Answer to Question #233680 in Molecular Physics | Thermodynamics for Unknown346307

Question #233680

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.


1
Expert's answer
2021-09-06T11:34:54-0400

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"


In conclusion, (i) the rate of heat loss per m2 of the tank surface area is 795.304 W/m2, and (ii) the temperature of the outside surface of the tank is 94.53 °C.


Reference:

  • Sears, F. W., & Zemansky, M. W. (1973). University physics.

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