Producing Dry Saturated Steam in a Fire Tube Boiler

[electr]

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.
(b) The overall heat transfer coefficient based on the outside area of the
tubes is given as 54 W m–2 K–1. Determine the area of heat transfer
required to perform this duty.
(c) The tubes within the boiler are to be 25 mm inside diameter with a
wall thickness of 3 mm. The average flue gas velocity through the
tubes to maintain the overall heat transfer coefficient value and to
minimise pressure losses is to be more than 22 m s–1 and less than
28 m s–1.
Assuming that the average density of the flue gases is 1.108 kg m–3,
calculate:
(i) the minimum and maximum number of tubes in each pass
(ii) the overall length of tubes at each of these numbers of tubes
(iii) the minimum number of tube passes in each case, if the length
of a boiler tube is to be less than 5 metres.

mostly i would like to confirm if the answers are correct or not

Homework Equations

The Attempt at a Solution

a) 0.9 x 50.000 x 1.15 x (1600 - 200 )= mc x 2015
mc = 35955.33 kg/h

b)mg = 50.000 kg / h ===> mg =50.000/3600= 13.889 kg/s
Δtem=Δτ1 - Δτ2 / ln (Δτ1/Δτ2)
= 20-1420 / ln (20 / 1420)
= 328.43
then
54 x A x 328.43 = 13889 x (1,15 x10^3) x (1600-200)
=> A = 1260.833
A = 1260.833 x 0,9
A = 1134.02 m^2

c (i) mg = 50.000 kg / h ===> mg =50.000/3600= 13.889 kg/s
mg = n (Π/4 χ d^2 x v) x ρ (ρ is density)
==>n = mg / (π/4 x .d^2 x ρ χ vmax)
nmin = 13.889 / π/4 χ 0.025^2 χ 1.108 χ 28
nmin = 912,48 aprox 913
for max
nmax = 13.899 / π/4 χ 0.025^2 χ 1.108 χ 22
nmax = 1161,338 aprox 1162
c (ii) A = n x d x l
d = 25 +(2 x 3)
d = 0.031 m
l= A/nmin x π χ d
l = 1134.02 / 913 x π x 0.031
l1 = 12.09 m
for max
lmax =1134.02 / 1162 x π x 0.031
l2 = 9,50 m
c(iii)
for min m(min)= l1/5 ==>m(min)12,09/5
m(min) = 2,418 passes (aprox 2 passes)
for max m(max)=l2/5 ==>m(max) = 9,50/5
m(max)=1,9 passes (aprox 2 passes)

if its not correct i would like your advise ,thank you,(i know i didnt write the equations,but like i said i m here just to check my answer if they are correct)

 

[CWatters]

Only had time to check a) and I made it 39,950kg/h.

 

[electr]

still i got 35955.33,you did something different?

 

[Chestermiller]

I didn't check you arithmetic, but your methodology is correct. The only change I would make would be to round the 2.4 tube passes up to 3 passes, rather than rounding down to 2 passes. Otherwise you won't be able to achieve the desired amount of heat transfer.

 

[CWatters]

still i got 35955.33,you did something different?

My bad. I forgot the 0.9.

 

[electr]

thank you both for your help