You need to siphon water from a clogged sink. The sink has an area of 0.60 m2 an

Question

You need to siphon water from a clogged sink. The sink has an area of 0.60 m2 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 the figure (Figure 1) . The siphon tube has a diameter of 1.9 cm . Ignore viscosity.

1. The tube in the pail is about 4.0 cm below the surface of the liquid in the pail. 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.

2. Estimate how long it will take to empty the sink.

Explanation / Answer

Solution: From the question we have

The sink has an area of 0.60 m2 and is filled to a height of 4.0 cm

siphon tube rises 45 cm above the bottom of the sink and then descends 85 cm to a pail

The siphon tube has a diameter of 1.9 cm

1. The tube in the pail is about 4.0 cm below the surface of the liquid in the pail. 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.

Set the the water going into the tube as point 1 and the water coming out as point 2. Bernoulli says they must be equal.

z1 + P1/pg + v1^2/2g = z2 + P2/pg + v2^2/2g

( using P for pressure and p for density)
z1 - z2 = 0.59 m
Assume the tubes are barely under water, so
P1 =0
P2 =0
They say that v1 = 0

So what is left
z1 + P1/pg + v1^2/2g = z2 + P2/pg + v2^2/2g
z1 - z2 = v2^2/2g
v^2 = 2g*(z1-z2)

v = sqrt (2g *(z1 - z2))

Notice that this gives the same function as the "exit velocity" in the answer above.

Is is said "Assuming that the water enters the siphon tube with almost zero velocity,". Well the diameter of the tube does not change, so the velocity at any point in the tube must be the same. So if the velocity is almost zero going in then it must be almost zero coming out. They should have never included that sentence. Then you could chose the surface of the sink as point 1 and the end of the hose in the pail as point 2. Anyway, enough ranting.

v = sqrt (2g *(z1 - z2))
v = sqrt (2*9.81 m/s^2(0.59 m))
v = 3.04 m/s answer

2. Estimate how long it will take to empty the sink.
Area of tube (At) = (pi/4)*d^2
Flow rate (Q) = v * At
Flow rate (Q) * time to drain (t) = Volume of sink (VOL)
Time to drain (t) = Volume of sink (VOL) / Flow rate (Q)
VOL = surface area * depth

So t = (surface area * depth) / (v * At)
t = (surface area * depth) / (v * (pi/4)*d^2 )
t = (0.60 m^2 * .04 m) / (3.04 m/s * (pi/4)*(0.019m)^2 )
t = 278.5 sec or 4.61minutes answer

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