A venturi meter is used to measure the how speed of a flu d in a pipe. The meter

Question

A venturi meter is used to measure the how speed of a flu d in a pipe. The meter is connected between two sections of the pipe (the figure); the cross-sectional area A the entrance and exit of the meter matches the pipe s cross sectional area Between the entrance and exit, the fluid Rows from the pipe with speed V and then through 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 Delta p m the fluid's pressure, which causes a height difference h of the liquid in the two arms of the manometer. (Here Delta p means pressure in the throat minus pressure in the pipe.) Let A equal 3 o. Suppose the pressure p_1 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 p_2 at a equal to zero, Compute the corresponding volume Row rate if the diameter at A is 6.0 cm. The phenomenon that occurs at a when p_2 falls to nearly zero is known as cavitation. Please assume that the fluid Is water. The water vaporizes into small hubbies.

Explanation / Answer

at point 1,

cross-sectional area is A1 =3*a

pressure is p1=2.1 atm

fluid speed is v1=V

diameter of pipe is d1=6cm

====> A1=pi*r1^2

A1=pi*(d1/2)^2

A1=pi*(0.06/2)^2

A1=2.83*10^-3 m^2

point 2.

cross-sectional area is A2= a =A1/3

A2=9.43*10^-3 m^2

pressure is p2

fluid speed is v2=v

a)

use,

A1*v1=A2*v2

3*a*V=a*v

====> V1=v/3

speed at point 1 is V1=v/3

b)

p1-p2=rho/2*(v2^2-v1^2)

(2.1*1.013*10^5)-0=1000/2*(v^2-(v/3)^2)

====> v=21.88 m/sec

c)

volume flow rate Q=A1*V1=A2*V2

Q=2.83*10^-3*(21.88/3)

Q =2.064*10^-2 m^3/sec

1) v1=v/3 ====> v1=7.3 m/sec

2) v2=21.88 m/sec

3) Q =2.064*10^-2 m^3/sec

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