Chapter 9, Problem 45P

The inside diameters of the larger portions of the horizontal pipe depicted in Figure P9.45 are 2.50 era. Water flows to the right at a rate of 1.80 × 10⁻⁴ m³/s. Determine the inside diameter of the constriction.

Figure P9.45

To determine

The inside diameter of the constriction.

Answer to Problem 45P

The inside diameter of the constriction is $1.47\text{cm}$.

Explanation of Solution

Section1:

To determine: The flow speed of water inside the larger section.

Answer: The flow speed of water inside the larger section is $0.367\text{m/s}$.

Explanation: The flow speed of water inside of the larger section is calculated using continuity equation as follows. $A_1{v}_1=\text{flow}\text{rate}\to v_1=\text{flow}\text{rate}/A_1=4\left(\text{flow}\text{rate}\right)/\pi d_1^2$.

Given info: The flow speed of water is $1.80\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}$ and the inside diameter of the larger section of the pipe is $2.50\text{cm}$.

The formula for the flow speed of water inside the larger section of the water is,

$v_1=\frac{4\left(\text{flow}\text{rate}\right)}{\pi d_1^2}$.

• $d_1$ is inside diameter of the larger section.

Substitute $1.80\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}$ for $\text{flow}\text{rate}$ and $2.50\text{cm}$ for $d_1$ to find $v_1$.

$\begin{array}{c}{v}_1=\frac{4\left(1.80\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}\right)}{\left(3.14\right){\left[\left(2.50\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\right]}^2}\\ =0.367\text{m/s}\end{array}$

Thus, the flow speed of water inside the larger section is $0.367\text{m/s}$.

Section2:

To determine: The cross sectional area of the constriction.

Answer: The cross sectional area of the constriction is $1.70\times {10}^{-4}{\text{m}}^{\text{2}}$.

Explanation: We use Bernoulli’s equation for the flow speed of water in the region of that constriction such that $v_2^2=v_1^2+\left(2/\rho \right)\left(P_1-P_2\right)$ by assuming that $y_1=y_2$. The pressure difference between these two regions would be $P_1-P_2=P_{\text{atm}}+\rho g{h}_1-P_{\text{atm}}+\rho g{h}_2=\rho g\left(h_1-h_2\right)$ which going to be substituted in the expression of the flow speed of the water in constriction region is $v_2=\sqrt{v_1^2+2g\left(h_1-h_2\right)}$.

Now we arrive that the area of constriction is $A_2=\text{flow}\text{rate}/v_2=\text{flow}\text{rate}/\left(\sqrt{v_1^2+2g\left(h_1-h_2\right)}\right)$.

Given info: The flow speed of water in the larger section of the pipe is $0.367\text{m/s}$, acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$, the water level in the left most stand pipe is $10.0\text{cm}$, the water level in the right most stand pipe is $5.00\text{cm}$, and the flow rate is $1.80\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}$.

The formula for the area of constriction is,

$A_2=\frac{\text{flow}\text{rate}}{\sqrt{v_1^2+2g\left(h_1-h_2\right)}}$

• $v_1$ is flow speed of water in the larger section of the pipe.
• $g$ is acceleration due to gravity.
• $h_1$ is water level in the left most stand pipe.
• $h_2$ is water level in the right most stand pipe.

Substitute $1.80\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}$ for $\text{flow}\text{rate}$, $0.367\text{m/s}$ for $v_1$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, $10.0\text{cm}$ for $h_1$, and $5.00\text{cm}$ for $h_2$ to find $A_2$.

$\begin{array}{c}{A}_2=\frac{1.80\times {10}^{-4}{\text{m}}^{\text{3}}\text{/s}}{\sqrt{{\left(0.367\text{m/s}\right)}^2+2\left(9.80{\text{m/s}}^{\text{2}}\right)\left(10.0\text{cm}-5.00\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}}\\ =1.70\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Thus, the cross sectional area of the constriction is $1.70\times {10}^{-4}{\text{m}}^{\text{2}}$.

Section3:

To determine: The inside diameter of the constriction.

Answer: The inside diameter of the constriction is $1.47\text{cm}$.

Explanation: The diameter of the constriction is calculated from the expression of the area of the constriction as follows. $d_2=\sqrt{4A_2/\pi }$.

Given info: The area of the constriction is $1.70\times {10}^{-4}{\text{m}}^{\text{2}}$.

The formula for the diameter of the constriction is,

$d_2=\sqrt{\frac{4A_2}{\pi }}$

• $A_2$ is area of the constriction.

Substitute $1.70\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A_2$ to find $d_2$.

$\begin{array}{c}{d}_2=\sqrt{\frac{4\left(1.70\times {10}^{-4}{\text{m}}^{\text{2}}\right)}{3.14}}\\ =1.47\times {10}^{-2}\text{cm}\\ =1.47\text{cm}\end{array}$

Thus, the inside diameter of the constriction is $1.47\text{cm}$.

Conclusion:

Therefore, the inside diameter of the constriction is $1.47\text{cm}$.