Question

(Chap 3- Heat conduction in cylinders and spheres) An 8-m-internal diameter spherical tank made of 1.5-cm-thick stainless steel (k = 15W/m-K) is used to store iced water at 0°C. The tank is located in a room whose temperature is 25°C. The walls of the room are also at 25°C. The outer surface of the tank is black (emissivity1), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are 80W/m2-K and 10W/m2-K, respectively. Determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0 C.that melts during a 24-h period. The heat of fusion of water at atmospheric pressure is hy 333.7 kJ/kg. 2-2 Ied water (Hint: Start with the assumption that the temperature of outer surface is at 5°C and check with such assumption in the last step) (25pt)

Answer 1

Assumptions:

1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time.

2 Heat transfer is one-dimensional since there is thermal symmetry about the midpoint.

3 Thermal conductivity is constant

Properties;

The thermal conductivity of steel is given to be k = 15 W/m⋅°C. The heat of fusion of water at 1 atm is h_if= 333.7 kJ/kg. The outer surface of the tank is black and thus its emissivity $\varepsilon =1$

(a) The inner and the outer surface areas of sphere are  

$A_{i}=\pi D_{i}^{2}=\pi \times 8^{2}=201.06m^{2}$

$A_{0}=\pi D_{0}^{2}=\pi \times 8.03^{2}=202.57m^{2}$

here,

We assume the outer surface temperature T₂ to be 5°C after comparing convection heat transfer coefficients at the inner and the outer surfaces of the tank. With this assumption, the radiation heat transfer coefficient can be determined from

S247 7

$h_{rad}=1(5.67\times 10^{-8}) ((273+5)^{2}+(273+25)^{2}))(273+5)$

The individual thermal resistances are

$R_{conv,i}=\frac{1}{h_{i}A}=\frac{1}{80\times 201.06}=0.000062^{o}C/W$

$R_{1}=R_{sphere}=\frac{r_{2}-r_{1}}{4\pi Kr_{1}r_{2}}=\frac{(4.015-4)}{4\pi \times 15\times 4.015\times 4}=0.000005^{o}C/W$

$R_{conv,o}=\frac{1}{h_{o}A}=\frac{1}{10\times 202.57}=0.000494^{o}C/W$

$R_{rad,o}=\frac{1}{h_{rad}A}=\frac{1}{5.424\times 202.57}=0.000910^{o}C/W$

$\frac{1}{R_{eqv}}=\frac{1}{R_{conv,0}}+\frac{1}{R_{rad}}=\frac{1}{0.000494}+\frac{1}{0.000910}\rightarrow R_{eqv}=0.000320^{o}C/W$

$R_{total}=R_{conv,i}+R_{1}+R_{eqv}=0.000062+0.000005+0.000320=0.000387^{o}C/W$

Then the steady rate of heat transfer to the iced water becomes

$\dot{Q}=\frac{T_{\infty ,1}-T_{\infty ,2}}{R_{total}}=\frac{25-0}{0.000387}=\textbf{64,600W}$

b) The total amount of heat transfer during a 24-hour period and the amount of ice that will melt during this period are

$Q=\dot{Q\Delta t}=(64,600)(24\times 3600)=5.581 \times 10^{6}KJ$

$m_{ice}=\frac{Q}{h_{if}}=\frac{5.581\times 10^{6}}{333.7}=\textbf{16,730Kg}$

Check: The outer surface temperature of the tank is

$\dot{Q}=h_{conv+rad}A_{o}(T_{\infty 1}-T_{s})$

$T_{s}=T_{\infty 1}-\frac{\dot{Q}}{h_{conv+rad}A_{o}}$

$T_{s}=25-\frac{64600}{10+5.424\times (202.57)}=4.3^{o}C$

which is very close to the assumed temperature of 5°C for the outer surface temperature used in the evaluation of the radiation heat transfer coefficient. Therefore, there is no need to repeat the calculations