In a double pipe counter-flow heat exchanger, 10,000 kg/hr of oil [ $C_{p} = 2.09 kJ/kg-K$ ] is cooled from 80 °C to 50 °C by water [ $C_{p} = 4.18 kJ/kg-K$ ] of flow rate 8000 kg/hr entering at 25 °C. What will be the outlet temperature of water?

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SJVN ET Mechanical 2019 Official Paper

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Answer 1

Concept:

In a heat exchanger, heat lost by hot fluid = heat gained by cold fluid.

Given:

Oil: $ m_o = 10000 \, \text{kg/hr}, \, C_{p,o} = 2.09 \, \text{kJ/kg·K}, \, T_{in,o} = 80^\circ C, \, T_{out,o} = 50^\circ C $
Water: $ m_w = 8000 \, \text{kg/hr}, \, C_{p,w} = 4.18 \, \text{kJ/kg·K}, \, T_{in,w} = 25^\circ C $

Step 1: Heat lost by oil

$ Q = m_o \times C_{p,o} \times (T_{in} - T_{out}) = 10000 \times 2.09 \times 30 = 627000 \, \text{kJ/hr} $

Step 2: Heat gained by water

$ Q = m_w \times C_{p,w} \times (T_{out} - 25) = 8000 \times 4.18 \times (T_{out} - 25) $

$ 627000 = 33440 \times (T_{out} - 25) \Rightarrow T_{out} - 25 = \frac{627000}{33440} = 18.75 $

$ \Rightarrow T_{out} = 25 + 18.75 = {43.75^\circ C} $