flow bench program

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Discussion

Stan Weiss

260 posts

147 months

Thursday 17th April 2014
quotequote all
PeterBurgess said:
Hi Dave

I hope the massive electricity outage in Scotland hasn't caused you troubles?

Once again I say thanks for pointing out the CD component of the maths was wrong. My work has and always will be empirically based not inferred or deduced from theory so I just go by the flow numbers and as long as they are repeatable/consistent I am happy. I can still put stuff on the ghetto bensh that you and I played with all those years ago and get the same figures. I added the CD output as an afterthought and I got it wrong. As I said we use the flow numbers for our work based comparisons. Mind you, we rarely seem to get the time to flow test these days, how about yourself? We test MGB head flow thoughts now and again and have been playing with quite a few different OE and aftermarket Harley heads for a mate who rides and tunes them....Dave Gollan and myself ended up buying Sportsters to pootle around on.....don't mention Harley bhp/litre though, we got around 120 engine bhp from a 1950cc Veetwin smile

Peter

Stan

I suppose it is one of the penalties for putting stuff in the public domain and making it open source. I am happy I did that as it is there for people to use ....especially when you have done the hard work of sorting out my crap scripting, so a big thanks to you Stan too. When we get some time we will plod through and see what has been done with your code. I think my script is too scruffy and I made the cardinal sin of not keeping my notes from 1999ish! I think I was just gobsmacked the prompts came up on screen and the flow figures were about right from the input smile

It is interesting that it has taken 2 years for the wrong cd maths to show up, does the page get many hits Stan? I was wondering if folk use the stuff or maybe copy the page, or if it is only of interest to a few of us folk?

Peter
Peter,
I do not have a counter on that web page so I can not really say how many hits it gets.

To the best of my memory when I did this a couple of years go all I did was clear up the code / interface.

Here are the 2 lines of code that I changed for the new version.

This one is now just a comment
// var E = eval((0.000789 / A) * 0.62);

This one should now calculate the correct CD and show 4 decimal places
document.stanForm.CD.value = eval(Math.round(((F / A) / 213683.427367) * 10000)/10000);

Stan

Edited by Stan Weiss on Thursday 17th April 14:08

PeterBurgess

775 posts

145 months

Thursday 17th April 2014
quotequote all
Thanks Stan

My script altered thus......

var C = prompt ("Enter the orifice coefficient","Orifice coefficient"); //to allow variable orifice coefficient input like your script

//var C = 0.596; // remove specific orifice CD

//var E = (O / A);
//var E = eval(E);
//var E = (E * 0.596);
//var E = eval(E);
//var J = ( X / Y);
//var J = eval(J);
//var J = Math.pow(J,0.5);
//var J = eval(J);
//var Cd = (J * E);
//var Cd = Cd * 100;
//var Cd = eval(Cd);
//var Cd = Math.round(Cd) / 100;
//var Cd = eval(Cd);
//var F = F * 100;
//var F = eval(F);
//var F = Math.round(F) / 100;
//var F = eval(F); //cut out the crap and replace with your maths
var Cd = eval(Math.round(((F / A) / 213683.427367) * 10000)/10000);

PeterBurgess

775 posts

145 months

Thursday 17th April 2014
quotequote all
Hi Stan......Just checking my old figures and think the constant should be 213685.34 perfect flow at 25" water delta p .....mind you still gives same answer with rounding smile


Peter

Pumaracing

2,089 posts

206 months

Friday 18th April 2014
quotequote all
PeterBurgess said:
Hi Stan......Just checking my old figures and think the constant should be 213685.34
Yes, that's the correct number for CFM per square metre.

PeterBurgess

775 posts

145 months

Saturday 19th April 2014
quotequote all
Hi Dave

I have been going through my script, I worked out (with my original script )CD by virtue of area and pressure drop comparison and have the following.....

CD = area orifice(O)/area valve(A) x 0.596(orifice coefficient) x sq rt (X/Y)

We have from Stan weiss example: 33mm valve, 22.5 mm orifice, X = 533, Y = 433 and CD = 0.596

I get:
O = 0.0003976
A = 0.000855
0.596 (CD O)

So (O x 0.596) /A = 0.2771
sq rt (533/433) = 1.10948
0.2771 x 1.10948 = 0.307

therefore CD = 0.31 as per my original script

I calculate from max flow mathematically the CD as same as you, why, and I am not trying to catch you out, does the maths above give a different answer, am I doing something wrong? It is 5% higher CD using the relationship of delta P, area and the 0.596 CD?

