Steam produced heat transfer heat exchangers

[Mitch1]

Homework Statement

Dry saturated steam at a temperature of 180oC is to be produced in a fire tube boiler from the cooling of 50 000 kg h–1 of flue gases from a pressurised combustion process. The gases enter the tubes of the boiler at 1600oC and leave at 200oC. The feed water is externally preheated to 180oC before entering the boiler.

The mean specific heat capacity of the flue gases is 1.15 kJ kg–1 K–1. The latent heat of vaporisation of the water at 180oC is 2015 kJ kg–1. Feed water temperature = 180oC.

Determine the amount of steam produced per hour, if the total heat loss is 10% of the heat available for steam raising.

Homework Equations

Mass steam= heat available / heat required to produce 1kg steam
Qmc=heat transfer rate/ Cpc(tc2-tc1)
Heat transfer rate=Qmh hfg

The Attempt at a Solution

Heat required would be 2257 from steam table at 100C evaporation
Heat available
Qmh hfg
50000*2015=10075x10^4 kW
Ht rate=10% of 10075x10^4 = 90675x10^3
Mass steam=heat available / heat req 1kg
Mass steam = 90675x10^3/2257
=40175 kg h-1
Can anyone check over this as I am unsure if it is correct
Thanks

 

[SteamKing]

Homework Statement

Dry saturated steam at a temperature of 180oC is to be produced in a fire tube boiler from the cooling of 50 000 kg h–1 of flue gases from a pressurised combustion process. The gases enter the tubes of the boiler at 1600oC and leave at 200oC. The feed water is externally preheated to 180oC before entering the boiler.

The mean specific heat capacity of the flue gases is 1.15 kJ kg–1 K–1. The latent heat of vaporisation of the water at 180oC is 2015 kJ kg–1. Feed water temperature = 180oC.

Determine the amount of steam produced per hour, if the total heat loss is 10% of the heat available for steam raising.

Homework Equations

Mass steam= heat available / heat required to produce 1kg steam
Qmc=heat transfer rate/ Cpc(tc2-tc1)
Heat transfer rate=Qmh hfg

The Attempt at a Solution

Heat required would be 2257 from steam table at 100C evaporation

Why? The feed water is being supplied already pre-heated to 180 C

Heat available
Qmh hfg
50000*2015=10075x10^4 kW

The units here are not kW. 1 W = 1 J/s, and 1 hour ≠ 1 second

Show the units here and carry thru with the multiplication. That's how to avoid mistakes.

Ht rate=10% of 10075x10^4 = 90675x10^3
Mass steam=heat available / heat req 1kg
Mass steam = 90675x10^3/2257=40175 kg h-1

As mentioned above, the feed water is already pre-heated to 180 C when it enters the heat exchanger. The problem statement tells you that the latent heat of vaporization is 2015 kJ/kg for water at 180 C.

The enthalpy of steam at 100 C is irrelevant here.

 

[Mitch1]

Why? The feed water is being supplied already pre-heated to 180 C

The units here are not kW. 1 W = 1 J/s, and 1 hour ≠ 1 second

Show the units here and carry thru with the multiplication. That's how to avoid mistakes.

As mentioned above, the feed water is already pre-heated to 180 C when it enters the heat exchanger. The problem statement tells you that the latent heat of vaporization is 2015 kJ/kg for water at 180 C.

The enthalpy of steam at 100 C is irrelevant here.

Thanks for your reply
Ah so if I work out the correct units and carry it through I will divide it by 2015 which is at 180C ? Does the rest Seem ok?

 

[SteamKing]

Thanks for your reply
Ah so if I work out the correct units and carry it through I will divide it by 2015 which is at 180C ? Does the rest Seem ok?

The heat supplied by the flue gas seems to have been overlooked. The gas enters at a temp. of 1600 C and leaves at a temp. of 200 C. The specific heat capacity of the flue gas is 1.15 kJ/kg-K. You've got to work out how much heat is available to be transferred from the gas to the feed water to make steam.

