NATURAL GAS PROBLEM

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NATURAL GAS PROBLEM

Postby aquarians » Tue Oct 18, 2005 7:29 pm

well i am trying to convert BTU/SCF in KJ/KG....i myself gave it a try.Basically this thing is concerned to Natural Gas.

i found out the GCV of natural gas from the N.G bill which is 939 BTU/SCF?

939 btu/scf x 1.055 KJ/1 BTU x 1 scf/0.0283 scu. m (*s is being used for standard.)

ANSWER COMES OUT TO BE = 35,005 KJ/SCU. M

I ASSUMED NATURAL GAS = 100% METHANE
=> MOLECULAR WEIGHT OF METHANE = 16 G/MOL = 0.016 KG/MOL

AT STANDARD CONDITIONS i.e at 1 atm pressure and 15 degree centigrade temperaute; i calculated the volume using IDEAL GAS EQUATION
V= (R x T)/P and it came out to be 21.721 SCU. M/MOL.

CALCULATED THE DENSITY AFTERWARDS,

= 0.016/21.721 = 0.00073661 KG/SCU. M

NOW FOR 1 SCU. M ----> MASS WILL BE= DENSITY X VOLUME
COMES OUT TO BE = 0.00073661 KG

THUS GCV OF NATURAL GAS =
35,005 KJ/SCU. M x (1 SCU. M/0.00073661 KG) = 47522400.2 KJ/KG

Now you can see first my calculations are based on assumption, second i am not sure whether i am doing it right or not?

can anyone please help me out in finding the way of calculating GCV of natural gas at standard conditions in KJ/KG? plus it would be highly appreciated if somebody let me know what is NET CALORIFIC VALUE (NCV) of natural gas at standard conditions? **STANDARD CONDITIONS ARE 1 ATM. AND 15 DEGREE CENTRIGRADE.

Your earlier response will be highly apprciated.

Thanking you and Regards,
aquarians
 
Posts: 3
Joined: Mon Oct 10, 2005 9:24 pm

Postby Guest » Tue Oct 18, 2005 11:33 pm

You are on the right track, but there are three issues with your calculation:
1) Somewhere you slipped three decimal places in molar volume. At 0 degrees C, molar volume is 22.4 L, not m^3, a 1000:1 difference.
2) Another minor error in molar volume, at 15 degrees C it should be larger than 22.4 L, by ratio (273.15 + 15)/273.15.
3) Natural gas is mostly, but not 100% methane. Methane is around 890 BTU/SCF. To the degree the energy is higher, traces of ethane, butane are adding more to the energy and weight, than non-burning impurities are subtracting from the energy. You'd need a detailed gas analysis for high accuracy, but you are in the ballpark. I'd use the 890 figure for methan, and proceed with the corrections above. The impurities add both weight and energy, but in about the right proportion.

(1) has to be corrected, as you are 1000X off. The others are minor to the accuracy you can hope to achieve without gas analysis.
Guest
 

Postby aquarians » Wed Oct 19, 2005 6:07 pm

ok i got one thing clear and thankyou for that..........the volume should be
V = 22.4 x (288.2/273.2)
=> V= 23.6 Litres

Again i am stuck......that is the volume of Methane at standard conditions..RIGHT?

Mr. Guest can You please show me the calculations? that would really be nice of you.

Secondly i want to know the NET CALORIFIC VALUE OF NATURAL GAS IN KJ/KG?

NOTE: If You consider Methane in your calculations thats fine too? i wanna learn how it should be solved?

What if i provide you the natural gas composition sheet..i mean the components percentages and may be their relative densities THEN ANY WAY OUT?

Looking forward to hear from you soon? views from other people are also awaited and offcourse welcomed....

thanking you and Regards,
aquarians
 
Posts: 3
Joined: Mon Oct 10, 2005 9:24 pm

Postby Guest » Wed Oct 19, 2005 7:32 pm

opppsss sorry i misunderstood....26.3 L thats the volume and should i be doing the rest of calculations as i did above? or what else?

second just wana make sure molecular weight of methane its 16 g/mol not 16 kg/mol.??

just came into my mind.....if i write an equation considering methane like

CH4 + 02 -------> CO2 + H20

USING this equation can we find out the NCG of methane? as the relation is.....if i am not wrong

NCV = GCV - m (lambda) *dont know how to put the sign of lambda??

thanking you and Regards,
Guest
 

Postby Guest » Wed Oct 19, 2005 10:06 pm

Anonymous wrote:opppsss sorry i misunderstood....26.3 L thats the volume and should i be doing the rest of calculations as i did above? or what else?

second just wana make sure molecular weight of methane its 16 g/mol not 16 kg/mol.??

just came into my mind.....if i write an equation considering methane like

CH4 + 02 -------> CO2 + H20

USING this equation can we find out the NCG of methane? as the relation is.....if i am not wrong

NCV = GCV - m (lambda) *dont know how to put the sign of lambda??

thanking you and Regards,


A mole is normally a gram-mole, a number of grams equal to the numerical value of atomic mass for the molecule, ie 16 g for methane.

Some texts also define pound-moles and kilogram-moles, but I think they just add confusion. Your natural gas is both heavier and more energy dense than methane because of ethane and propane imputities.

I would redo your calculations using 896 BTU/SCF for methane (From an old handbook of chemistry and physics). and use the molar volume of 23.6 L or 0.0236 m^3 per mole (16 g or 0.016 kg).

Your equation needs to be balanced
CH4 + 2 02 -------> CO2 + 2 H20
so total number of each C, H, and O atoms are conserved from left to right side.
Guest
 

Postby aquarians » Thu Oct 20, 2005 12:02 am

okies well i think i am almost there.....

as it is advised by you to consider methane, i did calculations which are as follows....please check them.

