Answer:
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Explanation:
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A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o
f 2.5. How much power is required to drive the unit? For the same COP, how high a high temperature would a Carnot heat pump have, assuming the same low T?
1 answer:
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Answer:
Explanation:
a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:
b) For a transient, 1-D, constant with energy generation
suppose T = f(x)
Then; the equation can be expressed as:
where;
= heat generated per unit volume
= Thermal diffusivity
c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:
where;
The radial directional term = and the axial directional term is
d) The heat equation for a wire going through a furnace is:
since;
the steady-state is zero, Then:
'
e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:
Answer:
The convective coefficient is 37.3 W/m²K.
Explanation:
Use Newton’s law of cooling to determine the heat transfer coefficient. Assume there is no heat transfer from the ends of electric resistor. Heat is transferred from the resistor curved surface.
Step1
Given:
Diameter of the resistor is 2 cm.
Length of the resistor is 16 cm.
Current is 5 amp.
Voltage is 6 volts.
Resistor temperature is 100°C.
Room air temperature is 20°C.
Step2
Electric power from the resistor is transferred to heat and this heat is transferred to the environment by means of convection.
Power of resistor is calculated as follows:
P=VI
P= 30 watts.
Step3
Newton’s law of cooling is expressed as follows:
Here, h is the convection heat coefficient and is the exposed surface area of the resistor.
Substitute the values as follows:
h = 37.3 W/m²K.
Thus, the convective coefficient is 37.3 W/m²K.