Chapter 8, Problem 159P

Water is siphoned from a reservoir open to the atmosphere by a pipe with diameter D. total length L. and friction factor f as shown in Fig. 8-159. Atmospheric pressure is 99.27 kPa. Minor losses may be ignored. (a) Verify the following equation for the ratio of exit velocity for the siphon with nozzle to the e>it velocity for the siphon without the nozzle. V_D/V_C
   $V_D/V_C=\sqrt{f\left(\frac{L}{D}\right)+1/\left(f\frac{L}{D}\frac{d^4}{D^4}+1\right)}$

Calculate its value for L = 28 m. h₁ = 1.8 m, h₂ = 12m. D = l2 cm, d = 3 cm, and f= 0.02.
(b) Show that static pressure at B in the siphon with the nozzle is greater and the velocity is smaller with respect to the siphon without the nozzle. Explain how this situation can be useful with respect to flow physics.
(e) Determine the maximum value of h2 in the piping system to avoid cavitation. The density and vapor pressure of water are p = 1000 kg/m³ and P_v = 4.25 kPa, and L₁ = 4 in.

To determine

(a)

The verification of relation $\frac{V_D}{V_C}=\sqrt{\left(\frac{f\frac{L}{D}+1}{f\frac{L}{D}\frac{d^4}{D^4}+1}\right)}$.

The value of $\frac{V_D}{V_C}$.

Answer to Problem 159P

The relation $\frac{V_D}{V_C}=\sqrt{\left(\frac{f\frac{L}{D}+1}{f\frac{L}{D}\frac{d^4}{D^4}+1}\right)}$ is verified.

The value of $\frac{V_D}{V_C}$ is $2.357$.

Explanation of Solution

Given information:

The length of pipe is $28\text{m}$, the value of $h_1$ is $1.8\text{m}$, the value of $h_2$ is $12\text{m}$, diameter of pipe is $12\text{cm}$, diameter of nozzle is $3\text{cm}$, friction factor is $0.02$ and atmospheric pressure is $99.27\text{kPa}$.

Write the expression for the Bernoulli Equation.

  $\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{v_2^2}{2g}+Z_2+h_f$  ......(I)

Here, pressure at point $1$ is $P_1$, velocity at point $1$ in $v_1$, density of the fluid is $\rho$, gravitational acceleration is $g$, elevation of point $1$ is $Z_1$, pressure at point $2$ is $P_2$ and elevation of point $2$ is $Z_2$.

Write the expression for conservation of mass in nozzle.

  $\begin{array}{l}\frac{\pi }{4}{D}^2{V}_C=\frac{\pi }{4}{d}^2{V}_D\\ D^2{V}_C=d^2{V}_D\\ V_C=\frac{d^2{V}_D}{D^2}\end{array}$   ....... (II)

Here, diameter of nozzle is $d$, diameter of pipe is $D$, velocity through nozzle is $V_D$ and velocity through pipe is $V_C$.

Write the expression for head loss.

  $h_f=\frac{fL{V}_C^2}{2gD}$   ....... (III)

Here, head loss is $h_f$, friction factor is $f$ and length of pipe is $L$.

Calculation:

Substitute $\frac{fL{V}_C^2}{2gD}$ for $h_f$, $P_{atm}$ for $P_1$, $P_{atm}$ for $P_2$, $0$ for $v_1$, $h_2$ for $Z_1$, $V_C$ for $v_2$, $0$ for $Z_2$ in Equation (I).

  $\begin{array}{l}\frac{P_{atm}}{\rho g}+\frac{{\left(0\right)}^2}{2g}+h_2=\frac{P_{atm}}{\rho g}+\frac{V_C^2}{2g}+0+\frac{fL{V}_C^2}{2gD}\\ h_2=\frac{V_C^2}{2g}+\frac{fL{V}_C^2}{2gD}\\ h_2=\frac{V_C^2}{2g}\left(1+\frac{fL}{D}\right)\\ V_C=\sqrt{\frac{2g{h}_2}{\left(1+\frac{fL}{D}\right)}}\end{array}$   ....... (IV)

Substitute $P_{atm}$ for $P_1$, $P_{atm}$ for $P_2$, $0$ for $v_1$, $h_2$ for $Z_1$, $V_D$ for $v_2$, $0$ for $Z_2$ and $\frac{fL{V}_C^2}{2gD}$ for $h_f$ in Equation (I).

