You need to siphon water from a clogged sink. The sink has an area of 0.40 m 2 a

Question

You need to siphon water from a clogged sink. The sink has an area of 0.40m2 and is filled to a height of 4.0 cm. Your siphon tube rises 50 cm above the bottom of the sink and then descends 100 cm to a pail as shown in the figure(Figure 1) . The siphon tube has a diameter of 1.8cm

1. Assuming that the tube is narrow compared to the sink, calculate the velocity of the water when it enters the pail. The tube in the pail is about 4.0 cm

below the surface of the liquid in the pail.

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

Explanation / Answer

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

(I am using P for pressure and p for density)

z1 - z2 = 0.54 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.

This is the part I don't like. They say "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))

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 )

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