Hot liquid water at 1MPa and 165°C is to be cooled to 70°C at 1MPa in a shell-tu

Question

Hot liquid water at 1MPa and 165°C is to be cooled to 70°C at 1MPa in a shell-tube heat exchanger by using air. Liquid water flows through tube section of heat exchanger. The air enters the heat exchanger at 100kPa and 28°C with volume flow rate 780m3/min and leaves the heat exchanger at 92kPa and 127°C. *The mass flow rate of water if the specific heat of water at constant pressure Cp,water = 4.2kJ/kg K = 0.046kg/s.

Determine heat transfer rate from water to air. Consider air as an ideal gas with Cp,air = 1.02kJ/(Kg K) and R=0.287kJ/(kg K).

Explanation / Answer

This is a "tricky" question with both an academic answer and a practical answer. The practical answer is to treat as a constant pressure problem, but for a thermo course the main point is to grasp the concepts which requires solving for ideal gas enthalpy with changing temperature and pressure.

Step 1 - Calculate enthalpy change of air from given data.

delta h (ideal gas) = (p2v2 - p1v1)(k/k-1)

To get v2 solve first for air mass flow then use the ideal gas equation.

mass flow = (100E3 N/m2 * 780 m3 / 0.287E3 N m/Kg K * 301 K) = 902.9 Kg/min = 903

exit volume = m R T /P = 1126.6 m3/min = 1127

Plugging into delta h and remembering that Pa = N/m2,

delta h, air = (92 KPa*1127 m3/min - 100 kPa * 780 m3/min) (1.4 / 0.4) =89.6 megaJoule/min

In contrast the constant pressure solution would give 91.1

Step 2 - Calculate mass flow of water

mass flow = q/Cp delta T = 89600 kJ/min / (4.2 kJ/kg K * 95 K) = 225 kg/min

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