  $\begin{array}{l}\frac{P_{atm}}{\rho g}+\frac{{\left(0\right)}^2}{2g}+h_2=\frac{P_{atm}}{\rho g}+\frac{V_D^2}{2g}+0+\frac{fL{V}_C^2}{2gD}\\ h_2=\frac{V_D^2}{2g}+\frac{fL{V}_C^2}{2gD}\\ h_2=\frac{1}{2g}\left(V_D^2+\frac{fL{V}_C^2}{D}\right)\end{array}$   ....... (V)

Substitute $\frac{d^2{V}_D}{D^2}$ for $V_C$ in Equation (V).

  $\begin{array}{l}{h}_2=\frac{1}{2g}\left(V_D^2+\frac{fL{\left(\frac{d^2{V}_D}{D^2}\right)}^2}{D}\right)\\ h_2=\frac{1}{2g}\left(V_D^2+\frac{fL}{D}\frac{d^4{V}_D^2}{D^4}\right)\\ h_2=\frac{V_D^2}{2g}\left(1+\frac{fL}{D}\frac{d^4}{D^4}\right)\\ V_D=\sqrt{\frac{2g{h}_2}{\left(1+\frac{fL}{D}\frac{d^4}{D^4}\right)}}\end{array}$   ...... (VI)

From Equation (IV) and (VI).

  $\begin{array}{l}\frac{V_D}{V_C}=\frac{\sqrt{\frac{2g{h}_2}{\left(1+\frac{fL}{D}\frac{d^4}{D^4}\right)}}}{\sqrt{\frac{2g{h}_2}{\left(1+\frac{fL}{D}\right)}}}\\ \frac{V_D}{V_C}=\sqrt{\left(\frac{1+\frac{fL}{D}}{1+\frac{fL}{D}\frac{d^4}{D^4}}\right)}\end{array}$   ....... (VII)

Substitute $0.02$ for $f$, $28\text{m}$ for $L$, $3\text{cm}$ for $d$ and $12\text{cm}$ for $D$ in Equation (VII).

  $\begin{array}{c}\frac{V_D}{V_C}=\sqrt{\left(\frac{1+\frac{\left(0.02\right)\left(28\text{m}\right)}{\left(12\text{cm}\right)}}{1+\frac{\left(0.02\right)\left(28\text{m}\right)}{\left(12\text{cm}\right)}\frac{{\left(3\text{cm}\right)}^4}{{\left(12\text{cm}\right)}^4}}\right)}\\ =\sqrt{\left(\frac{1+\frac{\left(0.02\right)\left(28\text{m}\right)}{\left(12\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}}{1+\frac{\left(0.02\right)\left(28\text{m}\right)}{\left(12\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}\frac{{\left(\left(3\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]\right)}^4}{{\left(\left(12\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]\right)}^4}}\right)}\\ =\sqrt{5.55}\\ =2.357\end{array}$

Conclusion:

The relation $\frac{V_D}{V_C}=\sqrt{\left(\frac{f\frac{L}{D}+1}{f\frac{L}{D}\frac{d^4}{D^4}+1}\right)}$ is verified.

The value of $\frac{V_D}{V_C}$ is $2.357$.

To determine

(b)

The static pressure at $B$ in case of with nozzle is greater than static pressure at $B$ in case of without nozzle or not.

The velocity at $B$ in case of with nozzle is less than velocity at $B$ in case of without nozzle or not.

Answer to Problem 159P

The static pressure at $B$ in case of with nozzle is greater than static pressure at $B$ in case of without nozzle.

