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THERMO Spoken Here! ~ J. Pohl © ~ 2017 | www.ThermoSpokenHere.com (E3340-231) |

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

**Prove** the velocity of the water at X = X_{R}(t*) is equal to (A_{L}/A_{R})v_{L}(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 **X _{L}(t*)**) and on the right by a plane at

(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, **A _{L}** and

(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.