Chapter 3, Problem 3.127P

To determine

(a)

The mass flow rate.

Answer to Problem 3.127P

The mass flow rate is $5.26\text{kg}/\text{s}$.

Explanation of Solution

Given information:

The diameter at section (1) is $12\text{cm}$ . The diameter of the jet is $4\text{cm}$ . The pressure at the centerline of section (1) is $110\text{kPa}$ . The pressure at the exit of the jet is $101.325\text{kPa}$.

The Figure-(1) shows the section (1) and the section (2) for the given system.

Figure-(1)

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

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=0\end{array}$ …… (I)

Here, the pressure at inlet is $p_1$, the velocity at inlet is $V_1$, the height of the inlet from the datum is $z_1$, the pressure at outlet is $p_2$, the velocity at outlet is $V_2$, the height of the outlet from the datum is $z_2$, the density of water is $\rho$, and the acceleration due to gravity is $g$.

Since the section (1) and the section (2) are at the same level, so the difference between the datum at the two sections is zero.

$z_1-z_2=0$

Substitute $0$ for $z_1-z_2$ in Equation (I).

$\begin{array}{l}\left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+0=0\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)=0\end{array}$ …… (II)

Write the expression for the flow rate at section (1).

$Q=A_1{V}_1$ …… (III)

Here, the flow rate is $Q$, the area at section (1) is $A_1$ and the velocity at section (1) is $V_1$.

Write the expression for the area at section (1).

$A_1=\frac{\pi }{4}{d}_1^2$

Here, the diameter at section (1) is $d_1$.

Substitute $\frac{\pi }{4}{d}_1^2$ for $A_1$ in Equation (III).

$Q=\frac{\pi }{4}{d}_1^2{V}_1$

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 (IV).

$Q=\frac{\pi }{4}{d}_2^2{V}_2$ …… (V)

Since the flow rate at both the sections is equal, it satisfies the continuity equation.

Substitute $\frac{\pi }{4}{d}_1^2{V}_1$ for $Q$ in Equation (V).

$\begin{array}{l}\frac{\pi }{4}{d}_1^2{V}_1=\frac{\pi }{4}{d}_2^2{V}_2\\ d_1^2{V}_1=d_2^2{V}_2\\ V_1=\frac{d_2^2}{d_1^2}{V}_2\end{array}$

Substitute $\frac{d_2^2}{d_1^2}{V}_2$ for $V_1$ in Equation (II).

$\begin{array}{l}\left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{{\left(\frac{d_2^2}{d_1^2}{V}_2\right)}^2}{2g}-\frac{V_2^2}{2g}\right)=0\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{d_2^4}{d_1^4}\frac{V_2^2}{2g}-\frac{V_2^2}{2g}\right)=0\\ \left(\frac{p_1-p_2}{\rho g}\right)+\frac{V_2^2}{2g}\left(\frac{d_2^4}{d_1^4}-1\right)=0\end{array}$ …… (VI)

Write the expression for mass flow rate.

$\dot{m}=\rho Q$ …… (VII)

Calculation:

Substitute $110\text{kPa}$ for $p_1$, $101.325\text{kPa}$ for $p_2$, $12\text{cm}$ for $d_1$, $4\text{cm}$ for $d_2$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in Equation (VI).

$\begin{array}{l}\left\{\frac{\left(110\text{kPa}\right)-\left(101.325\text{kPa}\right)}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^2\right)}\right\}+\frac{V_2^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\left(\frac{{\left(4\text{cm}\right)}^4}{{\left(12\text{cm}\right)}^4}-1\right)=0\\ \left\{\frac{\left(8.675\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^2\right)}\right\}+V_2^2\left(-0.05034{\text{s}}^2/\text{m}\right)=0\\ \left(0.88607388\text{N}\cdot \text{m}/\text{kg}\right)=\left(0.05034{\text{s}}^2/\text{m}\right)V_2^2\\ V_2=\sqrt{\frac{\left(0.88607388\text{N}\cdot \text{m}/\text{kg}\right)}{\left(0.05034{\text{s}}^2/\text{m}\right)}}\end{array}$

Further solve the above expression.

$\begin{array}{c}{V}_2=\sqrt{\frac{\left(0.88607388\text{N}\cdot \text{m}/\text{kg}\right)}{\left(0.05034{\text{s}}^2/\text{m}\right)}}\\ =\sqrt{17.60{\text{m}}^{\text{2}}/{\text{s}}^2}\\ =4.195\text{m}/\text{s}\end{array}$

Substitute $4.195\text{m}/\text{s}$ for $V_2$ and $4\text{cm}$ for $d_2$ in Equation (V).

$\begin{array}{c}Q=\frac{\pi }{4}{\left(4\text{cm}\right)}^2\left(4.195\text{m}/\text{s}\right)\\ =\frac{\pi }{4}\left(16{\text{cm}}^2\right)\left(\frac{1{\text{m}}^2}{10000{\text{cm}}^2}\right)\left(4.195\text{m}/\text{s}\right)\\ =0.00527{\text{m}}^3/\text{s}\end{array}$

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0.00527{\text{m}}^3/\text{s}$ for $Q$ in Equation (VII).

$\begin{array}{c}\dot{m}=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.00527{\text{m}}^3/\text{s}\right)\\ =5.26\text{kg}/\text{s}\end{array}$

Conclusion:

The mass flow rate is $5.26\text{kg}/\text{s}$.

To determine

(b)

The height of the fluid in stagnation tube.

Answer to Problem 3.127P

The height of the fluid in stagnation tube is $0.917\text{m}$.

Explanation of Solution

Given information:

The diameter at section (1) is $12\text{cm}$ . The diameter of the jet is $4\text{cm}$ . The pressure at the centerline of section (1) is $110\text{kPa}$ . The pressure at the exit of the jet is $101.325\text{kPa}$.

Write the expression for the velocity of the jet.

$V_2=\sqrt{2g\left(H-\frac{d_2}{2}\right)}$ …… (VIII)

Here, the height of the water in stagnation tube is $H$.

Calculation:

Substitute $4\text{cm}$ for $d_2$, $4.195\text{m}/\text{s}$ for $V_2$ and $9.81{\text{m}/\text{s}}^2$ for $g$ in Equation (VIII).

$\begin{array}{l}\left(4.195\text{m}/\text{s}\right)=\sqrt{2\left(9.81{\text{m}/\text{s}}^2\right)\left\{H-\frac{\left(4\text{cm}\right)}{2}\right\}}\\ \left(4.195\text{m}/\text{s}\right)=\sqrt{\left(19.62{\text{m}/\text{s}}^2\right)\left\{H-\left(2\text{cm}\right)\right\}}\end{array}$

Square on both the sides in above expression.

$\begin{array}{l}{\left(4.195\text{m}/\text{s}\right)}^2={\left[\sqrt{\left(19.62{\text{m}/\text{s}}^2\right)\left\{H-\left(2\text{cm}\right)\right\}}\right]}^2\\ \left(17.59{\text{m}}^2/{\text{s}}^2\right)=\left(19.62{\text{m}/\text{s}}^2\right)\left\{H-\left(2\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right\}\\ H-\left(0.02\text{m}\right)=0.8969\text{m}\\ H=0.917\text{m}\end{array}$

Conclusion:

The height of the fluid in stagnation tube is $0.917\text{m}$.