The velocity at $B$ in case of with nozzle is less than velocity at $B$ in case of without nozzle.

Explanation of Solution

Calculation:

Substitute $\frac{f\left(L-L_1\right)V_C^2}{2gD}$ for $h_f$, $P_{B,\text{without}\text{nozzle}}$ for $P_1$, $P_{atm}$ for $P_2$, $V_B$ for $v_1$, $\left(h_1+h_2\right)$ for $Z_1$, $V_C$ for $v_2$, $0$ for $Z_2$ in Equation (I).

  $\begin{array}{l}\frac{P_{B,\text{without}\text{nozzle}}}{\rho g}+\frac{V_B^2}{2g}+\left(h_1+h_2\right)=\frac{P_{atm}}{\rho g}+\frac{V_C^2}{2g}+0+\frac{f\left(L-L_1\right)V_C^2}{2gD}\\ \frac{P_{B,\text{without}\text{nozzle}}}{\rho g}=\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}+\frac{V_C^2}{2g}-\frac{V_B^2}{2g}-\left(h_1+h_2\right)\\ P_{B,\text{without}\text{nozzle}}=\rho g\left(\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\frac{V_B^2}{2g}-\left(h_1+h_2\right)+\frac{V_C^2}{2g}\right)\end{array}$   ....... (VIII)

Substitute $\frac{f\left(L-L_1\right)V_C^2}{2gD}$ for $h_f$, $P_{B,\text{with}\text{nozzle}}$ for $P_1$, $P_{atm}$ for $P_2$, $V_B$ for $v_1$, $\left(h_1+h_2\right)$ for $Z_1$, $V_D$ for $v_2$, $0$ for $Z_2$ in Equation (I).

  $\begin{array}{l}\frac{P_{B,\text{with}\text{nozzle}}}{\rho g}+\frac{V_B^2}{2g}+\left(h_1+h_2\right)=\frac{P_{atm}}{\rho g}+\frac{V_D^2}{2g}+0+\frac{f\left(L-L_1\right)V_C^2}{2gD}\\ \frac{P_{B,\text{with}\text{nozzle}}}{\rho g}=\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}+\frac{V_D^2}{2g}-\frac{V_B^2}{2g}-\left(h_1+h_2\right)\\ P_{B,\text{with}\text{nozzle}}=\rho g\left(\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}+\frac{V_D^2}{2g}-\frac{V_B^2}{2g}-\left(h_1+h_2\right)\right)\end{array}$   ....... (IX)

Substitute $\frac{D^2{V}_C}{d^2}$ for $V_D$ in Equation (IX).

  $\begin{array}{c}{P}_{B,\text{with}\text{nozzle}}=\rho g\left(\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\frac{V_B^2}{2g}-\left(h_1+h_2\right)+\frac{{\left(\frac{D^2{V}_C}{d^2}\right)}^2}{2g}\right)\\ =\rho g\left(\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\frac{V_B^2}{2g}-\left(h_1+h_2\right)+\frac{V_C^2}{2g}\frac{D^4}{d^4}\right)\end{array}$   ...... (X)

From Equation (VIII) and (X), since $D>d$, therefore $P_{B,\text{with}\text{nozzle}}>P_{B,\text{without}\text{nozzle}}$.

Since the pipe diameter is greater than nozzle diameter, therefore ratio of pipe diameter to nozzle diameter will be greater than one. Thus the static pressure at $B$ in case of with nozzle is greater than static pressure at $B$ in case of without nozzle.

From equation (VIII).

  $\begin{array}{l}\frac{P_{B,\text{without}\text{nozzle}}}{\rho g}=\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\frac{V_{B,\text{without}\text{nozzle}}^2}{2g}-\left(h_1+h_2\right)+\frac{V_C^2}{2g}\\ \frac{V_{B,\text{without}\text{nozzle}}^2}{2g}=\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\left(h_1+h_2\right)+\frac{V_C^2}{2g}-\frac{P_{B,\text{without}\text{nozzle}}}{\rho g}\\ V_{B,\text{without}\text{nozzle}}^2=2g\left(\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\left(h_1+h_2\right)+\frac{V_C^2}{2g}-\frac{P_{B,\text{without}\text{nozzle}}}{\rho g}\right)\end{array}$   ...... (XI)

From equation (X).

