You are assigned to design a heating system for a swimming pool that is 2 m deep

Question

You are assigned to design a heating system for a swimming pool that is 2 m deep, 20m long, 20 m wide. The client desires that the heating system be large enough to raise the water temperature from 20 degree C 30 degree C in 2 hours. The rate of heat loss from water to air at the outdoor design conditions is determined to be 900 W/m2, and the heater must be able to maintain the pool at 30 degree C at those conditions. Heat losses to the ground can be disregarded. It is requested to have an electric heater with its efficiency of 90 percent. Please recommend the heater size in term of kW and prepare answers for possible questions that your client could raise to you.

Explanation / Answer

Swimming pool is 2m deep ,20m wide,20 m long

volume=2*20*20 m^3= 800 m^3

Density=1000kg/m^3

The mass=1000*800=8*10^5 kg

Specific heat of water=4.18kj/kg*k

Total power=8*10^5*4.18*(10)=33.40 *10^6 kj in 2 hours

=33.40*10^6/7200=33.40/72*10^4 kwatt=4644.44kwatt

Power loss=900 *20*20=360000watt=360kilowatt

Total power required=5004.44kilowatt

Efficiency 90% so power of heater=5560.48 kilowatt[Answer]

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