Answer:
The rate of condensation of the steam in the condenser is 1.53 kg/s
Step-by-step 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