Answer to Question #232811 in Molecular Physics | Thermodynamics for Unknown346307

Question

. A 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 for 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

Answer 1

Thickness of mild steel tank wall

$L=12 \;mm = 0.012 \;m$

Temperature of water

$t_{hf}= 95 \;°C$

Temperature of air

$t_{cf}= 15 \;°C$

Thermal conductivity of mild steel

$k=50 \;W/m°C$

Heat transfer coefficients :

Hot fluid (water)

$h_{hf}= 2850 \;W/m^2°C$

Cold fluid (air)

$h_{cf}= 10 \;W/m^2°C$

(i) Rate of heat loss per m² of the tank surface area, q :

Rate of heat loss per m² of tank surface,

$q=UA(t_{hf}-t_{cf})$

The overall heat transfer coefficient, U is found from the relation

$\frac{1}{U}= \frac{1}{h_{hf}}+ \frac{L}{k} + \frac{1}{h_{cf}}= \frac{1}{2850}+ \frac{0.012}{50} + \frac{1}{10} \\ = 0.0003508 + 0.00024 + 0.1 = 0.1006 \\ U= \frac{1}{0.1006}=9.94 \;W/m^2°C \\ q= 9.94 \times 1 \times (95-15) = 795.2 \;W/m^2$

(ii) Temperature of the outside surface of the tank, t₂ :

We know that

$q=h_{cf} \times 1 \times (t_2-t_{cf}) \\ 975.2=10(t_2-15) \\ t_2=\frac{795.2}{10}+15 = 94.52 \;°C$