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Chapter 6: Problem 74

Water flows out of a kitchen faucet of \(1.25 \mathrm{cm}\) diameter at the rate of \(0.1 \mathrm{L} / \mathrm{s}\). The bottom of the sink is \(45 \mathrm{cm}\) below the faucet outlet. Will the cross-sectional area of the fluid stream increase, decrease, or remain constant between the faucet outlet and the bottom of the sink? Explain briefly. Obtain an expression for the stream cross section as a function of distance \(y\) above the sink bottom. If a plate is held directly under the faucet, how will the force required to hold the plate in a horizontal position vary with height above the sink? Explain briefly.

Short Answer

Expert verified
The cross-sectional area of the fluid stream will decrease as you move down along the sink, and can be modeled with \(A_2 = \pi \left( \frac{d}{2} \right)^2 \times \frac{Q}{\sqrt{(2g(45-y)/10^3)} }\). The force required to hold the plate in a horizontal position will increase as the plate moves down along the sink.

Step by step solution

01

Understand the Conservation of Mass Principle for Fluid Flow

The conservation of mass principle applied to fluid flow states that the volume flow rate (i.e., the product of velocity and cross-sectional area) is constant, providing there are no leaks or sources along the fluid stream. This principle is applied here to determine the change in cross-sectional area of the stream as we move down the sink.
02

Set up an Expression for Cross-sectional Area

Using the conservation of mass principle, when fluid is within a free fall, the equation to follow is \(A_1v_1 = A_2v_2\) where \(A_1\) and \(A_2\) are the cross-sectional areas at the faucet and at a distance \(y\) from the sink bottom respectively, and \(v_1\) and \(v_2\) are the velocities at these points. Since the volume flow rate at the faucet (where the velocity is \(v_1\)) is given as \(0.1 L/s = 0.1 \times 10^{-3} m^3/s\), we can express \(v_1\) as \(Q/A_1\), where Q is the volume flow rate and \(A_1\) is the cross-sectional area at the faucet, given by \(\pi (d/2)^2\), with d being the faucet diameter.
03

Obtain an Expression for the Velocity \(v_2\)

The velocity at a distance \(y\) from the bottom, \(v_2\), is governed by the principle of conservation of mechanical energy which states that the sum of the kinetic and potential energies per unit volume of a fluid is constant along a streamline in the absence of viscous losses. Since the faucet is exposed to atmosphere and also the bottom of sink is open to atmospheric pressure, \(v_1 = \sqrt{(2g(45-y)/10^3)}\), where g is gravitational acceleration (9.81 \(m/s^2\)).
04

Determine the Cross-sectional Area \(A_2\)

Substituting the expressions for \(v_1\) and \(v_2\) from Steps 2 and 3 respectively into the conservation of mass equation, we obtain \(A_2\) as a function of \(y\) from the sink bottom. Hence \(A_2 = A_1 \times \frac{v_1}{v_2} = \pi \left( \frac{d}{2} \right)^2 \times \frac{Q}{\sqrt{(2g(45-y)/10^3)} }\). This implies that the cross-sectional area of the fluid stream decreases as you move down along the sink, as \(y\) increases.
05

Analyzing the Force on the Plate

The force required to hold the plate in a horizontal position under the faucet is the force exerted by the water hitting the plate. This force is equal to the rate of change of momentum of the water, which can be given as \(F = mv\), where m is the rate of fluid mass hitting the plate and v is the fluid velocity. The rate of fluid mass hitting the plate, or mass flow rate, is constant (as it equals the rate water flows out of the faucet). As you move the plate down the sink, velocity increases, hence the force required to hold the plate in a horizontal position increases as you move the plate down along the sink.

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Most popular questions from this chapter

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