507 p k nag solution, Exam questions for Mechanical Engineering. Birla Institute of Technology and Science
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507 p k nag solution, Exam questions for Mechanical Engineering. Birla Institute of Technology and Science

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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

IntroductionBy: S K MondalChapter 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 mρ

∀→

⎛ ⎞= ⎜ ⎟∀⎝ ⎠

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

mρ ∀→ ∀

⎛ ⎞= ⎜ ⎟∀⎝ ⎠

Page 3 of 265

IntroductionBy: S K MondalChapter 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 asPatm = 101.325 kN/m2

Page 4 of 265

IntroductionBy: S K MondalChapter 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

IntroductionBy: S K MondalChapter 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

IntroductionBy: S K MondalChapter 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

IntroductionBy: S K MondalChapter 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 10pv = × , 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

IntroductionBy: S K MondalChapter 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 gρ

h

dh

p = h gρ

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

TemperatureBy: S K MondalChapter 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

TemperatureBy: S K MondalChapter 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

TemperatureBy: S K MondalChapter 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

2at btε = +

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

TemperatureBy: S K MondalChapter 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 = 550o 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 = -50o 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 tα 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 0R , &α β then 2

0(1 500 500 )R R α β= + × + ×

Page 14 of 265

Work and Heat TransferBy: S K MondalChapter 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 TransferBy: S K MondalChapter 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

3m

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 TransferBy: S K MondalChapter 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 TransferBy: S K MondalChapter 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 TransferBy: S K MondalChapter 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 TransferBy: S K MondalChapter 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 TransferBy: S K MondalChapter 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 TransferBy: S K MondalChapter 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 TransferBy: S K MondalChapter 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 TransferBy: S K MondalChapter 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 ma 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 TransferBy: S K MondalChapter 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

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