Question

Cold water (Cp = 4.18 KJ/kg*K) enters a heat exchanger at 15 degree C at a rate of 0.5 kg/s, where it is heated by hot air (Cp = 1.0 KJ/kg*K) that enters the heat exchanger at 50 degree C at a rate of 1.8 kg/s. The maximum possible heat transfer rate in this heat exchanger is:

A. 66.8 kW
B. 51.1 kW
C. 80.0 kW
D. 63.0 kW
E. 73.2 kW

Answer 1

The maximum possible heat transfer rate in the heat exchanger is 54.0 kW.

Step-by-step explanation:

The maximum possible heat transfer rate in the heat exchanger can be calculated using the formula Q = mcΔT, where Q is the heat transfer rate, m is the mass flow rate, c is the specific heat capacity, and ΔT is the temperature difference.

For the cold water, the heat transfer rate is Qc = mcΔTc = (0.5 kg/s)(4.18 KJ/kg*K)(50-15) = 62.7 kW.

For the hot air, the heat transfer rate is Qh = mhΔTh = (1.8 kg/s)(1.0 KJ/kg*K)(50-15) = 54.0 kW.

The maximum possible heat transfer rate in the heat exchanger is given by Q = min(Qc, Qh) = min(62.7 kW, 54.0 kW) = 54.0 kW.