Heat Recovery from Boiler Exhaust Gases

[Chestermiller]

Thank you for pointing that out.
Q = U * Area * Delta T
750W = 9.5 W/m²K * (3.14*Diameter*Length)*(T₂ - T₁)
750 W = 9.5 * (3.14*0.1*L)* (70-20)
5m = L
I feel as if I'm making a mistake here... I couldn't quite get the temperature. What do I do with the rest of temperatures? Help please?

Yes. See post #28, which I have corrected because I originally used the wrong heat load. It shows that the length of the pipe will have to be between 5 m and 10.9 m. See my next post to see how to calculated it precisely. The temperature is going to be changing along the pipe, and this has to be taken into account.

Chet

 

[Chestermiller]

Chet,
Since the water starts to condense at 49 deg C, does this mean there' s no water condensing in the system? To which then I use
H=MC(T−25)

You can for the portion of the pipe where T > 49, but, in the region where 43<T<49, you can't because water is condensing. The water condensation contribution is very important.

I should have been more explicit in what H actually is. Using your notation, H(T) is given by:

H(T) = m * ([2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (T - 25) + 2.25 (f h_v(T)+(1-f) h_l(T)-104.8))

where f is the fraction of water that is vapor, h_v(T) is the enthalpy of saturated water vapor, and h_l(T) is the enthalpy of saturated liquid water.

If so, what do I take for x to find H? If I was supposed to find x and not H, using the formula quoted and given above, what do I take for H? I don't really understand what H stands for. Is it the 750Watt?

I presume that the: H(T(x+\Delta x))-H(T(x)
where the temperature is the temperature between the increments, perhaps like this?
H (70) - H(20)
H (65) - H(20)
and so on until H (43) - H(20)?

No. In the first increment of pipe length, if you are using 5 degree increments, then you have H(70)-H(65). You are going to be adding up all the small incremental lengths $\Delta x$ to get the total length of the pipe. The following equation gives you the formula for calculating the incremental lengths:
\Delta x = \frac{H(T(x))-H(T(x+\Delta x))}{\pi DU(\frac{(T(x)+T(x+\Delta x))}{2}-20)}
So, for the first increment, for example, you get:
\Delta x = \frac{H(70)-H(65)}{\pi DU(\frac{(70+65)}{2}-20)}
Similarly for the second increment, you get:
\Delta x = \frac{H(65)-H(60)}{\pi DU(\frac{(65+60)}{2}-20)}
Etc.
You then add up all the increments in length.

You don't have to use 5C temperature increments. And the temperature increments don't all have to be equal either. So, when you are getting close to 43 C, for example, you can use a 2 C increment from 45 to 43.

I think your spreadsheet is already set up to calculate H(T), so setting up the integration should be pretty easy.

Chet

 

[nightingale]

You can for the portion of the pipe where T > 49, but, in the region where 43<T<49, you can't because water is condensing. The water condensation contribution is very important.

I should have been more explicit in what H actually is. Using your notation, H(T) is given by:

H(T) = m * ([2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (T - 25) + 2.25 (f h_v(T)+(1-f) h_l(T)-104.8))

where f is the fraction of water that is vapor, h_v(T) is the enthalpy of saturated water vapor, and h_l(T) is the enthalpy of saturated liquid water.

No. In the first increment of pipe length, if you are using 5 degree increments, then you have H(70)-H(65). You are going to be adding up all the small incremental lengths $\Delta x$ to get the total length of the pipe. The following equation gives you the formula for calculating the incremental lengths:
\Delta x = \frac{H(T(x))-H(T(x+\Delta x))}{\pi DU(\frac{(T(x)+T(x+\Delta x))}{2}-20)}
So, for the first increment, for example, you get:
\Delta x = \frac{H(70)-H(65)}{\pi DU(\frac{(70+65)}{2}-20)}
Similarly for the second increment, you get:
\Delta x = \frac{H(65)-H(60)}{\pi DU(\frac{(65+60)}{2}-20)}
Etc.
You then add up all the increments in length.

You don't have to use 5C temperature increments. And the temperature increments don't all have to be equal either. So, when you are getting close to 43 C, for example, you can use a 2 C increment from 45 to 43.

I think your spreadsheet is already set up to calculate H(T), so setting up the integration should be pretty easy.

