Applied Thermodynamics
for Engineering Technologists
Solutions Manqal
Contents
Introduction
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Introduction and the First Law of Thermodynamics
The Working Fluid
Reversible and Irreversible Processes
The Second Law
The Heat Engine Cycle
Mixtures
Combustion
Steam Cycles
Gas Turbine Cycles
Nozzles and Jet Propulsion
Rotodynamic Machiner,v
Positive Displacement Machines
Reciprocating Internal-combustion Engines
Refrigeration and Heat Pumps
Psychrometry and Air-conditioning
Heat Transfer
The Sources, Use, and Management of Energy
Vil
1
11
20
34
58
68
87
112
125
141
153
183
200
216
236
258
306
V
Introduction
The objectives of this Solutions Manual are to make the fifth edition of the Applied
Thermodynamics text-book more useful and accessible to students and to assist lecturers
by adding to the available range of study aids. With the increasing emphasis on directed
study a Solutions Manual has an important role to play.
After studying a particular chapter of the text-book, students are advised to make
their own attempt at the problems before turning to the Solutions Manual; this is a
study aid, not a crutch. Solutions are given in a reasonably full version but in some of
the more difficult problems reference back to the fifth edition of the text-book may be
necessary.
.
Many of the problems can be solved accurately and quickly using simple computer
programs but since the emphasis in the text-book is on the physical understanding of
practical problems the solutions in the manual have been written to illustrate the
principles involved and not to demonstrate computing techniques; students may
develop some of the solutions further through their own computer programs. One
example of this is in the problems on two-dimensional and transient heat conduction
in Chapter 16.
The problems have been reproduced from the fifth edition of the text-book, and all
page, table and figure numbers in the Solutions Manual refer to the fifth edition.
It is hoped that users of the fifth edition of the text-book will find this Solutions
Manual a welcome and useful complement to the book.
TDE
1993
vii
1
Introduction and the First Law of
Thermodynamics
1.1
Solution
A certain fluid at 10 bar is contained in a cylinder behind a piston, the initial volume
being 0.05 m 3 • Calculate the work done by the fluid when it expands reversibly :
(i) at constant pressure to a final volume of 0.2 m 3 ;
(ii) according to a linear law to a final volume of0.2 m 3 and a final pressure of2 bar;
(iii) according to a law pV = constant to a final volume of 0.1 m 3 ;
(iv) according to a law pv 3 = constant to a final volume of 0.06 m3 ;
(v) according to a law, p = (A/V 2 ) - (B/V), to a final volume of 0.1 m 3 and a final
pressure of 1 bar, where A and B are constants.
Sketch all process~s on a p-v diagram.
( i)
Work input
= 10xl05 x(0.05
-
=
150 000
"'
ie
Work output = + 150 000
(ii)
Work input
rn
"'
Work output
rn
= 1osx{(2x0.15)
= - 90 000 N
ie
0.2)
=+
-
10)}
m
"'
90 000
+ 0.15(2
2
rn
(iii) Work input = 1osx10xO.OSxln(O.OS/O.l)
=
-
34 700
"'
rn
"'
ie
Work output = + 34 700
(iv)
1
Work input = 10x10 5 x( o. OS )3x(
0.06 2
2
= - 7640
Work output = + 7640 N m
(v)
10 =
A
_B_
0.05
1
)
0.05 2
N m
ie
0.05 2
m
and
__
1 = _A
0. 1 2
_JL_
0.1
1
Applied Thermodynamics
1.1
Hence,
A= 0.04, and B = 0.3
Therefore,
0.3
p =
v2
V
Work input = 1osxo.04( :.l._+
0.05
- 1osx0.3xln(o.os/o.1)
= - 0.04xl0 6 + 0.020Bxl06
= - 19 200 Nm
Work output=+ 19 200 Nm
ie
10
L____:1k-_ _ _ _ __.(~il:....__ _ _ _l2.
O,OS
1.2
Solution
o.o,
0-1
0 ,2
1 kg of a fluid is compressed reversibly according to a law pv = 0.25, where P is in
3
bar and v is in m / kg. The final volume is ¼of the initial volume. Calculate the work
done on the fluid and sketch the process on a p-v diagram.
