THERMOFLUID: EXAMPLES AND PROBLEMS

THERMOFLUID

EXAMPLES AND PROBLEMS

Sylvester Gindan

THERMOFLUID
EXAMPLES AND PROBLEMS

FIRST EDITION

Published by:

POLITEKNIK KOTA KINABALU
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© Politeknik Kota Kinabalu
First Edition, 2021

All right reserved. No part of this publication may be reproduced or transmitted in any form or
by any means, electronic or mechanical including photocopy, recording, recording, or any
information storage and retrieval system, without permission in writing from Politeknik Kota
Kinabalu.

THERMOFLUID: EXAMPLES AND PROBLEMS

eISBN
BAR CODE eISBN

Preface

Praise to God for the successful implementation of the book Thermofluid.

The book was published as a reference and guide for all students studying a Diploma in
Mechatronics Engineering (DEM) who take courses DJM20053 Thermofluid. This
course is a core course for DEM and offered in semester 2 study in the programme.

This book contains five main topics based on the syllabus prescribed by the Ministry of
Higher Education for teaching DJM20053 Thermofluid in polytechnics. I would like to
give an appreciation to the parties that are involved in helping to complete and finish
this book.

This book have been written by Mr. Sylvester Gindan which are lecturer at Mechanical
Engineering Department, Polytechnic Kota Kinabalu for over a decade of experiences.

Lastly, I would like to take this opportunity to express our gratitude to our family,
colleagues and students who continuously supporting us with ideas, knowledges and
feedback throughout the process.

Sylvester Gindan

Notation

ρ mass density ( ) n polytropic index
γ adiabatic index
s relative density; specific gravity c velocity ( )
ω specific weight ( ) η efficiency
⅀ total
v specific volume ( )
ṁ mass flow rate ( )
g gravity acceleration ( )

P pressure ( )

z height (m)
F force (N)
A area ( )
w weight (N)
m mass (kg)
H energy (J)
T temperature (°C)

u internal energy ( )

h enthalpy ( )

s entropy ( )
x dryness fraction
R gas constant ( )

M molecular weight ( )
n number of moles
cv specific heat capacity at constant volume ( )

cp specific heat capacity at constant pressure ( )
Q heat (J)
W work (J)

ii

Synopsis
Thermofluids provides students to the basic concepts of thermodynamics and fluids mechanics
into one integrated course. This course emphasize on concepts of conceptual principles in
thermofluids, fluids applications, properties of pure substances, first and second law of
thermodynamics. This course also provides knowledge and understanding of theory, concepts
and application of principles to solve problems related to thermofluids processes.

Learning Outcomes
Upon completion of this course, the students will be able to :

1. Apply the fundamental concepts of thermodynamics and fluid mechanics to solve the
related problem

2. Perform appropriately experiments according to the Standard Operating Procedures
3. Demonstrate ability to work in team to complete assigned tasks

iii

TABLE OF CONTENT

TITLE PAGE
Preface i
Notation ii
Synopsis iii
Learning Outcomes iii
Table of Content iv

CHAPTER 1: CONCEPTUAL PRINCIPLES IN THERMOFLUIDS 1
1.1 Introduction 1
1.2 Examples 2
1.3 Problems
Formula

CHAPTER 2: FLUID APPLICATIONS 4
2.1 Introduction 4
2.2 Examples 11
2.3 Problems
Formula

CHAPTER 3: PROPERTIES OF PURE SUBSTANCES 17
3.1 Introduction 17
3.2 Examples 24
3.3 Problems
Formula

CHAPTER 4: FIRST LAW OF THERMODYNAMICS AND ITS PROCESSES

4.1 Introduction 28

4.2 Examples 28

4.3 Problems 34

Formula

CHAPTER 5: SECOND LAW OF THERMODYNAMICS 38
5.1 Introduction

iv

5.2 Examples 38
5.3 Problems 40
Formula
86
Reference

v

CHAPTER 1
CONCEPTUAL PRINCIPLES IN THERMOFLUIDS

1.1 Introduction
This topic covers basic concepts in Thermodynamics and Fluid Mechanics. The
students are expected to understand characteristic of fluids and solve unit
conversion. The student should be able to classify the fundamental concept of
thermodynamics such as the principles of a system, boundary and surrounding.

1.2 Examples
1) What is the weight, in N, of an object with a mass of 200 kg at a location where g
= 9.6 m/s2.
W = mg = 200 x 9.6 = 1920 N

2) A tank is filled with oil whose density is ρ = 850 kg/m3. If the volume of the tank
is v = 2 m3, determine the amount of mass m in the tank.
ρ=
m = ρv = 850 x 2 = 1700 kg

3) Determine the mass and the weight of the air contained in a room whose
dimensions are 6m x 6m x 8m. Assume the density of the air is 1 kg/m3.
ρ=

m = ρv = 1 (6 x 6 x 8) = 288 kg
W = mg = 288 x 9.81 = 2825.28 N

4) A lunar exploration module weights 4000 N at a location where g = 9.8 m/s2.
Determine the weight of this module in Newtons when it is on the moon where g =
1.64 m/s2.

W = mg
m = = = 408.16 kg

W = mg = 408.16 x 1.64 = 669.39 N

5) Calculate the specific weight, specific volume and density of methane at 100°C
and 120 N/m2 absolute. Given gas constant, R = 96.3 J/kgK.

i. Density, ρ = = = 3.341 x 10-3 kg/m3

ii. Specific weight, ω = ρg = 3.341 x 10-3 x 9.81 = 0.033 N/m3

-1-

iii. Specific volume, v = = = 299.31 m3/kg

6) At 90°C and 30 N/m2 absolute, the volume per unit weight of a certain gas was
11.4 m3/kg. Determine its gas constant and the density.

i. R = = = 0.942 J/kg.K

ii. ρ = = = 0.088 kg/m3

7) If the density and velocity of a liquid is 835 kg/m3 and 10 m/s, find its specific
volume and mass flow rate. Assume the pipe area is 10 m2.
i. Specific volume, v = = = 1.2 x 10-3 m3/kg

ii. Mass flow rate, ṁ = ρAv = 835 x 10 x 10 = 83500 kg/s

8) Water flows through a 200 mm diameter pipe at a velocity of 2 m/s. Determine the
volume flow rate, weight flow rate and mass flow rate.

i. Volume flow rate, Q = Av = v = (2) = 0.0628 m3/s

ii. Weight flow rate, w = ṁg = 1000 x 0.0628 x 9.81 = 616.36 N/s
iii. Mass flow rate, ṁ = ρAv = 1000 x 0.0628 = 62.83 kg/s

1.3 Problems

1. Calculate the specific weight of fluid and specific gravity of fluid if the weight is 10 N
and the volume is 500 cm3. [Dec. 2014]

