TITLE.PM5

(Ann) #1
VAPOUR POWER CYCLES 591

dharm
\M-therm\Th12-4.pm5

= 838.46 kJ/kg
(WT)total = 11.06 × 84.92 + 838.46 = 1777.67 kJ/kg
Neglecting pump work,

ηηηηηoverall =

1777.67
3853.26 = 0.4613 or 46.13%. (Ans.)

ADDITIONAL / TYPICAL EXAMPLES

Example 12.23. The following data relate to a regenerative steam power plant generating
22500 kW energy, the alternator directly coupled to steam turbine :
Condition of steam supplied to the steam turbine ... 60 bar, 450°C
Condenser vacuum ... 707.5 mm
Pressure at which steam is bled from the steam turbine ... 3 bar
Turbine efficiency of each portion of expansion ... 87 per cent
Boiler efficiency ... 86 per cent
Alternator efficiency ... 94 per cent
Mechanical efficiency from turbine to generator ... 97 per cent
Neglecting the pump work in calculating the input to the boiler, determine :
(i)The steam bled per kg of steam supplied to the turbine.
(ii)The steam generated per hour if the 9 percent of the generator output is used to run the
pumps.
(iii)The overall efficiency of the plant.
Solution. The schematic arrangement of the steam power plant is shown in Fig. 12.38 (a),
while the conditions of the fluid passing through the components are represented on T-s and h-s
diagrams as shown in Figs. 12.38 (b) and (c). The conditions of the fluid entering and leaving the
pump are shown by the same point as the rise in temperature due to pump work is neglected.
Given : Power generated = 22500 kW ;
p 1 = 60 bar ; t 1 = 450°C ; p 2 (= p 2 ′) = 3 bar ;


p 3 (= p 3 ′) = 760 707 5
760

−. × 1.013 = 0.07 bar ; η
turbine (each portion) = 87% ;
ηboiler = 86% ; ηalt. = 94%, ηmech. = 97%
l Locate point 1 corresponding to the conditions : p 1 = 60 bar ; t 1 = 450°C on the h-s chart
(Mollier chart).
From h-s chart ; we find : h 1 = 3300 kJ/kg.
l Draw vertical line through point 1 till it cuts the 3 bar pressure line, then locate point 2.
∴ h 2 = 2607 kJ/kg

Now, ηturbine = 0.87 = hh
hh

12
12

−′

or 0.87 =^3300
3300 2607

−′ 2

h

∴ h 2 ′ = 2697 kJ/kg
l Locate the point 2 on the h-s chart as enthalpy and pressure are known and then draw a
vertical line through the point 2 till it cuts the 0.07 bar pressure line and then locate the
point 3.
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