TITLE.PM5

290 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-4.pm5

W

Q 3

Q 1

Source-1

500 K

Source-2

400 K

Source-3

300 K

Q 2

Heat

engine

Fig. 5.46
(i) The direction of heat flow from source 1 is known as given in the problem. Assume that
the quantities of heats Q 2 and Q 3 are taken from heat sources and their directions are arbitrarily
chosen.
For the cyclic operation of the engine
dQ
T

F

H

I

z K = 0

∴

Q

T

Q

T

Q

T

1

1

2

2

3

3

+− = 0
and Q 1 + Q 2 – Q 3 = W
∴^1500500 +− 400 QQ^23300 = 0 ...(i)
and 1500 + Q 2 – Q 3 = 200 ...(ii)
Solving eqns. (i) and (ii), we get
Q 2 = – 1600 kJ/min and Q 3 = – 300 kJ/min.
The above values indicate that the direction of Q 2 and Q 3 are reversed. Since Q 2 should be
+v and Q 3 also must be +ve but both are –ve therefore, their assumed directions should be reversed.
The arrangement is shown in Fig. 5.47.

Fig. 5.47