Chapter 3, Problem 3.121P

To determine

(a)

The pressure at section (2).

Answer to Problem 3.121P

The pressure at section (2) is $169354.67\text{Pa}$.

Explanation of Solution

Given information:

The diameter of the tube at inlet and outlet in which the fluid is flowing is $10\text{cm}$, and $6\text{cm}$ respectively, the difference in the level of manometric fluid in two limbs of the manometer is $8\text{cm}$, the temperature of the fluid is $20\text{°C}$, and the pressure at section (1) is $170\times {10}^3\text{Pa}$ . Specific gravity of the red oil is $0.827$ and the fluid inside the tube is ${\text{CO}}_{\text{2}}$.

Concept Used:

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

Figure-(1)

Write the expression for specific gas constant for ${\text{CO}}_{\text{2}}$.

$R=\frac{R\text{'}}{M}$ ..... (I)

Here, universal gas constant is $R\text{'}$, specific gas constant for $C{O}_2$ is $R$, and molecular weight of $C{O}_2$ is $M$.

Write the expression for ideal gas equation.

$\begin{array}{l}{P}_1={\rho }_{C{O}_2}RT\\ \frac{P_1}{{\rho }_{C{O}_2}}=RT\\ {\rho }_{C{O}_2}=\frac{P_1}{RT}\end{array}$ ..... (II)

Here, density of the $C{O}_2$ is ${\rho }_{C{O}_2}$, the temperature of the fluid is $T$, the specific gas constant is $R$, and the atmospheric pressure is $p$.

Write the expression for the hydrostatic formula for section (A) and section (B).

$\begin{array}{l}{P}_1+{\rho }_{C{O}_2}gh=P_2+{\rho }_{Oil}gh\\ P_1-P_2={\rho }_{Oil}gh-{\rho }_{C{O}_2}gh\\ P_1-P_2=\left({\rho }_W\times S{G}_{Oil}-{\rho }_{C{O}_2}\right)gh\end{array}$ ..... (III)

Here, density of water is ${\rho }_W$, the density of $C{O}_2$ is ${\rho }_{C{O}_2}$, the specific gravity of oil is $S{G}_{Oil}$, the acceleration due to gravity is $g$, the pressure at section (A) is $P_1$, the pressure at section (B) is $P_2$, the elevation of section (B) from section (A) is $h$.

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

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

Here, pressure at section (1) is $P_1$, 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 $V_1$ in Equation (I).

$V_2=\sqrt{2\left(\frac{P_1-P_2}{\rho }\right)}$ ..... (IV)

Write the expression for discharge/flow rate.

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

Here, the cross-section area of the tube at exit is $A_2$, and the velocity of the fluid at exit of the tube is $V_2$.

Write the expression for cross section area.

$A_2=\frac{\pi }{4}{d}_2{}^2$ ..... (VI)

Here, diameter of the tube at exit is $d_2$.

Calculation:

Substitute $8314\text{J}/\text{kmol}\cdot \text{°K}$ for $R\text{'}$, $44.01\text{kg}/\text{kmol}$ for $M$ in Equation (I).

$\begin{array}{c}R=\frac{8314\text{J}/\text{kmol}\cdot \text{K}}{44.01\text{kg}/\text{kmol}}\\ \approx 189\text{J}/\text{kg}\cdot \text{K}\end{array}$

Substitute $170\times {10}^3\text{kPa}$ for $P_1$, $189\text{J}/\text{kg}\cdot \text{K}$ for $R$ and $293\text{K}$ for $T$.

$\begin{array}{c}{\rho }_{C{O}_2}=\frac{170\times {10}^3\text{Pa}}{189\text{J}/\text{kg}\cdot \text{K}\times 293\text{K}}\\ =\frac{170000\text{Pa}}{189\text{J}/\text{kg}\cdot \text{K}\times 293\text{K}}\times \frac{101350\text{N}/{\text{m}}^{\text{2}}}{1\text{atm}}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ =\frac{170000\text{kg}}{55377{\text{m}}^{\text{3}}}\\ =3.0698\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $170\times {10}^3\text{Pa}$ for $P_1$, $0.827$ for $S{G}_{Oil}$, $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_W$, $3.0698\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{C{O}_2}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $0.08\text{m}$ for $h$ in Equation (III).

$\begin{array}{l}{P}_1-P_2=\left(998\text{kg}/{\text{m}}^{\text{3}}\times 0.827-3.0698\text{kg}/{\text{m}}^{\text{3}}\right)\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 0.08\text{m}\\ 170\times {10}^3\text{Pa}-P_2=822.276\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 0.08\text{m}\\ 170\times {10}^3\text{Pa}-645.322\text{N}/{\text{m}}^{\text{2}}\times \frac{\text{1}\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}=P_2\\ P_2=169354.67\text{Pa}\end{array}$

Conclusion:

Thus, the pressure at section (2) is $169354.67\text{Pa}$.

To determine

(b)

The volume flow rate of the gas.

Answer to Problem 3.121P

The volume flow rate of the gas is $208.512{\text{m}}^{\text{3}}/\text{h}$.

Explanation of Solution

Given information:

The diameter of the tube in which the 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 $1\text{atm}$

Calculation:

Substitute $170\times {10}^3\text{Pa}$ for $P_1$, $645.322\text{Pa}$ for $P_2$, and $3.0698\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{C{O}_2}$ in Equation (IV).

$\begin{array}{c}{V}_2=\sqrt{2\times \frac{170\times {10}^3\text{Pa}-169354.67\text{Pa}}{3.0698\text{kg}/{\text{m}}^{\text{3}}}\times \frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}}\\ =\sqrt{2\times \frac{645.322\text{N}/{\text{m}}^{\text{2}}}{3.0698\text{kg}/{\text{m}}^{\text{3}}}}\\ =\sqrt{420.43{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ \approx 20.5044\text{m}/\text{s}\end{array}$

Substitute $0.06\text{m}$ for $d_2$ in Equation (VI).

$\begin{array}{c}{A}_2=\frac{\pi }{4}\times {\left(0.06\text{m}\right)}^2\\ =\frac{0.0113}{4}{\text{m}}^{\text{2}}\\ \approx 2.825\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Substitute $2.825\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_2$ and $20.5044\text{m}/\text{s}$ for $V_2$ in Equation (V).

$\begin{array}{c}Q=2.825\times {10}^{-3}{\text{m}}^{\text{2}}\times 20.5044\text{m}/\text{s}\\ =\left(57.92\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\\ \approx 208.512{\text{m}}^{\text{3}}/\text{h}\end{array}$

Conclusion:

Thus, the volume flow rate of the gas is $208.512{\text{m}}^{\text{3}}/\text{h}$.