Calculate steam raised per hour

[Andy86]

Homework Statement

A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume. It is to be fed to the combustion chamber in 10% excess air at 25ºC, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90ºC.

(i) If 5% of the heat available for steam production is lost to the atmosphere, determine the amount of steam raised per hour when the total flow of flue gases is 1400 kmol h–1.

Water Inlet temp = 90°C (Given in question)
Steam temp @ 5bar = 152°C (from steam tables)

Flue Gas inlet = 1997°C (Based on flame temp)
Flue Gas outlet = 300°C (Given in question)

Latent heat of steam @ 5 bar = 2108KJ/KG (from steam tables)

TOT flow of flue gas = 1440 kmol h-1

Homework Equations

Mass flow rate (H) * Heat Capacity * (ΔH) = Mass flow rate (C) * Heat Capacity (C) * (ΔC)

Rearrange to make "Mass flow rate (C)" the subject>

Mass flow rate (C) = (Mass flow rate (H) * Heat Capacity * (ΔH)) / (Heat Capacity (C) * (ΔC))

The Attempt at a Solution

[/B]
Mass flow rate (C) = (1400 * Heat Capacity) * (1997-300) / (4.2 * (288-90))

To attempt this question I need the heat capacity of the fuel gas mix, I am not sure how to calculate this and once I do am I using the correct equation??

Thanks
Andy

 

[BvU]

I need the heat capacity of the fuel gas mix

Why ? Isn't this question now all about CO2 and water ?

 

[Andy86]

Sorry heat capacity of flue gas

 

[BvU]

Well, you could http://www.kostic.niu.edu/350/_350-posted/350Chengel7th/Appendix1Udated.pdf

 

[jim hardy]

They gave you a lot of little facts to string together.

Flue gas is a mix of air(10%excess), N2, CO2 and water. N2 i'd probably treat like air.
Specific heats of all are easy to look up
from given fuel mix you know how much carbon and hydrogen are there so i think you can come up with fraction of each gas

and they gave you ΔT
and flow rate in moles which should translate to mass

it's a several step problem to come up with ΔH for flue gas side, and two or three steps for steam side.

Think in little steps.

 

[Andy86]

Thanks BvU & Jim, is this lot any closer?

Thanks
Andy


Heat capacities of the flue gases at 1997°C / 2270K

Co2 = 1.3865 KJ/KG
H2O = 2.41267 KJ/KG
O2 = 1.1999 KJ/KG
N2 = 1.2985 KJ/KG

Mass = Nmols * Molecular Mass

Co2 = 3.9mols * 44 = 171.6g
H2O = 4.75mols * 18 = 85.5g
O2 = 0.6mols * 32 = 19.2g
N2 = 25.9mols * 28 = 725.2g

Specific heat of flue gases

3.9 mols Co2 = 1.3865 /1000 * 171.6g = 0.24 KJ/KG
4.75 molsH2O = 2.41267 /1000 * 85.5g = 0.21 KJ/KG
0.6 mols O2 = 1.1999 /1000 * 19.2g = 0.02 KJ/KG
25.9 mols N2 = 1.2985 /1000 * 725.2g = 0.94 KJ/KG

TOTAL = 1.41KJ/KG


Allow for 5% heat loss > 1.41 KJ/KG * 0.95
= 1.34 KJ/KG

TOTAL mass = 1001.5g > 1.0015KG
TOTAL flue gas mols = 35.15 mols > 0.03515 Kmol

1001.5KG / 35.15mols = 28.49 grams per mol of flue gas

1400 kmol h^-1 (given in question)

mass h^-1 = 0.02849KG * 1400 Kmol
Flue gas mass= 39.9 KG / H

Mass flow rate (C) = (Mass flow rate (H) * Heat Capacity * (ΔH)) / (Heat Capacity (C) * (ΔC))
= 39.9 KG / H * 1.34 KJ/KG * (1997-300) / (4.2 * 152 - 90)
= 360 KG / HOUR

 

[jim hardy]

If you want someone to check your work you need to show it step by step.

