# Heat Recovery from Boiler Exhaust Gases

## Homework Statement

The exhaust gases from a natural gas-fired central heating boiler leave the boiler at a temperature of 70 deg C. The oxygen content of the exhaust gases is 9% by volume (dry). The rate of energy input (Gross calorific value) to the boiler from natural gas is 16kW. The gross calorific value of natural gas is 55MJ/kg. For the purposes of all calculations assume that natural gas is 100% Methane (CH4). The pressure in the combustion chamber of the boiler and throughout the heat recovery system is 1 bar.

The exhaust gases are blown down a plastic pipe situated in a room are then exhausted to atmosphere, as shown in the diagram attached. The warm pipe provides a heat input to the room of 0.75kW. The pipe is 100mm in diameter and is made up of a number sections 1m in length.

The overall heat transfer coefficient (U) for heat transfer from the exhaust gases through the pipe to the room is 9.5 W/m^2K. The room temperature is 20 deg C.

Calculate:
1. The temperature to which the exhaust gases must be cooled in order to recover 0.75kW of heat.
2. The number of 1m length of pipe needed to provide the heat input of 0.75kW to the room.

## Homework Equations

Hints:
The best way to calculate the final temperature of the exhaust gases is to consider cooling down to a different temperatures and identifying (by iterating between the nearest values) which temperature gives 0.75kW of heat recovery.

To calculate the heat recovery from the exhaust gases as they cool through a temperature difference of Delta T, use Q = mCpDeltaT for the dry products of combustion and Q = mDeltah for the water vapour, where appropriate enthalpy values are taken from tables. (A temperature interval of about 5 deg C is appropriate).

As the exhaust gas temperature varies along the pipe, the best way to calculate the length needed is to consider each temperature interval, assume a mean gas temperature, then calculate the length of the pipe needed to give the heat transfer between the exhaust gas and the room as the gas cools over the temperature interval.

## The Attempt at a Solution

CH4 + x 02 + 3.76x N2 = a CO2 + b H20 + c 02 + d N2

a = 1
b = 2
x = 2 + c
d = 3.76x = 3.76(2 + c) = 7.52 + 3.76c

9% oxygen, dry
0.09 = c / (a+c+d)
0.09 = c/(1+c+7.52 + 3.76c)
c = 1.3417 moles

and thus x = 3.417 moles

Reactant
mass (Molar Mass * moles)
CH4 1 moles 16 * 1 = 16kg

Products
mass (Molar Mass * moles) mass/kg fuel
CO2 1 mole 44*1 = 44kg 44/16 = 2.75kg/kg fuel
H2O 2 moles 18*2 = 36 kg 36/16 = 2.25kg/kg fuel
O2 1.3417 moles 32*2 = 42.9344kg 42.934/16 = 2.6834kg/kg fuel
N2 12.565 moles 28*12.565 = 351.82 kg 351.82/16 = 22kg/kg fuel

I am completely clueless on how to continue afterwards. The hint told me to make an assumptions of the final temperature that will allow a heat recovery of 0.75 kW.

By using the formula:

Q = ∑(mass of dry product*Cp* ΔT) + mass of H2O * enthalpy of steam at 70 deg C - (mass of H20 water * enthalpy of water at assumed Temp + mass of H2O vapour * enthalpy of steam at assumed temperature)

I found the mass of water= 0.0985kg and the water vapour= 0.110864kg.
I work on excel spread sheet, taking the assumptions of the temperature from 0 deg to 35 deg, all give me the answer of more than 0.75 kW, the required output. At 1.5 deg, the heat recovery is found to be 7510kW, an outrageous amount compared to the required 0.75kW. Could you please help?

#### Attachments

• Picture.png
54.1 KB · Views: 767

BvU
Homework Helper
How many kg/s of CH4 do you feed the boiler ? How many kg/s of exhaust gas does that yield ?

Then: if the room temperature is 20 degrees, what is the minimum exhaust temperature to the outside world ?

