Chapter 3, Problem 3.134P

To determine

(a)

The head loss.

Answer to Problem 3.134P

The head loss is $15\text{m}$.

Explanation of Solution

Given information:

The flow rate is $40{\text{m}}^3/\text{h}$ . The temperature of the water is $40°\text{C}$ . The height of the water level in the left side above the throat is $25\text{m}$ and the height of the water level in the right side above the throat is $10\text{m}$.

The Figure-(1) show three sections of constant flow rate in the system.

Figure-(1)

Here, the section (1) is for the top of the water level in the left region, the section (2) is at the throat and the section (3) is at the top of the water level in the right region.

Write the expression for the Bernoulli’s equation between the section (1) and section (3) of the given system.

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_3}{\rho g}+\frac{V_3^2}{2g}+z_3+h_f\\ \left(\frac{p_1-p_3}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_3^2}{2g}\right)+\left(z_1-z_3\right)=h_f\\ \left(\frac{p_1-p_3}{\rho g}\right)+\left(\frac{\left(V_1-V_3\right)\left(V_1+V_3\right)}{2g}\right)+\left(z_1-z_3\right)=h_f\end{array}$ ...(I)

Here, the pressure at section (1) is $p_1$, the velocity at section (1) is $V_1$, the height of the section (1) from the datum is $z_1$, the pressure at section (3) is $p_3$, the velocity at section (3) is $V_3$, the height of the section (3) from the datum is $z_3$, the density of water is $\rho$, the head loss due to friction is $h_f$ and the acceleration due to gravity is $g$.

Since the fluid at section (1) and section (3) is stationary, so the difference between the velocities at these sections is zero.

$V_1-V_3=0$

Since the pressure at the section (1) and section (3) is atmospheric, so the difference between the pressures is zero.

$p_1-p_3=0$

Substitute $0$ for $p_1-p_3$ and $0$ for $V_1-V_3$ in Equation (I).

$\begin{array}{l}\left(\frac{0}{\rho g}\right)+\left(\frac{0\times \left(V_1+V_3\right)}{2g}\right)+\left(z_1-z_3\right)=h_f\\ h_f=\left(z_1-z_3\right)\end{array}$ ...(II)

Calculation:

Substitute $25\text{m}$ for $z_1$ and $10\text{m}$ for $z_2$ in Equation (II).

$\begin{array}{c}{h}_f=\left\{\left(25\text{m}\right)-\left(10\text{m}\right)\right\}\\ =15\text{m}\end{array}$

Conclusion:

The head loss is $15\text{m}$.

To determine

(b)

The constriction diameter $D$ that causes cavitation.

Answer to Problem 3.134P

The constriction diameter $D$ that causes cavitation is $24.8\text{mm}$.

Explanation of Solution

Given information:

The diameter at section (1) is $5\text{cm}$ . The diameter at section (3) is $8\text{cm}$ . The atmospheric pressure is $100\text{kPa}$ . The height of the water level in the left side above the throat is $25\text{m}$ and the height of the water level in the right side above the throat is $10\text{m}$.

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system.

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+{h^{\prime }}_f\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)={h^{\prime }}_f\end{array}$ ...(III)

Here, the pressure at section (2) is $p_2$, the velocity at section (2) is $V_2$, the height of the section (2) from the datum is $z_2$ and the head loss at the throat due to cavitation is ${h^{\prime }}_f$.

Write the expression for the head loss due to cavitation.

${h^{\prime }}_f=\frac{h_f}{2}$

Write the expression for the flow rate at section (2).

$Q=A_2{V}_2$ ...(IV)

Here, the flow rate is $Q$, the jet area at section (2) is $A_2$ and the velocity at section (2) is $V_2$.

Write the expression for the area at section (2).

$A_2=\frac{\pi }{4}{D}_2^2$

Here, the diameter at section (2) is $D_2$.

Substitute $\frac{\pi }{4}{D}_2^2$ for $A_2$ in Equation (V).

$\begin{array}{l}Q=\frac{\pi }{4}{D}_2^2{V}_2\\ V_2=\frac{Q}{\frac{\pi }{4}{D}_2^2}\end{array}$

Substitute $\frac{Q}{\frac{\pi }{4}{D}_2^2}$ for $V_2$ and $\frac{h_f}{2}$ for ${h^{\prime }}_f$ in Equation (III).

$\begin{array}{l}\left(\frac{p_1-p_2}{\rho g}\right)+\left\{\frac{V_1^2}{2g}-\frac{{\left(\frac{Q}{\frac{\pi }{4}{D}_2^2}\right)}^2}{2g}\right\}+\left(z_1-z_2\right)=\frac{h_f}{2}\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left\{\frac{V_1^2}{2g}-\frac{8Q^2}{{\pi }^2{D}_2^{4}g}\right\}+\left(z_1-z_2\right)=\frac{h_f}{2}\end{array}$ ...(V)

Calculation:

Since the fluid at section (1) is stationary, so the velocity at this section is zero.

$V_1=0$

Since the pressure at the section (1) is atmospheric so,

$\begin{array}{l}{p}_1=p_{atm}\\ p_1=101325\text{Pa}\end{array}$

Refer to the property table to obtain the vapor pressure of water at $40°\text{C}$ as,

$p_2=7375\text{Pa}$

Substitute $40{\text{m}}^3/\text{h}$ for $Q$, $0$ for $V_1$, $101325\text{Pa}$ for $p_1$, $7375\text{Pa}$ for $p_2$, $25\text{m}$ for $z_1$, $10\text{m}$ for $z_2$

$998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $15\text{m}$ for $h_f$ in Equation (V).

$\begin{array}{l}\left[\begin{array}{l}\left\{\frac{\left(101325\text{Pa}\right)-\left(7375\text{Pa}\right)}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right\}+\\ \left\{\frac{0}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}-\frac{8{\left(40{\text{m}}^3/\text{h}\right)}^2}{{\pi }^2{D}_2^4\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right\}\\ +\left\{\left(25\text{m}\right)-\left(0\text{m}\right)\right\}\end{array}\right]=\frac{\left(15\text{m}\right)}{2}\\ \left[\begin{array}{l}9.60\text{m}+\left\{0-\frac{8{\left\{\left(40{\text{m}}^3/\text{h}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\right\}}^2}{{\pi }^2{D}_2^4\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right\}\\ +\left(25\text{m}\right)\end{array}\right]=7.5\text{m}\\ \left(9.60\text{m}\right)-\frac{\left(\text{1}\text{.0201}\times {\text{10}}^{-5}{\text{m}}^{\text{2}}/\text{s}\right)}{D_2^4}+\left(25\text{m}\right)=\left(7.5\text{m}\right)\\ D_2^4=\frac{\left(\text{1}\text{.0201}\times {\text{10}}^{-5}{\text{m}}^5\right)}{\left(27.1\text{m}\right)}\end{array}$

Further solve the above expression.

$\begin{array}{c}{D}_2=\sqrt[4]{\frac{\left(\text{1}\text{.0201}\times {\text{10}}^{-5}{\text{m}}^5\right)}{\left(27.1\text{m}\right)}}\\ =\sqrt{3.764\times {10}^{-7}{\text{m}}^4}\\ =\left(0.0248\text{m}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =24.8\text{mm}\end{array}$

Conclusion:

The constriction diameter $D$ that causes cavitation is $24.8\text{mm}$.