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Chapter 4: Problem 175

The tank shown rolls with negligible resistance along a horizontal track. It is to be accelerated from rest by a liquid jet that strikes the vane and is deflected into the tank. The initial mass of the tank is \(M_{0}\). Use the continuity and momentum equations to show that at any instant the mass of the vehicle and liquid contents is \(M=M_{0} V /(V-U) .\) Obtain a general expression for \(U\) / \(V\) as a function of time.

Short Answer

Expert verified
The velocity ratio of the fluid being ejected into the tank, to the velocity of the tank \(U/V\) as a function of time is \(1 - e^{-t}\).

Step by step solution

01

Understand the given information

We have a tank initially at rest with an initial mass \(M_{0}\). The tank is being accelerated by a fluid jet impacting a vane and being deflected into the tank. It is also mentioned that fluid outflow and tank's velocity, \(U\) and \(V\) respectively, are different at any given instance.
02

Apply momentum equation for the system

As per conservation of momentum, the rate of incoming momentum minus the rate of outgoing momentum equals the rate of change of momentum of the system (i.e.,\(M*V\)). Therefore, the momentum equation for the system is given by \(0 - (\rho * A * U^2) = d(M*V)/dt\), where \(\rho\) is the fluid density, \(A\) is the cross-sectional area of the jet and \(U\) is the velocity of the fluid being ejected into the tank.
03

Apply the continuity equation

The continuity equation implies the mass flow in equals the mass flow out. Since there is no outflow here, we have \(M = M_{0} + \rho * A * U * t\), where \(t\) is time.
04

Substituting

We substitute the continuity equation into the momentum equation to get \(d(M*V)/dt = -\rho * A * U^2.\) Now, divide both sides by \(\rho * A * U\), and substituting \(M\) from the continuity equation to get \(d(V)/(V - U) = - dt.\) Solving this differential equation we find \(V - U = k * e^{-t},\) where \(k\) is a constant.
05

Determining the constant

At \(t=0\), \(V=0\), thus we find \(k=U\). So, our expression for the velocity ratio as the function of time becomes \(V / U = 1 - e^{-t}.\)

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