 

[Mitch1]

The heat supplied by the flue gas seems to have been overlooked. The gas enters at a temp. of 1600 C and leaves at a temp. of 200 C. The specific heat capacity of the flue gas is 1.15 kJ/kg-K. You've got to work out how much heat is available to be transferred from the gas to the feed water to make steam.

Hi again,
To find out this is this the spec heat*temp difference
I'm not that clued up on this subject
Once this is found is this part of the heat available

 

[SteamKing]

Hi again,
To find out this is this the spec heat*temp difference
I'm not that clued up on this subject
Once this is found is this part of the heat available

Yes. You are transferring the heat from the flue gas to the feed water to turn it into steam. Remember to include the 10% heat loss in your calculations of steam production.

 

[Mitch1]

Yes. You are transferring the heat from the flue gas to the feed water to turn it into steam. Remember to include the 10% heat loss in your calculations of steam production.

So it would be 1400*1.15*.9 for the ten percent heat loss which gives 1449 kW I'm assuming?
So is this the heat available? Then we divide it by 2015 the enthalpy at 180C ?
I believe I am missing a step out as this seems too low of a value

 

[SteamKing]

So it would be 1400*1.15*.9 for the ten percent heat loss which gives 1449 kW I'm assuming?
So is this the heat available? Then we divide it by 2015 the enthalpy at 180C ?
I believe I am missing a step out as this seems too low of a value

Again, you need to show and carry thru the units in these calculations. You are just assuming what units appear in the results.

You've also neglected to include the flow rate of the flue gases in your heat calculations, which is why the result appears so low.

For the heat available:

H = eff. * flow * c_p * ΔT, which has units

H = kg/hr * kJ/kg-K * K

H = kJ/hr, which is not the same units as kW, so stop calling the heat rate kW

 

[Mitch1]

Again, you need to show and carry thru the units in these calculations. You are just assuming what units appear in the results.

You've also neglected to include the flow rate of the flue gases in your heat calculations, which is why the result appears so low.

For the heat available:

H = eff. * flow * c_p * ΔT, which has units

H = kg/hr * kJ/kg-K * K

H = kJ/hr, which is not the same units as kW, so stop calling the heat rate kW

So, (1600-200)*1.15*.9*50,000=72450000kj/hr
7245x10^4/2015=35955kg/hr ?

 

[SteamKing]

This looks better.

 

[Mitch1]

This looks better.

Finally,
Thanks for your help!

 

[lycee]

On the same question: The overall heat transfer coefficient based on outside area of the tubes is given as 54 Wm K .Determine the area of heat transfer required to perform this duty

 

[insightful]

On the same question: The overall heat transfer coefficient based on outside area of the tubes is given as 54 Wm K .Determine the area of heat transfer required to perform this duty

Are these units correct?

 

[lycee]

Should be Wm^-2 K^-1

 

[insightful]

Should be Wm^-2 K^-1

Good. Now, what's the equation for the heat transfer in a heat exchanger?

 

[lycee]

I'm new to this forum and not sure yet how to write down the symbols, but heat transfer= Phi = U A (LMTD).So finding A would be phi/U(LMTD).Finding the LMTD is proving difficult as I'm not sure of the outlet temp of the feed water.

 

[SteamKing]

I'm new to this forum and not sure yet how to write down the symbols, but heat transfer= Phi = U A (LMTD).So finding A would be phi/U(LMTD).Finding the LMTD is proving difficult as I'm not sure of the outlet temp of the feed water.

It's not clear what you are trying to analyze here.

The original problem involved a fire tube boiler. The conditions at which the boiler produced steam were not disclosed in the OP. The only thing you know for certain is that the feed water entered the boiler at 180° C.

