mCpdT and UAdT
mCpdT and UAdT
(OP)
Can anyone explain the difference between mCpdT and UAdT..
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS Come Join Us!Are you an
Engineering professional? Join Eng-Tips Forums!
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail. Posting GuidelinesJobs |
|
RE: mCpdT and UAdT
RE: mCpdT and UAdT
mCpdT is a singleside heat transfer equation. It accounts for the heat capacity of the fluid and the enthalpy change that is occurring.
RE: mCpdT and UAdT
mCpdT or mH (latent heat change) is heat balance equation.
UAdT is heat transfer equation.
You are starting from a given heat balance (duty), and then you choose particular heat transfer application where this heat exchange can be executed in a most effective way.
http://antwrp.gsfc.nasa.gov/apod/astropix.html
RE: mCpdT and UAdT
I seek your comments on the thread below:
http://www
RE: mCpdT and UAdT
Your exchangers are being coated with coke or salt causing pressure drop and lower heat transfer rates, U.
RE: mCpdT and UAdT
Hi dcasto,
I have a series of unknowns- thermal conductivity, viscosity, outlet temperature, fouling factor.
"Your exchangers are being coated with coke or salt causing pressure drop and lower heat transfer rates, U."
Is there any way of quantifying the decrease in heat transfer rate?
what could be the best way to predict the performance of the heat exchgr?
http
RE: mCpdT and UAdT
From the design the exchange had a U of 50 and know when you run the calc its only 40, you have basically lost 20% of your exchanger. You can do lost work analysis and all that, but I keep it simple. The lost work will tell you how much energy you lost and when the economics are to shutdown and clean. At todays energy prices, if you get under 90% eff, shutdown and clean.
RE: mCpdT and UAdT
1) Design,
2) Rating, and
3) Evaluation.
In a design problem, you know what it is you need your equipment to do and you proceed to "size" your equipment, presumably before buying the equipment.
In a rating problem, you know what equipment you have and you are attempting to see if it is able to perform under different conditions.
In an evaluation problem, both the equipment and its expected performance are known and you are determining how closely the equipment matches expectations. Ideally, you would have historical data to compare against. Especially useful would be data pertaining to the equipment when new and clean (best possible performance). rgrokkam, you can track degradation of your exchanger's performance by seeing how U falls with time. This happens for the reasons stated in some of the other posts.
Doug
RE: mCpdT and UAdT
Excellent summary - a great reference for when somebody asks me the difference. Great to explain exchanger performance, rating, and finally design parameters.
RE: mCpdT and UAdT
RE: mCpdT and UAdT
I have a situation. I need to make sure that a particular cooling unit / Equipment is suitable for a process or not. I am bit new with this type of assignments! I am sure it won't be very challenging for the experts but I really wish to get some help with this one. So here is the problem.
I have a vessel. Working volume for the process wil be 20 L. Starting temp. 37C End temp for the vessel content needs to be at the 8 C. There is a time restriction. Vessel need to go from 37C to 8 C in 30 minutes.
Now I used Q - mCpdT equation to find out the amount of power required.
here is the calculation
M = mass of vessel content. I assumed to be water for the calculation purpose. Working volume is going to be 20 L =
M = 20 Kg
Cp = I used Cp of water at 40 C found from
http://ww
Cp = 4.179 KJ/Kg/K
dT = 29 K
So Q = 2423.82 KJ
Time 30 minutes
So Rate of cooling heat removal = 1.346566667 KJ/sec
1 KJ 238.8459 Cal
Rate 321.6219274 Cal/ sec
1 Watt 0.2390585 cal /sec
Power 1345.369135 W
1.345369135 KW
Now I know that the equipment specification is 0.5 KW at -20 C
Does this mean that I can conclude that Equipment won't be able to meet the cooling requirement in a given timeframe.
Please suggest and comment on this calculation.
Mant Thanks in Advance
JP
RE: mCpdT and UAdT
JP: you should have started a new thread altogether. Anyway,
just referring to the calculation. 1 kJ/s = 1 kW by definition, therefore there is no need of so many transformations.
RE: mCpdT and UAdT
I did realize that few minutes ago! What a waste of time. :)
I will start new thread as well.
Thanks again
RE: mCpdT and UAdT
Thanks for your inputs
Rgds,
rgrokkam