Work input = 1xo.2sx1osxln(4) = 34 660 N m
2
pv: O,Z5
V
2
_1_)
0.1
Introduction and the first law of thermodynar11ics
1.3
3
0.05 m of a gas at 6.9 bar expands reversibly in a cylinder behind a piston according
to the law pv1. 2 = constant, until the volume is 0.08 m 3 • Calculate the work done by
the gas and sketch the process on a p- V diagram.
Solution
Final pressure= 6.9x(0.05/0.08)1.2
= 3.926 bar
Work input= 10 5 x{(3.926x0.08) (1.2 -
(6.9x0.05)}
1)
= - 15 480 Nm
ie
Work output=+ 15 480 Nm
-.
d
L
:s
..
a..
"'
o-os
o-oa
Vo t ...... --/ c,...,,
1.4
Solution
1 kg of a fluid expands reversibly according to a linear law from 4.2 bar to 1.4 bar;
the initial and final volumes are 0.004 m 3 and 0.02 m 3 • The fluid is then cooled
reversibly at constant pressure, and finally compressed reversibly according to a law
pv = constant back to the initial conditions of 4.2 bar and 0.004 m 3 • Calculate the
work done in each process and the net work of the cycle. Sketch the cycle on a p-v
diagram.
V3 = 4.2x0.004/1.4 = 0.012 m3 /kg
Fdr process 1 to 2:
Work input= -lOS{l.4(0.02 - 0.004)
+ 0.5(0.02 - 0.004)(4.2 - 1.4)}
= - 4480 Nm
For process 2 to 3:
Work i llPU t = l.4xl05x(0.02 - 0.012)
= 1120
N m
For process 3 to 1:
Work input = 4.2x1osxo.004xln(0.04/0.012)
= 1845 N m
3
Applied Thermodynamics
1.4
Therefore,
Net work input= - 4480 + 1120 + 1845
= - 1515 N
ie
Net work output = + 1515 Nm
t,4
1---1-------'-----~z
0•004
1.5
Solution
m
0.020
A fluid at 0. 7 bar occupying 0.09 m 3 is compressed reversibly to a pressure of 3.5 bar
according to a law pv" = constant. The fluid is then heated reversibly at constant volume
until the pressure is 4 bar; the specific volume is then 0.5 m 3 /kg. A reversible expansion
according to a law pv 2 = constant restores the fluid to its initial state. Sketch the
cycle on a p-v diagram and calculate:
(i) the mass of fluid present;
(ii) the value of n in the first process;
(iii) the net work of the cycle.
(i)
V3 = v2 = 0.5 m3/kg
V1
ie
v1
2
= 4x0.5 2 /0.7
= 1.195 m3/kg
Also, V1 = 0.09 m3 (given), therefore,
Mass of fluid= 0.09/1.195 = 0.0753 kg
(ii)
For the process 1 to 2:
3.5/0.7 = (1.195/0.5)n
ie
4
n = 1.847
Introduction and the first law of thermodynamics
1.5
(iii)
For the process 1 to 2:
Work input
= 0.0753xl05{(3.5x0.5)
(0.7xl.195)}
1
1. 847 -
= 8121 Nm
For the process 3 to 1:
Work input
= 0.0753xl05{(0.7xl.195)
(4x0.5)}
2 - 1
=
8761 Nm
Net work input= 8121 - 8761 = - 640 Nm
ie
Net work output=+ 640 Nm
4.01------'
J.~1-------'2:...t.
o., 1 - - - - - - - 1 - - - - - - - - 3 1 !
o.S
1.6
Solution
'\1j
A fluid is heated reversibly at a constant pressure of 1.05 bar until it has a specific
volume of 0.1 m 3 /kg. It is then compressed reversibly according to a law pv = constant
to a pressure of 4.2 bar, then allowed to expand reversibly according to a law
pv1. 1 = constant, and is finally heated at constant volume back to the initial conditions.
The work done in the constant pressure ·process is - 515 N m, and the mass of fluid
present is 0.2 kg. Calculate the net work of the cycle and sketch the cycle on a p-v
diagram.
