Question

A boiling water reactor had the following characteristics:

System pressure 1,090 psia

Saturated steam production rate 1.5x10^7 lb/hr

Average exit void fraction 38.5%

Slip ratio 1.92

Feed water temperature 285oF

Determine the following:

1) The enthalpy of subcooling.

2) Degree of subcooling

3) Total heat generated by the reactor

Question 2 (20%) A boiling water reactor has the following characteristics: System pressure Saturated steam production rate 1.5 x 10 lb/hr Average exit void fraction Slip ratio Feedwater temperature 1,090 psia 38.5% 1.92 285 oF Determine the following: 1) The enthalpy of subcooling 2) Degree of subcooling. 3) Total heat generated by the reactor.

Answer 1

$ubshhde tale inabove cquation 1.5 x lo Theu fore, the enthalpo SLub teo \SwbG.aling 15b6-285 hea Genea 1- 0-35$

Given that system pressure (P_system) = 1,090 P sia Saturated steam production rate (r) = 1.5 times 10^7 lb/hr Average exit void fraction (n) = 38.5% Slip ratio (s) = 1.92 Feed water temperature (T_F) = 285 degree F Assume, SP. heat of water (P = 1.020BTU/lbM - degree F Consider the boiling curvature of water From above table, the saturation temperature of water at 1090 Psia is 556 degree F 1) The enthalpy of subcooling. Ans To Calculate the enthalpy of subcooling h = M C_P [T_sat -T_F] Here, C_P is the specific heat of water at constant pressure. \r\n Substitute values in above equation. h = 1.5 times 10^7/60 times 60 times 1.020 [556 - 285] h = 1151750[BTU/sec Therefore, the enthalpy of subcooling is 1151750 BTU/sec 2. Degree of subcooling Sol To Calculate degree of subcooling T_subcooling = [T_sat - T_F] Substitute values in above equation T_subcooling = [556 - 285]