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? 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?
thanks. :)
1. Volume flow rate = (Area)(Velocity) = pi*2^2/4 V = pi cm^2* 35 cm/s = 110 cm^3/s
Amount filled = V = 10,000 cm^3 = Q t
t = V/Q = 10,000 cm^2/(110 cm^3/s)
= ? s
2. Use the incompressible continuity equation V*A = constant to get the velocity at the narroqw end. Velocity will be 100 times higher at the narrow end. Then apply Bernoulli's equation for the pressure change.
P + (1/2)*(density)*V^2 = constant
P2 - P1 = (1/2)(density)(V1^2 - V2^2)
thanks for your help. :)
1.) To find the time it will take to fill a 10 liter container, we need to calculate the volume flow rate of the water coming out of the pipe.
First, we need to find the cross-sectional area of the pipe. The diameter of the pipe is given as 2 cm, which means the radius is 1 cm (or 0.01 m). The cross-sectional area can be calculated using the formula: A = πr^2.
A = π(0.01 m)^2
A = π(0.0001 m^2)
A ≈ 0.000314 m^2
Now, we need to find the volume flow rate. The formula for volume flow rate is Q = A * v, where Q is the volume flow rate, A is the cross-sectional area, and v is the speed of the water.
Q = 0.000314 m^2 * 0.35 m/sec
Q = 0.0001099 m^3/sec
Since 1 liter is equal to 0.001 m^3, we can convert the volume flow rate to liters per second:
Q = 0.0001099 m^3/sec * (1 liter / 0.001 m^3)
Q = 0.1099 liter/sec
Now, we can calculate the time it will take to fill the 10 liter container:
Time = volume / flow rate
Time = 10 liters / 0.1099 liter/sec
Time ≈ 91.02 seconds
Therefore, it will take approximately 91.02 seconds to fill a 10 liter container.
2.) To find the pressure at the narrow end of the pipe segment, we can use the principle of conservation of mass and Bernoulli's equation.
According to the principle of conservation of mass, the mass flow rate at any point along the pipe remains constant. Therefore, we can write the equation:
A1 * v1 = A2 * v2
Where A1 and v1 are the cross-sectional area and velocity at the larger end, and A2 and v2 are the cross-sectional area and velocity at the narrow end.
Given that A1 = 50 cm^2, A2 = 0.5 cm^2, and v1 = 0.04 m/sec, we can rearrange the equation to solve for v2:
v2 = (A1 * v1) / A2
v2 = (50 cm^2 * 0.04 m/sec) / 0.5 cm^2
v2 = 4 m/sec
Now we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a container:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where P1 and P2 are the pressures at the larger and narrow ends, ρ is the density of the fluid, g is the acceleration due to gravity, and h1 and h2 are the heights at the larger and narrow ends (assuming they are at the same level).
Given P1 = 1.2 × 10^5 Pa, v1 = 0.04 m/sec, v2 = 4 m/sec, and assuming h1 = h2, ρgh1 and ρgh2 can be canceled out.
P2 = P1 + (1/2)ρ(v1^2 - v2^2)
Since we are not given the density of the fluid, we cannot directly calculate the pressure at the narrow end without additional information.
Please provide the density of the fluid to proceed with the calculation.
1.) To find how long it will take to fill a 10 liter container, we need to calculate the volume flow rate of water through the pipe and then divide it by the volume of the container.
The volume flow rate can be found using the formula:
Q = A * V
where Q is the volume flow rate, A is the cross-sectional area of the pipe, and V is the velocity of the water.
First, let's calculate the cross-sectional area of the pipe. The diameter is given as 2 cm, which means the radius is 1 cm or 0.01 m.
So, the area (A) is:
A = π * r^2
A = 3.14 * (0.01)^2
A = 0.000314 m^2
Now, we can calculate the volume flow rate:
Q = 0.000314 * 0.35
Q = 0.0001099 m^3/sec
Since the volume of the container is given in liters, we need to convert it to cubic meters for consistency.
1 liter = 0.001 cubic meters.
So, 10 liters = 0.01 cubic meters.
Now, let's calculate the time it takes to fill the container:
Time = Volume / Volume flow rate
Time = 0.01 / 0.0001099
Time ≈ 91.01 seconds
Therefore, it will take approximately 91.01 seconds to fill the 10-liter container.
2.) To find the pressure at the narrow end of the segment, we can use the principle of conservation of mass and Bernoulli's equation.
According to the principle of conservation of mass, the mass flow rate of fluid remains constant throughout the pipe. The mass flow rate is given by:
m = ρ * A * V
where m is the mass flow rate, ρ is the density of the fluid, A is the cross-sectional area, and V is the velocity.
Since the mass flow rate remains constant, we can find the cross-sectional area at the narrow end of the segment:
A_narrow = (m / (ρ * V_narrow))
Given that the cross-sectional area at the larger end of the pipe is 50 cm^2, the mass flow rate (m) can be calculated using:
m = ρ * A_large * V_large
where A_large is 50 cm^2 (or 0.005 m^2 as the area should be in square meters), and V_large is 0.04 m/sec.
Now, let's calculate the mass flow rate:
m = ρ * A_large * V_large
m = ρ * 0.005 * 0.04
m = 0.0002 * ρ
Next, we need to find the cross-sectional area at the narrow end. The velocity at the narrow end is given as 0.04 m/sec, the same as at the large end since the pipe is horizontal and the fluid is incompressible.
So, the cross-sectional area at the narrow end is:
A_narrow = m / (ρ * V_narrow)
A_narrow = (0.0002 * ρ) / (ρ * 0.04)
A_narrow = 0.0002 / 0.04
A_narrow = 0.005 m^2
Now that we have the cross-sectional area at the narrow end, we can find the pressure at that point using Bernoulli's equation:
P_narrow + (1/2) * ρ * V_narrow^2 = P_large + (1/2) * ρ * V_large^2
Since the fluid is at rest at both ends, V_narrow and V_large are both zero in this case.
Therefore, the pressure at the narrow end (P_narrow) can be calculated by rearranging the equation:
P_narrow = P_large
Given that the pressure at the larger end of the pipe is 1.2x10^5 Pa, the pressure at the narrow end of the segment is also 1.2x10^5 Pa.
So, the pressure at the narrow end of the segment is 1.2x10^5 Pa.