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Chapter 15: Problem 117

The nozzle discharges water at a constant rate of \(2 \mathrm{ft}^{3} / \mathrm{s} .\) The cross-sectional area of the nozzle at \(A\) is \(4 \mathrm{in}^{2}\) and at \(B\) the cross-sectional area is 12 in \(^{2}\). If the static gauge pressure due to the water at \(B\) is \(2 \mathrm{lb} / \mathrm{in}^{2},\) determine the magnitude of force which must be applied by the coupling at \(B\) to hold the nozzle in place. Neglect the weight of the nozzle and the water within it. \(\gamma_{w}=62.4 \mathrm{lb} / \mathrm{ft}^{3}\)

Short Answer

Expert verified
To solve this problem, one has to calculate the velocities at the two sections, the pressures at section \(A\), the force due to pressure at both sections, and finally, calculate the net force required to hold the nozzle in place

Step by step solution

01

Find the Velocity at A and B

Using the definition of flow rate which is \(Q=Av\), where \(Q\) is the flow rate, \(A\) is the cross-sectional area, and \(v\) is the velocity. The areas are given in square inches, so these need to be converted into square feet. Therefore, the area at A is \(A_{A} = \frac{4}{144} \, ft^{2}\) and the area at B is \(A_{B} = \frac{12}{144} \, ft^{2}\). Since the incompressible fluid flows at a constant rate, then the velocities of the fluid at \(A\) and \(B\) can be given as \(v_{A} = \frac{Q}{A_{A}}\) and \(v_{B} = \frac{Q}{A_{B}}\) respectively.
02

Apply Bernoulli's Equation and Solve for Pressure at A

Applying Bernoulli's equation from point A to point B which is \(p_{A} + \frac{1}{2} \gamma v_{A}^{2} = p_{B} + \frac{1}{2} \gamma v_{B}^{2}\). Where \(p\) is the pressure and \(\gamma\) is the weight density of the water. By rearranging the equation, you can solve for \(p_{A}\) which is \(p_{A} = p_{B} + \frac{1}{2} \gamma v_{B}^{2} - \frac{1}{2} \gamma v_{A}^{2}\). The pressure is given in \(lb / in^{2}\), therefore, these need to be converted into \(lb / ft^{2}\). The density of the water \(\gamma_{w}\) is given, hence the pressure at \(A\) can be calculated.
03

Calculate The Force at A and B

The force due to pressure at any section can be found as \(F = p \cdot A\). Therefore, the force at section A is given as \(F_{A} = p_{A} \cdot A_{A}\) and the force at section B is given as \(F_{B} = p_{B} \cdot A_{B}\).
04

Find the Net Force

The net force required to hold the nozzle in place is the difference between the force at \(A\) and \(B\). Therefore, \(F_{net} = F_{A} - F_{B}\), ensuring the direction of forces is respected. The calculated \(F_{net}\) is the force required to keep the nozzle in place.

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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 a branch of physics that studies the behavior of fluids (liquids and gases) and the forces on them. It is essential for understanding a variety of real-world scenarios, from simple plumbing to complex aerodynamics. A fundamental principle of fluid mechanics is that fluids can exert pressure and can flow with varying speeds, depending on their environment and the forces acting upon them.

In practice, fluid mechanics allows us to predict how fluids will move and interact with their surroundings, which is crucial for designing systems involving anything from water pipes to jet engines.
Flow Rate
Flow rate is a term used in fluid mechanics to describe the volume of fluid that passes through a given surface per unit of time. It is usually denoted as Q and is measured in units such as cubic feet per second (ft鲁/s) or liters per minute (L/min). The concept of flow rate is central to understanding fluid movement and is used to design systems that can handle the transport of fluids at various rates, ensuring that pipes, channels, and pumps are appropriately sized.

For example, a nozzle discharging water at a constant rate is directly related to how much water is being moved through the system and is a critical factor in calculating forces that act within and upon the system.
Pressure and Forces in Fluids
Pressure in fluids is a measure of the force exerted by a fluid per unit area, often described in the unit pounds per square inch (psi) or Pascals (Pa). It is an important concept in understanding how fluids interact with their containers and any objects they encounter.

This force arises from the movement and collision of molecules within the fluid. The pressure can vary significantly within a fluid system due to changes in velocity and elevation 鈥� a principle captured by Bernoulli's equation. Engineers and scientists use this concept to analyze and design systems such as dams, water treatment plants, and even blood circulation systems. In solving problems, they calculate the forces exerted by the fluid to ensure structural integrity and proper functionality of systems.
Velocity of Fluid
The velocity of a fluid refers to the speed and direction of the fluid particles in motion. In fluid mechanics, the understanding of fluid velocity is crucial as it impacts the flow rate and pressure within a system. The velocity can be altered by changes in cross-sectional area of the flow path, obstructions, and other forces acting on the fluid.

In the example of a nozzle discharging water, the velocity of the fluid changes from one section of the nozzle to another. This is described by the Continuity Equation, which in simple terms states that the product of the cross-sectional area and the fluid velocity at any two points along a streamline must be constant for an incompressible fluid, assuming the flow rate remains constant.

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Most popular questions from this chapter

An earth satellite of mass \(700 \mathrm{kg}\) is launched into a free-flight trajectory about the earth with an initial speed of \(v_{A}=10 \mathrm{km} / \mathrm{s}\) when the distance from the center of the earth is \(r_{A}=15 \mathrm{Mm}\). If the launch angle at this position is \(\phi_{A}=70^{\circ},\) determine the speed \(v_{B}\) of the satellite and its closest distance \(r_{B}\) from the center of the earth. The earth has a mass \(M_{e}=5.976\left(10^{24}\right)\) kg. Hint: Under these conditions, the satellite is subjected only to the earth's gravitational force, \(F=G M_{e} m_{s} / r^{2},\) Eq. \(13-1 .\) For part of the solution, use the conservation of energy.

The 20 -kg crate is lifted by a force of \(F=\left(100+5 t^{2}\right) \mathrm{N},\) where \(t\) is in seconds. Determine how high the crate has moved upward when \(t=3\) s, starting from rest.

A man kicks the 150 -g ball such that it leaves the ground at an angle of \(60^{\circ}\) and strikes the ground at the same elevation a distance of \(12 \mathrm{m}\) away. Determine the impulse of his foot on the ball at \(A\). Neglect the impulse caused by the ball's weight while it's being kicked.

For a short period of time, the frictional driving force acting on the wheels of the \(2.5-\mathrm{Mg}\) van is \(F_{D}=\left(600 t^{2}\right) \mathrm{N}\) where \(t\) is in seconds. If the van has a speed of \(20 \mathrm{km} / \mathrm{h}\) when \(t=0,\) determine its speed when \(t=5 \mathrm{s}\).

Two boxes \(A\) and \(B\), each having a weight of \(1601 b\), sit on the 500 -lb conveyor which is free to roll on the ground. If the belt starts from rest and begins to run with a speed of \(3 \mathrm{ft} / \mathrm{s},\) determine the final speed of the conveyor if (a) the boxes are not stacked and \(A\) falls off then \(B\) falls off, and (b) \(A\) is stacked on top of \(B\) and both fall off together.
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