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Chapter 4: Problem 33

As a valve is opened, water flows through the diffuser shown in Fig. \(\mathrm{P} 4.33\) at an increasing flowrate so that the velocity along the centerline is given by \(\mathbf{V}=u \hat{\mathbf{i}}=V_{0}\left(1-e^{-\alpha}\right)(1-x / \ell) \hat{\mathbf{i}},\) where \(u_{0}, c,\) and \(\ell\) are constants. Determine the acceleration as a function of \(x\) and \(t .\) If \(V_{0}=10 \mathrm{ft} / \mathrm{s}\) and \(\ell=5 \mathrm{ft},\) what value of \(c\) (other than \(c=0\) ) is needed to make the acceleration zero for any \(x\) at \(t=1 \mathrm{s} ?\) Explain how the acceleration can be zero if the flowrate is increasing with time.

Short Answer

Expert verified
The acceleration is given by the time derivative of the given velocity function. By setting this expression to zero and solving for c, it is found that c has to be a certain value for the acceleration to be zero at \(t=1s\) for any x, excluding the trivial solution of \(c=0\). In the context of this problem, the acceleration can be zero while the flow rate increases because the velocity (and hence the acceleration) pertains to the speed of the flow, whereas the flow rate pertains to the volume of water flowing, which can increase even if the speed of the flow remains constant due to an increase in the cross-sectional area of the flow path.

Step by step solution

01

Find the acceleration as a function of x and t

Velocity \(V\) of an object is defined as the rate of change of displacement with time. Acceleration \(a\) is defined as the rate of change of velocity with time. So, to find the acceleration, differentiate the velocity function with respect to time (t). It will result in the equation for acceleration. Thus, \(a = \frac{dV}{dt}\).
02

Substitute the given values into the acceleration equation

For \(V=V_{0}\left(1-e^{-\alpha}\right)(1-x / \ell)\), substitute the given values \(V_{0}=10 \mathrm{ft} / \mathrm{s}\) and \(\ell=5 \mathrm{ft}\) into the equation. Then, differentiate this equation with respect to time (t).
03

Find a non-zero value of c for which the acceleration is zero

Having obtained an expression for acceleration that includes c, set this expression equal to zero and solve for c. This will give us the value of c that results in zero acceleration at any x, for \(t=1s\).
04

Explain how acceleration can be zero while flow rate increases

Zero acceleration means that the velocity of the fluid is not changing with time. Even if the flow rate (the volume of fluid flowing through per unit of time) is increasing, this can happen if the increase in flow rate is due to the increase in the area the fluid flows through (as in the case of the diffuser) and not due to an increase in the flow velocity.

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