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Chapter 14: Problem 38

A horizontal pipe \(10.0 \mathrm{cm}\) in diameter has a smooth reduction to a pipe \(5.00 \mathrm{cm}\) in diameter. If the pressure of the water in the larger pipe is \(8.00 \times 10^{4} \mathrm{Pa}\) and the pressure in the smaller pipe is \(6.00 \times 10^{4} \mathrm{Pa},\) at what rate does water flow through the pipes?

Short Answer

Expert verified
Using the calculated velocity and the cross-sectional area of the smaller pipe, the flow rate of the water through the pipes can be determined.

Step by step solution

01

Setup Bernoulli's equation

Apply Bernoulli's equation which is given as \(P + \frac{1}{2} \rho v^{2} = constant\) where \(P\) is the pressure, \(\rho\) is the density of the fluid and \(v\) is the velocity of fluid. Since the fluid is incompressible, \(\rho\) remains constant. Therefore we can write Bernoulli's equation for two points 1 and 2 in the fluid in the following way: \(P_1 + \frac{1}{2} \rho v_1^{2} = P_2 + \frac{1}{2} \rho v_2^{2}\). The pressure in the larger pipe is given as \(P_1 = 8.00 \times 10^{4} \mathrm{Pa}\) and the pressure in the smaller pipe is given as \(P_2 = 6.00 \times 10^{4} \mathrm{Pa}\). The velocity of the fluid in the large pipe is what we want to find.
02

Setup the equation of continuity

The equation of continuity states that the mass flow rate must be constant throughout the pipe, since no fluid is being lost or gained. This can be written in terms of the areas of the two pipes (\(A_1\) and \(A_2\)) and the velocities through them (\(v_1\) and \(v_2\)), as \(A_1 v_1 = A_2 v_2\). The areas can be expressed as \(A=\pi r^{2}\), where \(r\) is the radius of the pipe. So the equation of continuity now becomes \(v_1 = \frac{A_2}{A_1} v_2 = \left(\frac{r_2}{r_1}\right)^2 v_2\). The diameters of the pipes are given, therefore, we can find the radii as \(r_1 = 5 \mathrm{cm}\) and \(r_2 = 2.5 \mathrm{cm}\). Replace \(v_1\) in Bernoulli's equation from the equation of continuity.
03

Solve for the velocity in the smaller pipe

Substitute the known values including the densities into Bernoulli's equation and solve for the unknown velocity \(v_2\). The density of water is typically \(\rho = 1000 \mathrm{kg/m^3}\). After substituting all the values and solving the equation, we should have the velocity of the fluid in the smaller pipe.
04

Determine the rate of water flow through the pipes

The flow rate of the water can be calculated using the product of the cross-sectional area of the pipe and the velocity of the fluid. This can be expressed as \(Q=A_2 v_2\), where \(Q\) is the flow rate, \(A_2\) is the cross-sectional area of the pipe, and \(v_2\) is the velocity of the fluid. The cross-sectional area \(A_2\) can be calculated using the formula \(\pi r^2\) with \(r\) as the radius of the smaller pipe. After substituting the values of \(A_2\) and \(v_2\) into the equation, we can solve for \(Q\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fluid Mechanics
Fluid mechanics is the branch of physics concerned with the behavior of fluids (liquids, gases, and plasmas) and the forces on them. At the core of fluid mechanics are the laws that govern the statics and dynamics of fluids. In practice, this field applies to numerous aspects of our daily lives, including water flow in pipes, the lift of an airplane wing, and weather patterns in the atmosphere.

Concerning this exercise, the aspect of fluid mechanics in focus is the flow of an incompressible fluid鈥攚ater鈥攖hrough a constricted pipe. The principles guiding the behavior of the water in this context are encapsulated by Bernoulli's principle and the equation of continuity. These principles assert that, within a closed system or a streamline flow, certain properties of the fluid remain consistent. For instance, changes in the pipe's cross-sectional areas are compensated by changes in the fluid's pressure and velocity, ensuring the flow rate remains constant.
Equation of Continuity
The equation of continuity is a fundamental principle of fluid mechanics stating that for any incompressible fluid, the flow rate must remain constant from one cross-section of a pipe to another. This is because the mass of fluid entering a system must equal the mass leaving it over any period, assuming there's no accumulation of mass within the system.

Based on the concept of conservation of mass, the equation of continuity can be mathematically represented as \( A_1v_1 = A_2v_2 \), where \( A \) is the cross-sectional area and \( v \) is the velocity of the fluid at two different points 1 and 2 along the pipe. If the area decreases, as with the narrowing pipe in the exercise, the fluid's velocity must increase to maintain a constant flow rate. This relationship is crucial for understanding how changes in a pipe's diameter affect a fluid's flow properties.
Flow Rate
Flow rate, often symbolized by \( Q \), is a measure of the volume of fluid that passes through a given surface per unit time. It's a crucial concept in various engineering and scientific disciplines where fluid behavior is studied or utilized. Flow rate is directly relevant in systems like pipelines, rivers, and even in the circulatory system of living organisms.

In the exercise, the flow rate is determined using the area of the pipe and the fluid velocity at that point. It's commonly expressed in cubic meters per second (m鲁/s) or liters per minute (L/min). For a pipe with a circular cross-section, the flow rate is calculated with \( Q = A_2v_2 \), where \( A \) represents the cross-sectional area and \( v \) is the fluid velocity. The area \( A \) can be found using the formula \( \(A=\) \(\pi r^2\) \) when the pipe diameter is known. Analyzing flow rate helps in designing systems that optimize fluid transport while avoiding issues like pipe erosion or bursts due to high velocity.

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Most popular questions from this chapter

A thin spherical shell of mass \(4.00 \mathrm{kg}\) and diameter \(0.200 \mathrm{m}\) is filled with helium (density \(=0.180 \mathrm{kg} / \mathrm{m}^{3}\) ). It is then released from rest on the bottom of a pool of water that is \(4.00 \mathrm{m}\) deep. (a) Neglecting frictional effects, show that the shell rises with constant acceleration and determine the value of that acceleration. (b) How long will it take for the top of the shell to reach the water surface?

In \(1983,\) the United States began coining the cent piece out of copper-clad rinc rather than pure copper. The mass of the old copper penny is \(3.083 \mathrm{g},\) while that of the new cent is 2.517 g. Calculate the percentage of sinc (by volume) in the new cent. The density of copper is \(8.960 \mathrm{g} / \mathrm{cm}^{3}\) and that of zinc is \(7.133 \mathrm{g} / \mathrm{cm}^{3} .\) The new and old coins have the same volume.

Review problem. A uniform disk of mass \(10.0 \mathrm{kg}\) and \(\mathrm{ra}\) dius \(0.250 \mathrm{m}\) spins at 300 rev/min on a low-friction axle. It must be brought to a stop in 1.00 min by a brake pad that makes contact with the disk at average distance \(0.220 \mathrm{m}\) from the axis. The coefficient of friction between pad and disk is \(0.500 .\) A piston in a cylinder of diameter \(5.00 \mathrm{cm}\) presses the brake pad against the disk. Find the pressure required for the brake fluid in the cylinder.

The four tires of an automobile are inflated to a gauge pressure of \(200 \mathrm{kPa}\). Each tire has an area of \(0.0240 \mathrm{m}^{2}\) in contact with the ground. Determine the weight of the automobile.

A swimming pool has dimensions \(30.0 \mathrm{m} \times 10.0 \mathrm{m}\) and a flat bottom. When the pool is filled to a depth of \(2.00 \mathrm{m}\) with fresh water, what is the force caused by the water on the bottom? On each end? On each side?
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