Sample Problem Statement
Calculate log meantemperaturedifference for the followingshell and tube heat exchangerfor two possible arrangements (i) counter-current and (ii) co-current flow patterns.Hot fluid in temperature TH1 = 1000CHot fluid out temperature TH2 = 900C Cold fluid out temperature TC2 = 500CCold fluid in temperature TC1 = 300C
Solution
As per the definition and equation for Log Mean Temperature Difference (LMTD),
ΔT1→ thetemperaturedifference between hot and cold fluids at one end of the heat exchanger
ΔT2→ the temperature difference between hot and cold fluids at the other end of the heat exchanger (i) For counter current heat exchanger,
ΔT1= TH1 - TC2 = 100 - 50 = 500C (At one end hot fluid enters and cold fluid exits.)
ΔT2= TH2 - TC1 = 90 - 30 = 600C
(在不her end cold fluid enters and hot fluid exits.)
by definition given above, LMTD for counter current flow = (60-50) / ln(60/50) = 10 / 0.1823 = 54.850C.
This can be also verified quickly inEnggCyclopedia's LMTD calculator. (ii)
For co-current heat exchanger,
ΔT1= TH1 - TC1 = 100 - 30 = 700C (At first end hot and cold fluids enter the heat exchanger.)
ΔT2= TH2 - TC2 = 90 - 50 = 400C
by definition given above, LMTD for counter current flow = (70-40) / ln(70/40) = 30 / 0.5596 = 53.610C. This can be also verified quickly inEnggCyclopedia's LMTD calculator.
Remarks
It can be readily noticed that for co-current heat exchanger the logarithmic mean temperature difference is lower compared to counter-current heat exchanger. This suggests that for same fluids, to achieve the same heat transfer, a co-current heat exchanger always requires more heat transfer area than the counter-current type exchanger. In this example, a simple heat exchanger with single shell pass and single tube pass is considered. For multiple shell and tube passes, the LMTD calculated has to be multiplied by a correction factor to account for geometric changes.EnggCyclopedia's LMTD correction factor calculatorcan be readily used for this correction factor calculation.