2. Fill in the blanks with suitable names for the closed system in the figure below.
[Dec. 2014]

3. Define the following terms;
i. Mass density
ii. Specific weight [Jun 2015]

4. Define internal energy and state the relevant equation. [Jun 2015]
5. Define fluids. [Dec. 2015]
6. A glass bottle with a volume of 100 cm3 full with fluid has relative density 1.25. If the

total mass is 301.7 gram and mass density of glass bottle is 2450 kg/m3, determine
i. Glass bottle mass
ii. Glass bottle volume [Dec. 2015]
7. A system is allowed to do work amounting to 500 kNm whilst heat energy amounting
to 800 kJ is transferred into it. Find the change of internal energy and state whether it
is an increase or decrease. [Jun 2015]
8. Define the following terms :
i. System
ii. Boundary [Jun 2016]
9. Mass of a fluid is 150 kg and its volume is 5000 cm3. Calculate the following :
i. Mass density
ii. Specific weight

-2-

iii. Specific gravity
iv. Specific volume [Jun 2016]
10. Define the following terms:
i. Fluid statics
ii. Fluid dynamics
iii. System
iv. Boundary
v. Surrounding [Dec. 2016]
11. Given mass of a fluid is 150 kg and its volume is 5000 cm3. Calculate the following:
i. Mass density
ii. Specific weight
iii. Specific gravity
iv. Specific volume [Dec. 2016]
12. Sketch and label system, surrounding and boundary. Then, give the definition of
surroundings. [Jun 2017]
13. The oil consist of 5.6 m3 volume and 46800 N weighs. Calculate :
i. Mass density, ρ
ii. Relative density, s [Jun 2017]
14. Define the following terms:
i. Fluid
ii. Mass density
iii. Specific weight
iv. Relative density
v. Specific volume [Dec. 2017]
15. A Bourdon pressure gauge which attached to a boiler which is at sea level with a
reading pressure of 10 bars. If atmospheric pressure is 1.013 bars, calculate the
pressure in that boiler in kN/m2. [Dec 2017]

Formula Density, ρ = = , kg/m3
i.
ii. Specific gravity, s = =
iii.
iv. Specific weight, ω = = = , N/m3
v.
Specific volume, v = = = , m3/kg

Mass flow rate, ṁ = ρAv , kg/s

-3-

CHAPTER 2
FLUID APPLICATIONS

2.1 Introduction
This topic covers the relationship between pressure and depth, Pascal’s Law,
manometer, continuity equation and Bernoulli equation. This topic also explains
the methods in solving problems involving pressure measurement, buoyancy and
pressure.

2.2 Example
1) A diver is working at a depth of 20 m below the surface of the sea. How
much greater is the pressure intensity at this depth than at the surface? Take
into consideration specific weight of water is 10000 N/m3.
P = ɷh = ρgh = 10000 x 9.81 x 20 = 200 kN/m2

2) A force, F of 500 N is applied to the smaller cylinder of a hydraulic jack.
The area, a of a small piston is 20 cm2 while the area, A of a larger piston is
200 cm2. What mass can be lifted on the larger piston?
F = 500 N; a = 20 cm2; A = 200 cm2
P1 = P2

W= = = 5000 N

m = = = 509.7 kg

What mass can be lifted on the larger piston if the larger piston is 0.65 m
below the smaller piston? Assume the liquid is water.
P1 = P2 + P

+ ωh
– ωh }A
– (1000 x 9.81)(0.65)} 200 x 10-4
.47 N

-4-

= = 496.68 kg

What mass can be lifted on the larger piston if the small piston is 0.40 m
below the larger piston? Assume the liquid is water.
P2 = P1 + P

+ ωh
+ ωh }A
+ (1000 x 9.81)(0.40)} 200 x 10-4
N
= = 517.68 kg

3) Find the height of a water column which is equivalent to the pressure of 2
N/m2. ( Take into consideration specific weight of water, water = 1000
kg/m2 x 9.81 m/s2 )
P = ωh

h= = = 2.04 x 10-4 m water

4) A U-tube manometer similar to that shown is used to measure the gauge
pressure of water (mass density ρ = 1000 kg /m3). If the density of mercury
is 13.6 × 103 kg /m3, what will be the gauge pressure at A if h1 = 0.45 m
and D is 0.7 m above BC.

PB = PC PD = Patm
PA + Pwater = PD + Pmercury
PA + ωwaterh1 = ωmercuryh2

-5-

PA = ωmercuryh2 – ωwaterh1
PA = 13.6 x 103 x 9.81 x 0.7 – 1000 x 9.81 x 0.45
PA = 88976.7 N/m2
5) A U tube manometer measures the pressure difference between two points
A and B in a liquid. The U tube contains mercury. Calculate the difference
in pressure if h = 1.5 m, h2 = 0.75 m and h1 = 0.5 m. The liquid at A and B
is water ( ω = 9.81 × 103 N/m2) and the specific gravity of mercury is 13.6.

PC = PD
PA + Pwater = PB + Pwater + Pmercury
PA + ωwaterh = PB + ωwater(h2 – h1) + ωmercuryh1
PA – PB = ωwater(h2 – h1) + ωmercuryh1 - ωwaterh
PA – PB = 9.81 x 103 (0.75 – 0.5) + 13.6 x 9810 x 0.5 – 9.81 x 103 x 1.5
PA – PB = 54445.5 N/m2
6) The top of an inverted U tube manometer is filled with oil of specific
gravity, soil=0.98 and the remainder of the tube with water whose specific
weight of water, ω= 9.81×103 N/m2. Find the pressure difference in N/m2
between two points A and B at the same level at the base of the legs where
the difference in water level h2 is 75 mm.

-6-

PD = PC

PA - Poil - Pwater = PB - Pwater
PA – ωoilh2 – ωwaterh1 = PB – ωwater(h2 + h1)
PA – PB = – ωwater(h2 + h1) + ωoilh2 + ωwaterh1
PA – PB = – ωwaterh2 + ωoilh2
PA – PB = - 9.81 x 103 x 75 x 10-3 + 0.98 x 1000 x 9.81 x 75 x 10-3
PA – PB = -14.715 N/m2

7) A pressure tube is used to measure the pressure of oil (mass density 640
kg/m3) in a pipeline. If the oil rises to a height of 1.2 above the centre of the
pipe, what is the gauge pressure in N/m2 at that point?
P = ωh = ρgh
P = 640 x 9.81 x 1.2 = 7534.08 N/m2

8) If the area A1 = 10  10-3 m2 and A2 = 3  10-3 m2 and the upstream mean
velocity, v1=2.1 m/s, calculate the downstream mean velocity.
Q1 = Q2
A1v1 = A2v2

V2 = = (2.1) = 7 m/s

9) A pipe is split into 2 pipes which are BC and BD. The following
information is given:
diameter pipe AB at A = 0.45 m
diameter pipe AB at B = 0.3 m
diameter pipe BC = 0.2 m
diameter pipe BD = 0.15 m
Calculate:

-7-

i. discharge at section A if vA = 2 m/s
ii. velocity at section B and section D if velocity at section C = 4 m/s

i. QA = AAvA (2) = 0.318 m3/s
QA = vA =

ii. QA = QB = 0.318 m3/s = 4.48 m/s
0.318 = ABvB
VB = =

QB = QC + QD
QB = ACvC + ADvD

0.318 = (4) + (vD)

0.318 = 0.126 + 0.018 vD
VD = 10.67 m/s

10) Water flows through a pipe 36 m from the sea level as shown in figure.
Pressure in the pipe is 410 kN/m2 and the velocity is 4.8 m/s. Calculate
total energy of every weight of unit water above the sea level.

