250 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-2.pm5

∴ Q

Q

1

4

= T

T

TT

TT

1

4

43

12

−

−

F

HG

I

KJ

= −

−

F

HG

I

KJ

1350

350

350 278

1350 350

= 0.278

or Q 4 =

Q 1

0278.

= 3.6 Q 1

and Q 2 =

T

T

(^2) Q
1
× 1 =
350
1350 1
Q = 0.259 Q 1
∴ Q 4 + Q 2 = (3.6 + 0.259) Q 1 = 100
∴ Q 1 =^100
36 0259..+
= 25.9 kJ. (Ans.)
CLAUSIUS INEQUALITY
+Example 5.18. 300 kJ/s of heat is supplied at a constant fixed temperature of 290°C to a
heat engine. The heat rejection takes place at 8.5°C. The following results were obtained :
(i)215 kJ/s are rejected.
(ii)150 kJ/s are rejected.
(iii)75 kJ/s are rejected.
Classify which of the result report a reversible cycle or irreversible cycle or impossible
results.
Solution. Heat supplied at 290°C = 300 kJ/s
Heat rejected at 8.5°C : (i) 215 kJ/s, (ii) 150 kJ/s, (iii) 75 kJ/s.
Applying Clausius inequality to the cycle or process, we have :
(i)
δQ
cycle∑T^ = + − +
300
290 273
215
8 5 273.
= 0.5328 – 0.7637 = – 0.2309 < 0.
∴ Cycle is irreversible. (Ans.)
(ii) δQ
cycle∑T

• −


• 
  

  300
  290 273
  150
  8 5 273.
  = 0.5328 – 0.5328 = 0
  ∴ Cycle is reversible. (Ans.)
  (iii)
  δQ
  cycle∑T 8.

  
  


• 
  

  −

  
  


• 
  

  300
  290 273
  75
  5 273
  = 0.5328 – 0.2664 = 0.2664 > 0.
  This cycle is impossible by second law of thermodynamics, i.e., Clausius inequality. (Ans.)
  Example 5.19. A steam power plant operates between boiler temperature of 160°C and
  condenser temperature of 50°C. Water enters the boiler as saturated liquid and steam leaves the
  boiler as saturated vapour. Verify the Clausius inequality for the cycle.
  Given : Enthalpy of water entering boiler = 687 kJ/kg.
  Enthalpy of steam leaving boiler = 2760 kJ/kg
  Condenser pressure = 0.124 × 105 N/m^2.