For the process 1 to 2:
Work input=
=
515 Nm= - 515/0.2
2575 Nm/kg
5
r
Applied Thermodynamics
-
-2575 = 1 . osx1osx(v1
ie
1.6
Vt
= 0.0755 m3 /kg =
0.1)
V4
For the process 2 to 3:
Work input = 1.osx1osxo.1xln(l.OS/4 . 2)
= - 14 556 Nm
Also,
V3
= 1.0Sx0.1/4.2 = 0.025 m3/kg
For the process 3 to 4:
p4
= 4.2/(0.0755/0.025) 1
· 7
= 0.642 bar
Work input
= 1os{(0.642x0.0755) 1. 7
-
(4.2x0.025}l
1
= - 8075.6 Nm
Then,
Net work input= 0.2x(14 556 - 8075.6 - 2575)
= 781 Nm
0 .1
1.7
Solution
In an air compressor the compression takes place at a constant internal energy and
50 kJ of heat are rejected to the cooling water for every kilogram of air. Calculate
the work input for the compression stroke per kilogram of air.
The change in internal energy is zero,
Q
ie
6
therefore ,
+ W = 0
Work input= -
Q
= -(-SO)
= SO kJ/kg
Introduction and the first law of thermodynamics
1.8
In the compression stroke of a gas engine the work done on the gas by the piston
is 70 kJ /kg and the heat rejected to the cooling water is 42 kJ /kg. Calculate the
change of specific internal energy stating whether it is a gain or a loss.
Solution
W = 70 kJ/kg
and
Q
= - 42 kJ/kg
Therefore,
Increase in internal energy=
Q
+ W
= - 42 + 70
= 28 kJ/kg
1.9
Solution
A mass of gas at an initial pressure of 28 bar, and with an internal energy of 1500 kJ,
is contained in a well-insulated cylinder of volume 0.06 m 3 • The gas is allowed to
expand behind a piston until its internal energy is 1400 kJ; the law of expansion is
pv 2 = constant. Calculate:
(i) the work done;
(ii) the final volume;
(iii) the final pressure.
(i)
0
ie
Work input= 1400 - 1500 = - 100 kJ
(ii)
The work input is also given by,
ie
- 100xl03 = p2V2 - (28xl05x0.06)
2 - 1
+ W = U2 - U1
p2V2 - p1V1
n-1
Therefore,
p2V2 = 68 000 Nm
Also,
p2V2 2
ie
= 28xl0 5 x(0.06) 2
= 10 080 N m4/kg 2
Final volume= 10 080/68 000 = 0.148 m3
(iii) Final pressure= 68 000/0.148xl05 = 4.59 bar
1.10
The gases in the cylinder of an internal combustion engine have a specific internal
energy of 800 kJ /kg and a specific volume of 0.06 m 3 /kg at the beginning of expansion.
The expansion of the gases may be assumed to take place according to a reversible
law, pv1.s = constant, from 55 bar to 1.4 bar. The specific internal energy after
expansion is 230 kJ /kg. Calculate the heat rejected to the cylinder cooling water per
kilogram of gases during the expansion stroke.
7
Applied Thermodynamics
v 2 = o.o6x(SS/1.4) 111
1.10
·
= 0.693 m3 /kg
5
Then,
- (SSxO.o6i)
work input = 10sx{fl.4X0.693)
1.5 - 1
= - 466 000 Nm/kg
= - 466 kJ/kg
Therefore,
ie
1.11
Solution
= 2 30
=-
Q
+ W
Q
= - 570 - (-466) = - 104 kJ/kg
- 800
570 kJ/kg
Heat rejected= 104 kJ/kg
A steam turbine receives a steam flow of 1.35 kg/ s and the power output is 500 kW.
The heat loss from the casing is negligible. Calculate:
(i) the change of specific enthalpy across the turbine when the velocities at entrance
and exit and the difference in elevation are negligible;
(ii) the change of specific enthalpy across the turbine when the velocity at entrance is
60 m/ s, the velocity at exit is 360 m/ s, and the inlet pipe is 3 m above the exhaust
pipe.
