Overall Heat Transfer Co-efficient Calulation based on Trials

Great day Readers.....!!!
This day i wanna demonstrate the way of determining the value of Overall Heat Transfer Co-efficient based on Practical Heating and Cooling trials, Which will return us the perfect value for an heat transfer equipment.

Previously i've demonstrated the Calculation of Overall Heat Transfer Co-efficient based on Properties, which seems to be somewhat difficult when you donno the properties of the Jacket fluids exactly, but this time i'll make it somewhat easy for you as this includes a pure practical approach.

For many Engineers the Overall Heat Transfer Co- efficient calculation remains as a nightmare, because the calculation will involves many terms and conditions and also many formulas, but here i'd like to use minimum formulae, which were well known by all engineer's.

Those well known formulae were,

1. M x Cp x dT - which is used to calculate the heat transfer to a fluid or a matter from source or heat transfer from fluid or a matter to sink.

2. U x A x LMTD - Which defines the overall heat transfer in between to matter of state separated by a material boundary.

May be right now, you came to know how i can derive a relation for calculating the Overall Heat Transfer Co-efficient here, deriving a relation is much more simpler, but incorporating the values i the formulae for calculation is some what sensitive stage in this.

First of all you need to know, how to take a Perfect trial .....!!

Taking trials is so easy, as we simply use water as mass in a batch heat exchanger, i'll demonstrate a simple case study now,
Consider i need to calculate the overall heat transfer coefficient for water as vessel mass and steam/hot water as jacket fluid, let the reactor volume be 5KL and then charge 3KL water into it, and start heating it to 70°C from RT [30°C], then just note down the time taken for heating the 3KL mass, use hot water at 90°C for heating as utility, and it has taken 2 hrs[say].

then try to incorporate the figures in the following relation,

U = M x Cp x dT / A x LMTD,

Here you need to know what is the heat transfer area of a 5KL reactor, then it is too simple to calculate it, follow this post

How To Find Reactor Heat Transfer Area ? 

so, now Heat Transfer Area of the 5KL reactor is 12.55Sq.m [L/D ration is considered as 1.15],

IF JACKET UTILITY IS HOT WATER

LMTD = (88-30)-(90-75)/Ln(88-30)/(90-75) = 31.85°C
 
U = 3000*1*1*(75-30)/(12.55* 2*  31.85) = 168.86KCal/hr.Sq.m.°C.

So, this is for water Vs Hot water at 90°C,

IF JACKET UTILITY IS STEAM

Now i'll Calculate for water vs Steam [ 110°C],

For the both cases only LMTD will vary, everything remains the same.

LMTD = (110-30) - (110-75)/Ln(110-30)/(110-75) =  54.87°C

U = 3000*1*1*(75-30)/(12.55*2*54.87) = 98.02KCal/Sq.m.hr.°C.

That's it, Here you need to know that U value vary depending on the Jacket utility temperature [T] , and the Temperature differences [dT], Time taken for the Heat Transfer [ t ], Heat Transfer Area [ A ],

So, my request to all the readers is that don't blindly go with thumb, if you do so the consequences will be severe to face.

 Some Typical U values, which can meet practical cases by some deviation, shown below,

Thanks for spending your time reading all these, Say cheers if you understood the above,
Any queries feel free to ask us, We will respond you,

Comments are most appreciated.....!!! 
 

 About The Author

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Hi! I am Ajay Kumar Kalva, Currently serving as the CEO of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.
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