Total energy per unit weight:
-8-

H=z+ +

H = 36 + + = 78.96 J/N

11) A bent pipe labeled MN measures 5 m and 3 m respectively above the

datum line. The diameter M and N are both 20 cm and 5 cm. The water
pressure is 5 kg/cm2. If the velocity at M is 1 m/s, determine the pressure at
N in kg/cm2.

ZM = 5m; ZN = 3m; ΦM = 20cm; ΦN = 5cm; PM = 5 kg/cm2; VM = 1 m/s; PN
=?
HM = HN

ZM + + = ZN + +

= (ZM – ZN) + + - ……………….(1)

QM = QN
AMvM = ANvN

VN = = (1) = 16 m/s

PM = x x = 490.5 kN/m2

ωwater = ρwater.g = 1000 x 9.81 = 9810 N/m3

From Eq. (1) + -
= (5 – 3) + - } (9810)

PN = {(5 – 3) + + -9-
PN = 382 620 N/m2

PN = = 39003.06 x = 3.900306

12) A horizontal venturi meter measures the flow of oil of specific gravity 0.9

in a 75 mm diameter pipe line. If the difference of pressure between the full
bore and the throat tapping is 34.5 kN/m2 and the area ratio, m is 4,

calculate the rate of flow, assuming a coefficient of discharge is 0.97.

Soil = 0.9; Ф1 = 75 mm; P1 – P2 = 34.5 kN/m2; m = 4; cd = 0.97

Qactual = Qtheory. Cd ……...(1)

Qtheory = A1v1 ……………(2)

A1 = = = 0.00441 m2

For horizontal venturi;

V1 = √ ………..……(3)

ωoil = soil.ωwater = 0.9 x 1000 x 9.81 = 8829 N/m3

H= = = 3.92 m

m= =4
From Eq. (3)

V1 = √ = 2.26 m/s

From Eq. (2)
Qtheory = 0.00441 x 2.26 = 0.00999 m3/s
From Eq. (1)
Qactual = 0.00999 x 0.97 = 0.00969 m3/s

13) A vertical venturi meter measures the flow of oil of specific gravity 0.82
and has an entrance of 125 mm diameter and throat of 50 mm diameter.
There are pressure gauges at the entrance and at the throat, which is 300
mm above the entrance. If the coefficient for the meter is 0.97 and pressure
difference is 27.5 kN/m2, calculate the actual discharge in m3/s.
Soil = 0.82; Ф1 = 125 mm; Ф2 = 50 mm; z2 – z1 = 300 mm; cd = 0.97;
P1 – P2 = 27.5 kN/m2
Qactual = Qtheory. Cd ……...(1)

- 10 -

Qtheory = A1v1 ……………(2)

A1 = = = 0.0123 m2

For horizontal venturi;

V1 = √ ………..……(3)

ωoil = soil.ωwater = 0.82 x 1000 x 9.81 = 8044.2 N/m3

H= = = 3.42 m

m= = = = 6.25
From Eq. (3)

V1 = √ = 1.268 m/s

From Eq. (2)
Qtheory = 0.0123 x 1.268 = 0.0156 m3/s
From Eq. (1)
Qactual = 0.0156 x 0.97 = 0.0151 m3/s

2.3 Problems
1. A pipe is split into 2 namely BC and BD as shown in the figure below. The
following information is given: [Dec. 2014]
Diameter pipe AB at A = 0.45m
Diameter pipe at AB at B = 0.3m
Diameter pipe BC = 0.2m
Diameter pipe BD = 0.15m
Calculate:
i. Discharge at section A if vA = 2 m/s
ii. Velocity at section B and section D if velocity at section C = 4 m/s

2. Explain the concept of Hydraulic Jack. [Jun 2015]
- 11 -

3. A force, P of 500 N is applied to the smaller cylinder of a hydraulic jack. The area,
a of a small piston is 20 cm2 while the area, A of a larger piston is 200 cm2. What
mass can be lifted on the larger piston ? [Jun 2015]

4. The main pipe A with a diameter of 50mm flows oil with the flow rate of QA =
2QB. The pipe then split into 2 pipes, where pipe B has a diameter of 95mm and
velocity of 2 m/s. Pipe C has a flow velocity of 0.6 m/s. Calculate: [Dec. 2015]
i. Discharge in pipe B, A and C
ii. Diameter of pipe C

5. A venturi tube tapers has a 350 mm diameter of the entrance and 120 mm diameter
of the throat given discharge coefficient, cd 0.98. A differential mercury U – tube
gauge is used to connect the entrance and the throat. The meter is used to measure
the flow of water fills the leads to the U – tube and is in contact with the mercury.
Calculate the discharge when the difference of the level in the U – tube is 45 mm.
[Jun 2016]

6. State Pascal’s Law. [Jun 2016]
7. A force, F of 800 N is applied to the smaller cylinder of a hydraulic jack. The area,

a of small piston is 20 cm2 while the area, A of a larger piston is 200 cm2. Identify
the mass which can be lifted on the larger piston. [Jun 2016]
8. Figure below shows a simple U – tube manometer which contains mercury to
measure water pressure. Given specific gravity of mercury, smercury = 13.6, h1 = 0.3
m and h2 = 0.5 m. Calculate the gauge pressure at A. Take Patm = 101.325 kPa.
[Jun 2016]

- 12 -

9. A rectangular pontoon has 6 m width (B), 12 m length (l) and 1.5 m draught (D)
fresh water (density 1000 kg/m3). Calculate [Dec 2016]
i. The weight of the pontoon
ii. Its draught in sea water (density 1025 kg/m3)
iii. The load (in kN) that can be supported by the pontoon in fresh water if the
maximum draught permissible is 2 m.

10. Define the following terms: [Dec 2016]
i. Solid phase
ii. Liquid phase

11. Inclined venture meter measures the flow of oil of specific gravity 0.82 and has an
entrance of 125 mm diameter and throat of 50 mm diameter. There are pressure
gauges at the entrance and at the throat, which is 300 mm above the entrance. If
the coefficient for the meter is 0.97 and pressure difference is 27.5 kN/m2,
calculate the actual discharge in m3/s. [Dec 2016]

12. A force, (F) at 650 N is applied to hydraulic jack system. The area of the small
piston is 15 cm2 and the larger piston is 150 cm2. Calculate the load, W. If: [Dec
2016]
i. The pistons are locate at the same level.
ii. The large piston is 0.65 m below the small piston.
Consider the mass density of the liquid jack is 1000 kg/m3.

