A hot water coil with an overall conductance U-500 W/m-C is designed to heat a flow...

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A hot water coil with an overall conductance U-500 W/m²-C is designed to heat a flow of 1000L/s of air from 25°C to 35°C. The design water flow rate is 1.5 L/s, and the water enters at 60°C. Use following air and water properties for the calculation: Air Water Properties Cp, kJ/kg-K p, kg/m³ 1.005 4.185 1.164 983.2 Determine the heat transfer rate and outlet water temperature. b) Calculate the heat transfer area for the design using the ε-NTU method. Use the following relationship for the e-NTU calculation: 1 - e-Ntu(1-C¹) ε = [1 -0.0411C*(1-e-0.414Ntu)] - C*e-Ntu(1-C) UA where C* = min and Ntu Cmax Cmin A hot water coil with an overall conductance U-500 W/m²-C is designed to heat a flow of 1000L/s of air from 25°C to 35°C. The design water flow rate is 1.5 L/s, and the water enters at 60°C. Use following air and water properties for the calculation: Air Water Properties Cp, kJ/kg-K p, kg/m³ 1.005 4.185 1.164 983.2 Determine the heat transfer rate and outlet water temperature. b) Calculate the heat transfer area for the design using the ε-NTU method. Use the following relationship for the e-NTU calculation: 1 - e-Ntu(1-C¹) ε = [1 -0.0411C*(1-e-0.414Ntu)] - C*e-Ntu(1-C) UA where C* = min and Ntu Cmax Cmin

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Given overall conductance heat transfer coefficient U 500 Wmc density of water Su 9832 kgm 3 density ... View the full answer