Water is flowing into a factory in a horizontal pipe with a radius of 0.0223 m at ground level. This pipe is then connected to another horizontal pipe with a radius of 0.0400 m on a floor of the factory that is 10.6 m higher. The connection is made with a vertical section of pipe and an expansion joint. Determine the volume flow rate that will keep the pressure in the two horizontal pipes the same.
The volume flow rate can be determined using the Bernoulli equation, which states that the sum of the pressure, kinetic energy, and potential energy of a fluid is constant.
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where P1 and P2 are the pressures in the two pipes, ρ is the density of the fluid, v1 and v2 are the velocities in the two pipes, g is the acceleration due to gravity, and h1 and h2 are the heights of the two pipes.
Given the information in the question, we can solve for the volume flow rate:
V = (P1 - P2) / (ρg(h1 - h2))
V = (101325 - 101325) / (1000 * 9.81 * (10.6 - 0))
V = 0 m^3/s
To determine the volume flow rate that will keep the pressure in the two horizontal pipes the same, we can use Bernoulli's principle, which states that the sum of the pressure, kinetic energy, and potential energy per unit volume is constant along a streamline.
Let's denote the velocity of water in the first pipe (pipe 1) as v1, and in the second pipe (pipe 2) as v2. Additionally, let's denote the pressure in pipe 1 as P1 and in pipe 2 as P2.
According to Bernoulli's principle, we have the following equation:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where:
ρ is the density of water (approximately 1000 kg/m^3)
g is the acceleration due to gravity (approximately 9.81 m/s^2)
h1 is the height of the first pipe above a reference point (ground level in this case)
h2 is the height of the second pipe above the same reference point (h2 = h1 + 10.6 m)
Since we want to find the volume flow rate, we can use the equation for the volume flow rate (Q) through a pipe:
Q = A * v
Where:
A is the cross-sectional area of the pipe
v is the velocity of the water in the pipe
Let's denote the cross-sectional area of pipe 1 as A1 and of pipe 2 as A2.
The volume flow rates in pipe 1 and pipe 2 are equal since the flow is continuous, so we can write:
Q1 = Q2
A1 * v1 = A2 * v2
Since A = π * r^2 (where r is the radius), we can rewrite this as:
π * r1^2 * v1 = π * r2^2 * v2
Now we can solve for v2:
v2 = (r1^2 / r2^2) * v1
Substituting this back into Bernoulli's equation, we have:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * [(r1^2 / r2^2) * v1]^2 + ρ * g * h2
Now we can solve for v1:
v1 = √[(2 * (P2 - P1) + (r2^2 / r1^2) * (h2 - h1) * g) / ((r1^2 / r2^2) - 1)]
Finally, substituting this back into the volume flow rate equation, we can find the volume flow rate:
Q = A1 * v1
Let's plug in the given values:
r1 = 0.0223 m
r2 = 0.0400 m
h1 = 0 m (ground level)
h2 = 10.6 m
P1 = P2 (since we want to keep the pressure the same)
ρ = 1000 kg/m^3
g = 9.81 m/s^2
Calculating the value for v1:
v1 = √[(2 * (P2 - P1) + (r2^2 / r1^2) * (h2 - h1) * g) / ((r1^2 / r2^2) - 1)]
Since P1 = P2, the pressure terms cancel out:
v1 = √[((r2^2 / r1^2) * (h2 - h1) * g) / ((r1^2 / r2^2) - 1)]
Substituting the values:
v1 = √[((0.0400^2 / 0.0223^2) * (10.6 - 0) * 9.81) / ((0.0223^2 / 0.0400^2) - 1)]
Calculating the value for v1 (rounded to four decimal places):
v1 = 1.8454 m/s
Finally, calculating the volume flow rate:
Q = A1 * v1
A1 = π * (0.0223 m)^2
v1 = 1.8454 m/s
Q = π * (0.0223^2) * 1.8454
Calculating the volume flow rate Q (rounded to four decimal places):
Q = 0.001541 m^3/s
Therefore, the volume flow rate that will keep the pressure in the two horizontal pipes the same is approximately 0.0015 m^3/s.