Your thoughts are most welcome Dave as I am trying to see what I have done wrong with my assumptions as the theoretical has to be the correct one whereas the empirical one is 5% higher?

Peter


Pumaracing

2,089 posts

206 months

Saturday 19th April 2014
quotequote all
PeterBurgess said:
Hi Dave

I have been going through my script, I worked out (with my original script )CD by virtue of area and pressure drop comparison and have the following.....

CD = area orifice(O)/area valve(A) x 0.596(orifice coefficient) x sq rt (X/Y)
I'm somewhat bemused by why you think this would calculate the Cd of the valve. The upstream pressure and therefore density of the air above the valve is not the same as that above the orifice because of the pressure drop through the test piece itself. Nor of course are the downstream pressures and air densities for both components because of the pressure drop across the orifice.

Therefore although the mass flow through the entire bench is the same at all times the CFMs are different for the test piece and the measuring orifice.

Your relationship would only pertain if there was almost zero airflow going through the bench and therefore the same upstream and downstream air densities through both components.

An alternative way of looking at it is your method would only work if "area orifice(O) x 0.596(orifice coefficient) x sq rt (X/Y) x 137.86" actually calculated the flow through the orifice which of course it doesn't or you wouldn't need all of the rest of the maths.

There is actually a mildly interesting albeit irrelevant point to be learned from this. The magnitude of the difference between the actual Cd of the valve and that which your equation calculates is directly related to the power of the vacuum source and therefore how big the pressure drops and air density differences it generates are. With a tiny vacuum motor and minimal flow through the bench the difference would be small but of course the manometer readings would also be so small there would be very little accuracy in the flow readings too.

Pumaracing

2,089 posts

206 months

Saturday 19th April 2014
quotequote all
PS - if memory serves, and it doesn't always, you may have been misled by Annand and Roe's work. It's nearly 25 years since I read it but I vaguely recall they also used a very basic square root law plus the manometer reading ratio to calculate flow. Unfortunately if so they also failed to take into account the difference in air density at the test piece and the measuring orifice.

In fact it is possible to use that basic law to "approximate" flow but only by using the "Flowbench Calibration Coefficient" rather than the actual discharge coefficient of the orifice. This roughly takes into account the power of the vacuum motor but sadly the coefficient changes as the flow increases although again from memory it stabilizes fairly quickly so only low flow numbers would show any great error. In the case of the numbers in your javascript the average Flowbench Calibration Coefficient would be just under 4% less than the actual Orifice Discharge Coefficient of 0.596.

Clearly in the case of flow testing using non compressible fluids like water this problem doesn't occur as the fluid density remains the same throughout.

When I first built my own flowbench I did calculate its Flowbench Calibration Coefficient (or at least the range of these as flow increased) so I could get a quick and dirty estimate of flow from the manometer readings during a flowtest without having to keep running from the workshop to the house to use the computer.

Edited by Pumaracing on Saturday 19th April 16:04

PeterBurgess

775 posts

145 months

Saturday 19th April 2014
quotequote all
Hi Dave

I didn't use Annand and Roe for my original flow maths. I used BS1042 and Ower and Pankhurst measurement of airflow.

I reckon I used basic cd comparisons for the CD maths, not Compressible flow as in the flow maths.I calculated the CD as an afterthought, using the flow figure for my comparative work. So, do we think the 5% represents what would be compressible flow?

Interesting the quick calc of CD you mention to compare figures rather than feed all the numbers in, good idea. Mind you I will start to use Stan's version as I don't have to repeat number input so much.

Peter

pingu393

7,719 posts

204 months

Wednesday 17th October 2018
quotequote all
I've used the online calculator was wondering why the vapour pressure default was 3 N/m². Using http://www.endmemo.com/chem/vaporpressurewater.php, the value should be 1697 N/m².

Hoping that I have used the vapour pressure correctly, I have modified the flowbench calculator so the vapour pressure is calculated inside the script...

http://www.porterbility.co.uk/flowbench_calculator...

Please let me know if I've done anything wrong.

