Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in figure below, the pressure is P1 = 1.75X104 Pa and the pipe diameter is 6.00 cm. At another point y = 0.250 m higher, the pressure is P2 = 1.20X104 Pa and the pipe diameter is 3.00 cm. Find the speed of flow (A) in the lower section and (B) in the upper section. (C) Find the volume flow rate through the pipe.
thanks, it was so helpful. not being sarcastic.
To solve this problem, we can use the principle of conservation of energy for fluid flow, also known as Bernoulli's equation. Bernoulli's equation states that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume of a fluid is constant along a streamline. Using this equation, we can solve for the speed of flow in each section and the volume flow rate through the pipe.
Step 1: Convert the given pressure values to SI units.
P1 = 1.75 × 10^4 Pa
P2 = 1.20 × 10^4 Pa
Step 2: Convert the given pipe diameters to radii.
r1 = 6.00 cm / 2 = 3.00 cm = 0.03 m
r2 = 3.00 cm / 2 = 1.50 cm = 0.015 m
Step 3: Apply Bernoulli's equation between the two points.
P1 + 1/2 ρv1^2 + ρgh1 = P2 + 1/2 ρv2^2 + ρgh2
Step 4: Simplify the equation using the given information.
Since the two points are at the same height (y = 0.250 m), the potential energy term cancels out.
P1 + 1/2 ρv1^2 = P2 + 1/2 ρv2^2
Step 5: Solve for the speed of flow in the lower section (v1).
v1^2 = (2(P2 - P1))/ ρ
Step 6: Calculate the density of water.
The density of water (ρ) is approximately 1000 kg/m^3.
Step 7: Plug in the values and calculate v1.
v1^2 = (2(1.20 × 10^4 Pa - 1.75 × 10^4 Pa))/(1000 kg/m^3)
v1^2 = -0.55 m^2/s^2
Since the velocity squared is negative, this means there is no real flow in the lower section of the pipe. However, it is possible that the equation was not set up correctly or there is an error in the given values.
Step 8: Calculate the speed of flow in the upper section (v2).
v2^2 = v1^2 + 2g(h2 - h1)
v2^2 = 0 + 2(9.8 m/s^2)(0.250 m)
v2^2 = 4.9 m^2/s^2
Step 9: Take the square root of v2^2 to find v2.
v2 = √4.9 m^2/s^2
v2 ≈ 2.21 m/s
The speed of flow in the upper section of the pipe is approximately 2.21 m/s.
Step 10: Calculate the volume flow rate through the pipe.
The volume flow rate (Q) is defined as the product of the cross-sectional area (A) and the velocity (v).
Step 11: Calculate the cross-sectional area of each section.
A1 = πr1^2 = π(0.03 m)^2 ≈ 0.00283 m^2
A2 = πr2^2 = π(0.015 m)^2 ≈ 0.00071 m^2
Step 12: Calculate the volume flow rate.
Q = A1v1 = A2v2
Q = (0.00283 m^2)(2.21 m/s)
Q ≈ 0.0063 m^3/s
The volume flow rate through the pipe is approximately 0.0063 m^3/s.
To find the speed of flow in the lower section (A), we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a horizontal flow. The equation is given as:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where:
P1 and P2 are the pressures at the lower and higher points respectively,
v1 and v2 are the velocities at the lower and higher points respectively,
ρ is the density of the fluid (in this case, water),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
h1 and h2 are the heights at the lower and higher points respectively.
In this case, the lower point (point 1) is given by P1 = 1.75 * 10^4 Pa and the pipe diameter is 6.00 cm. The higher point (point 2) is located at a height of 0.250 m above the lower point and has a pressure given by P2 = 1.20 * 10^4 Pa.
To solve for the speed of flow in the lower section (A):
1. Calculate the height difference between the two points: h1 = 0 m (reference) and h2 = 0.250 m.
2. Convert the pipe diameter at point 1 to radius: r1 = 6.00 cm / 2 = 0.03 m.
3. Use the formula A = π * r^2 to find the cross-sectional area of the pipe at point 1: A1 = π * (0.03 m)^2.
4. From the equation above, rearrange it to solve for v1:
P1 - P2 = ρ * (v2^2 - v1^2) / 2
v1 = sqrt((2 * (P1 - P2)) / ρ) = sqrt((2 * (1.75 * 10^4 - 1.20 * 10^4)) / ρ)
5. Calculate the density of water: ρ = 1000 kg/m^3.
6. Substitute the known values into the equation to find the speed of flow in the lower section (A) at point 1.
To find the speed of flow in the upper section (B), we can use the principle of continuity, which states that the volume flow rate is constant in a steady flow. Mathematically, it can be expressed as:
A1 * v1 = A2 * v2
Where:
A1 and A2 are the cross-sectional areas of the pipe at the lower and higher points respectively,
v1 and v2 are the velocities at the lower and higher points respectively.
From the previous steps, we already have the values for A1 and v1. The diameter at the upper point (point 2) is given as 3.00 cm, so the radius at point 2 will be r2 = 3.00 cm / 2 = 0.015 m. Using the formula A = π * r^2, we can calculate A2. Substituting the known values into the equation, we can solve for v2.
Finally, to find the volume flow rate through the pipe (C), we can use the formula:
Q = A1 * v1
Where:
Q is the volume flow rate,
A1 is the cross-sectional area of the pipe at the lower point (which we already calculated),
v1 is the speed of flow in the lower section (which we found earlier).
Substituting the known values into the equation will give us the volume flow rate through the pipe.
p1 + rho g h1 + (1/2) rho v1^2 = p1 + rho g h1 + (1/2) rho v1^2
and v2 A2 = v1 A1 = Q , because Q, vol flow, is constant
density = rho = 10^3 kg/m^3
g = 9.8
p1 = 1.75*10^4 N/m^2
r1 = 0.030 m so A1 = pi (.03)^2 = 2.83*10^-3 m^2
h1 = 0, select as base height
p2 = 1.20*10^4
r2 = 0.015 m so A2 = pi (0.015)^2 = 0.707*10^-3 m^2
h2 = 0.250 m
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v2 = v1 (A1/A2) = v1 (2.83/.707) = 4 v1
so
1.75*10^4 + 0 + 500 v1^2 = 1.2*10^4 + 10^3 * 9.8*0.25 + 500*16v1^2
solve for v1, then go back for the rest
v2 = 4 v1
Q = A1 v1 = A2 v2