- #1

- 29

- 0

## 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)

Last edited: