing in a pipeline is
**

**measured with a mercury manometer, shown in
Figure. Some steam condenses into water. Estimate
the steam pressure in kPa. Take the density of
mercury as 3 313.6 10 kg/m**× **, density of water as 103
kg/m3, the barometer reading as 76.1 cmHg, and g as
9.806 m/s2.
**

**Solution:**

2o Hg H O p 0.50 g 0.03 g p+ × ρ × = × ρ × +

3 3or p 0.761 13.6 10 9.806 0.5 13.6 10 9.806 0.03 1000 9.806 Pa.

167.875 kPa = × × × + × × × − × × =

**Q1.9 A vacuum gauge mounted on a condenser reads 0.66 mHg. What is the
absolute pressure in the condenser in kPa when the atmospheric
pressure is 101.3 kPa?
**

(**Ans. **13.3 kPa)
**Solution:** Absolute = atm. – vacuum
= 3 3101.3 – 0.66 × 13.6 × 10 × 9.81 × 10 kPa−
= 13.24 kPa

Page 9 of 265

Page 10 of 265

** Temperature****By: S K Mondal****Chapter 2
**

** 2. ** Temperature
Some Important Notes

Comparison of Temperature scale

** Relation:** C 0
100 0

− −

= F 32 212 32

− −

= K 273 373 273

− −

= 0 80 0 ρ −

− = −

− x 10

30 10

Questions with Solution P. K. Nag
**Q2.1 The limiting value of the ratio of the pressure of gas at the steam point and at
**

**the triple point of water when the gas is kept at constant volume is found to be
1.36605. What is the ideal gas temperature of the steam point?
**

(**Ans. **100°C)
**Solution:
**

t

p 1.36605 p

=

( )v t

p273.16 p

273.16 1.36605 373.15 C

∴ θ = ×

= × = °

**Q2.2 In a constant volume gas thermometer the following pairs of pressure
**

**readings were taken at the boiling point of water and the boiling point
of sulphur, respectively:
**

** Water b.p. 50.0 100 200 300
Sulphur b.p. 96.4 193 387 582
**

** The numbers are the gas pressures, mm Hg, each pair being taken with
the same amount of gas in the thermometer, but the successive pairs
being taken with different amounts of gas in the thermometer. Plot the
ratio of Sb.p.:H2Ob.p. against the reading at the water boiling point, and
extrapolate the plot to zero pressure at the water boiling point. This
**

o1 0 0 C o F 2 1 2 373K 3 0 c m

o 0 C o3 2 F 2 7 3 K 1 0 c m

C F K

o 8 0

o 0

Boiling Point

Test Temperature

Freezing Point

x

Page 11 of 265

** Temperature****By: S K Mondal****Chapter 2
**

**gives the ratio of Sb.p. : H2Ob.p. On a gas thermometer operating at zero
gas pressure, i.e., an ideal gas thermometer. What is the boiling point of
sulphur on the gas scale, from your plot?
**

(**Ans. **445°C)
**Solution : **Water b.p. 50.0 100 200 300
Sulphur b.p. 96.4 193 387 582

Ratio b.p b.p

S W

= 1.928 1.93 1.935 1.940

1

2

1

2

2

T 100 C 373K T ? p 1.926 p T 373 1.926 718K 445 C

∴ = ° =

=

=

∴ = × = = °

**
1
**

**.9
26
**

**E
xt
**

**ra
po
**

**la
ti
**

**ng
**

**0 50 100 200 300**

**Q2.3 The resistance of a platinum wire is found to be 11,000 ohms at the ice point,
**

**15.247 ohms at the steam point, and 28.887 ohms at the sulphur point. Find the
constants A and B in the equation
**

**2
0 (1 ) R R At Bt**= + +

** And plot R against t in the range 0 to 660°C.
Solution:
**

**x
**

**y
**

**(3271, 1668628)
**

**R
**

**t
**

**11
**

**36.595
**

**0 660°C
**

{ }

( )

