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Set of ten problems on hydrodynamics

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SOLUTION
Basic theory

Continuity equation : Assuming the liquid to be incompressible, rate of flow of liquid at any cross section is same. Thus, av = constant (a1v1 = a2v2 = a3v3) where a = area of cross section, v = velocity of liquid at that cross section. ........(1)

Bernoulli's theorem : For streamline flow of a perfect liquid in a pipe, the sum of pressure energy, potential energy and the kinetic energy per unit mass is constant throughout the flow. This can be expressed by the following equation : P/ρ + gh + ½ v2 = Constant ....(2)

Alternative forms of the above equation are :

Multiplying (2) by ρ we get : P + ρgh + ½ ρv2 = Constant ......(3)

Dividing (2) by g we get : P/ρg + h + ½ (v2)/g= Constant ....(4)

Torricelli's theorem : The velocity (v) of liquid emerging out of an orifice in a container will be the same as a freely falling object would acquire after falling through a distance equal to the vertical distance (h) between the free surface of the liquid and the orifice.
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V = √2gh (g = acceleration due to gravity) .........(5)

Problem 013

16m

Flow of water = 1.9 x 10-3 m3/min

Using Torricelli's theorem we calculate velocity of water coming out of the hole as:
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Solution Summary

Solutions to these ten problems illustrate different principles of hydrodynamics such as continuity equation, Bernoulli's theorem, Torricelli's theorem, siphon.

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