Physics presentation: ideal fluid flow

When this pump or vacuum system is started, air begins to flow through this hose, which has a slight crimp, and as a result this hose begins to totally collapse. Now don't you say it's because of suction, because as you know, physics don't suck. (It blows.) (Video link: "hose collapse.")

In a previous presentation we investigated the behavior of static fluids, now we'll consider dynamic fluids--more specifically, ideal fluid flow. First we'll define what we mean by "ideal" fluids, then we'll see how two conservation laws are applied simultaneously to flowing ideal fluids.

Ideal fluids fall under a restrictive class of fluids--let's take a look at the characteristics that set ideal fluids apart from, say, real fluids.

Ideal fluids are incompressible, which water is to some extent. Note that while air is not incompressible, there are situations where we can make the crude approximation that it is.

An ideal fluid should undergo laminar flow, where the adjacent particles flow smoothly past each other, as opposed to turbulent flow, where particles swirl around in a chaotic manner. Like many fluids water can undergo both laminar and turbulent flow, so we'll restrict our attention to certain conditions where water undergoes laminar flow.

Lastly, an ideal fluid should be non-viscous, that is, flow without appreciable frictional losses, as opposed to a viscous fluid that, well, looks and is literally, "gooey." (Video link: "Viscosity.")

Streamlines are a way to visualize ideal fluid flow. Here air is undergoing incompressible, laminar, non-viscous flow over the front end of a car in this wind tunnel, where the streamlines are smoothly conforming to the contours of the car and to adjacent streamlines. Note the rear of the car, where the streamlines are turbulent, and indicate the presence of non-ideal fluid flow. (Video link: "Mercedes-Benz SLS AMG Developement and Testing Wind tunnel.")

The first conservation law for ideal fluid flow follows from its incompressible nature.

Even if a pipe changes radius, the incompressibility of an ideal fluid means that the same volume flowing in one end must equal to the same volume coming out the other end in the same amount of time. "Stuff in, stuff out."

This conservation of volume flow rate can be reinterpreted in terms of cross-sectional areas and fluid speeds in the continuity equation. A large-area section of pipe will have a slower fluid speed than a small-area section of the same pipe, which will have a faster fluid speed. Note that the product of cross-sectional area and fluid speed at any section of a pipe results in the volume flow rate.

For a pipe with a constant cross-sectional area, the volume flow rate ∆V/∆t through the pipe is constant. To convince yourself of this you'll need a friend to watch this animation with you. Every time you see fluid particles entering the pipe, say "in." "In. In. In..." Keep doing that. Convince your friend to say "out" every time fluid particles are leaving the pipe. "Out. Out. Out..." If the two of you do this correctly, each time you say "in," your friend immediately follows-up by saying "out." This means that rate of fluid volume going in (represented here by three dots) must continuously be equal to the rate of fluid volume going out. "Stuff in, stuff out," right?

Then from the continuity equation:

A₁·v₁ = A₂·v₂,

since the cross-sectional areas of where the fluid goes in and where it goes out are the same (A₁ = A₂), then v₁ = v₂, so the speed of the fluid flowing through this pipe must be constant as well.

For a pipe with increasing cross-sectional area, the volume flow rate ∆V/∆t through the pipe is also constant. Let's do the same "in and out" exercise as above. Every time you see fluid particles entering the narrow end of the pipe, say "in," while your friend says "out" every time fluid particles are leaving the wider end of the pipe. "In. Out. In. Out. In. Out..." As before, since each time you say "in," your friend immediately follows-up by saying "out," so the rate of fluid volume going in the narrow end (represented here by three dots) must continuously be equal to the rate of fluid volume going out the wider end. "Stuff in, stuff out," right?

Then from the continuity equation:

A₁·v₁ = A₂·v₂,

since the cross-sectional area of where the fluid goes in is smaller than the cross-sectional area of where it goes out (A₁ < A₂), then v₁ > v₂, so the speed of the fluid flowing through this pipe slows down.

Conversely, for a pipe with decreasing cross-sectional area, the speed of the fluid through this pipe must have a corresponding increase, such that it flows more quickly through the narrow portion of the tube.