0

2 100 0

4

3

R 11.000 R R 1 A 100 B 100

or 15.247 11.000 1100A 11 10 B or 3.861 10 A 100B ... i−

= Ω

= + × + ×

= + + ×

× = +

( ) ( ) ( )

{ } ( )

( ) ( )

2

-3

7

3

3 7 2

3 7 2

2

660

28.887 11.00 445 11A 445 11B 3.6541×10 = A + 445B ... ii

equation ii i gives. B 6 10

A 3.921 10 R 11 1 3.921 10 t 6 10 t

or Y 11 1 3.921 10 t 6 10 t

or t 3271 4 37922 Y 1668628 R 36.595

−

−

− −

− −

= + × + ×

−

= − ×

= ×

= + × − ×

= + × − ×

− = − × × −

=

Page 12 of 265

** Temperature****By: S K Mondal****Chapter 2
**** Q2.4 when the reference junction of a thermocouple is kept at the ice point
**

**and the test junction is at the Celsius temperature t, and e.m.f. e of the
thermocouple is given by the equation
**

**2 at bt**ε = +

** Where a = 0.20 mV/deg, and b = - 5.0 × 10-4 mV/deg2
(a) Compute the e.m.f. when t = - l00°C, 200°C, 400°C, and 500°C, and
**

**draw graph of **ε ** against t in this range.
(b) Suppose the e.m.f. **ε

**is taken as a thermometric property and that a**

**temperature scale t* is defined by the linear equation.
**

** t* = a'**ε

**+**

*b'*** And that t* = 0 at the ice point and t* = 100 at the steam point. Find
the numerical values of a' and b' and draw a graph of **ε

**against**

*t*.*** (c) Find the values of t* when t = -100°C, 200°C, 400°C, and 500°C, and
draw a graph of t* against t. **

** (d) Compare the Celsius scale with the t* scale.
Solution: **Try please

**Q2.5 The temperature**

*t*on a thermometric scale is defined in terms of a**property K by the relation
**

*t *= *a *ln *K *+ *b
*

** Where a and b are constants.
The values of K are found to be 1.83 and 6.78 at the ice point and the
**

**steam point, the temperatures of which are assigned the numbers 0 and
100 respectively. Determine the temperature corresponding to a
reading of K equal to 2.42 on the thermometer.
**

(**Ans. **21.346°C)
**Solution: **t = a ln x + b
0 = a x ln 1.83 + b … (i)
100 = a x ln 6.78 + b … (ii)
Equation {(ii) – (i)} gives

*

6.78a ln 100 1.83

or a 76.35 b a ln 1.83

46.143 t 76.35 lnk 46.143

t 76.35 ln2.42 46.143 21.33 C

⎛ ⎞⋅ ⋅ =⎜ ⎟ ⎝ ⎠

= ∴ = − ×

= − ∴ = −

∴ = × − = °

**Q2.6 The resistance of the windings in a certain motor is found to be 80 ohms
**

**at room temperature (25°C). When operating at full load under steady
state conditions, the motor is switched off and the resistance of the
windings, immediately measured again, is found to be 93 ohms. The
windings are made of copper whose resistance at temperature t°C is
given by
**

Page 13 of 265

** Temperature****By: S K Mondal****Chapter 2
**

**t 0R = R [1 + 0.00393 t]**

** Where R0is the resistance at 0°C. Find the temperature attained by the
coil during full load.
**

(**Ans. **70.41°C)
**Solution: ** R25 = R0 [1 + 0.00393 × 25]

[ ]0 80R 72.84

1 0.00393 25 ∴ = = Ω

+ ×

{ }93 72.84 1 0.00393 t or t 70.425 C ∴ = + ×

= °

**Q2.7 A new scale N of temperature is divided in such a way that the freezing
**

**point of ice is 100°N and the boiling point is 400°N. What is the
temperature reading on this new scale when the temperature is 150°C?
At what temperature both the Celsius and the new temperature scale
reading would be the same?
**

