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.
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
To solve this problem, we can apply the principle of continuity of flow, Bernoulli's equation, and the equation for volume flow rate.
(a) To find the speed v2 of the gas in the larger original pipe, we can use the principle of continuity, which states that the volume flow rate at any point in an incompressible fluid is constant. Therefore, the volume flow rate in the larger original pipe is equal to the volume flow rate in the Venturi meter.
Using the equation for volume flow rate Q = A * v, where Q is the volume flow rate, A is the cross-sectional area of the pipe, and v is the speed of the fluid, we can set up the equation:
Q1 = Q2
A1 * v1 = A2 * v2
Here, A1 is the cross-sectional area of the Venturi meter (0.0400 m^2) and A2 is the cross-sectional area of the larger original pipe (0.0700 m^2). We need to solve for v2.
Since the Venturi meter is a constriction, we can assume that the fluid speed v1 at the entrance to the Venturi meter is negligible. Therefore, v1 ≈ 0.
Substituting the given values, we have:
(0.0400 m^2) * 0 = (0.0700 m^2) * v2
v2 = 0 m/s
Therefore, the speed of the gas in the larger original pipe is 0 m/s.
(b) To find the volume flow rate Q of the gas, we can use the equation for volume flow rate Q = A * v, where Q is the volume flow rate, A is the cross-sectional area of the pipe, and v is the speed of the fluid.
Using the given values, we have:
Q = A2 * v2
Q = (0.0700 m^2) * 0 m/s
Q = 0 m^3/s
Therefore, the volume flow rate of the gas is 0 m^3/s.
To find the speed v2 of the gas in the larger original pipe, we can make use of Bernoulli's equation. Bernoulli's equation relates the pressure, density, and velocity of a fluid at different points of a flow. It states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline.
We can express Bernoulli's equation as:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Since the pipe is horizontal, the potential energy term ρgh1 and ρgh2 can be neglected. Also, the pressure difference ∆P = P2 - P1 is given as 180 Pa.
Now, the Venturi meter is designed such that the velocity at the narrow section (A1) is higher than the velocity at the larger section (A2). Therefore, the velocity at the narrow section (A1) can be assumed to be the maximum velocity (v1).
The cross-sectional areas are given as follows:
A1 = 0.0400 m^2
A2 = 0.0700 m^2
Using Bernoulli's equation, we can rewrite it as:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
Substituting the given values:
P1 + (1/2)(1.80 kg/m^3)v1^2 = P2 + (1/2)(1.80 kg/m^3)v2^2
Since the pressure difference ∆P = P2 - P1 is given as 180 Pa, we can substitute P2 - P1 = 180 Pa into the equation:
(1/2)(1.80 kg/m^3)v1^2 = 180 Pa + (1/2)(1.80 kg/m^3)v2^2
Simplifying the equation, we can rearrange it to solve for v2:
(1/2)(1.80 kg/m^3)v2^2 = (1/2)(1.80 kg/m^3)v1^2 - 180 Pa
Now, we can solve for v2 by substituting the given values and solving the equation algebraically.