A heat pump is proposed that uses 50 degree F ground water to heat a house to 70 Degree F. The groundwater experiences a temperature drop of 12 degree F. If the house require 75,000 Btu/hr, Calculate the minimum mass flux of the groundwater and minimum HP required for the pump
cool water is going to make house hotter? Somehow, I dont get this.
To calculate the minimum mass flux of the groundwater and the minimum heat pump (HP) required for the pump, we can use the equation:
Q = m * Cp * ΔT
Where:
Q = Heat required by the house (75,000 Btu/hr)
m = Mass flux of the groundwater (unknown, to be calculated)
Cp = Specific heat capacity of water (1 Btu/lb·°F)
ΔT =Temperature drop of the groundwater (12 °F)
First, let's convert the heat requirement from Btu/hr to Btu/s:
Q = 75,000 Btu/hr * (1 hr/3600 s) = 20.83 Btu/s
Now, we can rearrange the equation to solve for the mass flux (m):
m = Q / (Cp * ΔT)
m = 20.83 Btu/s / (1 Btu/lb·°F * 12 °F)
Dividing the Btu units will cancel out, leaving us with the mass flux in lb/s:
m = 20.83 / (1 * 12) lb/s
m ≈ 1.736 lb/s
So, the minimum mass flux of the groundwater required is approximately 1.736 lb/s.
Now let's calculate the minimum heat pump (HP) required. To determine the HP, we need to calculate the heat transfer rate of the groundwater, which is equal to the heat requirement of the house (Q):
HP = Q = 75,000 Btu/hr
Therefore, the minimum HP required for the heat pump is 75,000 Btu/hr.