U Value of Half Pipe Jacket
U Value of Half Pipe Jacket
(OP)
I am trying to determine the overall heat transfer coefficient of our half pipe jacket for a reactor for a HAZAN. We can get the following data minute by minute:
Jacket Flow Rate
Jacket Inlet Temperature
Jacket Outlet Temperature
Mass Temperature (would just be water for the test)
So we would do this for a cooling cycle from 70 C to 50 C with just water in the reactor. I am going to calculate the total heat transfer area of the jacket. So here's my calculation methodology:
From data, we can get total heat transfered from reactor to jacket by:
Q=(m)(Cp)(deltaT) (1)
Using the average heat capacity of water in the delta T range.
Then could find U by
U=Q/(A*deltaT) (2)
Now for heat exchangers, I know you use the log mean temperature difference. For the reactor, would I just use the deltaT for the jacket in equation 2 like I used in equation 1? Or would I figure out in the minute rannge the temperature decrease in the reactor mass and use that delta T to get the log mean temperature difference?
So time 0 minute I would have a mass temperature
At Time 1 minute I would have a delta T in the reactor, a inlet jacket temperature, jacket flow rate, and outlet jacket temperature.
Cal calculate log mean temp difference, heat transferred, and U for that period. Or is just using delta T in the jacket suffice without using mass temperature?
Also to get just one U value, would I average all the U values calculated in this timeframe (should be close hopefully)?
Please review and give any good past practices for this type of calculation.
Thanks
Jacket Flow Rate
Jacket Inlet Temperature
Jacket Outlet Temperature
Mass Temperature (would just be water for the test)
So we would do this for a cooling cycle from 70 C to 50 C with just water in the reactor. I am going to calculate the total heat transfer area of the jacket. So here's my calculation methodology:
From data, we can get total heat transfered from reactor to jacket by:
Q=(m)(Cp)(deltaT) (1)
Using the average heat capacity of water in the delta T range.
Then could find U by
U=Q/(A*deltaT) (2)
Now for heat exchangers, I know you use the log mean temperature difference. For the reactor, would I just use the deltaT for the jacket in equation 2 like I used in equation 1? Or would I figure out in the minute rannge the temperature decrease in the reactor mass and use that delta T to get the log mean temperature difference?
So time 0 minute I would have a mass temperature
At Time 1 minute I would have a delta T in the reactor, a inlet jacket temperature, jacket flow rate, and outlet jacket temperature.
Cal calculate log mean temp difference, heat transferred, and U for that period. Or is just using delta T in the jacket suffice without using mass temperature?
Also to get just one U value, would I average all the U values calculated in this timeframe (should be close hopefully)?
Please review and give any good past practices for this type of calculation.
Thanks





RE: U Value of Half Pipe Jacket
Good luck,
Latexman
RE: U Value of Half Pipe Jacket
LN(T1-t1/T2-t1)=U*A*time/MC
Where:
T1 = intial temperature o the batch
t1 = temperature of vaporizing medium
T2 = batch temperature at "time"
M = mass in the reactor
C = specific heat of fluid in the reactor
So from these assumptions, can I use the data in my first post to rearrange to get U? Is an isothermal process a good assumption in this case?
The jacket fluid is water and water is in the jacket.
Please review and comment.
Thanks
RE: U Value of Half Pipe Jacket
There is a "Plant Notebook" article in the ChE issue of May 1989 titled How to predict batch-reactor heating and cooling by John McEwan, and another in the ChE issue of September 1996, titled Estimating Batch reactor heating or cooling time by V.V. Mahajani, both articles dealing with jacketed reactors, probably providing answers to your query.
RE: U Value of Half Pipe Jacket
No, use the “Coil-in-Tank or Jacketed Vessel, Nonisothermal Cooling Medium equation”:
LN((T1 – t1)/(T2 – t1) = wc/MC((K2 – 1)/K2)θ
K2 = exp(UA/wc)
In Kern, he left out that this also applies to a Jacketed Vessel (compare to the Nonisothermal Heating Medium equation on the previous page).
The equation can be rearranged to get U, you just have to go thru K2 to get it. Plot LN((T1 – t1)/(T2 – t1) versus θ in Excel. Regress the data to get the slope and use that to calculate U.
Good luck,
Latexman