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In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).
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Answer Posted / kang chuen tat (malaysia - pen
Answer 12 : Q = q(1) + q(2) + q(3) = [ UA (TI – TB) ] + [ UA (TB – TC) ] + [ UA (TC – TF) ] = UATI + (UATB – UATB) + (UATC – UATC) – UATF = UATI – UATF = UA (TI – TF) (Proven). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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