Chet

Thank you, thank you very much Chet! I have finally understood! I'll always be indebted to you.
I'll attempt to finish the work tomorrow after I finished one of my exam. Once again, thank you!

 

[nightingale]

You can for the portion of the pipe where T > 49, but, in the region where 43<T<49, you can't because water is condensing. The water condensation contribution is very important.

I should have been more explicit in what H actually is. Using your notation, H(T) is given by:

H(T) = m * ([2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (T - 25) + 2.25 (f h_v(T)+(1-f) h_l(T)-104.8))

where f is the fraction of water that is vapor, h_v(T) is the enthalpy of saturated water vapor, and h_l(T) is the enthalpy of saturated liquid water.

No. In the first increment of pipe length, if you are using 5 degree increments, then you have H(70)-H(65). You are going to be adding up all the small incremental lengths $\Delta x$ to get the total length of the pipe. The following equation gives you the formula for calculating the incremental lengths:
\Delta x = \frac{H(T(x))-H(T(x+\Delta x))}{\pi DU(\frac{(T(x)+T(x+\Delta x))}{2}-20)}
So, for the first increment, for example, you get:
\Delta x = \frac{H(70)-H(65)}{\pi DU(\frac{(70+65)}{2}-20)}
Similarly for the second increment, you get:
\Delta x = \frac{H(65)-H(60)}{\pi DU(\frac{(65+60)}{2}-20)}
Etc.
You then add up all the increments in length.

You don't have to use 5C temperature increments. And the temperature increments don't all have to be equal either. So, when you are getting close to 43 C, for example, you can use a 2 C increment from 45 to 43.

I think your spreadsheet is already set up to calculate H(T), so setting up the integration should be pretty easy.

Chet

Chet,

I have done the calculations, my final result shows the sum of H(T) to be 12.3kW and the length needed to be roughly 18.5meter, above the maximum length you have given which is 10.9m.

Here is an example of how I calculate H(T) before condensation of water at 70deg C:
H(T) = m * ([2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (70 - 25) + 2.25 (hg at 70 - hf at 25))
H(T) = 0.000333 * ( 27.85 * (70 - 25) + 2.25 * (2626.3 - 104.8))
H(T) = 2.3 kW

Using the same formula H(T) at 60 deg C is 2.25kW.
Delta X = [(2.3 - 2.25)*1000] / [3.14 * 9.5 * 0.1 * {(70+60)/2 - 20}]
Delta X = 0.372m

Here is an example of how I calculate H(T) after condensation of water at 45deg C:
H(T) = m * {[2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (45 - 25) + 2.25 (hg at 45) - [(1-f) * Hf at 25 + f * Hg at 25]}
H(T) = 0.000333 * { 27.85 * (45 - 25) + 2.25 * 2581 - [0.4687 * 104.8 + 1.78 * 2546.6]}
H(T) = 0.59 kW

Could you please help confirm if I have made mistakes? Thank you so much. My spreadsheet is attached.

 

[BvU]

In column J you are condensing all the H₂O vapour at every step. That's not the idea: you only cool down the vapour from 70 to 65 degrees.

And remember: the whole pipe is supposed to extract only 0.75 kW from the exhaust gas

[edit after post #36 from nightingale] My mistake I see what you do and the column I should have looked at is P, and it's fine up to the onset of condensation. I won't interrupt any more and watch you and Chet at work :)

 

[nightingale]

In column J you are condensing all the H₂O vapour at every step. That's not the idea: you only cool down the vapour from 70 to 65 degrees.

And remember: the whole pipe is supposed to extract only 0.75 kW from the exhaust gas

I couldn't quite capture what you meant by "in column J you are condensing all the vapour at every step". Do you mean I should not input the [mass of water * (hg at 70 - hf at 25)] ? Or I should change it to [mass of water * (hg at 70 - hf at 65)] or perhaps [mass of water * (hg at 70 - hg at 65)]?

Yes, I am well aware that my calculations showed the heat recovery of 12kW in total instead of the required 0.75kW, but I really couldn't pinpoint what I did wrong. My basic in this field was very limited as I have never studied this subject until 4 months ago. Could you please help?

