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), thf=95°Ct_{h_f} = 95°C

Temperature of cold fluid (air), tcf=15°Ct_{c_f} = 15°C

Thermal conductivity of mild steel, k=50Wm°Ck= 50 \cfrac{W}{m°C}

Heat transfer coefficients: Hot fluid (water), hhf=2850Wm2°Ch_{h_f} = 2850 \cfrac{W}{m^2°C}

Cold fluid (air), hcf=10Wm2°Ch_{c_f} = 10 \cfrac{W}{m^2°C}


First, we use the relation to find the rate of heat loss per m2 of the tank surface area asqA=U(thftcf)\cfrac{q}{A} = U(t_{h_f} – t_{c_f}) .


Before we proceed, we have to find the overall heat transfer coefficient, U, and we use the relation 1U=1hhf+Lk+1hcf\frac{1}{U}=\frac{1}{h_{h_f} }+\frac{L}{k}+\frac{1}{h_{c_f} }


We substitue and find:


1U=12850Wm2°C+0.012m50Wm°C+110Wm2°CU=10.100590877m2°CW=9.9413Wm2°C\cfrac{1}{U}=\cfrac{1}{2850 \frac{W}{m^2°C} }+\cfrac{0.012\,m}{50 \frac{W}{m°C}}+\cfrac{1}{10 \frac{W}{m^2°C}} \\U=\cfrac{1}{0.100590877\frac{m^2°C}{W}}=9.9413\frac{W}{m^2°C}


After that we can calculate the rate of heat loss per m2 of the tank surface:

qA=(9.9413Wm2°C)(9515)°C    qA=795.304Wm2\cfrac{q}{A} = (9.9413\frac{W}{m^2°C})(95 – 15)°C \\\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:


qA=hcf(t2tcf)    t2=q/Ahcf+tcf\cfrac{q}{A} = h_{c_f} (t_{2} – 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:

t2=795.304Wm210Wm2°C+15°C=94.53°Ct_{2} =\cfrac{795.304 \frac{W}{m^2}}{10 \frac{W}{m^2°C}} + 15°C=94.53 °C


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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