(**Ans. **550°N, – 50°C.)
**Solution:**
150 0

100 0 − −

= 100
400 100
*N *−

−

or N = 550*o N *
let N= C for ox

then 0
100 0
*C *−

− = 100

400 100
*N *−

−

or 100

*x * = 100
300

*x *−

or x = 100

3
*x *−

or 3 x = x -100
or 2 x = -100
or x = -50*o *C

**Q2.8 A platinum wire is used as a resistance thermometer. The wire resistance was
**

**found to be 10 ohm and 16 ohm at ice point and steam point respectively, and
30 ohm at sulphur boiling point of 444.6°C. Find the resistance of the wire at
500°C, if the resistance varies with temperature by the relation.
**

**2
0 (1 ) R R t t**α β= + +

(**Ans. **31.3 ohm)
** Solution:**
2010 (1 0 0 )*R *α β= + × + ×

2
016 (1 100 100 )*R *α β= + × + ×

2
030 (1 444.6 444.6 )*R *α β= + × + ×

Solve 0*R *, &α β then
2

0(1 500 500 )*R R *α β= + × + ×

Page 14 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

** 3. ** Work and Heat Transfer
Some Important Notes

**-ive
W
**

**+ive
W
**

**-ive
Q
**

**+ive
Q
**

Our aim is to give heat to the system and gain work output from it.

So heat input → +ive (positive) Work output → +ive (positive)

− = =∫ ∫ v ff

i f i v i

W pdV pdv

d Q = du + dW

= − +∫ ∫ f f

f i i i

dQ u u dW

− − += ∫ v f

i f f i v i

Q u u pdV

Questions with Solution P. K. Nag
**Q3.1 (a) A pump forces 1 m3/min of water horizontally from an open well to a closed
**

**tank where the pressure is 0.9 MPa. Compute the work the pump must do
upon the water in an hour just to force the water into the tank against the
pressure. Sketch the system upon which the work is done before and after
the process.
**

(**Ans. **5400 kJ/h)
**(b)If the work done as above upon the water had been used solely to raise the
**

**same amount of water vertically against gravity without change of
pressure, how many meters would the water have been elevated? **

(**Ans. **91.74 m)
** (c)If the work done in (a) upon the water had been used solely to accelerate
**

**the water from zero velocity without change of pressure or elevation, what
velocity would the water have reached? If the work had been used to
accelerate the water from an initial velocity of 10 m/s, what would the final
velocity have been?
**

(**Ans. **42.4 m/s; 43.6 m/s)
**Solution:** (a) Flow rate 1m3/hr.
Pressure of inlet water = 1 atm = 0.101325 MPa
Pressure of outlet water = 0.9 MPa

Page 15 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

( ) 33 Power pv

1 m0.9 0.101325 10 kPa s60 kJ13.31 s

∴ = Δ

= − × ×

=

(b) So that pressure will be 0.9 MPa

∴ ρ =

× =

×

6

h g 0.9 MPa 0.9 10or h = m 91.743 m

1000 9.81

( )

( )

2 2 2 1

2 2 2 1

2 2 2 1

1(c) m V V pv where m v 2

1or V V p 2

por V V 2

− = Δ = ρ

ρ − = Δ

Δ − =

ρ

2 22 1 por V V 2 Δ= +

ρ

( ) 62

2

2 0.9 0.101325 10 10

1000 V 41.2 m / s.

× − × = +

=

**Q3.2 The piston of an oil engine, of area 0.0045 m2, moves downwards 75 mm,
**

**drawing in 0.00028 m3 of fresh air from the atmosphere. The pressure in the
cylinder is uniform during the process at 80 kPa, while the atmospheric
pressure is 101.325 kPa, the difference being due to the flow resistance in the
induction pipe and the inlet valve. Estimate the displacement work done by
the air finally in the cylinder.
**

