Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.85 105 Pa and the pipe radius is 2.80 cm. At the higher point located at y = 2.50 m, the pressure is 1.28 105 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 in m/s
(c) Find the volume flow rate through the pipe in m^3/s
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To solve this problem, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a flowing system. The equation is given as:
P + 1/2ρv^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
h is the height of the fluid
(a) Find the speed of flow in the lower section:
At the lower point:
Pressure, P1 = 1.85 × 10^5 Pa
Radius, r1 = 2.80 cm = 0.028 m
We need to find the velocity, v1.
(b) Find the speed of flow in the upper section:
At the higher point:
Pressure, P2 = 1.28 × 10^5 Pa
Radius, r2 = 1.20 cm = 0.012 m
Height, h = 2.50 m
We need to find the velocity, v2.
(c) Find the volume flow rate through the pipe:
The volume flow rate is given by:
Q = Av
Where:
A is the cross-sectional area of the pipe
v is the velocity of the fluid
We need to find the volume flow rate, Q.
Let's solve each part step-by-step:
(a) Find the speed of flow in the lower section:
Using Bernoulli's equation, we can equate the pressures at the two points in the pipe:
P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2
Since the pipe is horizontal, the heights cancel out:
P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2
Solving for v1:
1/2ρv1^2 = P2 - P1 + 1/2ρv2^2
v1^2 = (2(P2 - P1))/ρ + v2^2
v1 = sqrt((2(P2 - P1))/ρ + v2^2)
Given:
P1 = 1.85 × 10^5 Pa
P2 = 1.28 × 10^5 Pa
ρ = density of water = 1000 kg/m³ (approximately)
Substituting the values, we have:
v1 = sqrt((2(1.28 × 10^5 - 1.85 × 10^5))/(1000) + v2^2)
Calculating the numerical value will give you the speed of flow in the lower section.
(b) Find the speed of flow in the upper section:
Using the same equation as above, but substituting the appropriate values, we have:
v2 = sqrt((2(P1 - P2))/(ρ) + v1^2)
Calculating the numerical value will give you the speed of flow in the upper section.
(c) Find the volume flow rate through the pipe:
The cross-sectional area of the pipe, A, is given by:
A = πr^2
For the lower section:
r1 = 0.028 m
For the upper section:
r2 = 0.012 m
The volume flow rate, Q, can be calculated as:
Q = A1v1 = A2v2
Substituting the values, we have:
Q = πr1^2v1 = πr2^2v2
Calculating the numerical value will give you the volume flow rate through the pipe.
To answer these questions, we can use Bernoulli's equation, which describes the relationship between pressure, velocity, and height in a fluid flow.
Bernoulli's equation is given by:
P1 + (1/2)ρv1^2 + ρgy1 = P2 + (1/2)ρv2^2 + ρgy2
Where:
P1 and P2 are the pressures at points 1 and 2
v1 and v2 are the velocities at points 1 and 2
ρ is the density of the fluid
g is the acceleration due to gravity
y1 and y2 are the heights at points 1 and 2
Let's solve the questions step by step:
(a) To find the speed of flow in the lower section, we can use the equation:
P1 + (1/2)ρv1^2 + ρgy1 = P2 + (1/2)ρv2^2 + ρgy2
Since the lower point is the reference point, we can set y1 = 0 and P1 = 1.85 x 10^5 Pa.
Also, we know that ρ and g are constants.
P2 + (1/2)ρv2^2 + ρgy2 = P1
Substituting the given values:
1.28 x 10^5 Pa + (1/2)ρv2^2 + ρgy2 = 1.85 x 10^5 Pa
Simplifying the equation, we get:
(1/2)ρv2^2 + ρgy2 = 1.85 x 10^5 - 1.28 x 10^5
(b) To find the speed of flow in the upper section, we can use the same equation as in part (a), but with different values for P2 and y2.
Simplifying the equation for the upper section, we get:
(1/2)ρv2^2 + ρgy2 = P2 - P1
(c) To find the volume flow rate through the pipe, we can use the equation:
Q = A1v1 = A2v2
Where Q is the volume flow rate, A1 and A2 are the respective cross-sectional areas at points 1 and 2, and v1 and v2 are the velocities at points 1 and 2.
Since the pipe is constricted, the cross-sectional areas change, but we can calculate them using the given radii of the pipe.
A1 = πr1^2
A2 = πr2^2
Substituting the values and the previously calculated velocities, we can find the volume flow rate Q.