The exercise instruction to consider the water in the exhaust from the boiler to be in the vapour phase at 70 degrees and 1 Bar is clear, but not very realistic, I would say.

Last edited:
• nightingale
Chestermiller
Mentor
As BvU indicated, you need to start out by determining the mass flow rate through the exhaust pipe. You are given the heat transfer rate from the exhaust pipe to the room(0.75 kW). From this, you can calculate the exit temperature from the exhaust pipe. Don't forget to take into account the heat of condensation for the part of the water vapor that condenses (probably most of it by the exit). Once you know the exit temperature, you need to calculate the part of the heat load occurring over 5C increments within the pipe. From this, you can use the overall heat transfer coefficient and the room air temperature to determine the length of each section required to bring about each 5 C increment. You then add these up to get the overall length.

Chet

• nightingale
Chestermiller
Mentor
How many kg/s of CH4 do you feed the boiler ? How many kg/s of exhaust gas does that yield ?

Then: if the room temperature is 20 degrees, what is the minimum exhaust temperature to the outside world ?

The exercise instruction to consider the water in the exhaust from the boiler to be in the vapour phase at 70 degrees and 1 Bar is clear, but not very realistic, I would say.
Why do you think this? What is the partial pressure of the water vapor in the exhaust? What is the equilibrium vapor pressure of water at 70C?

Chet

• nightingale
BvU
Homework Helper
Why do you think this? What is the partial pressure of the water vapor in the exhaust? What is the equilibrium vapor pressure of water at 70C?

Chet
Thanks for putting me right, Chet. Oh boy, what a beginner's mistake. Please disregard altogether, nightingale.

But do ask yourself how many kg/s of CH4 the boiler needs to generate 16 kW.

• nightingale
Chestermiller
Mentor
But do ask yourself how many kg/s of CH4 the boiler needs to generate 16 kW.
This is not as straightforward a calculation as it seems. Nightingale has to do an overall open system first law heat balance on the boiler, and has to take into account the enthalpy of the exhaust stream exiting the boiler (which will be significant). Also, the 55 MJ/kg is the heat of combustion at 25 C, assuming liquid water is part of the product stream. Nightingale also has to make some kind of assumption regarding the temperature of the feed air and methane to the boiler (20 C for each is reasonable).

It will be interesting to see how nightingale does this calculation.

Chet

• nightingale
BvU
Homework Helper
Time for a tweet from Nightingale, then !

• nightingale
Thank you Chet and BvU.

I found that the fraction of water vapour can be calculated from= 2 moles/ 16.91 (total moles in the product) * 100% = 11.827%.
The pressure of the system is 1 bar, thus partial pressure of water vapour in the flue gases is 0.11827* 1 bar= 0.11827 bar, and from the steam tables, at 0.11827 bar, the water vapour will start to condense at 49.1 deg C.

Both of you advised me to obtain the mass flow rate:
I assume the inlet air to be 25°C.
Q = m (mass flow rate of CH4)* [∑m (mass of dry products) * Cp *(T1 - T2) + mH2O * (hg at 70°C - hf at 25°C)]
Q = m * [2.75* 0.871 (CO2) +2.6834*0.923 (O2) +22*1.04 (N2)] * (70 - 25) + 2.25 * (2626.3-104.8)
16 = m *1248*5673.375
m = 2.26*10^(-6) kg/s

Am I going in the right way?
Please bear with me, I took this subject as an extra module and I am struggling to understand. Thank you very much!

Last edited:
BvU
Homework Helper
That would be a cheap gas bill :)

Isn't the Q you calculate this way the heat needed to bring the exhaust products from 25 to 70 degrees ?
What happened to the 55 MJ/kg ?
And why does a + sign change to a * suddenly ?
And is the 16 really 16 ?

• nightingale
Chestermiller
Mentor
Hi nightingale,

Nice job on getting the dew point.

