Question 5 Flying Circus of Physics A venturi meter is used to measure the flow

Question

Question 5 Flying Circus of Physics 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 y. A manometer connects the wider portion of the meter to the narrower portion. The change in the fluids speed is accompanied by a change tip in the fluid?s pressure, which causes a height difference h of the liquid in the two arms of the manometer. (Here tip means pressure in the throat minus pressure in the pipe.) Let A equal 5.a. Suppose the pressure P1 at A is 2.3 atm. Compute the values of (a) the speed Vat A and (b) the speed y 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 8.0 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. (a) Number Units (b) Number Units (c) Number Units

Explanation / Answer

a) Apply Continuty equation

A1*v1 = A2*v2

5*a*V = a*v

==> 5*V = v

Now use Bernoulli's equation,

P1 + 0.5*rho*V1^2 = P2 + 0.5*rho*V2^2

P1 - P2 = 0.5*rho*(V2^2 - V1^2)

delta P = 0.5*rho*((5*V)^2 - V^2)

delta P = 12*rho*V^2

==> V = sqrt(delta P/(12*rho))

= sqrt(2.3*1.013*10^5/(12*1000))

= 4.4 m/s

b) v = 5*V

= 5*4.4

= 22 m/s

C) Volume flowrate = A*V

= (pi*d^2/4)*V

= (pi*0.08^2/4)*4.4

= 0.022 m^3/s

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