- #1
Andy86
- 47
- 2
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
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(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 can't 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 into (iii) using L=5m and leaving N as the variable?? SOS