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 (Fig. 14 - 55); 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 Ap in the fluid's pressure, which causes a height difference h of the liquid in the two arms of the manometer. (Here Ap means pressure in the throat minus pressure in the pipe.) Let A equal 5 a. Suppose the pressure p1 at A is 2.4 atm. Compute the values of (a) the speed V at A and 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 7.5 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. FIG. 14 - 55 Problems 67 and 68.

Explanation / Answer

(a)    from the theory we know that the equation ofcontinuity as    A1 v1 = a2v2    from the given    p = p1          = 1.80atm    so we get    v = {2 p / [((A /a)2 - 1)]}       = {2 (1.01 x105 Pa) / [(1000 kg / m3)((5a /a)2 - 1)]}       = .......... m / s (b)    again using equation of continuity we get    V = (A / a) v        = (5a / a) v        = ......... m /s (c)    the rate of flow is    A v = ( / 4) (5.0 x 10-4m2) (v)           =......... m3 / s    from the given    p = p1          = 1.80atm    so we get    v = {2 p / [((A /a)2 - 1)]}       = {2 (1.01 x105 Pa) / [(1000 kg / m3)((5a /a)2 - 1)]}       = .......... m / s (b)    again using equation of continuity we get    V = (A / a) v        = (5a / a) v        = ......... m /s (c)    the rate of flow is    A v = ( / 4) (5.0 x 10-4m2) (v)           =......... m3 / s

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