= l.35(h2
(i)
W
ie
h1 - h2
(ii)
W
-
h1)
=-
500 kW
= 500/1.35 = 370
kJ/kg
= l.35{(h2 - h1) + w2-C12i + (Z2 - Zt )g}
2
=
500 kW
Therefore,
500
I
I
=
+ {3602 - 602) + (-3x9.807)
l · 35
2xl03
370 + 63 -
0.029
103
=
433 kJ/kg
(Note that the energy
decrease due to the change in
height is negligible.)
1.12
8
A steady flow of steam enters a cond
..
and a velocity of 350 m/s The cond
enser with a specific enthalpy of 2300 kJ/ kg
·
ensate leaves the
d
.
.
0 f 160 kJ/
kg and a velocity of 70 m/ C
con enser with a specific entha1PY
th
per kilogram of steam condensed.
s. alculate e heat transfer to the cooling fluid
Introduction and the first law of thermodynamics
Solution
Q
= (h2 - ht)+ (C2 2
C1 2
-
)
+ (Z2 - Z1)g
2
= (160 - 2300) + (7~ 2
350 2
-
2xl0 3
)
+
0
= - 2199 kJ/kg
ie
1.13
Heat rejected= 2199 kJ/kg
A turbine operating under steady-flow conditions receives steam at the following
state: pressure, 13.8 bar; specific volume 0.143 m 3 /kg, specific internal energy
2590 kJ/kg, velocity 30 m/s. The state of the steam leaving the turbine is as follows:
pressure 0.35 bar, specific volume 4.37 m 3 /kg, specific internal energy 2360 kJ /kg,
velocity 90 m/s. Heat is rejected to the surroundings at the rate of 0.25 kW and the
rate of steam flow through the turbine is 0.38 kg/s. Calculate the power developed
by the turbine.
Solution
Q +
w
= (u2 - u1) + (p2v2 - p1v1) + (C2 2
-
C1
2
)
2
= (2360 - 2590) + 10S(0.35x4.37 - 13.Bx0.143)
103
+ (90 2
-
30 2
)
2xl0 3
ie
Q +
W = - 270.79 kJ/kg
Therefore.
Power input= (-270.79x0.38) -
= ie
1.14
(-0.25)
102.7 kW
Power output= 102.7 kW
A nozzle is a device for increasing the velocity of a steadily flowing fluid. At the iniet
to a certain nozzle the specific enthalpy of the fluid is 3025 kJ /kg and the velocity
is 60 m/ s. At the exit from the nozzle the specific enthalpy is 2790 kJ /kg. The nozzle
is horizontal and there is a negligible heat loss from it. Calculate:
(i) the velocity of the fluid at exit;
.
.
•
2
(ii) the rate of flow of fluid when the mlet area 1s 0.1 m and the specific volume at
inlet is 0.19 m 3 /kg;
(iii) the exit area of the nozzle when the specific volume at the nozzle exit is 0.5 m 3 /kg.
9
Applied Thermodynamics
1.14
Solution
+ W
=
(h2 - h1) + {C2 2
( i)
Q
ie
c 2 2 = 60 2
-
=O
= 688 m/s
Mass flow rate= C1A1/v1 = 60x0.1/0.19
= 31.6 kg/s
(iii) Ex~t area, A2 = 31.6x0.5/688
= 0.0229 m2
10
2
C1 )/2
2xl0 3 x(2790 - 3025) = 47.36x1o 4
Final velocity, C2
(ii)
-
2
The Working Fluid
2.1
Table 2.4 Data for
Problem 2.1
Complete Table 2.4 (p. 48) using steam tables. Insert a dash for irrelevant items, and
interpolate where necessary.
p
t
V
(bar)
(°C)
(m 3 /kg)
90
20
5
34
3
15
130
38.2
2.3
44
X
Degree of
superheat
2.361
h
u
(kJ /kg)
(kJ /kg)
2799
0.3565
188
81.3
200
250
2400
0.9
0.85
0.152
1.601
297
300
420
3335
0.8
0.95
The completed table is given on p. 50 as Table 2.6.
Solution
The completed table is given in the text-book as
Table 2.6.
Line 1:
at t
= 90
°C , v
= 2.361
m3 /kg
= Vg,
and hence the
steam is dry saturated.
Line 2:
at p
= 20
bar, h
= 2799
kJ/kg
= hg,
and hence the
steam is dry saturated with t = 212.4 °C.
