Question

You need to siphon water from a clogged sink. The sink has an area of and is filled to a height of 4.0 cm. Your siphon tube rises 45 cm above the bottom of the sink and then descends 85 cm to a pail as shown in Fig. 10–59. The siphon tube has a diameter of 2.3 cm. (a) Assuming that the water level in the sink has almost zero velocity, use Bernoulli’s equation to estimate the water velocity when it enters the pail. (b) Estimate how long it will take to empty the sink. Ignore viscosity.

Short answer

(a) The water velocity when the siphon enters into the pail is\(2.66\;\;{\rm{m/s}}\).

(b) The time taken to empty the sink is \(13.75\;\;{\rm{s}}\).

Step-by-step solution

(a) Expression of Bernoulli’s equation and application to find the velocity.

Bernoulli's equationis given by,

\(\frac{P}{{\rho g}} + \frac{{{v^2}}}{{2g}} + h = {\rm{ constant }}\)

Here\(\frac{P}{{\rho g}}\)is the pressure head,\(\frac{{{v^2}}}{{2g}}\)is the velocity head, and\(h\)is the height.

Applying Bernoulli's equation at the endpoints of the siphon tube, we get

\(\frac{{{P_1}}}{{\rho g}} + \frac{{v_1^2}}{{2g}} + {h_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{v_2^2}}{{2g}} + {h_2}\)

Here, subscript 1 denotes the position at the base level of the clogged sink, and subscript 2 represents the position at the end of the siphon tube inside the pail. Since the pressure heads at both the ends of the siphon are equal to atmospheric pressure, Bernoulli's equation becomes,

\(\frac{{v_1^2}}{{2g}} + {h_1} = \frac{{v_2^2}}{{2g}} + {h_2}\)

The velocity inside the sink can be assumed to be zero. Therefore,\({v_1} = 0\).Now, Bernoulli's equation becomes,

\(\begin{array}{c}\frac{{v_2^2}}{{2g}}{\rm{ }} = {h_1} - {h_2}\\ = \Delta h\end{array}\)

Solving for\({v_2}\), we get

\({v_2} = \sqrt {2g\Delta h} \)

The change in height is given by,

\(\Delta h{\rm{ }} = 85\;\;{\rm{cm}} - (45\;\;{\rm{cm}} + 4\;\;{\rm{cm}}) = 36\;\;{\rm{cm}}\)

Substituting,\(36\;\;{\rm{cm}}\)for\(\Delta h\)and\(9.8\;\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\)for\(g\)in the above expression,

\(\begin{array}{c}{v_2}{\rm{ }} = \sqrt {2\left( {9.8\;\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)(36\;\;{\rm{cm}})\left( {\frac{{1\;\;{\rm{m}}}}{{100\;\;{\rm{cm}}}}} \right)} \\ = \sqrt {7.056\;\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\ = 2.66\;\;{\rm{m/s}}\end{array}\)

Therefore, the water velocity when the siphon enters into the pail is \(2.66\;\;{\rm{m/s}}\).

(b) Calculating the time taken.

The time taken to empty the tank is given by,

\(\begin{array}{c}{\rm{Time }} = \frac{V}{Q}\\ = \frac{{{A_1}{h_1}}}{{{A_s}{v_s}}}\\ = \frac{{{A_1}{h_1}}}{{\left( {\frac{\pi }{4}{d^2}} \right){v_s}}}\end{array}\)

Here\(V\)is the volume of the liquid in the tank,\(Q\)is the volumetric flow rate,\({A_1}\)is the area of the tank,\({h_1}\)is the liquid height,\({A_s}\)is the area of the siphon tube, and\({v_s}\)is the velocity of the liquid inside the siphon tube.

Substituting,\(0.38\;\;{{\rm{m}}^2}\)for\({A_1}\), \(0.04\;{\rm{m}}\)for\({h_1}\), \(0.023\;\;{\rm{m}}\)for\(d\), and\(2.66\;\;{\rm{m/s}}\)for\({v_{\rm{s}}}\)in the above expression, we get

\(\begin{array}{c}{\rm{ Time }} = \;\frac{{\left( {0.38\;\;{{\rm{m}}^2}} \right)(0.04\;\;{\rm{m}})}}{{\left( {\frac{\pi }{4}{{(0.023\;\;{\rm{m}})}^2}} \right)(2.66\;\;{\rm{m/s}})}}\\ = \;13.75\;\;{\rm{s}}\end{array}\)

Therefore, it will take \(13.75\;\;{\rm{s}}\) to empty the sink.