 

[Chestermiller]

The problem is with the water terms in your equation for H(T) at temperatures lower than the dew point 49C. It should be:

2.25(f *hg (T) + (1-f) * hf(T) - hf(25))

Chet

 

[nightingale]

The problem is with the water terms in your equation for H(T) at temperatures lower than the dew point 49C. It should be:

2.25(f *hg (T) + (1-f) * hf(T) - hf(25))

Chet

Thank you Chet. However, I have changed my calculations and still I got minus 10.3 as the result.

At 45 deg C.
H(T) = m * {[2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (45 - 25) + 2.25(f *hg (T) + (1-f) * hf(T) - hf(25))
H(T) = 0.000333 * { 27.85 * (45 - 25) + 2.25 * [1.78 * 2546.6 + 0.4687 * 188 - 104.8] }
H(T) = 3.57kW.
Done to the same to 43 deg C.

I'll give the list of H(T) I have calculated:
at 70 C ;H(T) = 2.3 kW
at 65 C ;H(T) = 2.25 kW
at 60 C ;H(T) = 2.2 kW
at 55 C ;H(T) = 2.148 kW
at 50 C ;H(T) = 2.094 kW
at 45 C ;H(T) = 3.57 kW
at 43 C ;H(T) = 3.15 kW

From 75 to 65: Δx = 0.372m
From 65 to 60: Δx = 0.417m
From 60 to 55: Δx = 0.473m
From 55 to 50: Δx = 0.555m
From 50 to 45: Δx = -18m (This is because the H(T) at 45 C is larger than the H(T) at 50 C)
From 45 to 43: Δx = 5.822m

Did I make another mistake? Thank you.

 

[Chestermiller]

Where did that 1.78 and that 0.4687 come from?

Chet

 

[nightingale]

Where did that 1.78 and that 0.4687 come from?

Chet

They are the mass fraction of water, they add up to 2.25kg, the original mass of water.

At 45, the saturated water vapour pressure is 0.096bar.
Thus, the moles of water vapour is 1.583mol and moles of water is 0.4166. (Total moles of water is 2 mol).

Then I find the mass of water by:
(2-1.583)/2 * 2.25 = 0.4687 kg

Then I find the mass of vapour by:
1.583/2 * 2.25 = 1.78 kg

Ah I see, If I do this, there is no need to multiply 2.25 into the formula anymore, isn't it? So the formula becomes:
At 45 deg C.
H(T) = m * {[2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (45 - 25) +(f *hg (T) + (1-f) * hf(T) - hf(25))
H(T) = 0.000333 * { 27.85 * (45 - 25) + [1.78 * 2546.6 + 0.4687 * 188 - 104.8] }
H(T) = 1.69kW.
Done to the same to 43 deg C.

Thus,
at 70 C ;H(T) = 2.3 kW
at 65 C ;H(T) = 2.25 kW
at 60 C ;H(T) = 2.2 kW
at 55 C ;H(T) = 2.148 kW
at 50 C ;H(T) = 2.094 kW
at 45 C ;H(T) = 1.689 kW
at 43 C ;H(T) = 1.494 kW

and the delta x
From 75 to 65: Δx = 0.372m
From 65 to 60: Δx = 0.417m
From 60 to 55: Δx = 0.473m
From 55 to 50: Δx = 0.555m
From 50 to 45: Δx = 4.928m
From 45 to 43: Δx = 2.731m
The total Δx = 9.47m... Above the min length and under the max length. Great, thanks Chet!

I got the answer! But Chet, something that bothers me is that the H(T) adds up to 14.18 kW. When I only need 0.75kW heat recovery. Any insight whether this is correct or not please?

 

[Chestermiller]

I got the answer! But Chet, something that bothers me is that the H(T) adds up to 14.18 kW. When I only need 0.75kW heat recovery. Any insight whether this is correct or not please?

First of all, it's 2.3-1.494 = 0.806 kW, which suggests that your value of 43 degrees may be a little low. I'm guessing about 43.8 C. Please recheck your overall heat balance. You must have that H(70)-H(T_exit ) must be equal to 0.75 kW.

Look at those large changes in enthalpy rate as soon as the water vapor starts condensing. Make a graph of H vs T to see this visually.

Chet

 

[nightingale]

First of all, it's 2.3-1.494 = 0.806 kW, which suggests that your value of 43 degrees may be a little low. I'm guessing about 43.8 C. Please recheck your overall heat balance. You must have that H(70)-H(T_exit ) must be equal to 0.75 kW.

