Chapter 20, Problem 100RQ

(a)

To determine

The rate of heat transfer in steady operation for the water in the tank is still.

Explanation of Solution

Given:

The outer diameter of the spherical tank $\left(d_o\right)$ is $30.0\text{cm}$.

The surface temperature of vessel $\left(T_s\right)$ is $30°\text{C}$.

The outside temperature of vessel $\left(T_{\infty }\right)$ is $20°\text{C}$.

The circulation velocity of water in tank $\left(V\right)$ is $20\text{cm}/\text{s}$.

Calculation:

Calculate the film temperature $\left(T_f\right)$ using the relation.

    $\begin{array}{c}{T}_f=\frac{T_s+T_{\infty }}{2}\\ =\frac{30°\text{C}+20°\text{C}}{2}\\ =25°\text{C}\end{array}$

Refer Table A-15 “Properties of saturated water”.

Obtain the following values of different properties corresponding film temperature $25°\text{C}$ to the as follows:

  $\begin{array}{l}\rho =997\text{kg}/{\text{m}}^{\text{3}}\\ \mu =0.891\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\\ k=0.607\text{W}/\text{m}\cdot \text{K}\\ \mathrm{Pr}=6.14\end{array}$

  $\begin{array}{l}v=8.937\times {10}^{-7}{\text{m}}^{\text{2}}/\text{s}\\ \beta =0.247\times {10}^{-3}{\text{K}}^{-1}\end{array}$

Calculate the characteristics length $\left(L_c\right)$ is using the relation.

    $\begin{array}{c}{L}_c=d_o\\ =30.0\text{cm}\end{array}$

Calculate the Reynolds number $\left(\text{Ra}\right)$ using the relation.

    $\begin{array}{c}\text{Ra}=\frac{g\beta \left(T_s-T_{\infty }\right){\left(d_o\right)}^3}{v^2}\mathrm{Pr}\\ =\frac{\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.247\times {10}^{-3}{\text{K}}^{-1}\right)\left(\begin{array}{l}30°\text{C}\\ -20°\text{C}\end{array}\right){\left(30.0\text{cm}\right)}^3}{{\left(8.937\times {10}^{-7}{\text{m}}^{\text{2}}/\text{s}\right)}^2}\left(6.14\right)\\ =\frac{\left(2.419\times {10}^{-3}\text{m}/{\text{s}}^2\cdot \text{K}\right)\left[\begin{array}{l}\left(30°\text{C}+\text{273}\right)\text{K}\\ -\left(20°\text{C}+\text{273}\right)\text{K}\end{array}\right]{\left(30\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^3}{{\left(8.94\times {10}^{-7}{\text{m}}^2/\text{s}\right)}^2}\left(6.14\right)\\ =5.029\times {10}^9\end{array}$

Calculate the Nusselt number $\left(\text{Nu}\right)$ using the relation.

    $\begin{array}{c}\text{Nu}=2+\frac{0.589{\left(\text{Ra}\right)}^{1/4}}{{\left[1+{\left(0.467/\mathrm{Pr}\right)}^{9/16}\right]}^{4/9}}\\ =2+\frac{0.589{\left(5.029\times {10}^9\right)}^{1/4}}{{\left[1+{\left(0.467/6.14\right)}^{9/16}\right]}^{4/9}}\\ =2+\frac{156.85}{1.098}\\ =144.8\end{array}$

Calculate the area of heat transfer $\left(A\right)$ using the relation.

    $\begin{array}{c}A=\pi d_o^2\\ =\pi {\left(30.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\\ =0.2827{\text{m}}^2\end{array}$

Calculate the heat transfer coefficient of convection $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{d_o}Nu\\ =\frac{0.607\text{W}/\text{m}\cdot \text{K}}{30.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}\left(144.8\right)\\ =293\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the heat transfer by convection $\left({\dot{Q}}_{cv}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{cv}=hA\left(T_s-T_{\infty }\right)\\ =\left(293\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.2827{\text{m}}^{\text{2}}\right)\left[\left(30°\text{C}+\text{273}\right)\text{K}-\left(20°\text{C}+273\right)\text{K}\right]\\ =828.3\text{W}\end{array}$

Thus, the of heat transfer in steady operation for the water in the tank is still is $828.3\text{W}$.

