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A cold liquid enters a counter flow heat exchanger at 15 deg at a rate of 8 kg/s. A hot stream of the same liquid enters the heat exchanger at 75 deg at 2 kg/s. Assuming the specific heat of the fluid as 4 kJ/kg.°C, determine the maximum heat transfer rate.
1. 960 kW
2. 240 kW
3. 1920 kW
4. 480 kW

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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,

c = 8 kg/s Tci = 15°C, Cpc = 4 kJ/kg°C,

For hot fluid, 

h = 2 kg/s, Thi = 75°C, Cph = 4 kJ/kg°C

Now let us find

c × Cpc = 8 × 4 = 32 kW/°C

n × Cph = 2 × 4 = 8 kW/°C

So the maximum heat transfer rate = 8 × (75 - 15) = 480 kW

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