Chapter 8, Problem 136P

Shell-and-tube heat exchangers with hundred of tubes housed in a shell are commonly used in practice for heat transfer between two fluids. Such a heat exchanger used in au active solar liot-watet system transfers heat froiui a water-atnifreeze solution flowing through the shell and the solar collector to fresh water flowing through the tubes at an average temperature of 64ƯC al a ¡‘ale of I S L’s. The heat exchangei’ contains 80 biass tubes I ciii in immer diameter and 1.5 in in length. Disregaidiiig inkt. exit, and header losses. detennine tite pressure drop across a single itibe and the pmnping power required by the rube-side finid of the heat exchaiiger.
After operating a long time. I-tutu-thick scale builds up on the inner surfaces with an equivalent rougluiess of 0.4 mm.
For the saine pwnplng power input. detennine tIme percent reductiomi in the flow rate of water through the tubes.

To determine

The pressure drop across a single tube.

The plumbing power required by the tube side fluid of the heat exchanger.

The percent reduction in the flow rate of water through the tubes.

Answer to Problem 136P

The pressure drop across a single tube is $8.93\text{kPa}$.

The plumbing power required by the tube side fluid of the heat exchanger is $0.133\text{kW}$.

The percent reduction in the flow rate of water through the tubes is $54\%$.

Explanation of Solution

Given information:

The average temperature of water flowing through the tubes is $60\text{°C}$, rate of flow of water is $15\text{L}/\text{s}$, the number of tube contained in the heat exchanger is $80$, inner diameter of tube is $1\text{cm}$ the length of tube is $1.5\text{m}$, and the buildup scale thickness on the inner surface is $1\text{mm}$, equivalence roughness is $0.4\text{mm}$.

Write the expression for the average velocity.

  $V=\frac{V_1}{A}$   ....... (I)

Here, volume flow rate of fluid is $V_1$, and cross-sectional area of tube is $A$.

Write the expression of total area of tubes.

  $A=n\left(\frac{\pi }{4}{d}^2\right)$   ....... (II)

Here, the number of tube is $n$ diameter of tube is $d$.

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{\rho Vd}{\mu }$   ....... (III)

Here, the density of fluid is $\rho$, the average velocity is $V$, diameter of tube is $d,$ and dynamic viscosity is $\mu$.

Write the expression for the Colebrook equation.

  $\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\left(\frac{\epsilon }{d}\right)}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$   ....... (IV)

Here, the friction factor is $f$, the roughness of tube is $\epsilon$, the diameter of tube is $d$, and the Reynold's number is $\mathrm{Re}$.

Write the expression for the pressure drop.

  $\Delta P=f\frac{L}{d}\frac{\rho V^2}{2}$   ....... (V)

Here, the friction factor is $f$, length of the tube is $L$, diameter of the tube is $d$, density of fluid is $\rho$ and velocity of flow is $V$.

Write the expression for the pumping power.

  $W_{pump}=V_1\Delta P$   ....... (VI)

Here, volume flow rate of fluid is $V_1$ and the pressure drop is $\Delta P$.

Write the expression of percentage reduction in flow rate.

  $P_{\text{reduction}}=\left(\frac{V_1-{V^{\prime }}_1}{V_1}\right)\times 100$   ....... (VII)

Here, the given flow rate is $V_1$ and the calculated flow rate is ${V^{\prime }}_1$.

Substitute $80$ for $n$ and $1\text{cm}$ for $d$ in Equation (II).

  $\begin{array}{c}A=80\left(\frac{\pi }{4}{\left(1\text{cm}\right)}^2\right)\\ =20\pi \times {\left(\left(1\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =20\pi \times 0.0001{\text{m}}^{\text{2}}\\ =0.00628{\text{m}}^2\end{array}$

Substitute $15\text{L}/\text{s}$ for $V_1$, $0.00628{\text{m}}^2$ for $A$ in Equation (I).