11
Applied Thermodynamics
2.1
Line 3:
at p = 5 bar. v = 0 _3565 m3/kg which is less than
Vg •
= 151.8 °c;
and hence the steam is wet with t
the dryness fra ction, x = 0.3565/0.3748 = 0.951,
then,
=
h
= 640
ht + Xhfg
+ 0.95lx2109
= 2646 kJ/kg
=
u = Uf + XUfg
639 + 0.951(2562
-
639)
= 2471 kJ/kg.
Line 4:
at t = 188 °c, u = 2400 kJ/kg which is less than u 9
= 2588 kJ/kg and hence the steam is wet with P = 12
= u/(Ug
bar; x
- Uf)
= 2400/(2588
-
797)
= 0.895,
and hence, v = 0.895x0.1632 = 0.1461 m3 /kg, and
h
= 798
+ 0.895xl986
= 2576
kJ/kg.
Line 5:
the steam is wet with t
= 240.9
3
0.0529 m /kg, h = 1042 + 0.9xl761
= 1038
and u
+ 0.9(2603 -
1038)
= 0.9x0.05875
°c; v
= 2627 kJ/kg,
= 2447
kJ/kg.
Line 6:
the steam is wet with p
=
2 75
-
= 0.5
bar; v
= 0.85x3.239
3
m /kg, h = 340 + 0.8Sx2305 = 2300 kJ/kg, and
u = 340 + 0.85(2483 - 340) = 2165 kJ/kg.
Line 7:
at P = 3 bar
133.5
•
t
= 200 °C Which is greater than tg
oc and hence the steam 1.·s
superheat tables hand
Line 8:
at P = 15 bar, v
than
12
Vg
= 0.1317
superheated; from
u can be read off.
=
0 152 3
·
m /kg Which is greater
m
9 a nd hence the steam is
3/k
=
The working fluid
2.1
superheated at t = 250 °C; from superheat tables h
and u can be read off.
Line 9:
at P = 130 bar, h = 3335 kJ/kg which is greater
than hg and hence the steam is superheated with t =
500 °C; from superheat tables v and u can be read
off.
Line 10:
at t
= 250 •c,
= 1.601
v
m3/kg which is greater
than v 9 and hence the steam is superheated at p =
1.5 bar; from superheat tables hand u can be read
off.
Line 11:
the steam is wet with x = 0.8; at p= 38.2 bar
values of t 9
,
v9
,
ht, ht 9
,
Uf
and u 9 can be found
by interpolating between the values at 38 bar and
40 bar and the values of v, hand u found using the
dryness fraction; for example,
Vg = 0.05246 - (38.2 - 38)(0.05246 - 0.04977)
(40
- 38)
= 0.05219 m3/kg
ie
v
=
0.8x0.05219 = 0.04175 m3/kg
Line 12:
the steam is wet with x = 0.95; as for line 11, the
values can be found by interpolating, in this case
between t = 295 •c and t = 299.2 °C.
Line 13:
at p = 2.3 bar, t = 300 °C which is greater than t 9
and hence the steam is superheated; in superheat
tables values are tabulated for P = 2 bar and p = 3
13
Applied Thermodynamics
2.1
bar, therefore it is necessary to interpolate
between these va 1 ues at t = 300 °C; for example,
V
= 1 _31 6 - o.3(1.316 - 0.8754) = 1.184 m3/kg
Line 14:
at P = 44 bar, t= 420 °c which is greater than t 9
and hence the Stea m is superheated; in superheat
tables values are tabulated for pressures of 40 bar
and 50 bar at temperatures of 400 °C and 450 °c
therefore a double interpolation is required; for
example,
at p = 40 bar and t
V
= 420 °C,
= 0.0733 + 20(0.08 - 0.0733) = 0.076 ml/kg
50
at p
= 50
V
ie
2.2
Solution
bar and t
= 420
°C,
= 0.0578 + 20(0.0632
50
0.0578) = 0.06 ml/kg
v = 0.076 - 0.4(0.076 - 0.06) = 0.0696 m3 /kg
A vessel of volume 0.03 m 3 contains dry saturated steam at 17 bar. Calculate the
mass of steam in the vessel and the enthalpy of this mass.