**(Ans. **27 J)
**Solution : **Volume of piston stroke
= 0.0045 × 0.075m3
= 0.0003375m3
∴ ΔV = 0.0003375 m3
as pressure is constant
= 80 kPa
So work done = pΔV
= 80 × 0.0003375 kJ
= 0.027 kJ = 27 J

**Final volume = 3.375×10-4
**

**Initial volume = 0
**

3*m
*

**Q3.3 An engine cylinder has a piston of area 0.12 m3 and contains gas at a
**

**pressure of 1.5 MPa. The gas expands according to a process which is
represented by a straight line on a pressure-volume diagram. The final
pressure is 0.15 MPa. Calculate the work done by the gas on the piston if
the stroke is 0.30 m.
**

** (Ans. **29.7 kJ)
**Solution: **Initial pressure ( 1p ) = 1.5 MPa
Final volume (V1) = 0.12m2 × 0.3m

Page 16 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

= 0.036 m3 Final pressure ( 2p ) = 0.15 MPa As initial pressure too high so the volume is neglected.

Work done = Area of pV diagram

( )

( )

1 2

3

1 p p V 2 1 1.5 0.15 0.036 10 kJ 2 29.7 kJ

= + ×

= + × ×

=

**Vneg.
**

**0.36 m3
**

**p
**

**0.15 MPa
**

**1.5 M aP
**

**
Q3.4 A mass of 1.5 kg of air is compressed in a quasi-static process from 0.1
**

**MPa to 0.7 MPa for which pv = constant. The initial density of air is 1.16
kg/m3. Find the work done by the piston to compress the air.
**

(**Ans. **251.62 kJ)
**Solution:****For quasi-static process
**

Work done pdV [ given pV C= =∫ 2

1

v

1 1 1 1 2 2 v

2 1 1 1 1

1

1 1 2 1 1

2 2 1

1

31 1

1

2

dVp V p V pV p V C V

V p Vp V l n p V V

p p Vp V ln p p V

0.10.1 1.2931 ln MJ given p 0.1 MPa 0.7

m 1.5251.63 kJ V m 1.16

p 0.7 MPa

= ∴ = = =

⎛ ⎞ = ∴ =⎜ ⎟

⎝ ⎠ ⎛ ⎞

= ∴ =⎜ ⎟ ⎝ ⎠

= × × =

= = = ρ

=

∫

**Q3.5 A mass of gas is compressed in a quasi-static process from 80 kPa, 0.1 m3
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.
**

(**Ans. **–11.83 kJ)
**Solution:** Given initial pressure ( )1p = 80kPa
Initial volume ( )1V = 0.1 m3

Page 17 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

Final pressure ( )2p = 0.4 MPa = 400 kPa Final volume ( )2V = 0.03 m3 As p-V relation npV = C

[ ]

( )

n n 1 1 2 2

e

1 1 2 2

1 2 2 1

1 2

2 1

2

1

1

2

1 1

p V p V taking log both side ln p n ln V ln p n ln V or n ln V ln V ln p ln p

V por n ln ln V p

400p lnln p 1.6094480or n 1.3367 1.34 V 0.1 1.20397ln lnV 0.03

p VWork done W

∴ =

+ = +

− = −

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎛ ⎞⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠= = = ≈ ≈ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∴ = 2 2 p V

n 1 80 × 0.1 400 × 0.03 11.764 kJ

1.34 1

− −

− = = −

−

**Q3.6 A single-cylinder, double-acting, reciprocating water pump has an
indicator diagram which is a rectangle 0.075 m long and 0.05 m high. The
indicator spring constant is 147 MPa per m. The pump runs at 50 rpm.
The pump cylinder diameter is 0.15 m and the piston stroke is 0.20 m.
Find the rate in kW at which the piston does work on the water.
**