13. A pipe, AB splits into BC and BD. Pipe BC is 60mm in diameter. The discharge
in pipe AB is 8 liter/s. The diameter of pipe A is 75 mm and the diameter of pipe
B is 50 mm. The velocity in pipe BD is 1.5 m/s and the discharge in pipe BC is
half than discharge in pipe BD. Calculate the, [Jun 2017]
i. Discharge in pipe A, B, BC and BD
ii. Velocity at pipe A, B and BC
iii. Diameter of BD

- 13 -

14. State the function of piezometer and barometer. [Jun 2017]
15. A u-tube differential manometer is connected to two pipes at A and B as figure

below. Pipe A contains oil with specific gravity 0.92 and pipe B is carrying water.
The u-tube contains mercury. Find the pressure different between A and B. If the
pressure at point A is 125 kN/m2, find the pressure at point B. [Jun 2017]

16. A load on a hydraulic lift will rise by pouring oil from a thin tube. The required
height of oil in the tube is required to increase the weight is to be determined.
Calculate: [Jun 2017]
i. Gauge pressure, Pgauge in the fluid under the load
ii. Required oil height, h if the density of oil is given ρoil = 780 kg/m3

17. A pipe is split into two; namely BC and BD, as shown in figure below. The
following information is given: [Dec 2017]
Diameter pipe AB at A = 0.45m
Diameter pipe AB at B = 0.3m
Diameter pipe BC = 0.2m
Diameter pipe BD = 0.15m
Determine:
- 14 -

i. Discharge rate at section A if VA = 2 m/s
ii. Velocity at section B and section D if velocity at section C = 4 m/s

18. Define the following terms: [Dec 2017]

i. Solid phase
ii. Liquid phase
iii. Gas phase

19. Inclined venture meter measures the flow rate of oil with specific gravity 0.82, at

the pipe diameter is 125 mm and the throat is 50 mm. The pressure gauges consist

of difference level at the entrance and at the throat is 300 mm where throat

more over then entrance. If the coefficient for the meter is 0.97 and

pressure difference is 27.5 kN/m2, calculate the actual discharge in m3/s. [Dec

2017]

20. A force, F 650 N is applied to hydraulic jack system. The area of small and larger
piston are 15 cm2 and 150 cm2. Calculate the load, W if : [Dec 2017]

i. The piston are the same level
ii. The large piston is 0.65m below than the smaller piston
iii. The small piston is 0.40m below than the larger piston

Consider the mass density, ρwater of the liquid in side the jack is 1000
kg/m3

Formula
i. Pressure, P = ρgh, N/m2
ii. Buoyancy force, Fb = ρgv, N
iii. Flow rate, Q = Av, m3/s

iv. Continuity equation law : Qin = Qout

v. Area, A = m2

vi. Total energy, HT = z + + , Nm @ Joule

vii. Bernoulli theorem : Hin = Hout
viii. Actual flow rate, Qactual = Qtheory . cd , m3/s

- 15 -

ix. Theory flow rate, Qtheory = Ainvin , vin. horizontal = √ ,

vin. incline = √ , m3/s
x. Pressure head, H = ,m
xi. Area ratio, m =

- 16 -

CHAPTER 3
PROPERTIES OF PURE SUBSTANCES

3.1 Introduction
This topic enables the students to review the properties of pure substances, analyze
the state of steam using the properties of pure substances and relate the ideal gas
model to pure substances.

3.2 Examples
1) For a steam at 20 bar with a dryness fraction of 0.9, calculate the
i. Specific volume
ii. Specific enthalpy
iii. Specific internal energy
iv. Specific entropy
v. Sketch the P – v diagram showing the location point of the dryness
i. V = vf + xvfg vf ≈ 0
V = xvg = 0.9 (0.09957) = 0.0896 m3/kg
ii. h = hf + xhfg = 909 + 0.9 (1890) = 2610 kJ/kg
iii. u = uf + xufg = uf + x(ug – uf) = 907 + 0.9 (2600 – 907)
= 2430.7 kJ/kg
iv. s = sf + xsfg = 2.447 + 0.9 (3.893) = 5.9507 kJ/kg.K
v.

2) Steam at pressure 165 bar has specific internal energy of 2380 kJ/kg. Show
that the steam is saturated wet steam and determine
i. Dryness fraction
ii. Specific volume
iii. Specific enthalpy

- 17 -

iv. Specific entropy
i. u = uf + xufg = uf + x (ug – uf)

x= = = 0.949

ii. v = vf + xvfg = xvg = 0.949 (0.00884) = 0.008389 m3/kg
iii. h = hf + xhfg = 1670 + 0.949 (895) = 2519.36 kJ/kg
iv. s = sf + xsfg = 3.777 + 0.949 (1.437) = 5.141 kJ/kg

3) Calculate the specific internal energy of superheated steam at pressure of
45 bar and 380°C temperature.
U = ? if P = 45 bar and T = 380°C

350 380 400
40 2828 a 2921
45 b U d
50 2810 c 2907

a = 2883.8 kJ/kg

b = 2819 kJ/kg

c = 2868.2 kJ/kg

- 18 -

d = 2914 kJ/kg

U = 2876 kJ/kg

4) The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar,

what is the value of dryness fraction?

u = uf + xufg

x= = = 0.6

5) Determine the specific volume, specific enthalpy and specific internal
energy of wet steam at 32 bar if the dryness fraction is 0.92.
i. v = vf + xvfg = xvg = 0.92 (0.06246) = 0.057 m3/kg
ii. h = hf + xhfg = 1025 + 0.92 (1778) = 2660.76 kJ/kg
iii. u = uf + xufg = uf + x (ug – uf) = 1021 + 0.92 (2603 – 1021)
= 2476.44 kJ/kg

6) Find the dryness fraction, specific volume and specific internal energy of
steam at 105 bar and specific enthalpy 2100 kJ/kg.

i. h = hf + xhfg

x= = = 0.52

ii. v = vf + xvfg = xvg = 0.52 (0.01696) = 8.8192 x 10-3 m3/kg
iii. u = uf + xufg = uf + x (ug – uf) = 1414 + 0.52 (2537 – 1414)

= 1997.96 kJ/kg
7) Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Calculate :

i. Temperature
ii. Degree of superheat
iii. Specific volume

- 19 -

iv. Specific internal energy

At 160 bar, hg = 2582 kJ/kg. Ini bermakna stim dalam keadaan panas

lampau kerana h = 3139 kJ/kg lebih tinggi berbanding hg = 2582 kJ/kg.

i. T = 450°C P = 160 bar dan h = 3139 kJ/kg

ii. Degree of superheat = T – Ts = 450 – 347.3 = 102.7°C

iii. v = 1.702 x 10-2 m3/kg P = 160 bar dan h = 3139 kJ/kg

iv. h = u + pv
u = h – pv = 3139 – (160 x 102 x 1.702 x 10-2) = 2866.68 kJ/kg

8) A quantity of a certain perfect gas is heated at a constant temperature
(isothermal process) from an initial state of 0.15 m3 and 3 bar to a final
state of 1.6 bar. Calculate the final volume of the gas.

Pv = mRT
= mR

T1 = T2 for isothermal process

P1v1 = P2v2

V2 = = x 0.15 = 0.281 m3

9) A quantity of gas at 0.45 m3 and 350 oC undergoes a constant pressure
process that causes the volume of the gas to decreases to 0.25 m3.
Calculate the temperature of the gas at the end of the process.

- 20 -

Pv = mRT
= mR
P1 = P2 for constant pressure process
=

T2 = = x (350 + 273) = 346.11 K

10) A rigid vessel with constant volume 2 m3 contains air at 8 bar and
70°C. The vessel is cooled until the temperature of the air drops to
30°C. Calculate:
i. Mass of air in the vessel.
ii. Final pressure
iii. Heat removed during the process
For related calculation, standard values of Cp, Cv, R, etc of air must be
referred from the Steam Table.

