Water (η = 1.00E-3 Pa*s) is flowing through a horizontal pipe with a volume flow rate of 0.0120 m3/s. As the figure below shows, there are two vertical tubes that project from the pipe. Assume that H = 0.0421 m and L = 0.706 m. Calculate the radius of the horizontal pipe.
Use Poiseuille's Law, and sub rho*g*h for (P2-P1)
To calculate the radius of the horizontal pipe, we can use the principles of fluid dynamics and Bernoulli's equation.
First, let's understand the setup:
- Water is flowing through a horizontal pipe with two vertical tubes protruding from it.
- The length of the vertical tubes is labeled as L = 0.706 m.
- The height difference between the top of the pipe and the top of the water level in the vertical tubes is labeled as H = 0.0421 m.
- The volume flow rate of water through the pipe is given as Q = 0.0120 m^3/s.
- The dynamic viscosity of water is given as η = 1.00E-3 Pa*s.
Now, let's use Bernoulli's equation to solve for the radius of the horizontal pipe.
The equation is:
P₁ + ½ρv₁² + ρgh₁₁ = P₂ + ½ρv₂² + ρgh₂₂
Where:
P₁ and P₂ are the pressures at two different points in the fluid.
ρ is the density of the fluid (which we can assume to be the density of water, approximately 1000 kg/m³).
v₁ and v₂ are the velocities at points 1 and 2.
h₁₁ and h₂₂ are the heights at points 1 and 2.
In this case, we can simplify Bernoulli's equation by canceling out the height terms since they are at the same horizontal level (the water surface and the top of the pipe). Therefore, the equation becomes:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
Next, let's consider the points 1 and 2:
- Point 1 is inside the horizontal pipe where the water is flowing with a velocity v₁.
- Point 2 is in the vertical tube where the water has come to rest, and its velocity is now 0.
Since the water is flowing in a horizontal pipe, we can assume that the pressures at points 1 and 2 are the same (i.e., P₁ = P₂), so the equation further simplifies to:
½ρv₁² = ½ρv₂²
Now, we have two equations:
Q = Av₁ (equation 1)
and
0 = Av₂ (equation 2)
where A is the cross-sectional area of the pipe at points 1 and 2.
We can rearrange equation 1 to solve for v₁:
v₁ = Q/A
Similarly, rearranging equation 2 gives us:
v₂ = 0
Now, let's find the values of v₁ and v₂.
Substituting v₂ = 0 in the simplified Bernoulli's equation:
½ρv₁² = 0
Since v₂ is 0, we can cancel out the velocity terms, simplifying the equation further to:
0 = 0
This means that the pressure at point 1 is equal to the pressure at point 2.
Now, let's calculate the cross-sectional area A of the pipe.
From equation 1,
Q = Av₁
Solving for A:
A = Q / v₁
Using the given values:
- Q = 0.0120 m³/s (volume flow rate)
- v₁ = Q / A (from the previous equation)
Substituting the given values:
A = 0.0120 / v₁
Now, let's calculate the viscosity and density of water:
- η = 1.00E-3 Pa*s (dynamic viscosity)
- ρ = 1000 kg/m³ (density of water)
We know that dynamic viscosity (η) is equal to the product of viscosity (μ) and density (ρ):
η = μ * ρ
Therefore, we can rearrange the equation to solve for viscosity (μ):
μ = η / ρ
Substituting the given values:
μ = (1.00E-3) / (1000)
Now, we have all the information we need to calculate the radius of the horizontal pipe.
To do so, we can follow these steps:
1. Calculate the viscosity (μ) using the equation mentioned above.
2. Calculate the cross-sectional area (A) of the pipe using the equation A = Q / v₁.
3. Calculate the radius (r) of the pipe using the equation A = πr².