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 2.0 m. Find the velocity, vm (in m/s), and the volumetric flow rate, Q (in m3/s), of the water in the main.
vm=
Q=
vm = 2.213594362
Q = 0.278168471
Can you plz give the formula??
wrong
vm : 0.39597
Q : 0.04972
To find the velocity, vm, and the volumetric flow rate, Q, of the water in the main using the Venturi flow meter, we can use the principle of the Bernoulli equation.
The Bernoulli equation states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of fluid is constant along a streamline.
First, we need to find the velocity, v1, at the larger diameter of the water main. We can use the equation of continuity, which states that the product of the cross-sectional area and the velocity at any point in an incompressible fluid flow is constant.
A1 * v1 = A2 * v2
Given:
Diameter of the water main (D1) = 40.0 cm = 0.4 m
Diameter of the constriction (D2) = 10.0 cm = 0.1 m
Difference in water level (h) = 2.0 m
Calculating the cross-sectional area of the water main A1:
A1 = π * (D1/2)^2
A1 = π * (0.4/2)^2
A1 = π * (0.2)^2
A1 ≈ 0.12566 m^2
Calculating the cross-sectional area of the constriction A2:
A2 = π * (D2/2)^2
A2 = π * (0.1/2)^2
A2 = π * (0.05)^2
A2 ≈ 0.00785 m^2
Applying the equation of continuity:
A1 * v1 = A2 * v2
0.12566 * v1 = 0.00785 * v2
v1 = (0.00785 * v2) / 0.12566
Next, we can use the Bernoulli equation to relate the velocities and the pressure difference between the two vertical pipes.
Bernoulli equation:
P1 + 0.5 * ρ * v1^2 + ρ * g * h1 = P2 + 0.5 * ρ * v2^2 + ρ * g * h2
Where:
P1 = pressure at the larger diameter of the water main
P2 = pressure at the constriction
ρ = density of water (approximately 1000 kg/m^3)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h1 = height difference between the water main and the constriction
h2 = 0 (as both vertical pipes are open at the top)
Rearranging the equation:
P1 - P2 = 0.5 * ρ * (v2^2 - v1^2) + ρ * g * h1
As both vertical pipes are open at the top, the pressure difference (P1 - P2) can be approximated as zero.
0 = 0.5 * ρ * (v2^2 - v1^2) + ρ * g * h1
Simplifying the equation:
0.5 * ρ * (v2^2 - v1^2) = -ρ * g * h1
Now, we can substitute the value of v1 into the equation:
0.5 * ρ * (v2^2 - [(0.00785 * v2) / 0.12566]^2) = -ρ * g * h1
Simplifying further:
0.5 * ρ * (v2^2 - [(0.00785^2 * v2^2) / 0.12566^2]) = -ρ * g * h1
Dividing by ρ:
0.5 * (v2^2 - [(0.00785^2 * v2^2) / 0.12566^2]) = -g * h1
Simplifying and rearranging:
0.5 * v2^2 - 0.00785^2/0.12566^2 * v2^2 = -2 * g * h1
Solving for v2:
0.5 * v2^2 - 0.00785^2/0.12566^2 * v2^2 = -2 * 9.8 * 2
0.5 * v2^2 - 0.00785^2/0.12566^2 * v2^2 = -39.2
0.5 * v2^2 - 0.0000392 * v2^2 = -39.2
0.5 * v2^2 - 0.0000392 * v2^2 = -39.2
0.5 * v2^2 - 3.872e-9 * v2^2 = -39.2
(0.5 - 3.872e-9) * v2^2 = -39.2
v2^2 = -39.2 / (0.5 - 3.872e-9)
v2^2 = -39.2 / 0.499999996128
v2^2 ≈ -78.408
Since velocity cannot be negative in this context, there seems to be an error in the given problem. Please check the values provided for the difference in water level (h) and ensure that it is correct.