  $\begin{array}{c}V=\frac{\left(15\text{L}/\text{s}\right)}{\left(0.00628{\text{m}}^2\right)}\\ =\frac{\left(15\text{L}/\text{s}\right)\times \left(\frac{1{\text{m}}^{\text{3}}/\text{s}}{1000\text{L}/\text{s}}\right)}{\left(0.00628{\text{m}}^2\right)}\\ =\frac{0.015{\text{m}}^{\text{3}}/\text{s}}{0.00628{\text{m}}^2}\\ =2.38\text{m}/\text{s}\end{array}$

Refer table 3E "Properties of water" to find the dynamic viscosity of water corresponding to $60°$ temperature as $0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$.

Substitute, $983.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2.38\text{m}/\text{s}$ or $V$, $1\text{cm}$ for $d$ and $0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (III).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(2.38\text{m}/\text{s}\right)\times \left(1\text{cm}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(2.38\text{m}/\text{s}\right)\times \left(1\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =50112.5\\ \approx 50112\end{array}$

Refer Table 8-2 "Equivalence roughness values for new commercial pipes" to find the value of roughness is $1.5\times {10}^{-6}\text{m}$ corresponding to copper or brass pipes.

Substitute $1.5\times {10}^{-6}\text{m}$ for $\epsilon$, $1\text{cm}$ for $d$ and $50112$ for $\mathrm{Re}$ in Equation (IV).

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\left(\frac{1.5\times {10}^{-6}\text{m}}{1\text{cm}}\right)}{3.7}+\frac{2.51}{50112\sqrt{f}}\right)\\ =-2\log \left(\frac{\left(\frac{1.5\times {10}^{-6}\text{m}}{\left(1\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)}\right)}{3.7}+\frac{2.51}{50112\sqrt{f}}\right)\\ =-2\log \left(0.405\times {10}^{-4}+\frac{0.5\times {10}^{-4}}{\sqrt{f}}\right)\\ f=0.0214\end{array}$

Substitute $0.0214$ for $f$, $1.5\text{m}$ for $L$

  $983.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2.38\text{m}/\text{s}$ for $V$ and $1\text{cm}$ for $d$ in Equation (V).

  $\begin{array}{c}\Delta P=0.0214\frac{\left(1.5\text{m}\right)}{\left(1\text{cm}\right)}\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right){\left(2.38\text{m}/\text{s}\right)}^2}{2}\\ =0.0107\left(\frac{\left(1.5\text{m}\right)\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\left(5.6644{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}{\left(1\text{cm}\right)\times \left(\frac{1\text{ m}}{100\text{cm}}\right)}\right)\\ =\left(89.39\times 100\frac{\text{kg}}{\text{m}\cdot {\text{s}}^{\text{2}}}\right)\left(\frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ \approx 8.93\text{kPa}\end{array}$

Substitute $15\text{L}/\text{s}$ for $V_1$ and $8.93\text{kPa}$ for $\Delta P$ in Equation (VI).

  $\begin{array}{c}{W}_{pump}=\left(15\text{L}/\text{s}\right)\times \left(8.93\text{kPa}\right)\\ =\left(15\text{L}/\text{s}\right)\times \left(8.93\text{kPa}\right)\left(\frac{\text{1}{\text{m}}^{\text{3}}/\text{s}}{\text{1000}\text{L}/\text{s}}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\\ =\left(133.95\text{Pa}\cdot {\text{m}}^{\text{3}}/\text{s}\right)\left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{1\text{Pa}}\right)\left(\frac{1\text{W}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}\right)\left(\frac{1\text{kW}}{1000\text{W}}\right)\\ \approx 0.133\text{kW}\end{array}$

Substitute $80$ for $n$ and $0.8\text{cm}$ for $d$ in Equation (II).

  $\begin{array}{c}A=80\left(\frac{\pi }{4}{\left(0.8\text{cm}\right)}^2\right)\\ =20\pi \times {\left(\left(0.8\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =20\pi \times 0.000064{\text{m}}^{\text{2}}\\ =0.004021{\text{m}}^2\end{array}$

Substitute $V^{\prime }$ for $V_1$ and $0.004021{\text{m}}^2$ for $A$ in Equation (I).

