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A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 \(\mathrm{cm}^{3} / \mathrm{s}\) . At one point in the pipe, where the radius is \(4.00 \mathrm{cm},\) the water's absolute pressure is \(2.40 \times 10^{5}\) Pa. At a second point in the pipe, the water passes through a constriction where the radius is 2.00 \(\mathrm{cm} .\) What is the water's absolute pressure as it flows through this constriction?

Short Answer

Expert verified
The absolute pressure in the constricted section is 1.92脳10鈦 Pa.

Step by step solution

01

Understand the Given Information

We have a pipe with two different sections. The first section has a radius of 4.00 cm and a water flow rate of 7200 cm鲁/s while the absolute pressure is 2.40脳10鈦 Pa. The second section has a radius of 2.00 cm, and we need to find the pressure at this point.
02

Calculate the Flow Velocity in the First Section

Using the formula for volumetric flow rate, \(Q = A \cdot v\), where \(A\) is the cross-sectional area and \(v\) is the velocity, we can calculate the velocity in the first section. The area \(A_1 = \pi \cdot (4.00)^2 \). Therefore, \(v_1 = \frac{Q}{A_1} = \frac{7200}{\pi \cdot (4.00)^2} \). Calculate \(v_1\).
03

Calculate the Flow Velocity in the Second Section

The flow rate \( Q \) remains constant, so use the same formula \(Q = A \cdot v\). The area \(A_2 = \pi \cdot (2.00)^2\). Therefore, \(v_2 = \frac{7200}{\pi \cdot (2.00)^2} \). Calculate \(v_2\).
04

Apply Bernoulli's Equation

Now, apply Bernoulli鈥檚 equation, which is useful for incompressible fluids: \(P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2\), where \(P_1\) is the pressure in the first section, and \(P_2\) is the pressure in the second section. Solve for \(P_2\) using the known \(P_1\), \(v_1\), and \(v_2\). Assume water density, \(\rho = 1000\, \mathrm{kg/m^3}\).
05

Solve the Equation for P_2

Substitute the computed velocities and the known pressures into Bernoulli's equation to solve for \(P_2\). Since the pipe is horizontal, the height terms cancel. Rearrange to find \(P_2:\)\[P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2)\].Compute \(P_2\) using the given pressure \(P_1\), and calculated \(v_1\) and \(v_2\).

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is a branch of physics that focuses on the movement of liquids and gases. Understanding how fluids behave when in motion helps solve practical problems in many fields, such as engineering and meteorology. It involves various principles and equations, with Bernoulli's Principle being a key player. Bernoulli's Principle is especially important for understanding how pressure changes in moving fluids. This principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or potential energy. This is vital when analyzing problems like the one we're solving for the sprinkler system. Fluid dynamics helps us calculate the changes in pressure as water flows from one section of the pipe to another.
Volumetric Flow Rate
Volumetric flow rate, often denoted as \( Q \), is the quantity of three-dimensional space that a fluid occupies as it moves through a given area per unit time. In other words, it tells us how much fluid is passing through a section of a pipe or channel at any given moment. In the case of the sprinkler system, the given volumetric flow rate of 7200 \( \mathrm{cm}^3/\mathrm{s} \) lets us calculate the speed of the water in different sections of the pipe. The flow rate remains constant for incompressible fluids, which ensures calculations are consistent across the varying pipe diameters. Calculating the velocity involves using the cross-sectional area of the pipe and applying the formula \( Q = A \cdot v \), where \( A \) represents the area and \( v \) the velocity.
Pressure Calculation
Pressure calculation within fluid systems helps determine the force exerted by the fluid per unit area. In the sprinkler problem, we need to calculate the pressure difference between two points inside the pipe. This involves using Bernoulli's equation, \[P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2\], where \( P_1 \) and \( P_2 \) are the pressures at the first and second points respectively, and \( v_1 \) and \( v_2 \) are the velocities. By knowing the pressure at the first point and calculating the velocities, we can solve for \( P_2 \), the pressure at the narrowest part of the pipe. This approach is essential in designing systems like sprinklers, where consistent water pressure ensures effective operation.
Sprinkler System
A sprinkler system is an effective way to distribute water across a wide area, ideal for uses like golf courses and agriculture. These systems depend crucially on understanding fluid flow and pressure changes. In our problem, water is transported through pipes with varying diameters, requiring precise calculations to ensure uniform spraying. Different segments of the pipe will influence the speed and pressure of the water, which ultimately affects how it is dispersed to the surrounding area. By applying Bernoulli鈥檚 Principle, we can design sprinkler systems that balance pressure and speed, ensuring even coverage across the landscape.
Incompressible Fluid Flow
Incompressible fluid flow refers to the behavior of a fluid whose density remains constant throughout its motion. Most liquids, like water, are considered incompressible under normal conditions. This characteristic simplifies calculations in physics and engineering since the mass flow rate is conserved. In the context of our sprinkler problem, this assumption means that the volumetric flow rate stays the same across different sections of the pipe. By treating the fluid as incompressible, we straightforwardly apply Bernoulli鈥檚 equation, facilitating the calculation of pressures and velocities through the system. This allows for a precise understanding of how the fluid will behave as it travels from one point to another in the pipe, which is useful for ensuring an efficient and effective sprinkler system.

