Question 1

An evaporative cooling unit needs to pump 100 litres of air per second through a pipe of radius 10 cm and length 10 m. How much of a pressure difference must be applied to the air to produce this flow rate? The viscosity of air is 1.8x10-5 Pa s.

Accepted Answer

Using Poiseuille's law, we have:

Q = (pi*r^4 / 8*eta*L) * ∆P

Where Q is the flow rate, r is the radius of the pipe, eta is the viscosity of air, L is the length of the pipe, and ∆P is the pressure difference.

Rearranging for ∆P, we get:

∆P = (8*eta*L*Q) / (pi*r^4)

Substituting the given values, we get:

∆P = (8*1.8x10^-5 * 10 * 100) / (pi*0.1^4)

∆P ≈ 0.46 Pa

Question 2

Water flows down a pipe of radius 5 cm at a speed of 5 cm s-1. What is the Reynolds number of this flow and will it be turbulent? The density of water is 1000 kg m-3 and the viscosity of water is 0.001 Pa s.

Accepted Answer

The Reynolds number (Re) is calculated using the formula Re = (density * velocity * diameter) / viscosity. Converting the radius to meters (0.05 m), the diameter is 0.1 m. Re = (1000 kg/m³ * 0.05 m/s * 0.1 m) / 0.001 Pa·s = 5000, indicating turbulent flow.

Question 3

Imagine that you have been given the job of designing a fountain to be the centrepiece of a new shopping mall on the Gold Coast. The fountain will consist of a single upward jet of water, launched from a pipe of radius 2 mm and length 1.2 m. The pipe is fed by recycled water, pushed by a pump which produces a pressure difference of 50 000 Pa. How high will the water in the fountain go? The viscosity of water is 0.001 Pa s, and g = 9.8 m s-2.

Accepted Answer

The correct answer is D.The height of the water in the fountain can be calculated using the Bernoulli equation:$$P_1 + \frac{1}{2} ho v_1^2 + ho g h_1 = P_2 + \frac{1}{2} ho v_2^2 + ho g h_2$$where:* $P_1$ is the pressure at the bottom of the pipe* $v_1$ is the velocity of the water at the bottom of the pipe* $ho$ is the density of water* $g$ is the acceleration due to gravity* $h_1$ is the height of the water at the bottom of the pipe* $P_2$ is the pressure at the top of the pipe* $v_2$ is the velocity of the water at the top of the pipe* $h_2$ is the height of the water at the top of the pipeIn this case, we have:* $P_1 = 50 000 Pa$* $v_1 = 0 m/s$* $ho = 1000 kg/m^3$* $g = 9.8 m/s^2$* $h_1 = 0 m$* $P_2 = 0 Pa$* $v_2 = 0 m/s$Substituting these values into the Bernoulli equation, we get:$$50 000 Pa + 0 + 1000 kg/m^3 	imes 0 m/s^2 	imes 0 m = 0 + 0 + 1000 kg/m^3 	imes 9.8 m/s^2 	imes h_2$$Solving for $h_2$, we get:$$h_2 = 5.1 m$$Therefore, the water in the fountain will go 5.1 m high.

Question 4

A natural gas pipeline with a diameter 0.250 m delivers 1.55 cubic meters of gas per second. What is the flow speed of the gas in the pipeline?

Accepted Answer

The answer of A natural gas pipeline with a diameter...

Question 5

Imagine that you are working at an outback mine. In part of the on-sight mineral separation plant, a slurry consisting of a mixture of mineral dust and water flows from one part of the plant to another down a pipe of diameter 10 cm. The flow needs to remain turbulent, as otherwise the mineral dust will settle to the bottom of the pipe and block it. If the density of the flurry is 2000 km m-3 and its viscosity is 0.001 Pa s, what is the minimum velocity the slurry must flow at to ensure it remains turbulent?

Accepted Answer

The answer of Imagine that you are working at an...

Question 6

A water pipe must deliver at least 10 litres per second to a water tank on a property on the outskirts of a country town. The pipe will have to be 2 km long, and is fed by the town water supply which delivers a pressure difference of 100 000 Pa. What is the minimum radius this pipe needs? The viscosity of water is 0.001 Pa s.

Accepted Answer

We can use the Poiseuille's law to solve this problem:

Q = (πr^4ΔP)/(8ηl)

Where Q is the flow rate, r is the radius of the pipe, ΔP is the pressure difference, η is the viscosity of the fluid and l is the length of the pipe.

We want to find the minimum radius, so we need to rearrange this equation to solve for r:

r = [(8ηlQ)/(πΔP)]^(1/4)

Substituting the given values, we get:

r = [(8 x 0.001 x 2000 x 10)/(π x 100000)]^(1/4)

r = 0.027 m = 2.7 cm

Therefore, the minimum radius this pipe needs is 2.7 cm, which is option D.

Question 7

Filtered air is pumped down a tube of diameter 50 cm into a hospital operating theatre. The idea is that a laminar flow over the patient will prevent contamination. What is the maximum safe speed at which the air can be pumped which will ensure that the flow is not turbulent? The density of air is 1.2 kg m-3 and its viscosity is 1.8x10-5 Pa s.

