Question

12-3 A 250-MW vapor-dominated hydrothermal powerplant uses well steam that is saturated at 450 psia at shutoff. The steam is throttled to a turbine inlet pressure of 140 psia. A direct-contact condenser operatęs at a pressure of 5 psia with a cooling-water inlet temperature of 48°F. The turbine polytropic efficiency is 0.80 and the turbine-generator combined mechanical-electrical efficiency is 0.90. Calculate (a) the steam mass flow rate, in pounds mass per hour, (b) the condenser cooling water mass flow rate, in pounds mass per hour. (c) the cycle thermal efticiency and heat rate, in Blus per kilowatt hour, and (d) the difference between well flow and disposal flow, in pounds mass per hour, if the cooling tower losses by evaporation and drift are 2 percent.

Answer 1

Refer to the given problem. Consider the given data: Warw = 250 MW power of hydrothermal powerplant 17 po= 0.80 ngen = 0.90 Steam saturated at 450 psia, P =450 psia Steam is throttle at inlet pressure at 140 psia, that is, P2 = 140 psia Condenser operate at pressure of 5psia, that is, P, = 5 psia T: = 48 °F Find the power of the plant in BTU/hr. Wactual = 250 MW =(250 x 3412141.16 BTU/hr) (::1 MW = 3412141.16 BTU/hr) =853035250 BTU/hr =853.035x10° BTU/hr Thus, Ww=853.035 x 10 BTU/hr Consider the expression for the ideal output power of the plant. W Substitute 18 = 0.90, poly = 0.80, and Wacua=853.035x10 BTU/hr. 853.035 x10 BTU/hr Wideal = (0.90).(0.80) _853.035 x 10° (0.90).(0.80) = 1184.7 x 10 BTU/hr Thus, Wided = 1184.7 x 10 BTU/hr

Consider the following properties: s, = 1.474 BTU/lbm-R At a 140 psia that is P, = 140 psia. s = s, =1.474 BTU/Ibm-R For this condition, using interpolation method, the enthalpy of steam is given as, h= 437.3 BTU/lbm, and h = h =1110.8 BTU/lbm At exit of turbine, at 5 psia that is P, = 5 psia. S = sz=1.474 BTU/lbm-R For this condition, the enthalpy is given as, h = 900.7 BTU/ibm h, = 130.1 BTU/lbm Tar=162.2 °F (a) Consider the expression for the rate of mass flow of steam. Wide al = myteam (ha - h) Or, Misscom (h-h) Substitute Wided = 1184.7 x 10 BTU/hr , h=h. = 1110.8 BTU/lbm, and m.team 1184.7x10 BTU/hr (1110.8 BTU/lbm-900.7 BTU/Ibm) *

1184.7x10 (1110.8-900.7) = 5.638 x10 lbm/hr Therefore, the rate of mass flow of steam is m.com = 5.638 x 10 lbm/hr. (b) Using energy balance to condenser. Consider the expression for the mass flow rate of cooling heater. m.gam (1 - 1,)=mac,(T. -T) Or, C, (TEX - Ton) Substitute ha = 900.7 BTU/lbm, h, =130.1 BTU/ibm , Tigr = 162.2 °F, mean = 5.638x10lbm/hr , and T: = 48 °F. 5.638 x10 lbm/hr ( 900.7 - 130.1) 1) (c, =1) 1 162.2 - 48 5.638x10° 900.7–130.1) 1.002 1 1 62.2 - 48 mooi = 38.04x10 lbm/hr Therefore, the condenser cooling water mass flow rate is m = 38.04x10 lbm/hr. (c) Consider the expression for the plant thermal efficiency. Wanted WITH = 2 meteam (ha -h;)

Substitute m.com = 5.638 x 10° lbm/hr and, and W = 853.6 x 100 BTU/hr. IT 5.638 x 10' (1110.8-437.3) nx = 0.2247 NA = 22.47% Consider the expression for the difference between well flow and the disposal flow. $=(.001 – m, eam) Substitute mood = 38.04x106 lbm/hr , and mean = 5.638 x 100 lbm/hr. 8=(38.04x10 lbm/hr - 5.638 x100 lbm/hr) 8=(38.04x10® – 5.638x10°) 5 = 32.40 x100 lbm/hr between well flow and the disposal flow is Therefore, the difference 8 = 32.40 x 10° lbm/hr.

Hope it helps!