Answer:
b. 1655.7 KJ/kg ( net work produced )
c. 2324.86 KJ/kg
d. 0.25539 --- 25.5%
e. 3980.63 KJ/kg
Explanation:
Given:-
Condenser exit parameters:
P1 = 50 KPa , saturated liquid
Boiler exit / Turbine exit parameters:
P3 = 2 MPa
T3 = 1100°C
Solution:-
- Adiabatic and reversible processes for pump and turbine are to be applied
- Assume changes in elevation heads within the turbomachinery to be negligible.
- Assume steady state conditions for fluid flow and the use of property tables will be employed.
Isentropic compression of water in pump:
Pump inlet conditions : Pump exit to Boiler pressure:
P1 = 50 KPa, sat liquid P2 = P3 = 2 MPa
h1 = 340.54 KJ/kg s2 = s1 = 1.0912 KJ/kg.K
s1 = 1.0912 KJ/kg.K h2 = 908.47 KJ/kg
- Apply energy balance for the pump and determine the work input ( Win ) required by the pump:
Win = h2 - h1
Win = 908.47 - 340.54
Win = 567.93 KJ/kg
Isentropic expansion of steam in turbine:
Turbine inlet conditions : Turbine exit to condenser pressure:
P3 = 2MPa, T3 = 1100°C P4 = P1 = 50 kPa
h3 = 4889.1 KJ/kg s4 = s3 = 8.7842 KJ/kg.K .. superheated
s3 = 8.7842 KJ/kg.K h4 = hg = 2665.4 KJ/kg
- Apply energy balance for the turbine and determine the work output ( Wout ) produced by the turbine:
Wout = h3 - h4
Wout = 4889.1 - 2665.4
Wout = 2223.7 KJ/kg
- The net work-output obtained from the cycle ( W-net ) is governed by the isentropic processes of pump and turbine.
W_net = Wout - Win
W_net = 2223.7 - 567.93
W_net = 1655.77 KJ/kg ... Answer
- The fraction of work generated by turbine is used to operate the pump. The a portion of Wout is used to drive the motor of the pump. The pump draws ( Win ) amount of work from pump. The ratio of work extracted from turbine ( n ) would be:
n = Win / Wout
n = 567.93 / 2223.7
n = 0.25539 ... Answer ( 25.5 % ) of work is used by pump
- The amount of heat loss in the condenser ( consider reversible process ). Apply heat balance for the condenser, using turbine exit and condenser exit conditions:
Ql = h4 - h1
Ql = 2665.4 - 340.54
Ql = 2324.86 KJ/kg ... Answer
- The amount of heat gained by pressurized water in boiler ( consider reversible process ). Apply heat balance for the boiler, using pump exit and boiler exit conditions:
Qh = h3 - h2
Qh = 4889.1 - 908.47
Qh = 3980.63 KJ/kg ... Answer