At 17 bar, v 9
= 0.1167 m3/kg, therefore,
Mass of steam= 0.03/0.1167 = 0.257 kg
Also, h 9
= 2795 kJ/kg, therefore,
H = mh = 0.257x2795 = 718 kJ
2.3
Solution
Steam at 7 bar and 250 °C enters a pipeline and flows along it at constant pressure.
If the steam rejects heat steadily to the surroundings, at what temperature will droplets
of water begin to form in the vapour? Using the steady-flow energy equation, ao<l
neglecting changes in velocity of the steam, calculate the heat rejected per kilogram of
steam flowing.
Water droplets will begin to form at the saturation
temperature corresponding to 7 bar.
14
The working fluid
2.3
ie
= 165 ·c
t
From superheat tables. h 1 = 2955 kJ/kg, and for dry
saturated steam at 7 bar. h 2 = 2764 kJ/kg, then.
Q = 2764 -
ie
2.4
2955 = -
191 kJ/kg
Heat rejected= 191 kJ/kg
0.05 kg of steam at 15 bar is contained in a rigid vessel of volume 0.0076 m 3 • What
is the temperature of the steam? If the vessel is cooled, at what temperature will the
steam be just dry saturated? Cooling is continued until the pressure in the vessel is
11 bar; calculate the final dryness fraction of the steam, and the heat rejected between
the initial and the final states.
Solution
= 0.0076/0.05 = 0 ~152
V
m3 /kg
Hence the steam is superheated since v > v 9
;
from
superheat tables at p = 15 bar and v = 0.152 m3/kg
t
= 250 ·c.
When cooling takes place at constant volume the
steam is dry saturated when v = v 9
= 0.152 m3 /kg.
Hence interpolating,
t
=
191.6 -
=
191.4
(0.1520 - 0.1512)(191.6 -188)
(0.1632 - 0.1512)
·c
At 11 bar and v = 0.152 m3 /kg the steam is wet with
a dryness fraction of, x = 0.152/0.1774 = 0.857,
and therefore,
u2
= 780 + 0.857(2586 - 780) = 2327.4 kJ/kg
Initially at p = 15 bar and v = 0.152 m3 /kg, from
superheat tables,
u1
= 2697 kJ/kg
Hence,
Q
ie
= 2327.4 - 2697 = - 369.6 kJ/kg
Heat rejected= 369.6x0.05 = 18.5 kJ
15
Applied Thermodynamics
2.5
. ·ven in ref. 2.1, calculate:
Using the tables for ammond1a gt 'fie volume of ammonia at 0.7177 bar, dryness fraction
(i) the specific enthalpy an spect
0 9·
.
lume of ammonia at 13
saturated;
. '
'fl
th 1 y and spec16c vo
(ii) the spec1 c en a P
• t 7 529 bar 30 °C.
(iii) the specific enthalpy of ammonta a .
,
cc
(i)
Solution
h = 0 + o.9x1390 = 1251 kJ/kg
V
0.9xl.552
:
=
1.397 m3/kg
(ii)
°c and 14 •c,
Interpolating between 12
h
= 1457
kJ/kg, and v
= 0.1866
m3/kg
(iii)
The vapour is superheated hence interpolating,
h = 1459.5 + (30 - 16)(1591.7 - 1459.5)
so
= 1496.5 kJ/kg
2.6
Using the property values for refrigerant HF A 134a given in Table 2.5, calculate:
(i) the specific enthalpy and specific volume of HF A 134a at -8 °C, dryness fraction 0.85;
(ii) the specific enthalpy of HFA 134a at 5.7024 bar, 35 °C.
