Chapter 3, Problem 3.120P

To determine

(a)

The volume flow rate in the tube if the fluid in the tube is gasoline.

Answer to Problem 3.120P

The volume flow rate in the tube if the fluid in the tube is gasoline is $0.5{\text{ft}}^3/\text{s}$.

Given information:

The diameter of the tube in which fluid is flowing is $3\text{in}$, the difference in the level of manometric fluid in two limbs of the manometer is $1\text{in}$, the temperature of the fluid is $20\text{°C}$ at a pressure of $1\text{atm}$.

Concept Used:

The following figure shows the two sections where we apply the Bernoulli equation.

Figure-(1)

Write the expression for Bernoulli equation at section (1) and section (2).

$\frac{P_1}{\rho }+\frac{V_1{}^2}{2}+g{z}_1=\frac{P_2}{\rho }+\frac{V_2{}^2}{2}+g{z}_2$

Substitute $z_1=z_2$ in above equation.

$\begin{array}{l}\frac{P_1}{\rho }+\frac{V_1{}^2}{2}=\frac{P_2}{\rho }+\frac{V_2{}^2}{2}\\ \frac{V_1{}^2}{2}=\frac{V_2{}^2}{2}+\frac{P_2}{\rho }-\frac{P_1}{\rho }\\ V_1=\sqrt{\frac{V_2{}^2}{2}+2\left(\frac{P_2-P_1}{\rho }\right)}\end{array}$ …… (I)

Here, pressure at section (1) is $P_1$, the pressure at section (2) is $P_2$, the velocity of the jet at section (1) is $V_1$, and the velocity of the jet at section (2) is $V_2$.

Substitute $0$ for $V_2$ in Equation (I).

$V_1=\sqrt{2\left(\frac{P_2-P_1}{\rho }\right)}$ …… (II)

Write the expression for finding the pressure difference by using the manometer reading.

$p=\left({\rho }_{Hg}-\rho \right)gh$ …… (III)

Here, the pressure difference between the limbs of manometer is $p$, the acceleration due to gravity is $g$, the difference in the elevation level of the manometric fluid is $h$, the density of the manometric fluid is ${\rho }_{Hg}$.

Write the expression for the flow rate.

$Q=A_1{V}_1$ …… (IV)

Here, cross section area of the tube is $A_1$, and the velocity of the fluid inside the tube is $V_1$.

Write the expression for cross section area.

$A_1=\frac{\pi }{4}{d}_1{}^2$ …… (V)

Here, diameter of the tube is $d_1$.

Calculation:

Substitute $26.34\text{slug}/{\text{ft}}^{\text{3}}$ for ${\rho }_{Hg}$, $1.32\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, and $1/12\text{ft}$ for $h$ in Equation (III).

$\begin{array}{c}p=\left(26.34\text{slug}/{\text{ft}}^{\text{3}}-1.32\text{slug}/{\text{ft}}^{\text{3}}\right)\times 32.2\text{ft}/{\text{s}}^{\text{2}}\times 1/12\text{ft}\\ =25.02\text{slug}/{\text{ft}}^{\text{3}}\times 32.2\text{ft}/{\text{s}}^{\text{2}}\times 1/12\text{ft}\times \frac{\text{1}\text{lbf}\cdot {\text{s}}^{\text{2}}/\text{ft}}{1\text{slug}}\\ =\frac{805.644\text{lbf}/{\text{ft}}^{\text{2}}}{12}\\ \approx 67.13\text{lbf}/{\text{ft}}^{\text{2}}\end{array}$

Substitute $67.13\text{lbf}/{\text{ft}}^{\text{2}}$ for $P_2-P_1$, and $1.32\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$.

$\begin{array}{c}{V}_1=\sqrt{2\times \frac{67.13\text{lbf}/{\text{ft}}^{\text{2}}}{1.32\text{slug}/{\text{ft}}^{\text{3}}}}\\ =\sqrt{2\times \left(\frac{67.13\text{lbf}/{\text{ft}}^{\text{2}}}{1.32\text{slug}/{\text{ft}}^{\text{3}}}\right)\left(\frac{1\text{slug}}{1\text{lb}\cdot {\text{fs}}^{\text{2}}/\text{ft}}\right)}\\ =\sqrt{101.71{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ \approx 10.1\text{ft}/\text{s}\end{array}$

Substitute $3/12\text{ft}$ for $d_1$ in Equation (V).

$\begin{array}{c}{A}_1=\frac{\pi }{4}\times {\left(\frac{3}{12}\text{ft}\right)}^2\\ =\frac{0.785}{16}{\text{ft}}^{\text{2}}\\ \approx 0.049{\text{ft}}^{\text{2}}\end{array}$

Substitute $0.049{\text{ft}}^{\text{2}}$ for $A_1$ and $10.1\text{ft}/\text{s}$ for $V_1$ in Equation (IV).