(b)

To determine

The rate of heat transfer in steady operation for the water in the tank is still and the buoyancy force caused by the difference in water density is assumed to be negligible.

Explanation of Solution

Calculation:

If the buoyancy force becomes zero, the convective current becomes negligible means $\left(\beta =0\right)$ so the heat transfer will be from conduction. For this condition the Rayleigh number is $\left(\text{Ra}=0\right)$ and Nusselt number is $\left(\text{Nu}=2\right)$.

Calculate the convective heat transfer $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{d_o}\left(\text{Nu}\right)\\ =\frac{0.607\text{W}/\text{m}\cdot \text{K}}{30.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}\left(2\right)\\ =4.05\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the heat transfer by conduction $\left(Q_{cond}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{cond}=hA\left(T_s-T_{\infty }\right)\\ =\left(4.05\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.2827{\text{m}}^2\right)\left[\left(30°\text{C}+\text{273}\right)\text{K}-\left(20°\text{C}+273\right)\text{K}\right]\\ =11.4\text{W}\end{array}$

Thus, the rate of heat transfer in steady operation for the water in the tank is still and the buoyancy force caused by the difference in water density is assumed to be negligible is $11.4\text{W}$.

(c)

To determine

The rate of heat transfer in steady operation for the water in the tank is circulated at an average velocity of $20\text{cm}/\text{s}$.

Explanation of Solution

Calculation:

This condition is forced convection.

Refer Table A-15 “Properties of saturated water”.

Obtain the following values of different properties corresponding temperature $20°\text{C}$ to the as follows:

  $\begin{array}{l}\rho =998\text{kg}/{\text{m}}^{\text{3}}\\ {\mu }_s{}_{\left(30°\text{C}\right)}=0.798\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\\ {\mu }_{\infty }=1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\\ k=0.598\text{W}/\text{m}\cdot \text{K}\end{array}$

  $\begin{array}{l}\mathrm{Pr}=7.01\\ v=1.004\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}\end{array}$

Calculate the Reynolds number $\left(\mathrm{Re}\right)$ using the relation.

    $\begin{array}{c}\mathrm{Re}=\frac{V{d}_o}{v}\\ =\frac{\left(0.2\text{m}/\text{s}\right)\left(30.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{\left(1.004\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =\frac{\left(0.06{\text{m}}^{\text{2}}/\text{s}\right)}{\left(1.004\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =59760\end{array}$

Calculate the Nusselt number $\left(\text{Nu}\right)$ using the relation.

    $\begin{array}{c}\text{Nu}=2+\left[0.4{\left(\mathrm{Re}\right)}^{0.5}+0.06{\left(\mathrm{Re}\right)}^{2/3}\right]{\mathrm{Pr}}^{0.4}{\left(\frac{{\mu }_{\infty }}{{\mu }_{s\left(30°\text{C}\right)}}\right)}^{1/4}\\ =2+\left[0.4{\left(58760\right)}^{0.5}+0.06{\left(59760\right)}^{2/3}\right]{\left(7.01\right)}^{0.4}{\left(\frac{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}{0.798\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\right)}^{1/4}\\ =2+\left[0.4{\left(58760\right)}^{0.5}+0.06{\left(58760\right)}^{2/3}\right]\left(1.255\right)\\ =439.1\end{array}$

Calculate the convective heat transfer $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{d_o}\left(\text{Nu}\right)\\ =\frac{0.598\text{W}/\text{m}\cdot \text{K}}{30.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}\left(439.1\right)\\ =\frac{0.598\text{W}/\text{m}\cdot \text{K}}{0.3\text{m}}\left(439\right)\\ =875.3\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the heat transfer by forced convection $\left(Q_{fcon}\right)$ using the relation.

    $\begin{array}{c}{\dot{Q}}_{fcon}=hA\left(T_s-T_{\infty }\right)\\ =\left(875.3\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(0.2827{\text{m}}^2\right)\left[\left(30°\text{C}+\text{273}\right)\text{K}-\left(20°\text{C}+273\right)\text{K}\right]\\ =2474\text{W}\end{array}$

Thus, the rate of heat transfer in steady operation for the water in the tank is circulated at an average velocity of $20\text{cm}/\text{s}$ is $2474\text{W}$.