The second conservation law for ideal fluid flow follows from its laminar, non-viscous nature, as energy per volume density will be conserved if there are no losses to dissipative turbulence and frictional losses. (Incompressibility matters here too, such that the volume term in the energy per volume will be conserved.)

Bernoulli's equation is the extension of the static fluid relationship between pressure and gravitational potential energy per volume changes, to ideal fluid flow, with the addition of a translational kinetic energy per volume term. All three terms have equivalent units of Pa or J/m³, such that they can transfer to/from each other, as long as the net balance of exchanges is zero.

For an ideal fluid flowing through a horizontal pipe with a widening cross-sectional area, does the kinetic energy density term (1/2)·ρ·∆(v²) term increase, decrease, or have no change? Let's refer back to the continuity equation discussion above and recall that while the volume flow rate doesn't change, the speed changes, where the fluid slows down travels through this pipe. This means that the kinetic energy density will decrease (as it depends on the square of the speed), and this term will be negative.

Does the gravitational potential energy density term ρ·g·∆y term increase, decrease, or have no change? Since the center of the pipe has no change in height (even though the cross-sectional areas are different, they are still "horizontally aligned" with each other), there is no change in the gravitational potential energy density term, and this term will be zero.

Then as a result, does the pressure of the ideal fluid flowing through this pipe increase, decrease, or have no change? Note the steps in determining the changes (if any) in pressure for this ideal flowing fluid--first we apply the continuity equation to determine the change in speeds (if any), which tells us the change (if any) in the kinetic energy density. We then look at the change in height of the centerline of the pipe (if any), which tells us the change (if any) in the gravitational potential energy density. Then we can look at Bernoulli's equation:

0 = ∆P + ρ·g·∆y + (1/2)·ρ·∆(v²),

and look at the increases (+) or decreases (–) (or no changes) of the terms we know so far:

0 = ∆P + (0) + (–).

In order to balance out this equation to equal zero on the left-hand side, the pressure of the fluid as it flows through this pipe must increase, making ∆P positive, such that:

0 = (+) + (0) + (–),

and both sides of Bernoulli's equation are balanced. So for this ideal fluid flowing through this widening pipe, the pressure will increase. This is why a weakened, enlarged blood vessel (an aneurysm) is dangerous, as blood flowing through this damaged, wider section will temporarily experience an increase in pressure as it slows down, and may widen the blood vessel even more and cause it to eventually rupture.

For an ideal fluid flowing through a horizontal pipe with a narrowing cross-sectional area, does the kinetic energy density term (1/2)·ρ·∆(v²) term increase, decrease, or have no change? (Refer back to the continuity equation discussion to determine this.) Does the gravitational potential energy density term ρ·g·∆y term increase, decrease, or have no change?

Then as a result, does the pressure of the ideal fluid flowing through this pipe increase, decrease, or have no change?

(Note that the pressure should decrease in the narrow portion of this pipe, which is why the crimped hose at the start of this presentation collapsed--as air flowed through the narrow crimped portion of the hose, its speed increased, which made the pressure decrease inside the hose, and the surrounding atmospheric pressure then flattened the crimped portion even further.)

For an ideal fluid flowing through a descending horizontal pipe with a constant cross-sectional area, does the kinetic energy density term (1/2)·ρ·∆(v²) term increase, decrease, or have no change? (Refer back to the continuity equation discussion to determine this.) Does the gravitational potential energy density term ρ·g·∆y term increase, decrease, or have no change?

Then as a result, does the pressure of the ideal fluid flowing through this pipe increase, decrease, or have no change?

Physics presentation: static fluids

In this scene from Man from Atlantis, Mark Harris (played by Patrick Duffy) is put into a water pressure chamber that simulates different depths, and watches as test canisters are crushed by pressures equivalent to 20,000 ft, 25,000 ft, and 30,000 ft below sea level.

On a more reality-based aside, consider tourists who float in the high salinity waters of the Dead Sea.

We'll discuss these two different aspects of static fluids here: pressure, and buoyancy.