$\begin{array}{c}Q=0.049{\text{ft}}^{\text{2}}\times 10.1\text{ft}/\text{s}\\ =0.4949{\text{ft}}^3/\text{s}\\ \approx 0.5{\text{ft}}^3/\text{s}\end{array}$

Conclusion:

Thus, the volume flow rate in the tube if the fluid in the tube is gasoline is $0.5{\text{ft}}^3/\text{s}$.

(b)

The volume flow rate in the tube if the fluid in the tube is nitrogen.

The volume flow rate in the tube if the fluid in the tube is nitrogen is $12.3{\text{ft}}^3/\text{s}$.

Given information:

The diameter of the tube in which fluid is flowing is $3\text{in}$, the difference in the level of manometric fluid in two limbs of the manometer is $1\text{in}$, the temperature of the fluid is $20\text{°C}$ at a pressure of $1\text{atm}$

Concept Used:

Write the expression for ideal gas equation.

$\begin{array}{l}p=\rho RT\\ \frac{p}{\rho }=RT\\ \rho =\frac{p}{RT}\end{array}$ …… (VI)

Here, the density of the fluid inside the tube is $\rho$, the temperature of the fluid is $T$, the specific gas constant is $R$, and the atmospheric pressure is $p$.

Write the expression for discharge/flow rate.

$Q=A_1{V}_1$

Here, the cross section area of the tube is $A_1$, and velocity of the fluid inside the tube is $V_1$.

Calculation:

Substitute $1\text{atm}$ for $p$, $297\text{J}/\text{kg}\cdot \text{K}$ for $R$ and $293\text{K}$ for $T$.

$\begin{array}{c}\rho =\frac{1\text{atm}}{297\text{J}/\text{kg}\cdot \text{K}\times 293\text{K}}\\ =\left(\frac{1\text{atm}}{297\text{J}/\text{kg}\cdot \text{K}\times 293\text{K}}\right)\left(\frac{101350\text{N}/{\text{m}}^{\text{2}}}{1\text{atm}}\right)\left(\frac{1\text{J}}{1\text{Nm}}\right)\\ =\left(\frac{101350\text{kg}}{87021{\text{m}}^{\text{3}}}\right)\left(\frac{1\text{kg}/{\text{m}}^{\text{3}}}{515.38\text{slug}/{\text{ft}}^{\text{3}}}\right)\\ \approx 0.00226\text{slug}/{\text{ft}}^{\text{3}}\end{array}$

Substitute $26.34\text{slug}/{\text{ft}}^{\text{3}}$ for ${\rho }_{Hg}$, $0.00226\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, and $1/12\text{ft}$ for $h$ in Equation (III).

$\begin{array}{c}p=\left(26.34\text{slug}/{\text{ft}}^{\text{3}}-0.00226\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(1/12\text{ft}\right)\\ =\left(26.337\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(1/12\text{ft}\right)\left(\frac{\text{1}\text{lbf}\cdot {\text{s}}^{\text{2}}/\text{ft}}{1\text{slug}}\right)\\ =\frac{848.05\text{lbf}/{\text{ft}}^{\text{2}}}{12}\\ \approx 70.7\text{lbf}/{\text{ft}}^{\text{2}}\end{array}$

Substitute $70.7\text{lbf}/{\text{ft}}^{\text{2}}$ for $P_2-P_1$, and $0.00226\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$.

$\begin{array}{c}{V}_1=\sqrt{2\times \frac{70.7\text{lbf}/{\text{ft}}^{\text{2}}}{0.00226\text{slug}/{\text{ft}}^{\text{3}}}}\\ =\sqrt{2\times \frac{67.13\text{lbf}/{\text{ft}}^{\text{2}}}{0.00226\text{slug}/{\text{ft}}^{\text{3}}}\times \frac{1\text{slug}}{1\text{lb}\cdot {\text{fs}}^{\text{2}}/\text{ft}}}\\ =\sqrt{62566.37{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ \approx 250.1\text{ft}/\text{s}\end{array}$

Substitute $0.049{\text{ft}}^{\text{2}}$ for $A_1$ and $250.1\text{ft}/\text{s}$ for $V_1$ in Equation (IV).

$\begin{array}{c}Q=0.049{\text{ft}}^{\text{2}}\times 250.1\text{ft}/\text{s}\\ =12.2549{\text{ft}}^3/\text{s}\\ \approx 12.3{\text{ft}}^3/\text{s}\end{array}$

Conclusion:

Thus, the volume flow rate in the tube if the fluid in the tube is nitrogen is $12.3{\text{ft}}^3/\text{s}$.