Pv = mRT
=m

i. = = = 16.25 kg

ii. P2 = = = 706.56 kN/m2

iii. Qout = mcvdT
Qout = mcv (T2 – T1)
Qout = 16.25 x 718 x (303 – 343) = -466.7 kJ

11) A mass of 1.5 kg of air at a pressure of 2 bar and a temperature at
50°C is compressed adiabatically to 7 bars pressure. Determine the
following
i. Final temperature

- 21 -

ii. Final volume
iii. Work done
iv. Heat supply
v. Change in internal energy
For air, take γ = 1.4, Cv = 718 J/kgK and R = 287 J/kgK

Pvγ = c
i. = { }

T2 = T1 { }

T2 = (50 + 273) { } = 462.17 K
ii. = { }

{} =

V2 = ………….(1)

{}

P1v1 = mRT1

V1 = = = 0.695 m3

From Eq. 1;

V2 = = = 0.284 m3

{ }

iii. W = = = -149.5 kJ

iv. Q = 0 adiabatic process
v. dU = mcvdT

- 22 -

U2 – U1 = mcv (T2 – T1)
U2 – U1 = 1.5 x 718 x (462.17 – 323) = 149.886 kJ

12) The combustion gases in a petrol engine cylinder are at 30 bar and 800oC
V2 8.5

before expansion. The gases expand through a volume ratio ( V1 ) of ( 1 )
and occupy 500 cm3 after expansion. When the engine is air cooled the
polytropic expansion index n = 2. What is the temperature and pressure of
the gas after expansion, and what is the work output?

={ } = 155.65 J
T2 = T1 { }
T2 = (800 + 273) { } = 126.24 K

={ }
= { } = 30 x { } = 0.415 bar
= = 5.88 x 10-5 m3
W= =

13) Two kilograms of a gas receive Q kJ as heat at constant volume process. If
the temperature of the gas increases by 100°C, and cv = 0.268 kJ/kgK of the
process, determine the heat flow.
M = 2 kg; cv = 0.268 kJ/kgK; T2 – T1 = 100°C = 373 K
Q – W = U2 – U1
mcv (T2 – T1) – 0 = U2 – U1

- 23 -

mcv (T2 – T1) – 0 = Q
2 (0.268)(373) = Q
Q = 199.928 kJ

3.3 Problems
1. Name TWO (2) phases of pure substance. [Dec. 2014]
2. The mass of 0.18 kg gas is at temperature of 15°C and pressure of 130
kN/m2. If the gas has a value of Cv = 720 J/kg.K, calculate the [Dec.
2014]
i. Gas constant if volume is 0.17 m3
ii. Molecular weight
iii. Specific heat at constant pressure
iv. Specific heat ratio
3. A rigid vessel with a constant of volume 1.2 m3 contains air at 7.5 bar and
65°C. The vessel is cooled until the temperature of the air drops to 32°C.
Calculate
i. Mass of air in the vessel
ii. Final pressure
iii. Heat removed during the process
For related calculation, standard values of Cp, Cv, R, etc of air must be
reffered from the Steam Table. [Dec. 2014]
4. Draw the P – v diagram of a pure substance. [Jun 2015]
5. For a steam at 20 bar with a dryness fraction of 0.9, calculate the [Jun
2015]
i. Specific volume
ii. Specific enthalpy
iii. Specific internal energy
iv. Sketch the P – v diagram showing the location point of the dryness

6. A mass of 0.25 kg of air at a pressure of 1.5 bars and a temperature at 50°C
is compressed adiabatically to 8 bars pressure. Determine the following
i. Final temperature
ii. Final volume
iii. Work done
- 24 -

iv. Heat supply
v. Change in internal energy
For air, take γ = 1.4, Cv = 0.718 kJ/kgK and R = 0.287 kJ/kgK [Jun 2015]
7. Give TWO (2) application of steam in industry. [Dec. 2015]
8. Steam at pressure 165 bar has specific internal energy of 2380 kJ/kg. Show
that the steam is saturated wet steam and determine [Dec. 2015]
i. Dryness fraction
ii. Specific volume
iii. Specific enthalpy
iv. Specific entropy
9. Calculate the specific internal energy of superheated steam at pressure of
45 bar and 380 °C temperature. [Dec. 2015]
10. An insulated cylinder has 140 kN/m2 pressure at 300 K temperature has
been expanded to 623 K. If the gas has a mass of 0.88 kg, determine the
final pressure and final volume of the gas. Given Cp = 0.865 kJ/kgK and
Cv = 0.445 kJ/kgK.[Dec. 2015]
11. Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Calculate : [Jun
2016]
i. Temperature
ii. Degree of superheat
iii. Specific volume
iv. Specific internal energy
12. An air contained in a pneumatic cylinder is initially at pressure 100 kN/m2,
volume of 0.012 m3 and temperature of 120 °C. The air is heated at
constant volume until the pressure is increase to 240 kN/m2. Assume
that air is a perfect gas. Calculate: [Jun 2016]
i. The mass of air
ii. The change of entropy
iii. Sketch the process on T – s diagram
13. The pressure of wet steam given is 36 bar and the dryness fraction is 0.62.
Calculate: [Dec. 2016]
i. Specific enthalpy
ii. Specific entropy
iii. Specific internal energy

- 25 -

14. A mass of 0.35 kg gas is at a temperature of 35°C, volume of 0.075 m3/kg
and pressure of 190 kN/m2. If the gas has a value of Cv = 720 J/kgK,
calculate; [Jun 2017]

i. Gas constant

ii. Molecular weight

iii. Specific heat at constant pressure

iv. Specific heat ratio

15. 1 kg of nitrogen (molecular weight 28) is compressed reversibly and

isothermally from 1.01 bar, 35°C to 4.2 bar. Calculate the work done and

the heat flow during the process. Assume nitrogen to be a perfect gas. [ Jun

2017]

16. A perfectly insulated horizontal nozzle has a steady stream of mass flow.

The following even was obtained as shown in Table 1. [Dec. 2017]

At the entry At the exit

Enthalpy, h1 = 3025 kJ/kg Enthalpy, h2 = 2790 kJ/kg

Pressure, P1 = 90 bar Pressure, P2 = 10 bar
Velocity, C1 = 60 m/s Specific volume, v = 0.5 m3/kg
Diameter, d1 = 357 mm
Specific volume, v = 0.19 m3/kg

From the above information, determine:
i. Velocity at the exit
ii. Mass flow rate
iii. Diameter of the exit

Formula
i. Dryness fraction, x =
ii. mtotal = mliquid + mvapor = mf + mg
iii. Specific volume, v = vf + xvfg, m3/kg
iv. Specific enthalpy, h = hf + xhfg, kJ/kg
v. Specific internal energy, u = uf + xufg, kJ/kg
vi. Specific entropy, s = sf + xsfg, kJ/kgK
vii. Ideal gas law, Pv = mRT

- 26 -

viii. Gas constant, R = = constant, kJ/kgK

ix. Universal gas constant, Ro = MR, kJ/kmolK M - molecular weight of the
gas, kg/kmol

x. Non-flow energy equation, Q – W = (U2 – U1)
xi. Heat flow in a constant volume process, Q12 = mcv(T2 – T1), kJ/kg
xii. Heat flow in a reversible constant pressure process Q = mcp(T2 – T1), kJ/kg

xiii. Constant pressure process, = = constant

xiv. Constant volume process, dQ = mCvdT, kJ/kg W=0
xv. Isothermal process, P1 v1 = P2v2 = constant
xvi. Polytrophic process, pvn = constant

xvii. Work = ,J

xviii. = { } = { }

xix. =
xx. Adiabatic process, Pv = C Q = 0
xxi. = { } = { }

xxii.  = ratio of specific heat = C p
Cv

xxiii. Work =

xxiv. cp- cv = R

xxv. cv =  R
1

- 27 -

CHAPTER 4
FIRST LAW OF THERMODYNAMICS AND ITS PROCESSES

4.1 Introduction
This topic explains the concept of the first law of thermodynamics. The students
should be able to apply the first law of thermodynamics such as on the closed and
open system and calculate power in the boiler, condenser, turbine, nozzle, throttle
and pump.

