LMTD – PARALLEL & COUNTER FLOW HEAT EXCHANGER CLASS NOTES
LMTD – PARALLEL & COUNTER
FLOW HEAT EXCHANGER
CLASS NOTES
In a heat exchanger, temperature
difference is variable all the length of the
heat exchanger. There is a difficulty in
handling variable temperature
difference. Efforts are made to find the
mean temperature difference. In case of
heat exchanger the mean temperature
difference comes in the logarithmic form. Hence
it is called logarithmic mean temperature
difference. Thus, LMTD is an equivalent
constant temperature difference in place
of variable temperature difference in
actual. It will give constant average rate
of heat transfer.
DERIVATION LMTD COUNTER FLOW HEAT EXCHANGER
Fig. Counter Flow Heat Exchanger
Assumptions used
1. U is constant all along the HEX.
2. Steady flow conditions.
3. Specific heats of hot and cold fluids are constant
4. Flow rates of hot and cold fluids are constant.
5.There is no heat loss to the surroundings. There is no heat gain from the surroundings.
6.There is no phase change of the either fluid.
7.Conduction in the tube and shell are negligible.
8. Kinetic as well potential energies of fluids are constant.
Fig. Temp Distribution for a Counter Flow HEX
Consider a small area dA with dQ. heat transfer and temperature differences on hot is dTh and cold side is dTc as shown in Fig.
dQ. =U dA ΔT =U dA (Th – Tc)=-m.h cph dth = — m.c cpc dtc
dth= – dQ. / m.h cph =– dQ. /Ch
dtc= – dQ. / m.c cpc =– dQ./Cc
where Ch = m.h cph = heat capacity of hot fluid
Cc = m.c cpc = heat capacity of cold fluid
dth — dtc =–d Q.(1/Ch — 1/Cc)
dθ = –d Q. (1/Ch — 1/Cc)
dθ = — U dA (Th – Tc) (1/Ch — 1/Cc)
dθ = — U dA θ (1/Ch — 1/Cc)
= –(1/Ch + 1/Cc)
ln( θ2/θ1) = — U A (1/Ch –1/Cc)
Q. =Ch(Th1 – Th2)= Cc(Tc1—Tc2 )
But 1/Ch =Th1 –Th2/ Q. and 1/Cc =Tc2 –Tc1/ Q.
ln( θ2/θ1) = — U A ( (Th1 –Th2/ Q.) — ( Tc2 –Tc1/ Q.) )
ln( θ2/θ1) = (U A/ Q. ) ( ( Th2 –Tc2) — ( Th1 –Tc1) )
ln( θ2/θ1) = –(U A/ Q. ) (θ1—θ2)
ln( θ2/θ1) = (U A/ Q. ) (θ2— θ1)
Q. = U A (θ2— θ1)/ ln( θ2/θ1)
Q. = U A θm
θm = LMTD = (θ2— θ1)/ ln( θ2/θ1)
LMTD = (θ1—θ2)/ ln( θ1/θ2)
LMTD = (θmax—θmin)/ ln( θmax/θmin)
If θ1=θ2=θ, then LMTD = θ
Then Q=UAθ
DERIVATION LMTD PARALLEL FLOW HEAT EXCHANGER
Fig. Parallel Flow Heat Exchanger
ASSUMPTIONS USED IN DERIVATION
1. U is constant all along the HEX.
2. Steady flow conditions.
3. Specific heats of hot and cold fluids are constant
4. Flow rates of hot and cold fluids are constant.
5.There is no heat loss to the surroundings. There is no heat gain from the surroundings.
6.There is no phase change of the either fluid.
7.Conduction in the tube and shell is negligible.
8. Kinetic as well potential energies of fluids are constant.
PROOF OF LMTD FOR PARALLEL FLOW HEAT EX-CHANGER
Fig. Temp variation in a parallel flow heat exchanger
Consider a small area dA with dQ. heat transfer rate. Temperature differences on hot fluid is dTh and cold fluid is dTc in length dx
dQ. =U dA ΔT =-m.h cph dth = m.c cpc dtc =U dA (Th – Tc)
dth= – dQ. / m.h cph = dQ. /Ch
dtc= – dQ/ m.c cpc =dQ/Cc
where Ch = m.h cph = heat capacity of hot fluid
Cc = m.c cpc = heat capacity of cold fluid
dth — dtc =– dQ. (1/Ch + 1/Cc)
dθ = — dQ. (1/Ch + 1/Cc)
dθ = — U dA (Th – Tc) (1/Ch + 1/Cc)
dθ = — U dA θ (1/Ch + 1/Cc)
dθ/θ = –UdA(1/Ch + 1/Cc)
ln( θ2/θ1) = — U A (1/Ch + 1/Cc)
But 1/Ch =Th1 –Th2/ Q. and 1/Cc =Tc1 –Tc2/ Q.
ln( θ2/θ1) = — U A ( Th1 –Th2/ Q. + Tc1 –Tc2/ Q. )
ln( θ2/θ1) = (U A/ Q. ) (( Th2 –Tc2) — ( Th1 –Tc1) )
ln( θ2/θ1) = (U A/ Q. ) (θ2— θ1)
Q. = U A (θ2— θ1)/ ln( θ2/θ1)
Q. = U A θm
θm = LMTD = (θ2— θ1)/ ln( θ2/θ1)
LMTD = (θmax—θmin)/ ln( θmax/θmin)
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