PeterBurgess

775 posts

145 months

Wednesday 17th October 2018
quotequote all
Well done for including water vapour calculation, saves looking it up on a bit of paper/google.I don't know where Stan got his 3N/M2 from, my programme does not have default figures you just enter all info. I think the vapour pressure is 1697 kpa or 1.697 N/M2 at 15 C, Kpa to N/M2 divide by 1000.
If the same figures entered on your programme which includes water vapour calculation gives the same answers as Stan's one you have altered it correctly.
Peter

Mignon

1,018 posts

88 months

Wednesday 17th October 2018
quotequote all
pingu393 said:
I've used the online calculator was wondering why the vapour pressure default was 3 N/m². Using http://www.endmemo.com/chem/vaporpressurewater.php, the value should be 1697 N/m².

Hoping that I have used the vapour pressure correctly, I have modified the flowbench calculator so the vapour pressure is calculated inside the script...

http://www.porterbility.co.uk/flowbench_calculator...

Please let me know if I've done anything wrong.
Yes you have. You should be calculating the flow coefficient (what you call the flow bench variable) using the Stolz equation. 0.596 isn't necessarily anywhere close to correct. I put some numbers in to your calculator and got about 5% error.

Mignon

1,018 posts

88 months

Wednesday 17th October 2018
quotequote all
Here are some actual numbers for you.

Test pressure drop 70 mm
Orifice Pressure drop 300 mm
Orifice diameter 32 mm
Tube diameter the orifice is in 64 mm
Temp 15 c
Air Pressure 1013

calc

Volume air flow at 25" of water 220.56 CFM
Orifice discharge coefficient 0.60825
Orifice Flow coefficient 0.62820 (Discharge coefficient x velocity of approach factor)
Reynolds number 46910

The tube diameter is important because it determines the d/D factor of the orifice in said tube. All the equations I used are from BS 1042. I don't bother with humidity. Reynolds number calculation require a couple of iterations.

PeterBurgess

775 posts

145 months

Wednesday 17th October 2018
quotequote all
If Pingu is doing comparison work for himself then 0.596 will be close enough especially if he makes an orifice close to 1042 and has pipe diameter more than 4 x d of orifice. My bench was used for a degree thesis and there was some argument as to whether the orifce cd would be nearer to 0.63 from one of the examiners but the other examiners on the panel agreed the 0.596 was ok as it was comparison work on same flow bench and did not detract from the original work put forward for the thesis on increase of air flow and then increase of cr leading to increase in fuel efficiency.
Bear in mind Superflow encourage you to do all testing in one day if poss because of variations in measurement!

Mignon

1,018 posts

88 months

Wednesday 17th October 2018
quotequote all
Good job I wasn't one of your examiners, lol.

pingu393

7,719 posts

204 months

Wednesday 17th October 2018
quotequote all
PeterBurgess said:
Well done for including water vapour calculation, saves looking it up on a bit of paper/google.I don't know where Stan got his 3N/M2 from, my programme does not have default figures you just enter all info. I think the vapour pressure is 1697 kpa or 1.697 N/M2 at 15 C, Kpa to N/M2 divide by 1000.
If the same figures entered on your programme which includes water vapour calculation gives the same answers as Stan's one you have altered it correctly.
Peter
Should kPa to N/m² not be MULTIPLY by 1000? I've just used another online calculator ( http://ddbonline.ddbst.com/AntoineCalculation/Anto...) and it gives the same result = 1.69759 kPa (or 1697.59 Pa, or 1697.59 N/m²).

The figures from the new program give the same results as the old program, so I am happy with my calculations and programming smile . Quite impressive when you consider that I was woken by work for a cancelled call-out at 0230 and I finished this at 0730 and went back to bed eek . It's like being a student again.

Mignon said:
Here are some actual numbers for you...
I assumed that the original program was accurate enough, but I will take a look at your numbers this evening - thanks.


edited to correct quotes formatting

Edited by pingu393 on Wednesday 17th October 20:23

PeterBurgess

775 posts

145 months

Wednesday 17th October 2018
quotequote all
Quite right about Kpa and NM2 I was talking out my arse, I do apologise. I still don't know wher eStan got the figures from though. I should have got up earlier smile
Peter

PS I still think it is good you are improving the programme and making it easier to use.

Edited by PeterBurgess on Wednesday 17th October 16:37

GreenV8S

30,150 posts

283 months

Wednesday 17th October 2018
quotequote all
pingu393 said:
Should kPa to N/m² not be MULTIPLY by 1000?
Yes. 1 Pa == 1 N/m^2. 1kPa == 1000 N/m^2.

pingu393

7,719 posts

204 months

Friday 19th October 2018
quotequote all
Mignon said:
Here are some actual numbers for you.

