Derivation of LMTD Page 1

Derivation of LMTD

As we know from 1st law of thermodynamics

Q = mcp(hc,o − hc,i)

So with the specific heat assumption that we made (and treating the fluids as incompressible), we can

write

Q = mcp∆T

We’ll also define Ci = micp,i as the heat capacity rate. We want to relate the inlet and outlet

temperatures,the U, and the A, to the rate of heat transferred, and we can do this as:

Q = U A ∆Tmean

We now need to find what this mean temperature is. To do this, we’ll look at a differential area of the

heat exchanger where a differential amount of heat is transferred.

Using this equation to examine the hot and cold streams separately yields:

δQ= −Ch dTh ; Ch = ( m cp)h (1)

δQ= ±Cc dTc ; Cc = ( m cp)c (2)

where ± or ∓ appears, the top sign designates parallel-flow, and the bottom sign designates

counterflow.

ch TTΔT

mΔTAUQ

Derivation of LMTD Page 2

From the following steps LMTD for Double pipe Heat Exchangers can be done

1. hhhhh dTCdTcmdQ

2. h

hC

dQdT

3. hcccc dTCdTcmdQ

4. c

cC

dQdT

5.

ch C

1

C

1TΔAdUT)(Δd

6.

2

1ch

ΔT

ΔT

dAC

1

C

1U

ΔT

T)(Δd2

1

7.

ch1

2

C

1

C

1UA

ΔT

ΔTln

8. )T(T

QC

oh,ih,

h

9. )T(T

QC

ic,oc,

c

10.

ch1

2

C

1

C

1UA

ΔT

ΔTln

11. ])T(T)T[(TQ

UA

ΔT

ΔTln ic,c,0oh,ih,

1

2

12. ])T(T)T[(TQ

UAoc,h,0ic,ih,

]ΔTT[Δ

Q

UA21

13. )T/ΔT(Δln

ΔTΔTUAQ

12

12

)T(TC oh,ih,h )T(Tcm Q oh,ih,hh