Is it wrong to derive Log mean temperature difference (LMTD) of a Heat Exchanger, this way?

Physics Asked by Harshit Rajput on May 9, 2021

I tried deriving the LMTD of a parallel flow double pipe heat exchanger by first finding the temperature profiles of the hot and cold fluid and then averaging them over the entire length of the Heat exchanger. Then I subtracted these average values hoping that it would give me the relation for LMTD, but I’m stuck. The relations for temperature variation and average temperature are in the picture.

enter image description here

enter image description here

EDIT: I have come one step closer to make the relation look like the one for LMTD given in books.

enter image description here

The relation would match with the relation of LMTD given in books if we assume the wall surface temperatures to be equal. Do books take that assumption while deriving LMTD?

Please Help.

One Answer

This is not analyzed correctly. $T_{s1}$ and $T_{s2}$ are not constant, and they need to be eliminated from the analysis all-together. If $dot{Q}$ is the rate of heat flow per unit length along the exchanger, then $$dot{Q}=rho_ih_i(T_h-T_{s1})=rho_oh_o(T_{s2}-T_c)=frac{(rho_i+rho_o)}{2}frac{k_W}{delta}(T_{s1}-T_{s2})$$where $delta$ is the wall thickness and $k_w$ is its thermal conductivity. If we eliminate $T_{s1}$ and $T_{s2}$ from these equations, we obtain: $$dot{Q}=Urho (T_h-T_c)$$Where $$frac{1}{Urho}=frac{1}{h_1rho_i}+frac{1}{frac{(rho_i+rho_o)}{2}frac{k_W}{delta}}+frac{1}{rho_oh_o}$$ Based on this, the heat balance equations for the heat exchanger become $$dot{m}_hC_hfrac{dT_h}{dx}=-Urho(T_h-T_c)$$and$$dot{m}_cC_cfrac{dT_c}{dx}=+Urho(T_h-T_c)$$ I leave it to you to complete the analysis and show that the overall heat load Q is given by:$$Q=(Urho x_f)Delta T_{LM}$$

Answered by Chet Miller on May 9, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP