# Fluid Dynamics

Mass Flow Rate | kg / s | Δmass / Δtime Δm |
---|---|---|

Volume Flow Rate | m^{3} / s | Area · velocity = A · v |

Equation of Continuity | A_{1} v_{1} = A_{2} v_{2}The equation of continuity is derived from the mass flow rates: Δm ρ Δv ρ / Δt ( ΔL ρ A ρ A A | |

Bernoulli's Equation | P P = Pressure at ... | |

Viscosity | When considering a viscous liquid between two surfaces, the following relationship is observed: η = ( F · ℓ ) / ( A · v ) A = Surface area of viscous liquid v = Speed of liquid between two plates F = Force to set the system in motion ℓ = Separation between the surfaces | |

Coefficient of Viscosity (η) | 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 | 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 R Q = Volume rate of flow (m Of note, the pressure gradient is ΔP / ℓ → (units of Pa m | |

Surface Tension (&gamma) | Surface tension is represented by γ, the lower-case gamma. | |

U-Shaped Apparatus | γ = ½ F / ℓ |

P_{1} + ½ ρ v_{1}^{2} + ρ g y_{1} = P_{2} + ½ ρ v_{2}^{2} + ρ g y_{2}

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}
P_{1} = ?
P_{2} = atmospheric pressure = 1000 Pascals
P_{G} = Gauge pressure = P_{1} - atmospheric pressure = P_{1} - P_{2}
y_{1} = 0
y_{2} = 20 meters
v_{1} = 0
v_{2} = 0

P_{1} + ½ · 1000 · 0^{2} + 1000 · 9.8 · 0 = P_{2} + ½ · 1000 · 0^{2} + 1000 · 9.8 · 20.0
P_{1} + 0 + 0 = P_{2} + 0 + 196000
P_{1} - P_{2} = P_{G} = 19600 Pascals needed in the water main

^{-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.

v_{1} A_{1} = v_{2} A_{2}

P_{1} + ½ ρ v_{1}^{2} + ρ g y_{1} = P_{2} + ½ ρ v_{2}^{2} + ρ g y_{2}

ρ = 1000 kg m^{-3}
P_{1} = 4 atm = 4 atm · 101325 Pa / atm = 405300 Pa
P_{2} = ?
v_{1} = 1.0 m s^{-1}
v_{2} = ?
y_{1} = 0
y_{2} = 30 meters
A_{1} = π · .04^{2} = 0.00502654825 m^{2}
A_{1} = π · .015^{2} = 0.000706858347 m^{2}

1.0 m s^{-1} · 0.00502654825 m^{2} = v_{2} · 0.000706858347 m^{2}
v_{2} = 7.11111112 m s^{-1} → The flow velocity is 7.1 m s^{-1} out of the faucet.

405300 + ½ · 1000 · 1.0^{2} + 405300 · 9.8 · 0 = P_{2} + ½ · 1000 · 7.11111112^{2} + 1000 · 9.8 · 30
405300 + 500 + 0 = P_{2} + 500 · 50.5679014 + 294000
P_{2} = 86516.0493 Pascals → The gauge pressure is 86500 Pa on the top floor.

^{5}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 m

^{2}, how fast must the air flow over the upper surface of the wing if the plane is to stay in the air?

P_{1} + ½ ρ v_{1}^{2} + ρ g y_{1} = P_{2} + ½ ρ v_{2}^{2} + ρ g y_{2}

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

P_{1} + ½ ρ v_{1}^{2} = P_{2} + ½ ρ v_{2}^{2}

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}.]
P_{1} = ?
P_{2} = ?
v_{1} = ?
v_{2} = 90 m s^{-1}

It is not possible with the information provided to derive P_{1} and P_{2}.
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: P_{2} - P_{1} = ( 5 · 10^{5} kg · 9.8 m s^{-1} ) / 250 m^{2} = 19600 Pascals

P_{1} + ½ · 1.22521 · v_{1}^{2} = P_{2} + ½ · 1.22521 · 90
0.612605 · v_{1}^{2} = P_{2} - P_{1} + 55.13445
0.612605 · v_{1}^{2} = 19600 + 55.13445
v_{1}^{2} = 32084.5152
v_{1} = 179.12151 m s^{-1} → 180 m s^{-1} is the air flow on the upper wing surface.

^{-3}, η = 0.30 Pa s) at a rate of 500 cm

^{3}s

^{-1}?

Q = π R^{4} ΔP / ( 8 η ℓ )

Q = 500 cm^{3} s^{-1} = 500 · 10^{-6} m^{3} 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

^{3}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 R^{4} 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 cm^{3} / min → 3.0 · 10^{-6} m^{3} / min = 5 · 10^{-8} m^{3} / 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 · 0.00025^{4} · ΔP / ( 8 · .004 · .020 )
3.2 · 10^{-11} = π · 0.00025^{4} · 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 = P_{needle} - P_{blood} or ΔP = P_{blood} - P_{needle}
However, consider if ΔP = P_{blood} - P_{needle} 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!

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)
Tube_{radius} = ½ · 1.50 cm · .01 m cm^{-1} = .0075 m
Rod_{radius} = ½ · 0.50 cm · .01 m cm^{-1} = .0025 m
Rod_{length} = 40.0 cm · .01 m cm^{-1} = .40 m
Rod_{weight} = 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 m^{2}
ℓ = Distance between two surfaces = Tube_{radius} - Rod_{radius} = .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}

^{-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}