 

[lycee]

If you refer to the first post from Mitch1 all the details of the first part are there.The second part of the question is asking for the area required to perform this duty when the overall heat coefficient is 54 Wm^-2 K^-1

 

[insightful]

I'm new to this forum and not sure yet how to write down the symbols, but heat transfer= Phi = U A (LMTD).So finding A would be phi/U(LMTD).Finding the LMTD is proving difficult as I'm not sure of the outlet temp of the feed water.

Good. You have a pool of water boiling at 180^oC producing steam at 180^oC. What might the water temperature be everywhere outside the tubes?

 

[lycee]

Good. You have a pool of water boiling at 180^oC producing steam at 180^oC. What might the water temperature be everywhere outside the tubes?

Of course it's obvious when you look at it logically.Thanks for the heads up.I can know find the LMTD and ultimately find the required area

 

[MCTachyon]

Hi,

I have a query to what lycee was trying to work out.

When working out the LMTD I keep getting a negative number. My working is as follows:

LMTD = (Tc1 -Tc2) / In[(Th-Tc1)/(Th-Tc2)]

Where I have:

Th = 180°C
Tc1 = 1600°C
Tc2 = 200°C

Subbing those values in gives me an answer of -328.41°C.

Do I have the values correct? Or even the equation?

Thank you in advance for any response.

 

[insightful]

Your LMTD equation is incorrect.

 

[SteamKing]

Hi,

I have a query to what lycee was trying to work out.

When working out the LMTD I keep getting a negative number. My working is as follows:

LMTD = (Tc1 -Tc2) / In[(Th-Tc1)/(Th-Tc2)]

Where I have:

Th = 180°C
Tc1 = 1600°C
Tc2 = 200°C

Subbing those values in gives me an answer of -328.41°C.

Do I have the values correct? Or even the equation?

Thank you in advance for any response.

You can't calculate a valid LMTD for a boiler since there is latent heat involved.

https://en.wikipedia.org/wiki/Logarithmic_mean_temperature_difference

 

[MCTachyon]

Your LMTD equation is incorrect.

Is the equation I want:

LMTD = (Tc1-Tc2) / [(In(Tc1) - In(Tc2))]

Therefore:

1400 / 2.07 = 676.33°C

*Source for that equation from Wiki page SteamKing posted.Edit: I've just looked again.. The equation on the Wiki page mentions temperature change at points A and B. Not set temperatures.

Lemme have some time with this today and I'll report back. The only equation I know for the LMTD is the one I originally stated.

 

[MCTachyon]

Having worked out correctly what Mich1 was trying to work out, I now have the values for:

q_m = 35955 kgh^-1
Φ = 72.45 x 10⁶kJh^-1

I am now trying to work out the area of heat transfer if the overall heat transfer coefficient based on the outside area of the tubes is given as:

U = 54 Wm^-2k^-1

I am now using the equation:

Φ = UA(LMTD)

Rearranged for A:

A = Φ / (U x LMTD) ..

So far so good?

 

[SteamKing]

Having worked out correctly what Mich1 was trying to work out, I now have the values for:

q_m = 35955 kgh^-1
Φ = 72.45 x 10⁶kJh^-1

I am now trying to work out the area of heat transfer if the overall heat transfer coefficient based on the outside area of the tubes is given as:

U = 54 Wm^-2k^-1

I am now using the equation:

Φ = UA(LMTD)

Rearranged for A:

A = Φ / (U x LMTD) ..

So far so good?

Yeah, except boilers don't work with LMTD because the fluid inside them is undergoing a phase change. The boiler furnishes the latent heat required to turn the feed water into saturated vapor, and this process occurs at constant temperature.

The opposite occurs in the condenser. The condenser removes the latent heat of vaporization from the incoming steam and turns it back into saturated liquid. This process occurs at constant temperature, also.

You can't analyze a boiler or a condenser like it was any other heat exchanger.

