Steam is to be condensed in the condenser of a steam power plant at a temperature of 50°C with cooling water from a nearby lake, which enters the tubes of the condenser at 18°C at a rate of 97 kg/s and leaves at 27°C. Determine the rate of condensation of the steam in the condenser. The heat of vaporization of water at 50°C is hfg = 2382.0 kJ/kg and specific heat of cold water is cP = 4.18 kJ/kg·°C.

The rate of condensation of the steam in the condenser is 1.53 kg/s

Explanation:

Given

Tcond = 50°C

mc = 97kg/s

Tc = 18°C

Th = 27°C

cP of water= 4.18 kJ/kg·°C

hfg = 2382.0 kJ/kg

The rate of heat add to the system is defined from applying  the energy balance on the system

Ein - Eout = ΔEsys = 0

Ein = Eout

Qin + mh1 = mh2

Qin = m(h2-h1)

Qin = mcP(Th - Tc)

Qin = 97*4.18(27-18) = 3649.14 kW

The rate of the condensation of the steam could be defined as the following

Qin = m(condensate)*hfg

m(condensate) = Qin/hfg

m(condensate) = 3649.14/2382 = 1.53 kg/s

answer:

yes that is right have a good day

answer:

c

explanation: