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mCpdT and UAdT
2

mCpdT and UAdT

mCpdT and UAdT

(OP)
Can anyone explain the difference between mCpdT and UAdT..

RE: mCpdT and UAdT

they both equal energy, the first is a physical measurement, the second is an emperical estimate.

RE: mCpdT and UAdT

UAdt is an exchanger design and performance equation.  It is a function of thermal performance, LMTD, exchanger area and configuration, etc.

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

you can mess with the  mCpdT equation all you want and get an exact answer (within measurement limits).  But the emperical UAdT is just that, you can try to optimize the equation in the linked post with the UAdT, but you have several series unknowns, fouling factors.

Your exchangers are being coated with coke or salt causing pressure drop and lower heat transfer rates, U.

RE: mCpdT and UAdT

(OP)
Thanks for your replies.
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://www.eng-tips.com/viewthread.cfm?qid=218659&page=1  

RE: mCpdT and UAdT

The simplist is temperature approach. Look at how close the temperatures are between the inlets and outlets on both ends of the exchangers.  If the design called for say 10 degrees F between the cold in and hot out on one end and 15 degrees between the cold out and the hot in and now its 14 and 20 degrees with the same rates, then there is fouling to calaculate how bad, use Q= UAdTlm.  

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

2
Practical engineering calculations can be summarized as falling into one of three catagories; namely:
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

Doug:

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

Thanks for the kudos. Star? :-)

RE: mCpdT and UAdT

Can I please ask one question!

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://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

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

Thanks 25362

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

(OP)
djack, dcasto-
Thanks for your inputs


Rgds,
rgrokkam

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