Question

Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in Figure P14.42, the pressure is ${\mathit{P}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.75}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\text{Pa}}$and the pipe diameter is${\mathit{P}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.20}\mathbf{\times }\mathbf{1}{\mathbf{\text{0}}}^{\mathbf{\text{4}}}\mathbf{\text{Pa}}$. At another point$\mathit{y}\mathbf{=}\mathbf{2}\mathbf{.5}\mathbf{\text{0m}}$higher, the pressure is$\mathbf{\text{6.00cm}}$and the pipe diameter is$\mathbf{\text{3.00cm}}$. Find the speed of flow (a) in the lower section and (b) in the upper section. (c) Find the volume flow rate through the pipe.

Short answer

(a) The speed of flow in the lower section is $\text{0.638}\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

(b) The speed of flow in the upper section is $\text{2.55}\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

(c) The volume flow rate through the pipe is $1.80\times {10}^{-3}\raisebox{1ex}{${\displaystyle {\text{m}}^{\text{3}}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

Step-by-step solution

A concept:

The flow rate (volume flux) through a pipe that varies in cross-sectional area is constant; that is equivalent to stating that the product of the cross-sectional area$\mathit{A}$and the speed$\mathit{v}$at any point is a constant.

This result is expressed in the equation of continuity for fluids:

${\mathit{A}}_{\mathbf{1}}{\mathit{V}}_{\mathbf{1}}\mathbf{=}{\mathit{A}}_{\mathbf{2}}{\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathbf{\text{constant}}$

Where, role="math" localid="1663767302281" ${\mathit{A}}_{\mathbf{1}}$is the initial area and role="math" localid="1663767464748" ${\mathit{A}}_{\mathbf{2}}$is the final area, role="math" localid="1663767407393" ${\mathit{V}}_{\mathbf{1}}$initial volume and${\mathit{V}}_{\mathbf{2}}$final volume.

(a) Determine the speed of flow in the lower section:

The mass flow rate and the volume flow rate are constant:

$$\begin{gathered} \rho A_1{V}_1=\rho A_2{V}_2 \\ \pi r_1^2{v}_1=\pi r_2^2{v}_2 \end{gathered}$$

Substituting, the values of radii, you get

$$\begin{gathered} {(3.00cm)}^2{v}_1={(1.50cm)}^2{v}_2 \\ {v}_2=4v_1 \end{gathered}$$

For ideal flow is define by,

$P_1+\rho g{y}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{y}_2+\frac{1}{2}\rho v_2^2$

$\begin{array}{rcl}1.75\times {10}^{\text{4}}\text{Pa}+0+\frac{1}{2}\left(1000\raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right){\left(v_1\right)}^2& =& 1.20\times {10}^4\text{Pa}+\left(1000\raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right)\left(9.8\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\right)\left(2.50\text{m}\right)+\\ & & \frac{1}{2}\left(1000\raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right){\left(v_2\right)}^2\end{array}$

$\begin{array}{rcl}\left(500\raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right){\left(4v_1\right)}^2-5\left(00\raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right){\left(v_1\right)}^2& =& 1.75\times {10}^{\text{4}}\text{Pa}-1.20\times {10}^4\text{Pa}\\ & & -\left(1000\raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right)\left(9.8\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\right)\left(2.50\text{m}\right)\end{array}$

Solving for $v_1$gives that gives,

$\begin{array}{rcl}{v}_1& =& \sqrt{\frac{3050{\displaystyle \text{Pa}}}{7500{\displaystyle \raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.}}}\\ & =& 0.638\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

(b) The speed of flow in the upper section:

From part (a), you have

$\begin{array}{rcl}{V}_2& =& 4V_1\\ & =& 4\left(0.638\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)\\ & =& 2.55\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

(c) The volume flow rate through the pipe:

The volume flow rate is,

$\begin{array}{rcl}\pi r_1^2{v}_1& =& 3.14\times {\left(0.030\text{0m}\right)}^2\left(0.638\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)\\ & =& 1.80\times {10}^{-3}\raisebox{1ex}{${\text{m}}^{\text{3}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

Hence, the volume flow rate through the pipe is $\begin{array}{rcl}& & 1.80\times {10}^{-3}\raisebox{1ex}{${\text{m}}^{\text{3}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$.