Chapter 9, Problem 123P

Water flows down a long, straight, inclined pipe of diameter D and length L (Fig. 9-123). There is no forced pressure gradient between points 1 and 2; in other words, the water flows through the pipe by gravity alone, and $P_1=P_2=P_{\text{atm}}$ . The flow is steady, fully developed, and laminar. We adopt a coordinate system in which x follows the axis of the pipe, (a) Use the control volume technique of Chap. 8 to generate an expression for average velocity V as a function of the given parameters, $p,g,D,\Delta z,\mu$ and L (b) Use differential analysis to generate an expression for V as a function of the given parameters. Compare with your result of part (a) and discuss, (c) Use dimensional analysis to generate a dimensionless expression for V as a function of the given parameters. Construct a relationship between your II's that matches the exact analytical expression.

FIGURE P9-123

To determine

(a)

The expression for average velocity as the function of $\rho$, $g$, $D$, $\mu$ and $L$ using control volume method.

Answer to Problem 123P

The expression for average velocity is $V=\frac{\left(\rho g\Delta z\right)D^2}{32\mu L}$.

Explanation of Solution

Given Information:

The pressure at the points 1 and 2 is same and equal to the atmospheric pressure and the velocity of the flow is $V$.

Write the energy equation for the pipe.

  $\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+z_2+h_f$   ...... (I)

Here, the kinetic energy correction factor at point 1 is ${\alpha }_1$, the kinetic energy correction factor at point 2 is ${\alpha }_2$, the pressure at inlet is $P_1$, the pressure at outlet is $P_2$, the density of the water is $\rho$, the acceleration due to gravity is $g$, the velocity at inlet is $V_1$, the velocity at outlet is $V_2$, the head at inlet is $z_1$, the head at outlet is $z_2,$ and the frictional loss is $h_f$.

Calculation:

Substitute $P_{atm}$ for $P_1$, $P_{atm}$ for $P_2$, $V$ for $V_1$, $V$ for $V_2$, $\alpha$ for ${\alpha }_1$ and $\alpha$ for ${\alpha }_2$ in equation (I).

  $\begin{array}{l}\frac{P_{atm}}{\rho g}+\alpha \frac{V^2}{2g}+z_1=\frac{P_{atm}}{\rho g}+\alpha \frac{V^2}{2g}+z_2+h_f\\ z_1=z_2+h_f\\ \Delta z=h_f\end{array}$   ...... (II)

Write the expression to calculate the frictional loss.

  $h_f=f\frac{L{V}^2}{2Dg}$   ...... (III)

Here, the length of the pipe is $L$ and the diameter of the pipe is $D$.

Write the expression for friction factor in laminar flow.

  $f=\frac{64\mu }{\rho VD}$

Substitute $\frac{64\mu }{\rho VD}$ for $f$ in Equation (III).

  $\begin{array}{c}{h}_f=\frac{64\mu }{\rho VD}\times \frac{L{V}^2}{2Dg}\\ =\frac{32LV\mu }{\rho D^{2}g}\end{array}$

Substitute $\frac{32LV\mu }{\rho D^{2}g}$ for $h_f$ in Equation (II).

  $\begin{array}{l}\Delta z=\frac{32LV\mu }{\rho D^{2}g}\\ V=\frac{\rho D^{2}g}{32L\mu }\end{array}$

Conclusion:

Thus, the expression for average velocity is $V=\frac{\left(\rho g\Delta z\right)D^2}{32\mu L}$.

To determine

(b)

The expression for average velocity as the function of $\rho$, $g$, $D$, $\mu ,$ and $L$ using differential method.

Answer to Problem 123P

The expression for average velocity is $V_{avg}=\frac{\left(\rho g\Delta z\right)D^2}{32\mu L}$.

Explanation of Solution

Given Information:

The pressure at the points 1 and 2 is same and equal to the atmospheric pressure.

Write the force balance equation in the direction of the flow for the volume element shown in Figure-(1).