The following attachment shows the calculations for a student-designed fire tube boiler:

http://u.osu.edu/joesportfolio/files/2015/01/Three-Phase-Water-Boiler-2253n42.pdf

 

[MCTachyon]

Yeah, except boilers don't work with LMTD because the fluid inside them is undergoing a phase change. The boiler furnishes the latent heat required to turn the feed water into saturated vapor, and this process occurs at constant temperature.

The opposite occurs in the condenser. The condenser removes the latent heat of vaporization from the incoming steam and turns it back into saturated liquid. This process occurs at constant temperature, also.

You can't analyze a boiler or a condenser like it was any other heat exchanger.

The following attachment shows the calculations for a student-designed fire tube boiler:

http://u.osu.edu/joesportfolio/files/2015/01/Three-Phase-Water-Boiler-2253n42.pdf

Thank you. I will go through this PDF and have another stab.

Ta,

 

[MCTachyon]

Right I've had a good go at this over weekend and haven't got any closer to answering.

There is an equation on the PDF I was playing with to try and pulling something together but think I am completely barking up the wrong tree.

The equation being:

Q = A [(T_air-T_w) / (1/h_air+t_tube/K_tube+1/h_w)]

We are not given the thickness of tubes to work this equation out though.

I have tried reading further into the thermal design of fire tube boilers on other sites and seem to have the same equations I was encountering before.

Sorry to have dragged this out. I spent so long on the LMTD route that my train of thought probably isn't on the right track at all.

 

[SteamKing]

Right I've had a good go at this over weekend and haven't got any closer to answering.

There is an equation on the PDF I was playing with to try and pulling something together but think I am completely barking up the wrong tree.

The equation being:

Q = A [(T_air-T_w) / (1/h_air+t_tube/K_tube+1/h_w)]

We are not given the thickness of tubes to work this equation out though.

I have tried reading further into the thermal design of fire tube boilers on other sites and seem to have the same equations I was encountering before.

Sorry to have dragged this out. I spent so long on the LMTD route that my train of thought probably isn't on the right track at all.

From what I remember about boiler design (I studied marine engineering at college), the characteristics of the tubes were important to the heat transfer occurring inside the boiler itself.

The original problem was never about the heat transfer through the tubes into the feed water, but about the gross heat energy which was available in the flue gasses, therefore, the right design information for the tubes was not provided.

 

[insightful]

You can't analyze a boiler or a condenser like it was any other heat exchanger.

I have specified dozens of condensers and boilers and used LMTD successfully. Since simple phase-change boiling or condensing cannot have a temperature cross, no correction to LMTD is needed.

Given the statement of the problem, I don't know of any other approach.

MCTachyon, you still haven't got the correct LMTD.

 

[MCTachyon]

I have specified dozens of condensers and boilers and used LMTD successfully. Since simple phase-change boiling or condensing cannot have a temperature cross, no correction to LMTD is needed.

Given the statement of the problem, I don't know of any other approach.

MCTachyon, you still haven't got the correct LMTD.

The only other equation I've got in my notes is:LMTD = (T_H1 - T_C2) - (T_H2 - T_C1) / In[(T_H1 - T_C2) / (T_H2 - T_C1)]

From what I can see from my notes you need 2 inlet and 2 outlet temperatures to use this equation through. I have the flue gases inlet and outlet temperatures (1600°C and 200°C) and the desired outlet temperature of the steam (180°C).

The question mentions "The feed water is externally preheated to 180°C before entering the boiler". Does this mean we can take T_H1 and T_H2 as 180°C to fill this equation?

 

[insightful]

Does this mean we can take TH1 and TH2 as 180°C to fill this equation?

Aren't these the cold side temperatures?

 

[MCTachyon]

Aren't these the cold side temperatures?

Blugh, my bad.

So taking T_C1 and T_C2 as 180 in the equation:

LMTD = (T_H1 - T_C2) - (T_H2 - T_C1) / In[(T_H1 - T_C2) / (T_H2 - T_C1)]

Gives:

LMTD = (1600 - 180) - (200 - 180) / In[(1600 - 180) / (200 - 180)]

= 1400 / 4.26

= 328.64°C

 

[insightful]

= 328.64°C

Bingo!

