Chapter 19, Problem 128RQ

(a)

To determine

The tube length using the appropriate Nusselt number relation for liquid metals.

Explanation of Solution

Given:

The flow rate of mercury $\left(\dot{m}\right)$ is $0.6\text{kg}/\text{s}$.

The diameter $\left(D\right)$ is $5\text{cm}$.

The inlet mean temperature $\left(T_i\right)$ is $100°\text{C}$.

The outlet mean temperature $\left(T_o\right)$ is $200°\text{C}$.

The maintained surface temperature of the tube surface $\left(T_s\right)$ is $250°\text{C}$.

Calculation:

Calculate $T_f$.

  $\begin{array}{c}{T}_f=\frac{100°\text{C}+200°\text{C}}{2}\\ =150°\text{C}\end{array}$

Refer to table A-20 “Properties of liquid metals” to obtain the following properties corresponding to the temperature of $150\text{°C}$:

  $\begin{array}{l}{c}_p=136.1\text{J}/\text{kg}\cdot \text{K}\\ k=10.0778\text{W}/\text{m}\cdot \text{K}\\ \mu =1.126\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\\ \mathrm{Pr}=0.0152\end{array}$

And the Prandtl number corresponding to the surface temperature of $250°\text{C}$ is ${\left(\mathrm{Pr}\right)}_s=0.0119$.

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

    $\begin{array}{c}\mathrm{Re}=\frac{4\dot{m}}{\pi D\mu }\\ =\frac{4\left(0.6\text{kg}/\text{s}\right)}{\pi \left(5\text{cm}\right)\left(1.126\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{4\left(0.6\text{kg}/\text{s}\right)}{\pi \left[\left(5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right]\left(1.126\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =13570\end{array}$

Calculate the Nusselt number $\left(Nu\right)$ by using the relation.

    $\begin{array}{c}Nu=4.8+0.0156{\mathrm{Re}}^{0.85}{\left(\mathrm{Pr}\right)}_s{}^{0.93}\\ =4.8+0.0156{\left(13570\right)}^{0.85}{\left(0.0119\right)}^{0.93}\\ =5.624\end{array}$

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

    $\begin{array}{c}h=\frac{kNu}{D}\\ =\frac{\left(10.0778\text{W}/\text{m}\cdot \text{K}\right)\left(5.624\right)}{5\text{cm}}\\ =\frac{\left(10.0778\text{W}/\text{m}\cdot \text{K}\right)\left(5.624\right)}{\left(5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =1134\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the length of the tube $\left(L\right)$ by using the relation.

    $\begin{array}{c}{L}_1=-\frac{\dot{m}{c}_p}{\pi Dh}\ln \left(\frac{T_s-T_o}{T_s-T_i}\right)\\ =-\frac{\left(0.6\text{kg}/\text{s}\right)\left(136.1\text{J}/\text{kg}\cdot \text{K}\right)}{\pi \left(5\text{cm}\right)\left(1134\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}\ln \left(\frac{250°\text{C}-100°\text{C}}{250°\text{C}-200°\text{C}}\right)\\ =-\frac{\left(0.6\text{kg}/\text{s}\right)\left(136.1\text{J}/\text{kg}\cdot \text{K}\right)}{\pi \left(5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(1134\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}\ln \left(\frac{250°\text{C}-100°\text{C}}{250°\text{C}-200°\text{C}}\right)\\ =0.504\text{m}\end{array}$

Thus, the tube length is $0.504\text{m}$.

(b)

To determine

The tube length using the appropriate Dittus–Boelter equation.

Explanation of Solution

Given:

The velocity of air $\left(V\right)$ is $25\text{m}/\text{s}$.

The mean temperature $\left(T\right)$ is $50°\text{C}$.

Calculation:

Calculate the Nusselt number $\left(Nu\right)$ by using the relation.

    $\begin{array}{c}Nu=0.023{\mathrm{Re}}^{0.8}{\left(\mathrm{Pr}\right)}^{0.4}\\ =0.023{\left(13570\right)}^{0.8}{\left(0.0152\right)}^{0.93}\\ =8.72\end{array}$

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

    $\begin{array}{c}h=\frac{kNu}{D}\\ =\frac{\left(10.0778\text{W}/\text{m}\cdot \text{K}\right)\left(8.72\right)}{5\text{cm}}\\ =\frac{\left(10.0778\text{W}/\text{m}\cdot \text{K}\right)\left(8.72\right)}{\left(5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =1758\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the length of the tube $\left(L\right)$ by using the relation.

    $\begin{array}{c}{L}_2=-\frac{\dot{m}{c}_p}{\pi Dh}\ln \left(\frac{T_s-T_o}{T_s-T_i}\right)\\ =-\frac{\left(0.6\text{kg}/\text{s}\right)\left(136.1\text{J}/\text{kg}\cdot \text{K}\right)}{\pi \left(5\text{cm}\right)\left(1758\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}\ln \left(\frac{250°\text{C}-100°\text{C}}{250°\text{C}-200°\text{C}}\right)\\ =-\frac{\left(0.6\text{kg}/\text{s}\right)\left(136.1\text{J}/\text{kg}\cdot \text{K}\right)}{\pi \left(5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(1758\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)}\ln \left(\frac{250°\text{C}-100°\text{C}}{250°\text{C}-200°\text{C}}\right)\\ =0.325\text{m}\end{array}$

Thus, the tube length is $0.325\text{m}$.

(c)

To determine

The comparison.

Explanation of Solution

Calculation:

Calculate the difference between the length $\left(\text{Difference}\right)$ using the relation.

    $\begin{array}{c}\text{Difference}=L_1-L_2\\ =0.504\text{m}-0.325\text{m}\\ =0.179\text{m}\end{array}$

Thus, the length of the pipe obtained in first case (part ‘a’) is $0.179\text{m}$ greater as compared to the length of the pipe obtained by second process.