(**Ans. **43.3 kW)
**Solution:** Area of indicated diagram ( )*da * = 0.075 × 0.05 m2 = 3.75 × 310− m2
Spring constant (k) = 147 MPa/m

Page 18 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

**Q3.7 A single-cylinder, single-acting, 4 stroke engine of 0.15 m bore develops
an indicated power of 4 kW when running at 216 rpm. Calculate the area
of the indicator diagram that would be obtained with an indicator
having a spring constant of 25 × 106 N/m3 . The length of the indicator
diagram is 0.1 times the length of the stroke of the engine.
**

(**Ans. **505 mm2)
**Solution:** Given Diameter of piston (D) = 0.15 m
I.P = 4 kW = 4 × 1000 W
Speed (N) = 216 rpm
Spring constant (k) = 25 × 106 N/m
Length of indicator diagram ( )dl = 0.1 × Stoke (L)
Let Area of indicator diagram = ( )*da *

∴ Mean effective pressure ( mp ) = d d

a k l

×

∴

[ ]m

d

d

d d

p LANand I.P. as 4 stroke engine 120

a k L A Nor I.P. l 120

I.P l 120or a k L A N

=

× × × = ×

× × =

× × ×

2

2

d

2 6 2

4 2

2

Darea AI.P 0.1L 120 4 4k L D N and l 0.1L

4 0.1 120 4 1000 m 25 10 0.15 216 5.03 10 m 503mm

−

⎡ ⎤π =× × × ⎢ ⎥= ⎢ ⎥× × π × × =⎢ ⎥⎣ ⎦

× × × × =

× × π × × = ×

=

**Q3.8 A six-cylinder, 4-stroke gasoline engine is run at a speed of 2520 RPM.
**

**The area of the indicator card of one cylinder is 2.45 × 103 mm2 and its
length is 58.5 mm. The spring constant is 20 × 106 N/m3 . The bore of the
cylinders is 140 mm and the piston stroke is 150 mm. Determine the
indicated power, assuming that each cylinder contributes an equal
power.
**

(**Ans. **243.57 kW)

**Solution: **dm
d

ap k l

= ×

3 2 3 2

3 2 2.45 10 mm N mm N 120 10 Pa N / m

58.5 mm 1000m m m 837.607 kPa

L 0.150 m

× × ⎛ ⎞= × × ∴ × ⇒ = ⎜ ⎟× ⎝ ⎠ = =

Page 19 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

2 2D 0.14A

4 4 N 2520

π π × = =

=

[ ]m

2

n 6 p LANI.P. n as four stroke

120 0.14 2520 6837.607 0.15 kW 4 120

243.696 kW

=

∴ = ×

π × × = × × ×

=

**Q3.9 A closed cylinder of 0.25 m diameter is fitted with a light frictionless
**

**piston. The piston is retained in position by a catch in the cylinder wall
and the volume on one side of the piston contains air at a pressure of 750
kN/m2.The volume on the other side of the piston is evacuated. A helical
spring is mounted coaxially with the cylinder in this evacuated space to
give a force of 120 N on the piston in this position. The catch is released
and the piston travels along the cylinder until it comes to rest after a
stroke of 1.2 m. The piston is then held in its position of maximum travel
by a ratchet mechanism. The spring force increases linearly with the
piston displacement to a final value of 5 kN. Calculate the work done by
the compressed air on the piston.
**

(**Ans. **3.07 kJ)
**Solution:** Work done against spring is work done by the compressed gas

φ** 0.25m
**

**1.2m **

120 5000Mean force 2

2560 N Travel 1.2m

Work Done 2560 1.2 N.m 3.072 kJ

+ =

= =

∴ = × =

** By Integration
** At a travel (x) force (Fx) = 120 + kx
At 1.2 m then 5000 = 120 + k × 1.2
∴ Fx = 120 + 4067 x

Page 20 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

[ ]

