Correct Answer - Option 4 : 480 kW
Concept:
Maximum heat transfer rate through a heat exchanger is given by
Qmax = (ṁ Cp)min × \(\left( {{T_{{h_i}}} - {T_{{c_i}}}} \right)\)
Using subscript c for cold fluid and h for hot fluid
(ṁCp)min is the minimum of ṁc × Cpc and ṁh × Cph
Cp is the specific heat of the fluid, and ṁ is the mass flow rate of the fluid
Thi – Hot fluid inlet temperature
Tci – cold fluid inlet temperature
Calculation:
Given,For cold fluid,
ṁc = 8 kg/s Tci = 15°C, Cpc = 4 kJ/kg°C,
For hot fluid,
ṁh = 2 kg/s, Thi = 75°C, Cph = 4 kJ/kg°C
Now let us find
ṁc × Cpc = 8 × 4 = 32 kW/°C
ṁn × Cph = 2 × 4 = 8 kW/°C
So the maximum heat transfer rate = 8 × (75 - 15) = 480 kW