# Fluid Dynamics

By Levi Clancy for Student Reader on
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Mass Flow Rate Volume Flow Rate kg / s Δmass / ΔtimeΔm1 / Δt = Δm2 / Δt m3 / s Area · velocity = A · v A1 v1 = A2 v2The equation of continuity is derived from the mass flow rates:Δm1 / Δt = Δm2 / Δtρ Δv1 / Δt = ρ Δv2 / Δtρ / Δt ( ΔL1 ΔA1 ) = ρ / Δt ( ΔL2 ΔA2 )ρ A1 ( ΔL1 / Δt ) = ρ A2 ( ΔL2 / Δt )ρ A1 v1 = ρ A2 v2A1 v1 = A2 v2 P1 + ½ ρ v12 + ρ g y1 = P2 + ½ ρ v22 + ρ g y2P = Pressure at ... When considering a viscous liquid between two surfaces, the following relationship is observed:η = ( F · ℓ ) / ( A · v )A = Surface area of viscous liquidv = Speed of liquid between two platesF = Force to set the system in motionℓ = Separation between the surfaces Pa s-1 or Poise A fluid has a characteristic coefficient of viscosity η, measured in N s m-2 (Pa s-1) or dyne · s · cm-2 (Poise, P). The two units are related as: 1 Pa s-1 = 10 Poise. Poiseuille's Equation concerns the flow of a viscous liquid through a cylindrical tube. Poiseuille's Equation was initially developed to study blood flow.Q = π R4 ΔP / ( 8 η ℓ )Q = Volume rate of flow (m3 s-1). R = Inner radius of the tube (meters)ΔP = Pressure difference between the two ends of the tube ( P1 - P2 )η = Coefficient of viscosity of the fluidℓ = Length travelled.Of note, the pressure gradient is ΔP / ℓ → (units of Pa m-1). Surface tension is represented by γ, the lower-case gamma. γ = ½ F / ℓ
Practice Problems
What gauge pressure in the water mains is necessary if a firehose is to spray water to a height of 20 meters?

P1 + ½ ρ v12 + ρ g y1 = P2 + ½ ρ v22 + ρ g y2

Let 1 refer to the water main, and 2 refer to the point where the water was reached 20 meters.

ρ = 1000 kg m-3
P1 = ?
P2 = atmospheric pressure = 1000 Pascals
PG = Gauge pressure = P1 - atmospheric pressure = P1 - P2
y1 = 0
y2 = 20 meters
v1 = 0
v2 = 0

P1 + ½ · 1000 · 02 + 1000 · 9.8 · 0 = P2 + ½ · 1000 · 02 + 1000 · 9.8 · 20.0
P1 + 0 + 0 = P2 + 0 + 196000
P1 - P2 = PG = 19600 Pascals needed in the water main

Water at a gauge pressure of 4 atm at street level flows into a high-rise building at a speed of 1.0 m s-1 through a pipe 8 cm in diameter. The pipe tapers down to 3 cm in diameter by the top floor, 30 meters above, where a faucet has been left open. Calculate the flow velocity and gauge pressure in the pipe on the top floor. Assume no branch pipes and ignore viscosity.

v1 A1 = v2 A2

P1 + ½ ρ v12 + ρ g y1 = P2 + ½ ρ v22 + ρ g y2

ρ = 1000 kg m-3
P1 = 4 atm = 4 atm · 101325 Pa / atm = 405300 Pa
P2 = ?
v1 = 1.0 m s-1
v2 = ?
y1 = 0
y2 = 30 meters
A1 = π · .042 = 0.00502654825 m2
A1 = π · .0152 = 0.000706858347 m2

1.0 m s-1 · 0.00502654825 m2 = v2 · 0.000706858347 m2
v2 = 7.11111112 m s-1 → The flow velocity is 7.1 m s-1 out of the faucet.

405300 + ½ · 1000 · 1.02 + 405300 · 9.8 · 0 = P2 + ½ · 1000 · 7.111111122 + 1000 · 9.8 · 30
405300 + 500 + 0 = P2 + 500 · 50.5679014 + 294000
P2 = 86516.0493 Pascals → The gauge pressure is 86500 Pa on the top floor.

An airplane has a mass of 5 · 105 kg, and the air flows past the lower surface of the wings at 90 m s-1. If the wing surface area on the bottom and top is each 250 m2, how fast must the air flow over the upper surface of the wing if the plane is to stay in the air?

P1 + ½ ρ v12 + ρ g y1 = P2 + ½ ρ v22 + ρ g y2

The height of the wings is not provided, and is generally negligible, so assume that Δy = 0
Bernoulli's Equation now simplifies to:

P1 + ½ ρ v12 = P2 + ½ ρ v22

Let 1 refer to the top surface and 2 refer to the bottom surface.

ρ = Air density = 1.22521 kg m-3 [However, Giancoli uses 1.29 kg m-3.]
P1 = ?
P2 = ?
v1 = ?
v2 = 90 m s-1

It is not possible with the information provided to derive P1 and P2.
However, consider that airplanes lift by an air pressure difference comparing the upper and lower wing surfaces.
The lower wing surface must experience higher air pressure than the upper wing surface.
This pressure must be sufficient to lift the entire mass of the plane.
Also, remember that Pressure = Force / Area
Therefore: P2 - P1 = ( 5 · 105 kg · 9.8 m s-1 ) / 250 m2 = 19600 Pascals

P1 + ½ · 1.22521 · v12 = P2 + ½ · 1.22521 · 90
0.612605 · v12 = P2 - P1 + 55.13445
0.612605 · v12 = 19600 + 55.13445
v12 = 32084.5152
v1 = 179.12151 m s-1 → 180 m s-1 is the air flow on the upper wing surface.