First, pressure--as a force per unit area density, then reinterpreted as a pressure per unit area density.

Pressure can be consider as the amount of force exerted over a certain area on a surface, whether by a macroscopic object, or by the random bombardment by atoms/molecules of gases or liquids. If you don't wear snowshoes, the force of your weight, distributed over the area of your feet will create a pressure that cannot be supported by soft, unpacked snow, and your feet will sink in.

However, the force of your weight distributed over a much larger snowshoe area will reduce the pressure exerted on the snow, and you will not sink in much, if at all.

From the definition of pressure as a force per area density, the unit of pressure is pascals (Pa), equivalent to N/m².

A more useful interpretation of pressure, especially in regards to fluids (gases and liquids) is to think of it as an energy per unit volume. Notice how the units of pascals equals N/m², and when both numerator and denominator by are multiplied by meters (m), these units become N/m² = (N·m)/(m³) = J/m³.

By interpreting pressure as a form of energy per unit volume, we can connect it to gravitational potential energy per unit volume ∆U_grav/V = m·g·∆y/V = (m/V)·g·∆y = ρ·g·∆y, which also has units of J/m³.

Pressure and gravitational potential energy per unit volume are then terms in an energy density "conservation" equation, and they are allowed to "exchange" Pa provided the fluid is static and there is no external work being put in or taken out of the fluid. In this form, then by picking two locations in the same static fluid, an increase or decrease in the ρ·g·∆y must have a corresponding decrease or increase in pressure (∆P).

A weather balloon that is partially filled at ground level will rise, and will seemingly inflate and eventually explode at very high elevations.

To explain what's going on here, we are going to analyze the static fluid that exists at both ground level and at a higher elevation: the air surrounding the balloon (i.e., the entire atmosphere), and not the contents of the balloon, which do not simultaneously exist at both locations (as it is "transported," and technically not a "static" fluid.)

Let's compare the air surrounding the balloon at ground level, and compare it to the air at the final higher elevation. Since the gravitational potential energy density depends on elevation, as the elevation of the balloon increases, the gravitational potential energy density of the surrounding air increases.

Looking at the energy density "conservation" equation for static fluids:

0 = ΔP + ρ·g·∆y,

Since the gravitational potential energy density of the air surrounding the balloon increases as it moves to higher elevations, then ρ·g·∆y is positive. In order to balance out this equation to equal zero on the left-hand side, the pressure of the air surrounding the balloon must have a corresponding decrease, making ΔP negative, such that:

0 = (–) + (+),

meaning that there is more air pressure at ground level than at a higher elevation. Essentially the pressure within the balloon remains constant, and because it is surrounded with lower pressure air at a higher elevation, the balloon will expand in size, and eventually "pop."

Notice the full-sized Styrofoam™ cup in the back, compared to other cups that were carried in the outside storage compartment of a submarine, where the air pockets inside these cups were collapsed by the surrounding water, effectively permanently shrinking the sizes of these cups. During this process, did the ρ·g·∆y increase or decrease? Did the water pressure surrounding the cups increase or decrease?

Second, buoyancy.

The buoyant force on an object depends on the density ρ of the fluid, and the volume of the object that is actually submerged in the fluid. While this is a simple definition, knowing the appropriate amount of volume to use in this equation is key to understanding buoyancy.

For a fully-submerged object, the volume used in calculating the buoyant force is the volume of the entire object, such that the buoyant force is given by:

F_B = ρ·g·V,

where ρ is the density of the surrounding fluid (water), and volume V is the entire volume of the diver, as he is fully submerged.

Here, since the object (the submerged diver) is floating underwater, Newton's first law applies, and the downwards weight force and the upwards buoyant force balance out.

For a partially-submerged object like this red ship, the volume used in calculating the buoyant force is not the volume of the entire object, but only the portion of the object that is actually submerged.

For the red ship, which Newton's law applies to its motion? How do the magnitudes of the downwards weight force and the upwards buoyant force compare? What fluid density should be put into the ρ in the buoyant force calculation?

Astronomy midterm question: summer coming when Scorpius rises in morning?

