In a steam power plant, the temperature of the burning fuel is 1100 °C, and cooling water is available at 15 °C. Steam leaving the boiler is at 2 MPa and 700 °C, and the condenser produces a saturated liquid at 50 kPa. The steam lines are well insulated. The turbine and pump operate reversibly and adiabatically. Some of the mechanical work generated by the turbine is used to drive the pump.
a. Draw a T-s diagram of this cycle.
b. What is the net work obtained in the cycle per kg steam generated in the boiler?
c. How much heat is discarded in the condenser per kg steam generated in the boiler?
d. What fraction of the work generated by the turbine is used to operate the pump?
e. How much heat is absorbed in the boiler per kg steam generated?

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

answer:

ya

explanation:

sure ill do it