mild steel tank of wall thickness 12 mm contains water at 95°C. The
thermal conductivity of mild steel is 50 W/m°C, and the heat transfer coefficients dfor the inside
and outside the tank are 2850 and 10 W/m2°C, respectively. If the atmospheric temperature is
15°C, calculate :
(i) The rate of heat loss per m2 of the tank surface area ;
(ii) The temperature of the outside surface of the tank.
The variables that we have are:
Thickness of mild steel tank wall, L = 12 mm = 0.012 m
Temperature of hot fluid (water), "t_{h_f} = 95\u00b0C"
Temperature of cold fluid (air), "t_{c_f} = 15\u00b0C"
Thermal conductivity of mild steel, "k= 50 \\cfrac{W}{m\u00b0C}"
Heat transfer coefficients: Hot fluid (water), "h_{h_f} = 2850 \\cfrac{W}{m^2\u00b0C}"
Cold fluid (air), "h_{c_f} = 10 \\cfrac{W}{m^2\u00b0C}"
First, we use the relation to find the rate of heat loss per m2 of the tank surface area as"\\cfrac{q}{A} = U(t_{h_f} \u2013 t_{c_f})" .
Before we proceed, we have to find the overall heat transfer coefficient, U, and we use the relation "\\frac{1}{U}=\\frac{1}{h_{h_f} }+\\frac{L}{k}+\\frac{1}{h_{c_f} }"
We substitue and find:
"\\cfrac{1}{U}=\\cfrac{1}{2850 \\frac{W}{m^2\u00b0C} }+\\cfrac{0.012\\,m}{50 \\frac{W}{m\u00b0C}}+\\cfrac{1}{10 \\frac{W}{m^2\u00b0C}}\n\\\\U=\\cfrac{1}{0.100590877\\frac{m^2\u00b0C}{W}}=9.9413\\frac{W}{m^2\u00b0C}"
After that we can calculate the rate of heat loss per m2 of the tank surface:
"\\cfrac{q}{A} = (9.9413\\frac{W}{m^2\u00b0C})(95 \u2013 15)\u00b0C\n\\\\\\implies \\cfrac{q}{A} = 795.304 \\frac{W}{m^2}"
After that, we can also establish another equation for the heat loss on the outside to find the temperature of the shell:
"\\cfrac{q}{A} = h_{c_f} (t_{2} \u2013 t_{c_f}) \\implies t_{2} =\\cfrac{q\/A}{h_{c_f}} + t_{c_f}"
We substitute and we can confirm the temperature of the outside:
"t_{2} =\\cfrac{795.304 \\frac{W}{m^2}}{10 \\frac{W}{m^2\u00b0C}} + 15\u00b0C=94.53 \u00b0C"
Reference:
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