Question

An adiabatic heat exchanger is used to heat cold water at 15°C entering at the rate of 10kg/sec by hot air at 90°C entering at the rate of 10 kg/sec. if the exit temperature of cold water IS 29°C then what will be the temperature of hot air at the exit. C_p for air = 1.008 KJ/Kg. C_p for water = 4.18 KJ/Kg: 

1. 26.54°C
2. 31.88°C
3. 58.65°C
4. 89.12°C

Answer 1

Correct Answer - Option 2 : 31.88°C

Concept:

By usimg equilibrium of heat exchanger, we know that,

Heat loss by Hot air = Heat gain by cold water

\({\left( {m{C_p}\Delta T} \right)_{Air}} = {\left( {m{C_p}\Delta T} \right)_{water}}\)

Calculation:

Given:

For air:

Ti = 90°C, To = 20°C, m = 10 kg/s, Cp = 1.008 kJ/kgK

For water:

Ti = 15°C, To = ?,  m = 10 kg/s, Cp = 4.18 kJ/kgK

⇒ 10 × 1.008 × (90 – T_f) = 10 × 4.18 × (29 – 15)

⇒ Tf = 31.6° C