Question

Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one, and reheat. Steam enters the first turbine stage at 12 MPa, 480°C, and expands to 2 MPa. Some steam is extracted at 2 MPa and fed to the closed feedwater heater. The remainder is reheated at 2 MPa to 440°C and then expands through the second-stage turbine to 0.3 MPa, where an additional amount is extracted and fed into the open feedwater heater operating at 0.3 MPa. The steam expanding through the third-stage turbine exits at the condenser pressure of 33 kPa. Feedwater leaves the closed heater at 210°C, 12 MPa, and condensate exiting as saturated liquid at 2 MPa is trapped into the open feedwater heater. Saturated liquid at 0.3 MPa leaves the open feedwater heater. Each turbine stage and the pumps have an isentropic efficiency of 80%. Determine for the cycle: (a) the heat transfer to the working fluid passing through the steam generator, in kj per kg of steam entering the first stage turbine. (b) the percent thermal efficiency. (c) the heat transfer from the working fluid passing through the condenser to the cooling water, in kj per kg of steam entering the first stage turbine.

Answer 1

Heat transfer from working condense is fluid through 11-y'-y") (hu-h5) mi E (1-0:162-0.123) (1975-5-15153) = 1304.14 KJ/Kg (And)

For any further clarification please comment below