Explanation of Solution

Given information:

The diameter of the tube in which fluid is flowing is $3\text{in}$, the difference in the level of manometric fluid in two limbs of the manometer is $1\text{in}$, the temperature of the fluid is $20\text{°C}$ at a pressure of $1\text{atm}$.

Concept Used:

The following figure shows the two sections where we apply the Bernoulli equation.

Figure-(1)

Write the expression for Bernoulli equation at section (1) and section (2).

$\frac{P_1}{\rho }+\frac{V_1{}^2}{2}+g{z}_1=\frac{P_2}{\rho }+\frac{V_2{}^2}{2}+g{z}_2$

Substitute $z_1=z_2$ in above equation.

$\begin{array}{l}\frac{P_1}{\rho }+\frac{V_1{}^2}{2}=\frac{P_2}{\rho }+\frac{V_2{}^2}{2}\\ \frac{V_1{}^2}{2}=\frac{V_2{}^2}{2}+\frac{P_2}{\rho }-\frac{P_1}{\rho }\\ V_1=\sqrt{\frac{V_2{}^2}{2}+2\left(\frac{P_2-P_1}{\rho }\right)}\end{array}$ …… (I)

Here, pressure at section (1) is $P_1$, the pressure at section (2) is $P_2$, the velocity of the jet at section (1) is $V_1$, and the velocity of the jet at section (2) is $V_2$.

Substitute $0$ for $V_2$ in Equation (I).

$V_1=\sqrt{2\left(\frac{P_2-P_1}{\rho }\right)}$ …… (II)

Write the expression for finding the pressure difference by using the manometer reading.

$p=\left({\rho }_{Hg}-\rho \right)gh$ …… (III)

Here, the pressure difference between the limbs of manometer is $p$, the acceleration due to gravity is $g$, the difference in the elevation level of the manometric fluid is $h$, the density of the manometric fluid is ${\rho }_{Hg}$.

Write the expression for the flow rate.

$Q=A_1{V}_1$ …… (IV)

Here, cross section area of the tube is $A_1$, and the velocity of the fluid inside the tube is $V_1$.

Write the expression for cross section area.

$A_1=\frac{\pi }{4}{d}_1{}^2$ …… (V)

Here, diameter of the tube is $d_1$.

Calculation:

Substitute $26.34\text{slug}/{\text{ft}}^{\text{3}}$ for ${\rho }_{Hg}$, $1.32\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, and $1/12\text{ft}$ for $h$ in Equation (III).

$\begin{array}{c}p=\left(26.34\text{slug}/{\text{ft}}^{\text{3}}-1.32\text{slug}/{\text{ft}}^{\text{3}}\right)\times 32.2\text{ft}/{\text{s}}^{\text{2}}\times 1/12\text{ft}\\ =25.02\text{slug}/{\text{ft}}^{\text{3}}\times 32.2\text{ft}/{\text{s}}^{\text{2}}\times 1/12\text{ft}\times \frac{\text{1}\text{lbf}\cdot {\text{s}}^{\text{2}}/\text{ft}}{1\text{slug}}\\ =\frac{805.644\text{lbf}/{\text{ft}}^{\text{2}}}{12}\\ \approx 67.13\text{lbf}/{\text{ft}}^{\text{2}}\end{array}$

Substitute $67.13\text{lbf}/{\text{ft}}^{\text{2}}$ for $P_2-P_1$, and $1.32\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$.

$\begin{array}{c}{V}_1=\sqrt{2\times \frac{67.13\text{lbf}/{\text{ft}}^{\text{2}}}{1.32\text{slug}/{\text{ft}}^{\text{3}}}}\\ =\sqrt{2\times \left(\frac{67.13\text{lbf}/{\text{ft}}^{\text{2}}}{1.32\text{slug}/{\text{ft}}^{\text{3}}}\right)\left(\frac{1\text{slug}}{1\text{lb}\cdot {\text{fs}}^{\text{2}}/\text{ft}}\right)}\\ =\sqrt{101.71{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ \approx 10.1\text{ft}/\text{s}\end{array}$

Substitute $3/12\text{ft}$ for $d_1$ in Equation (V).

$\begin{array}{c}{A}_1=\frac{\pi }{4}\times {\left(\frac{3}{12}\text{ft}\right)}^2\\ =\frac{0.785}{16}{\text{ft}}^{\text{2}}\\ \approx 0.049{\text{ft}}^{\text{2}}\end{array}$

Substitute $0.049{\text{ft}}^{\text{2}}$ for $A_1$ and $10.1\text{ft}/\text{s}$ for $V_1$ in Equation (IV).

