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 A2 = 0.0700 m^2. The gas has a density of ρ = 1.80 kg/m^3. 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.
To find the speed of the gas in the larger original pipe, we can use Bernoulli's equation, which relates the pressure, density, and speed of a fluid in a continuous flow system.
(a) Let's start by using Bernoulli's equation for the venturi meter:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Since the venturi meter is horizontal and at the same level, we can ignore the terms involving the height (h) for this case. Also, we assume that the velocity v1 in the smaller pipe is negligible compared to v2, so we can ignore the (1/2)ρv1^2 term.
The equation simplifies to:
P1 = P2 + (1/2)ρv2^2
We are given that P2 - P1 = 180 Pa, so we can rewrite the equation as:
P2 - (180 Pa) = P2 + (1/2)ρv2^2
Now solve for v2:
(1/2)ρv2^2 = -180 Pa
v2^2 = -360 Pa / ρ
Since density (ρ) is positive, we can ignore the negative sign:
v2^2 = 360 Pa / ρ
Substitute the given value ρ = 1.80 kg/m^3:
v2^2 = 360 Pa / (1.80 kg/m^3)
Simplify:
v2^2 = 200 m^2/s^2
Take the square root of both sides to find v2:
v2 = √(200 m^2/s^2)
v2 = 14.14 m/s
Therefore, the speed of the gas in the larger original pipe is 14.14 m/s.
(b) To find the volume flow rate Q of the gas, we can use the equation:
Q = A2 * v2
Substitute the given values:
Q = (0.0700 m^2) * (14.14 m/s)
Q = 0.988 m^3/s
Therefore, the volume flow rate of the gas is 0.988 m^3/s.