Heat Transfer through Evaporators

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Heat transfer in refrigerator evaporators involves complexities beyond Newton's law of cooling due to the refrigerant existing as a mixture of saturated liquid and gas. While Newton's equation can provide a simplified calculation, it may not yield accurate results without considering the boiling and condensing processes of the refrigerant. The heat convection equation requires knowledge of the log mean temperature difference (LMTD) and overall heat transfer coefficient, which can be challenging to determine. For a more accurate approximation, using the enthalpy equation is recommended, focusing on the energy added to the refrigerant, which can be simplified if the fluid remains relatively saturated. Ultimately, knowing the refrigerant flow rate and the change in enthalpy is crucial for effective heat transfer calculations.
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Dear all,

I am a bit stuck trying to find out how heat is transferred through an evaporator in a refrigerator. I was using Newton's law of cooling before I realized 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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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;

\dot{Q}=\dot{m}\Deltahfg
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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