TITLE.PM5

VAPOUR POWER CYCLES 599

dharm

\M-therm\Th12-4.pm5

Work done by the pump P 2 ,

WP 2 = vw 2 (1 – m 1 )(150 – 5) × 10^5 × 10–3 kJ/kg

=

1

1000

(1 – 0.179)(150 – 5) × 10^5 × 10–3 = 11.9 kJ/kg

Work done by pump P 3 ,

WP 3 = vw 3 × m 1 × (150 – 40) × 10^5 × 10–3

=

1

1000

× 0.179 (150 – 40) × 10^5 × 10–3 = 1.97 kJ/kg

Total pump work = WP 1 + WP 2 + WP 3

= 0.338 + 11.9 + 1.97 = 14.21 kJ/kg of steam supplied by boiler

∴ Net work done by the turbine per kg of steam supplied by the boiler,

Wnet = 1456.94 – 14.21 = 1442.73 kJ/kg

Heat of feed water extering the boiler

= (1 – m 1 ) × 1611 + m 1 × 1611 = 1611 kJ/kg

Heat supplied by the boiler per kg of steam,

Qs 1 = h 1 – 1610 = 3578 – 1610 = 1968 kJ/kg

Qs 2 = Heat supplied in the reheater

= (1 – m 1 )(h 3 – h 2 ) = (1 – 0.179)(3678 – 3140)

= 441.7 kJ/kg of steam supplied by the boiler

Qst (Total heat supplied) = Qs 1 + Qs 2 = 1968 + 441.7 = 2409.7 kJ/kg

∴ ηthermal =

W

Qst

net=1442 73

2409 7

.
.
= 0.5987 or 59.87%. (Ans.)
Example 12.27. Steam at 70 bar and 450°C is supplied to a steam turbine. After expanding to
25 bar in high pressure stages, it is reheated to 420°C at the constant pressure. Next ; it is expanded in
intermediate pressure stages to an appropriate minimum pressure such that part of the steam bled at
this pressure heats the feed water to a temperature of 180°C. The remaining steam expands from this
pressure to a condenser pressure of 0.07 bar in the low pressure stage. The isentropic efficiency of H.P.
stage is 78.5%, while that of the intermediate and L.P. stages is 83% each. From the above data,
determine :
(i)The minimum pressure at which bleeding is necessary.
(ii)The quantity of steam bled per kg of flow at the turbine inlet.
(iii)The cycle efficiency.
Neglect pump work. (Roorkee University)
Solution. The schematic arrangement of the plant is shown in Fig. 12.42 (a) and the proc-
esses are represented on T-s and h-s diagrams as shown in Figs. 12.42 (b) and (c) respectively.

(i) The minimum pressure at which bleeding is necessary :
It would be assumed that the feed water heater is an open heater. Feed water is heated to
180°C. So psat at 180°C ~− 10 bar is the pressure at which the heater operates.
Thus, the pressure at which bleeding is necessary is 10 bar. (Ans.)
From the h-s chart (Mollier chart), we have :
h 1 = 3285 kJ/kg ; h 2 = 2980 kJ/kg ; h 3 = 3280 kJ/kg ; h 4 = 3030 kJ/kg
h 3 – h 4 ′ = 0.83(h 3 – h 4 ) = 0.83(3280 – 3030) = 207.5 kJ/kg
∴ h 4 ′ = h 3 – 207.5 = 3280 – 207.5 = 3072.5 kJ/kg