How many gallons/day of water is required to cooling a 300kW load using a coolin

Question

How many gallons/day of water is required to cooling a 300kW load using a cooling tower? Assume the load will increase the water temperature by 20

Explanation / Answer

M = E + D + W = E + 0.003C E = CTcp/Hv = (20*4.184/2260)*C = (523/14125)C => M = (523/14125)C + 0.003C M = (784/19587)C Now, E = mcpT E = V*d*cpT ( mass = volume * density) P = dE/dt P = C*d*cp*T ( d - density of water) C = P/(d*cp*T) C = 300*10^3/(1*4.184*20) Liter/s Now, M = (784/19587)C M = (784/19587)*300*10^3/(1*4.184*20) Liter/s M = 143.4986208223747 Liter/s M = (143.4986208223747/3.78541178) Gallon/s M = (143.4986208223747/3.78541178)*3600 Gallon/day M = 136469.9705882353 Gallons/day = 136470 gallons/day (approx)

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