December 22, 2024, 02:41:23 PM
Forum Rules: Read This Before Posting


Topic: Heat Exchanger Single pass shell and tube Help Pleaseeeeeeeeeee  (Read 4812 times)

0 Members and 1 Guest are viewing this topic.

Offline mal

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
Heat Exchanger Single pass shell and tube Help Pleaseeeeeeeeeee
« on: December 07, 2008, 01:02:48 AM »
Hello everyone

I need your help guys to understand this concept. If you have a single pass shell and tube heat exchanger (counter current) between two fluids hot & cold what would be the result of the final(exit, outlet) temperature for both fluids if you increase the mass flow rates for both fluids( hot & cold)

those are the results that i got let me know if it does make sense

Outle(exit) Temperature for Hot fluid Increases (As) Mass flow rate for Hot fluid Increases & Uo Overall heat coeffecient increases.

Outle(exit) Temperature for Cold fluid Decreases (As) Mass flow rate for Cold fluid Increases & Uo Overall heat coeffecient increases

if it's right let me know, if not let me know why and what is the general rule or concept of the velocity or mass flow rate of the fluid Vs. Outlet Temperature of both of the fluids

Thanks in a million

You guys are a lifesaver

Offline technologist

  • Full Member
  • ****
  • Posts: 148
  • Mole Snacks: +6/-4
    • Chemical Professionals
Re: Heat Exchanger Single pass shell and tube Help Pleaseeeeeeeeeee
« Reply #1 on: December 11, 2008, 11:17:18 PM »
Dear Mal,
Your both the results may be correct depending on the ratio of increase in hot fluid flow Vs increase in Cold fluid flow.

What U need to do is to first make a balance for original condition for three things - 1. Hot fluid side, 2. Cold fluid side & 3. Geometry or thermo means U A LMTD.

All three must balance.

In your new condition.... Copy all the data in adjacent cell in excel. Change hot fluid flow & assume hot fluid outlet temp. make a Qh for hot fluid.

Put this Qh in place of Qc & back calculate Outlet T for cold fluid with increased cold fluid flow.
Now accordingly LMTD will change if U have put in a formula.
A is fixed. U can be changed based on individual change in fluid flows.....e.g. Break original U in hi & hio and then apply correction in hi & hio for higher flow using 0.8 rule.
Now combine new hi & hio to get new U.

All this should be in formula form...

Now using new U you will get final Q0 now this Q0 should be equal to Qh or Qc which are already eqaul.

So now just change (Or use goal seek) the Outlet T of hot fluid to equal Q0 & Qh. This is your final answer & will be the actual consition for rvised case.

Sponsored Links