Natural gas consumption

You can use ideal gas equation mentioned in physical chemistry

PV=nRT

V=volume in dm3 = ?

P=pressure in bar= 18 bar

n=number of moles of steam (water) =mass (gm)/18 =23690*1000/18 = 1316111.11

R=gas constant =0.0821

T=absolute temperature =103 C+273 =376 K

V =nRT/P =1316111.11*0.0821*376/18 = 2257101 dm3 = 2257.101 m3

80% consumption = V*0.80 = 1805.68 m3

30% consumption = V *0.30 = 677.13 m3

Good Luck

Ashraf

No, that is not correct. I don't have time to get into the details right now, but what you want to do is this.

Calculate the amount of heat required to produce your steam - you can do that with steam tables. You need to know a couple more parameters such as feed water inlet pressure and steam outlet temperature/superheat.

You will also need the efficiency of your boiler - that is how much heat is wasted when burning natural gas.

Then, using the heating value of burning natural gas - you will have your answer.

I am trying to calculate base on the effeciency of the boiler that the vendor guarantee of 99.477% calculated as per DIN 1942. Ncv = 10.35kw-hr/m3 with the Enthalpy of steam at 18 bar = 0.7767 and Enthalpy of Feed water at 103 degC = 0.1201 kw-hr/m3 Nominal

mFuel = m steam * [(0.7767 - 0.1201) / (10.35* 0.9477)

= 1585.83 m3/hr.

80% = 1487 m3/hr

30% = 476 m3/hr

Is my calculation correct?

It looks like you are on the right track - just a couple comments:

1. Your heating value for natural gas looks appropriate
2. The boiler efficiency seems a little high to me in respects to this type of calculation - in effect @ 99.477% this is saying that there is essentially no heat lost in the exhaust gas - it just doesn't seem right to me
3. I just VERY briefly looked at the steam tables, but I think your enthalpy values are a little off - I am getting 7.035 kwhr/m3 for saturated steam at 18 bar. I would review your numbers and conditions (I may be wrong because of some unit conversions there)
4. For the feed water - did you use the saturated water condition at 18 bar?

I would like to know what manufacture of a boiler that offers that effeciency. We normally install (at least for testing and start-up) a fuel flow meter. Because I am in an area that uses the "other set of numbers" I will need to convert your data but I think you are close except for the 99.477%.

More later

Miketheboilerguy

<...vendor guarantee of 99.477%...>

Whether it is <...correct...> or not, it is rather pointless.

Call you gas company they have tons of info!!!

I dont suppose it would be to easy to consult the meter that feeds your boiler as to gas consumption??

Or does your meter supply other factory equipment as well??

Garth CR4

Hi friend,

I will try to supply an approximation that you can track for your need.

Firstly the entalphy of steam @18bar is 2795 kJ/kg. (18 bar is absolute pressure here)

The mass of steam you've supplied is 23690 kg/hr. This yields that the required NET energy to reach this pont is 23690*2795 ≈ 66,2 MJ/hr (u can find these values from steam tables).

This is the net energy required but we will have many losses like heat tranfer. Say total losses are 10% then required energy will be approx 72 MJ/hr.

You will supply this by burning natural gas and the combustion energy of it is approx 54MJ/kg. This means you must burn approx 1,4kg/hr natural gas (≈2 m³ / hr). This is very coarse value, can be used as a starter only.

To find the exact values you must dig the "combustion" that is the world of themodynamics.

For any combustible gas u must use the chemical equation like:

CH₄ + 2O₂ → CO₂ + 2H₂O + heat. Here heat is the c_p*Δt. c_p of CH4 from thermadynamic tables and Δt is the temperature of combustible media after combustion, as to the starting temperature (say 25+273 ^oK).

Mass of reactants used with this equation are : C=12kg (why molar weight of carbon is 12gr/mol = 12kg/kmol); O₂ = 2*32kg; H₂ = 4kg.

Now we must speak a bit more about c_p. This value can be found in thermodynamic tables for different temperatures! You can start with all reactants at 25^oC (273+25 ^oK) but any combustion equation must be calculated more than one! Why many reactants can not complete the burning during 1st combustion and the half-burned byproducts are still exist (like CO, even H₂). You must at least a 2nd calculation with the same equation, taking c_p of temperature high ( say 1000 ^oK) this time.

2nd point with combustion is how oxygen is supplied? Possibly you do use the air, this is very common. Then you must include the nitrogen with above equation. For this reason a general form of equation is used:

C_aH_b + (a+(b/4)O₂ + 3,76(a+(b/4))N₂ = a*CO₂ + (b/2)*H₂O + 3,76(a+(b/4))N₂

N₂ does nothing but there exists and has a heat content.

As a result what you study is not steam side but the combustion itself that has huge content and we have not enough room here for this.

Feri,

Good comments.

But it looks like you missed out on subtracting the enthalpy of the feed water - a pretty significant contribution.

Cheers

Hello Guest,

Thank you.

It must be because of concentrating on combustion...

regards

In reply 2 & 3, I made an estimate of gas consumption at 80% and 30% steam rate of 23690 KG/HR, (18 BAR/ - FW 103 C) using a typical boiler efficiency of 80% and the natural gas heating value of 1000 BTU/SCFT. Ofcourse, I converted all the data into British Units and converted back to M³/HR. This works out to 1503 M³/HR and 564 M³/HR (for 80 and 30%respectively). In this calc 1 LB of steam needs 1.27 SCFT of Natural Gas. You can correct this (rule of thumb)suitably for different boiler effriciency other than 80% and heating value of NG other than 1000 BTU/SCFT.