A horizontal water pipe goes from a large diameter to a small diameter and then back to the first diameter as shown in the figure below. The level of water (8cm in the larger tube and 4 cm in the constricted tube) in the small vertical tubes provides us with information about the water pressure in the two different sizes of pipe. The inside diameter of the larger pipe is 2.60 cm and water travels through both sizes of pipe with a volume flow rate of 2.00 10-4 m3/s. Determine the inside diameter of the smaller pipe.
I have attempted many times by combining Bernoulli's Equation with the equation of continuity to no avail. Please help.
The answer is 1.30 cm. To solve this problem, you need to use the equation of continuity, which states that the flow rate (Q) is equal to the product of the cross-sectional area (A) and the velocity (V) of the fluid. Since the flow rate is given, you can solve for the velocity of the fluid in the two pipes. Then, you can use Bernoulli's equation to calculate the pressure in each pipe. Finally, you can use the pressure difference between the two pipes to calculate the inside diameter of the smaller pipe.
To determine the inside diameter of the smaller pipe, we can apply Bernoulli's equation and the equation of continuity.
First, let's review Bernoulli's equation:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Where:
P₁ and P₂ are the pressures at points 1 and 2 (in our case, the larger and smaller pipe),
ρ is the density of water (~1000 kg/m³),
v₁ and v₂ are the velocities of water at points 1 and 2, respectively,
g is the acceleration due to gravity (~9.8 m/s²),
h₁ and h₂ are the heights of the water levels in the smaller vertical tubes.
From the equation of continuity, we know that the volume flow rate (Q) is constant along the pipe:
A₁v₁ = A₂v₂
Where:
A₁ and A₂ are the cross-sectional areas of the larger and smaller pipe, respectively.
Now, let's begin solving the problem step-by-step:
Step 1: Convert the given values to SI units.
Inside diameter of the larger pipe (D₁) = 2.60 cm = 0.0260 m
Water level in the larger vertical tube (h₁) = 8 cm = 0.08 m
Water level in the smaller vertical tube (h₂) = 4 cm = 0.04 m
Volume flow rate (Q) = 2.00 × 10⁻⁴ m³/s
Step 2: Calculate the cross-sectional areas of the pipes.
A₁ = (π/4)D₁²
A₂ = (π/4)D₂²
Step 3: Use the equation of continuity to find the velocity of water in the smaller pipe.
v₁ = Q / A₁
v₂ = Q / A₂
Step 4: Substitute the values into Bernoulli's equation and simplify.
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Assuming the pressures at both points are atmospheric (P₁ = P₂ = Patm), we can simplify the equation further:
½ρv₁² + ρgh₁ = ½ρv₂² + ρgh₂
Step 5: Substitute the expressions for v₁, v₂, A₁, and A₂ into the simplified Bernoulli's equation.
½ρ(Q/A₁)² + ρgh₁ = ½ρ(Q/A₂)² + ρgh₂
Step 6: Solve for D₂, the inside diameter of the smaller pipe.
Simplify the equation and solve for D₂:
(0.5ρ(Q/A₁)² + ρgh₁) - (0.5ρ(Q/A₂)² + ρgh₂) = 0
D₂ = sqrt((2gh₁ - 2gh₂ + Q² (A₂² - A₁²) / (A₁²)))
Substitute the given values into the equation to find D₂.
To solve this problem, you can use the principles of Bernoulli's equation and the equation of continuity. Let's break it down step by step:
1. Bernoulli's equation relates the pressure, velocity, and height of a fluid in motion. It can be stated as P + (1/2)ρv^2 + ρgh = constant, where P is the pressure, ρ is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the height of the fluid.
2. The equation of continuity states that the volume flow rate of a fluid is constant throughout a pipe or tube, assuming no loss of fluid. Mathematically, it can be expressed as A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the two sections of the pipe, and v1 and v2 are the velocities of the fluid in each section.
Now let's apply these principles to the given problem:
1. Determine the cross-sectional area of the larger pipe (A1):
- The inside diameter of the larger pipe is given as 2.60 cm. We need to convert it to meters: 2.60 cm = 0.026 m.
- The radius of the larger pipe is half the diameter: r1 = 0.026 m / 2 = 0.013 m.
- The cross-sectional area of the larger pipe is A1 = π(r1)^2 = π(0.013 m)^2.
2. Use the equation of continuity to relate the velocities in the two sections:
- The volume flow rate is given as 2.00 * 10^(-4) m^3/s.
- Since A1v1 = A2v2, we can rearrange it to v2 = (A1/A2) * v1.
- Substitute the known values: v2 = (A1/A2) * v1 = (π(0.013 m)^2) / A2 * v1.
3. Apply Bernoulli's equation to the upper and lower points in the pipe:
- At the larger pipe, P1 = atmospheric pressure, v1 = 0 (water is at rest), h1 = 0.
- At the smaller pipe, P2 = atmospheric pressure, h2 = 8 cm = 0.08 m.
- The velocities v1 and v2 are related through the equation of continuity as mentioned in step 2.
With these equations set up, you can solve for the unknown variable, which is the inside diameter of the smaller pipe (A2). Plug the necessary values into the equations and solve for A2.