*ack after a long gap, and i got many queries asking about LMTD, and how to decide the LMTD factor, so today i gonna explain you exactly what does LMTD mean and how does it vary and what factors does have influence over LMTD.*

**B**

*What is LMTD ?*LMTD means

__L__ogarthemic

__M__ean

__T__emperature

__D__ifference, there are many ways to calculate Mean value, i.e., Average value of the temperatures of fluid streams like Arthematic mean, Geometric mean, Harmonic mean etc, but those all were limited for a certain extent, while coming to the case where two fluid streams were involved, it is difficult to calculate the mean of those, for such cases there is a need to develop one expression which can define the exact mean value, so there our LMTD was born,

The Basic expression which defines LMTD is,

LMTD = ( ( T1 - T2 ) - ( T3 - T4 ) ) / Log [ ( T1 - T2 ) / ( T3 - T4 ) ],

So, with an example i'll explain you the way of calculating the LMTD value,

**Calculation :**Let us we are having two process fluid where heat transfer taking place, hot fluid inlet is having 80 deg C and out let if having 70 deg C, and the cold fluid is having inlet temperature of 25 deg C and outlet temperature of 45 deg C, so for this we need to calculate the LMTD,

Giving notations for the temperatures, Th1 = 90 deg C, Th2 = 75 deg C,

Tc1 = 25 deg C, Tc2 = 35 deg C,

LMTD = ( ( Th1 - Tc2 ) - ( Th2 - Tc1 ) ) / Log [ ( Th1 - Tc2 ) / ( Th2 - Tc1 ) ]

= ( ( 90 - 35 ) - ( 75 - 25 ) ) / Log [ ( 90 - 35 ) / ( 75 - 25 ) ]

= ( 55 - 50 ) / Log [ 55 / 50 ]

= 5 / 0.04 = 125

So our LMTD value is 125 deg C, for Counter current flow,

Now calculating for Parallel flow,

LMTD = ( ( Th1 - Tc1 ) - ( Th2 - Tc2 ) ) / Log [ ( Th1 - Tc1 ) / ( Th2 - Tc2 ) ]

= ( ( 90 - 25 ) - ( 75 - 35 ) ) / Log [ ( 90 - 25 ) / ( 75 - 35 ) ]

= ( 65 - 40 ) / Log [ 65 / 40 ] = 119

So our LMTD value in Parallel flow is 119 deg C.

So by now you gotta clear picture of calculating the LMTD. and also one thing you can observe is the LMTD value is somewhat higher in Counter flow than parallel flow, so the Heat transfer profile is better in counter flow.

**Factors influencing LMTD :**The expression of Heat transfer rate is given in the expression,

Q = U x A x LMTD,

The value of Q will increase if LMTD increases,

LMTD = Q / [ U x A ],

So from the above form it is clear that if Heat Transfer Area increases then the LMTD will decrease, and same is the case of Overall Heat Transfer Co-efficient also.

Coming to LMTD correction factor, you need to know some thing, when actually there is need for LMTD correction factor.

**When does it require ?**

Usually our condensers in process industry will be Shell & Tube heat exchangers, where the concept of passes will arise, usually there will be two passes for cold fluid and single pass for hot fluid, simply a 1,2-shell & tube heat exchanger, for effective heat transfer rate the hot fluid will be passed through shell and cold fluid on tube side, as our main motto is to remove heat from hot fluid to cold fluid and let our hot fluid condense, so there wont be any problem if our hot fluid exchanges heat with the shell to atmosphere, but we need to use the utility energy very effectively,

So, while coming across these cases, our cold fluid will be passing through one counter flow and one parallel flow with respect to hot fluid and we can calculate either parallel flow LMTD or Counter flow LMTD, but together cannot, due tot these kind of circumstances the concept of LMTD correction factor arises,,

**LMTD Correction Factor :**Overall LMTD = LMTD of Counter flow x Correction factor ( F ),

Usually many of the engineers will considers the Value of F as 0.9, but actually there is a way to calculate the value of F theoretically, i'll demonstrate it below,

R = ( Ta - Tb ) / ( tb - ta )

P = ( tb - ta ) / ( Ta - ta )

Ta , Tb - Shell in & out temperatures,

ta , tb - Tube in & out temperatures.

If R =! 1,

Î± = [ ( 1 - ( RxP ) ) / ( 1 - P ) ] ^ ( 1 / N ), S = ( Î± - 1 ) / ( Î± - R ),

F = Q x Ln [ ( 1 - S ) / ( 1 -RS ) ] / [ ( R - 1 ) Ln [ ( 2 - S( R+1 - Q ) ) / ( 2 - S(R+1 + Q ) ) ]

Where, Q = [ ( R^2 +1 )^0.5 ],

if R = 1,

S = P / (N - ( N- 1 ) P ),

F = 1.414 S /[ ( 1 - S ) Ln [ ( 2 - 0.59S ) / ( 2 - 3.41S ) ]

That's it..........!!! You're done,

If you understood the above calculation part, then Say cheers,

If any doubt got arised then simply ping me, i'll be there to resolve your queries........!!!

Comments are most appreciated.........!!!

**A**bout The Author

Thank you for sharing such a useful information. It was so useful. Please visit

ReplyDeleteIntrviewHow to calculate LMTD for Batch reactor??

ReplyDeleteThanks

Dear Khan,

DeleteGo ahead by considering initial and final temperatures of reactor as cold fluid and the jacket inlet&outlet temperatures as hot fluid.

Regards,

AJAY K

Is there a correction factor for that?

DeleteHello Mr. AJAY KUMAR

DeleteI am trying to estimate the evaporation capacity of a well stirred batch reactor / half-pipe jacket coil/. I am using saturated steam at 3 barG/ specific heat = 2163 kj/kg.K, Temp.=133 deg.C./ and in the reactor I have to evaporate 2000 L or 2979 kg Chloroform /INLET temp.= 25 deg.C., OUTLET temp.= boiling point= 61,2 deg.C, cP=0.95kj/kg.K.25deg.C, Latent heat,lambda = 245 kj/kg.k.25 deg.C/. I have estimated the required Q for the heating and for the evaporation processes = 231,16 kW, the steam consumption Msteam= 384,79 kg./.

I need to find out:

1. Evaporation capacity for 1 hr., kg/h;

2. Time for evaporation for the whole quantity of Chloroform, hrs;

3. To size a counter flow shell and tube heat exchanger, sq.m.

I apreciate your help and mostly for LMTD for the batch reactor.

Thanks for the quick reply Ajay.

ReplyDeletewhat is N?

ReplyDeleteDear Sri,

DeleteN is no. of passes on tube side,

I've seen your comments, which are awaiting for approval, i'll add the requested content in main post itself, don worry.

Regards,

AJAY K