TITLE.PM5

292 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-4.pm5

(∆ S)working fluid in (^) HE = 0
∴ (∆ S)universe = (∆ S)system + (∆ S)reservoir = – 1054.7 + 20508 10 125

. ×−^3 W

Since (∆ S)universe ≥ 0

∴ – 1054.7 + 205.08 10

125

×−^3 W

≥ 0

or – 1054.7 + 1640.64 – 125 W ≥ 0 or 585.94 – 125 W ≥ 0

or 585.94 ≥ 125 W or 125 W ≤ 585.94

i.e., W(max.) = 585.94 × 125 = 73.24 kJ. (Ans.)

Example 5.46. In an insulated duct air is flowing steadily. The pressure and temperature
measurements of the air at two stations A and B are given below :
Station Pressure Temperature
A 140 kPa 60 °C
B 110 kPa 15 °C
Establish the direction of the flow of the air in the duct.
For air assume that :
cp = 1.005 kJ/kg K

h = cp T and (^) Tv =
0.287
p
where p, v and T are pressure (in kPa), volume (in m^3 /kg) and temperature (in K) respectively.
Solution. From property relation,
Tds = dh – vdp
ds = dhT – vdpT
For two states at A and B the entropy changes of the system is given by
ds
cdT
T
dp
s p
s p
T
T
p
p
A
B
A
B
A
B
zzz=−0 287.
∴ sB – sA = 1.005 loge
T
T
B
A – 0.287 loge^
110
140
= 1.005 loge
15 273
60 273

• 
  


• 
  

  F
  H
  I
  K – 0.287 loge^
  110
  140
  = – 0.1459 + 0.0692 = – 0.0767 kJ/kg K
  (∆s)system = – 0.0767 kJ/kg K
  Since the duct is insulated, (∆ s)surroundings = 0
  ∴ (∆ s)universe = – 0.0767 kJ/kg K.
  This is impossible.
  So, the flow is from B to A. (Ans.)
  Example 5.47. 3 kg of water at 80°C is mixed with 4 kg of water at 15°C in an isolated
  system. Calculate the change of entropy due to mixing process.