1.2

x 0

1.2

0 1.22

0 2

W F dx

120 4067x dx

x120x 4067 2

1.2120 1.2 4067 J 2

144 2928.24 J 3072.24J 3.072 kJ

∴ =

= +

⎡ ⎤ = + ×⎢ ⎥

⎣ ⎦

= × + ×

= + = =

∫

∫

**Q 3.l0 A steam turbine drives a ship’s propeller through an 8: 1 reduction gear.
**

**The average resisting torque imposed by the water on the propeller is
750 × 103 mN and the shaft power delivered by the turbine to the
reduction gear is 15 MW. The turbine speed is 1450 rpm. Determine (a)
the torque developed by the turbine, (b) the power delivered to the
propeller shaft, and (c) the net rate of working of the reduction gear.
**

(**Ans. **(a) *T = *98.84 km N, (b) 14.235 MW, (c) 0.765 MW)

**Solution:** Power of the propeller = Power on turbine shaft

The net rate of working of the reduction gear = (15 – 14.235) MW = 0.7647 MW

**Q 3.11 A fluid, contained in a horizontal cylinder fitted with a frictionless leak
proof piston, is continuously agitated by means of a stirrer passing
through the cylinder cover. The cylinder diameter is 0.40 m. During the
stirring process lasting 10 minutes, the piston slowly moves out a
distance of 0.485 m against the atmosphere. The net work done by the
fluid during the process is 2 kJ. The speed of the electric motor driving
the stirrer is 840 rpm. Determine the torque in the shaft and the power
output of the motor.
**

(**Ans. **0.08 mN, 6.92 W)

Page 21 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

**Solution: **Change of volume = A L

π = ×

π× = ×

=

2

2 3

3

d L 4

0.4 0.485 m 4

0.061 m

As piston moves against constant atmospheric pressure then work = Δdone p V

**0.485m
**

φ** = 0.4m
M
**

101.325 0.061 kJ 6.1754 kJ

= × =

Net work done by the fluid = 2 kJ ∴ Net work done by the Motor = 4.1754 kJ There for power of the motor

34.1754 10 W

10 60 ×

= ×

6.96 W PTorque on the shaft W

=

=

6.96 60 2 840

0.0791mN

× =

π ×

=

**Q3.12 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.
**

(**Ans. **17.95 kW)

Page 22 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

**Solution:
**

( ) 2

1 dInitial volume V L 4

π = ×

( ) 2

30.15 0.25 m 4

π × = ×

30.00442 m= ( )1Initial pr essure p 101.325 kPa.=

( ) 312 VFinal volume V 0.000884 m 5

= =

1.2 1.21 1 2 2p V p V=

Or 1.2

1 1 2 1.2

2

p Vp 699.41 700 kPa V

= = ≈

( )Work done / unit stroke unit cylinder W−

[ ]1 1 2 2 1.2 p V p V

1.2 1 ⎛ ⎞= × −⎜ ⎟−⎝ ⎠

101.325 0.00442 700 0.000884 1.2

1.2 1 × − ×⎛ ⎞= ×⎜ ⎟−⎝ ⎠

( )-ive work, as work done on the system

W 500 2 1.2Power kW 60

× × × =

= 17.95 kW

**Q3.13 Determine the total work done by a gas system following an expansion
process as shown in Figure.
**

(**Ans. **2.253 MJ)