Table 2.5 Data for
Problem 2.6
Superheat
values degree
of superheat
20K
Saturation values
Solution
tg
Pg
Vg
hr
(QC)
(bar)
(m 3 /kg)
(kJ/kg)
-10
-5
20
2.0051
2.4371
5.7024
0.098
0.081
0.036
86.98
93.46
126.92
hg
h
(kJ/kg)
288.86
291.77
306.22
308.64
312.05
328.93
(i )
Interpolating,
at -10 °c,
h
= 86.98 + 0.85(288.86 - 86.98)
= 258.58 kJ/kg
16
The working fluid
2.6
at -5 °c,
h = 93.46 + 0.85(291.77 - 93.46)
= 262.02
Therefore at -8 °c,
h
= 258.58 + 0.4(262.02 - 258.58)
= 259.96 kJ/kg
at -10 °c
V
at -5
= 0.85x0.098 = 0.0833 m3 /kg
·c,
V
= 0.85x0.081 = 0.06885 m3/kg
Therefore at -8
V
oc,
= 0.0833 - 0.4(0.0833 - 0.06885)
= 0.0775 m3 /kg
(ii)
Interpolating,
h = 306.22 + (35 - 20)(328.93 - 306.22)
20
= 323.25 kJ/kg
2.7
The relative molecular mass of carbon dioxide, CO 2 , is 44. In an experiment the
value of y for CO 2 was found to be 1.3. Assuming that CO 2 is a perfect gas, calculate
the specific gas constant, R, and the specific heat capacities at constant pressure and
constant volume, cP and CvR
Solution
= 8.3143/44 =
0.189 kJ/kg K
=
Cp - Cv
Also,
Cp/Cv = 1. 3
Therefore,
l.3Cv - Cv = 0.189
ie
cv
= 0.189/0.3 =
0.63 kJ/kg K
Cp = l.3x0.63 = 0.819 kJ/kg K
17
Applied Thermodynamics
2.8
5 3
Calculate the internal energy and enthalpy of 1 kg _of_ air occupying o.o m at 20 bar.
5
If the internal energy is increased by 120 kJ as the air is compressed to obar, calculate
the new volume occupied by 1 kg of the air.
=
T
Solution
pV/mR
=
2ox1osxo.05/lx287.l
=
348.3 K
Therefore,
250.1 kJ/kg
u = cvT = 0.71Bx348.3 =
350.1 kJ/kg
h = Cpl = l.005x348.3 =
ie
U2
kJ/kg
= 250.l + 120 = 370.l
T2
= 370.1/0.718 = 515.5 K
Therefore,
= lx287.lx515.5/50xl0 5 = 0.0296
v2
2.9
3
m
Oxygen, 0 2 , at 200 bar is to be stored in a steel vessel at 20 °C. The capacity of the
vessel is 0.04 m 3 . Assuming that 0 2 is a perfect gas, calculate the mass of oxygen
that can be stored in the vessel. The vessel is protected against excessive pressure by
a fusible plug which will melt if the temperature rises too high. At what temperature
must the plug melt to limit the pressure in the vessel to 240 bar? The molar mass of
oxygen is 32 kg/kmol.
Solution
R
= 8.3143/32 = 0.26
kJ/kg K
Mass of oxygen= 2oox1osxo.04/0.26x293
= 10.5
kg
At constant volume,
T2 = T1p2/p1
ie
2.10
Solution
limiting temperature= 351.6 -
K
273 = 78. 6 °C
Whe~ a c~rtain perfect gas is heated at constant pressure from 15 cc to 95 oc the heat
required
1s 1136 kJ /kg · Wh en t he same gas 1s
· heated at constant volume' between
h
t e same temperatures the heat required is 808 kJ /kg. Calculate c c y R and the
mo1ar mass of the gas.
P' v• '
At constant pressure,
Q
ie
-
Cp (
t2
- ti)
= 1136/(95
Cr
At constant volume,
Q
ie
18
= 293x240/200 = 351.6
Cv
=
- 15) = 14.2 kJ/kg K
- t1)
= 808/(95 - 15)
Cv (
t2
=:
10.1 kJ/kg K
The working fluid
2.10
Then,
isentropic index= cp/cv = 14.1/10.1 = 1.405
R = Cp - Cv = 14.2 - 10.1 = 4.1 kJ/kg K
molar mass= 8.3143/4.1 = 2.028 kg/kmol
2.11
Solution
In an air compressor the pressures at inlet and outlet are 1 bar and 5 bar respectively.
The temperature of the air at inlet is 15 °C and the volume at the beginning of
compression is three times that at the end of compression. Calculate the temperature
of the air at outlet and the increase of internal energy per kg of air.
