/*! 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 96 Water is discharged at a flow ra... [FREE SOLUTION] | 91Ó°ÊÓ

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Water is discharged at a flow rate of \(0.3 \mathrm{m}^{3} / \mathrm{s}\) from a narrow slot in a 200 -mm-diameter pipe. The resulting horizontal twodimensional jet is \(1 \mathrm{m}\) long and \(20 \mathrm{mm}\) thick, but of nonuniform velocity, the velocity at location (2) is twice that at location (1). The pressure at the inlet section is \(50 \mathrm{kPa}\) (gage). Calculate (a) the velocity in the pipe and at locations (1) and (2) and (b) the forces required at the coupling to hold the spray pipe in place. Neglect the mass of the pipe and the water it contains.

Short Answer

Expert verified
The velocities in the pipe, at location (1) and location (2) can be calculated using the continuity equation. Moreover, to calculate the force required to hold the pipe in place, we can use impulse-momentum principle.

Step by step solution

01

Calculate the Velocity in the Pipe

From the continuity equation \(A_{1}V_{1} = A_{2}V_{2}\), where \(A\) is the cross-sectional area and \(V\) is the velocity. Given the flow rate (Q) is \(0.3 \mathrm{m}^{3} / \mathrm{s}\), we can express \(V_{1}\) as \(Q / A_{1}\). Hence the velocity in the pipe (\(V_{1}\)) can be calculated by taking the ratio of the flow rate to the cross sectional area of the pipe, which can be determined by the formula for the area of a circle \(\pi (D/2)^{2}\), where D is the diameter of the pipe.
02

Calculate the Velocity at location (1)

From the continuity equation \(A_{1}V_{1} = A_{3}V_{3}\), where \(A_{3} = (1000mm*20mm)\) is the area of the jet and \(A_{1}\) and \(V_{1}\) are from the previous step, we can compute the velocity at location (1) (\(V_{3}\)) by the equation \(V_{3} = A_{1}V_{1} / A_{3}\).
03

Calculate the Velocity at location (2)

The problem states that the velocity at location (2) is twice that at location (1). Thus, the velocity at location (2) is \(2V_{3}\).
04

Calculate the Force Needed to Hold the Pipe

Based on impulse momentum principle, the force required to hold the pipe can be obtained by the equation \(F = \rho Q (V_{2} - V_{1})\), where \(\rho\) is the density of water. Now can use this equation to find the force required.

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