As far as the heat balance on the boiler is concerned, there are a few improvements needed. Since your mass balances were done on the basis of 1 kg-mole of methane, it would be easier to base the heat balance on the molar flow rate of methane to the reactor. So, let m represent the inlet molar flow rate of methane in kg-mole/sec. You can then easily calculate the rate of sensible heat exiting the boiler. In this case, you would directly use the moles of the various species in the exit stream calculated from your mass balance, and you would use molar heat capacities, rather than mass heat capacities. Also, in the case of the water, you would multiply the enthalpies from your steam table by 18. If your mass flow rate m is in kg-moles/sec of methane fed to the reactor, then the sensible heat rate of your exit stream will be kW.

As BvU indicated, you omitted the heat of combustion from the boiler heat balance. That should be on the right side of the equation, and you have to convert that to kJ/kg-mole (multiply by 16) and you also have to multiply it by the inlet molar flow rate of methane m. The 16 kW is subtracted from this, since it is being transferred from the combustion gases to make the steam.

Chet

• nightingale
What happened to the 55 MJ/kg ?
And why does a + sign change to a * suddenly ?
And is the 16 really 16 ?

Yes, thank you for pointing that out.
Does this mean I have to convert 55MJ/kg into watt and then subtract 16 from it to obtain the Q?

As far as the heat balance on the boiler is concerned, there are a few improvements needed. Since your mass balances were done on the basis of 1 kg-mole of methane, it would be easier to base the heat balance on the molar flow rate of methane to the reactor. So, let m represent the inlet molar flow rate of methane in kg-mole/sec. You can then easily calculate the rate of sensible heat exiting the boiler. In this case, you would directly use the moles of the various species in the exit stream calculated from your mass balance, and you would use molar heat capacities, rather than mass heat capacities. Also, in the case of the water, you would multiply the enthalpies from your steam table by 18. If your mass flow rate m is in kg-moles/sec of methane fed to the reactor, then the sensible heat rate of your exit stream will be kW.

I'm confused on how do I change the base of my calculations into molar flow rate of methane. It was:
Q = m (mass flow rate of CH4)* [∑m (mass of dry products/kg fuel of Methane) * Cp *(T1 - T2) + mH2O * (hg at 70°C - hf at 25°C)]
Q = m * [2.75* 0.871 (CO2) +2.6834*0.923 (O2) +22*1.04 (N2)] * (70 - 25) + 2.25 * (2626.3-104.8)
16 = m *1248+5673.375
m = 2.3*10^(-3) kg/s.... which is wrong because I did not take into account the heat of combustion.

Is it like this then?
Q = m (molar flow rate of CH4)* [∑m (moles of dry products*Molar of products) * Cp *(T1 - T2) + moles H2O * molarity* (hg at 70°C - hf at 25°C)]
Q = m * [1*44* 0.871 (CO2) +1.3417*32*0.923 (O2) +12.565*28*1.04 (N2)] * (70 - 25) + 2.25 * 18* (2626.3-104.8)

But then again by doing that, the result will just be the mass (kg) of the products. I'm really confused, could you please help?

As BvU indicated, you omitted the heat of combustion from the boiler heat balance. That should be on the right side of the equation, and you have to convert that to kJ/kg-mole (multiply by 16) and you also have to multiply it by the inlet molar flow rate of methane m. The 16 kW is subtracted from this, since it is being transferred from the combustion gases to make the steam.

What is this heat of combustion? Is it a parameter I have to find? Or is it just 16kW * molar flow rate methane * 16? and thus:
Q = m (molar flow rate of CH4)* [∑m (moles of dry products*Molar of products) * Cp *(T1 - T2) + moles H2O * molarity* (hg at 70°C - hf at 25°C)]+ 16*16*m
55* 10^3 = m * [1*44* 0.871 (CO2) +1.3417*32*0.923 (O2) +12.565*28*1.04 (N2)] * (70 - 25) + 2.25 * 18* (2626.3-104.8) + 16*16*m

Thank you

Chestermiller
Mentor

55* 10^3(16)m = m * ([1*44* 0.871*44 (CO2) +1.3417*32*0.923 (O2) +12.565*28*1.04 (N2)] * (70 - 25) + 2.25 * 18* (2626.3-104.8)) + 16*10^3

The corrections are indicated in bold. Note the parenthesis include the water term.

