Chapter 3, Problem 3.110P

To determine

The new bolt force and the head loss.

Answer to Problem 3.110P

The new bolt force is $1344\text{lbf}$.

The head loss is $7.43\text{ft}$.

Explanation of Solution

Given information:

The diameter of nozzle at section 1 is $12\text{in}$, the diameter of nozzle at section 2 is $6\text{in}$ and the velocity at section 2 is $56\text{ft/s}$.

Write the expression for area of section 1.

$A_1=\frac{\pi D_1^2}{4}$ …… (I)

Here, diameter at section 1 is $D_1$ and the area at section 1 is $A_1$.

Write the expression for area at section 2.

$A_2=\frac{\pi D_2^2}{4}$ …… (II)

Here, diameter at section 2 is $D_2$ and the area at section 2 is $A_2$.

Write the expression for velocity at section 1 using continuity equation.

$V_1=V_2\left(\frac{A_2}{A_1}\right)$ …… (III)

Here, velocity at section 1 is $V_1$ and the velocity at section 2 is $V_2$.

Write the expression for Bernoulli’s Equation at inlet and outlet without losses.

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}$ …… (IV)

Here, pressure at inlet is $P_1$, pressure at outlet is $P_2$ and the density of liquid is $\rho$.

Write the expression for mass flow rate.

$\dot{m}=A_1{V}_1\rho$ ……. (V)

Write the expression for force on bolt using momentum principle.

$F_{bolts}=P_{1gage}{A}_1-\dot{m}\left(V_2-V_1\right)$ …… (VI).

Write the expression for Bernoulli’s Equation considering head loss.

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+h_f$ ……. (VII)

Here, the head loss is $h_f$.

Calculation:

Substitute $12\text{in}$ for $D_1$ in equation (I)

$\begin{array}{c}{A}_1=\frac{\pi {\left(12\text{in}\right)}^2}{4}\\ =\frac{\pi \left(144{\text{in}}^2\right)}{4}\\ =\left(113.097{\text{in}}^2\right){\left(\frac{1\text{ft}}{12\text{in}}\right)}^2\\ =0.78539{\text{ft}}^2\end{array}$

Substitute $6\text{in}$ for $D_2$ in equation (II)

$\begin{array}{c}{A}_2=\frac{\pi {\left(6\text{in}\right)}^2}{4}\\ =\frac{\pi \left(36{\text{in}}^2\right)}{4}\\ =28.2743{\text{in}}^2{\left(\frac{1\text{ft}}{12\text{in}}\right)}^2\\ =0.196349{\text{ft}}^2\end{array}$

Substitute $113.097{\text{in}}^2$ for $A_1$

$28.2743{\text{in}}^2$

$A_2$ and $56\text{ft}/\text{s}$ for $V_2$ in equation (III).

$\begin{array}{l}113.097{\text{in}}^2{V}_1=\left(28.2743{\text{in}}^2\right)\left(56\text{ft}/\text{s}\right)\\ V_1=\frac{1583.3608\text{ft}/\text{s}\cdot {\text{in}}^2}{113.097{\text{in}}^2}\\ V_1=14\text{ft}/\text{s}\end{array}$

Substitute $14\text{ft}/\text{s}$ for $V_1$, $56\text{ft}/\text{s}$ for $V_2$, $32.2{\text{ft}/\text{s}}^2$ for $g$ and $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$ and $15\text{lbf}/{\text{in}}^{\text{2}}$ for $P_2$ in equation (IV).

$\begin{array}{l}\frac{P_1}{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2{\text{ft}/\text{s}}^2\right)}+\frac{{\left(14\text{ft}/\text{s}\right)}^2}{2\left(32.2{\text{ft}/\text{s}}^2\right)}=\left[\begin{array}{l}\frac{\left(15\text{lbf}/{\text{in}}^{\text{2}}\right){\left(\frac{12\text{in}}{1\text{ft}}\right)}^2}{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2{\text{ft}/\text{s}}^2\right)}\\ +\frac{\left(56\text{ft}/\text{s}\right)}{2\left(32.2{\text{ft}/\text{s}}^2\right)}\end{array}\right]\\ \frac{P_1}{1.94\text{slug}/{\text{ft}}^{\text{3}}}=2583.4\\ P_1=2583.4\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\\ P_1=5012\text{lbf}/{\text{ft}}^2\end{array}$

Substitute $14\text{ft}/\text{s}$ for $V_1$, $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$ and $0.78539{\text{ft}}^2$ for $A_1$ in Equation (V).

$\begin{array}{c}\dot{m}=\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(0.78539{\text{ft}}^2\right)\left(14\text{ft}/\text{s}\right)\\ =21.33\text{lbf/s}\end{array}$

Substitute $5012{\text{lbf/ft}}^{\text{2}}$ for $P_{1gage}$, $21.33\text{lbf/s}$ for $\dot{m}$, $56\text{ft}/\text{s}$ for $V_2$

$0.78539{\text{ft}}^2$ for $A_1$, $14\text{ft}/\text{s}$ for $V_1$ in Equation (VI).

$\begin{array}{c}{F}_{bolts}=\left(5012\text{lbf}/{\text{ft}}^{\text{2}}-15\text{lbf}/{\text{in}}^{\text{2}}\right)\left(0.78539{\text{ft}}^2\right)-21.33\text{lbf/s}\left(\begin{array}{l}56\text{ft}/\text{s}\\ -14\text{ft}/\text{s}\end{array}\right)\\ =\left(5012\text{lbf}/{\text{ft}}^{\text{2}}-\left(15\text{lbf}/{\text{in}}^{\text{2}}\right){\left(\frac{12\text{in}}{1\text{ft}}\right)}^2\right)\left(0.78539{\text{ft}}^2\right)-21.33\text{lbf/s}\left(\begin{array}{l}56\text{ft}/\text{s}\\ -14\text{ft}/\text{s}\end{array}\right)\\ =1344\text{lbf}\end{array}$

Substitute $14\text{ft}/\text{s}$ for $V_1$, $56\text{ft}/\text{s}$ for $V_2$, $32.2{\text{ft}/\text{s}}^2$ for $g$ and $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$ and $15\text{lbf}/{\text{in}}^{\text{2}}$ for $P_2$ in equation (IV).

$\begin{array}{l}\frac{5012{\text{lbf/ft}}^{\text{2}}}{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2{\text{ft}/\text{s}}^2\right)}+\frac{{\left(14\text{ft}/\text{s}\right)}^2}{2\left(32.2{\text{ft}/\text{s}}^2\right)}=\left(\begin{array}{l}\frac{\left(15\text{lbf}/{\text{in}}^{\text{2}}\right){\left(\frac{12\text{in}}{1\text{ft}}\right)}^2}{\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2{\text{ft}/\text{s}}^2\right)}\\ +\frac{\left(56\text{ft}/\text{s}\right)}{2\left(32.2{\text{ft}/\text{s}}^2\right)}\end{array}\right)\\ 83.3\text{ft}+h_f=90.73\text{ft}\\ {\text{h}}_f=90.73\text{ft}-83.3\text{ft}\\ h_f=7.43\text{ft}\end{array}$

Conclusion:

The bolt force is $1344\text{lbf}$.

The head loss is $7.43\text{ft}$.