A venturi meter is used to measure the flow speed of a fluid in a pipe. The mete

Question

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (the figure); the cross-sectional area A of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed V and then through a narrow ''throat'' of cross-sectional area a with speed v. A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change ?p in the fluid's pressure, which causes a height difference h of the liquid in the two arms of the manometer. (Here ?p means pressure in the throat minus pressure in the pipe.) Let A equal 5

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (the figure); the cross-sectional area A of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed V and then through a narrow ''throat'' of cross-sectional area a with speed v. A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change ?p in the fluid's pressure, which causes a height difference h of the liquid in the two arms of the manometer. (Here ?p means pressure in the throat minus pressure in the pipe.) Let A equal 5·a. Suppose the pressure p1 at A is 2.1 atm. Compute the values of (a) the speed V at A and (b) the speed v at a that make the pressure p2 at a equal to zero. (c) Compute the corresponding volume flow rate if the diameter at A is 4.5 cm.The phenomenon that occurs at a when p2 falls to nearly zero is known as cavitation. Please assume that the fluid is water. The water vaporizes into small bubbles.

Explanation / Answer

a)

v = ?{2 ?p / [?((A /a)2 - 1)]}

      = ?{2 (2.1 x 1.01 x105 Pa) / [(1000 kg/m3) ((5a /a)2 - 1)]}

      = 4.2 m / s

b)

According to the equation of continuity we have

aV = Av

V = (A / a) v

= (5a / a) v

= (5a / a) (4.2 m/s)

= 21 m/s

c)

Volume flow rate is given by

   A v = ? (2.25 x 10-2 m)^2 (4.2 m/s)

          = 6.68 * 10^-3 m3/s

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