Chapter 3, Problem 3.126P

To determine

(a)

The volume flow rate from the tank for no loss.

Answer to Problem 3.126P

The volume flow rate from the tank for no loss is $0.2{\text{ft}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

The liquid in the tank is kerosene at $20\text{°C}$ temperature. The air pressure inside the tank is $P_1=20\text{lbf}/{\text{in}}^{\text{2}}$, and the pressure outside the tank is $P_2=14.7\text{lbf}/{\text{in}}^{\text{2}}$ . Height of the fluid element is $5\text{ft}$, diameter at exit is $1\text{in}$, and no head losses.

Concept Used:

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

Figure-(1)

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

$\frac{P_1}{\rho g}+\frac{V_1{}^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{V_2{}^2}{2g}+z_2+h_f$

Substitute $0$ for $z_2$, and $0$ for $V_1$ and $K\frac{V_2{}^2}{2g}$ for $h_f$ in above equation.

$\begin{array}{l}\frac{P_1}{\rho g}+z_1=\frac{P_2}{\rho g}+\frac{V_2{}^2}{2g}+K\frac{V_2{}^2}{2g}\\ \frac{P_1-P_2}{\rho g}+z_1=\frac{V_2{}^2}{2g}(K+1)\end{array}$ ..... (I)

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

Write the expression for the volume flow rate for the tank.

$Q=A{V}_2$ ..... (II)

Here, area of the hole is $A_{hole}$, velocity of jet is $V_{jet}$.

Write the expression for the area of the hole.

$A_{jet}=\frac{\pi }{4}{d}^2$ ..... (III)

Here, the diameter of the hole is $d$.

Calculation:

Substitute $1\text{in}$ for $d$ in Equation (III).

$\begin{array}{c}{A}_{hole}=\frac{\pi }{4}\times {\left(1\text{in}\right)}^2\\ =\frac{\pi }{4}\times 1{\text{in}}^{\text{2}}\\ =\text{0}\text{.7854}{\text{in}}^{\text{2}}\end{array}$

Refer to table A.3, “Properties of common liquids” to obtain the value of density of kerosene at $1\text{atm}$ pressure and $20\text{°C}$ as $804\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $5\text{ft}$ for $z_1$, $20\text{lbf}/{\text{in}}^{\text{2}}$ for $P_1$, $14.7\text{lbf}/{\text{in}}^{\text{2}}$ for $P_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $0$ for $K$ in Equation (I).

$\begin{array}{l}\frac{20\text{lbf}/{\text{in}}^{\text{2}}-14.7\text{lbf}/{\text{in}}^{\text{2}}}{804\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}+5\text{ft}=\frac{V_2{}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}(0+1)\\ \left(\frac{5.3\text{lbf}/{\text{in}}^{\text{2}}}{7887.24\text{kg}/{\text{m}}^{\text{2}}{\text{×s}}^{\text{2}}}\right)\left(\frac{4.45\text{kg×m}/{\text{s}}^{\text{2}}}{1\text{lbf}}\right)\left(\frac{144{\text{in}}^{\text{2}}}{0.09{\text{m}}^{\text{2}}}\right)+5\text{ft}=\frac{V_2{}^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ \left(\frac{3396.24}{709.85}\text{m}\right)\left(\frac{\text{1}\text{ft}}{.3\text{m}}\right)+5\text{ft}=\left(\frac{V_2{}^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{0.3\text{m}}{1\text{ft}}\right)\\ \frac{V_2{}^2}{65.4\text{ft}/{\text{s}}^{\text{2}}}=15.9\text{ft}+5\text{ft}\end{array}$

$\begin{array}{l}\frac{V_2{}^2}{65.4\text{ft}/{\text{s}}^{\text{2}}}=20.9\text{ft}\\ V_2{}^2=1366.86{\text{ft}}^2/{\text{s}}^{\text{2}}\\ V_2=\sqrt{1366.86{\text{ft}}^2/{\text{s}}^{\text{2}}}\\ V_2=36.9\text{ft}/\text{s}\end{array}$

Substitute $\text{0}\text{.7854}{\text{in}}^{\text{2}}$ for $A$, and $36.9\text{ft}/\text{s}$ for $V_2$ in Equation (II).

