Gas Calculation

Take the ratio of absolute temperatures and multiply.

Take the ratio of absolute pressures and multiply.

That will get close enough for all practical purposes.

http://en.wikipedia.org/wiki/Ideal_gas_equation

You say it's a mass flowmeter. If it's reading m³, it will have been calibrated to some standard temperature and pressure, pressure almost certainly 1 atmosphere. You need to find the calibration temperature. If it's 0°C, you're already in Nm³. If it's something else, convert as #1.

Codey

Dear Muhmmad

Volume at NTP = 11678 x [273.15/(273.15+35)] x [(1.013+0.8)/1.013] = 18,526.6 Nm3

I have calculated it for you. For more accurate result you need to include compressibility factor which is in this case I assume to be equal to 1.0 (normally valid for low pressure system).

Hope you are happy with the answer

Best regards

No, we were told it is a mass flowmeter. One such is a thermal hotwire type. Please read my #2. If the mass flow stays constant, the reading does not change with pressure and temperature. It can be set up to read mass flow or volume flow. OP said m³, so volume flow in this case. It can be calibrated to read flow corrected to any chosen pressure and temperature, but in practice it is likely to be calibrated to one of the standard conditions. Most common are normal (0°C, 1 atm), or standard (15°C, 20°C or 70°F depending on your preference, and 1 atm). If Muhmmad doesn't know what it's calibrated to, he needs to find out.

If it's calibrated to normal, 11678 m³ is already correct. If it's one of the standards, multiply by 273.15/(273.15+T) where T = 15, 20 etc, but not the actual gas temperature. Just noticed he said ambient temperature 35°C. That's even less relevant than the gas temperature.

Codey

Dear all

Thanks for showing concern and helping me to resolve the Natural Gas calculation issue.Hopefully this will help me lot and I will follow the same method for calculation in future.In case any problem I will come to you again and will get the soloution, Inshahallah.Thanks once again for your concern and support.

Best Regard

Basharat

Hang on, you've had conflicting advice from me and other posters. We can't all be right. I believe my approach is correct, if it is a mass flowmeter as you said in original post. Can you confirm and give more details of the flowmeter?

Codey

Dear Mr Cody

Thanks for the concern.Working detail of the meter I am mentioning here. This is all mentioned from manufacturer, (Incase an total burner control, there is a gas flow meter with analogue output signals in the corresponding gas pipe.The mass flow meter measure the individual amounts of gas in the burners and control them with the control valve.

Calculation from operation flow(m3b/h) to (m3n/h) will be the electric signal from gas mass meters, the pressure, and the temperature sensor.)the calculation of this meter is doubtful that's why I made request for calculation because me and my collegues calculated many time with the formulation(P1v1/T1=P2V2/T2)

P1V1/T1 = P2V2/T2

P1= pressure at normal condition.i.e..1 atm or 1.013 bar A

T1= Temperature at normal condition.i.e..288.5 K.( 15.5 +273.0)

V1= Volume at normal condition in M3/h or M3/min.....

P2= pressure at standard condition.i.e..1 atm or 1.013 bar A

T2= Temperature at standard condition.i.e..273.0 K. 273 is absolute zero.

V2= Volume at standard condition in M3/h or M3/min.....

With this formulation ,the result is not matching with soft ware calcultion and this situation is confusing.

Today reading at mass flow meter is, 11883 m3

Temperature is indicating 38 C

Pressure is . 0.8 bar

Hope this all information is enough to find the result.I am greatful for your concern.Wating for your advise.

Basharat

Muhmmad

You still haven't given details of the flowmeter. What manufacturer, what model, what does it look like, etc? What are the calibration conditions?

If it is a mass flow meter you don't need to measure the temperature and pressure to correct to normal (or standard) conditions (as I've said a couple of times!). You need T & P if it's e.g. a differential pressure type meter.

You say "Calculation from operation flow(m3b/h) to (m3n/h) will be the electric signal from gas mass meters, the pressure, and the temperature sensor". What is m3b/h? I assume m3n/h means normal (usually written Nm³/h).