**Solution: **Area under AB
= (0.4 – 0.2) × 50 × 510 J
= 610 W 1 MJ=

Page 23 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

**A B
**

**C
**

**bar
p
**

**0.2 0.4 0.8
V m1 3
**

**50
**

**pV = c1.3
**

1 1 2 2

5 5

Area under BC p V p V

n 1 50 10 0.4 20.31 10 0.8 W

1.3 1 1.251MJ

− =

− × × − × ×

= −

=

5 B B

3 B

3 C

1.3 5 1.3 B B

C 1.3 1.3 C

5

Here p p 50 bar 50 10 Pa V 0.4m V 0.8m

p V 50 10 0.4p V 0.8

20.31 10 Pa

⎡ ⎢ = = = ×⎢ ⎢ = ⎢

=⎢ ⎢

× ×⎢ = =⎢ ⎢ ⎢ = ×⎣

Total work = 2.251MJ
**Q3.14 A system of volume V contains a mass m of gas at pressure p and
**

**temperature T. The macroscopic properties of the system obey the
following relationship:
**

**2
ap + (V b) = mRT
**

**V
**⎛ ⎞ −⎜ ⎟
⎝ ⎠

**Where a, b, and R are constants.
Obtain an expression for the displacement work done by the system
**

**during a constant-temperature expansion from volume V1 to volume V2.
Calculate the work done by a system which contains 10 kg of this gas
expanding from 1 m3 to 10 m3 at a temperature of 293 K. Use the values
**

**4 2 315.7 10 Nm , 1.07 10 m a b **−= × = ×

**, and R= 0.278 kJ/kg-K.**(

**Ans.**1742 kJ)

**Solution:** As it is constant temp-expansion then
( ) ( ) ( )2

ap V b constant mRT k as T constant V

⎛ ⎞+ − = =⎜ ⎟ ⎝ ⎠

Page 24 of 265

** Work and Heat Transfer****By: S K Mondal****Chapter 3**

( ) ( ) ( )

( )

( )

( )

1 1 2 22 2 1 2

2

1 2

2 2 1

2

2 1

2

1 2 1

2 1 12

1 1 2

a ap V b p V b k V V

constant kaW p dV p V V b

k a k adV or p V b V V b V

a 1 1kln V b dv c V V V

V b 1 1k ln a V b V V

V ba 1 1p V b ln a V V b V

⎛ ⎞ ⎛ ⎞ ∴ + − = + − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞= ∴ + =⎜ ⎟ −⎝ ⎠

⎛ ⎞= − = −⎜ ⎟− −⎝ ⎠

⎡ ⎤= − + − = +⎢ ⎥⎣ ⎦

⎛ ⎞ ⎛ ⎞− = + −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

⎛ ⎞ − = + − + −⎜ ⎟ −⎝ ⎠

∫

∫

∫

1V ⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

( ) ( ) ( )2 ap V b constant mRT k as T constant

V ⎛ ⎞+ − = =⎜ ⎟ ⎝ ⎠

Given m = 10 kg; T = 293 K; R = 0.278 kJ/kg. K ∴ Constant k = 10 × 293 × 0.278 kJ = 814.54 kJ a = 15.7 × 10 Nm4; b = 1.07 × 10-2m3 ⇒ V2 = 10m3, V1 = 1m3

( ) ( )

2

2 10 1.07 10 1 1W 814.54 ln a 1 1.07 10 10 1

1883.44 a 0.9 kJ 1883.44 157 0.9 kJ

1742.14kJ

−

−

⎛ ⎞− × ⎛ ⎞∴ = + −⎜ ⎟ ⎜ ⎟− × ⎝ ⎠⎝ ⎠ = − ×

= − ×

=

**Q3.15 If a gas of volume 6000 cm3 and at pressure of 100 kPa is compressed
**

**quasistatically according to pV2= constant until the volume becomes
2000 cm3, determine the final pressure and the work transfer.
**

(**Ans. **900 kPa, – 1.2 kJ)
**Solution: **Initial volume ( 1v ) = 6000 cm3
= 0.006 m3
Initial pressure ( )1p = 100 kPa

Final volume ( 2v ) = 2000 cm3 = 0.002 m3 If final pressure ( )2p

( )

( ) ×

∴ = = = 22

1 1 2 2 2

2

100 0.006p Vp 900 kPa V 0.002

Page 25 of 265