Question

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 cm${}^3$/s. At one point in the pipe, where the radius is 4.00 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 cm. What is the water's absolute pressure as it flows through this constriction?

Short answer

The water's absolute pressure at the constriction is approximately 2.25 x 10^5 Pa.

Step-by-step solution

Understanding the Given Values

We are given the flow rate of water as 7200 cm${}^3$/s, which we can convert to m${}^3$/s by dividing by 1000000. So, the flow rate is 0.0072 m${}^3$/s. The initial radius $r_1$ is 4.00 cm (0.04 m) and the initial pressure $P_1$ is 2.40 \times 10^5 Pa. The radius at the constriction $r_2$ is 2.00 cm (0.02 m). We need to find the pressure $P_2$ at the constriction.

Applying the Continuity Equation

The Continuity Equation for fluid dynamics is $A_1{v}_1=A_2{v}_2$, where $A$ is the cross-sectional area and $v$ is the velocity. The area $A$ is $\pi r^2$. Calculate $A_1=\pi (0.04{)}^2=0.0050265$ m${}^2$ and $A_2=\pi (0.02{)}^2=0.001257$ m${}^2$. The volumetric flow rate $Q=A_1{v}_1=A_2{v}_2=0.0072$ m${}^3$/s. Therefore, $v_1=\frac{0.0072}{0.0050265}=1.432$ m/s and $v_2=\frac{0.0072}{0.001257}=5.727$ m/s.

Using Bernoulli’s Equation

Bernoulli's Equation states $P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$. Since the height $h$ remains constant, the equation simplifies to $P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$. We are given $P_1=2.40\times {10}^5$ Pa, and we assume the density $\rho$ of water is 1000 kg/m${}^3$. Substitute the velocities: $P_2=P_1+\frac{1}{2}\rho (v_1^2-v_2^2)$.

Calculating the Absolute Pressure at the Constriction

Substitute $v_1=1.432$ m/s and $v_2=5.727$ m/s into Bernoulli's equation: $P_2=2.40\times {10}^5+\frac{1}{2}\times 1000\times ({1.432}^2-{5.727}^2)$. Calculate the pressure difference: $\mathrm{\Delta }P=1000\times \frac{1}{2}\times ((1.432{)}^2-(5.727{)}^2)$. This equals $-15048.806$ Pa. Therefore, $P_2=2.40\times {10}^5-15048.806=2.24951\times {10}^5$ Pa.

Key concepts

Continuity Equation

When dealing with fluid flow in pipes, the Continuity Equation is a crucial concept. It helps us understand how the fluid behaves when it moves from one section of a pipe to another, especially when there is a change in the pipe's diameter. The Continuity Equation states that the mass flow rate of a fluid must remain constant throughout a streamline, assuming there is no accumulation of fluid in the pipe. This concept is presented mathematically as $A_1{v}_1=A_2{v}_2$, where:

• $A$ represents the cross-sectional area of the pipe.
• $v$ represents the velocity of the fluid.

Hence, given a fluid with a constant density, the product of area and velocity is conserved. This means that if a pipe narrows, the fluid velocity must increase to maintain the same volumetric flow rate. Understanding this equation allows engineers and scientists to predict how changes in a pipe's shape will affect the flow speed of the fluid.

Fluid Dynamics

Fluid dynamics is the study of fluids in motion. It examines the forces and energies involved when fluids like water and air flow. In the context of our sprinkler problem, understanding fluid dynamics allows us to analyze how water moves through pipes of varying diameters. Key aspects include:

• Velocity, which is how fast the fluid moves through the pipe.
• Pressure, which is the force exerted by the fluid per unit area.
• Flow rate, which is the volume of fluid flowing per unit time.

These aspects all interact under the principles of fluid dynamics, governed by equations like the Continuity Equation and Bernoulli's Equation. Appreciating how these elements interplay helps in designing systems for water distribution, like our sprinkler system, ensuring efficient delivery and operation.

Volumetric Flow Rate

The volumetric flow rate is a measure of how much volume of fluid passes through a given cross-sectional area in a certain amount of time. In fluid dynamics, it’s often denoted as $Q$ and calculated using the formula:$$Q=A\times v$$where $A$ is the cross-sectional area and $v$ is the velocity of the fluid. For the golf course sprinkler scenario, the constant volumetric flow rate of 7200 cm${}^3$/s (or 0.0072 m${}^3$/s) means that regardless of how the pipe's diameter changes, the same volume of water passes through every second. This concept is foundational in applications where the supply of fluid needs to be consistent, ensuring devices like sprinklers operate evenly and as intended across different sections of a piping network.

Pressure Calculation

Calculating pressure differences in fluid systems is essential to understanding how systems function under varying conditions. Using Bernoulli's Equation, which is $P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$, we can find the pressure at different points in a pipe when changes in velocity occur. In this equation:

• $P_1$ and $P_2$ are pressures at two points along the pipe.
• $v_1$ and $v_2$ are velocities at these points.
• $\rho$ is the fluid's density.

For our example, we used initial conditions at one section of the pipe and solved for pressure at a constricted section. This pressure balance is vital in design to ensure safe and efficient fluid system operations, particularly where pressures and velocities change due to varying pipe diameters.