•The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with
some third body, then they are also in equilibrium with each other.

Zeroth law

First law •The First Law of Thermodynamics states that heat is a form of energy, and thermodynamic processes
are therefore subject to the principle of conservation of energy. This means that heat energy cannot
be created or destroyed. It can, however, be transferred from one location to another and converted
to and from other forms of energy.

•The Second Law of Thermodynamics is about the quality of energy. It states that as energy is
transferred or transformed, more and more of it is wasted. The Second Law also states that there is

Second law a natural tendency of any isolated system to degenerate into a more disordered state.

Third law •The Third Law of Thermodynamics is concerned with the limiting behavior of systems as the
temperature approaches absolute zero. Most thermodynamics calculations use only entropy
differences, so the zero point of the entropy scale is often not important. However, we discuss the
Third Law for purposes of completeness because it describes the condition of zero entropy.

Figure 4.1 Law of thermodynamics

4.2 Examples
1) The specific internal energy of a fluid is increased from 120 kJ/kg to 180
kJ/kg during a constant volume process. Determine the amount of heat
energy required to bring about this increase for 2 kg of fluid.
The non flow energy equation is
Q – W = U2 – U1
For a constant volume process
W=0
and the equation becomes

- 28 -

Q = U2 – U1
Q = 180 – 120
= 60 kJ/kg
Therefore for 2 kg of fluid
Q = 60 x 2 = 120 kJ
120 kJ of heat energy would be required.

2) 2.25 kg of fluid having a volume of 0.1 m3 is in a cylinder at a constant
pressure of 7 bar. Heat energy is supplied to the fluid until the volume
becomes 0.2 m3. If the initial and final specific enthalpies of the fluid
are 210 kJ/kg and 280 kJ/kg respectively, determine
a) the quantity of heat energy supplied to the fluid
b) the change in internal energy of the fluid
Data: p = 7.0 bar; V1 = 0.1 m3 ; V2 = 0.2 m3
a) Heat energy supplied = change in enthalpy of fluid
Q = H2 – H1
= m( h2 - h1 )
= 2.25( 280 – 210 )
= 157.5 kJ
b) For a constant pressure process
W = P(V2 – V1)
= 7 x 105 x ( 0.2 – 0.1)
= 7 x 104 J
= 70 kJ

Applying the non-flow energy equation
Q – W = U2 – U1
gives
U2 – U1 = 157.5 – 70

= 87.5 kJ

3) The figure below shows a certain process, which undergoes a complete
cycle of operations. Determine the value of the work output for a complete
cycle, Wout.

- 29 -

Non flow energy equation : dU = Q – W
We assume no internal energy so
0=∑ ∑
∑∑
Qin + Qout = Win + Wout
+10 + (-3) = -2 + Wout
Wout = 9 kJ

4) A system is allowed to do work amounting to 500 kNm whilst heat energy
amounting to 800 kJ is transferred into it. Find the change of internal
energy and state whether it is an increase or decrease.

Non flow energy equation : dU = Q – W
U2 – U1 = Q – W
U2 – U1 = +800 – (+500) = 300 kJ (increase)

5) The specific internal energy of a fluid is increased from 120 kJ/kg to 180
kJ/kg during a constant volume process. Determine the amount of heat
energy required to bring about this increase for 2 kg of fluid.
For close system (constant volume):
Non flow energy equation : dU = Q – W
U2 – U1 = Q – W W = 0 (constant volume)
U2 – U1 = Q
180 – 120 = Q

- 30 -

60 kJ/kg = Q
For 2 kg of fluid : Q = 60 x 2 = 120 kJ

6) During a complete cycle operation, a system is subjected to the following:
Heat transfer is 800 kJ supplied and 150 kJ rejected. Work done by the
system is 200 kJ. Calculate the work transferred from the surrounding to
the system.
Qin = 800 kJ; Qout = -150 kJ; Wout = 200 kJ
∑ =∑
800 + (-150) = Win + 200
Win = 450 kJ

7) A close system undergoes a process in which there is a heat transfer of 200
kJ from the system to the surroundings. The work done from the system to
the surroundings is 75 kJ. Calculate the change of internal energy and state
whether it is an increase or decrease.
Qout = -200 kJ; Wout = 75 kJ
dU = Q – W
dU = -200 – 75
dU = -275 kJ

8) A thermodynamic system undergoes a process in which its internal energy
decreases by 300 kJ. If at the same time, 120 kJ of work is done on the
system, find the heat transferred to or from the system.
dU = -300 kJ; Win = -120 kJ
dU = Q – W
-300 = Q – (-120)
Q = -420 kJ

9) The internal energy of a system increases by 70 kJ when 180 kJ of heat is
transferred to the system. How much work is done by the gas?
dU = 70 kJ; Q = 180 kJ
dU = Q – W
70 = 180 – W

- 31 -

W = 110 kJ

10) During a certain process, 1000 kJ of heat is added to the working fluid
while 750 kJ is extracted as work. Determine the change in internal energy
and state whether it is increased of decreased.
Qin = 1000 kJ; Wout = 750 kJ
dU = Q – W
dU = 1000 – 750
dU = 250 kJ

11) If the internal energy of a system is increased by 90 kJ while the system
does 125 kJ of work to the surroundings, determine the heat transfer to or
from the system.
dU = 90 kJ; Wout = 125 kJ
dU = Q – W
90 = Q – 125
Q = 215 kJ

12) Fluid with specific enthalpy of 2800 kJ/kg enters a horizontal nozzle with
negligible velocity at the rate of 14 kg/s. At the outlet of the nozzle, the
specific enthalpy and specific volume of the fluid are 2250 kJ/kg and 1.25
m3/kg respectively. Assuming an adiabatic flow, determine the required
outlet area of the nozzle.
h1 = 2800 kJ/kg; C1 = 0; ṁ = 14 kg/s; h2 = 2250 kJ/kg; v2 = 1.25 m3/kg;
Q=0
C2 = √
C2 = √
C2 = 33.17 m/s
ṁ=

14 =
A2 = 0.53 m2

- 32 -

13) Fluid enters a condenser at the rate of 30 kg/min with a specific enthalpy of
2100 kJ/kg, and leaves with a specific enthalpy of 205 kJ/kg. Determine the
rate of heat energy loss from the system.
ṁ = 30 kg/min; h1 = 2100 kJ/kg; h2 = 205 kJ/kg
Q = ṁ (h2 – h1)
Q = (205 – 2100)

