1. A Venturi meter is a device for measuring the speed of a fluid within a pipe.

Question

1. A Venturi meter is a device for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed v2 through a horizontal section of pipe whose cross-sectional area is A2 = 0.0674 m2. The gas has a density of = 1.30 kg/m3. The Venturi meter has a cross-sectional area of A1 = 0.0162 m2 and has been substituted for a section of the larger pipe. The pressure difference between the two sections is P2 - P1 = 119 Pa. Find (a) the speed v2 of the gas in the larger original pipe and (b) the volume flow rate Q of the gas.

2. A liquid is flowing through a horizontal pipe whose radius is 0.0321 m. The pipe bends straight upward through a height of 3.30 m and joins another horizontal pipe whose radius is 0.0572 m. What volume flow rate will keep the pressures in the two horizontal pipes the same?

Explanation / Answer

1)Figure is missing for this one

2)

r = 0.0321 m ; h = 3.3 m ; r' = 0.0572 m

We know from Bernaulli's equation:

P1 + 1/2 rho v1^2 + rho g y1 = P2 + 1/2 rho v2^2 + rho g y2

For P1 = P2 ; y2 - y1 = 3.3 m

1/2 rho v1^2 = 1/2 rho v2^2 + rho g (y2 - y1) (1)

Volume flow rate is goven by:

Q = A v

A1v1 = A2v2 => v1 = A2v2/A1

putting the value of V1 from above is (1)

1/2 rho (A2v2/A1)^2 = 1/2 rho v2^2 + rho g(y2 - y1)

1/2 [(A2/A1)^2 - 1] v2^2 = h (y2 - y1)

v2^2 = 2 g (y2 - y1)/[(A2/A1)^2 - 1]

v2 = sqrt [2 g (y2 - y1)/[(A2/A1)^2 - 1]]

v2 = sqrt [ 2 x 9.8 x 3.3/(pi x 0.0572^2/ pi x 0.0321^2) - 1]

v2 = 5.45 m/s

Q = A2v2

Q2 = 3.14 x 0.0572^2 x 5.45 = 0.056 m^3/s

Hence, Q2 = 0.056 m^3/s

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.