Answer:
The log mean temperature difference is:
ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)
Explanation:
To evaluate the equivalent average temperature difference between two fluids we consider a parallel-flow double-pipe heat exchanger (see attached diagram). The temperature of the hot and cold fluids is large at the inlet of the heat exchanger and decreases exponentially toward the outlet.
We can assume that the outer surface of the heat exchanger is well insulated and that heat transfer only occurs between the two fluids. We can also assume negligible kinetic and potential. The energy balance on each fluid can be written as the rate of heat loss from the hot fluid is equal to the rate of heat gained by the cold fluid in any section of the heat exchanger:
Q = -m,h×c,ph×dT,h (1)
where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the hot fluid, dT,h= differential temperature of the hot fluid
Q = m,c×c,pc×T.c (2)
where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the cold fluid, dT,h= differential temperature of the cold fluid
The temperature of the hot fluid change is negative and is added to make Q positive. Solving equations 1 and 2 in terms of dT:
dT.h = - Q/(m,h×c,ph)
dT.c = Q/(m,c×c,pc)
and taking the difference:
dT,h-dT,c= d(T,h - T,c) = -Q(1/(m,h×c,ph) + 1/(m,c×c,pc)) (3)
The heat transfer rate in the differential section of the heat exchanger can be expressed as:
Q = U(T,h-T,c)×dA,s (4)
where U=overall heat transfer coefficients, dA,s = differential sectional area. Substitute equation 4 into 3:
d(T,h - T,c)/(T,h - T,c) = -U×dA,s×(1/(m,h×c,ph) + 1/(m,c×c,pc)) (5)
Integrating equation 5:
㏑((T,h out - T,c out)/(T,h in - T,c in)) = -U×A,s×(1/(m,h×c,ph) + 1/(m,c×c,pc)) (6)
The first law of thermodynamics requires the rate of heat transfer from hot and cold fluid to be equal.
Q= m×c, pc×(T, c out-T, c in) (7)
Q= m×c, ph×(T,h out-T, h in) (8)
Solve equations 7 and 8 for m,c×c, pc and m,h×c, ph and substituting into equation 6:
Q = U×A,s×ΔT,lm
Where the log mean temperature difference is:
ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)