What must be the pressure difference between the two ends of a 2.0 km section of pipe, 20.0 cm in diameter, if it is to transport oil (ρ = 800 kg m-3, η = 0.30 Pa s) at a rate of 500 cm3 s-1?

Q = π R4 ΔP / ( 8 η ℓ )

Q = 500 cm3 s-1 = 500 · 10-6 m3 s-1
R = .100 m = 10-1 m
ℓ = 2000 m
ΔP = ?
η = 0.30 Pa s

500 · 10-6 = π · ( 10-1 )4 · ΔP / ( 8 · 0.30 · 2000 )
ΔP = 7639.43727 Pascals → The pressure difference is 7640 Pascals

A patient is to be given a blood transfusion. The blood is to flow through a tube from a raised bottle to a needle inserted in the vein. The inside diameter of the 20 mm long needle is 0.50 mm, and the required flow rate is 3.0 cm3 of blood per minute. How high should the bottle be placed above the needle? Assume the blood pressure is 100 Torr above atmospheric pressure and look up ρ and η for blood.

Q = &pi R4 ΔP / ( 8 η ℓ )

Pressure at a depth = ρ g h

ρ of blood ≈ 1050 kg m-3 (according to Giancoli)
η of blood ≈ .004 Pa s (according to Giancoli)
Q = 3.0 cm3 / min → 3.0 · 10-6 m3 / min = 5 · 10-8 m3 / s
R = ½ · 0.50 mm *rarr; ½ · 0.00050 = 0.00025 m
ℓ = 20 mm → .020 m
ΔP = Pressure at needle - Pressure of blood
Needle Pressure = ρ g h → 1050 · 9.8 · h = 10290 · h
Blood Pressure = 78 Torr → 78 · 133.322368 Pascal = 10399.1447 Pa

5 · 10-8 = π · 0.000254 · ΔP / ( 8 · .004 · .020 )
3.2 · 10-11 = π · 0.000254 · ΔP
2607.59459 = ΔP

2607.59459 = 10290 · h - 10399.1447
13006.7393 = 10290 · h
1.26401742 = h

The bottle must be 1.3 meters above the patient's arm.

At first I was unclear whether ΔP = Pneedle - Pblood or ΔP = Pblood - Pneedle
However, consider if ΔP = Pblood - Pneedle were used.
Then, there would have been a negative value for h.
Having the bottle below the vein would cause blood to flow out of the patient!
So remember to use some intuition to help solve ambiguous parts of a problem.
And if in doubt, work it out!

A long vertical hollow tube with an inner diameter of 1.50 cm is filled with SAE 10 motor oil. A rod with a diameter of 0.50 cm, a length of 40.0 cm and a weight of 100 grams is dropped vertically through the oil in the tube. What is the maximum velocity of the rod as it falls?

Right away it is clear that this is a problem that involves viscosity.

η = ( F · ℓ ) / ( A · v )

The equation relates the flow velocity of a fluid between two surfaces.
How does the equation tell us the velocity of a rod travelling through a cylinder?
Well, imagine if there is a vertical sandwich of two surfaces and oil.
Then the oil is moving subject to the force of gravity → its flow velocity is v.
Or, you can view it as its velocity is v relative to the velocity (0) of the surfaces.
So what if the oil and one surface are held in place?
Then the other surface moves with velocity v relative to the rest of the apparatus.

v = ( F · ℓ ) / ( η · A )

η = 200 · 10-3 Pa · s (according to Giancoli)
Tuberadius = ½ · 1.50 cm · .01 m cm-1 = .0075 m
Rodradius = ½ · 0.50 cm · .01 m cm-1 = .0025 m
Rodlength = 40.0 cm · .01 m cm-1 = .40 m
Rodweight = 100 grams · .001 kg g-1 = .1 kg

Based on the values provided in the problem, it is easy to deduce what to plug into the equation.
But the reasoning is as follows:
The surface area of the rod and the inner surface of the cylinder provides two surfaces.
The cylinder and the rod have different surface areas; one is greater in diameter than the other.
However, the equation only has a variable for one surface area.
Which surface area to use?
Imagine if the two surface areas were flattened out.
The cylinder's surface area would be a large rectangle and the rod would be a narrower rectangle.
You would use the smaller surface area to solve the problem.
Because it is only that surface area that matches the conditions of the equation:
two surfaces sandwiching a layer of oil.
The larger surface area would give you those conditions, plus an area with just one surface and oil.
Thus you use the smaller surface area.

A = 2 · π · .0025 · .40 = 0.00628318531 m2
ℓ = Distance between two surfaces = Tuberadius - Rodradius = .0075 - .0025 = .005 m
F = Force of gravity on rod = .1 kg · 9.8 m s-2 = .98 N

v = ( .98 · .005 ) / ( 200 · 10-3 · 0.00628318531 ) = 2.59953074 m s-1 = 2.6 m s-1

A u-shaped apparatus has a movable wire of length 6.6 cm. If the force needed to move the wire is is 2.0 · 10-3, calculate the surface tension of the enclosed fluid.

γ = ½ F / ℓ

F = 2.0 · 10-3 m
ℓ = 6.6 m → .066 m

γ = ½ · .002 / .066 = ½ · 0.0303030303 = 1.5 · 10-2 N m-1