$\begin{array}{c}Q=0.049{\text{ft}}^{\text{2}}\times 10.1\text{ft}/\text{s}\\ =0.4949{\text{ft}}^3/\text{s}\\ \approx 0.5{\text{ft}}^3/\text{s}\end{array}$

Conclusion:

Thus, the volume flow rate in the tube if the fluid in the tube is gasoline is $0.5{\text{ft}}^3/\text{s}$.

To determine

(b)

The volume flow rate in the tube if the fluid in the tube is nitrogen.

Answer to Problem 3.120P

The volume flow rate in the tube if the fluid in the tube is nitrogen is $12.3{\text{ft}}^3/\text{s}$.

Explanation of Solution

Given information:

The diameter of the tube in which fluid is flowing is $3\text{in}$, the difference in the level of manometric fluid in two limbs of the manometer is $1\text{in}$, the temperature of the fluid is $20\text{°C}$ at a pressure of $1\text{atm}$

Concept Used:

Write the expression for ideal gas equation.

$\begin{array}{l}p=\rho RT\\ \frac{p}{\rho }=RT\\ \rho =\frac{p}{RT}\end{array}$ …… (VI)

Here, the density of the fluid inside the tube is $\rho$, the temperature of the fluid is $T$, the specific gas constant is $R$, and the atmospheric pressure is $p$.

Write the expression for discharge/flow rate.

$Q=A_1{V}_1$

Here, the cross section area of the tube is $A_1$, and velocity of the fluid inside the tube is $V_1$.

Calculation:

Substitute $1\text{atm}$ for $p$, $297\text{J}/\text{kg}\cdot \text{K}$ for $R$ and $293\text{K}$ for $T$.

$\begin{array}{c}\rho =\frac{1\text{atm}}{297\text{J}/\text{kg}\cdot \text{K}\times 293\text{K}}\\ =\left(\frac{1\text{atm}}{297\text{J}/\text{kg}\cdot \text{K}\times 293\text{K}}\right)\left(\frac{101350\text{N}/{\text{m}}^{\text{2}}}{1\text{atm}}\right)\left(\frac{1\text{J}}{1\text{Nm}}\right)\\ =\left(\frac{101350\text{kg}}{87021{\text{m}}^{\text{3}}}\right)\left(\frac{1\text{kg}/{\text{m}}^{\text{3}}}{515.38\text{slug}/{\text{ft}}^{\text{3}}}\right)\\ \approx 0.00226\text{slug}/{\text{ft}}^{\text{3}}\end{array}$

Substitute $26.34\text{slug}/{\text{ft}}^{\text{3}}$ for ${\rho }_{Hg}$, $0.00226\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, and $1/12\text{ft}$ for $h$ in Equation (III).

$\begin{array}{c}p=\left(26.34\text{slug}/{\text{ft}}^{\text{3}}-0.00226\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(1/12\text{ft}\right)\\ =\left(26.337\text{slug}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(1/12\text{ft}\right)\left(\frac{\text{1}\text{lbf}\cdot {\text{s}}^{\text{2}}/\text{ft}}{1\text{slug}}\right)\\ =\frac{848.05\text{lbf}/{\text{ft}}^{\text{2}}}{12}\\ \approx 70.7\text{lbf}/{\text{ft}}^{\text{2}}\end{array}$

Substitute $70.7\text{lbf}/{\text{ft}}^{\text{2}}$ for $P_2-P_1$, and $0.00226\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$.

$\begin{array}{c}{V}_1=\sqrt{2\times \frac{70.7\text{lbf}/{\text{ft}}^{\text{2}}}{0.00226\text{slug}/{\text{ft}}^{\text{3}}}}\\ =\sqrt{2\times \frac{67.13\text{lbf}/{\text{ft}}^{\text{2}}}{0.00226\text{slug}/{\text{ft}}^{\text{3}}}\times \frac{1\text{slug}}{1\text{lb}\cdot {\text{fs}}^{\text{2}}/\text{ft}}}\\ =\sqrt{62566.37{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ \approx 250.1\text{ft}/\text{s}\end{array}$

Substitute $0.049{\text{ft}}^{\text{2}}$ for $A_1$ and $250.1\text{ft}/\text{s}$ for $V_1$ in Equation (IV).

$\begin{array}{c}Q=0.049{\text{ft}}^{\text{2}}\times 250.1\text{ft}/\text{s}\\ =12.2549{\text{ft}}^3/\text{s}\\ \approx 12.3{\text{ft}}^3/\text{s}\end{array}$

Conclusion:

Thus, the volume flow rate in the tube if the fluid in the tube is nitrogen is $12.3{\text{ft}}^3/\text{s}$.