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 = 0.0769 m2. The gas has a density of ρ = 1.30 kg/m3. The Venturi meter has a cross-sectional area of A1 = 0.0107 m2 and has been substituted for a section of the larger pipe. The pressure difference between the two sections is P2 - P1 = 101 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 find the speed of the gas in the larger original pipe (v2), we can use Bernoulli's equation, which states that the total energy of a fluid flowing through a pipe is constant along a streamline.
The equation is given as:
P1 + 0.5ρv1^2 + ρgh1 = P2 + 0.5ρv2^2 + ρgh2
Where P1 and P2 are the pressures at points 1 and 2 respectively, v1 and v2 are the speeds of the fluid at points 1 and 2, ρ is the density of the fluid, g is the acceleration due to gravity, and h1 and h2 are the heights of the fluid at points 1 and 2 respectively.
In this case, since the section 2 is the larger original pipe, we have the following information:
P1 = P2 + 101 Pa (given)
A1 = 0.0107 m^2 (given)
A2 = 0.0769 m^2 (given)
ρ = 1.30 kg/m^3 (given)
We can rearrange Bernoulli's equation to solve for v2:
P1 - P2 = 0.5ρv2^2 * (A1/A2)^2
101 Pa = 0.5 * 1.30 kg/m^3 * v2^2 * (0.0107 m^2 / 0.0769 m^2)^2
Now let's solve for v2:
101 Pa = 0.5 * 1.30 kg/m^3 * v2^2 * 0.0185
101 Pa = 0.012925 kg/m s^2 * v2^2
v2^2 = 101 Pa / 0.012925 kg/m s^2
v2^2 = 7819.31 m^2/s^2
v2 ≈ 88.42 m/s
Therefore, the speed of the gas in the larger original pipe is approximately 88.42 m/s.
To find the volume flow rate Q of the gas, we can use the equation:
Q = A2 * v2
Now let's substitute the values:
Q = 0.0769 m^2 * 88.42 m/s
Q ≈ 6.85 m^3/s
Therefore, the volume flow rate of the gas is approximately 6.85 m^3/s.
To solve this problem, we can use the Bernoulli's equation, which states that the pressure difference (P2 - P1) between two points in a fluid flow is related to the speed difference (v2 - v1) through the equation:
P2 - P1 = (1/2) ρ (v1^2 - v2^2)
(a) To find the speed v2 of the gas in the larger pipe, we can rearrange the equation as follows:
P2 - P1 = (1/2) ρ (v1^2 - v2^2)
101 Pa = (1/2) * 1.30 kg/m^3 * (0 - v2^2)
101 Pa = -0.65 kg/m^3 * v2^2
Now we can solve for v2:
v2^2 = -(101 Pa) / (-0.65 kg/m^3)
v2^2 = 155.38 m^2/s^2
v2 = √(155.38) m/s
v2 ≈ 12.47 m/s
(b) To find the volume flow rate Q of the gas, we can use the equation:
Q = A2 * v2
Substituting the given values:
Q = 0.0769 m^2 * 12.47 m/s
Q ≈ 0.954 m^3/s
Therefore, the speed v2 of the gas in the larger original pipe is approximately 12.47 m/s and the volume flow rate Q of the gas is approximately 0.954 m^3/s.