GCV of methane = 896 BTU/SCF or 33,402 KJ/SCU. M

Molecular weight = 16 g/ mol

volume as advised by you as i recalculated also = 23.6 CU. M/Kmol

33,402 KJ/SCU. M x 23.6 CU. M/Kmol x 1Kmol/1000 mol x 1 mol/16 g

GCV at standard conditions COMES OUT TO BE = 49.267 KJ/G or 49,267 KJ/KG.

As,

NCG = GCV - m x latent heat of vaporization of water

latent heat of vaporization of water is at standard conditions = 2442 KJ/KG (i am not sure of this value please verify)

EQUATION: CH4 + 2O2 -----> CO2 + 2H2O

from equation it is seen that m=mass of water is 2 KG (not sure though..needs verification)

therefore,
NCV = 49,267 - 2 x 2442 = 44,383 KJ/KG

please check and verify if its all correct....kindly show your work if you find some correction.

Thanking you and Regards,
aquarians
 
Posts: 3
Joined: Mon Oct 10, 2005 9:24 pm

Postby Guest » Thu Oct 20, 2005 9:17 am

I should have been clearer that the 896 BTU/SCF is a "net" or lower heating value rating for methane, so you don't need to account for the energy of condensing the water vapor. The gross or higher heating valu is 995 BTU/SCF

As for the equation, the coeficients are moles, so 1 mole of CH4 (16 g) produces 2 moles of H2O (36 g). 1 kg of CH4 produces 2.25 kg of H2O.

My source also gives values for BTU/lb-mol (1 lb-mol of CH4 is 16 lb). The values are:
HHV: 384,000 BTU/lb-mol
LHV: 346,000 BTU/lb-mol
Divide by 16 to get BTU/lb

These can be converted directly to kJ/kg as 1 BTU/lb = 2.326 kJ/kg. I get 50,300 kJ/kg for LHV or net.
Guest
 

Postby Guest » Sat Oct 22, 2005 6:12 am

thankz for Your help....really kind of you.

can You tell me the book name + Author name........from where you got the NCV and GCV of methane in BTU/lb-mol

secondly......if it is possible to show me how to convert BTU/SCF in KJ/KG........KINDLY SHOW ME THE WORK ?

and the formula i wrote above,

NCV = GCV - mass of water x latent heat of vaporization of water

is this right?


Thanking you and Regards,
Guest
 

Postby Guest » Sat Oct 22, 2005 8:23 am

My source is the Handbook of Chemistry and Physucs, published by the Chemical Rubber Publishing Company (CRC). They also publish a math handbook. My data is from the 43rd edition (1961). Those tables do not apear in the 70th edition (1989). I don't know what year they dropped the fuels data.

The difference between HHV and LHV should just be the heat of vaporization of water resulting from combustion; however, I've never done that calculation.

My data points are at 60 degrees F, so molar volume is 23.7 L by the method from above.

Simple chain calculation and units cancellation

896 BTU/SCF CH4 x 1.05506 kJ/BTU x 1 ft^3/28.3168 L
x 23.7 L at "standard"/1 mol x 1mol/0.016 kg CH4 =
49450 kJ/kg for LHV

For HHV, same process gives 54,900 kJ/kg.

That does not agree exactly with the value derived from the BTU/lb-mol figure. The ideal gas equation is for ideal gas, not real gas which may differ by 1-2%. There are tables of "compressibility factor" which is the ratio of real/ideal gas behavior, material specific and a weak function of temperature and pressure. It is probably most of the error. Write all those "units fractions" above and below a horizontal bar, instead of with a slash. The unit cancellation will be a lot clearer.
Guest
 

Postby Guest » Sat Oct 22, 2005 4:46 pm

:wink: thankz....well i think that will satisfy my Boss....i will discuss this with him........

once again thankz alot for your help....if i may know what do u do? i mean ur profession. i'm a chemical engineer (graduate) ...not much experience.....but am on my way.

thanking you and Regards,
Guest
 

Postby Guest » Sat Oct 22, 2005 6:01 pm

Retired electrical engineer & engineering manager. But I worked in an automotive company, so I know a little more about combustion than the average electrical engineer.
Guest
 

Postby Guest » Sat Oct 22, 2005 7:06 pm

Anonymous wrote::wink: thankz....well i think that will satisfy my Boss....i will discuss this with him........


If you need any great precision, I recommend getting a chemical analysis of the natural gas composition, and using this "heat of combustion analyzer." It will accept either weight or mole fraction percentages for input and give either Customary or metric output. For 100% methane, it gives numbers very much like what we have discussed above.

Link: http://www.processassociates.com/proces ... hc_gas.htm
Guest
 

Postby Guest » Sun Oct 23, 2005 4:21 pm

thankYou Sir....that was really kind of You once again.
i've not discussed it with my Boss yet....but i will give You the feedback as soon as i discuss.

After making this problem clear.....i would like to discuss with you about pipe sizing? its thickness....etc. and selection of pumps..

hoping you will help me in this matter also.

nice knowing You Sir,

keep in touch

Thanking You and Regards,
Guest
 

Postby Guest » Sun Oct 23, 2005 6:17 pm

Glad I could help.

I probably can't help much on piping and pumps. My work outside my own field of engineering (electrical) is pretty limited to things that matter to automobiles, and not 100% of that. I doubt our fuel lines and pumps generalize well to what you are concerned with.

But there are a lot of people here with different experiences. Hopefully someone else will step up to the plate.
Guest
 


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