Chapter 19, Problem 82P

To determine

The tube surface temperature necessary to heat water.

The tube surface temperature necessary to heat engine oil.

The tube surface temperature necessary to heat liquid mercury.

Explanation of Solution

Given:

The diameter of tube is $\left(D\right)$ is $25\text{mm}$.

The length of tube is $\left(L\right)$ is $15\text{m}$.

The inlet temperature $\left(T_i\right)$ of fluid is $50{}^{\text{o}}\text{C}$.

The outlet temperature $\left(T_e\right)$ of fluid is $150{}^{\text{o}}\text{C}$.

The mass flow rate $\left(\dot{m}\right)$ is $0.01\text{kg}/\text{s}$.

Calculation:

Calculate the average temperature.

    $\begin{array}{c}{T}_{avg}=\frac{T_i+T_e}{2}\\ =\frac{50°\text{C}+150°\text{C}}{2}\\ =100°\text{C}\end{array}$

For water:

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

Obtain the following properties of water corresponding to the temperature of $100°\text{C}$ as follows:

$\rho =957.9\text{kg}/{\text{m}}^3$

$k=0.679\text{W}/\text{m}\cdot \text{K}$

$\mathrm{Pr}=1.75$

$\mu =0.282\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$

Calculate the Reynolds number for water.

    $\begin{array}{c}\mathrm{Re}=\frac{4\dot{m}}{\pi D\mu }\\ =\frac{4\times 0.01\text{kg}/\text{s}}{\pi \left(25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\left(0.282\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =1806.01\end{array}$

The value of Reynolds number is less than $2000$. Therefore the flow is laminar flow.

Calculate the hydrodynamic entry length for water.

    $\begin{array}{c}{L}_H=0.05\mathrm{Re}D\\ =0.05\times 1806.01\times 25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\\ =0.05\times 1806.01\times 0.025\text{m}\\ \text{=2}\text{.2575}\text{m}\end{array}$

Calculate the thermal entry length for water.

    $\begin{array}{c}{L}_T=\text{Pr}\times L_H\\ =1.75\times \text{2}\text{.2575}\text{m}\\ =3.9506\text{m}\end{array}$

The value of hydrodynamic entry length and thermal entry length is less than $15\text{m}$. Therefore the flow is fully developed flow.

The value of Nusselt number for fully developed laminar flow is $3.66$.

    $Nu=3.66$

Calculate the heat transfer coefficient.

    $\begin{array}{c}h=Nu\left(\frac{k}{D}\right)\\ =3.66\left(\frac{0.679\text{W}/\text{m}\cdot \text{K}}{25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}}\right)\\ =99.41\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate the surface area of tube.

    $\begin{array}{c}{A}_s=\pi DL\\ =\pi \left(25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\left(15\text{m}\right)\\ =1.178{\text{m}}^2\end{array}$

Calculate the exit temperature of fluid.

    $\begin{array}{c}{T}_s=\frac{T_e-T_i\text{exp}\left[\frac{-h{A}_s}{\dot{m}{c}_p}\right]}{1-\text{exp}\left[\frac{-h{A}_s}{\dot{m}{c}_p}\right]}\\ =\frac{150°\text{C}-\left(50°\text{C}\right)\text{exp}\left[\frac{-\left(99.41\text{W}/{\text{m}}^2°\text{C}\right)\times \left(1.178{\text{m}}^2\right)}{\left(0.01\text{kg}/\text{s}\right)\left(4217\text{J}/\text{kg}\cdot \text{K}\right)}\right]}{1-\text{exp}\left[\frac{-\left(99.41\text{W}/{\text{m}}^2°\text{C}\right)\times \left(1.178{\text{m}}^2\right)}{\left(0.01\text{kg}/\text{s}\right)\left(4217\text{J}/\text{kg}\cdot \text{K}\right)}\right]}\\ =\frac{150°\text{C}-\left(50°\text{C}\right)\text{exp}\left[-2.7769\right]}{1-\text{exp}\left[-2.7769\right]}\\ =156.6°\text{C}\end{array}$

Thus, the tube surface temperature is $156.6°\text{C}$.

For engine oil:

Refer Table A-19 “Properties of liquids”.

Obtain the following properties of the engine oil corresponding to the temperature of $100°\text{C}$ as follows:

$k=0.1367\text{W}/\text{m}\cdot \text{K}$

$c_p=2220\text{J}/\text{kg}\cdot \text{K}$

$\mathrm{Pr}=279.1$

$\mu =0.01718\text{kg}/\text{m}\cdot \text{s}$

Calculate the Reynolds number for engine oil.

    $\begin{array}{c}\mathrm{Re}=\frac{4\dot{m}}{\pi D\mu }\\ =\frac{4\times 0.01\text{kg}/\text{s}}{\pi \left(25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\left(0.01718\text{kg}/\text{m}\cdot \text{s}\right)}\\ =29.64\end{array}$

The value of Reynolds number is less than $2000$. Therefore the flow is laminar flow.

Calculate the hydrodynamic entry length for engine oil.

