Solutions to Exercises in Applied Thermodynamics
Set 3
1.
General equation of continuity for an extensive property
of an open system with
n
streams transporting
i
into the system is
n
=
i
i
=1
The rate of change
,
of the extensive property is given by
n
=
i
i
=1
Applications
i)
Conservation of mass
The T

elbow of hot and co
ld water shower serves as the mixing chamber for the hot
and cold water streams. Under steady flow conditions the net mass flow will be zero.
Hot water in
Cold water in
(ii) Conservation of matter in a combustion chamber
In a combustion cham
ber, under steady flow conditions, net inflow of matter is zero.
Also conservation can be applied to each chemical element.
iii)
Conservation of electric charge
The junction of an electric circuit cannot store electric charge and conservation of
electric charge is stated as Kirchoff’s law (which, in this case, is
i
1
+
i
2
+
i
3
+
i
4
= 0
).
i
4
i
3
i
2
i
1
Products of combustion out
Air in
Fuel in
Warm water to
shower
Open system
2.
Properties common to two systems in equilibrium
i) Simple thermodynamic systems: temperature
ii) Electrically charged bodies: electri
c potential
iii) Fluids in separate vessels connected by a valve: pressure
3
p
i
= 300 kPa
V
i
= 150 l
p
o
=105 kPa
Given that theprocess is fully rtesisted,
FIG.Q3:ABSOLUTE PRESSURE VS VOLUME
By graphical interpolation of F
igure 1 to give 6 equal intervals of
V
at 100 litres each,
Table Q3a: Absolute pressure

volume relationship
p
/ kPa
405
245
185
155
115
85
70
V
/
l
150
250
350
450
550
650
750
Using Simpson’s rule, for any even number (
2n)
of equal width strips width
x
, area
A
is
given by,
A
=
x
(
y
1
+ 4
y
2
+2
y
3
+ . . . + 4
y
2n
+
y
2n+1
) / 3
Work done,
W
= area under
p

V
curve in Figure Q3
=
–
100 [405+ 4(245+ 155+ 85) + 2(185+115) + 70] / 3
=
–
100 500 J =
–
100 kJ
The result could be obtained with the tabulated values
, ignoring
V
= 200 litres to give
Table 3b and very nearly the same result as above
Table Q3b: Absolute pressure

volume relationship
p
/ kPa
405
210
155
95
70
V
/
l
150
300
450
600
750
Work done,
W
= area under
p

V
curve in Figure Q3
=
–
150 [405+ 4(210+
95) + 2(155) + 70] / 3
=
–
100 250 J =
–
100 kJ
Shaft work =
W
s
=
W
–
p
o
(
V
1
–
V
2
) =
–
100 kJ
–
105 (150

750)/1000 =
–
37 kJ
4.
The expression is valid only for fully resisted expansion of a closed system.
(a)
i)
The expression cannot be used for a p
artly resisted expansion.
ii) The expression cannot be used for an un

resisted expansion.
iii) The expression can be used for a fully resisted expansion.
(b)
i)
Friction does not affect the process within the closed system comprising the gas.
Thus,
the expression can be used to find the work done on the system.
ii)
The expression cannot be used to find the work done on the system comprising
the surroundings, as friction occurs within the surroundings and affects the work
transmitted across the pist
on.
(c)
The expression is not for open systems and therefore cannot be used for any flow
process.
5
For an ideal gas,
pV
=
mRT
For the process,
pV
n
=
K
Therefore, work done in process 1

2 =
=
=
(
p
2
V
2
–
p
1
V
1
) / (
n

1)
For a constant pressure process,
p
= constant. Therefore
n
= 0.
p
2
=
p
1
=
p
W
1

2
=
(
p
2
V
2
–
p
1
V
1
) / (
n

1)
= p (V
1

V
2
)
For a constant volume process,
V
= constant. Therefore
n
=
.
W
1

2
= 0
For const
ant temperature process,
n
= 1. The formula,
W
1

2
=
(
p
2
V
2
–
p
1
V
1
) / (
n

1)
cannot be used, and the result is obtained by integration as
W
1

2
=
mRT
ln
(V
1
/V
2
)
6
Cyclic process
: There is no net change in the system boundary during a cycle. So th
ere is
no work done in displacing the matter in the surroundings. Thus the use of either gauge
pressure or absolute pressure will give the same result.
Non

cyclic process
: Here the use of gauge pressure will give the shaft work or useful work
and the use
of absolute pressure the work interaction across the system boundary.
Set 5
1
The boundary interactions including the unaccounted heat interaction rate
Q
x
are as
identified in the figure above. W
s
and W
s
are, resp
ectively the work output rates to the
shaft and the dynamo, Q
in
is
the heat supply rate, and Q
c
is the heat removal rate by the
cooling water.
Rate of work transfer through shaft = W
s
=
= 40*2
*4000/60 W
= 16.8 kW
Power delivered to dynamo =
W
d
= 0.5 kW (by data)
Total work interaction rate
= 16.8 kW + 0.5 kW =
17.3 kW
Ratio of shaft power to heat input
= 17.3 / 68
=
0.254
(or 2
5.4%)
Applying the first law of thermodynamics to the system,
Q
in
–
Q
c
–
Q
x
–
W
s

