Chapter 19, Problem 132RQ

(a)

To determine

The length of the tube.

Explanation of Solution

Given:

The mass flow rate $\stackrel{\cdot }{\left(\text{m}\right)}$ is $1500\text{kg}/\text{hr}$.

The diameter $(D)$ of tube is $10\text{mm}$.

The initial temperature $\left(T_i\right)$ is $10°\text{C}$.

The final temperature $\left(T_e\right)$ is $40°\text{C}$.

The tube wall temperature $(T)$ is $49°\text{C}$.

Calculation:

Calculate the bulk mean fluid temperature $\left(T_m\right)$ using the relation.

    $\begin{array}{c}{T}_m=\left[\frac{T_e+T_i}{2}\right]\\ =\left[\frac{40°\text{C}+\text{10}°\text{C}}{2}\right]\\ =25°\text{C}\end{array}$

Refer table $\text{A-15}$ “Properties of saturated water”.

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

$\begin{array}{l}\rho =997\text{kg}/{\text{m}}^3\\ k=0.607\text{W}/\text{m}\cdot \text{K}\\ \mu =0.891\times {10}^{-3}{\text{m}}^2/\text{s}\\ c_p=4180\text{J}/\text{kg}\cdot \text{K}\\ \mathrm{Pr}=6.14\end{array}$

Calculate the rate of heat transfer $(\stackrel{.}{Q})$ using the relation.

    $\begin{array}{c}\dot{Q}=\dot{m}{c}_p(T_e-T_i)\\ =\left(1500\text{kg}/\text{h}\right)\left(4180\text{J}/\text{kg}\cdot \text{K}\right)\left(\left(40°\text{C}+273\right)\text{K}-\left(10°\text{C}+273\right)\text{K}\right)\\ =\left\{\left(1500\text{kg}/\text{h}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\right\}\left(4180\text{J}/\text{kg}\cdot \text{K}\right)\times 30\text{K}\\ =52250\text{W}\end{array}$

Calculate the velocity of water $(V)$ using the relation.

  $\begin{array}{c}V=\frac{\stackrel{.}{m}}{\rho A_c}\\ =\frac{\left(1500\times \frac{1}{60\times 60}\right)\text{kg}/\text{s}}{\left(997\text{kg}/{\text{m}}^3\right)\pi \left({\frac{\left(\text{10}\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}{4}}^2\right)}\\ =5.321\text{}\text{m}/\text{s}\end{array}$

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

    $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{\left(997\text{kg}/{\text{m}}^3\right)\left(5.321\text{m}/\text{s}\right)\left(10\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}{0.891\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =59540\end{array}$

The Reynolds number is greater than 10000 therefore flow is turbulent flow.

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

    $\begin{array}{c}\text{Nu}=0.023{\mathrm{Re}}^{0.8}{\mathrm{Pr}}^{0.4}\\ =0.023\times {\left(59540\right)}^{0.8}{\left(6.14\right)}^{0.4}\\ =313.9\end{array}$

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

    $\begin{array}{c}\text{Nu}=\frac{hD}{k}\\ h=\frac{k}{D}\text{Nu}\\ h=\frac{0.607\text{W}/\text{m}\cdot \text{K}}{\left(10\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}\left(313.9\right)\\ h=19054\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

Calculate the logarithmic mean temperature difference $\left(\Delta T_{\text{lm}}\right)$ using the relation.

  $\begin{array}{c}\left(\Delta T_{\text{lm}}\right)=\frac{T_i-T_e}{\text{ln}\left(\frac{T-T_e}{T-T_i}\right)}\\ =\frac{\left(\left(10°\text{C}+273\right)\text{K}-\left(40°\text{C}+273\right)\text{K}\right)}{\text{ln}\left(\frac{(49°\text{C})-(40°\text{C})}{\left(49°\text{C}\right)-\left(10°\text{C}\right)}\right)}\\ =20.46\text{K}\end{array}$

Calculate the surface area $\left(A_s\right)$ using the relation.

    $\begin{array}{c}\stackrel{.}{Q}=h{A}_s\Delta T_{\text{lm}}\\ 52250\text{W}=\left(19054\text{W}/{\text{m}}^2\cdot \text{K}\right)A_s\left(20.46\text{K}\right)\\ A_s=0.134{\text{m}}^2\end{array}$

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

    $\begin{array}{c}{A}_s=\pi DL\\ L=\frac{A_s}{\pi D}\\ =\frac{0.134{\text{m}}^2}{\pi \left(10\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}\\ =4.26\text{m}\end{array}$

Thus, the length of the tube is $4.26\text{m}$.

(b)

To determine

The outlet temperature of water.

Explanation of Solution

Given:

If the tube length is doubled.

Calculation:

Calculate the outlet temperature $\left(T_e\right)$ of water by using relation.

    $\begin{array}{c}\stackrel{.}{m}{c}_p(T_e-T_i)=h{A}_s\Delta T_{\text{lm}}\\ \stackrel{.}{m}{c}_p(T_e-T_i)=h{A}_s\frac{T_i-T_e}{\text{ln}\left(\frac{T_s-T_e}{T_s-T_i}\right)}\\ \left[\begin{array}{l}\left(1500\times \frac{1}{60\times 60}\right)\text{kg}/\text{s}\times \\ \left(4180\text{J}/\text{kg}\cdot \text{K}\right)(T_e-10°\text{C})\end{array}\right]=\left[\begin{array}{l}\left(19054\text{W}/{\text{m}}^2\text{K}\right)\left(2\times 0.134{\text{m}}^2\right)\times \\ \left[\frac{(10°\text{C}-T_e)}{\text{ln}\left(\frac{49°\text{C}-T_e}{49°\text{C}-10°\text{C}}\right)}\right]\end{array}\right]\\ 1741.66\text{J}/\text{s}\left(T_e-10\right)=5106.47\left[\frac{(10°\text{C}-T_e)}{\text{ln}\left(\frac{49°\text{C}-T_e}{39°\text{C}}\right)}\right]\end{array}$

Use trial error method to solve the above equation.

Assume $T_e=45°\text{C}$

$\begin{array}{c}1741.66\text{J}/\text{s}\left(45°\text{C}-10°\text{C}\right)=5106.47\left[\frac{(10°\text{C}-45°\text{C})}{\text{ln}\left(\frac{49°\text{C}-45°\text{C}}{39°\text{C}}\right)}\right]\\ 11.93\ne 15.36\end{array}$

Since, the left hand side and right hand is not equal therefore, this assumption is wrong.

Assume $T_e=46°\text{C}$.

    $\begin{array}{c}1741.66\text{J}/\text{s}\left(46°\text{C}-10°\text{C}\right)=5106.47\left[\frac{(10°\text{C}-46°\text{C})}{\text{ln}\left(\frac{49°\text{C}-46°\text{C}}{39°\text{C}}\right)}\right]\\ 12.27\ne 14.03\end{array}$

Since, the left hand side and right hand is not equal therefore, this assumption is wrong.

Assume $T_e=47°\text{C}$

    $\begin{array}{c}1741.66\text{J}/\text{s}\left(47°\text{C}-10°\text{C}\right)=5106.47\left[\frac{(10°\text{C}-47°\text{C})}{\text{ln}\left(\frac{49°\text{C}-47°\text{C}}{39°\text{C}}\right)}\right]\\ 12.61=12.46\end{array}$

Since, the left hand side and right hand is equal therefore, this assumption is correct.

Thus, the outlet temperature of the water is $47°\text{C}$.