Chapter 3, Problem 3.185P

To determine

The reading of mercury manometer.

Answer to Problem 3.185P

The reading of mercury manometer is $3.92\text{ft}$.

Explanation of Solution

Given information:

The temperature of Kerosine is $20°\text{C}$ . The flow rate of Kerosine is $2.3{\text{ft}}^3/\text{s}$ . The pump power to flow is $8\text{hp}$ . The head loss between the section (1) and section (2) is $8\text{ft}$.

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

$\begin{array}{l}\frac{p_1}{{\rho }_{k}g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{{\rho }_{k}g}+\frac{V_2^2}{2g}+z_2+h_f-h_p\\ \left(\frac{p_1-p_2}{{\rho }_{k}g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f-h_p\end{array}$ …… (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 $\rho$, the head loss due to friction is $h_f$, the head loss due to pump is $h_p$ and the acceleration due to gravity is $g$.

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

$\begin{array}{l}Q=A_1{V}_1\\ V_1=\frac{Q}{A_1}\end{array}$ …… (II)

Here, 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 (II).

$\begin{array}{c}{V}_1=\frac{Q}{\frac{\pi }{4}{D}_1^2}\\ V_1=\frac{4Q}{\pi D_1^2}\end{array}$ …… (III)

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

$\begin{array}{l}Q=A_2{V}_2\\ V_2=\frac{Q}{A_2}\end{array}$ …… (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).

$\begin{array}{c}{V}_2=\frac{Q}{\frac{\pi }{4}{D}_2^2}\\ V_2=\frac{4Q}{\pi D_2^2}\end{array}$ …… (V)

Write the expression for the pump power.

$\begin{array}{l}P={\gamma }_{k}Q{h}_p\\ h_p=\frac{P}{{\gamma }_{k}Q}\end{array}$ …… (VI)

Here, the specific density of kerosene is ${\gamma }_k$, the pump power is $P$ and the pump head is $h_p$.

Write the expression for the weight density of kerosene.

${\gamma }_k={\rho }_{k}g$

Here, the density of kerosene is ${\rho }_k$ and the acceleration due to gravity is $g$.

Write the expression for the weight density of mercury.

${\gamma }_m={\rho }_{m}g$

Here, the density of kerosene is ${\rho }_m$ and the acceleration due to gravity is $g$.

Substitute ${\gamma }_k$ for ${\rho }_{k}g$ in Equation (I).

$\left(\frac{p_1-p_2}{{\gamma }_k}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f$ …… (VII)

Write the expression for the pressure difference.

$p_2-p_1=\left({\rho }_m-{\rho }_k\right)gh-{\rho }_{k}g\left(z_2-z_1\right)$ …… (VIII)

Substitute ${\rho }_{k}g$ for ${\gamma }_k$ and ${\rho }_{m}g$ for ${\gamma }_m$ in Equation (VIII).

$p_2-p_1=\left({\gamma }_m-{\gamma }_k\right)h-{\gamma }_k\left(z_2-z_1\right)$ …… (IX)

Calculation:

Refer to the property table for liquids to obtain the weight density of kerosene as,

${\gamma }_k=50.23\text{lbf}/{\text{ft}}^{\text{3}}$

Refer to the property table for liquids to obtain the weight density of mercury as,

${\gamma }_m=846.5\text{lbf}/{\text{ft}}^{\text{3}}$

Substitute $8\text{hp}$ for $P$, $50.23\text{lbf}/{\text{ft}}^{\text{3}}$ for ${\gamma }_k$ and $2.3{\text{ft}}^3/\text{s}$ for $Q$ in Equation (VI).

$\begin{array}{c}{h}_p=\frac{\left(8\text{hp}\right)}{\left(50.23\text{lbf}/{\text{ft}}^{\text{3}}\right)\left(2.3{\text{ft}}^3/\text{s}\right)}\\ =\frac{\left(8\text{hp}\right)\left(\frac{550\text{ft}\cdot \text{lbf}/\text{s}}{1\text{hp}}\right)}{\left(50.23\text{lbf}/{\text{ft}}^{\text{3}}\right)\left(2.3{\text{ft}}^3/\text{s}\right)}\\ =38.1\text{ft}\end{array}$

Substitute $2.3{\text{ft}}^3/\text{s}$ for $Q$ and $3\text{in}$ for $D_1$ in Equation (III).

