SECOND LAW OF THERMODYNAMICS AND ENTROPY 247dharm
/M-therm/th5-2.pm5Example 5.15. Two Carnot engines work in series between the source and sink tempera-
tures of 550 K and 350 K. If both engines develop equal power determine the intermediate tem-
perature.
Solution. Fig. 5.17 shows the arrangement of the system.
Temperature of the source, T 1 = 550 K
Temperature of the sink, T 3 = 350 K
Intermediate temperature, T 2 :
The efficiencies of the engines HE 1 and HE 2 are given by
η 1 =
W
Q 1 =TT
T12
1−
=
W
QW 2 + ...(i)η 2 = W
Q 2
= TT
T23
2− = W
QW 3 + ...(ii)
From eqn. (i), we getW = (Q 2 + W)
TT
T12
1F −
HGI
KJ∴ W
TT
T
1 12
1− −
F
HGI
KJL
NM
MO
QP
P= Q
TT(^2) T
12
1
F −
HG
I
KJ
∴ W
T
T
2
1
F
HG
I
KJ
= Q
TT
(^2) T
12
1
F −
HG
I
KJ
∴ W = Q TT
(^2) T
12
2
F −
HG
I
KJ
...(iii)
From eqn. (ii), we get
W = Q
TT
(^2) T
23
2
F −
HG
I
KJ
...(iv)
Now from eqns. (iii) and (iv), we get
T 1 – T 2 = T 2 – T 3
2 T 2 = T 1 + T 3 = 550 + 350
∴ T 2 = 450 K
Hence intermediate temperature = 450 K. (Ans.)
Example 5.16. A Carnot heat engine draws heat from a reservoir at temperature T 1 and
rejects heat to another reservoir at temperature T 3. The Carnot forward cycle engine drives a
Carnot reversed cycle engine or Carnot refrigerator which absorbs heat from reservoir at tem-
perature T 2 and rejects heat to a reservoir at temperature T 3. If the high temperature T 1 = 600 K
and low temperature T 2 = 300 K, determine :
(i)The temperature T 3 such that heat supplied to engine Q 1 is equal to the heat absorbed
by refrigerator Q 2.
(ii)The efficiency of Carnot engine and C.O.P. of Carnot refrigerator.
Solution. Refer Fig. 5.18.
Temperature, T 1 = 600 K
Temperature, T 2 = 300 K
Fig. 5.17