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Proper way of sizing a steam Trap 1

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shup0739

Mechanical
Feb 26, 2003
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Hello All,

I am trying to size a steam trap for a main steam header.
I have a 165 psig steam header and would like to put a trap at the low point. I've done some calculations and my load comes out to less than 1/2 lb/hr. I'm not sure if that really makes sense, I was anticapating more but I might have made a mistake in my calculation can someone please tell me what I'm doing wrong...here is my calculations:


A:External Area of Pipe, ft^2: 1.178
U:Heat Transfer Factor, Btu/ft^2/F/hr=0.343636364
Delta T:Temperature Diff(steam temp - air temp), F=304.9992
E: 1 minus Efficiency of Insullation= 0.1
H: Latent Heat of Steam, Btu/lb = 844.9
S: Safety Factor(End of Header)= 3

Q :Condensate Load, lbs/hr= A*U*(DELTA T)*E*S/H = .04383884

This is telling me that I will anticapate about 4 % of a pound of condensate. which comes out to about a pound per day.

Thats not much at all.

I have crane tech 410 but I can't seem to find anything different.


Regards,

Sam
 
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Perhaps you calculation is ok for steady state.
However, upon startup, transient conditions should be taken into account. The metal of the piping will heat up to approximately the steam temp from cold contitions.
Also the insulation will heat up.
The mc(delta t) of the metal pipe and insulation will result in more condensate.
 
As a rule of thumb, use the steady state flow x 2 or 3 to give the start up rate. I would only expect a 1/2" trap for a simple drain anyway.

To do the calc accurately you would need the heat loss of the pipe and the heat to raise the mass of steel through ambient to operating conditions in the allotted warm up time.

This takes a while to do, but thankfully, steam traps are flexible in their sizing and the 'rule of thumb' usually suffices.

Friar Tuck of Sherwood
 
shup0379, Check with the Armstrong Machine Works and get a copy of "Steam-A-Ware" sizing and selection software. They may even have it on their website. Sizes and selects, all in one neat little package.

Hope this helps.
saxon
 
In addition to the comments already posted, I think you're forgetting to multiply the external area by the length of the pipe. 1.178 ft^2 is for 1 ft of 4" pipe.
 
shup0739:

What Saxon is advising is exactly what you should do. sailoday28 also brings up the practical, empirical fact that initial startup has to be considered because it will be the biggest instantaneous condensate load on the trap.

This is a classic heat transfer example covered in detail in Kern's "Process Heat Transfer", example 2.5; page 19. You are using the wrong terminology and algorithm to determine the amount of condensate formed. (There is no "U" involved in the analytical calculation; there are film coefficients that are in play - not an overall heat transfer coefficient). You have to calculate the heat loss from the steam main, taking into account all the conduction and convection factors such as insulation type, thickness, metal wall, wind effect, etc. Once the heat loss is calculated the conversion to condensate formed is a snap. But, on a more practical basis, Saxon brings out the fact that steam trap companies have already resolved the easiest and fastest manner of doing this calculation - either with tables or a "slide-rule". Now, they may even have a quick, PC program to do the routine calculation. Since you depend on the steam trap manufacturer for the equipment, I would also recommend you follow their recommendations on what load to size the trap for - and which type of trap is recommended by them - given all the basic data that you have on the application. It's the most practical and cost-effective way to resolve steam trap applications.

I hope this experience helps.

Art Montemayor
Spring, TX
 
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