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Thread: Regulation of cooling capacity
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03-02-2002, 09:20 PM #6
Based on your figures of 15ft of 5/16 copper tube we can calculate that the surface area of the tube is
A = Pi x D x L
Where
15ft =4.57m
5/16bore = 0.01m o.d.
so A = 3.141592654 x 0.01 x 4.57 = 0.143m2
The inlet temp of the vapour into the shell is stated @ 78.5deg C and the outlet temp is required @ 60deg C.
Assuming the inlet temperature of the cooling water entering the tube is 6deg C and leaving @ 35deg C we can calculate that the LMTD would equate to 44.61deg C.
The U value of copper is 390w/m2.K
So: the heat transfer would be Q = U A dT
Q= 390 x 0.143 x 44.61 = 248W
Trying to convert a domestic fridge rated at or about 500W to do this load directly would take some engineering! but it would be possible to create a secondary water tank cooled by the fridge circuit with a little pipework evaporator submerged in the tank and then a small domestic circulator feeding the tube/shell exchanger. A hand control valve would be required in the secondary circuit to set up the cooling water mass flow rate and an immersion stat in the tank to control the fridge circuit.
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