  $\begin{array}{l}\frac{P_{B,\text{with}\text{nozzle}}}{\rho g}=\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\frac{V_{B,\text{with}\text{nozzle}}^2}{2g}-\left(h_1+h_2\right)+\frac{V_C^2}{2g}\frac{D^4}{d^4}\\ \frac{V_{B,\text{with}\text{nozzle}}^2}{2g}=\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\left(h_1+h_2\right)+\frac{V_C^2}{2g}\frac{D^4}{d^4}-\frac{P_{B,\text{with}\text{nozzle}}}{\rho g}\\ V_{B,\text{with}\text{nozzle}}^2=2g\left(\frac{P_{atm}}{\rho g}+\frac{f\left(L-L_1\right)V_C^2}{2gD}-\left(h_1+h_2\right)+\frac{V_C^2}{2g}\frac{D^4}{d^4}-\frac{P_{B,\text{with}\text{nozzle}}}{\rho g}\right)\end{array}$   ....... (XII)

From Equation (XI) and (XII), since $P_{B,\text{with}\text{nozzle}}>P_{B,\text{without}\text{nozzle}}$, therefore $V_{B,\text{with}\text{nozzle}}<V_{B,\text{without}\text{nozzle}}$.

Since the static pressure at $B$ in case of with nozzle is greater than static pressure at $B$ in case of without nozzle, therefore The velocity at $B$ in case of with nozzle is less than velocity at $B$ in case of without nozzle.

Conclusion:

The static pressure at $B$ in case of with nozzle is greater than static pressure at $B$ in case of without nozzle.

The velocity at $B$ in case of with nozzle is less than velocity at $B$ in case of without nozzle.

To determine

(c)

The maximum value of $h_2$.

Answer to Problem 159P

The maximum value of $h_2$ is $16.35\text{m}$.

Explanation of Solution

Given information:

The density of water is $1000\text{kg}/{\text{m}}^{\text{3}}$, vapor pressure is $4.25\text{kPa}$ and $L_1$ is $4\text{m}$.

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $12\text{m}$ for $h_2$, $0.02$ for $f$, $28\text{m}$ for $L$ and $12\text{cm}$ for $D$ in Equation (IV).

  $\begin{array}{c}{V}_C=\sqrt{\frac{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(12\text{m}\right)}{\left(1+\frac{\left(0.02\right)\left(28\text{m}\right)}{\left(12\text{cm}\right)}\right)}}\\ =\sqrt{\frac{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(12\text{m}\right)}{\left(1+\frac{\left(0.02\right)\left(28\text{m}\right)}{\left(12\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}\right)}}\\ =\sqrt{41.548{\text{m}}^2/{\text{s}}^{\text{2}}}\\ =6.445\text{m}/\text{s}\end{array}$

Substitute $6.445\text{m}/\text{s}$ for $V_C$, $4.25\text{kPa}$ for $P_B$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1.8\text{m}$ for $h_1$, $0.02$ for $f$, $28\text{m}$ for $L$, $4\text{m}$ for $L_1$, $V_C$ for $V_B$, $99.27\text{kPa}$ for $P_{atm}$ and $12\text{cm}$ for $D$ in equation (VIII).

  $\begin{array}{l}\left(9810\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\left(\left(1.8\text{m}\right)+h_2\right)=\left(83076.05\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)+\left(95020\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\\ \left(1.8\text{m}\right)+h_2=\left(18.15\text{m}\right)\\ h_2=16.35\text{m}\end{array}$

Conclusion:

The maximum value of $h_2$ is $16.35\text{m}$.