 

[MCTachyon]

So we now just add to the equation:

Φ = U*A*LMTD

Therefore:

A = Φ / U*LMTD

A = 72.45 x 10⁶Jh^-1 / (54 Wm^-2k^-1 * 328.64°C)

Balancing units gives

(72.45x10⁶) / 3600 = 20125 Js^-1

Completing:

A = 20125 / 1762.56

A = 11.42m²Close? Or have I missed something out?

 

[SeaofEnergy]

I am confused by this question as well.

When i calculate the area as per your last post MCtachyon, i end up with 1.13m² ( i had a slightly different temp result than you - and 54x328.64 doesn't equal what you calculated its 17746.56)

But when i use this to calculate the number of tubes i end up with n= 1160 for the max and n=913 for the min.

This seems like a really high number of tubes, this also leads me to overall length of 0.1m and 0.12 m respectively.

I know this can't be right because the question directly after requires that the tubes be divided into passes less than 5 m long so the overall length must be greater than what i have calculated.

Any help with this would be greatly appreciated!

 

[insightful]

Show your work for getting phi (total heat transferred).

 

[SeaofEnergy]

Φ =q_mhC_ph (T_H1-T_H2)
= 50000 x 1.15(1600-200)
= 80.5x10⁶ J/hr
with 10% heat loss = 80.5x10⁶ x 0.9
=724.5 x 10⁵ J/hr

Calculated ΔT as per MCtachyons post above (slightly different answer as i didn't round the 4.26 number)

Φ=UAΔT
724.5x10⁵ = 54 x A(328.4)
A= 724.5x10⁵/3600 / 54 x 328.4
= 1.13m²

q_mh = ρAfu
so for u=22m/s
Af= 50000/3600 / 1.108 x 22
=0.569 kg s^-1

Af= πd²n / 4 where n=number of tubes
n= 4Af / πd²
= 4 x 0.569 / π(0.025)²
=1159.15 = 1160 tubes.

I have also tried the following equation which gives me an answer of 1160.7

n= q_mh / ρπd²uCan anyone see where I am going wrong?

 

[SteamKing]

Φ =q_mhC_ph (T_H1-T_H2)
= 50000 x 1.15(1600-200)
= 80.5x10⁶ J/hr
with 10% heat loss = 80.5x10⁶ x 0.9
=724.5 x 10⁵ J/hr

Calculated ΔT as per MCtachyons post above (slightly different answer as i didn't round the 4.26 number)

Φ=UAΔT
724.5x10⁵ = 54 x A(328.4)
A= 724.5x10⁵/3600 / 54 x 328.4
= 1.13m²

q_mh = ρAfu
so for u=22m/s
Af= 50000/3600 / 1.108 x 22
=0.569 kg s^-1

Af= πd²n / 4 where n=number of tubes
n= 4Af / πd²
= 4 x 0.569 / π(0.025)²
=1159.15 = 1160 tubes.

I have also tried the following equation which gives me an answer of 1160.7

n= q_mh / ρπd²uCan anyone see where I am going wrong?

Don't you want to calculate the surface area of the tubes, rather than the cross sectional area? Isn't the heat transferred thru the circumference of each tube?

 

[insightful]

Φ =qmhCph (TH1-TH2)
= 50000 x 1.15(1600-200)
= 80.5x106 J/hr
with 10% heat loss = 80.5x106 x 0.9
=724.5 x 105 J/hr

Check your units closely here.
Why not add units to all equations??
Also, I would not deduct the 10% loss here for gross heat transferred.

 

[MCTachyon]

I am confused by this question as well.

When i calculate the area as per your last post MCtachyon, i end up with 1.13m² ( i had a slightly different temp result than you - and 54x328.64 doesn't equal what you calculated its 17746.56)

I rushed this as I was on lunch in work when I was posting. See edited post for actual working.

