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Chapter 4: Problem 36

A constant-density fluid flows through a converging section having an area \(A\) given by \\[ A=\frac{A_{0}}{1+(x / \ell)} \\] where \(A_{0}\) is the area at \(x=0 .\) Determine the velocity and acceleration of the fluid in Eulerian form and then the velocity and acceleration of a fluid particle in Lagrangian form. The velocity is \(V_{0}\) at \(x=0\) when \(t=0\)

Short Answer

Expert verified
The velocity and acceleration of the fluid in Eulerian form are \(V(x) = V_{0}(1 + x/l)\) and \(a = 0\) respectively. The velocity and acceleration in Lagrangian form are \(V(x) = V_{0}(1 + x/l)\) and \(a = V_{0}^2/l\) respectively.

Step by step solution

01

Find velocity in Eulerian form

Begin by applying the continuity equation \(A_1V_1=A_2V_2\) that states the flow rate is constant. Let's apply this principle between two segments: one at location \(x\) with area \(A\) and velocity \(V\), and the other at \(x=0\) with area \(A_{0}\) and velocity \(V_{0}\). So, the equation becomes \(A_0V_0 = AV\). Substituting the given equation \(A=A_0/(1+x/l)\) in the above equation, we get the velocity as a function of the position \(x\), \(V(x) = V_0*(1 + x/l)\).
02

Find acceleration in Eulerian form

Acceleration is the time derivative of velocity. As fluid moves, the observer sees different particles at the same location \(x\) with different velocities at different times. So, in Eulerian form, we calculate acceleration as \(a = dV/dt\). However, since \(V = V_0*(1 + x/l)\), this becomes zero as there is no time dependence.
03

Find velocity in Lagrangian form

In Lagrangian form, we look at the velocity of a single particle of fluid as it moves along the flow. Since we are watching the same fluid particle at different times, its velocity will vary with time and space. This is the same as the velocity we derived in the Eulerian form, \(V(x) = V_0*(1 + x/l)\), because the observer in Eulerian form is seeing the exact trajectory of the fluid particle.
04

Find acceleration in Lagrangian form

Acceleration is the time derivative of velocity. Note that, from the Lagrangian perspective, we treat \(x\) as a function of time as the same particle is followed. So, acceleration is obtained by the total derivative of velocity, i.e., \(a = d(V)/dt = DV_0/dt = V_0*dx/dt*(1/l) = V(dV/dx)\). Substituting the value of \(V\) from the Eulerian form, we get \(a = V_0^2/l\).

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Most popular questions from this chapter

The velocity components of \(u\) and \(v\) of a two-dimensional flow are given by \\[ u=a x+\frac{b x}{x^{2} y^{2}} \quad \text { and } \quad v=a y+\frac{b y}{x^{2} y^{2}} \\] where \(a\) and \(b\) are constants. Calculate the acceleration.

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