A Venturi flow meter is used to measure the the flow velocity of a water main. The water main has a diameter of 30.0 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 3.0 m.
Find the velocity, vm (in m/s)of the water in the main.
The volumetric flow rate, Q (in m3/s) of the water in the main.
volume flow is the same in the big section as the small.
pi Rbig^2 Vbig = pi Rsmall^2 Vsmall
so
Vsmall = Vbig (15^2/5^2) = 9 Vbig
then Bernoulli
(1/2)rho Vsmall^2 = (1/2) rho Vbig^2 +rho g h
2 g h = (Vsmall^2 - Vbig^2)
2 (9.81)(3) = 81Vbig^2 - 1 Vbig^2
= 80 Vbig^2
so
Vbig^2 = .736 m^2/s^2
Vbig = .858 m/s in big main
Q = pi (.15^2) .858 = .06 m^3/sec
thanks
I am not getting the correct answer, can you plz explain a bit better? Thankx
To find the flow velocity of the water in the main, we can use the Bernoulli's equation and the principle of continuity.
1. Bernoulli's equation:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where:
P1 and P2 are the pressures at two different points in the flow (in this case, the pressure at the top of the water column in each vertical pipe)
ρ is the density of the fluid (in this case, water)
v1 and v2 are the velocities at two different points in the flow (in this case, the velocity of the water in the main and the velocity in the constriction)
g is the acceleration due to gravity (9.8 m/s^2)
h1 and h2 are the heights of the water column in each vertical pipe
2. The principle of continuity:
A1 * v1 = A2 * v2
Where:
A1 and A2 are the cross-sectional areas of the flow at two different points (in this case, the area of the water main and the area of the constriction)
Now let's solve for the velocity in the main (vm):
Using the principle of continuity, we get:
π * r1^2 * vm = π * r2^2 * v2
Since the diameter is given, we can calculate the radius:
r1 = 0.5 * 30.0 cm = 15.0 cm = 0.15 m
r2 = 0.5 * 10.0 cm = 5.0 cm = 0.05 m
So the equation becomes:
π * (0.15 m)^2 * vm = π * (0.05 m)^2 * v2
The π terms on either side cancel out, simplifying the equation to:
(0.15 m)^2 * vm = (0.05 m)^2 * v2
Now let's calculate the velocity in the constriction (v2):
Using Bernoulli's equation, we can compare the two vertical pipes:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
The pressure at the top of both pipes is atmospheric pressure, which we can assume cancels out from both sides of the equation. Also, since the top of both pipes are open, the height terms (h1 and h2) cancel out. Thus, we get:
1/2 * ρ * v1^2 = 1/2 * ρ * v2^2
Now we can solve for v2:
v2^2 = v1^2 * (r1/r2)^2
Plugging in the known values:
v2^2 = vm^2 * (0.15 m / 0.05 m)^2
v2^2 = 9 * vm^2
Now we can substitute this value of v2^2 into the equation for the principle of continuity:
(0.15 m)^2 * vm = (0.05 m)^2 * 9 * vm^2
Simplifying the equation, we get:
0.0225 * vm = 0.0025 * 9 * vm^2
0.0225 = 0.0025 * 9 * vm
Now, solve for vm:
vm = 0.0225 / (0.0025 * 9)
vm ≈ 0.1 m/s
So, the flow velocity of the water in the main (vm) is approximately 0.1 m/s.
To find the volumetric flow rate (Q) of the water in the main, we can use the equation:
Q = A1 * vm
Where A1 is the cross-sectional area of the water main and vm is the flow velocity we just found.
Plugging in the values, we get:
Q = π * (0.15 m)^2 * 0.1 m/s
Calculating this expression, we find:
Q ≈ 0.0707 m^3/s
Therefore, the volumetric flow rate of the water in the main (Q) is approximately 0.0707 m^3/s.