Chet

• nightingale

55* 10^3(16)m = m * ([1*44* 0.871*44 (CO2) +1.3417*32*0.923 (O2) +12.565*28*1.04 (N2)] * (70 - 25) + 2.25 * 18* (2626.3-104.8)) + 16*10^3

I don't understand why do I have to multiply the molar mass of CO2 twice and why is the 16kW multiplied by 10^3? I thought that since all the heat capacities and masses are in kg, then the unit should be kW, which makes it like this:

55* 10^3(16)m = m * ([1*44* 0.871 (CO2) +1.3417*32*0.923 (O2) +12.565*28*1.04 (N2)] * (70 - 25) + 2.25 * 18* (2626.3-104.8)) + 16

Could you please elaborate further? Thank you

Chestermiller
Mentor
I don't understand why do I have to multiply the molar mass of CO2 twice and why is the 16kW multiplied by 10^3? I thought that since all the heat capacities and masses are in kg, then the unit should be kW, which makes it like this:

55* 10^3(16)m = m * ([1*44* 0.871 (CO2) +1.3417*32*0.923 (O2) +12.565*28*1.04 (N2)] * (70 - 25) + 2.25 * 18* (2626.3-104.8)) + 16

Could you please elaborate further? Thank you
Oops. I missed the 44 in the equation you wrote. Of course, you are right. It should only be there once. Also, somehow I though that the 16 kW was 16 MW.

Chet

• nightingale
Oops. I missed the 44 in the equation you wrote. Of course, you are right. It should only be there once. Also, somehow I though that the 16 kW was 16 MW.

Chet

Thank you Chet,
I was able to find the mass flow rate with your help and obtain 3.33 * 10^(-4) kg/s. I checked with the teacher, who confirmed the flow rate value. I try to find the answer for the 2nd part:
2. The temperature to which the exhaust gases must be cooled in order to recover 0.75kW of heat:

I first assume that the temperature is cooled to around 5 degC and thus took the the Cp, hf and hg based on this. I will recalculate a more precise one later. My first iteration:
Q = m (mass flow rate of CH4)* [∑m (mass of dry products/kg fuel of Methane) * Cp *(T1 - T2) + mH2O * hg at 70°C -(mass of H2O water *hf + mass of vapour *hg)]
0.75kW = 0.000333 *(27.752 (70- Temperature of exhaust) + 5923.35 - 277.85)
2252 = 27.752 (70 - T) + 5645.5
-3393 = 27.752 (70 - T)
-122.3 = 70 - T
T = 192 deg C

BvU
Homework Helper
Check your math. Brackets have a purpose. It might help if you consistently hang on to units as well.

• nightingale
Please disregard my answer above. I realized my mistake, where the moles of water and water vapour varies greatly with the temperature and thus my answer will not be valid.

BvU
Homework Helper
Yes, it will depend on the temperature. With your spreadsheet you can solve that by iterating by hand (or by using the solver).
And do keep track of what is what (dimensions) and of the brackets (m H2O/kg CH4 , not just m H2O)

• nightingale
Thank you very much Chet and BvU, I was finally able to find the temperature that will give me 0.75kW heat recovery! Which I found was around 43 degC from the excel spreadsheet (attached if you would like to see).

My last question:
How do I find the number of 1m lengths of pipe to provide the heat input of 0.75kW to the room? I don't know where to start, any hints please?

#### Attachments

• Efficiency Coursework.xlsx
13.6 KB · Views: 254
BvU
Homework Helper
"Don't know where to start" is too easy. PF guidelines.
You have a lot of info at hand. And ther are the original hints. If you really have no idea: start with 1 chunk of pipe of 1 m length and set up a heat balance.

Chestermiller
Mentor
In post #16, you indicated 3.33x10^-4 kg/sec. Did you mean kg-moles/sec? I got something like 2.1x10^-4 kg-moles/sec. My calculator battery is dead right now, so I can't check my calculations, but can you please double check yours.