 

[SeaofEnergy]

Don't you want to calculate the surface area of the tubes, rather than the cross sectional area? Isn't the heat transferred thru the circumference of each tube?

As far as I am aware the A in the Φ=UAΔT equation gives you the surface area of the heat transfer. Which is what I'm after no?

Check your units closely here.
Why not add units to all equations??
Also, I would not deduct the 10% loss here for gross heat transferred.

That's what Steamking basically confirmed was ok to Mitch earlier? i.e Mitch said 'So, (1600-200)*1.15*.9*50,000=72450000kj/hr'

Steamking said, 'that looks much better' all i did was separate out the x0.9 stage.

Any help beyond these pretty vague clues? I have looked at the units over and over before i came on here to try and get some help.

 

[MCTachyon]

So we now just add to the equation:

Φ = U*A*LMTD

Therefore:

A = Φ / U*LMTD

A = 72.45 x 10⁶Jh^-1 / (54 Wm^-2k^-1 * 328.64°C)

Balancing units gives

(72.45x10⁶) / 3600 = 20125 Js^-1

Completing:

A = 20125 / 17746.56

A = 1.13m²Close? Or have I missed something out?

*Edited from post 35.

 

[SteamKing]

As far as I am aware the A in the Φ=UAΔT equation gives you the surface area of the heat transfer. Which is what I'm after no?

Yes, but in order to calculate the number of tubes which gives this surface area, you don't use the cross-sectional area of each tube like you did in your calculations:

Af= πd²n / 4 where n=number of tubes
n= 4Af / πd²
= 4 x 0.569 / π(0.025)²
=1159.15 = 1160 tubes.

Can anyone see where I am going wrong?

For an individual tube, SA ≠ π d² / 4

 

[insightful]

Φ =qmhCph (TH1-TH2)
= 50000 x 1.15(1600-200)
= 80.5x106 J/hr
with 10% heat loss = 80.5x106 x 0.9
=724.5 x 105 J/hr

Should this be kJ/hr??

 

[insightful]

A note on the "10% loss."

Since we don't know exactly where the heat loss occurs, it is prudent to consider it to occur after the heat is transferred through the tubes. (Prudent because it results in 10% greater surface area and is thus more conservative.)
That is why I would not incorporate the loss when calculating phi.

 

[SeaofEnergy]

Yes, but in order to calculate the number of tubes which gives this surface area, you don't use the cross-sectional area of each tube like you did in your calculations:For an individual tube, SA ≠ π d² / 4

My materials state this as the route to determine number of tubes.

q_mh=ρAfu

where
q_mh - mass flowrate of the liquid/gas = 50000/3600 kJ/s
ρ is the density of the liquid/gas = 1.108 kg m^-3
Af is the flow area - (unknown calculated as 0.569 m² using this equation)
u is the velocity of the liquid/gas = 22 ms^-1

then

Af= πd²n / 4

where
d is the internal diameter of the tubes = 0.025m
n is the number of tubes.

or alternatively:

n= 4q_mh / ρπd²u
The problem i am having is that i cannot get these equations to return a viable answer. There must be something i am missing here but i still cannot see what it is. I have tried recalculating this with the revised surface area of 1134.8 m² but still have 1160 tubes at max and 913 tubes at min with 65 and 85 passes respectively when restricted to 5m length passes. I am new to heat exchangers but this seems like an unrealistically high amount of tubes/passes?

Should this be kJ/hr??

Yes your right. Cheers!

 

[insightful]

I have tried recalculating this with the revised surface area of 1134.8 m2 but still have 1160 tubes

Use these numbers. How long would a single tube bundle be to give the required surface area?

 

[SeaofEnergy]

Thanks I have got there now. Just needed a nudge in the right direction!Appreciate the help folks! :)

 

[MCTachyon]

surface area of 1134.8 m²

Is this surface area correct for his part of the question?