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Chapter 4: Problem 38

A fluid particle flowing along a stagnation streamline, as shown in Video \(V 4.9\) and Fig. \(P 4.38,\) slows down as it approaches the stagnation point. Measurements of the dye flow in the video indicate that the location of a particle starting on the stagnation streamline a distance \(s=0.6 \mathrm{ft}\) upstream of the stagnation point at \(t=0\) is given approximately by \(s=0.6 e^{-0.5 y},\) where \(t\) is in seconds and \(s\) is in feet. (a) Determine the speed of a fluid particle es a function of time, \(V_{\text {particle }}(t),\) as it flows along the streamline. (b) Determine the speed of the fluid as a function of position along the streamline, \(V=V(s)\) (c) Determine the fluid acceleration along the streamline as a function of position, \(a_{s}=a_{s}(s)\)

Short Answer

Expert verified
The speed of a fluid particle as a function of time is \(V_{particle}(t) = -0.3e^{-0.5t}\). The speed of the fluid as a function of position along the streamline is \(V(s) = -0.5s\). The fluid acceleration along the streamline as a function of position is \(a_s(s) = -0.5\).

Step by step solution

01

Express Particle Displacement as Function of Time

According to the provided equation, the position or displacement of the particle at any given time t is given by \(s(t) = 0.6 e^{-0.5t}\).
02

Find Particle Speed as Function of Time

The speed of the particle is the rate of change of displacement with respect to time, which is simply the first derivative of the displacement equation. So, take derivative of the displacement equation to find: \(V_{particle}(t) = -0.3e^{-0.5t}\).
03

Find Fluid Speed as Function of Position

The speed of the fluid as a function of the position along the streamline can be obtained by rewriting the speed function in terms of s using \(s(t) = 0.6 e^{-0.5t}\) and updating t with this expression in \(V_{particle}(t)\). So, \(V(s) = -0.3e^{ln(0.6/s - ln(0.6))} = -0.3s/0.6 = -0.5s\).
04

Calculate Fluid Acceleration as Function of Position

The acceleration is simply the rate of change of the speed function with respect to position, so take the derivative of the speed function with respect to s. This gives \(a_s(s) = -0.5\).

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Most popular questions from this chapter

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.

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