  ${\left(2\pi rdrP\right)}_x-{\left(2\pi rdrP\right)}_{x+dx}+{\left(2\pi rdx\tau \right)}_r-{\left(2\pi rdx\tau \right)}_{r+dr}-W_x=0$   ...... (IV)

Here, the fluid weight in the direction of the flow is $W_x$, the radius of the volume element is $r$, the thickness of the volume element is $dr$, the length of the volume element is $dx$, the shear force on the volume element is ${\tau }_r$.

Write the equation for horizontal component of weight of the fluid.

  $W_x=W\sin \alpha$

Here, the angle of inclination is $\alpha$.

Substitute $W\sin \alpha$ for $W_x$ in Equation (IV).

  ${\left(2\pi rdrP\right)}_x-{\left(2\pi rdrP\right)}_{x+dx}+{\left(2\pi rdx\tau \right)}_r-{\left(2\pi rdx\tau \right)}_{r+dr}-W\sin \alpha =0$   ...... (V)

Write the expression to calculate the weight of the fluid for the volume element.

  $W=\rho gV$   ...... (VI)

Here, the density of the fluid is $\rho$, the acceleration due to gravity is $g,$ and the volume of the element is $V$.

Calculation:

The figure below represents the free body diagram of the pipe and forces acting on the pipe.

  

  Figure-(1)

Write the expression to calculate the volume of the element.

  $V=2\pi r\times dr\times dx$

Substitute $2\pi r\times dr\times dx$ for $V$ in Equation (VI).

  $W=\rho g\left(2\pi rdrdx\right)$

Substitute $\rho g\left(2\pi rdrdx\right)$ for $W$ in Equation (V).

  $\left[\begin{array}{l}{\left(2\pi rdrP\right)}_x-{\left(2\pi rdrP\right)}_{x+dx}+{\left(2\pi rdx\tau \right)}_r\\ -{\left(2\pi rdx\tau \right)}_{r+dr}-\rho g\left(2\pi rdrdx\right)\sin \alpha \end{array}\right]=0$   ...... (VII)

Divide the equation (V) by $\left(2\pi drdx\right)$.

  $\begin{array}{l}\left[\begin{array}{l}\frac{{\left(2\pi rdrP\right)}_x-{\left(2\pi rdrP\right)}_{x+dx}+{\left(2\pi rdx\tau \right)}_r-{\left(2\pi rdx\tau \right)}_{r+dr}}{\left(2\pi drdx\right)}\\ -\frac{\rho g\left(2\pi rdrdx\right)\sin \alpha }{2\pi drdx}\end{array}\right]=0\\ \frac{P_x-P_{x+dx}}{dx}+\frac{{\left(\tau \right)}_r}{dx}-\frac{{\left(\tau \right)}_{r+dr}}{dx}-\rho g\sin \alpha =0\\ -\frac{dP}{dx}+\frac{{\left(\tau \right)}_r}{dx}-\frac{{\left(\tau \right)}_{r+dr}}{dx}-\rho g\sin \alpha =0\end{array}$   ...... (VIII)

Substitute $\left(-\mu \frac{du}{dr}\right)$ for $\tau$ in equation (VIII)

  $\begin{array}{l}-\frac{dP}{dx}+\frac{{\left(\left(-\mu \frac{du}{dr}\right)\right)}_r}{dx}-\frac{{\left(\left(-\mu \frac{du}{dr}\right)\right)}_{r+dr}}{dx}-\rho g\sin \alpha =0\\ \frac{{\left(\left(-\mu \frac{du}{dr}\right)\right)}_r}{dx}-\frac{{\left(\left(-\mu \frac{du}{dr}\right)\right)}_{r+dr}}{dx}=\frac{dP}{dx}+\rho g\sin \alpha \\ \frac{\mu }{r}\left(\frac{du}{dr}\right)=\frac{dP}{dx}+\rho g\sin \alpha \\ rdr=\frac{\mu du}{\frac{dP}{dx}+\rho g\sin \alpha }\end{array}$   ...... (IX)

Since the pressure at point 1 and point 2 is same, therefore substitute zero for $\frac{dP}{dx}$ in equation (IX).

  $\begin{array}{l}rdr=\frac{\mu du}{0+\rho g\sin \alpha }\\ rdr=\frac{\mu du}{\rho g\sin \alpha }\end{array}$   ...... (X)

Integrate the equation (X).

