Question

Water at 200 Fahrenheit is pumped from a storage tank at the rate of 50 gallons/minute. The motor for the pump supplies work at the rate of 2 hp. The water goes through a heat exchanger, giving up heat at the rate of 40,000 BTU/minute(Q Net In), and is delivered to a second storage tank at an elevation 50 feet above the first tank. What is the temperature of the water delivered to the second tank?

Answer 1

Temperature of water = 200 F = 93.33 C

Flowrate of water = 50 gal/min

1 gal = 0.00378 m³

50 gal/min = 50*0.00378 = 0.189 m³/min

Flow rate of water =  0.189 m³/min = 0.189/60 m³/sec = 0.00315 m³/sec

Power supplied from from pump = 2 hp = 2*746 = 1492 watts

Net elevation = 50 ft = 15.24 m

Power to be supplied for pumping water = Qgh

Q = 0.00315 m³/sec ; = 1000 kg/m³; g = 9.81 m/sec²; h = 15.24 m

Power to be supplied for pumping water = 0.00315*1000*9.81*15.24 = 470.938 Watts

Excess energy supplied to water = 1492 - 470.938 = 1021.06 watts (or Joules/sec)

Case 1: Assuming the excess energy from pump is not converted to thermal energy of water

Thermal energy lost by water in heat exchanger = 40000 BTU /min = 40000*1055 Joules/min

mCpT = 40000*1055 Joules/min

m =  v = 1000 kg/m³ * 0.189 m³/min = 189 Kg/min

Cp = 4180 J and T = (T_in - T_out) = 93.33 - T

189*4180*(93.33 - T ) = 40000*1055

T = 39.91 C = 103.838 F

Temperature of water deilvered to second tank in case 1 = 103.838 F

Case 2: Assuming the excess energy from pump is converted to heat energy of water

Increase in thermal energy of water = excess energy from pump

mCpT = 1021.06 watts

Note that the right habd side unit is Watts (Joules /sec). Thus the mass flow rate should be in Kg.sec.

m =  v = 1000 kg/m³ * 0.00315 m³/sec = 3.15 Kg/sec

The temperature of water leaving exchanger = 39.91 C or 103.838 F

Cp = 4180 J and T = T - 39.91

3.15*4180*(T - 39.91) = 1021.06

T = 39.987 C = 103.976 F

Temperature of water deilvered to second tank in case 2 = 103.976 F