Should not 188 (C) be the inlet temperature?

Hello,

I agree to you, the case on which i am working is much wider than the data i have presented and this is just an example

Still i have made the changes, as per your suggestions, so could you help me on how to calculate it.

see attachment

Hello Sir,

The situation is that the Outlet Temperatures of both the stream are at same temperatures, so how to calculate LMTD for the case. 

For counterflow:

LMTD=[(188-68) - (68-42)] / ln[(188-68) / (68-42)]

Please Note sir it is a Parallel configuration as per your suggestion earlier

I think the solution that you have given is for Counter flow arrangement

Regards

Parallel co-flow is unrealistic in such a case. It would require an infinite heat transfer surface. Formally calculated LMTD (limit) is equal to 0 in this case.

Hello Sir,

I agree that the case would be unrealistic but the value of LMTD cannot be zero as it would result in no heat transfer.

Is there any special case to find out LMTD for this?

see attachment

For a Parallel Arrangement will it not be ?

ΔT'≔188-42 = 146

ΔT'' = 68-68 = 0

ΔT" = 68 - 68 = 0

LMTD tends to 0 as ΔT" tends to 0 (see the calculation in attachment LMTD-2.pdf).

Why do not you apply counter-current configuration?

Its a case study on which i am working and it has the similar data and configuration 

so i need to find LMTD for the case

Dear Divyesh

I am going through a experimental research paper where in the following temperatures have been extracted and trying to design a methodology to solve it. Also the mass flow rate does not make any role for calculating LMTD 

Regards

Hello,

I tried putting the same case in HTRI and it gave me a LMTD = 0.6.

Though manually if we try it should be 0.

What would be the method that HTRI took into account?