Regarding the length of the pipe, if H(T) is the overall enthalpy flow rate of the stream moving down the exhaust pipe, what is the differential heat balance on the pipe between location x and location x + dx? If the water weren't condensing, you could integrate this equation analytically, but, since water is condensing, you need to solve it numerically.

BTW, nice job on that spreadsheet.

Chet

In post #16, you indicated 3.33x10^-4 kg/sec. Did you mean kg-moles/sec? I got something like 2.1x10^-4 kg-moles/sec. My calculator battery is dead right now, so I can't check my calculations, but can you please double check yours.

55* 10^3 *16*m = m * [1*44* 0.871 (CO2) +1.3417*32*0.923 (O2) +12.565*28*1.04 (N2)] * (70 - 25) + 2 * 18* (2626.3-104.8) + 16*16

I realized that this is the same as using mass of each products/kg of fuel [Fuel of CH4 =16kg]

55* 10^3*m = m * [2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (70 - 25) + 2.25 (2626.3-104.8) + 16
and thus m = 3.33 * 10(-4)... at this point I don't really know what the unit is.

Did I make a mistake? Thank you.

-3
"Don't know where to start" is too easy. PF guidelines.
You have a lot of info at hand. And ther are the original hints. If you really have no idea: start with 1 chunk of pipe of 1 m length and set up a heat balance.

I'm sorry BvU.

Regarding the length of the pipe, if H(T) is the overall enthalpy flow rate of the stream moving down the exhaust pipe, what is the differential heat balance on the pipe between location x and location x + dx? If the water weren't condensing, you could integrate this equation analytically, but, since water is condensing, you need to solve it numerically.

Chet, I don't really understand what you're trying to tell me. I'm sorry.

The heat balance I know that utilized the heat transfer coefficient U is
Q = U * Area * (T2-T1)

The T1 is the room temperature is it not? Diameter is 100mm. Since the room temperature is 20 deg C, thus taking the temperature interval by 10 deg starting from
70 deg:
Q = U * Area * (70 - 20)
Q = 9.5 * (7.85*10-3)*50
Q = 3.72875kW
Which is absolutely incorrect, but I don't understand what I'm looking for. The hint told me to:
"As the exhaust gas varies along the pipe, the best way to calculate the length needed is to consider each temperature interval, assume a mean gas temperature and then calculate the length of pipe needed to give the heat transfer between the exhaust gas and the room as the gas cools over the temperature interval".

I don't understand what is the mean gas temperature and what do I need it for. Why do I have to consider each temperature interval? Am I trying to find the Q or the Area?

BvU
Homework Helper
Can you remember my advice about the units ? the 9.5 has units attached.
Area = 0.00785 what kind of things ? a 1 m section of pipe ?

 Ah, the cross sectional area, m2. But the heat transfer takes place in a direction perpendicular to that! So you want the pipe wall area !

Last edited:
Chestermiller
Mentor
The differential heat balance equation should be:
$$\frac{dH}{dx}=-\pi DU(T-20)$$
where T is the local temperature at axial location x along the exhaust pipe and H is a function of T (the enthalpy flow along the tube). If the water were not condensing, H(T) would be given by: ##H=MC(T-25)## where M is the molar flow rate and C is the heat capacity per mole. You integrate the equation until the temperature is 43 degrees.

You approximate the above equation in finite difference form by:
##H(T(x+\Delta x))-H(T(x))=-\pi DU(\frac{(T(x+\Delta x)+T(x)}{2}-20)\Delta x##
where ##\Delta x ## is the incremental distance along the exhaust pipe. Solving for ##\Delta x##, we get:

##\Delta x = -\frac{H(T(x+\Delta x))-H(T(x))}{\pi DU(\frac{(T(x+\Delta x)+T(x)}{2}-20)}##

So you specify 5C increments on T, evaluate the right hand side, and this gives you the ##\Delta x## increment for that temperature change. Your spreadsheet is already practically set up to do this.

