A 6.6 diameter horizontal pipe gradually narrows to 3.6 . When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 30.0 and 26.0 , respectively. What is the volume rate of flow? Answer in m^3/s
12
To find the volume rate of flow of water through the pipe, we can make use of Bernoulli's equation which relates the pressure, height, and velocity of a fluid flowing in a pipe.
Bernoulli's equation is given by:
P1 + 1/2 ρv1^2 + ρgh1 = P2 + 1/2 ρv2^2 + ρgh2
Where:
P1 and P2 are the gauge pressures at points 1 and 2 respectively,
ρ is the density of water,
v1 and v2 are the velocities at points 1 and 2 respectively,
g is the acceleration due to gravity,
h1 and h2 are the heights at points 1 and 2 respectively.
In this case, the two points are the sections where the pipe diameter narrows.
Since the pipe is horizontal, the height difference (h1-h2) can be assumed to be negligible. Therefore, the equation can be simplified to:
P1 + 1/2 ρv1^2 = P2 + 1/2 ρv2^2
Substituting the given values:
P1 = 30.0 psi (converted to Pa) = 206842 Pa
P2 = 26.0 psi (converted to Pa) = 179231 Pa
ρ = density of water = 1000 kg/m^3
Now, let's find the velocities v1 and v2.
To find the velocity at a given point, we can use the equation:
v = Q/A
Where:
Q is the volume rate of flow,
A is the cross-sectional area of the pipe.
The area can be found using the formula for the area of a circle:
A = π * r^2
Given:
diameter of the larger section = 6.6 m
radius of the larger section = 6.6/2 = 3.3 m
diameter of the narrower section = 3.6 m
radius of the narrower section = 3.6/2 = 1.8 m
We can determine the velocity at each section by dividing the volume rate of flow by the cross-sectional area.
Let's calculate the velocities:
v1 = Q/A1
v2 = Q/A2
Where A1 is the cross-sectional area of the larger section and A2 is the cross-sectional area of the narrower section.
A1 = π * (3.3)^2
A2 = π * (1.8)^2
Now, we can rewrite Bernoulli's equation as:
P1 + 1/2 ρ(v1^2) = P2 + 1/2 ρ(v2^2)
Simplifying further:
(v1^2 - v2^2)/2 = (P2 - P1)/ρ
Substituting the known values:
(v1^2 - v2^2)/2 = (179231 - 206842)/1000
Now, solve for v1^2 - v2^2, which represents the squared difference in velocities.
v1^2 - v2^2 = ((179231 - 206842)/1000) * 2
Finally, find the volume rate of flow Q:
Q = v1 * A1
Substituting v1 = √((v1^2 - v2^2) + v2^2) and A1 = π * (3.3)^2, we get:
Q = √((v1^2 - v2^2) + v2^2) * π * (3.3)^2
Simplifying and substituting the values of v1^2 - v2^2:
Q = √(((((179231 - 206842)/1000) * 2) + v2^2)) * π * (3.3)^2
Calculate the value of Q to find the volume rate of flow through the pipe in m^3/s.