Looks to me like you're slapping up equations and expecting them to work. They only work if you've got every term entered correctly.

This process cools flue gas, transferring energy to the water . I think you've got that.
Now do the footwork to get the right numbers in your formulas.

water side is easy, it comes in as liquid at 90 degrees and leaves as 5 bar steam presumably dry saturated.

So your question becomes two questions:
How much energy comes out when you cool a kg of flue gas from 1997 to 300 degrees?
How much energy goes into turn a kg of 90 degree water into 5 bar steam ? Hint - a lot more than 4.2*152-90.

From those two numbers and mass flow rate of either fluid , you can calculate flowrate of the other.

corrections welcome.

old jim

 

[Andy86]

If you want someone to check your work you need to show it step by step.

Looks to me like you're slapping up equations and expecting them to work. They only work if you've got every term entered correctly.

This process cools flue gas, transferring energy to the water . I think you've got that.
Now do the footwork to get the right numbers in your formulas.

water side is easy, it comes in as liquid at 90 degrees and leaves as 5 bar steam presumably dry saturated.

So your question becomes two questions:
How much energy comes out when you cool a kg of flue gas from 1997 to 300 degrees?
How much energy goes into turn a kg of 90 degree water into 5 bar steam ? Hint - a lot more than 4.2*152-90.

From those two numbers and mass flow rate of either fluid , you can calculate flowrate of the other.

corrections welcome.

old jim

Thanks for the constructive criticism >>>
Correction...

How much energy comes out when you cool a kg of flue gas from 1997 to 300 degrees?

Mass = 28.9 g per mol of flue gas
Flow rate = 1400 Kmol / hour

Mass flow rate = 28.9g * (1400 * 1000)
= 39, 886 KG / hour
= 39, 886 / 3600
= 11.08 KG / Sec

q = Cp * M * ΔT
q = (1.41 KG / KJ) * (11.08 KG / Sec) * (1997 - 300)
q = 7.36 KW / S
q = 7.36 * 3600
q = 26, 511 KW / H

How much energy goes into turn a kg of 90 degree water into 5 bar steam ?

q = Cp * M * ΔT

Rearranged for M

M = q / Cp * ΔT
M = 26511 / 4.2 * 62
M = 101.8 KG / H

 

[jim hardy]

Always cross-check a calculation to see if you're using the right algorithm.

How much energy goes into turn a kg of 90 degree water into 5 bar steam ?

http://www.engineeringtoolbox.com/saturated-steam-properties-d_457.html

Really should use a subcooled water table for h_liquid

 

[Andy86]

Always cross-check a calculation to see if you're using the right algorithm.http://www.engineeringtoolbox.com/saturated-steam-properties-d_457.html
View attachment 113577Really should use a subcooled water table for h_liquid

Apologies Jim, I am not sure what point your are trying to make? The question asks me for the Mass of the steam over an hour. Using the information I have I came up with M = 101.8 KG / H. Have I used an incorrect value somewhere?> I thought the heat capacity of water was constant @ 4.2 KJ /KG up to around 125°C. Since the water is fed @ 90°C I used 4.2 KJ/KG. Should I have used a different figure? I see you have very kindly highlighted 1.99 KJ/KG?? Is this the figure I should have used for Cp?

How does my figure of 108.1 KG / H look?

 

[jim hardy]

I thought the heat capacity of water was constant @ 4.2 KJ /KG up to around 125°C. Since the water is fed @ 90°C I used 4.2 KJ/KG.

Sounds reasonable for heating water from 90 degrees to boiling point

I'm more accustomed to BTU's,

MCΔT is for simply heating a substance. A single BTU will heat a pound of water a single degree F
What about a phase change? Takes ~1000 BTU's to boil a pound of water into a pound of steam...
How about in SI units?

