582 ENGINEERING THERMODYNAMICS

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\M-therm\Th12-3.pm5

Heat supplied = h 1 – hf 2

= 3250 – 173.9 [hf 2 = 173.9 kJ/kg at 0.08 bar]

= 3076.1 kJ/kg

Thermal efficiency =

Work done

Heat supplied

=

1080

3076.1

= 0.351 or 35.1%. (Ans.)

Exhaust steam condition, x 2

= 0.83 (From Mollier chart). (Ans.)

Second case. Refer Fig. 12.30 (b).

(a)

Turbine

Condenser

Reheater

1

2 3

h

s

32 bar

(^1) 410 Cº 3 5.5. bar400 Cº
2
x = 0.935 4
0.08 bar
(b)
Fig. 12.30
From Mollier chart :
h 1 = 3250 kJ/kg ;
h 2 = 2807 kJ/kg ;
h 3 = 3263 kJ/kg ;
h 4 = 2426 kJ/kg.
Work done = (h 1 – h 2 ) + (h 3 – h 4 ) = (3250 – 2807) + (3263 – 2426) = 1280 kJ/kg
Heat supplied = (h 1 – hf 4 ) + (h 3 – h 2 )
= (3250 – 173.9) + (3263 – 2807) = 3532 kJ/kg
Thermal efficiency =
Work done
Heat supplied

1280
3532
= 0.362 or 36.2%. (Ans.)
Condition of steam at the exhaust,
x 4 = 0.935 [From Mollier chart]. (Ans.)
Example 12.21. (a) How does erosion of turbine blades occur? State the methods of
preventing erosion of turbine blades.
(b) What do you mean by TTD of a feed water heater? Draw temperature-path-line dia-
gram of a closed feed water heater used in regenerative feed heating cycle.