A horizontal pipe 10cm has asmootion to apipe 5 cm in diameter if the pressure of water in the large pipe is 80000pa and the pressure in the smaller pipe is 60000pa at what rate does water flow through the pipes
velocity in small pipe is four times velocity in big pipe
p + 1/2 rho v^2 is constant and rho is constant
8*10^4 + .5 rho v^2 = 6*10^4 + .5 rho (16)v^2
density of water = rho is about 10^3 kg/m^3
so
8*10^4 + .5*10^3 v^2 = 6^10^4 +8*10^3 v^2
7.5 *10^3 v^2 = 2*10^4
v^2 = (20/7.5)
solve for v in meters/s in large pipe
then volume flow rate = pi r^2 v
note use meters for radius 0.10
To find the rate at which water flows through the pipes, we can apply Bernoulli's equation, which relates the flow of a fluid to its pressure and cross-sectional area. Bernoulli's equation is given by:
P1 + (1/2)ρv1² + ρgh1 = P2 + (1/2)ρv2² + ρgh2
Where:
P1 and P2 are the initial and final pressures in the pipes respectively,
ρ is the density of the fluid (water in this case),
v1 and v2 are the velocities of the fluid in the large and small pipes respectively,
g is the acceleration due to gravity, and
h1 and h2 are the heights of the fluid in the large and small pipes respectively (we can assume they are at the same height since the pipe is horizontal).
The terms (1/2)ρv1² and (1/2)ρv2² represent the kinetic energy of the fluid, and ρgh1 and ρgh2 represent the potential energy of the fluid due to its height.
In this case, we have:
P1 = 80000 Pa (pressure in the large pipe)
P2 = 60000 Pa (pressure in the small pipe)
v1 = ? (velocity in the large pipe)
v2 = ? (velocity in the small pipe)
Since the pipes are horizontal, the heights of the fluid in both pipes, h1 and h2, can be assumed to be the same and therefore can be canceled out in Bernoulli's equation.
Simplifying Bernoulli's equation, we get:
P1 + (1/2)ρv1² = P2 + (1/2)ρv2²
Now, we need to find the velocities v1 and v2. To do this, we can use the relationship between velocity and cross-sectional area, given by the equation:
A1v1 = A2v2
Where A1 and A2 are the cross-sectional areas of the large and small pipes respectively.
Since the small pipe has a diameter of 5 cm, its radius is 2.5 cm (or 0.025 m). Therefore, its cross-sectional area A2 is given by:
A2 = π(0.025²) = 0.0019635 m²
The large pipe has a diameter of 10 cm, so its radius is 5 cm (or 0.05 m). Therefore, its cross-sectional area A1 is:
A1 = π(0.05²) = 0.0078539 m²
Now we can substitute the values of A1v1 and A2v2 into Bernoulli's equation:
P1 + (1/2)ρ(A1v1)² = P2 + (1/2)ρ(A2v2)²
Solving for v1 and v2, we can rewrite Bernoulli's equation as:
v1² = ((2(P2 - P1)) / ρ) + (A2 / A1)²(v2²)
And applying the equation for A2v2 = A1v1, we get:
v2 = (A1 / A2)v1
Substituting the expression for v2 into the previous equation, we have:
v1² = ((2(P2 - P1)) / ρ) + ((A2 / A1)²((A1 / A2)v1)²)
v1² = ((2(P2 - P1)) / ρ) + ((A1² / A2²)v1²)
Bringing all the terms involving v1 to one side:
v1² - ((A1² / A2²)v1²) = ((2(P2 - P1)) / ρ)
Simplifying further, we get a quadratic equation in terms of v1:
((A2² - A1²) / A2²)v1² = ((2(P2 - P1)) / ρ)
Now we can substitute the known values into this equation and solve for v1.
Note: In this explanation, we have derived the equation to find the velocity v1. To find the rate at which water flows through the pipe, you would need to multiply the velocity by the cross-sectional area of the pipe (A1).