Smart questions
Smart people
 Find A ForumFind An Expert
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Remember Me

Are you an
Engineering professional?
Join Eng-Tips now!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### LINK TO THIS FORUM!

Add Stickiness To Your Site By Linking To This Professionally Managed Technical Forum.
Just copy and paste the
code below into your site.

#### Partner With Us!

"Best Of Breed" Forums Add Stickiness To Your Site

#### Feedback

"...These forums are an excellent source and example of the way people can help each other..."

#### Geography

Where in the world do Eng-Tips members come from?

# Heat exchanger flow rate

 Forum Search FAQs Links Jobs Whitepapers MVPs
 RS82 (Mechanical) 4 May 12 19:58
 Hello, could you someone explain to me the effect of increasing the process fluid flow rate, in the tube, in regards to the heat transfer to the cooling water in the shell?What happens if I increase or decrease the flow rate in the tube by keeping the cooling water flow rate constant?  What happens if the cooling water flow rate in increased or decreased with the process fluid flow rate constant?I was reading a heat exchanger chart which shows the outlet temperature with respect to the process fluid flow rate..it showed the inlet temp as 500F and with a flow rate of 2GPM, the fluid cooled down to 300F..and with 1GPM it went from 500F to 200F..by increasing the process fluid flow rate, is the heat transfer rate reduced?
 Latexman (Chemical) 4 May 12 21:25
 Q = U A ΔTLM = w CP Δt Then you have two choices.  U is constant or U changes with the flow rate.Sounds very textbookish to me. Good luck,Latexman
 IRstuff (Aerospace) 5 May 12 3:03
 "it showed the inlet temp as 500F and with a flow rate of 2GPM, the fluid cooled down to 300F..and with 1GPM it went from 500F to 200F..by increasing the process fluid flow rate, is the heat transfer rate reduced? "Really?  How did you come to that conclusion?  Perhaps you're just being sloppy with terminology.  Since you say nothing about your cold side, I'll assume that nothing changed.  If nothing changed, then the transfer RATE is constant.  What's changed on the hot side is that you decreased the POWER = Joule heat*flow rate.  Decreased power results in a larger temperature drop
 Nivrah (Mechanical) 6 May 12 21:05
 RS82,changing the process fluid flow rate will change its exit temperatures.the overall heat transfer coefficient is dependent on both the cooling water and process water flow rates since it is a function of the convection heat transfer on the shell side and tube side. Any heat transfer textbook will show u a basic understanding on how and why it depends on these flowrates.when you change the process fluid flow rates, the amount of heat transferred from the cooling water to the process fluid does not change assuming that your inlet temperatures stay constant. What happens is that your U, exit temperatures and hence ΔTLM will adjust to match the "same" heat transferred.On the other hand, your overall heat transfer area remains a constant.
 rmw (Mechanical) 9 May 12 21:46
 Heat transfer is a function of the three "t's", time, temperature and turbulence.  Velocity interacts with two of them in a major way and one in a lesser way.Let's assume for a moment that the temperature rise of your cooling water is constant regardless of velocity - not true, but hold temperature for a moment.Your Hx will have a "U" value at a given velociy with units expressed in heat units per unit of tube surface area per time unit - Btu's per hr per square foot per degF in English units.  What is happening on each side of the tube will influence the "U" value.  For now we are holding the process side constant.  Therefore, if you increase  the cooling water velocity to infinity, your fluid won't stay in the Hx long enough to absorb any heat.  Obviously you wouldn't run it that fast, but that extreme is given to make a point.On the other hand, as the velocity increases, (to a point) the turbulence increases, so you get better heat transfer at the higher reynolds numbers.Now add the temperature back in because it is important to the conduction through the tube wall and the ovreall LMTD of the Hx.  Slow the fluid down so much that it heats up to its exit temperature too soon and little to no heat transfer occurs in the last few feet/meters of the tube(s).So it is a balancing act. Picking the velocity that will give you the best heat transfer for the surface area available.rmw
 RS82 (Mechanical) 10 May 12 6:21
 Thanks folks..so if I would like to increase the heat transfer of the process fluid through the tubes, I should increase the flow rate to ensure it is turbulent flow..this way the exit temp will lower as opposed to slowing the flow rate down which creates a laminar flow which depends on the thermal conductivity of fluid..did I get this right?
 Latexman (Chemical) 10 May 12 7:38
 Somewhat, the flow regime does not depend on thermal conductivity of fluid.  What year in school are you?  It appears you have not had fluid flow or heat transfer yet.  Replies can accomodate students if we know.  Otherwise, we assume you have at least a BS. Good luck,Latexman
 willard3 (Mechanical) 21 May 12 8:21
 From forum guidelines:Academic related posts are discouraged. If you post topics related to undergraduate course-work or projects, your posts will be red flagged and marked for deletion. Students generally have vast resources available to them (libraries, professors, etc), and need to learn how to use those sources instead of relying on the Eng-Tips.com fora (it is similar to having someone else do your homework for you; not a good idea.)
 zekeman (Mechanical) 2 Jun 12 0:38
 "when you change the process fluid flow rates, the amount of heat transferred from the cooling water to the process fluid does not change assuming that your inlet temperatures stay constant. What happens is that your U, exit temperatures and hence ΔTLM will adjust to match the "same" heat transferred."By what magic do you get this??
 25362 (Chemical) 2 Jun 12 5:12
 On the process fluid side:Before: Q1 = w1Cp1ΔT1After: Q2 = w2Cp2ΔT2Now replacing the symbols with numbers one gets the following:Q1 = (2×200) Cp1Q2 = (1×300) Cp2Assuming Cp1 ≈ Cp2 → Q1 = (400/300) Q2 Then, to understand what happened, see Latexman's answer, for the overall equation. Q = U×A×LMTDBy keeping A constant and changing the others, since Q dimished, it means U×LMTD decreased.

#### Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!