  $\begin{array}{c}V=\frac{V^{\prime }}{0.004021{\text{m}}^2}\\ =\frac{248.68\times V^{\prime }}{{\text{m}}^2}\\ =248.68\times V^{\prime }/{\text{m}}^2\end{array}$   ....... (VIII)

Substitute, $983.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.8\text{cm}$ for $d$ and $0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (III).

  $\mathrm{Re}=\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\times V\times \left(0.8\text{cm}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}$   ....... (IX)

Substitute $0.0004\text{m}$ for $\epsilon$, $0.8\text{cm}$ for $d$ in Equation (IV).

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\left(\frac{0.0004\text{m}}{0.8\text{cm}}\right)}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ =-2\log \left(\frac{\left(\frac{0.0004\text{m}}{\left(0.8\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)}\right)}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ =-2\log \left(\frac{0.05}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ =-2\log \left(0.014+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\end{array}$   ....... (X)

Substitute $1.5\text{m}$ for $L$, $0.8\text{m}$ for $d$ and $983.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in Equation (V).

  $\begin{array}{c}\Delta P=f\frac{\left(1.5\text{m}\right)}{\left(0.8\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)V^2}{2}\\ =f\times 93.75\times \left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\times V^2\\ =\left(92184.37\times f\times V^2\text{kg}/{\text{m}}^{\text{3}}\right)\left(\frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =92.1844\times f\times V^2\text{kPa}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}\end{array}$   ....... (XI)

Substitute $0.133\text{kW}$ for $W_{\text{pump}}$ in Equation (VI).

  $\begin{array}{l}0.133\text{kW}=\left(V^{\prime }\right)\times \Delta P\\ \left(0.133\text{kW}\right)\left(\frac{1000\text{W}}{1\text{kW}}\right)\left(\frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)=\left(V^{\prime }\right)\times \Delta P\\ \left(133\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}\right)\left(\frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)=\left(V{V^{\prime }}_1{}^{\text{'}}\right)\times \Delta P\\ \Delta P=\left(\frac{0.133}{V^{\prime }}\frac{{\text{m}}^{\text{3}}}{\text{s}}\text{kPa}\right)\end{array}$..(XII)

There are five Equations (VIII), (IX), (X), (XI) and (XII) with five unknowns $V$, $V^{\prime }$, $f$, $\mathrm{Re}$ and $\Delta P$.

Use the trial and error method to find the values.

Assume $2\text{m}/\text{s}$ for $V$.

Substitute $2\text{m}/\text{s}$ for $V$ in Equation (VIII).

  $\begin{array}{l}2\text{m}/\text{s}=\frac{V^{\prime }}{0.004021{\text{m}}^2}\\ V^{\prime }=\left(2\text{m}/\text{s}\right)\times \left(0.004021{\text{m}}^2\right)\\ V^{\prime }=0.008042{\text{m}}^3/\text{s}\end{array}$

Substitute $2\text{m}/\text{s}$ for $V$ in Equation (IX).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(2\text{m}/\text{s}\right)\times \left(0.8\text{cm}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(2\text{m}/\text{s}\right)\times \left(0.8\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{\left(15.73\text{kg}/\text{m}\cdot \text{s}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ \approx 33689\end{array}$

Substitute $33689$ for $\mathrm{Re}$ in Equation (X).

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2\log \left(0.014+\frac{2.51}{33689\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2\log \left(0.014+\frac{0.000074}{\sqrt{f}}\right)\\ f=0.0726\end{array}$

Substitute $2\text{m}/\text{s}$ for $V$ and $0.0726$ for $f$ in Equation (XI).

  $\begin{array}{c}\Delta P=92.1843\times 0.0726\times {\left(2\text{m}/\text{s}\right)}^2\text{kPa}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}\\ =\left(6.692\text{kPa}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}\right)\times \left(4{{\text{m}}^2/\text{s}}^2\right)\\ =26.77\text{kPa}\end{array}$

Substitute $26.77\text{kPa}$ for $\Delta P$, in Equation (XII).