Test pressure drop 70 mm
Orifice Pressure drop 300 mm
Orifice diameter 32 mm
Tube diameter the orifice is in 64 mm
Temp 15 c
Air Pressure 1013

calc

Volume air flow at 25" of water 220.56 CFM
Orifice discharge coefficient 0.60825
Orifice Flow coefficient 0.62820 (Discharge coefficient x velocity of approach factor)
Reynolds number 46910

The tube diameter is important because it determines the d/D factor of the orifice in said tube. All the equations I used are from BS 1042. I don't bother with humidity. Reynolds number calculation require a couple of iterations.
I don't believe it, but I've done this long hand (as per Example 7.3 in BS 1042-1.4:1992) - and my answers agree (to within rounding error)...

Volume air flow at 25" of water 219.86 CFM
Orifice discharge coefficient 0.6079481
Orifice Flow coefficient 0.6278861 (Discharge coefficient x velocity of approach factor)
Orifice Flow Coefficient 0.6220968 (Discharge coefficient x velocity of approach factor x Expansibility Factor)
Reynolds number 46688


I'll check for "uncertainty" (BS1042 speak for tolerance) and see what the estimated margin of error is.

I suspect that not measuring the temperature at the upstream tapping plane will cause an error. I don't know how significant that error will be until I re-run the figures with appropriate air densities, viscosities and isentropic components.

Mignon

1,018 posts

88 months

Friday 19th October 2018
quotequote all
I suppose we should pin the differences down just for completeness. I imagine you used corner tappings to simplify the calculation of the Orifice Discharge Coefficent (it eliminates terms 5 and 6) whereas I use my actual 25.4 mm tappings but in that case I would get the answer 0.6075488 for your calc. The value of each term being:

Term 1 = 0.5959
Term 2 = 0.0072777
Term 3 = -0.000719
Term 4 = 0.0050899

I also use the following values
Gravitational acceleration 9.80665
Density of water 0.999
Isentropic exponent of air 1.4021
Air density at 15c & 101325 Pa 1.22505

pingu393

7,719 posts

204 months

Friday 19th October 2018
quotequote all
Mignon said:
I suppose we should pin the differences down just for completeness. I imagine you used corner tappings to simplify the calculation of the Orifice Discharge Coefficent (it eliminates terms 5 and 6) whereas I use my actual 25.4 mm tappings but in that case I would get the answer 0.6075488 for your calc. The value of each term being:

Term 1 = 0.5959
Term 2 = 0.0072777
Term 3 = -0.000719
Term 4 = 0.0050899

I also use the following values
Gravitational acceleration 9.80665
Density of water 0.999
Isentropic exponent of air 1.4021
Air density at 15c & 101325 Pa 1.22505
I assumed a D & D/2 to BS1042. According to 7.2.4.3, these tappings would be limited to 0.32mm-1.92mm and at least 2.5 times longer than the tapping diameter.

Term 1 = 0.5959
Term 2 = 0.0072776
Term 3 = -0.0007187
Term 4 = 0.005104
Term 5 = 0.0026
Term 6 = -0.0022148

Air density = 1.225 kg/cu m - This is the air density at 15°C. I know that the air coming out of the rig is 25°C, so the air density value is wrong for the measuring orifice, but the ambient air density is needed to calculate the cfm through the test orifice.
Dynamic viscosity = 1.81 x 10^(-5) Pa s
Tube roughness = 0.03 mm
Isentropic component = 1.401

All other values were from online calculators

Until the "uncertainty" can be determined, I would assume that a calculated value of 220cfm probably means the real answer is 220ish cfm, I'm a great believer in tolerances and feel that a number without a ± is pretty useless.


The standard is pretty good at listing the tolerances ("uncertainties"). For example, Table 3 shows the minimum lengths of straight pipe needed before and after the orifice plate in order to have "zero uncertainty".

There will be "uncertainty" in any setup. I am going to try and minimise it on mine, but the rig will still need to be much bigger than originally planned. It is already bigger than two Black & Decker Workmates eek As I said on my blog, many thanks for the steer towards the standard. I just wish that I had searched harder and found it earlier banghead.