Steam produced heat transfer heat exchangers

[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?

 

[SeaofEnergy]

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

I originally had 1.13m² but that was due to a units error, the 724.5 x 10⁵ is in KJ/hr which means when it comes to calculating the Area it is 724.5 x 10⁸ as the coefficient is in W m^-2 K^-1. This balances the units. I think that's correct anyway.