$\begin{array}{c}{V}_1=\frac{4\left(2.3{\text{ft}}^3/\text{s}\right)}{\pi {\left(3\text{in}\right)}^2}\\ =\frac{4\left(2.3{\text{ft}}^3/\text{s}\right)}{\pi {\left\{\left(3\text{in}\right)\left(\frac{1\text{ft}}{12{\text{in}}^2}\right)\right\}}^2}\\ =46.86\text{ft}/\text{s}\end{array}$

Substitute $2.3{\text{ft}}^3/\text{s}$ for $Q$ and $6\text{in}$ for $D_2$ in Equation (III).

$\begin{array}{c}{V}_2=\frac{4\left(2.3{\text{ft}}^3/\text{s}\right)}{\pi {\left(6\text{in}\right)}^2}\\ =\frac{4\left(2.3{\text{ft}}^3/\text{s}\right)}{\pi {\left\{\left(6\text{in}\right)\left(\frac{1\text{ft}}{12{\text{in}}^2}\right)\right\}}^2}\\ =11.71\text{ft}/\text{s}\end{array}$

Substitute $46.86\text{ft}/\text{s}$ for $V_1$, $11.71\text{ft}/\text{s}$ for $V_2$, $38.1\text{ft}$ for $h_p$, $8\text{ft}$ for $h_f$, $0\text{ft}$ for $z_1$, $5\text{ft}$ for $z_2$, $50.23\text{lbf}/{\text{ft}}^{\text{3}}$ for ${\gamma }_k$ and $32.2\text{ft}/{\text{s}}^2$ for $g$ in Equation (VII).

$\begin{array}{l}\left\{\frac{p_1-p_2}{\left(50.23\text{lbf}/{\text{ft}}^{\text{3}}\right)}\right\}+\left\{\frac{{\left(46.86\text{ft}/\text{s}\right)}^2}{2\left(32.2\text{ft}/{\text{s}}^2\right)}-\frac{{\left(11.71\text{ft}/\text{s}\right)}^2}{2\left(32.2\text{ft}/{\text{s}}^2\right)}\right\}+\left\{\left(0\text{ft}\right)-\left(5\text{ft}\right)\right\}=\left(8\text{ft}\right)-\left(38.1\text{ft}\right)\\ \frac{p_1-p_2}{\left(50.23\text{lbf}/{\text{ft}}^{\text{3}}\right)}=\left(-57.06\text{ft}\right)\\ \left(p_1-p_2\right)=-2866.12\text{lbf}/{\text{ft}}^2\\ \left(p_2-p_1\right)=2866.12\text{lbf}/{\text{ft}}^2\end{array}$

Substitute $2866.12\text{lbf}/{\text{ft}}^2$ for $p_2-p_1$, $0\text{ft}$ for $z_1$, $5\text{ft}$ for $z_2$, $50.23\text{lbf}/{\text{ft}}^{\text{3}}$ for ${\gamma }_k$, $846.5\text{lbf}/{\text{ft}}^{\text{3}}$ for ${\gamma }_m$ and $32.2\text{ft}/{\text{s}}^2$ for $g$ in Equation (IX).

$\begin{array}{l}\left(2866.12\text{lbf}/{\text{ft}}^2\right)=\left[\begin{array}{l}\left(846.5\text{lbf}/{\text{ft}}^{\text{3}}\right)-\left(50.23\text{lbf}/{\text{ft}}^{\text{3}}\right)h-\\ \left(50.23\text{lbf}/{\text{ft}}^{\text{3}}\right)\left(5\text{ft}-0\text{ft}\right)\end{array}\right]\\ \left(2866.12\text{lbf}/{\text{ft}}^2\right)=\left(796.27\text{lbf}/{\text{ft}}^3\right)h-251.15\text{lbf}/{\text{ft}}^2\\ h=\frac{\left(3117.27\text{lbf}/{\text{ft}}^2\right)}{\left(796.27\text{lbf}/{\text{ft}}^3\right)}\\ h=3.92\text{ft}\end{array}$

Conclusion:

The reading of mercury manometer is $3.92\text{ft}$.