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Most popular questions from this chapter

An incompressible fluid with density \(\rho\) is in a horizontal test tube of inner cross-sectional area \(A .\) The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed \omega. Gravitational forces are negligible. Consider a volume element of the fluid of area \(A\) and thickness \(d r^{\prime}\) a distance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho \omega^{2} r^{\prime} d r^{\prime}\) . (b) If the surface of the fluid is at a radius \(r_{0}\) where the pressure is \(p_{0},\) show that the pressure \(p\) at a distance \(r \geq r_{0}\) is \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .\) (c) An object of volume \(V\) and density \(\rho_{\mathrm{ob}}\) has its center of mass at a distance \(R_{\mathrm{cmob}}\) from the axis. Show that the net horizontal force on the object is \(\rho V \omega^{2} R_{\mathrm{cm}},\) where \(R_{\mathrm{cm}}\) is the distance from the axis to the center of mass of the displaced fluid. (d) Explain why the object will move inward if \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{ob}} R_{\mathrm{cmob}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}} .\) (e) For small objects of uniform density, \(R_{\mathrm{cm}}=R_{\mathrm{cmob}}\) . What happens to a mixture of small objects of this kind with different densities in an ultracentrifuge?

A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa. (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.

An object of average density \(\rho\) floats at the surface of a fluid of density \(\rho_{\text { fluid. }}\) (a) How must the two densities be related? (b) In view of the answer to part (a), how can steel ships float in water? (c) In terms of \(\rho\) and \(\rho\) fluid, what fraction of the object is submerged and what fraction is above the fluid? Check that your answers give the correct limiting behavior as \(\rho \rightarrow \rho_{\text { fluid }}\) and as \(\rho \rightarrow 0 .\) (d) While on board your your your cousin Throckmorton cuts a rectangular piece (dimensions \(5.0 \times 4.0 \times 3.0 \mathrm{cm}\) out of a life preserver and throws it into the ocean. The piece has a mass of 42 g. As it floats in the ocean, what percentage of its volume is above the surface?

It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is 0.0154 \(\mathrm{kg} / \mathrm{m}^{3}\) (although this varies with temperature). Suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 5.00 g. We inflate them with a very light gas whose mass we can neglect. (a) What should be the radius and mass of these balloons so they just hover above the surface of Mars? (b) If we released one of the balloons from part (a) on earth, where the atmospheric density is \(1.20 \mathrm{kg} / \mathrm{m}^{3},\) what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down? (c) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

A swimming pool is 5.0 \(\mathrm{m}\) long, 4.0 \(\mathrm{m}\) wide, and 3.0 \(\mathrm{m}\) deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h,\) and integrate this over the end of the pool.) Do not include the force due to air pressure.

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