Heat exchanger question

  • Thread starter Andy86
  • Start date
  • #1
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Homework Statement


flue gases = 50,000kg h^-1
surface area = 113.57m^2
Do = 25 + (3*2) = 31mm

The tubes within a boiler are 25mm inside diameter with a wall thickness of 3mm. The flue gas velocity through the tubes is to maintain the overall heat transfer coefficient value and minimise pressure losses to be more than 22ms^-1 and less than 28ms^-1

Assuming average density of the flue gases is 1.108kgm^-3

(i) calculate the minimum and maximum number of tubes in each pass
(ii) overall length of tubes at each of these numbers of tubes
(iii) the minimum number of tube passes in each case if the boiler tube is to be less than 5M

Homework Equations


number of tubes = (4 * qmc(s)) / (density * π * Di * V)
A= π * Do * L * N
L = A / π * Do * N
A= π * D * L * N
rearranged
N= (A/L) / (π*Do)


The Attempt at a Solution


[/B]
(i)

qmc= 50,000/3600
=13.88kg s^-1

minimum velocity @ 22ms^1

number of tubes = (4 * qmc(s)) / (density * π * Di * V)
= (4 x 13.88) / (1.108 * π * (25*10^-3)*22)
=1160 tubes

minimum velocity @ 28ms^1

number of tubes = (4 * qmc(s)) / (density * π * Di * V)
= (4 x 13.88) / (1.108 * π * (25*10^-3)*28)
=912 tubes

(ii)

A= π * Do * L * N

rearranged

L = A / π * Do * N

1160 tubes
L = 113.57 / π * (31*10^-3) * 1160
L= 1.005M

912 tubes
L = 113.57 / π * (31*10^-3) * 912
L= 1.279M

(iii) COMPLETELY LOST!!! been barking up the following tree but cant make it work!! some help would be greatly appreciated!!

calculate number of tubes required in each case based on a tube length of 5M

A= π * D * L * N
rearranged
N= (A/L) / (π*Do)

22 ms^-1

=(113.57/5)/(π*(31*10^-3)
= 233.2
= 234 tubes @ 5m

26 ms^-1

=(113.57/5)/(π*(31*10^-3)
= 233.2
= 234 tubes @ 5m

I was thinking of working out the area for that number of tubes (in each case) ie 912 and 1160 based on the tube lengths calculated in (ii) then plugging the new area in to (iii) using L=5m and leaving N as the variable?? SOS
 

Answers and Replies

  • #2
21,627
4,911
It looks like only one pass is required. There must be something wrong with the problem statement.
 
  • #3
47
2
http://imgur.com/a/D5qqN

That is the original question paper ^^^
 
Last edited:
  • #5
47
2
I have just checked it. I got ;

heat transfer duty = qmH * Hfg
= (3598.5 * 2015) / 3600
= 2014.1KW

Area m^2 = heat transfer duty / U * LMTD
= (2014.1*10^3) / (54 * 328.4)
= 113.57m^2
 
  • #7
47
2
Im not entirely sure where that calc comes from (im not doubting it BTW), if that was the case the area would be;

Area m^2 = heat transfer duty / U * LMTD
= (22400*10^3) / (54 * 328.4)
= 1263.1m^2

seems a bit excessive but I could be wrong?
 
  • #8
21,627
4,911
Im not entirely sure where that calc comes from (im not doubting it BTW)

That's the rate of heat loss from the flue gas, which gets transferred to the liquid water to make steam.
 
  • #9
47
2
ok, so if you follow it through what surface area do you get?
 
  • #11
47
2
ok great, so if you follow the question through to where I was struggling;

(iii) the minimum number of tube passes in each case if the boiler tube is to be less than 5M

Do you have any thoughts on how to tackle this?
 
  • #12
21,627
4,911
ok great, so if you follow the question through to where I was struggling;

(iii) the minimum number of tube passes in each case if the boiler tube is to be less than 5M

Do you have any thoughts on how to tackle this?
The total area is ##\pi D L n##, where n is the number of tubes and L is the length of each tube. So, first find the length of each tube. The tube bank has to fit into a space no longer than 5 meters. If L is greater than 5 meters, you need to "fold" the tubes once or more so that there are enough tube passes to accommodate the total tube length within the allowable space.
 
  • #13
47
2
The total area is ##\pi D L n##, where n is the number of tubes and L is the length of each tube. So, first find the length of each tube. The tube bank has to fit into a space no longer than 5 meters. If L is greater than 5 meters, you need to "fold" the tubes once or more so that there are enough tube passes to accommodate the total tube length within the allowable space.

Ok great, ill have a look in the morning its very late here now (UK), thanks for your help Chester! Hopefully I will have some sensible answers tomorrow!!
 
  • #14
47
2
3ci)

Di = 25mm
D0= 31mm
Min V = 22ms^-1
Max V= 28ms^-1

qMc = 50000/3600

Heat load = 50000/3600 * 1150 * (1600-200)
=24000

Area = 22400*10^3 / (54*328.4)
= 1263.1M^2

Min number of tubes in each

22 ms^-1

N = 4*13.88 / 1.108 * π * (25*10^-3)^2 * 22
N = 1160 Tubes @ 22 m s ^-1

28 ms^-1

N = 4*13.88 / 1.108 * π * (25*10^-3)^2 * 28
N = 912 Tubes @ 22 m s ^-1

3cii) Overall length of tubes in each case

A = π * D * L * N

Re arranged

L = A / π * D * N
= 1263 / π * (31*10^-3) * 1160
= 11.18M @ 1160 Tubes

L = A / π * D * N
= 1263 / π * (31*10^-3) * 912
= 14.22M @ 1160 Tubes

3ciii) Minimum number of passes in each case

V = 22 ms^-1

L = 11.18M @ 1160
= 11.18 / 5
= 2.236 > 3 Passes required

V = 28 ms^-1

L = 14.22M @ 912
= 14.22 / 5
= 2.844 > 3 Passes required

Hopefully this is somewhere near!!!???
 
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