    $\begin{array}{c}{L}_H=0.05\mathrm{Re}D\\ =0.05\times 29.64\times 25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\\ =\text{0}\text{.037}\text{m}\end{array}$

Calculate the thermal entry length for engine oil.

    $\begin{array}{c}{L}_T=\text{Pr}\times L_H\\ =279.1\times \text{0}\text{.037}\text{m}\\ =10.34\text{m}\end{array}$

Calculate the heat transfer coefficient for engine oil.

    $\begin{array}{c}h=Nu\left(\frac{k}{D}\right)\\ =3.66\left(\frac{0.1367\text{W}/\text{m}\cdot \text{k}}{25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}}\right)\\ =20.01\text{W}/{\text{m}}^2°\text{C}\end{array}$

Calculate the exit temperature of fluid.

    $\begin{array}{c}{T}_s=\frac{T_e-T_i\text{exp}\left[\frac{-h{A}_s}{\dot{m}{c}_p}\right]}{1-\text{exp}\left[\frac{-h{A}_s}{\dot{m}{c}_p}\right]}\\ =\frac{150°\text{C}-\left(50°\text{C}\right)\text{exp}\left[\frac{-\left(20.01\text{W}/{\text{m}}^2°\text{C}\right)\times \left(1.178{\text{m}}^2\right)}{\left(0.01\text{kg}/\text{s}\right)\left(2220\text{J}/\text{kg}\cdot \text{K}\right)}\right]}{1-\text{exp}\left[\frac{-\left(20.01\text{W}/{\text{m}}^2°\text{C}\right)\times \left(1.178{\text{m}}^2\right)}{\left(0.01\text{kg}/\text{s}\right)\left(2220\text{J}/\text{kg}\cdot \text{K}\right)}\right]}\\ =\frac{150°\text{C}-\left(50°\text{C}\right)\text{exp}\left[-1.06126\right]}{1-\text{exp}\left[-1.06126\right]}\\ =202.9°\text{C}\end{array}$

Thus, the tube surface temperature is $202.9°\text{C}$.

For liquid mercury:

Refer Table A-20 "Properties of liquid metals”.

Obtain the following properties of liquid mercury corresponding to the temperature of $100°\text{C}$ as follows:

$k=9.467\text{W}/\text{m}\cdot \text{K}$

$c_p=137.1\text{J}/\text{kg}\cdot \text{K}$

$\mathrm{Pr}=0.018$

$\mu =1.245\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$

Calculate the Reynolds number for liquid mercury.

    $\begin{array}{c}\mathrm{Re}=\frac{4\dot{m}}{\pi D\mu }\\ =\frac{4\times 0.01\text{kg}/\text{s}}{\pi \left(25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\left(1.245\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =409.07\end{array}$

The value of Reynolds number is less than $2000$. Therefore the flow is laminar flow.

Calculate the hydrodynamic entry length for liquid mercury.

    $\begin{array}{c}{L}_H=0.05\mathrm{Re}D\\ =0.05\times 409.07\times 25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}\\ =0.05\times 1806.01\times 0.025\text{m}\\ =\text{0}\text{.511}\text{m}\end{array}$

Calculate the thermal entry length for liquid mercury.

    $\begin{array}{c}{L}_T=\text{Pr}\times L_H\\ =0.018\times \text{0}\text{.511}\text{m}\\ =0.0092\text{m}\end{array}$

Calculate the heat transfer coefficient for liquid mercury.

    $\begin{array}{c}h=Nu\left(\frac{k}{D}\right)\\ =3.66\left(\frac{9.467\text{W}/\text{m}\cdot \text{k}}{25\text{mm}\frac{{10}^{-3}\text{m}}{1\text{mm}}}\right)\\ =1385.98\text{W}/{\text{m}}^2°\text{C}\end{array}$

Calculate the exit temperature of fluid.

    $\begin{array}{c}{T}_s=\frac{T_e-T_i\text{exp}\left[\frac{-h{A}_s}{\dot{m}{c}_p}\right]}{1-\text{exp}\left[\frac{-h{A}_s}{\dot{m}{c}_p}\right]}\\ =\frac{150°\text{C}-\left(50°\text{C}\right)\text{exp}\left[\frac{-\left(1385.98\text{W}/{\text{m}}^2°\text{C}\right)\times \left(1.178{\text{m}}^2\right)}{\left(0.01\text{kg}/\text{s}\right)\left(137.1\text{J}/\text{kg}\cdot \text{K}\right)}\right]}{1-\text{exp}\left[\frac{-\left(1385.98\text{W}/{\text{m}}^2°\text{C}\right)\times \left(1.178{\text{m}}^2\right)}{\left(0.01\text{kg}/\text{s}\right)\left(137.1\text{J}/\text{kg}\cdot \text{K}\right)}\right]}\\ =\frac{150°\text{C}-\left(50°\text{C}\right)\text{exp}\left[-1190.87\right]}{1-\text{exp}\left[-1190.87\right]}\\ =150°\text{C}\end{array}$

Thus, the tube surface temperature is $150°\text{C}$.