Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is$\mathbf{0}\mathbf{.}\mathbf{070}{\mathbf{m}}^{\mathbf{2}}$, and the magnitude of the fluid velocity is 3.50m/s . (a) What is the fluid speed at points in the pipe where the cross-sectional area is (a) $\mathbf{0}\mathbf{.}\mathbf{105}{\mathbf{m}}^{\mathbf{2}}$and (b)$\mathbf{0}\mathbf{.}\mathbf{047}{\mathbf{m}}^{\mathbf{2}}$? (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

• Cross-sectional area at point 1,$A_1=0.070m^2$
• Cross-sectional area at point 2,${\mathrm{A}}_2=0.105{\mathrm{m}}^2$
• The velocity of the fluid at point 1,${\mathrm{v}}_1=3.50\mathrm{m}/\mathrm{s}$

Mathematicians use "transport equations" or "continuity equations" to describe the motion of a given quantity in terms of time.

When dealing with huge amounts, it is best to use a method like the preserved quantity since it is both straightforward and effective.

$A_1{v}_1=A_2{v}_2$………………(1)

Where A and v are the area of the cross-section of pipe and v is the velocity of the fluid.

Evaluate fluid speed at cross-section using equation (1),

$$\begin{gathered} v_2=\left(\frac{\left(0.70m^2\right)\left(3.50\mathrm{m}/\mathrm{s}\right)}{0.105{\mathrm{m}}^2}\right) \\ =2.34\mathrm{m}/\mathrm{s} \end{gathered}$$

Evaluate fluid speed at cross-section using equation (1),

To evaluate discharge, we rearrange equation (1) as,

$$\begin{gathered} Rv=A_1{v}_1 \\ Rv=\frac{V}{1}=A_1{v}_1 \\ V=A_1{v}_{1}t \end{gathered}$$

Substitute the values in the above equation,

$$\begin{gathered} V=\left(0.70m^2\right)\left(3.50\mathrm{m}/\mathrm{s}\right)\left(1.00\mathrm{hr}\right)\left(\frac{3600\mathrm{s}}{1\mathrm{hr}}\right) \\ =882{\mathrm{m}}^3 \end{gathered}$$

Thus, the volume of water discharged from the pipe is $882{\mathrm{m}}^3$.