Chet

Last edited:
Chestermiller
Mentor
55* 10^3 *16*m = m * [1*44* 0.871 (CO2) +1.3417*32*0.923 (O2) +12.565*28*1.04 (N2)] * (70 - 25) + 2 * 18* (2626.3-104.8) + 16*16

I realized that this is the same as using mass of each products/kg of fuel [Fuel of CH4 =16kg]

55* 10^3*m = m * [2.75* 0.871 (CO2) +2.6834 *0.923 (O2) +22 *1.04 (N2)] * (70 - 25) + 2.25 (2626.3-104.8) + 16
and thus m = 3.33 * 10(-4)... at this point I don't really know what the unit is.

Did I make a mistake? Thank you.
No. I confirm this as the mass flow rate kg/s of CH4 fed to the boiler. I got the kg-mole flow rate of CH4 fed to the boiler as 2.1x10^-5 kg-mole/s, in agreement with your result. Incidentally, the first equation should not have a second 16 multiplying the first 16.

Chet

Chestermiller
Mentor
You can get upper and lower bounds to the length of the tube by recognizing that the temperature difference is always going to be less than 70 - 20 = 50 C, and greater than 43 - 20 = 23 C. You can use such results to compare with the detailed result you get when you integrate the heat balance equation.

##0.1\pi (9.5)(50) L_{min}=750##

##0.1\pi (9.5)(23)L_{max}=750##

Lmin = 5.0 m

Lmax = 10.9 m

Chet

Last edited:
Can you remember my advice about the units ? the 9.5 has units attached.
Area = 0.00785 what kind of things ? a 1 m section of pipe ?

 Ah, the cross sectional area, m2. But the heat transfer takes place in a direction perpendicular to that! So you want the pipe wall area !

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?

The differential heat balance equation should be:
$$\frac{dH}{dx}=-\pi DU(T-20)$$
where T is the local temperature at axial location x along the exhaust pipe and H is a function of T (the enthalpy flow along the tube). If the water were not condensing, H(T) would be given by: ##H=MC(T-25)## where M is the molar flow rate and C is the heat capacity per mole. You integrate the equation until the temperature is 43 degrees.

You approximate the above equation in finite difference form by:
##H(T(x+\Delta x))-H(T(x))=-\pi DU(\frac{(T(x+\Delta x)+T(x)}{2}-20)\Delta x##
where ##\Delta x ## is the incremental distance along the exhaust pipe. Solving for ##\Delta x##, we get:

##\Delta x = -\frac{H(T(x+\Delta x))-H(T(x))}{\pi DU(\frac{(T(x+\Delta x)+T(x)}{2}-20)}##

So you specify 5C increments on T, evaluate the right hand side, and this gives you the ##\Delta x## increment for that temperature change. Your spreadsheet is already practically set up to do this.

Chet

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)

or should I use
dH/dx=−πDU(T−20)
H = ∫ −πDU(T−20) dx
H =−πxDU(T−20)?

Do I first find the H from the formulas above and the insert it into:

Δx=−H(T(xx))−H(T(x))πDU((T(xx)+T(x)2−20)\Delta x = -\frac{H(T(x+\Delta x))-H(T(x))}{\pi DU(\frac{(T(x+\Delta x)+T(x)}{2}-20)}

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 7500Watt?

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)?

Chestermiller
Mentor
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
Mentor
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 hv(T)+(1-f) hl(T)-104.8))

where f is the fraction of water that is vapor, hv(T) is the enthalpy of saturated water vapor, and hl(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 hv(T)+(1-f) hl(T)-104.8))

where f is the fraction of water that is vapor, hv(T) is the enthalpy of saturated water vapor, and hl(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!

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 hv(T)+(1-f) hl(T)-104.8))

where f is the fraction of water that is vapor, hv(T) is the enthalpy of saturated water vapor, and hl(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

#### Attachments

• Efficiency Coursework.xlsx
15.1 KB · Views: 146
BvU
[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 :)