Question

Water runs into a fountain, filling all the pipes, at a steady rate of $\mathbf{0}\mathbf{.}\mathbf{750}{\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{s}$. (a) How fast will it shoot out of a hole 4.50 cm in diameter? (b) At what speed will it shoot out if the diameter of the hole is three times as large?

Short answer

The speed of water when the diameter is 4.50 cm in diameter is,$V_1=471.4\mathrm{m}/\mathrm{s}$

The speed of water when the diameter is three times as large is,$V_2=52.4\mathrm{m}/\mathrm{s}$

Step-by-step solution

Identification of given data

• The rate of water flow is,$Q=0.750{\mathrm{m}}^3/\mathrm{s}$
• The diameter of hole is,$d=4.50\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=4.50\times {10}^{-2}\mathrm{m}$
• Therefore the radius of hole is,$r=\frac{d}{2}=2.25\times {10}^{-2}\mathrm{m}$

Significance of volume flow rate.

Volume flow rate is the product of flow velocity and cross-sectional vector area.

Volume flow rate can be given as,

Q=vA …………..(1)

Where, v and A are the velocity and the area of the cross-section.

Determine speed when diameter is 4.50 cm

The cross sectional area of the pipe can be evaluated by,

$$\begin{gathered} A_1={\mathrm{\pi r}}^2 \\ =\mathrm{\pi }\times {\left(2.25\times {10}^{-2}\mathrm{m}\right)}^2 \\ =1.59\times {10}^{-3}{\mathrm{m}}^2 \end{gathered}$$

The volume flow rate can be evaluated by using equation (1),

$v\frac{Q}{A}$

Substittute the values in above expression will give,

$$\begin{gathered} {\mathrm{V}}_1=\frac{0.750{\mathrm{m}}^3/\mathrm{s}}{1.591\times {10}^{-3}{\mathrm{m}}^2} \\ =471.4\mathrm{m}/\mathrm{s} \end{gathered}$$

The speed of water when the diameter is 4.50 cm in diameter is,$v_1=471.4m/s$