Please someone help..A horizontal pipe 10.0cm in diameter has a smooth reduction to a pipe 5.0cm in diameter. If the pressure of the water in the large pipe is 80000Pa and the pressure in the small pipe is 60000Pa at what rate does water flow through the pipe?
Well, well, well, looks like we have some flowy business here! Now, let's dive right in.
To find the rate at which water flows through the pipe, we can make use of the famous Bernoulli's principle. According to Mr. Bernoulli, the flow rate is inversely proportional to the cross-sectional area of the pipe. So, let's do some cross-sectional math!
The cross-sectional area of a pipe is given by the formula: A = πr², where "r" is the radius of the pipe.
In the large pipe, the diameter is 10.0 cm (or 0.1 m) which means the radius (r) is 0.05 m. So, the cross-sectional area of the large pipe (A₁) would be π(0.05)².
In the small pipe, the diameter is 5.0 cm (or 0.05 m) which means the radius (r) is 0.025 m. So, the cross-sectional area of the small pipe (A₂) would be π(0.025)².
Now, we can find the ratio of the flow rates (Q₁/Q₂) using the ratio of the cross-sectional areas (A₁/A₂).
Using the formula, Q₁/Q₂ = A₂/A₁, and plugging in the values, we have:
Q₁/Q₂ = (π(0.025)²) / (π(0.05)²)
Simplifying this, we get:
Q₁/Q₂ = (0.025)² / (0.05)²
Q₁/Q₂ = 0.025² / 0.05²
Q₁/Q₂ = 1/4
So, the rate at which water flows through the pipe is 1/4. That means, for every 1 unit of flow in the large pipe, there will be 0.25 units of flow in the small pipe.
I hope that floats your boat!
To find the rate at which water flows through the pipe, we can use Bernoulli's equation, which states that the total pressure at any point in an ideal fluid flow system is the sum of the static pressure, dynamic pressure, and potential energy per unit volume.
Bernoulli's equation can be written as:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 = pressure in the large pipe = 80000 Pa
P2 = pressure in the small pipe = 60000 Pa
ρ = density of water
v1 = velocity of water in the large pipe
v2 = velocity of water in the small pipe
g = gravitational acceleration
h1 = height of the large pipe
h2 = height of the small pipe
Since the pipes are horizontal, the height component (ρgh) can be ignored.
Let's assume that the fluid is incompressible, so the density (ρ) remains constant. Also, we'll assume that both pipes are at the same height, so the height component cancels out.
Therefore, the equation becomes:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
We can rewrite it as:
(1/2)ρv1^2 - (1/2)ρv2^2 = P2 - P1
Plugging in the given values:
(1/2)ρv1^2 - (1/2)ρv2^2 = 60000 - 80000
(1/2)ρv1^2 - (1/2)ρv2^2 = -20000
Since the flow is smooth, we can assume that the kinetic energy per unit volume is constant:
(1/2)ρv1^2 = (1/2)ρv2^2
v1^2 = v2^2
Taking the square root of both sides:
v1 = v2
This means that the velocity of water in the large pipe (v1) is the same as the velocity of water in the small pipe (v2).
Therefore, the rate at which water flows through the pipe is the same in both pipes.
To determine the rate of water flow through the pipe, you can use the principle of continuity, which states that the mass flow rate is constant at different points along a streamline (assuming incompressible fluid and steady flow).
The mass flow rate (ṁ) can be calculated using the equation:
ṁ = ρAv
Where:
ṁ is the mass flow rate
ρ is the density of the fluid
A is the cross-sectional area of the pipe
v is the velocity of the fluid
Since the density of water is typically around 1000 kg/m³, we can assume this value for ρ.
To find the velocities at each section, we can use Bernoulli's equation, which states that the total energy of a fluid particle is conserved along a streamline. Bernoulli's equation can be written as:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂,
Where:
P₁ and P₂ are the pressures at the two points
v₁ and v₂ are the velocities at the two points
h₁ and h₂ are the heights at the two points
ρ is the density of the fluid
g is the acceleration due to gravity
Since the pipe is horizontal, we can ignore the height terms in Bernoulli's equation.
Using the given pressures, we can rewrite Bernoulli's equation as:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
We can rearrange this equation to solve for v₂:
v₂ = sqrt((P₁ - P₂) * (2/ρ))
Now that we have the velocity at the smaller pipe (v₂), we can use the principle of continuity to find the velocity at the larger pipe (v₁).
Since the cross-sectional area of the smaller pipe (A₂) is known (calculated using the diameter), we can find the cross-sectional area of the larger pipe (A₁) using the equation:
A₁ = π(r₁)²
Where r₁ is the radius of the larger pipe (half the diameter).
Now that we have the cross-sectional areas of both pipes, we can find the velocity at the larger pipe (v₁) using the principle of continuity equation:
A₁v₁ = A₂v₂
Finally, we can substitute the known values into the equation and solve for v₁:
v₁ = (A₂v₂) / A₁
Once you have calculated v₁ (velocity at the larger pipe), you can multiply it with the cross-sectional area (A₁) to find the flow rate (Q = A₁v₁).
Solution
rho water = 1000 kg/m^3
P1 + (1/2) rho V1^2 = P1 + (1/2) rho V2^2
and
V1 A1 = V2 A2 so V2 = V1(100/25) = 4 V1
8*10^4 +(1/2)(10^3) V1^2 = 6*10^4 + (1/2)(10^3)(16 V1^2)
solve for V1
then volume flow rate = pi R1^2 V1
or
mass flow rate = 1000 pi R1^2 V1
note R1 = 0.10 meters