Your formula P1V1/T1 = P2V2/T2 is only correct if you know it is reading accurately in standard (15.5°C) m³/h and you want to convert to Nm³/h.

Do you mean the signals from the pressure and the temperature sensor are fed into the controls and used to calculate normal flow? (assuming it's a ΔP type) or do they give readings which you are using to calculate manually? If they go into the controls the reading should be correct. There would be no need for conversion as you've asked for. It would be a case of making sure everything is working right, and the formulas in the controls are correct.

If T & P do not go into the controls, the flowmeter may have been calibrated to read normal (or standard) when the actual T & P are at estimated average values. That minimises the error due to T & P variations, and if they don't vary too much the result could be near enough in practice. But you would need to know what those values are in order to check.

A final point - why do you want to measure gas flow? Is it to check gas usage, or for boiler control? The boilers I've seen just modulate the gas control valve based on water output temperature, no need to measure or control the flow to a setpoint. Your boiler might be different.

Codey

Dear sir

The detail of flowmeter is mentioned here.

Q max = 1000 m3/h

Q min = 50 m3/h

P max = 10 Bar

T = -20 c to + 60 c 1 m3 ^ 1 1mp

ELESTER

ii : Yes signals from pressure and the temperaturesensor are fed tocontrols and used to calculate NM3.That reading is confusing.Actually this Glass plant and we need to calculate cals/kgs.and melting efficiency is very important.

This is Natural Gas and Molicular weight is 19.5

Hopefully this information will help to sort out the problem.

Today reading at flowmeter is 12167 m3 Temperature is 36 c and pressure is 0.8 bar

Best regard

Basharat

Need more information on Elster. According to web they make various types of gas meter, but from a quick look I think they all measure actual m³. BK diaphragm type, QA turbine type and RVG-ST rotary type. Is it one of these? Can you give more details?

The actual m³ reading is then (presumably) converted to Nm³ by the software, using formula you've given. So the flowmeter itself does not give mass flow (or, equivalently, normal flow), but the system as a whole does.

I'm not clear what the problem is. Do you have a direct reading from the meter, before conversion (actual m³) from which you calculate normal m³, and get different answer from the software's? If you do not know actual m³ you cannot calculate Nm³ so do not have a figure to compare with the software figure.

Codey

Dear Sir

This is QA turbine type flow meter.The model is Q75 and in the list of Q75 ,open QUANTROMETRO Q (P).this PDF file is carrying complete detail of this flow meter.

ii:- I am getting two readings , one on flow meter in M3 and other on pc in Nm3 after calculation by the soft ware.My problem is that the reading I am getting in Nm3 is not matching with formula calculation. Because of this fuel efficiency cannot be calculated correctly.This is the reason ,I am seeking help to find out the correct answer.Difference of soft ware calculation is varying from 1.5 to 1.7 time where as formula calculation is giving difference only 1.2 to 1.3 time.

Today reading is 12067 M3

Pressure is 0.8 bar

temperature is 37 c

Hopefully you will find the soloution.

Regards

Basharat

It looks to me like the software is correct. Conversion factor from actual to normal is 273.15/(273.15+35) x (1.013+0.8)/1.013 = 1.59 (for 35°C), as given by sabriahmad in #3 (after removing superfluous brackets! Right in the middle of your 1.5 to 1.7.

Please give your calculation which yields 1.2 to 1.3.

Codey

Dear Muhmmad

Would you please show your calculation.

Come on Muhmmad, we're still waiting for your calculation!

Codey

Dear Sir

Sorry for delay. What I calculated , is mentioned as under.Please find the mistake and correct it.Hopefully this time my problem will resolve.

Best regard

V1/V2 = P2 X T2 / P1 x T2

P1 = Pressure at normal condition = 0.8 bar

T1 = Temperature at normal condition = 273 + 15.5 = 288.5 Kelvin

V1 = volume at normal condition = 512 m3/h

P2 = pressure at standard condition = 1.013 bar

T2 = Temperature at stranded condition = 273 + 33 = 306 Kelvin

V2 = volume at standard condition = 512 M3/h

V1 / V2 = 1 M3

P2 x T 1 = 1.013 x 288.5 = 292.2505

P 1 x T 2 = 0.8 x 306 = 244.8

292.2505 / 244.8 = 1.19383374183

512 x 1.19383374183 = 611.242875816 NM3

In 24 hours.

