Answered

An industrial boiler consists of tubes inside of which flow hot combustion gases. Water boils on the exterior of the tubes. When installed, the clean boiler has an over all heat transfer coefficient of 300 W/m^2 . K. Based on experience, i is anticipated that the fouling factors on the inner and outer surfaces will increase linearly with time as Ra,t and Ryo-at where a, 2.5 x 10^-11 m2 K/W s and a,-1.0 x 10^-11 m^2 - K/W s for the inner and outer tube surfaces, respectively. If the boiler is to be cleaned when the overall heat transfer coeffi- cient is reduced from its initial value by 25%, how long after installation should the first cleaning be scheduled?

Answer :

NUHULAWAL20GO

Answer:

the first cleaning be scheduled 1.006 years after installation

Explanation:

 Given the data in the question;

U[tex]_{clean[/tex] = 300 W/m².K

first we determine the heat coefficient of the dirt surface;

overall heat transfer coefficient is reduced from its initial value by 25%

U[tex]_{dirt[/tex] = ( 1 - 25%) × U[tex]_{clean[/tex]

U[tex]_{dirt[/tex] = ( 1 - 0.25) × 300

U[tex]_{dirt[/tex] = 0.75 × 300

U[tex]_{dirt[/tex] = 225 W/m².K

next we find the inner fouling factor

[tex]R"_{f ,i[/tex] = [tex]a_it[/tex]

[tex]R"_{f ,o[/tex] = (2.5 × 10⁻¹¹)t

for the outer fouling water;

[tex]R"_{f ,o[/tex] = [tex]a_ot[/tex]

[tex]R"_{f ,o[/tex] = ( 1.0 × 10⁻¹¹ )t

now, we determine the total heat transfer coefficient

[tex]\frac{1}{U}[/tex] = [tex]R"_{f ,i[/tex] + [tex]R"_{f ,o[/tex]

we substitute

[tex]\frac{1}{U}[/tex] =  (3.5 × 10⁻¹¹)t

so the first cleaning duration after insulation will be;

[tex]\frac{1}{U} = \frac{1}{U_{dirt}} - \frac{1}{U{clean}}[/tex]

we substitute

(3.5 × 10⁻¹¹)t = [tex]\frac{1}{225} - \frac{1}{300}[/tex]

(3.5 × 10⁻¹¹)t = 0.001111

t = 0.001111 / (3.5 × 10⁻¹¹)

t = 31742857.142857 seconds

t = 31742857.142857 / 3.154 × 10⁷

t = 1.006 years

Therefore, the first cleaning be scheduled 1.006 years after installation

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