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A circular jet of water leaves a nozzle in a vertical upward direction with a velocity of \(20 \mathrm{ft} / \mathrm{s}\). The jet diameter is 1 inch. A large circular disk weighing 2 pounds is held in a horizontal position above the nozzle. Neglect thermal and friction effects. What is the distance between nozzle and disk?

Short Answer

Expert verified
The distance between the nozzle and the disk is 6.21 feet.

Step by step solution

01

Understand the problem

We need to find the distance between the nozzle and a disk suspended above it. Water jet from a nozzle impacts the disk. We use momentum principles considering the kinetic energy transfer from the jet to the disk.
02

Convert Units

Convert jet diameter from inches to feet: \( 1 \text{ inch} = \frac{1}{12} \text{ feet} \), thus the radius \( r = \frac{1}{24} \text{ feet} \). Convert weight from pounds to mass in slugs: \( m = \frac{2}{32} \text{ slugs} = \frac{1}{16} \text{ slugs} \).
03

Calculate Area of Water Jet

Calculate the cross-sectional area, \( A \), of the jet: \( A = \pi r^2 = \pi \left( \frac{1}{24} \right)^2 = \pi \times \frac{1}{576} \text{ square feet} \).
04

Calculate Volume Flow Rate

Use the formula for volumetric flow rate \( Q = A \times v \), where \( v = 20 \text{ ft/s} \). Thus, \( Q = \pi \times \frac{1}{576} \times 20 = \frac{20\pi}{576} \text{ cubic feet/second} \).
05

Calculate Force Exerted by Water Jet

Force \( F \) equals rate of change of momentum: \( F = Q \times \text{density of water}\times v \). Assume density \( \rho = 1.94 \text{ slugs/ft}^3 \). Calculate force: \( F = \frac{20\pi}{576} \times 1.94 \times 20\).
06

Use Equilibrium of Forces

The upward force from the water jet equals the weight of the disk for equilibrium: \( \frac{20\pi}{576} \times 1.94 \times 20 = 2 \text{ pounds}\).
07

Solve for Distance

Using energy conservation, as water rises, initial kinetic energy converts to potential energy: \( mgh = 0.5mv^2 \). Given \( v = 20 \text{ ft/s} \), solve \( gh = 0.5v^2 \). Use \( g = 32.2 \text{ ft/s}^2 \). Solve for \( h = \frac{0.5 \times 20^2}{32.2} \).
08

Final Calculation

Calculate \( h = \frac{0.5 \times 400}{32.2} = 6.21 \text{ feet} \). Thus, the distance between nozzle and disk is 6.21 feet.

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Key Concepts

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

Momentum Principles
Momentum principles are essential when analyzing the interaction between a water jet and a disk. Momentum refers to the quantity of motion an object has, and it is calculated as the product of mass and velocity. When the water jet strikes the disk, it exerts a force due to the change in momentum. This happens as the velocity of the water changes upon hitting the disk.
To calculate the force exerted by the water jet, we use the principle that the force is equal to the rate of change of momentum. This is calculated using the formula \( F = Q \times \rho \times v \), where \( Q \) is the volumetric flow rate, \( \rho \) is the density of water, and \( v \) is the velocity.
This principle helps us understand the equilibrium of forces needed to keep the disk suspended above the nozzle.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion, expressed as \( KE = 0.5 \times m \times v^2 \). In this exercise, the kinetic energy of the water jet is converted into work done to hold the disk at a specific height.
When the water leaves the nozzle, it has high kinetic energy due to its high velocity. As the water jet impacts the disk, part of this kinetic energy is transferred into a force that acts against the weight of the disk. This balancing force maintains the disk at equilibrium.
Understanding the transformation of kinetic energy into potential energy as the water moves upward is key in finding the distance the water can travel before being halted by the gravitational pull acting on the disk.
Volumetric Flow Rate
Volumetric flow rate represents the volume of fluid passing through a given area per unit of time. It is a crucial parameter when working with fluid dynamics as it helps quantify how much water comes out of the nozzle.
In our calculation, the volumetric flow rate \( Q \) can be determined using the formula \( Q = A \times v \), where \( A \) is the cross-sectional area of the water jet and \( v \) is its velocity. This gives us an indication of the water's capacity to perform work or exert force on the suspended disk.
Determining the volumetric flow rate is essential in applying the momentum principle and calculating the force exerted on the disk by the water jet's impact.
Energy Conservation
Energy conservation is a fundamental principle used to understand that energy cannot be created or destroyed; it can only be transformed from one form to another. In the exercise, it helps explain how the initial kinetic energy of the water jet changes as the water rises towards the disk.
As the water ascends, it slows down, and its kinetic energy is converted into potential energy. This transformation is described by the equation \( mgh = 0.5 \times m \times v^2 \), where \( h \) is the height the water can reach.
This principle allows us to calculate the maximum height or distance between the nozzle and the disk by setting the initial kinetic energy equal to the gravitational potential energy at the peak of the water's trajectory.

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Most popular questions from this chapter

Water is moving through a nozzle at a volume rate of flow of \(3 \mathrm{~m}^{3} / \mathrm{s}\). A pinhole leak in the nozzle exists. If the velocity at a downstream area of \(8 \mathrm{~cm}^{2}\) is \(12 \mathrm{~m} / \mathrm{s}\), how much fluid is lost every 10 seconds?

An \(8.95 \mathrm{~cm}\) outer diameter by \(9.0 \mathrm{~cm}\) long piston moves downward in a \(9.0 \mathrm{~cm}\) inner diameter cylinder. The downstroke velocity is \(40 \mathrm{~m} / \mathrm{s}\). SAE 30 oil at \(100^{\circ} \mathrm{C}\) provides lubrication between cylinder wall and piston. If the oil velocity profile is assumed to be linear, and the piston weight is negligible, find (a) the shear stress on the piston, (b) the force required to maintain the piston speed, (c) the energy equation for this problem. (d) If the oil were at \(10^{\circ} \mathrm{C}\) and the piston \(12 \mathrm{~cm}\) long, find the force required for the same speed.

A rocket-propelled craft flies at a speed of \(1700 \mathrm{ft} / \mathrm{s}\) in a westerly direction. The combustion gases exhaust aft of the craft through an area \(1 \mathrm{ft}^{2}\) at 15 psia. Ambient pressure is 11 psi. The exhaust gases leave the aircraft at a speed of \(300 \mathrm{ft} / \mathrm{s}\) relative to the ground in an easterly direction. (a) Determine the velocity of the exhaust gases relative to the aircraft. (b) Calculate the thrust of the gases on the aircraft.

A toy balloon of \(100 \mathrm{~g}\) mass is filled with air of density \(1.030 \mathrm{~kg} / \mathrm{m}^{3}\), by a small hose \(5 \mathrm{~mm}\) in diameter oriented in a vertical direction. The balloon is released, letting air escape through the hose. Neglecting friction, calculate the rate at which air escapes if the initial acceleration of the balloon is zero.

Air flows adiabatically through a supersonic wind tunnel as shown in Fig. P5.114. At the inlet to the tunnel a large reservoir exists. Therefore, \(\tilde{V}_{1} \cong 0\). The air temperature and pressure in the reservoir of the tunnel are \(T_{1}=1000^{\circ} \mathrm{R}\) and \(p_{1}=10 \mathrm{~atm} .\) The throat temperature is \(T_{2}=800^{\circ} \mathrm{R} .\) Calculate the velocity at the throat.

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