Calculate steam raised per hour

[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.

 

[Rogue]

Thanks for that, and for pointing out that it should be a multiplication.
Think the stoichiometry has jumbled my head somewhat.

 

[Rogue]

Something still isn't quite right somewhere.

Missing a step methinks.

 

[Rogue]

Right, I think I should be adding enthalpy into the system into my qmc equation.

But I'm not sure I understand why.

Why would you divide the heat load by the heat capacity of water x dT + system enthalpy?

 

[Rogue]

Right, I think I should be adding enthalpy into the system into my qmc equation.

But I'm not sure I understand why.

Why would you divide the heat load by the heat capacity of water x dT + system enthalpy?

Sorry, enthalpy of steam at 5bar.

I guess this tells you the energy required to produce the steam at 5bar?
Don't think I've factored this in before on other heat transfer equations though.

 

[Chestermiller]

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.

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

Total 35.21

I don't confirm these mole percents or your molecular weight. I get
11.06%
13.48%
1.78%
73.68%
And a molecular weight of 28.5 kg/kmole

 

[Rogue]

Sorry, enthalpy of steam at 5bar.

I guess this tells you the energy required to produce the steam at 5bar?
Don't think I've factored this in before on other heat transfer equations though.

Now, I think I understand.
It appears that most of my other work has involved heating a substance without change of state.
During the heating, when the water (liquid) turns to steam (water -gas), additional enthalpy is required for the change of state.

This being an additional draw on the heating system needs to be factored in.

 

[Chestermiller]

Check out my post #12 from a year ago. This has the correct results.

 

[Rogue]

I don't confirm these mole percents or your molecular weight. I get
11.06%
13.48%
1.78%
73.68%
And a molecular weight of 28.5 kg/kmole

Sorry Chester, my mistake. I did have these correct but it would appear I have transferred across onto here incorrectly.

 

[Chestermiller]

Sorry Chester, my mistake. I did have these correct but it would appear I have transferred across onto here incorrectly.

What about the molecular weight?

What do you get for the heat load now?

 

[Rogue]

Check out my post #12 from a year ago. This has the correct results.

Thanks Chester.
I don't doubt your answer or replies, but I'm not willing to just copy across from something.
I need to work through it and gain the understanding where I'm lacking (I have no one to teach me or advise).
I'd rather get it wrong on my own merits then blindly repeat something.

Whilst I appreciate it might be frustrating for yourself, my posts and your replies have helped me rethink my original attempt.

 

[Rogue]

What about the molecular weight?

What do you get for the heat load now?

Yeh, got that. Too many pages with scribbles.

My heat load is 101276973 after 5% heat loss.
The difference coming from my flame temp difference.

Thanks Chester

 

[Chestermiller]

Yeh, got that. Too many pages with scribbles.

My heat load is 101276973 after 5% heat loss.
The difference coming from my flame temp difference.

Thanks Chester

I got less than that. With the 5% heat loss, I get 90 MJ./hr

 

[Rogue]

I got less than that. With the 5% heat loss, I get 90 MJ./hr

My delta T was 1750 (2050-300), which is different to your original answer.

 

[Chestermiller]

My delta T was 1750 (2050-300), which is different to your original answer.

Where did the 2050 come from?

 

[Rogue]

Where did the 2050 come from?

This is determined by a previous question where flame temperature needs to be calculated.
There will always be some differences in answer to this question as it has to be interpolated from a graph drawn by the student.
Accuracy will obviously depend on each student to an extent.
Any rounding errors calculating enthalpy in the fuel gas will also affect this.

 

[Chestermiller]

This is determined by a previous question where flame temperature needs to be calculated.
There will always be some differences in answer to this question as it has to be interpolated from a graph drawn by the student.
Accuracy will obviously depend on each student to an extent.
Any rounding errors calculating enthalpy in the fuel gas will also affect this.

If this is the case, the heat load could have been determined much more accurately by neglecting the heating to the adiabatic flame temperature all together and, instead, simply using Hess' Law to determine the enthalpy change between reactants at 25 C and products at 300 C. In this way, one would only need to know the average heat capacities of the products over the range from 25 C to 300 C.

 

[Rogue]

If this is the case, the heat load could have been determined much more accurately by neglecting the heating to the adiabatic flame temperature all together and, instead, simply using Hess' Law to determine the enthalpy change between reactants at 25 C and products at 300 C. In this way, one would only need to know the average heat capacities of the products over the range from 25 C to 300 C.

Definitely a good point.
I think the question was worded in such a way to point us down the interpolation route.

I had not even considered what you said, suppose that comes with experience/confidence.