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# Expansion Flow

The sketch depicts water flowing from the left to the right without friction (we assume) through a horizontal pipe. A fixed "origin" is notated. Left of the origin the pipe diameter is d. An expansion is located at X = 0. The pipe dimeter thereafter is D. Suppose at the instant, t = t*, a point in the flow at X = XL(t*) is marked and observed to have the velocity vL(t*). Prove the velocity of the water at X = XR(t*) is equal to (AL/AR)vL(t*).

♦  The answer to this problem is obvious. We're interested in the method of solution; not the answer. Let the system be the water bounded on the left by a plane at XL(t*)) and on the right by a plane at XR(t*). Both planes are moving. The points were noted at a superscripted time, t*, (which is usually set at, t* = 0+). Since the points move with the water, the mass between them is constant (by physical constraint of the pipe) and the derivative of that the mass so bounded equals zero. ( The notation, t*, is used because an observation is not made at time, t, a variable. However, after a conclusion is made with regard to a t*, it can be said that since t* was arbitrary, the results apply for all t). (1) 1

The system mass equals the density times area integrated between the left and right boundaries. The boundaries move with the fluid. The integral is written - below left. (2) 2

The mass between the moving boundaries is constant. The derivative of the mass (integral above left) is written as above right. Since mass is constant, that integral equals zero. To continue, observing that the area of the the integrand, A, is discontinuous at the non-moving location, X = 0. We break the integral (above right) into two pieces (shown below left) then reverse the limits on first integral and identify the areas as, AL and AR. The rearranged integral is written below right. (3) 3

The densities and areas are constant so we perform the integrations. (4) 4

Next we differentiate. The result is: (5) 5

Recognizing dX/dt as linear speed (magnitude of velocity), a result obtained is: (6) 6

And our solution is: (7) 7

This is what we expected. The purpose of this was to show that the mathematics works.

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