# Effectiveness to calculate the outlet temperatures of a heat exchanger

#### JohnnyS

Hi, I am working this same question at the moment and I have got the same answers you did for the outlet temperatures and the heat transfer rate of TC2 = 66.25°C and TH2 = 37.5°C, and the transfer rate of 129.15kW.
I am now struggling with the correction factor part of the question. Following the equations given in the lessons which are as you mention above I am not getting to 0.59.

I get :
∆Tc = Φ / UA
129.15 / (0.36 x 14)
∆Tc = 25.625°C
Then

F=∆Tc/LMTD

LMTD = (TC2 - TC1) / In [(TH - TC1) / (TH - TC2)]
= (66.25 - 15) / In [(140 - 15) / (140 - 66.25)]
LMTD = 97.13°C

F=∆Tc/LMTD
25.625 / 97.13
F = 0.264
Please can someone point me in the right direction please as I am going round in circles.

#### BvU

Homework Helper
I suspect your LMTD expression -- It looks wrong and it should have four temperatures as argument, not three

By the way, if you start a new thread, you appear on the 'unanswered' list (Threads with no replies) which is seen by many helpers. Now only the participants in this thread get a signal.

#### Chestermiller

Mentor
The effectiveness factor is the ratio of the heat transfer rate to the maximum possible heat transfer rate. The maximum heat transfer rate is what you get if you have countercurrent flow. With countercurrent flow, you have different inlet and outlet temperatures than the ones given.

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