Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.85x10^5 Pa and the pipe radius is 2.80 cm. At the higher point located at y = 2.50 m, the pressure is 1.28X10^5 Pa and the pipe radius is 1.20 cm.
(a) Find the speed of flow in the lower section.
(b) Find the speed of flow in the upper section.
(c) Find the volume flow rate through the pipe.
Can u please show steps on how to solve that question
To find the speed of flow in each section and the volume flow rate through the pipe, we can apply the principles of Bernoulli's equation and the continuity equation for fluid flow.
(a) Speed of flow in the lower section:
1. First, we need to find the velocity in the lower section using Bernoulli's equation, which states that the total pressure energy is constant along a streamline:
P1 + 1/2 * rho * v1^2 + rho * g * y1 = P2 + 1/2 * rho * v2^2 + rho * g * y2
Here:
P1 = 1.85 x 10^5 Pa (pressure at the lower point)
P2 = ? (pressure at the higher point - unknown)
rho = density of water (assume 1000 kg/m^3 for most cases)
v1 = ? (velocity at the lower point - unknown)
v2 = ? (velocity at the higher point - unknown)
g = acceleration due to gravity (approximately 9.8 m/s^2)
y1 = 0 (height at the lower point - reference point)
y2 = 2.50 m (height at the higher point)
2. Simplifying Bernoulli's equation for the lower section:
P1 + 1/2 * rho * v1^2 = P2 + 1/2 * rho * v2^2 + rho * g * y2
3. Since the pressure at the higher point, P2, is given as 1.28 x 10^5 Pa, we can substitute the known values:
1.85 x 10^5 Pa + 1/2 * rho * v1^2 = 1.28 x 10^5 Pa + 1/2 * rho * v2^2 + rho * g * y2
4. Now, we need the radius of the pipe in meters to calculate the areas:
r1 = 2.80 cm = 0.028 m (radius at the lower point)
r2 = 1.20 cm = 0.012 m (radius at the higher point)
5. The next step is to find the velocity at each point using the equation for the cross-sectional area:
A1 = pi * r1^2 (area at the lower point)
A2 = pi * r2^2 (area at the higher point)
6. To calculate the velocity, we use the continuity equation, which states that the volume flow rate is constant along a streamline:
A1 * v1 = A2 * v2
7. Rearranging the continuity equation to solve for v1:
v1 = (A2 * v2) / A1
8. Substituting the expressions for A1 and A2:
v1 = (pi * r2^2 * v2) / (pi * r1^2)
v1 = (r2^2 * v2) / r1^2
Now we can proceed to find the values for v1, v2, and other quantities.
(b) Speed of flow in the upper section:
9. Using the same equation and substituting the values:
v2 = (r1^2 * v1) / r2^2
(c) Volume flow rate through the pipe:
10. The volume flow rate is given by Q = A * v, where A is the cross-sectional area and v is the velocity.
Q = A1 * v1 = A2 * v2
Now we can follow the steps and calculations described above to find the values for v1, v2, and Q.