Heat Transfer through Evaporators


by recreated
Tags: evaporators, heat, transfer
recreated
recreated is offline
#1
Aug25-13, 06:39 AM
P: 48
Dear all,

I am a bit stuck trying to find out how heat is transfered through an evaporator in a refrigerator. I was using Newton's law of cooling before I realised there are more complexities in the equation as the refrigerant is in a mixture of saturated liquid and saturated gas in the evaporator.

Newton's Law of cooling (heat transfer from a solid surface to a fluid):

QL=UL*AL*LMTD
where
QL= heat transfer
UL= overal heat transfer coefficient
AL= external surface area of the evaporator
LMTD= log mean temp different over the evaporator tube. i.e. The average temperature difference between the solid surface and the fluid (air in this case) across the evaporator inlet and outlet.

My Question

Is Newton's Law of cooling applied to refrigerator evaporators and/or condensers with this equation, for a simplified calculation? Or would the results be so far from correct that calculations must always take into consideration the boiling and condensing of the refrigerant?
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sanka
sanka is offline
#2
Aug26-13, 01:03 PM
P: 43
The heat convection equation will provide a good calculation of the heat transfer but it requires a knowledge of LMTD meaning you need to know the hot fluid inlet temp, hot fluid outlet temp, cold fluid inlet temp and cold fluid outlet temp. You also need to be able to calculate U, which requires the use of empirical correlations (for condensers anyway, not sure about evaporators). It can be difficult to accurately calculate all these terms.

If you are simply looking for a good approximation of the heat transfer, an energy rate equation (enthalpy eqn) should suffice. In the case of an evaporator, the working fluid (refrigerant) acts as if it has an infinite specific heat capacity and can be assumed as isothermal. When adding energy to this fluid, the enthalpy of vaporization is the major contributor and sensible heat can probably be neglected, provided the liquid is relatively saturated at inlet and outlet (isnt too subcooled or superheated) If this is the case, the energy eqn simplifies to the heat/energy added to the fluid;

[itex]\dot{Q}[/itex]=[itex]\dot{m}[/itex][itex]\Delta[/itex]hfg
All you need to know is the flow rate of the refrigerant and the change in enthalpy from inlet to outlet (get this in thermo tables for whatever fluid you are looking at).


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