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Sagot :
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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