$\begin{array}{c}Q=\text{0}\text{.7854}{\text{in}}^{\text{2}}\times 36.9\text{ft}/\text{s}\\ =\left(\text{0}\text{.7854}{\text{in}}^{\text{2}}\right)\left(\frac{1{\text{ft}}^{\text{2}}}{144{\text{in}}^{\text{2}}}\right)\left(36.9\text{ft}/\text{s}\right)\\ \approx 0.2{\text{ft}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

Thus, the volume flow rate from the tank for no loss is $0.2{\text{ft}}^{\text{3}}/\text{s}$.

To determine

(b)

The volume flow rate from the tank for pipe losses.

Answer to Problem 3.126P

The volume flow rate from the tank for pipe losses is $0.085{\text{ft}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

The pipe losses in the form of head losses are given by $h_f=4.5\frac{V_2{}^2}{2g}$.

Concept Used:

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

$\frac{P_1-P_2}{\rho g}+z_1=\frac{V_2{}^2}{2g}(K+1)$

Write the expression for volume flow rate for the tank.

$Q=A{V}_2$

Write the expression for the area of the hole.

$A_{jet}=\frac{\pi }{4}{d}^2$

Calculation:

Substitute $5\text{ft}$ for $z_1$, $20\text{lbf}/{\text{in}}^{\text{2}}$ for $P_1$, $14.7\text{lbf}/{\text{in}}^{\text{2}}$ for $P_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $4.5$ for $K$ in Equation (I).

$\begin{array}{l}\frac{20\text{lbf}/{\text{in}}^{\text{2}}-14.7\text{lbf}/{\text{in}}^{\text{2}}}{804\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}+5\text{ft}=\frac{V_2{}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}(4.5+1)\\ \left(\frac{5.3\text{lbf}/{\text{in}}^{\text{2}}}{7887.24\text{kg}/{\text{m}}^{\text{2}}{\text{×s}}^{\text{2}}}\right)\left(\frac{4.45\text{kg×m}/{\text{s}}^{\text{2}}}{1\text{lbf}}\right)\left(\frac{144{\text{in}}^{\text{2}}}{0.09{\text{m}}^{\text{2}}}\right)+5\text{ft}=\frac{5.5V_2{}^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ \left(\frac{3396.24}{709.85}\text{m}\right)\left(\frac{\text{1}\text{ft}}{.3\text{m}}\right)+5\text{ft}=\left(\frac{5.5V_2{}^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{0.3\text{m}}{1\text{ft}}\right)\\ \frac{V_2{}^2}{11.89\text{ft}/{\text{s}}^{\text{2}}}=15.9\text{ft}+5\text{ft}\end{array}$

$\begin{array}{l}\frac{V_2{}^2}{11.89\text{ft}/{\text{s}}^{\text{2}}}=20.9\text{ft}\\ V_2{}^2=248.501{\text{ft}}^2/{\text{s}}^{\text{2}}\\ V_2=\sqrt{248.501{\text{ft}}^2/{\text{s}}^{\text{2}}}\\ V_2=15.76\text{ft}/\text{s}\end{array}$

Substitute $\text{0}\text{.7854}{\text{in}}^{\text{2}}$ for $A$, and $36.9\text{ft}/\text{s}$ for $V_2$ in Equation (II).

$\begin{array}{c}Q=\text{0}\text{.7854}{\text{in}}^{\text{2}}\times 15.76\text{ft}/\text{s}\\ =\left(\text{0}\text{.7854}{\text{in}}^{\text{2}}\right)\left(\frac{1{\text{ft}}^{\text{2}}}{144{\text{in}}^{\text{2}}}\right)\left(15.76\text{ft}/\text{s}\right)\\ \approx 0.085{\text{ft}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

Thus, the volume flow rate from the tank for pipe losses is $0.085{\text{ft}}^{\text{3}}/\text{s}$.