112) Steam is to be condensed on the shell side of a heat exchanger at 30 oC. Cooling water enters the tubes at 15 oC at a rate of 63 kg/s and leaves at 22 oC. Assuming the heat exchanger to be well-insulated, determine (a)the rate of heat transfer (kJ/s) in the heat exchanger and (b) the rate of condensation (kg/s) of the steam.
a). The rate of heat transfer is given by Q = m Cp dT where m = mass flow of cooling water, Cp = the heat capacity of the cooling water, and dT = the difference between the outlet and inlet temperatures. Therefore, Q = (63 kg/s) x (4182 J/kg-K) x (22 - 15 deg C) = 1844262 J/s = 1844 kJ/s. b). The rate of condensation of steam is dependent on the steam pressure and the amount of superheat. Assuming saturated steam at 1 atm (1.01325 bar absolute), the heat balance is: Q = m * Cp * dT + m * Hv. Here m = the rate of steam condensation, Cp = heat capacity of water, dT = difference between the saturated steam temperature and the outlet condensate temperature, and Hv is the heat of vaporization of steam @ 1 atm. So Q = m * ( 4182 J/kg-K) * (100 - 30 deg C) + m * (2257000 J/kg). Solving for m yields 0.72 kg/s. Alternatively, you can use standard steam tables to obtain a more accurate answer.