T2 = T1x~x'.'{_i_ = 2e8x5xl = 480 K = 207 °C
Pl V2
lx3
T1) = 0.718(480 - 288)
= 138 kJ/kg
2.12
A quantity of a certain perfect gas is ·compressed from an initial state of 0.085 m 3 ,
1 bar to a final state of 0.034 m 3 , 3.9 bar. The specific heat at constant volume is
0. 724 kJ /kg K, and the specific heat at constant pressure is 1.020 kJ /kg K. The observed
temperature rise is 146 K. Calculate the specific gas constant, R, the mass of gas
present, and the increase of internal energy of the gas.
Solution
R = Cp - cv = 1 . 02 - O. 724 =
I2._
T1
=
~X'.Y'..2_
Pt
Vt
o. 296
kJ / kg K
= 3.9X0.034 = 1.56
0.085
Also,
T2 - T1
= 146 K
Therefore,
l.S6T1 - T1 = 146
ie
T1
= 146/0.56 = 261 K
Therefore,
mass of gas= p1V1/RT1
= lxl05x0.085/0.296xl03x~61
= 0.11 kg
and,
U2 - U1
= mcv(T2 - T1)
= O.llx0.724x(l46)
= 11. 63 kJ
19
Reversible and Irreversible
Processes
3.1
Solution
1 kg of air enclosed in a rigid container is initially at 4.8 bar and 150 °C. ~he container
is heated until the temperature is 200 °C. Calculate the pressure of the air finally and
the heat supplied during the process.
At constant volume,
= 5.37 bar
p2 = 4.8x473/423
Q
= cv(T2 - Tt) = 0.718(200 -150)
= 35.9 kJ/kg
3.2
Solution
A rigid vessel of volume 1 m 3 contains steam at 20 bar and 400 °C. The vessel is
cooled until the steam is just dry saturated. Calculate the mass of steam in the vessel,
the final pressure of the steam, and the heat rejected during the process.
From superheat tables at 20 bar and 400 °c,
v1
= O. 1511
m3 /kg
Therefore,
Mass of steam= 1/0.1511 = 6.62 kg
At
Vg
= 0.1511 m3 /kg,
P2
interpolating,
= 13 + (1/104) = 13.01 bar
From superheat tables,
u1 = 2946 kJ/kg
uz , =
Ug
at 13.01 bar = 2590 + (3/104)
= 2590.03 kJ/kg
Q
ie
= m(uz
-
Ul )
= - 2355 kJ
Heat rejected
= 6.62(2590.03 - 2946)
= 2355 kJ
20
I
J
Reversible and irreversible processes
3.3
Oxygen (molar mass 32 kg/kmol) expands reversibly in a cylinder behind a piston
at a_ c?~stant pressure of 3 bar. The volume initially is 0.01 m 3 and finally is 0.03 m 3 ;
the m1t1al temperature is 17 °C. Calculate the work input and the heat supplied during
the· expansion. Assume oxygen to be a perfect gas and take cP = 0.917 kJ/kg K.
Solution
Work input= 3xl0 5 x(O.Ol - 0.03)
= - 6000 Nm= - 6 kJ
At constant pressure,
T2 = 290x0.03/0.0l = 870 K
Also,
R
=
8314.3/32
=
259.82 J/kg K
mass of oxygen= 3xl05x0.01/259.82x290
= 0.0398 kg
Therefore,
= mcp(T2
Q
T1)
= 0.0398x0.917x(870 - 290)
= 21.18 kJ
3.4
Steam at 7 bar, dryness fraction 0.9, expands reversibly at constant pressure until the
temperature is 200 °C. Calculate the work input and heat supplied per unit mass of
steam during the process.
Solution
v1
= o. 9x0. 2728 = o. 2455 m3 /kg
At 7 bar and 200 °C the steam is superheated and
from superheat tables,
v2
= 0.3001 m3 /kg
Therefore,
Work input= 7xl0 5 x(0.2455 - 0.3001)
= - 38.2 kJ/kg
h1 = 697 + 0.9x2067 = 2557.3 kJ/kg
From superheat tables,
h2 = 2846 kJ/kg
Then,
Heat supplied= 2846 - 2557.3 = 288.7 kJ/kg
21