Solutions Manual for Advanced Engineering Thermodynamics – 4th Edition (Bejan, 2017) | All Chapters Covered

Covers All Chapters




SOLUTIONS MANUAL

,Chapter 1 Solutions



P1.1 This example shows how the First Law can be applied to individual processes and how these
can make up a cycle.

A cycle of events is shown in Fig A.10. It is made up of four processes, and the heat and work
associated with those processes is as given.




Fig A.10: Cycle of events made up of four processes.

Process 12: Q = +10J W = -18J
Process 23: Q = +100J W = 0J
Process 34: Q = -20J W = +70J
Process 41: Q = -10J W = +28J

Calculate the values of Un - U1, the net work, the net heat transfer and the heat supplied for the cycle.

Solution
This problem can be solved by applying the First Law to each of the processes in turn, when the change
in internal energy is dU   Q   W .

Process 12: dU12  U2  U1   Q12   W12  10  18  28J
Process 23: dU23  U3  U2   Q23   W23  100  0  100J
Process 34: dU34  U4  U3   Q34   W34  20  70  90J
Process 41: dU41  U1  U4   Q41   W41  10  28  38J

The change in internal energy relative to the internal energy at point 1, U1, is given by
dU1n  U n  U 1  U n  U n 1  U n 1  U n  2 ......U 2  U1
 dU n 1,n  dU n  2,n 1 ...... dU 12

Hence:
dU12  U 2  U1  28J
dU13  dU12  dU 23  28  100  128J
dU14  dU12  dU 23  dU 34  28  100  (90)  38J
dU11  dU12  dU 23  dU 34  dU 41  28  100  (90)  (38)  0J
The result dU11 = 0 confirms that the four processes constitute a cycle, because the net change of state
is zero.

The net work done in the cycle is
W   W12   W23   W34   W41
cycle
 18  0  70  28  80J
The positive sign indicates that the system does work on the surroundings.

The net heat supplied in the cycle is


Solutions Chapter 1 Page 1
 D E Winterbone Current edition: 13/02/2015

,Chapter 1 Solutions



Q   Q 12   Q23   Q34   Q41
cycle
 10  100  20  10  80J

Hence the net work done and net heat supplied are both 80J, as would be expected because the state of
the system has not changed between both ends of the cycle. It is also possible to differentiate between
the heat supplied to the system, Q > 0, and heat rejected from the system, Q < 0. In this case the
total heat supplied is
 Q   Q12   Q23
cycle
 10  100  110J
while the heat rejected is the sum of the negative heat transfer terms, viz.:
 Q   Q34   Q41 .
cycle
 20  10  30J

It should also be noted that the work done, W, is the difference between the heat supplied and that
rejected. This is an important point when considering the conversion of heat into work, which is dealt
with by the Second Law of Thermodynamics.
__________________________________________________________________________________
P1.2
There are two ways to solve this problem:
1. A simple one based on the p-V diagram
2. A more general one based on the p-V relationship




Work done = Area under lines.
Assuming a linear spring, then a linear (straight) line relationship joins the points.
1 105
Hence, Work done by spring = Area abca = Ws   0.5  4  3  100kJ
2 10
Work done by gas = Area adeca = Area abca + Area adbca
105
= 100  1.0  0.5 
 150kJ
103
More general method is to evaluate the work as W   pdV .
First find the relationship for the spring. If linear pg1  kV1  k ,and pg 2  kV2  k 
p g 2  p g1
k  8bar/m3 , and k   7bar
Thus V2  V1
giving pg  8V  7.


Solutions Chapter 1 Page 2
 D E Winterbone Current edition: 13/02/2015

, Chapter 1 Solutions


δWg  pg dV ,
2
2
 8V 2 
giving Wg    8V  7 dV    7V 
Work done by gas,
1  2 1
 4  2.25  1  7 1.5  1 102  150kJ


δWs   ps dV
2
Work done by spring, 2
V2 
Ws    8V  8  dV  8   V   100kJ
 2 
1  1
The benefit of the latter approach is that it can be applied in the case of a non-linear spring: it is more difficult to
use the simpler approach.
_________________________________________________________________________________________
P1.3
p1 1.5
Pressure of gas is proportional to diameter, i.e. p  d , giving p1  kd1, and hence k    5bar/m
d1 0.3
13
d3  6V 
Volume V  , thus d   
6   
Work done during process, W   pdV
3 d 2 d2
Working in terms of d dV  dd  dd
6 2
d2 k k  d 2 4 d14 
2 2
W   kd
2 1
dd  d 3
.dd    
1
2 2  4 4 
5 
0.334  0.34  105  738J
4 2  
This problem can also be solved in terms of V; however, it cannot be solved using a linear approximation. You
might have been close to the correct solution for P1.3, but it does not work for P1.4.
_________________________________________________________________________________________
P1.4
The problem is the same as P1.3, but the final diameter is 1m.
d2 k k  d 2 4 d14 
2 2
W   kd
2 1
dd  d 3
.dd    
2 2  4 4 
The work done is 1
5  4
 1.0  0.34  105  194759J
4 2  
Using the WRONG APPROACH, assuming a linear relationship, gives
1.5  5 
W
2
 
 d 23  d13 105  165580J
6
_________________________________________________________________________________________
P1.5
ts at 20bar = 212.4C
Initial conditions:
ug  2600kJ / kg ; hg  2799kJ / kg; vg  0.0995m3 / kg
p  20bar; t  500C
Final conditions:
ug  3116kJ / kg ; hg  3467kJ / kg; vg  0.1756m3 / kg
To evaluate the energy added use 1st Law for closed system
Q  dU  W  dU  pdV



Solutions Chapter 1 Page 3
 D E Winterbone Current edition: 13/02/2015