I don't see a heat of vaporization in your formula so i don't know what you are using for an algorithm.
Maybe it's in there just i missed it.
That's why it's important to work step by step defining your terms. Explain what each represents. Paint a word picture of your analysis.

 

[Chestermiller]

I would have done the calculation very differently.

Heat capacities of the flue gases at 1997°C / 2270K

Co2 = 1.3865 KJ/KG-K = 1.3865 x 44 = 61 kJ/kmol-K
H2O = 2.41267 KJ/KG-K = 2.41267 x 18 = 43.42 kJ/kmol-K
O2 = 1.1999 KJ/KG-K = 1.1999 x 32 = 38.97 kJ/kmol-K
N2 = 1.2985 KJ/KG-K = 1.2985 x 28 = 36.36 kJ/kmol-K

Mole Fractions of gases

Co2 = 3.9mols /35.15 mols = 0.111
H2O = 4.75mols /35.15 mols = 0.135
O2 = 0.6mols /35.15 mols = 0.017
N2 = 25.9mols/35.15 mols = 0.739

Heat Capacity of mixture = (0.111)(61)+(0.135)(43.42)+(0.017)(38.97)+(0.739)(36.36)= 40.17 kJ/kmol-K

Available heat = (1400)(40.17)(1997-300)=95,400,000 kJ/hr

Heat load after 5 % loss = 90,700,000 kJ/hr

Enthalpy change per kg of steam produced = 4.184 (152-90) + 2108 = 2367 kJ/kg

Rate of steam production = 90700000/2367 = 38300 kg/hr

 

[Andy86]

@Chestermiller

Thanks Chester- that completely blows my calculations out the water! Back to square 1 we go!

 

[jim hardy]

Note how Chester explained what he was doing in every line
and expressed heat capacity per K
and went straight to an answer using logical small steps
old jim

 

[Chestermiller]

@Chestermiller

Thanks Chester- that completely blows my calculations out the water! Back to square 1 we go!

Andy86: I didn't mean to discourage you. The point I was trying to make was that, rather than going back and forth between g, kg, moles, and kmol with the fuel gas, it would have been much easier to work entirely in terms of kmol (after all, the flow rate was already in terms of kmol). All you would need to have done would have been to multiply each heat capacity by the corresponding molecular weight to get the heat capacities per kmol. After that, everything could have been done in kmols.

 

[Andy86]

@Chestermiller I was only joking! Thanks for stepping in, I have now completed all my thermodynamics modules. VERY steep learning curve. Thanks for all the help! @jim hardy Thank you also for the guidance! Off to sit in a darkened room for 3 hours!

Thanks
Andy

 

[jim hardy]

I'm still learning the ropes in homework forums...

Chester's molar method for flue gas energy is way superior to mine. I went to mass because it's my "safe zone" . I'm a chemistry plodder, out of my element...

But i was pretty sure on water side you'd missed heat of vaporization
and wanted you to go single-stepwise rather than leap into MCΔT which doesn't account for phase change.

Enthalpy change per kg of steam produced = 4.184 (152-90)(heat feedwater to saturation temperature) + 2108(heat of vaporization) = 2367 kJ/kg

Glad you're seeing it clearly now.

Thanks @Chestermiller !

 

[Chillwind]

can I ask how you got these figures?

Heat capacities of the flue gases at 1997°C / 2270K

Co2 = 1.3865 KJ/KG
H2O = 2.41267 KJ/KG
O2 = 1.1999 KJ/KG
N2 = 1.2985 KJ/KG

 

[Rogue]

can I ask how you got these figures?

Heat capacities of the flue gases at 1997°C / 2270K

Co2 = 1.3865 KJ/KG
H2O = 2.41267 KJ/KG
O2 = 1.1999 KJ/KG
N2 = 1.2985 KJ/KG

I'm struggling with that one as well.