Q = -947.5 kW

14) Steam flows steadily into a turbine at 5350 kg/h and produces 2700 kW of
power. Steam properties for inlet and outlet of the turbine are shown in the
Table 1. If the potential energy can be neglected, determine:

i. The heat which is transferred to surrounding
ii. The area for the outlet vessel
Item Inlet Outlet

Velocity, C 325 m/s 109 m/s

Pressure, P 10 bar 1.4 bar

Internal Energy, U 3700 kJ/kg 2500 kJ/kg
Specific Volume, v 0.75 m3/kg 1.823/kg

ṁ = 5350 kg/h; W = 2700 kW; gz = 0

i. Q = (U2 – U1) + (P2V2 – P1V1) + +W

Q = (2500 – 3700) + ((1.4 x 102)(1.82) – (10 x 102)(0.75)) + +

Q = -1200 – 495.2 – 46.9 + 1816.82
Q = 74.72 kJ/kg
ii. ṁ =

=
A2 = 0.025 m2

- 33 -

15) A rotary pump draws 650 kg/hour of atmospheric air and delivers it at a
higher pressure. The specific enthalpy of air at the pump inlet is 350 kJ/kg
and that at the exit is 500 kJ/kg. The heat lost from the pump casing is 5200
W. Neglecting the changes in kinetic and potential energy, determine the
power required to drive the pump.
ṁ = 650 kg/h; h1 = 350 kJ/kg; h2 = 500 kJ/kg; Q = -5200 W; C = 0; gz = 0
W = (h1 – h2) + Q
W = (350 – 500) -

W = -150 – 28.8 = -178.8 kJ/kg
W = = 32.28 kW

16) A boiler operates at a constant pressure of 25 bar, and evaporates fluid at
the rate of 1250 kg/h. At entry of the boiler, the fluid has an enthalpy of
185 kJ/kg and on leaving the boiler, the enthalpy of the fluid is 2150 kJ/kg.
Determine the heat energy supplied to the boiler.
P = 25 bar; ṁ = 1250 kg/h; h1 = 185 kJ/kg; h2 = 2150 kJ/kg
Q = h2 – h1
Q = 2150 – 185 = 1965 kJ/kg
Q = (1965) = 682.3 kW

4.3 Problems
1. Fluid with a specific enthalpy of 2800 kJ/kg enters a horizontal nozzle with
negligible velocity at the rate of 14 kg/s. At the outlet from the nozzle the
specific enthalpy and specific volume of the fluid are 2250 kJ/kg and
1.25 m3/kg respectively. Assuming an adiabatic flow, determine the
required outlet area of the nozzle. [Dec. 2014]
2. Define potential energy. [Dec. 2014]
3. In a steady flow system, a substance flows at the rate of 4 kg/s. It enters at a
pressure of 620 kN/m2, a velocity of 300 m/s, internal energy 2100 kJ/kg
and specific volume 0.37 m3/kg. It leaves the system at a pressure of 130
kN/m2, a velocity of 150 m/s, internal energy 1500 kJ/kg and specific

- 34 -

volume 1.2 m3/kg. During its passage through the system the substance
has a loss by heat transfer of 30 kJ/kg to the surroundings. Determine the
power of the system in kilowatts, stating whether it is from or to the
system. Neglect any change in potential energy. [Dec. 2014]
4. Fluid enters a condenser at the rate of 35 kg/min with a specific enthalpy
of 2200 kJ/kg, and leaves with a specific enthalpy of 255 kJ/kg. Determine
the rate of heat energy loss from the system. [Jun 2015]

5. Explain briefly the First Law of Thermodynamics. [Jun 2015]

6. Steam flows steadily into a turbine at 5400 kg/h and produces 2500 kW of

power. Steam properties for inlet and outlet of the turbine are shown in the

Table 1. If the potential energy can be neglected, determine:

i. The heat which is transferred to surrounding

ii. The area for the outlet vessel [Jun 2015; Dec. 2016]

Item Inlet Outlet

Velocity, C 330 m/s 119 m/s
Pressure, P
Internal Energy, U 10 bar 1.4 bar
Specific Volume, v
3770 kJ/kg 2550 kJ/kg
0.65 m3/kg 1.82 m3/kg

7. State TWO (2) device that use the principle of steady flow system.
[Dec. 2015]

8. In a steady flow open system, a substance flows at rate 5 kg/s. It enters the
system at velocity of 250 m/s and 2750 kJ/kg of specific enthalpy. It leaves
the system at velocity of 100 m/s and 1050 kJ/kg. The heat loss to the
surroundings as the steam passes through the system is 4 kW. Calculate the
power generates by the system and state whether it is from or to the system.
Neglect any changes in potential energy. [Dec. 2015]

9. State TWO (2) differences between non flow process and flow process.
[Jun 2016]

10. A boiler operates at a constant pressure of 20 bar, and evaporates fluid at
the rate of 1200 kg/h. At entry of the boiler, the fluid has an enthalpy of
175 kJ/kg and on leaving the boiler, the enthalpy of the fluid is 2100 kJ/kg.
Determine the heat energy supplied to the boiler. [Jun 2016]

11. State the following terms:

- 35 -

i. First Law of Thermodynamics
ii. Open System and Close System [Dec. 2016]
12. Define the heat supply, work done and change of internal energy value for
the system shown in figure below. [Dec. 2016]

Q2, Q3, W3 System Q1, W1, W2

Q1 = 15.5 kJ, Q2 = 147 kJ, Q3 = 0.68 MJ
W1 = 125 kJ, W2 = 0.23 MJ, W3 = 68 kJ
13. In a steam power plant, the heat supply to boiler is 2100 kJ/kg. The
plant produces 1800 kJ/kg turbine work out. The steam then goes to
condenser to change it phase to water by removing 1200 kJ/kg heat.
The water then pump into the boiler. Define the work in by the
water pump. [Jun 2017]
14. In a certain steam plant, the turbine produces 1000 kW. The heat supplied
to the steam in the boiler is 2800 kJ/kg, the heat rejected by the steam to the
cooling water in the condenser is 2100 kJ/kg and the required feed pump
work to pump the condensate back into the boiler is 5 kW. Determine
i. Change in heat
ii. Change in work
iii. Steam flow rate [Jun 2017]
15. State THREE (3) devices that use the principle of steady flow system and
explain ONE (1) device that you have listed. [Dec. 2017]
16. A perfectly insulated horizontal nozzle has a steady stream of mass flow.
The following even was obtained as shown in table below.

