fouling factor and oversurface
fouling factor and oversurface
(OP)
I'd like to know the formula that permit to correlate the value of heat transfer coefficient in case of an heat exchanger with and without fouling, with the value of the fouling factor.





RE: fouling factor and oversurface
I'm not quite sure I understand what you're wanting - are you trying to figure out how to compare a clean to a fouled heat exchanger? Or do you just need the formula for finding the fouling factor in terms of Q?
Patricia Lougheed
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RE: fouling factor and oversurface
usually, when I ask for an heat exchanger, I indicate these values: obviously flow and temperature of the fluids, and fouling for both the sides.
I want to understand how correlate the value of overall fouling (e.g. 0,0002 m2°C/W) with the excess of surface: I usually calculate it as the difference between the inverse of the transmission coefficient for the clean and dirty heat exchanger : 1/kdirty- 1/k clean, where the 2 heat coefficients are known, and then I recalculate (from the temperature difference) the surface of the heat exchanger: this let me to know the consequences (in terms of surfaces and then money) of the chosen of the value of fouling.
My question is, therefore this: the equation above reported, is correct?
or does exist another formula to relation fouling and heat transfer coefficient for dirty and clean HX?
Thank a lot for Your attention, and have a good day
Gabriele
RE: fouling factor and oversurface
HT coefficients in Btu/(h.ft2.oF) :
tubes' inside film HT coefficient : 250
tube metal wall: 3500
tubes' outside (shell-side) film HT coefficient: 50
resistances in (h.ft2.oF/Btu)
external fouling factor: 0.001
internal fouling factor: 0.002
The various resistances would show to be:
1/250 + 1/3500 + 1/50 + 0.001 + 0.002 = 0.0272857
The calculated value for U dirty would show to be:
1/0.0272857 = 36.6. The U clean = 41.1
Now, if one tabulates the various resistances as follows:
ri = 1/250 = 0.0040000 = 14.7 %
rw = 1/3500 = 0.0002857 = 1.0 %
ro = 1/50 = 0.0200000 = 73.3 %
fouling factors = 0.0030000 = 11.0%
total resistance= 0.0272857 = 100 %
One can see that by improving the outside (shell-side) HT film coefficient (having the highest resistance of 73.3%), for example, by adding baffles and paying for more friction drop, we can directly improve the efficiency of this particular HE.
I sincerely hope the exercise helps to grasp the importance of analysing the HT resistances.
RE: fouling factor and oversurface
A lot of work By E. Somerscales on fouling is also available.