Question
Download Solution PDFAn adiabatic heat exchanger is used to heat cold water at 15°C entering at the rate of 10 kg/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. Cp for air = 1.008 KJ/Kg. Cp for water = 4.18 KJ/Kg:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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 – Tf) = 10 × 4.18 × (29 – 15)
⇒ Tf = 31.6° CLast updated on May 28, 2025
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