Chapter 19, Problem 123RQ

(a)

To determine

The rate of heat transfer to the iced water in the tank.

Explanation of Solution

Given:

The temperature $\left(T_{s,in}\right)$ of air is $0°\text{C}$.

The temperature $\left(T_{surr}\right)$ of tank located outdoor is $30°\text{C}$.

The velocity $\left(v_l\right)$ of air is $25\text{km}/\text{hr}$.

The internal diameter $\left(d_i\right)$ of tank is $3\text{m}$.

The thickness $\left(t\right)$ of the tank is $1\text{cm}$.

The thermal conductivity of tank is $15\text{W}/\text{m}\cdot \text{K}$.

The surrounding surface temperature $\left(T_s\right)$ from tank is $25\text{°C}$.

The emissivity $\left(\epsilon \right)$ of the tank is $0.75$.

The heat of fusion $\left(h_{if}\right)$ of water is $333.7\text{kJ}/\text{kg}$.

The Stefan Boltzmann constant $\left(\sigma \right)$ is $5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

Calculation:

Refer to the table A-22 “properties of air at 1 atm pressure”.

Obtain properties of the air corresponding to the free stream temperature of $30°\text{C}$ as follows:

$k=0.02588\text{W}/\text{m}\cdot \text{K}$

$\mathrm{Pr}=0.7282$

$v=1.608\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$

${\mu }_{\infty }=1.872\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}$

${\mu }_s=1.729\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}$

Calculate the outer diameter by using the relation.

  $\begin{array}{c}D=d_i+\left(2\times t\right)\\ =3\text{m}+\left(2\times \left(1\text{cm}\times \frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ =3.02\text{m}\end{array}$

Calculate the Reynolds number by using the relation.

  $\begin{array}{c}\mathrm{Re}=\frac{v_{{}_l}D}{v}\\ =\frac{\left(25\text{km}/\text{hr}\times \frac{\frac{1000}{3600}\text{m}/\text{s}}{\text{1}\text{km}/\text{hr}}\right)\times 3.02\text{m}}{1.608\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =1.304\times {10}^6\end{array}$

Calculate the heat transfer coefficient by using the relation.

  $\begin{array}{c}h=\frac{k}{D}\times \left(\left(2\right)+\left(0.4{\left(\mathrm{Re}\right)}^{0.5}+0.06{\mathrm{Re}}^{\frac{2}{3}}\right){\mathrm{Pr}}^{0.4}{\left(\frac{{\mu }_{\infty }}{{\mu }_s}\right)}^{\frac{1}{4}}\right)\\ h=\left[\frac{0.02588\text{W}/\text{m}\cdot \text{K}}{3.02\text{m}}\times \left(\begin{array}{l}\left(2\right)+\left(\begin{array}{l}0.4{\left(1.304\times {10}^6\right)}^{0.5}\\ +0.06{\left(1.304\times {10}^6\right)}^{\frac{2}{3}}\end{array}\right)\times \\ {0.7282}^{0.4}{\left(\frac{1.872\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}}{1.729\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}}\right)}^{\frac{1}{4}}\end{array}\right)\right]\\ =9.05{\text{W}/\text{m}}^2\cdot \text{K}\end{array}$

Calculate the radius of sphere by using the equation.

  $\begin{array}{c}{R}_{sphere}=\frac{\frac{D}{2}-\frac{d_1}{2}}{4\pi k{r}_1{r}_2}\\ =\frac{\frac{3.02\text{m}}{2}-\frac{3\text{m}}{2}}{4\times \pi \times 0.02588\text{W}/\text{m}\cdot \text{K}\times \frac{3.02\text{m}}{2}\times \frac{3\text{m}}{2}}\\ =2.342\times {10}^{-5}\text{K}/\text{W}\end{array}$

Calculate the heat transfer through the tank by conduction by using the relation

  $\begin{array}{c}{\dot{Q}}_{throgh\tan k}={\dot{Q}}_{cond+rad}\\ \left(\frac{T_{s,out}-T_{s,in}}{R_{sphere}}\right)=h\pi D^2\left(T_{surr}-T_{s,out}\right)+\epsilon \pi D^2\sigma \left(T_s^4-T_{s,out}^4\right)\\ \left(\frac{T_{s,out}-\left(0°\text{C}+273\right)\text{K}}{2.342\times {10}^{-5}\text{K}/\text{W}}\right)=\left[\begin{array}{l}9.05\text{W}/\text{m}\cdot \text{K}\times \pi \times {\left(3.02\text{m}\right)}^2\left(\left(30°\text{C}+273\right)\text{K}-T_{s,out}\right)\\ +0.75\times \pi {\left(3.02\text{m}\right)}^2\times 5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^4\left(\begin{array}{l}{\left(\left(25°\text{C}+273\right)\text{K}\right)}^4-\\ {\left(\left(T_{s,ou}{}_t+273\right)\text{K}\right)}^4\end{array}\right)\end{array}\right]\\ T_{s,out}=\left(273.25\text{K}-273\right)°\text{C}\\ \text{=0}\text{.25}°\text{C}\end{array}$

Calculate the rate of heat transfer by using the relation.

  $\begin{array}{c}\dot{Q}=\frac{T_{s,out}-T_{s,in}}{R_{sphere}}\\ =\frac{\left(0.25°\text{C}+273\right)\text{K}-\left(0°\text{C}+273\right)\text{K}}{2.342\times {10}^{-5}\text{K}/\text{W}}\\ =10674.6\text{W}\end{array}$

Thus, the rate of heat transfer is $10674.6\text{W}$.

(b)

To determine

The amount of ice that melts during $24\text{hr}$.

Explanation of Solution

Calculation:

Calculate the amount of heat transfer during $24\text{hr}$ by using the relation.

  $\begin{array}{c}Q=\dot{Q}\Delta t\\ =\left(10674.6\text{J}/\text{s}\times \frac{\frac{1}{1000}\text{kJ}}{1\text{J}}\right)\times \left(24\text{hr}\times \frac{3600\text{s}}{1\text{hr}}\right)\\ =909880\text{kJ}\end{array}$

Calculate the amount of ice that melts during $24\text{hr}$ by using the relation.

  $\begin{array}{c}\dot{m}=\frac{Q}{h_{if}}\\ =\frac{909880\text{kJ}}{333.7\text{kJ}/\text{kg}}\\ =2726.64\text{kg}\\ \approx \text{2727}\text{kg}\end{array}$

Thus the amount of ice that melts during $24\text{hr}$ is $\text{2727}\text{kg}$.