Heat Exchanger Fouling Effect.

[murphymok]

Hi,

I am analyzing a steam to water shell/tube HX. When there is an increase in fouling, what changes should I expect on temperature and flow of both sides? Will the steam flow increase? Or will the temperature change?

Thanks.

 

[sailoday28]

Assuming the fouling is on the inside (water side) of the tubes, the fouling will act as an increased flow and thermal resistance.
Impact on cooling water side:
will depend upon where for example the system pump curve intersects the new system resistance curve.
Overall thermal impact will now have to consider new flow, and new overall heat transfer coefficient.
Will steam side flow change? Is the flow from a turbine and choked?.
Will steam side temp change? In general with increased fouling, yes.
Will cooling side temp change? Assuming lower cooling flow, yes. HOwever, if overall U=0. No change of temp on cooling water side.
Many variables, many assumtions on steam and water side can lead to many different solutions?

 

[murphymok]

The HX is used for domestic hot water heating. For my analysis, I am keeping the steam side constant (flow and temp) to make things simple. I am also keeping the inlet water constant. So the variables are the outlet water temp, water flow and u-value.

Right now, I am changing the u-value to simulate fouling. Do I keep the same Qwater and calculate the required water flow?

Thanks.

 

[rmw]

Go to

http://www.processassociates.com

select process tools, and use

http://www.processassociates.com/process/heat/hi_calc.htm

to get the tubeside coefficient, and use

http://www.processassociates.com/process/heat/ho_calc.htm

to get the shell side coefficient then use

http://www.processassociates.com/process/heat/u_calc.htm

to put various fouling factors in and watch the OHTC change accordingly.

You can put a shellside fouling factor representing the steam side in and hold it constant and vary the tube side or vice versa, or both at once, as you wish.

This should give you a clear picture of what is going on in your Hx.

rmw

 

[murphymok]

Thanks guys, great helps!

I am going to try the programs and see what my HXs are doing.

Thanks again.

 

[25362]

It is apparent that by keeping the steam flow rate constant murphymok tries to maintain the heat transfer rate (Btu/h)constant in the simulation.

Thus Q = constant = (mCp[Δ]T)[sub]water[/sub] = AU(LTD).

U (LTD) would have to remain constant since the area, A, is.

By increasing the water flow rate, m, (turbulent regime Reynolds number) one could estimate the new heat transfer coefficient by convection on the water side that would overcome, at least in part, the added resistance originated by fouling, if one knew the value of this resistance; all this, of course, at the expense of pressure drop 9pumping energy). Cp is assumed to remain approximately unchanged.

One could then estimate by use of iterations the value of water's outlet temperature that would satisfy both equations above.

Murphymok, is this what you are planning to do ?

 

[EdStainless]

Remember, your outlet steam temp will rise also.

Just as a tip, higher flow rate are not only better for heat transfer, but also help minimize fouling.
Just watch your material limitations.

= = = = = = = = = = = = = = = = = = = =
Corrosion never sleeps, but it can be managed.

http://www.trenttube.com/Trent/tech_form.htm

 

[25362]

To EdStainless. In general, for all purposes, as long as the steam pressure is kept constant, so would its condensing temperature.

Condensate sub-cooling is generally neglected as a factor in steam heaters where the condensate is removed (for example, by traps) immediately upon being formed. Any comment ?

 

[murphymok]

To 25362 and all,

I know all the perimeters on the steam side (pressure, flow and temp). But on the water side, I only know the temps. By keeping the steam side constant, I would like to see a change in flow rate or temp to justify the inefficiency of the HXs and how much it would have improved after cleaning it chemically.

Thx.

 

[sailoday28]

(Ts-tout)/(Ts-tin)=e^(UA/wc)
Based on constant specific heat and mean U
Ts=tsteam tout= temp cooling water out
tin= temp cooling water in
With Ts, and tin known, the only unknowns are U and w
If you vary the quantity U/w solve for tout
OR vary tout and solve for U/w
Since U is fouled and you are picking the fouling factor, calculate w or tout.

 

[sailoday28]

correction exponential should be negative.