Flow From a Faucet

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Unit: Fluids

Prerequisites

Bernoulli's Equation

Later Topics

None

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Part 1

A faucet with a circular opening of radius $r_0$ has a flow rate $Q_0$ at its opening. Let $x$ be the vertical distance measured downward from the opening of the faucet. Once the water exits the opening, the stream is in free fall with acceleration $g$ (taking downward as positive for $x$ ).

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Using Newton's second law applied to a fluid element in the stream, determine the pressure difference $\Delta P = P(x) - P_0$ between a point at distance $x$ below the faucet and the opening. Express your answer in terms of $\rho$ , $x$ , and $g$ .

Correct!

Solution:

Once the water exits the faucet, it forms a free-falling stream exposed to the atmosphere on all sides. Since the stream is open to the surrounding air, every point along the falling water column is at atmospheric pressure $P_0$ .

Applying Newton's second law to a short fluid element of length $\Delta x$ and cross-sectional area $A$ , its mass is

m = \rho A\Delta x.

Take downward as the positive $x$ direction. The forces on the element in the $x$ direction are the pressure force on the top face downward, the pressure force on the bottom face upward, and the weight downward:

\sum F_x = P(x)A - P(x+\Delta x)A + \rho A\Delta x\, g.

Because the stream is in free fall, the acceleration is $a_x=g$ , so Newton's second law gives

P(x)A - P(x+\Delta x)A + \rho A\Delta x\, g = (\rho A\Delta x) g.

Canceling the identical weight terms yields

P(x)A - P(x+\Delta x)A = 0,

P(x)=P(x+\Delta x).

Thus the pressure is constant along the free-falling stream. Since the stream is exposed to the atmosphere, that constant pressure is $P_0$ , meaning

P(x)=P_0.

Therefore,

\Delta P = P(x) - P_0 = 0.

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