/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 A flat plate orifice of 2 in. di... [FREE SOLUTION] | 91影视

91影视

A flat plate orifice of 2 in. diameter is located at the end of a 4 -in.-diameter pipe. Water flows through the pipe and orifice at $20 \mathrm{ft}^{3} / \mathrm{s}$. The diameter of the water jet downstream from the orifice is 1.5 in. Calculate the external force required to hold the orifice in place. Neglect friction on the pipe wall.

Short Answer

Expert verified
The external force required to hold the orifice in place can be found by using the Bernoulli's equation and equation of motion. After computing the velocities and pressures by using the given diameters and flow rate, the internal force exerted by water can be found. This will be the force required externally to hold the orifice in place, but in the opposite direction.

Step by step solution

01

Calculate velocity and pressure before and after passing through the orifice

Firstly we need to calculate the velocity \(v_1\) and \(v_2\) and pressure \(P_1\) and \(P_2\). The formula to calculate the velocity of water in a pipe is \(v_0 = Q/A\), where \(Q\) is the flow rate and \(A\) is the cross-sectional area of the flow. The cross-sectional area \(A = (\pi d^2)/4\). So the cross-sectional areas of the pipe and the orifice can be obtained as \(A_1 = (\pi (4^2))/4 = \pi\) in^2 and \(A_2 = (\pi (2^2))/4 = \pi/2\) in^2, respectively. The velocities \(v_1 = Q/A1\) and \(v_2 = Q/A2\) can now be calculated.
02

Calculate the internal force exerted by the water on the orifice

The force that the water exerts on the orifice internally is equal to the change in momentum per unit time. Considering that the friction on the pipe wall is negligible and that we have steady flow, we can write, by Newton's second law, that Force exerted = mass flow rate * (velocity downstream - velocity upstream). To obtain the mass flow rate, 蟻 * Q can be utilized, where 蟻 is the density of the fluid. Thus, the force exerted can be computed as \(F_i = \rho Q (v_2 - v_1)\). Considering a density of water \(\rho\) = 62.4 lb/ft^3.
03

Calculate the external force required to hold the orifice in place

The force required to hold the orifice in place will be equal to the internal force exerted by water on the orifice. The force will act in the direction opposing the force exerted by water. So external force \(F_{ext} = -F_i\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A rocket motor is used to accelerate a kinetic energy weapon to a speed of 3500 mph in horizontal flight. The exit stream leaves the nozzle axially and at atmospheric pressure with a speed of 6000 mph relative to the rocket. The rocket motor ignites upon release of the weapon from an aircraft flying horizontally at \(U_{0}=600 \mathrm{mph}\). Neglecting air resistance, obtain an algebraic expression for the speed reached by the weapon in level flight. Determine the minimum fraction of the initial mass of the weapon that must be fuel to accomplish the desired acceleration.

A cylindrical holding water tank has a \(3 \mathrm{m}\) ID and a height of \(3 \mathrm{m}\). There is one inlet of diameter \(10 \mathrm{cm},\) an exit of diameter \(8 \mathrm{cm},\) and a drain. The tank is initially empty when the inlet pump is turned on, producing an average inlet velocity of \(5 \mathrm{m} / \mathrm{s}\). When the level in the tank reaches \(0.7 \mathrm{m}\) the exit pump turns on, causing flow out of the exit; the exit average velocity is \(3 \mathrm{m} / \mathrm{s}\). When the water level reaches \(2 \mathrm{m}\) the drain is opened such that the level remains at \(2 \mathrm{m}\). Find (a) the time at which the exit pump is switched on, (b) the time at which the drain is opened, and (c) the flow rate into the drain \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\)

When a plane liquid jet strikes an inclined flat plate, it splits into two streams of equal speed but unequal thickness. For frictionless flow there can be no tangential force on the plate surface. Use this assumption to develop an expression for \(h_{2} / h\) as a function of plate angle, \(\theta\). Plot your results and comment on the limiting cases, \(\theta=0\) and \(\theta=90\).

\( \mathrm{A}\) disk, of mass \(M,\) is constrained horizontally but is free to move vertically. A jet of water strikes the disk from below. The jet leaves the nozzle at initial speed \(V_{0}\), Obtain a differential equation for the disk height, \(h(t),\) above the jet exit plane if the disk is released from large height, \(H .(\mathrm{You}\) will not be able to solve this ODE, as it is highly nonlinear!) Assume that when the disk reaches equilibrium, its height above the jet exit plane is \(h_{0}\) (a) Sketch \(h(t)\) for the disk released at \(t=0\) from \(H>h_{0}\) (b) Explain why the sketch is as you show it.

\( \mathrm{A}\) mass of \(5 \mathrm{lbm}\) is released when it is just in contact with a spring of stiffness 25 lbf/ft that is attached to the ground. What is the maximum spring compression? Compare this to the deflection if the mass was just resting on the compressed spring. What would be the maximum spring compression if the mass was released from a distance of 5 ft above the top of the spring?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.