A Venturi meter is a device for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed v2 through a horizontal section of pipe whose cross-sectional area is A2 = 0.0779 m2. The gas has a density of ρ = 1.30 kg/m3. The Venturi meter has a cross-sectional area of A1 = 0.0202 m2 and has been substituted for a section of the larger pipe. The pressure difference between the two sections is P2 - P1 = 128 Pa. Find (a) the speed v2 of the gas in the larger original pipe and (b) the volume flow rate Q of the gas.
To solve this problem, we can make use of Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a pipe.
The equation can be written as:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at the two sections of the pipe,
v1 and v2 are the velocities at the two sections of the pipe,
ρ is the density of the gas,
g is the acceleration due to gravity,
h1 and h2 are the heights of the two sections (considering they are horizontal, h1 and h2 will cancel out).
In this case, we know:
P2 - P1 = 128 Pa,
A1 = 0.0202 m^2,
A2 = 0.0779 m^2,
ρ = 1.30 kg/m^3.
The first step is to determine the speed v2 of the gas in the larger original pipe (a):
We can rearrange the Bernoulli equation and solve for v2:
P1 - P2 = (1/2)ρ(v2^2 - v1^2)
Since P2 - P1 = 128 Pa, we have:
128 = (1/2)(1.30 kg/m^3)(v2^2 - v1^2)
Now we need to find v1. We can use the equation of continuity, which states that the volume flow rate is constant in an incompressible fluid:
Q1 = Q2
A1v1 = A2v2
Substituting the given values:
(0.0202 m^2)v1 = (0.0779 m^2)v2
We can solve this equation for v1:
v1 = (0.0779/0.0202)v2
Now let's substitute this expression for v1 back into the Bernoulli equation:
128 = (1/2)(1.30 kg/m^3)(v2^2 - [(0.0779/0.0202)v2]^2)
Simplifying:
128 = (1/2)(1.30 kg/m^3)(v2^2 - 15.614v2^2)
128 = (1/2)(1.30 kg/m^3)(-14.614v2^2)
Now we can solve for v2:
v2^2 = -(2(128))/(1.3(-14.614))
v2^2 = 99.847
Taking the square root of both sides:
v2 = 9.992 m/s
Therefore, the speed of the gas in the larger original pipe is 9.992 m/s.
Now let's move on to finding the volume flow rate Q of the gas (b):
From the equation of continuity, we know that:
Q1 = Q2
A1v1 = A2v2
Substituting the given values:
(0.0202 m^2)([(0.0779/0.0202)(9.992)])
Simplifying:
Q = (0.0202 m^2)(30.862 m/s)
Q = 0.628 m^3/s
Therefore, the volume flow rate of the gas is 0.628 m^3/s.
To find the speed v2 of the gas in the larger original pipe and the volume flow rate Q of the gas, we can use Bernoulli's equation and the continuity equation.
(a) To find the speed v2 of the gas in the larger original pipe:
1. Start with Bernoulli's equation, which relates the pressure, velocity, and height of a fluid at two points in a streamline. In this case, it can be written as:
P1 + (0.5)ρ(v1)^2 + ρgh1 = P2 + (0.5)ρ(v2)^2 + ρgh2
2. Since the pipe is horizontal, the height difference (h) is the same at both points and can be ignored. Additionally, we can assume that the velocity v1 is negligible compared to v2 because the Venturi meter acts as a constriction, causing an increase in fluid velocity.
3. Rewrite Bernoulli's equation with these assumptions:
P1 = P2 + (0.5)ρ(v2)^2
4. We are given the pressure difference P2 - P1, which is equal to 128 Pa. Rearrange the equation to solve for v2:
P1 - P2 = (0.5)ρ(v2)^2
128 Pa = (0.5)(1.30 kg/m^3)(v2)^2
5. Solve for v2:
v2 = sqrt((2)(128 Pa) / (1.30 kg/m^3))
(b) To find the volume flow rate Q of the gas:
1. The volume flow rate Q is defined as the volume of fluid passing through a certain point per unit time. It can be calculated using the equation:
Q = A1 * v1 = A2 * v2
2. We are given the values for A1 and A2, as well as v2 (which we found in part a). Plug in the values to calculate Q:
Q = (0.0202 m^2) * v2
Now you can substitute the values into the equations and calculate the desired quantities.