  $\begin{array}{l}{\displaystyle \underset{r}{\overset{R}{\int }}rdr}={\displaystyle \underset{0}{\overset{u}{\int }}\frac{\mu du}{\rho g\sin \alpha }}\\ {\displaystyle \underset{r}{\overset{R}{\int }}rdr}=\frac{\mu }{\rho g\sin \alpha }{\displaystyle \underset{0}{\overset{u}{\int }}du}\\ u=\frac{R^2}{4\mu }\left(\rho g\sin \alpha \right)\left(1-\frac{r^2}{R^2}\right)\end{array}$

Write the expression for average velocity.

  $V_{avg}=\frac{2}{R^2}{\displaystyle \underset{0}{\overset{R}{\int }}urdr}$   ...... (XI)

Substitute $-\frac{R^2}{4\mu }\left(\rho g\sin \alpha \right)\left(1-\frac{r^2}{R^2}\right)$ for $u$ in equation (XI).

  $\begin{array}{c}{V}_{avg}=\frac{2}{R^2}{\displaystyle \underset{0}{\overset{R}{\int }}\left(\frac{R^2}{4\mu }\left(\rho g\sin \alpha \right)\left(1-\frac{r^2}{R^2}\right)\right)rdr}\\ =\frac{2\left(\rho g\sin \alpha \right)}{R^{2}4\mu }{\displaystyle \underset{0}{\overset{R}{\int }}{R}^2\left(1-\frac{r^2}{R^2}\right)dr}\\ =\frac{\left(\rho g\sin \alpha \right)R^2}{8\mu }\end{array}$   ...... (XII)

Substitute $\frac{D}{2}$ for R in equation (XII).

  $\begin{array}{c}{V}_{avg}=\frac{\left(\rho g\sin \alpha \right){\left(\frac{D}{2}\right)}^2}{8\mu }\\ =\frac{\left(\rho g\sin \alpha \right)\left(\frac{D^2}{4}\right)}{8\mu }\\ =\frac{\left(\rho g\sin \alpha \right)D^2}{32\mu }\end{array}$   ....... (XIII)

Write the expression to calculate the value of $\sin \alpha$.

  $\sin \alpha =\frac{\Delta z}{L}$

Here, the length of the pipe is $L$ and the head for the pipe is $\Delta z$.

Substitute $\frac{\Delta z}{L}$ for $\sin \alpha$ in Equation (XIII)

  $\begin{array}{c}{V}_{avg}=\frac{\left(\rho g\frac{\Delta z}{L}\right)D^2}{32\mu }\\ =\frac{\left(\rho g\Delta z\right)D^2}{32\mu L}\end{array}$

Conclusion:

The expression for average velocity of inclined pipe is $V_{avg}=\frac{\left(\rho g\Delta z\right)D^2}{32\mu L}$.

To determine

(c)

The dimensionless expression for velocity.

Answer to Problem 123P

The dimensionless equation for the velocity is $V=\frac{\frac{\sqrt{gD}}{32}\times \frac{\rho D\sqrt{gD}}{\mu }\times \frac{\Delta z}{D}}{\frac{L}{D}}.$

Explanation of Solution

Write the expression for the velocity in the pipe.

  $V=\frac{\left(\rho g\Delta z\right)D^2}{32\mu L}$   ...... (XIV)

Calculation:

Rearrange the equation (XIV) to obtain the dimensionless expression for velocity.

  $V=\frac{\frac{\sqrt{gD}}{32}\times \frac{\rho D\sqrt{gD}}{\mu }\times \frac{\Delta z}{D}}{\frac{L}{D}}$

Here, the first dimensionless parameter is $\frac{\sqrt{gD}}{32}$, the second dimensionless parameter is $\frac{\rho D\sqrt{gD}}{\mu }$, third dimensionless parameter is $\frac{\Delta z}{D}$ and the fourth dimensionless parameter is $\frac{L}{D}$.

Conclusion:

The dimensionless equation for the velocity is $V=\frac{\frac{\sqrt{gD}}{32}\times \frac{\rho D\sqrt{gD}}{\mu }\times \frac{\Delta z}{D}}{\frac{L}{D}}$.