Chapter 10, Problem 84GP

(a)

To determine

To Find: The velocity of water.

Answer to Problem 84GP

The velocity of water when it enters the pail is $3.25\text{ m/s}$ .

Explanation of Solution

Given:

Area $A=0.48{\text{ m}}^{\text{2}}$ .

  $\Delta h=Z_{\text{sink}}-Z_{\text{pail}}=100-50+4=54\text{ cm = }0.54\text{ m}$

Diameter of siphon tube $d=2\text{ cm = }0.02\text{ m}$ .

Radius of siphon tube $r=\frac{d}{2}=1\text{ cm }=0.01\text{ m}$ .

  $\begin{array}{l}{v}_{\text{sink}}=0\\ P_{\text{sink}}=P_{\text{pail}}\end{array}$

Formula used:

To determine the velocity of water when it enters the pail Bernoulli's equation is used,

  $\frac{P_{\text{sink}}}{\rho \times g}+\frac{v_{\text{sink}}^2}{2\times g}+Z_{\text{sink}}=\frac{P_{\text{pail}}}{\rho \times g}+\frac{v_{\text{pail}}^{\text{2}}}{2\times g}+Z_{\text{pail}}$

Here.

  $P$ is the pressure.

  $v$ is velocity of fluid.

  $Z$ is the distance from the reference.

  $\rho$ is density of fluid.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{Z}_{\text{sink}}=\frac{v_{\text{pail}}^2}{2g}+Z_{\text{pail}}\\ \frac{v_{\text{pail}}^2}{2g}=Z_{\text{sink}}-Z_{\text{pail}}\\ v_{\text{pail}}^{\text{2}}=2g\times \left(Z_{\text{sink}}-Z_{\text{pail}}\right)=2\times 9.8\times 0.54=10.584\\ v_{\text{pail}}=3.25\text{ m/s}\end{array}$

Conclusion:

Thus, the velocity of water when it enters the pail is $3.25\text{ m/s}$ .

(b)

To determine

To Find: The time required.

Answer to Problem 84GP

The time required to empty the sink is $18.8\text{ s}$ .

Explanation of Solution

Given:

Area $A=0.48{\text{ m}}^{\text{2}}$ .

Height $h=4\text{ cm }=0.04\text{ m}$ .

Diameter of siphon tube $d=2\text{ cm = }0.02\text{ m}$ .

Radius of siphon tube $r=\frac{d}{2}=1\text{ cm }=0.01\text{ m}$ .

Formula used:

Time required to empty the sink is calculated from the rate of flow and continuity equation,

  $\begin{array}{l}{Q}_{\text{pail}}={\left(A\times v\right)}_{\text{pail}}=\frac{V_{\text{sink}}}{t}\\ \text{Volume of sink, }{V}_{\text{sink}}=A\times h\end{array}$

Here,

  $A$ is the area through which fluid is flowing.

  $v$ is velocity of fluid.

  $V$ is volume of fluid.

  $h$ is the height.

Calculation:

Write the expression for volume.

  $V_{\text{sink}}=0.48\times 0.04=0.0192{\text{ m}}^{\text{3}}$

Write the expression for discharge.

  $Q_{\text{pail}}={\left(A\times v\right)}_{\text{pail}}=\frac{V_{\text{sink}}}{t}$

Calculate the time.

  $\begin{array}{l}\text{Time }t=\frac{V_{\text{sink}}}{{\left(A\times v\right)}_{\text{pail}}}=\frac{0.0192}{\pi \times r^2\times v_{\text{pail}}}=\frac{0.0192}{\pi \times {0.01}^2\times 3.25}\\ \text{Time }t=18.8\text{ s}\end{array}$

Conclusion:

Thus, the time required to empty the sink is $18.8\text{ s}$ .