Pressure drop accross a wire mesh screen calculation? 1

[kenVTME90]

How do you calculate the pressure drop accross a wire screen mesh with the following parameters?

Air Flow, 68inch/sec thru a stainless steel woven wire screen mesh, #6 mesh, (0.12inch square openings) with .047 wire which is 51.8% open.

The numbers we calculated using the ASHRAE methods seemed too high to believe. We then did a crude measurement with a sample of the screen and found the pressure differential to be much, much less than what the ASHRAE charts are saying we should expect.

Can anybody help me out?

thanks.

 

[ChasBean1]

Which ASHRAE charts are you referring to? This one's from 2003 Applications Hand Book, Ch. 52.5:

Q = 2610 * A * dP^.5

Q is flow in cfm
A is open area in ft2
dP is pressure drop across the restriction, in. w.c.

Using this would estimate a pressure drop across the screen of 0.063 in. w.c. What did you measure?

 

[kenVTME90]

Thanks for the reply.

is the 2610 a constant or derived from a chart or what?

 

[SAK9]

Not sure whether the ASHRAE formula takes into account the flow coefficient which can be any thing between 0.6~0.7

An SI version of the equation is:

Flow,M3/s=CvxA(m2)xSQRT(2xDelta P(pa)/Density(kg/m3)

Cv= flow coefficient(0.6~0.7)(this accounts for the area reduction across the aperture due to the formation of Vena Contracta.

 

[AbbyNormal]

I would just base it on what you measured. It will be proportional to the velocity head.

the 68 inch per second face velocity is about 340 fpm, velocity head of dry standard air is (340/4005)^2=0.007" WC

on your 'free area' maybe the velocity is 340/.51=667 fpm, velocity head is 0.028" WC

Some thing that is abrupt or turns air 90 degrees without vanes and maybe the loss will work out to something like 1.5 times the velocity head. Something where the air is not changing direction will be less than the velocity head.

Use your own empirical data

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[ChasBean1]

SAK9, the equation does account for a typical flow coefficient as you indicated. It starts with the orifice equation as:

Q = 776 C A [(2 dP)/rho)]^.5

and offers a typical C value of .65 and a density of .075 lbm/ft3 to simplify.

Granted, this is designed to model crack leakage for smoke systems to look at pressures across doors, etc. The linearity might start to fall apart with large leakage areas such as we have here (52% open).

So kenVTME90, to answer your question, the 2610 comes from the 776 (very helpful, right?)

Can you post what the actual measured value was?

 

[kenVTME90]

We had a really basic testing setup. Not at all calibrated.

We are machine designer type engineers, not HVAC type guys.

but, we used a blower with about twice as much flow as required and measured about 0.2 to 0.25 inch WC differential with a homemade manometer.

Couldn't have made a much cruder test setup than that.

I appreciate everybodies help. I ordered the 2003 ashrae handbook off amazon.

I have a contract where I have to show the flow thru this screen is going to meet spec. This flow part is a small part of the contract, but I still have to be able to prove it.

 

[stanlsimon]

I hope you have filters upstream of the screen otherwise the screen will act as a filter and the pressure drop will become a function of the elapsed time since the last screen cleaning exercise.

 

[willard3]

SMACNA "Duct System Design" also has calculations for exactly the question you are asking with tabularized coefficients.

 

[zekeman]

The ASHRAE clculations are probably based on a conservative estimate of potential blockage during in the application and should be considered in the design.

 

[ChasBean1]

If you double the flow and plug back in the numbers the equation produces .25 in. w.c. as an answer... not too bad.