Question

To a steam turbine comes 10 kg / s 10 bar steam at the temperature of 400 ° C. The steam leaves the turbine at a pressure of 10 kPa. The isentropic efficiency is 90%, and the other efficiency is a total of 87%.
a. Determine the electrical power supplied by the generator connected to the steam turbine.
b. The steam after the turbine is condensed into saturated liquid in a condenser. How much work must the pump supply to the liquid to raise the condensate pressure from 10 kPa to 10 bar?

Answer 1

3 (400°c) 10 bar 2 91 lokPa 4 L shown curve in. T-S plot steam expanded en turbine 3 to 4 0 but is entropic efficiency extraction value to Point 4 Reduse energy at KJ/kg kol kg 10 bar, 400c (Point - 3) enthalpy hat - 3264.4 S3= 7-467 entropy entropy for Point 4 ip pame as Point -3 Sq = S3 = 7:467 ks liegk Let Dryness fraction xq at Point 4 (10kPa) S4= Sf+XqSfg 7.467 = 0,649 + X4 (71502) = 0.9088 X 4 =

Enthalpy of Point 4 (10kPa) + X ha 19100 + (0-9088)x 2392-9 90% ha= 2366.53 kJ/kg given efficieny of turbine turbine hz - ha 2. actual work gain Ideal work t hiz - h4 hiz-hal = 0.gx (32644 - 2366-53) actual tubine worka 808.087 kJ/kg maso flow rate in= 10 Kg/sec total Actual turbine work = inx (hip-h41) = 8080.87 kW
( A) other total efficiency is given 87% . Overall efficiency of rankine cycle is nearly 30 to 40 % . So considering 87 % is loss in generator .

So work supply by generator

= efficiency × turbine actual work

= 0.87 × 8080.87 KW = 7030.357 KW

(B) pump work from Point 1 (10 kpa ) to Point 2 (10 bar )

V1f (m³/kg) = volume of saturated water at suction of pump (10 kpa)

Pump work = m × V1f × ( P2 - P1)

Pump work = 10 × 0.001011 × (10⁶ - 10⁴)

Pump work = 10008.9 W or 10.0089 KW

Please refer data steam table for above selected data.