W
d
= 0
Q
x
= 68

12.5

17.3
=
38.2 kW
2
Consider the rigid insulated sealed vessel shown above. There is no boundary interaction
Therefore there is no chan
ge energy content.
When the system is cooled there is a heat interaction with heat flow to the surroundings,
but no work. Therefore the energy content will decrease.
Q
in
= 68 kW
System boundary
Q
c
Q
x
W
s
W
d
partition system boundary
acid water mix
ture
Before breaking of partition
After breaking of partition
3
Data:
m
= 0.2 kg
R
= 287 J/kg K
= 1.4
p
1
= 800 kPa,
T
1
= 1000 K
Using ideal gas equation,
V
1
=
m R T
1
/
p
1
= 0.0717 m
3
= 71.7 litres
Table of properties of the air
State
p
/kPa
V
/
l
T
/ K
1
800
71.7
1000
2
100
407
709
(Numbers in red are directly obtained from the data. Numbers
in green are calculated
values).
C
v
=
R
/ (
–
1) = 717 J/kg K
C
p
=
R
/ (
–
1) = 1004 J/kg K
V
1
=
m R T
1
/
p
1
= 0.0717 m
3
= 71.7 litres
According to the law of expansion,
p
1
V
1
1.2
=
p
2
V
2
1.2
V
2
= (100/800)
1/1.2
V
1
= 407
l
Using ideal gas e
quation,
T
2
=
p
2
V
2
T
1
/
p
1
V
1
= 709 K
Work done on the system,
W
1

2
= (
p
2
V
2
–
p
1
V
1
) / (
n
–
1) =
m R
(
T
2
–
T
1
) / (
n
–
1)
= 0.2*287* (709

1000) / 0.2
=
–
83.3 kJ
Change in internal energy
U
=
m C
v
(
T
2
–
T
1
) = 0.2*
717 (709

1000) J =
–
41.7 kJ
By first law of thermodynamics,
U
=
Q
–
W
Heat interaction,
Q
=
–
41.7
–
(
–
83.3) =
41.6 kJ
pV
1.2
= constant
V
p
1
2
4
Data:
V
= 1 litre
R
= 287 J/kg K
= 1.4
p
1
= 100 kPa,
T
1
= 303 K
Usi
ng ideal gas equation,
m
=
p
1
V
1
/
R T
1
= 1.15*10

3
kg = 1.15 g
C
v
=
R
/ (
–
1) = 717 J/kg K
C
p
=
R
/ (
–
1) = 1004 J/kg K
Table of properties of the air
State
p
/kPa
V
/
l
T
/ K
1
100
1.0
303
2
201
0.607
370
(Numbers in red are directly obtaine
d from the
data. Numbers in green are calculated values).
V
1
–
V
2
= (
d
2
/4)*
x
, where
x
is the inward displacement of the piston.
V
2
= 1
–
(
*0.1*0.1*0.05 / 4 * 10
3
) litre
= 0.607
l
According to law of expansion,
p
1
V
1
=
p
2
V
2
p
2
= 100*
(1/0.607)
= 201 kPa
Using ideal gas equation,
T
2
=
p
2
V
2
T
1
/
p
1
V
1
= 367 K
Work done on the system,
W
1

2
= (
p
2
V
2
–
p
1
V
1
) / (
n
–
1) =
m R
(
T
2
–
T
1
) / (
n
–
1)
= 1.15*10

3
*287* (370

303) / 0.4
=
55 J
Change in
internal energy
U
2
–
U
1
=
m C
v
(
T
2
–
T
1
) = 1.15*10

3
* 717 (370

303) J = 55 J
By first law of thermodynamics,
U
2
–
U
1
=
Q
1

2
+
W
1

2
Heat interaction,
Q
1

2
=
0 kJ
(A small difference between
W
1

2
and
U
2
–
U
1
is possible
because of rounding errors in the
arithmetic.)
50 mm
pV
†
=
constant
V
p
2
1
position 2 position 1
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