1.)Water flows from a 2 cm diameter pipe, at a speed of 0.35m/sec. How long will it take to fill a 10 liter container?
2.)A horizontal segment of pipe tapers from a cross-sectional area of 50 cm^2 to 0.5 cm^2. The pressure at the larger end of the pipe is 1.2x 10^5 Pa and the speed is 0.04m/sec. What is the pressure at the narrow end of the segment?
1. time = Volume / (volume flow rate)
= V/(area*velocity)
2. Because of the continuity equation, V*Area = constant. Use that fact to get the narrow-end velocity. Then apply the Bernoulli equation for the pressure change.
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1.) To find the time it takes to fill a 10 liter container, we first need to find the volume of water that flows through the pipe per second.
The flow rate (Q) of water through a pipe can be calculated using the formula:
Q = A * V
Where:
Q = Flow rate (in cubic meters per second)
A = Cross-sectional area of the pipe (in square meters)
V = Velocity of the water (in meters per second)
Given that the diameter of the pipe is 2 cm, we can calculate the cross-sectional area (A) using the formula:
A = π * (d/2)^2
Where:
d = Diameter of the pipe
Plugging in the values, we have:
A = π * (2/2)^2
A = π * 1^2
A = π
Since the pipe is circular and we are given the diameter, we can use the value of π as it is.
Next, we can calculate the flow rate (Q) by multiplying the cross-sectional area (A) by the velocity (V):
Q = π * (2/2)^2 * 0.35
Q = π * 1^2 * 0.35
Q = 0.35π cubic meters per second
Since we want to find the time it takes to fill a 10 liter container, we need to convert the volume to liters and then divide it by the flow rate:
10 liters = 0.01 cubic meters (since 1 liter = 0.001 cubic meters)
Time = 0.01 / (0.35π) seconds
To get the final answer, we can round the result to an appropriate number of decimal places.
2.) To find the pressure at the narrow end of the pipe segment, we need to apply the principles of continuity and Bernoulli's principle.
According to the principle of continuity, the product of the cross-sectional area and the velocity of a fluid remains constant along a streamline. That is:
A1 * V1 = A2 * V2
Where:
A1 = Cross-sectional area of the larger end of the pipe
V1 = Velocity at the larger end of the pipe
A2 = Cross-sectional area of the narrow end of the segment
V2 = Velocity at the narrow end of the segment
Given that A1 = 50 cm^2, V1 = 0.04 m/sec, and A2 = 0.5 cm^2, we can solve for V2:
50 * 0.04 = 0.5 * V2
V2 = (50 * 0.04) / 0.5
V2 = 4 m/sec
Now, we can apply Bernoulli's principle, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline, assuming there is no energy loss due to friction or other factors.
Using Bernoulli's equation, we have:
P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2
Where:
P1 = Pressure at the larger end of the pipe
P2 = Pressure at the narrow end of the segment
ρ = Density of the fluid (assumed to be constant)
V1 = Velocity at the larger end of the pipe
V2 = Velocity at the narrow end of the segment
Given that P1 = 1.2 x 10^5 Pa and V2 = 4 m/sec, we can solve for P2:
P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2
P1 - P2 = (1/2) * ρ * V2^2 - (1/2) * ρ * V1^2
P2 = P1 - (1/2) * ρ * V2^2 + (1/2) * ρ * V1^2
Since the density of the fluid (ρ) is not given in the question, we cannot determine the exact pressure at the narrow end. However, we can still simplify the equation using the given values and express the pressure at the narrow end in terms of the given variables.