Look at those large changes in enthalpy rate as soon as the water vapor starts condensing. Make a graph of H vs T to see this visually.

Chet

Chet,

Thank you for the explanation.
I understand that my temperature is a bit off, so I try to find a more accurate temperature.
At 42.8 deg C, I found the heat recovery 0.754kW, the nearest to 0.75kW I could find.
At 43.5 deg C, the heat recovery is only 0.737kW.

If I took 42.8 deg C, I still have the difference of 0.806kW. Here is the H(T):
at 70 C ;H(T) = 2.3 kW
at 65 C ;H(T) = 2.25 kW
at 60 C ;H(T) = 2.2 kW
at 55 C ;H(T) = 2.148 kW
at 50 C ;H(T) = 2.094 kW
at 45 C ;H(T) = 1.689 kW
at 42.8 C ;H(T) = 1.5 kW

This happen after I have taken a more specific Cp (heat capacities) and h (enthalpy) for the products. I'm at my wits end as to why the difference is still 0.806kW.
I also made a graph as you advised, which shows that the heat recovery for 0.75kW occurs somewhere between 42.7 and 42.8 deg.

Do you think it will be acceptable if I take 0.806kW heat recovery as my final answer?

Once again, thank you Chet.

 

[Chestermiller]

Chet,

Thank you for the explanation.
I understand that my temperature is a bit off, so I try to find a more accurate temperature.
At 42.8 deg C, I found the heat recovery 0.754kW, the nearest to 0.75kW I could find.
At 43.5 deg C, the heat recovery is only 0.737kW.

If I took 42.8 deg C, I still have the difference of 0.806kW. Here is the H(T):
at 70 C ;H(T) = 2.3 kW
at 65 C ;H(T) = 2.25 kW
at 60 C ;H(T) = 2.2 kW
at 55 C ;H(T) = 2.148 kW
at 50 C ;H(T) = 2.094 kW
at 45 C ;H(T) = 1.689 kW
at 42.8 C ;H(T) = 1.5 kW

This happen after I have taken a more specific Cp (heat capacities) and h (enthalpy) for the products. I'm at my wits end as to why the difference is still 0.806kW.
I also made a graph as you advised, which shows that the heat recovery for 0.75kW occurs somewhere between 42.7 and 42.8 deg.

Do you think it will be acceptable if I take 0.806kW heat recovery as my final answer?

Once again, thank you Chet.

If it were me, there's no way I would quit until I found the difference between the two calculations. The two calculations are supposed to give exactly the same results. The only difference between the upper calculations and the lower calculation should be the enthalpy of the stream at 25 C.

What's the story with those heat capacities? One calculation assumes constant heat capacity, and the other appears to include temperature dependence of heat capacity. But the enthalpy change between two temperatures should not be Cp(T2-T1) if temperature dependence of heat capacity is taken into account. It should be the integral of CpdT from T1 to T2. This could be part of the problem.

Chet

 

[nightingale]

If it were me, there's no way I would quit until I found the difference between the two calculations. The two calculations are supposed to give exactly the same results. The only difference between the upper calculations and the lower calculation should be the enthalpy of the stream at 25 C.

What's the story with those heat capacities? One calculation assumes constant heat capacity, and the other appears to include temperature dependence of heat capacity. But the enthalpy change between two temperatures should not be Cp(T2-T1) if temperature dependence of heat capacity is taken into account. It should be the integral of CpdT from T1 to T2. This could be part of the problem.

Chet

Understood, true, I interpolate the Cp according to the temperature of the system. I understand to obtain the more precise answer, I will have to integrate the Cp. I think the teacher meant us to use only one single constant value for all the Cp. But since, I have already done the interpolation, I decided to let it go.

I recalculated the temperature, as always, you are right Chet, the temperature of the final exhaust gas is around 43.7 deg C and not 42.8 deg C, as I have thought before.

The difference of the H(T) between H(70) and H(43.7) is now 7.2 kW, and as the deadline for the submission was today, I have submitted this as my final answer.

Thank you very much for all the help, Chet!

Ps. Is it possible to edit my posts after I edit them? I saw that you are able to do it, but I couldn't seem to find the right button.

 

[Chestermiller]

I believe non-staff can edit their posts within a few hours after posting. There is an edit link in the stripe of options below each post. There is no time limit for access to the edit link for staff.

Chet