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.
Use the Carnot cycle H.C.O.P. and the required house heating rate to get the required heat transfer rate to ground water and the electrical input. Use Rankine temperatures for the HCOP.
The mass flow rate of ground water can be computed from the water temperature drop and the water heat transfer rate.
If you don't know what the HCOP is, do some Googling or reading of your course materials. Use Rankine temperatures, not Fahrenheit.
And please review my previous commentsabout your incorrect labeling of the subject
Well, it seems like we've got ourselves a heat pump that's ready to make some waves with that 50 degree F ground water! Now, let's get down to the calculations.
First, we need to find the heat energy required to warm up the house. The temperature difference is 70 - 50 = 20 degree F. Using the equation:
Q = m * Cp * ΔT
where Q is the heat energy, m is the mass flux, Cp is the specific heat capacity, and ΔT is the temperature difference.
We know that Q = 75,000 Btu/hr and ΔT = 20 degree F. The specific heat capacity of water is approximately 1 Btu/lb-°F.
So, substituting in the values, we find:
75,000 = m * 1 * 20
Simplifying this equation, we get:
m = 75,000 / 20
m ≈ 3750 lb/hr
So, the minimum mass flux of the groundwater required is approximately 3750 lb/hr.
Now, let's move on to finding the minimum horsepower (HP) required for the pump. To do this, we can use the equation:
HP = Q / (η * 2545)
where HP is the horsepower, Q is the heat energy, η is the efficiency of the heat pump, and 2545 is a conversion factor.
Since we're trying to find the minimum HP, let's assume an efficiency of 1 (or 100%). So, substituting Q = 75,000 Btu/hr into the equation, we get:
HP = 75,000 / (1 * 2545)
Simplifying this equation, we find:
HP ≈ 29.48 HP
So, the minimum HP required for the pump is approximately 29.48 HP.
Now, I must admit, these calculations are no laughing matter. But hey, at least we got to have a splash with some heat pump fun!
To calculate the minimum mass flux of the groundwater and minimum heat pump (HP) required for the pump, we can use the principle of energy conservation.
Step 1: Calculate the heat energy required:
The heat energy required by the house is given as 75,000 Btu/hr.
Step 2: Calculate the temperature rise required:
The house needs to be heated from 50°F to 70°F, which is a temperature rise of 20°F.
Step 3: Calculate the mass flow rate of the groundwater:
The heat energy required can be determined using the following equation:
Q = m x Cp x ΔT
where:
Q = Heat energy required (75,000 Btu/hr),
m = Mass flow rate of groundwater (unknown),
Cp = Specific heat capacity of water (1 Btu/lb°F),
ΔT = Temperature rise (20°F).
Substituting the values, we can solve for m:
75,000 = m x 1 x 20
m = 75,000 / (1 x 20)
m = 3,750 lb/hr
So, the minimum mass flow rate of the groundwater is 3,750 lb/hr.
Step 4: Calculate the minimum heat pump required:
The heat absorbed by the heat pump is equal to the heat energy required by the house. Therefore, the minimum heat pump capacity required is 75,000 Btu/hr.
Thus, the minimum mass flux of the groundwater is 3,750 lb/hr, and the minimum heat pump required for the heat pump is 75,000 Btu/hr.
To calculate the minimum mass flux of the groundwater, we first need to determine the heat transfer rate between the groundwater and the house.
The equation for heat transfer is Q = m * c * ΔT, where Q is the heat transfer rate, m is the mass flow rate of the groundwater, c is the specific heat capacity of water, and ΔT is the temperature difference.
Given:
Groundwater temperature before heating, T1 = 50 °F
House temperature, T2 = 70 °F
Temperature drop of groundwater, ΔT = 12 °F
Heat transfer rate required, Q = 75,000 Btu/hr
We need to convert the heat transfer rate from Btu/hr to energy units that match the specific heat capacity of water. 1 Btu (British thermal unit) is defined as the energy needed to raise the temperature of 1 pound of water by 1 °F.
To convert Btu/hr to watts (W), use the conversion factor: 1 Btu/hr ≈ 0.293071 W.
Therefore, the heat transfer rate in watts is Q = 75,000 Btu/hr * 0.293071 W/Btu ≈ 21,980 W.
Since the specific heat capacity of water is approximately c = 1 cal/g·°C or 4.184 J/g·°C, we need to convert the heat transfer rate to mass units in grams per second.
To do that, we use the formula: m = Q / (c * ΔT).
Now, calculating the mass flow rate:
m ≈ 21,980 W / (4.184 J/g·°C * (70 °F - 50 °F))
m ≈ 21,980 W / (4.184 J/g·°C * 20 °F)
m ≈ 262.94 g/s
So, the minimum mass flux of the groundwater required is approximately 262.94 grams per second.
To calculate the minimum power required for the heat pump, we need to determine the work input required for the heat pump.
The equation for work input to the heat pump is W = m * (h2 - h1), where W is the work input, m is the mass flow rate of the groundwater, h2 is the specific enthalpy of the groundwater after heating, and h1 is the specific enthalpy of the groundwater before heating.
To simplify the calculation, we can assume that the specific enthalpy of the groundwater remains constant during the heating process.
Therefore, the minimum power required for the heat pump is P = W.
P ≈ 262.94 g/s * (h2 - h1)
Since we are given that the temperature drop of the groundwater is 12 °F, we can calculate the change in specific enthalpy using the specific heat capacity of water, c.
Δh = c * ΔT
Now, substituting this back into the equation for power:
P ≈ 262.94 g/s * c * ΔT
Finally, substituting c = 1 cal/g·°C or 4.184 J/g·°C and ΔT = 12 °F:
P ≈ 262.94 g/s * 4.184 J/g·°C * (12 °F)
P ≈ 16,461 W
So, the minimum power required for the heat pump is approximately 16,461 watts.