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.cph dth = — m.cpc dtc

dth=  – dQ. / m.cph =– dQ. /Ch

dtc=  – dQ. / m.cpc =– dQ./Cc

where  Ch = m.cph = heat capacity of hot fluid

Cc = m.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( θ21)  =  — 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( θ21)  =  — U  A (  (Th1 –Th2/ Q.)  — ( Tc2 –Tc1/ Q.) )

ln( θ21)  =  (U  A/ Q. ) ( ( Th2 –Tc2)   — ( Th1 –Tc1) )

ln( θ21)  =  –(U  A/ Q. ) (θ1—θ2)

ln( θ21)  =  (U  A/ Q. ) (θ2— θ1)

Q.  = U A (θ2— θ1)/ ln( θ21)

Q.  = U A θm

θm = LMTD = (θ2— θ1)/ ln( θ21)

         LMTD = (θ1—θ2)/ ln( θ12)

         LMTD = (θmax—θmin)/ ln( θmaxmin)

If   θ12=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.cph dth =  m.cpc dtc =U dA (Th – Tc)

dth=  – dQ. / m.cph = dQ.  /Ch

dtc=  – dQ/ m.cpc =dQ/Cc

where  Ch = m.cph = heat capacity of hot fluid

Cc = m.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( θ21)  =  — U  A  (1/Ch + 1/Cc)

But 1/Ch =Th1 –Th2/ Q.    and   1/Cc =Tc1 –Tc2/ Q.

ln( θ21)  =  — U  A (  Th1 –Th2/ Q.       +  Tc1 –Tc2/ Q.    )

ln( θ21)  =  (U  A/ Q.    ) (( Th2 –Tc2)   — ( Th1 –Tc1) )

ln( θ21)  =  (U  A/ Q.    ) (θ2— θ1)

Q.    = U A (θ2— θ1)/ ln( θ21)

Q.      = U A θm

θm = LMTD = (θ2— θ1)/ ln( θ21)

 LMTD = (θmax—θmin)/ ln( θmaxmin)

 

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