The speed in the constricted Venturi section is:
v2' = v2*(.07/.04) = 1.75 v2
Bernoull's principle tells you that:
P2 - P1 = (1/2)ñ [(1.75v2)^2 -v2^2]
= 180
= (1/2)*ñ*2.0625 v2^2
(a) Solve for v2
(b) Volume flow rate = A*v2
A venturi meter is a device for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed v2 through a horizontal section of pipe whose cross-sectional area A^2 = 0.0700 m2. The gas has a density of ρ = 1.80 kg/m3. The Venturi meter has a cross-sectional area of A1 = 0.0400 m^2 and has been substituted for a section of the larger pipe. The pressure difference between the two sections is P2 - P1 = 180 Pa.
(a) Find the speed v2 of the gas in the larger original pipe.
(b) Find the volume flow rate Q of the gas.
1 answer
1 answer