  $\begin{array}{l}26.77\text{kPa}=\left(\frac{0.133}{V_1{}^{\text{'}}}\frac{{\text{m}}^{\text{3}}}{\text{s}}\text{kPa}\right)\\ V_1{}^{\text{'}}=\frac{0.133{\text{m}}^{\text{3}}/\text{s}}{26.77\text{kPa}}\text{kPa}\\ V_1{}^{\text{'}}=0.004968{\text{m}}^{\text{3}}/\text{s}\end{array}$

By substituting $2\text{m}/\text{s}$ for $V$, desired value is not obtained because the value of $V_1{}^{\text{'}}$ we get different from Equation (I) and Equation (II) so the value is not desirable.

Now use $1.714\text{m}/\text{s}$ for $V$.

Substitute $1.714\text{m}/\text{s}$ for $V$ in Equation (VIII).

  $\begin{array}{l}1.714\text{m}/\text{s}=\frac{V^{\prime }}{0.004021{\text{m}}^2}\\ V^{\prime }=\left(1.714\text{m}/\text{s}\right)\times \left(0.004021{\text{m}}^2\right)\\ V^{\prime }=0.00689{\text{m}}^3/\text{s}\end{array}$

Substitute $1.714\text{m}/\text{s}$ for $V$ in Equation (IX).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(1.714\text{m}/\text{s}\right)\times \left(0.8\text{cm}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(1.714\text{m}/\text{s}\right)\times \left(0.8\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{\left(13.483\text{kg}/\text{m}\cdot \text{s}\right)}{\left(0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ \approx 28871\end{array}$

Substitute $28871$ for $\mathrm{Re}$ in Equation (X).

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2\log \left(0.014+\frac{2.51}{28871\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2\log \left(0.014+\frac{0.0000869}{\sqrt{f}}\right)\\ f=0.0723\end{array}$

Substitute $1.714\text{m}/\text{s}$ for $V$ and $0.0723$ for $f$ in Equation (XI).

  $\begin{array}{c}\Delta P=92.1844\times 0.0723\times {\left(1.714\text{m}/\text{s}\right)}^2\text{kPa}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}\\ =\left(6.6649\text{kPa}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}\right)\times \left(2.9378{{\text{m}}^2/\text{s}}^2\right)\\ =19.58\text{kPa}\end{array}$

Substitute $19.58\text{kPa}$ for $\Delta P$, in Equation (XII).

  $\begin{array}{l}19.58\text{kPa}={V^{\prime }}_1\left(\frac{0.133}{V^{\prime }}\frac{{\text{m}}^{\text{3}}}{\text{s}}\text{kPa}\right)\\ {V^{\prime }}_1=\frac{0.133{\text{m}}^{\text{3}}/\text{s}}{19.58\text{kPa}}\text{kPa}\\ {V^{\prime }}_1=0.00689{\text{m}}^{\text{3}}/\text{s}\end{array}$

Now, By substituting $1.714$ for $V$ these gives similar value of ${V^{\prime }}_1$ from Equation (VIII) and Equation (XII).

The values of $V_1{}^{\text{'}}$ is $0.00689{\text{m}}^{\text{3}}/\text{s}$.

The value of $f$ is $0.0723$.

The value of $\mathrm{Re}$ is $28871$.

The value of $V$ is $1.714\text{m}/\text{s}$.

And the value of $\Delta P$ is $19.58\text{kPa}$.

Substitute $6.89\left({\text{m}}^{\text{3}}/\text{s}\right)$ for ${V^{\prime }}_1$ and $15{\text{m}}^{\text{3}}/\text{s}$ for $V_1$ in Equation (VII).

  $\begin{array}{c}{P}_{\text{reduction}}=\left(\frac{15{\text{m}}^{\text{3}}/\text{s}-6.89{\text{m}}^{\text{3}}/\text{s}}{15{\text{m}}^{\text{3}}/\text{s}}\right)\times 100\\ =\left(\frac{8.11{\text{m}}^{\text{3}}/\text{s}}{15{\text{m}}^{\text{3}}/\text{s}}\right)\times 100\\ =0.54\times 100\\ =54\%\end{array}$

Conclusion:

The pressure drop across a single tube is $8.93\text{kPa}$.

The plumbing power required by the tube side fluid of the heat exchanger is $0.133kW$.

The percent reduction in the flow rate of water through the tubes is $54\%$.