p + (1/2) rho v^2 = constant
rho = water density = about 1000 kg/m^3
p = rho g h
p big pipe + (1/2) rho vbig^2 = plittle pipe + (1/2)rho vlittle^2
p big pipe - p little pipe =1000(9.81)(4)
= 39240 Newtons/m^2 or Pascals
so
39240 = (1/2)(1000)(vlittle^2-vbig^2)
now what can we say about vbig and vlittle?
Same amount of water per second through both the big pipe and the little pipe
so
Vbig(pi Dbig^2/4)=Vlittle(pi*Dlittle^2/4)
or
vlittle = Vbig (40^2/10^2) =Vbig(16)
so now
2*39.240 = ([16vbig]^2-vbig^2)=255 vbig^2
so
vbig = .555 m/s
Q = .555 * pipe area = .555(pi*.4^2/4)
A Venturi flow meter is used to measure the the flow velocity of a water main. The water main has a diameter of 40.0 cm, and the constriction has a diameter of 10.0 cm. The two vertical pipes are open at the top, and the difference in water level between them is 4.0 m. Find the velocity, vm (in m/s), and the volumetric flow rate, Q (in m3/s), of the water in the main.
1 answer