/*! 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 174 The moving tank shown is to be s... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 174

The moving tank shown is to be slowed by lowering a scoop to pick up water from a trough. The initial mass and speed of the tank and its contents are \(M_{0}\) and \(U_{0}\) respectively. Neglect external forces due to pressure or friction and assume that the track is horizontal. Apply the continuity and momentum equations to show that at any instant \(U=U_{0} M_{0} / M .\) Obtain a general expression for \(U I U_{0}\) as a function of time.

Short Answer

Expert verified
The velocity \(U\) of the tank at any instant is given by \(U = \frac{M_{0}U_{0}}{M}\). The ratio \(U / U_{0}\) as a function of time is given by \(U / U_{0} = \frac{M_{0}}{M_{0} + \frac{dm}{dt} * t}\).

Step by step solution

01

Identify Important Parameters

The key parameters in the problem are the initial mass (\(M_{0}\)) and speed (\(U_{0}\)) of the tank, the velocity (\(U\)) of the tank at any instant, and the mass of the tank at the same instant (\(M\)). These indicate that the mass of the tank is not constant, but actually changes with time. We can assume that the rate at which the mass changes is constant and represented by \(dm/dt\). Given that no external forces are acting on the tank and neglecting friction, the momentum of the tank should be conserved.
02

Apply the Conservation of Momentum

Since the tank's momentum is conserved and changes only due to the mass picked up, we can write the conservation of momentum equation as \(M_{0}U_{0} = (M + \frac{dm}{dt} * dt) * U\), where \(\frac{dm}{dt}\) is the change in mass with respect to time and \(dt\) is a small time duration.
03

Rearrange the Equation to Solve for \(U\)

We rearrange the equation to solve for \(U\), the velocity of the tank at any instant. Doing this, we obtain the equation \(U = \frac{M_{0}U_{0}}{M + \frac{dm}{dt} * dt}\). However, as \(dt\) approaches zero (for any instant), the equation becomes \(U = \frac{M_{0}U_{0}}{M}\).
04

Derive the Expression for \(U / U_{0}\) as a Function of Time

Next, we have to derive an expression for the ratio \(U / U_{0}\) as a function of time. We know that \(U = \frac{M_{0}U_{0}}{M}\), therefore, \(U / U_{0} = \frac{M_{0}}{M}\). As \(M\) is increasing due to the addition of water, \(M = M_{0} + \frac{dm}{dt} * t\). Substituting this back into the equation for \(U / U_{0}\) we get \(U / U_{0} = \frac{M_{0}}{ M_{0} + \frac{dm}{dt} * t}\) this is the required expression.

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Most popular questions from this chapter

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A porous round tube with \(D=60 \mathrm{mm}\) carries water. The inlet velocity is uniform with \(V_{1}=7.0 \mathrm{m} / \mathrm{s}\). Water flows radially and axisymmetrically outward through the porous walls with velocity distribution \\[ v=V_{0}\left[1-\left(\frac{x}{L}\right)^{2}\right] \\] where \(V_{0}=0.03 \mathrm{m} / \mathrm{s}\) and \(L=0.950 \mathrm{m} .\) Calculate the mass flow rate inside the tube at \(x=L\)

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In a laboratory experiment, the water flow rate is to be measured catching the water as it vertically exits a pipe into an empty open tank that is on a zeroed balance. The tank is \(10 \mathrm{m}\) directly below the pipe exit, and the pipe diameter is \(50 \mathrm{mm} .\) One student obtains a flow rate by noting that after 60 s the volume of water (at \(4^{\circ} \mathrm{C}\) ) in the tank was \(2 \mathrm{m}^{3}\). Another student obtains a flow rate by reading the instantaneous weight accumulated of \(3150 \mathrm{kg}\) indicated at the \(60-\mathrm{s}\) point. Find the mass flow rate each student computes. Why do they disagree? Which one is more accurate? Show that the magnitude of the discrepancy can be explained by any concept you may have.
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