611.242875816 x 23.5 = 14364.2075816 NM3

Dear Muhmmad

Your calculation is incorrect. My observation is that you are confused with the definitions of NORMAL and STANDARD conditions.

NORMAL conditions is defined as P = 1.013 bara, T= 0C or 273K

STANDARD conditions P1 = 1.013 bara, T = 15.5 C or 288.5 K

Data recorded by the turbine flow meter is at ACTUAL conditions

Flow rate at ACTUAL conditions:

P1 = 0.8 barg = (0.8 + 1.013) bar

T1 = 33 C = (33 + 273) K

V1 = 512 m3/h

512 m3/h is the actual flow rate, sometimes we denote it by ACM/h or AM3/h

Flow rate at NORMAL conditions:

P2 = 1.013 bar

T2 = 0 C = (0+273) K

V2 = 512 X (0.8 + 1.013) X (0+273)/1.013/(33+273) = 817.5222 NM3/h

The ratio between NORMAL and ACTUAL conditions:

V2/V1 = 1.6

The total volume over 24 hrs = 817.5222 x 24 = 19,620.5 NM3

(I am wondering why you multiply by 23.5 hrs?)

Volume at STANDARD conditions

V3 = 817.5222 x (15.5 + 273) / 273 = 863.938 SM3/h

By the way, what is the software reading?

Dear Sir

Thanks lot for guiding me on right path , your calcultion is almost same what soft ware is calculating.I am sorry for dealay in answering.I was away for Eid.You asked me ,why I am calculating with 23.5 hrs, actually it is more less time because this is a glass plant and we are using regenrative system with two chambers and gas firing at one end is 20 minuts and then it reverse to other chamber for 20 minuts and during reversing their is no gas firing and that time is almost more then two minuts.In this case ,their is no gas consumption for two minuts after every 20 minuts.I am great full both of you for proper guidence.Hopefully you will help me again when I need you.

Best regards

Basharat

.... , your calcultion is almost same what soft ware is calculating...

Yes, I said that way back in #13

Codey

Dear Sir

Yes ,you are right.You indicated in your convesation #13 ,soft ware calculation is correct but the way I was calculating ,not letting me to digest that but your patient and coperation made my calculation correct too.I am great full.Thanks once again.

Best regard

Basharat

No problem, hope you had a pleasant eid

Codey

There are so many mistakes here I don't know where to start. But 2 points - you need to work in absolute pressure, and if your converting between actual and normal m³, there's no need to bring standard temperature (15.5°C) into it.

Please read other posts in detail and have a good think about it.

Codey

Salam Muhmmad,

If it is a mass flowmeter, then what is the reading in mass e.g kg/h? what is the gas molecular weight?If you know the mass flowrate, divide it by the molecular weight and multiply by 22.414 to arrive at Nm3/h.

The reading of 11883 m3/h is it actual volume at operating conditions (T=38C, P=0.8 barG) or at standard conditions (T=15.5 P=1 atm)? and you want to convert to volume at NTP (T=0C P=1 atm)?, you need to tell us more about your problem. Probably the manufacturer include molecular weight into the calculation? If that is so the equation should be:

P1V1/Z1M1T1 = P2V2/Z2M2T2

Where Z is the compressibility factor and M is the mol wt.

Perhaps you could give us a more clearer picture of the whole situations, so that we can help you once for all.

Wassalaam.

Sabri Ahmad

Dear Muhmmad

After going through your data, I notice that you are using T1 as 288.5K. By definition, normal conditions is referred to 0C or 273K and P 1 atm. You are using 288K and 1 atm respectively.

For turbine meter, the figure given by turbine is actual volume rate at operating conditions (e.g P=0.8 barg, T=35C). However, you are not using these data in your calculation.

The equation you use is correct but the figures for pressure and temperature are incorrect.

Please recheck your calculation.

Best regards