Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.70 105 Pa and the pipe radius is 2.80 cm. At the higher point located at y = 2.50 m, the pressure is 1.22 105 Pa and the pipe radius is 1.70 cm.
(a) Find the speed of flow in the lower section.
m/s
(b) Find the speed of flow in the upper section.
m/s
(c) Find the volume flow rate through the pipe.
m3/s
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Obviously your previous answer wasn't good enough seeing as many others keep posting the same question. Please consider revising.
To solve this problem, we can apply Bernoulli's equation, which relates pressure, velocity, and height in a fluid flow system.
(a) To find the speed of flow in the lower section, we first need to find the velocity using Bernoulli's equation:
P_1 + 1/2 * ρ * v_1^2 + ρ * g * h_1 = P_2 + 1/2 * ρ * v_2^2 + ρ * g * h_2
Where:
P_1 is the pressure at the lower point,
v_1 is the velocity at the lower point,
P_2 is the pressure at the higher point,
v_2 is the velocity at the higher point,
ρ is the density of water, and
g is the acceleration due to gravity.
Since the flow is steady and ideal, we can assume that the height of the liquid remains constant, so h1 = h2 = 2.50 m. Using this information, we can rearrange the equation to solve for v1:
P_1 + 1/2 * ρ * v_1^2 = P_2 + 1/2 * ρ * v_2^2
Substituting the given values:
P_1 = 1.70 * 10^5 Pa
P_2 = 1.22 * 10^5 Pa
Now we can solve for v1:
1.70 * 10^5 + 1/2 * ρ * v_1^2 = 1.22 * 10^5 + 1/2 * ρ * v_2^2
We can ignore the height term since it is the same in both sections.
Now, let's calculate the speed of flow in the lower section.
(b) To find the speed of flow in the upper section, we can use the same approach as in part (a), but with the given values for the higher point.
(c) Once we have the speeds of flow in both sections, we can calculate the volume flow rate using the equation:
Q = A1 * v1 = A2 * v2
Where:
Q is the volume flow rate,
A1 is the cross-sectional area at the lower point,
A2 is the cross-sectional area at the higher point,
v1 is the speed of flow in the lower section, and
v2 is the speed of flow in the upper section.
Now, let's calculate the speed of flow in the upper section and the volume flow rate.
To find the speed of flow in the lower section, you can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid within a pipe.
Bernoulli's equation states: P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
where:
P1 and P2 are the pressures at points 1 and 2,
ρ is the density of the fluid (water in this case),
v1 and v2 are the velocities of the fluid at points 1 and 2,
g is the acceleration due to gravity,
and h1 and h2 are the heights of points 1 and 2.
In this case, point 1 is at the lower section and point 2 is at the higher section.
(a) First, we need to find the speed of flow in the lower section.
Given:
Pressure at the lower point (P1) = 1.70 × 10^5 Pa
Pipe radius at the lower point (r1) = 2.80 cm = 0.028 m
Substituting the given values into Bernoulli's equation:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Since point 1 and point 2 have the same height (horizontal pipe), we can disregard the h terms.
Therefore, the equation simplifies to:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
Solving for v1, the speed of flow in the lower section:
v1 = sqrt((2(P2 - P1))/ρ)
Now we need to convert the given pressure difference to the SI unit, Pascal.
Pressure difference (P2 - P1) = 1.22 × 10^5 Pa - 1.70 × 10^5 Pa = -0.48 × 10^5 Pa
Substituting all the given values and calculated pressure difference into the equation for v1:
v1 = sqrt((2(-0.48 × 10^5))/ρ)
The density of water (ρ) is approximately 1000 kg/m^3.
Using the given information, we can calculate the speed of flow in the lower section.
(b) To find the speed of flow in the upper section, we can use the same equation as in part (a), but now we'll substitute the given pressure difference between the upper and lower points.
Given:
Pressure at the higher point (P2) = 1.22 × 10^5 Pa
Pipe radius at the higher point (r2) = 1.70 cm = 0.017 m
Now we can substitute the given values into the equation for v2:
v2 = sqrt((2(P2 - P1))/ρ)
(c) To find the volume flow rate through the pipe, we can use the equation:
Volume flow rate (Q) = Av
where:
A is the cross-sectional area of the pipe (πr^2) and
v is the speed of flow.
To calculate the cross-sectional area at each point, we will use the corresponding radii (r1 and r2) given.
Substituting the calculated values into the equation for volume flow rate:
Q = A1 * v1 = A2 * v2
Now you can substitute all the calculated values and solve for the volume flow rate through the pipe.
See my previous answer.
http://www.jiskha.com/display.cgi?id=1352505868