Accepted Answer

The maximum safe speed at which the air can be pumped without causing turbulence can be found using the formula for critical Reynolds number for laminar flow:

Re = (ρvd)/η

where:
ρ = density of air = 1.2 kg m^-3
v = maximum safe speed of air
d = diameter of tube = 50 cm = 0.5 m
η = viscosity of air = 1.8x10^-5 Pa s

The critical Reynolds number for laminar flow is around 2300. For laminar flow, Re < 2300.

Substituting the values:

Re = (1.2 x v x 0.5)/1.8x10^-5

Re = 33333.33v

For laminar flow, Re < 2300.

Therefore,

33333.33v < 2300

v < 0.069 m/s = 6.9 cm/s

Hence, the maximum safe speed at which the air can be pumped without causing turbulence is 6.9 cm s^-1. The answer is option C.

Question 8

Water flows through a pipe of radius 1 cm and length 1 m. The pressure difference is 100 Pa. What is the rate of flow, in litres per second? The viscosity of water is 0.001 Pa s.

Accepted Answer

The rate of flow can be calculated using Poiseuille's Law: Q = (π * r^4 * ΔP) / (8 * η * L). Substituting the given values: r = 0.01 m, ΔP = 100 Pa, η = 0.001 Pa s, L = 1 m, we get Q = (π * (0.01)^4 * 100) / (8 * 0.001 * 1) = 0.00039 m^3/s = 0.39 liters/s.

Question 9

Imagine that you work at a forensics laboratory and have been asked to identify a strange, highly viscous transparent fluid found at a crime scene. The fluid has a density of 2 g cm-3. To measure the density, you drop a lead sphere of radius 2mm into the fluid, and observe that it drops at a steady speed of 8.1 mm s-1. What is the viscosity of the unknown fluid? You may assume that Stokes' law applies and that g = 9.8 m s-2.

Accepted Answer

The answer of Imagine that you work at a forensics...

Question 10

A fountain sends a stream of water 20.0 m up into the air. If the base of the stream is 10.0 cm in diameter, what power is required to send the water to this height?

Accepted Answer

The answer of A fountain sends a stream of water...

Question 11

Imagine that you have been given the job of designing a fountain to be the centrepiece of a new shopping mall on the Gold Coast. The fountain will consist of a single upward jet of water, launched from a pipe of radius 5 mm and length 2 m. The pipe is fed by recycled water, pushed by a pump which produces a pressure difference of 100 000 Pa. What volume of water will come out of the pipe? The viscosity of water is 0.001 Pa s.

Accepted Answer

The volume flow rate can be calculated using the Hagen-Poiseuille equation for laminar flow through a pipe: Q = (π * r^4 * ΔP) / (8 * η * L). Substituting the given values (r = 0.005 m, ΔP = 100,000 Pa, η = 0.001 Pa s, L = 2 m) into the equation gives approximately 0.012 m³/s.

Question 12

Honey is around 3000 times more viscous than water, but has a similar density. How much faster could honey flow though a pipe compared to water, while still maintaining a laminar flow?

Accepted Answer

The answer of Honey is around 3000 times more viscous...

Question 13

The water supply to an old suburb in Perth is no longer adequate, and the water company is planning to put in a higher capacity supply. Which of the following options would give the biggest increase in the water flow into the suburb?

Accepted Answer

Increasing the diameter of the pipeline by 10% will increase the cross-sectional area by approximately 21% (since area is proportional to the square of the diameter). This will allow a larger volume of water to flow through the pipeline and into the suburb, resulting in the biggest increase in water flow. Re-routing the pipeline (option A) may reduce friction losses, but a 13% reduction in length is unlikely to have a significant impact. Likewise, increasing the pressure difference by 15% (option C) may increase flow, but not as much as increasing the diameter. Exposing the pipeline to the sun (option D) may slightly decrease viscosity, but this effect is not significant enough to make a significant difference in flow rate.

Question 14

You have been asked to modify the nozzle of a rocket engine. The existing rocket produces a laminar flow of exhaust gasses, but is not powerful enough to lift a new generation of communication satellites. You have been asked to advise on the best way to increase the thrust without causing the flow to become turbulent. The rocket thrust is proportional to the density of the exhaust, to its velocity, and to the square of the nozzle diameter. What is the best approach?

Accepted Answer

Increasing the diameter of the nozzle increases thrust due to the square relationship, and decreasing velocity by the same factor maintains laminar flow.

Question 15

Imagine that you have been given the job of designing a fountain to be the centrepiece of a new shopping mall on the Gold Coast. The fountain will consist of a single upward jet of water, launched from a pipe of radius 4 mm and length 1.5 m. The pipe is fed by recycled water, pushed by a pump which produces a pressure difference of 100 000 Pa. With what speed will the water come out of the pipe? The viscosity of water is 0.001 Pa s.

Accepted Answer

The speed of the water coming out of the pipe is given by the Torricelli equation:$$v = \sqrt{2gh}$$where:* v is the speed of the water (m/s)* g is the acceleration due to gravity (9.8 m/s^2)* h is the height of the water column (m)In this case, the height of the water column is equal to the length of the pipe, which is 1.5 m. Therefore, the speed of the water coming out of the pipe is:$$v = \sqrt{2gh} = \sqrt{2(9.8 m/s^2)(1.5 m)} = 4.2 m/s$$Therefore, the correct answer is E) 133 m s-1.

Quiz 14: Fluid Dynamics