At the entry At the exit

Enthalpy, h1 = 3025 kJ/kg Enthalpy, h2 = 2790 kJ/kg

Pressure, P1 = 90 bar Pressure, P2 = 10 bar
Velocity, C1 = 60 m/s Specific volume, v = 0.5 m3/kg

Diameter, d1 = 357 mm
Specific volume, v = 0.19 m3/kg

- 36 -

17. From the above information, determine
i. Velocity at the exit
Formula ii. Mass flow rate
i. iii. Diameter of the exit [Dec. 2017]
ii. A nozzle is device used to increase the velocity of a fluid.
i. Sketch TWO (2) types of nozzle.
iii. ii. State SIX (6) engineering systems applying the nozzle. [Dec. 2017]

Non flow energy equation, Q – W = U2 – U1
Steady flow energy equation,

gZ1 + u1 + P1v1 + + Q = gZ2 + u2 + P2v2 + + W
Specific enthalpy, h = u + Pv
Continuity equation, ṁ1 = ṁ2 =

- 37 -

CHAPTER 5
SECOND LAW OF THERMODYNAMICS

5.1 Introduction
This topic explains the second law of thermodynamics in relation to heat engine
and heat pump efficiency and also shows T – s diagram to explain the total flow of
heat due to entropy changes. The students should also be able to sketch T – s
diagram and calculate heat, work and entropy changes for the compression
processes and expansion of steam and gas.

5.2 Examples

1) An engine does 31 kJ of useful work. If it sinks 5 kJ of energy in the form

of heat to its cold reservoir,
i. Find the heat energy supplied to the engine’s hot reservoir.

ii. What is the efficiency of this engine?

i. Qin = Qout + W

Qin = 5 + 31 = 36 kJ

ii. ɳengine = = = 0.86 @ 86%

2) An engine has an efficiency of 55% and produces 450 J of work. Determine
the following;
i. The input heat
ii. The rejected heat
i. ɳengine =

Winput = Qin = = = 818.18 J

ii. Qin = Qout + Woutput
Qout = Qin - Woutput
Qout = 818.18 – 450 = 368.18 J

3) A heat engine draws 500 J of heat from its high temperature source and
discards 250 J of exhaust heat into its cold temperature reservoir during
each cycle.

- 38 -

i. How much work does this engine perform per cycle?

ii. What is its thermal efficiency?

i. Qin = Qout + Woutput

Woutput = Qin - Qout
Woutput = 500 – 250 = 250 J

ii. ɳengine = = = 0.5

4) A steam power plant operates between a boiler pressure 75 bar and a

condenser pressure 0.080 bar. Determine for these limits the

i. Boiler temperature

ii. Condenser temperature

iii. Carnot efficiency

iv. Turbine work

v. Pump work

vi. Work ratio

vii. Heat supplied

viii. Heat rejected

i. Tboiler = 290.5°C

ii. Tcondenser = 41.5°C

iii. ɳcarnot = ( )( ) = 0.442 @ 44.2%
()
=

iv. Wturbine = h1 – h2

h1 = hg at 75 bar = 2766 kJ/kg

S1 = Sg = S2 at 75 bar = 5.779 kJ/kg.K

S2 = Sf + x2Sfg at 0.080 bar

5.779 = 0.593 + x2 (7.634)

x2 = 0.68

h2 = hf + x2hfg at 0.080 bar

h2 = 174 + 0.68 (2402)

h2 = 1807.36 kJ/kg
Wturbine = 2766 – 1807.36 = 958.64 kJ/kg
v. Wpump = h4 – h3

h4 = hf at 75 bar = 1293 kJ/kg

S4 = Sf = S3 at 75 bar = 3.166 kJ/kg.K

- 39 -

S3 = Sf + x3Sfg at 0.080 bar
3.166 = 0.593 + x3 (7.634)
x3 = 0.34
h3 = hf + x3hfg at 0.080 bar
h3 = 174 + 0.34 (2402)
h3 = 990.68 kJ/kg
Wpump = 1293 – 990.68 = 302.32 kJ/kg

vi. Wratio = = = = 0.685

vii. Qin = h1 – h4 = 2766 – 1293 = 1473 kJ/kg
viii. Qout = h3 – h2 = 990.68 – 1807.36 = -816.68 kJ/kg

5) Sketch the Carnot cycle and explain the maximum possible efficiency
(Carnot efficiency) if the boiler temperature rise from 250°C until 500°C
and the condenser temperature still remain at 12°C.

ɳcarnot = =( )( ) = 0.455 @ 45.5%
()

If temperature rise to 500°C;

ɳcarnot = ( )( ) = 0.631 @ 63.1%
()
=

To achieve the maximum possible Carnot efficiency, the boiler temperature

need to be rise.

5.3 Problems
1) Illustrate a diagrammatic representation of heat engine. [Dec. 2014]
2) A steam power plant operates between a boiler pressure of 42 bar and a
condenser pressure of 0.035 bar. Determine for these limits the cycle
efficiency and the work ratio for a Carnot cycle using wet steam.
[Dec. 2014]

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3) A steam power plant operates between a boiler pressure of 38 bar and a
condenser pressure of 0.045 bar. If steam entry to the turbine with dry
saturated, calculate the work ratio for the Carnot cycle. [Jun 2015]

4) The reversed Carnot heat engine receives heat at temperature of 10°C and
rejects the heat at temperature of 32°C with heat transfer of 100 kW.
Determine the power required in kJ/h. [Jun 2015; Dec. 2016]

5) A steam power plant operates between a boiler pressure of 65 bar and a
condenser pressure of 0.035 bar. Calculate for these limits the cycle
efficiency and the work ratio for a Carnot cycle using wet steam.
[Dec. 2015]

6) Heat is transferred to a heat engine from a furnace at a rate of 70 MW. If
the rate of waste heat rejection to a nearby river is 35 MW, determine the
net work done and the thermal efficiency for this heat engine. [Jun 2016]

7) Explain briefly the principles of heat engine. [Jun 2016]
8) A Carnot heat engine receives heat from furnace at a rate of 70 MW. If the

rate of waste heat rejects heat at low temperature sink of 175°C, calculate
the thermal efficiency of the heat engine. [Jun 2016]
9) A steam power plant operates between a boiler pressure of 38 bar and a
condenser pressure of 0.045 bar. If steam entry to the turbine with dry
saturated, calculate for a Carnot cycle:
i. The Carnot efficiency
ii. The work ratio [Dec. 2016]
10) In a Carnot cycle operating between 307.2 bar and 99.6°C, the maximum
and minimum pressure are 95 bar and 1 bar. Define
i. Cycle efficiency
ii. Net work
iii. Gross work
Assume steam as the working fluid. [Jun 2017]
11) The reversed carnot heat engine receives heat at a temperature of 10°C and
produce 7.213 kW of power. Determine the heat transferred from the
system at temperature of 32°C. [Jun 2017]
12) A steam plant operates with Carnot cycle at a boiling pressure of 40 bar.
The steam condition expands in the turbine until the pressure decreased to
0.035 bar. Based on the power plant statement above:

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i. Find h1, h2, h3 and h4
ii. Calculate the heat required
iii. Calculate the work input [Dec. 2017]
13) A heat extracts energy from groundwater at 8°C and transfers it to water at
68°C to heat a building. Find
i. COP
ii. Electric power consumption if it supplies heat at the rate of 28 kW
[Dec. 2017]

Formula Wnet. out = Wout - Win
i. Wnet. out = Qin - Qout
ii. Thermal efficiency =
iii.
ɳth = =
iv.

v. Carnot efficiency = x 100%

vi. Carnot efficiency =

vii. Work ratio =

viii. COPHeat pump = COP : coefficient of performance
ix. COPRefrigerator =

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