Talk:Orifice plate

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[edit] Something seems wrong

The derivation of the orifice equation doesn't seem right somehow. I'll be back after I've had a chance to research it. Meanwhile, I cleaned up the math equations quite a bit (without changing them) and did some other very minor fixes in the text body of the article. - mbeychok 02:42, 1 May 2006 (UTC)

Yes I noticed that as well. Can't do any substantial cleanup while at work, but if someone could go through that would be appeciated. The left side of the equation remains unchanged, but the right side is modified by introducing non-unity factors such as C_d. —The preceding unsigned comment was added by 24.73.96.170 (talk) 14:02, 3 January 2007 (UTC).

[edit] Complete re-write and expansion

I have just completed re-writing this article and I believe it now a better article. I still intend to add a section on gas flows through orifices in the next few days. - mbeychok 22:53, 2 May 2006 (UTC)

Looks very good, Mbeychok. I just have one suggestion, to shorten the lead section per WP:LAYOUT#Lead_section. The equations would probably work better in their own section. Coming into the article cold, I'm used to a general lead section with details in later sections. Spalding 03:20, 15 July 2006 (UTC)

Done. Thanks for the suggestion. - mbeychok 04:25, 15 July 2006 (UTC)

[edit] Losses in orifice plate

Thank you for your useful article on the pressure drop across an orifice plate. It would be useful if you could add a section on the losses that occur across the orifice plate. It is frequently necessary to know the losses in a system and in some instances - such as in the outlet to a variable declining rate water filter - an orifice plate is used to increase the losses to prevent excessive flow. 217.15.119.158 13:40, 5 December 2006 (UTC)

Perhaps I don't understand your question, but Equation 2 in the article relates the mass flow to the pressure drop (P₁ - P₂) across the orifice for a liquid. By a simple re-arrangement of that equation, you could easily solve for the pressure drop, which is the pressure loss across the orifice, is it not? - mbeychok 19:28, 5 December 2006 (UTC)

Thank you for your response. The pressure change in Formula 2 is due to the change in velocity in the orifice arising from the the Bernouilli Equation. If you measure the pressure downstream of the vena contracta the pressure will recover but not to the same value as the upstream pressure because of energy losses in the device. It is these losses that I would like to be able to quantify. 217.15.119.158 11:26, 6 December 2006 (UTC)

I apologise for mis-understanding your question. I don't have one at hand, but any good textbook on fluid dynamics or fluid flow should have that information for you. - mbeychok 16:57, 6 December 2006 (UTC)

[edit] Reason for deleting external link added by Loughmsa

These are the reasons why I deleted the external link to a "Simple orifice flow calculation":

• The equation in that link is only good for air and only for one specific air density (which means only for one specific temperature and pressure).

• It is only good for the use of USA customary units, and we are now pretty much the only nation still using those units. The Wikipedia has a very large non-USA readership.

Basically, the linked calculation method was too limited and too simple for an article that is intended to explain orifice flow equations for use with any gas or any liquid. - mbeychok 01:46, 23 January 2007 (UTC)

[edit] external link not working

Flow through a sharp-edged orifice external link is not working please remove the same if it does not work for more. 59.180.94.55 10:49, 23 January 2007 (UTC)

I fixed that link and it now works. It requires that Shockwave be installed and it did that automatically ... for me at least. - mbeychok 17:05, 23 January 2007 (UTC)

[edit] Calculation of coefficient of discharge from Reynolds' number?

The article says that the coefficient of discharge can be calculated from the Reynolds' number (source given is Perry). If someone knows how to calculate it, or has a copy of Perry to hand, I would be very grateful if they would add this information to the article (I have googled extensively in vain). Cheers, 84.12.252.210 15:49, 4 July 2007 (UTC)

[edit] Orifice pressure for viscous fluids

If anyone would like to fill in orifice pressures for viscous fluids... after all, most of them are, and it becomes pretty important at small radii... Would be really nice!!! Cheers Michi zh 17:49, 26 September 2007 (UTC)

[edit] Measures vs. allows measurement

What distinction was the object of the last edit (09:16, 8 October 2007)?

The previous version says that an orifice plate "measures the rate". It was changed to say that it "allows the measurement of the rate".

It raises the question: If the orifice plate only _allows_ the measurement to be made, then what actually _does_ the measuring?

It also raises the question: _How_ does it _allow_ the measurement?

Perhaps the point is that only people can measure? Or perhaps that a connected differential pressure meter does a measurement which is ultimately interpreted as a flow rate?

Maybe the point is that the orifice plate is but one component in a total system which reports a flow rate.

The revised form is wordier. Maybe it is necessary to expand the reader's vision beyond this one component.