 

[Rogue]

I was going more down the route of converting 1400kmol/h to kg/h
Molecular mass of flue gas = 35.21g/kmol
=0.03521kg/kmol
$\frac{1400}{0.03521}$ = 39761.43kg/h

 

[Rogue]

I'm struggling with that one as well.

Looked it up haha

 

[Rogue]

I was going more down the route of converting 1400kmol/h to kg/h
Molecular mass of flue gas = 35.21g/kmol
=0.03521kg/kmol
$\frac{1400}{0.03521}$ = 39761.43kg/h

After this, I've tried factoring into the following with the specific heat cap of my flue gas being 1.53kj/kg

qmc= $\frac{qmh \times hfg}{cpc \times (Tc2-Tc1)}$

With hfg being my enthalpy in x0.95

Thus gives me an answer of 376000 (give or take) which is either completely incorrect or out by a factor of 10?

Please help...?

 

[Rogue]

Tried again by calculating heat load of the flue gas using my kmol/h to kg/h conversion.

Then applied the 5% heat loss and tried to calculate mass flowrate :

qmc = $\frac{Q}{(cpC (Tc2-Tc1))}$

Still leaving me out by a factor of 10 at 309253kg/h??

I'm now doubting my kmol/h to kg/h conversion.

Why won't my mass calculation pan-out?

 

[Rogue]

Yep.
My kmol/h to kg/h conversion is incorrect.
Units error.
Which would make it 39.76kg/h.

This would make my answer 390.253kg/h.

Too far the other way now? Scratching brain.

 

[Chestermiller]

I was going more down the route of converting 1400kmol/h to kg/h
Molecular mass of flue gas = 35.21g/kmol
=0.03521kg/kmol
$\frac{1400}{0.03521}$ = 39761.43kg/h

What were the mole fractions of CO2, H2O, O2, and N2 you calculated for the flue gas?

 

[Rogue]

What were the mole fractions of CO2, H2O, O2, and N2 you calculated for the flue gas?

Flue gas:
CO2 = 3.9. (11%)
H2O = 4.75. (13.5%)
O2 = 0.6. (0.6%)
N2 = 25.96. (25.96%)

Total 35.21

 

[Rogue]

So my next step, I tried to convert kmol/h to kg/h

So $\frac{1400}{35.21}$
= 39.76143kg/h

 

[Rogue]

This meant that my specific heat capacities of the flue gas don't need converting to kj/kmol, only fractionating according to the previously identified %'s.

So:
CO2 = 1.39 x 0.111 = 0.154
H2O = 2.93 x 0.135 = 0.396
O2. = 1.2 x 0.017 = 0.0204
N2. = 1.3 x 0.737 = 0.958

Total heat capacity of flue gas = 1.53kj/kg

 

[Chestermiller]

So my next step, I tried to convert kmol/h to kg/h

So $\frac{1400}{35.21}$
= 39.76143kg/h

Check the units

 

[Rogue]

Flue gas heat load:

Q = qmh x cph (Th1 - Th2)

= 39.76143 x 1.53 (2050-300)
= 106461.2288kj/h

x0.95 (for 5% heat loss) = 101138.1674kj/h

Then into qmc = $\frac{Q}{cpc \times (Tc2-Tc1)}$

 

[Rogue]

Check the units

Sorry Chester, I know you're trying to help, but you'll have to give me a bit more than that.
I've already adjusted the units once (another previous post) - I'm missing/not grasping something.

 

[Chestermiller]

kmoles x (kg/kmole) = kg

 

[Chestermiller]

Did you really think that the flow rate in a piece of industrial equipment was only 40 kg/hr?

 

[Rogue]

I noticed it was small, especially with my steam produced being so small.
But as I haven't done any mass and energy balance modules etc or much in the way conversion kmol to kg/h, I'm just trying best I can.

 

[Chestermiller]

Well, one thing you must make sure about is that the units correctly cancel. This will prevent you from mutiplying when you should be dividing, or dividing when you should be multiplying.