To be consistent, the next sentence would seem to need tweaking as well: "It uses ... Bernoulli's principle". Actually, the orifice plate itself doesn't do this -- a person interpreting a delta_P reading might Bernoulli's principle. More likely, they would use a precalculated chart.

Perhaps too much precision in language ultimately obscures the message by building a haystack around the needle?

I'm not suggesting undoing the last edit. I'm just trying to understand why the wording was changed and to explore how much anthropomorphizing is appropriate in technical writing. -- Ac44ck 16:01, 8 October 2007 (UTC)

I completely agree with you. So I changed the wording again to avoid the excessive precision in the language. - mbeychok 17:31, 8 October 2007 (UTC)

[edit] Redundancy is in the eye of the beholder

Good catch, mbeychok, on the squaring of the Cd term in the formula for permanent pressure drop.

As to whether the last equality is "needed", I would argue:

First:

It could be handy to have an alternate, equivalent form readily available when a deadline is looming.

The cost of having this additional information available is practically zero. The cost of botching a hurried transformation of the equation (by overlooking the need to square a factor) may more significant.

That is why I rearranged the pressure drop equation when I created this section of the article: to have the equivalent form readily available. Unhappily, I overlooked the need to square a factor when doing so -- and the mistake survived for months in a document which is subject to public scrutiny.

Second:

(Q/A1) = V1 = constant regardless of bore diameter if the flow rate and pipe size are known.

(Q/A2) = V2 varies with bore diameter and is generally of less interest to me.

I like to be able to see formulae which are more directly related to V1, as the last equality is.

Third:

The last equality can be handy in situations where the bore diameter is to be determined based on the desired pressure drop, known pipe size and flow rate.

Using the last equality, one can iterate for beta without caring about or having to keep track of a tentative bore diameter during the process.

--Ac44ck 22:07, 17 October 2007 (UTC)

Ac44ck, I would very much like to respond to your above comment ... but I must leave in a minute for a very important meeting. Please bear with me and I will respond when I return in about 4-5 hours. Thanks and best regards, - mbeychok 22:40, 17 October 2007 (UTC)

Ac44ck, thanks for waiting. I could easily check your first equality and find it to be correct except for the lack of squaring the Cd term.

However,I could not do the same with the second equality. Not only did it have the same lack of squaring the Cd term, I simply don't see how that second equality of yours ended up with a β⁴ in the denominator. That made me worry enough to ask myself, do we we really need it?

If you would correct the Cd squared errors in both equalities as well as explain to me how β⁴ ended up in the denominator of the second equality, then I would have no objection retaining your second equality. Again, best regards. - mbeychok 03:15, 18 October 2007 (UTC)

mbeychok, thanks for the further reply after your long meeting.

The transformation seemed a bit tricky to me. Perhaps it wouldn't have been so tricky (mumble) years ago in college, but transforming equations isn't something that I do every day now and I didn't want to lose the effort under a pile of papers. So I added the second equality to the article -- where someone else is doing the archiving, etc.

The denominator of the first equality contains the factor A₂².

The square-root of that factor can be rewritten as follows:

A₂ = A₁ * (A₂ / A₁) = A₁ * (d₂ / d₁)²

But: β = d₂ / d₁

So: A₂ = A₁ * β²

And:

A₂² = A₁² * β⁴

So the transformation to the second equality replaces A₂² in the denominator with A₁² * β⁴.

I considered dividing through by β⁴ -- changing the factor in the numerator to (1/β⁴ - 1) -- so there would be only one instance of β in the second equality. I might write it that way in a spreadsheet formula, but it seemed better to avoid stacking fractions in the printable form.

Thanks for finding that the Cd factor wasn't squared.

Best regards --Ac44ck 04:56, 18 October 2007 (UTC)

I am happy that we were able to settle this amicably. In an earlier response you said ... and the mistake survived for months in a document which is subject to public scrutiny. I assume that you have now corrected that document as well. Again, with best regards, Milt Beychok. -- mbeychok 05:16, 18 October 2007 (UTC)

[edit] formulae expansion factor

The formulae, stated for the expansion factor is used with nozzles and venturis, according to perry's handbook. The formulae for Y for orifices is different :

Y = 1 - [(1-r)/k](0.41+0.35Beta^4)

Comment plz. —Preceding unsigned comment added by 195.73.116.62 (talk) 10:08, 5 March 2008 (UTC)

I found the same mistake in the formula. The two formula gives quite different results. Using the current formula Y=0.20 and using the above formula, Y=0.71. Pls correct the formula in the article —Preceding unsigned comment added by 203.199.60.145 (talk) 04:58, 13 May 2008 (UTC)