Astronomy 210 Midterm 1, fall semester 2014
Cuesta College, San Luis Obispo, CA

An astronomy article[*] discusses how the constellation Scorpius can be used as a sign of the changing seasons:

I once heard Edwin Hubble remark, "Well, summer is on the way. I saw Scorpius rising this morning."

Discuss a plausible date and time for an observer in San Luis Obispo, CA to make this observation of Scorpius a signal that "summer is on the way." If there is no such plausible date and time, then explain why. Defend your answer by clearly explaining how you used your starwheel to do this, along with any assumptions that you may have made.

[*] G. Purdam; R.S. Richardson "Naked Eye Objects: Your Favorites?" The Griffith Observer, Vol. 44, No. 8 (August 1980), p. 23, books.google.com/books?id=Y_vxAAAAMAAJ.

Solution and grading rubric:

• p:
  Correct. Discussion arguing for plausible date/time for Scorpius rising in the morning to signal the start of summer includes the following:
  1. manipulates starwheel (planisphere) such that Scorpius is rising from the eastern horizon;
  2. chooses a plausible "morning" (but still dark) time (12 AM to 6 AM) to view stars;
  3. chooses a corresponding month that would plausibly be soon before the start of summer.
  However, if "rises in the morning" is strictly applied to sunrise or just before sunrise, then Scorpius is found to be rising during December/January, in which case student may either argue for plausibility (just past winter solstice, so days will get longer advancing towards the summer solstice) or implausibility (middle of winter).
• r:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Diagram and/or explanation has minor errors. Problems with one of the three discussion points in (p); may have Scorpius transiting or high overhead; may pick a daytime hour; or may pick a month that is not soon before the start of summer. Or intends "rises in the morning" to be taken as sunrise, but instead picks a time well after sunrise.
• t:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Problems with diagram or discussion; has only one of the three discussion points in (p).
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some related discussion of starwheel use, morning times, and seasons.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Section 70158
Exam code: midterm01neVs
p: 22 students
r: 15 student
t: 6 students
v: 2 students
x: 0 students
y: 0 students
z: 0 students

Section 70160
Exam code: midterm01s4Tn
p: 20 students
r: 2 students
t: 4 students
v: 6 students
x: 4 students
y: 0 students
z: 0 students

A sample "p" response (from student 7809), finding for plausibility:
A sample "p" response (from student 1770), arguing for implausibility:
A sample "r" response (from student 2125), placing Scorpius at its highest point in the sky:
A sample "r" response (from student 3566), placing Scorpius low over the west horizon instead of the east horizon:
A sample "t" response (from student 5656), placing Scorpius in its highest point in the sky at noon:

Astronomy midterm question: crescent moon setting just before dawn?

Astronomy 210 Midterm 1, fall semester 2014
Cuesta College, San Luis Obispo, CA

The following excerpt[*] describes the setting of a crescent moon:

[She] looked up and out, over the flat land with no intervening towers or hills, and saw the crescent moon...disappearing beyond the edge of the world. It would be dawn soon...

Discuss whether or not this description is plausible, and how you know this. Support your answer using a diagram showing the positions of the sun, the moon, Earth, and an observer on Earth.

[*] Pamela Dean, "Cousins," from Firebirds Rising: An Anthology of Original Science Fiction and Fantasy, Sharon November (ed.), Firebird (2005), p. 473.

Solution and grading rubric:

• p:
  Correct. Complete diagram (with the sun, moon, and observer on Earth), and discusses how this scenario is impossible using at least one of two arguments:
  1. waxing crescent moon sets at 9 PM, and waning crescent moon sets at 3 PM, and thus neither crescent moon sets "soon" before sunrise;
  2. none of the phases that would set "soon" before sunrise (first quarter, waxing gibbous, and full moon; at 12 AM, 3 AM, and 6 AM respectively) are crescent phases.
• r:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Diagram and/or explanation has minor errors. Typically demonstrates that one of the crescent phases cannot set "soon" before sunrise, but does not eliminate the other crescent phase.
• t:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Problems with either diagram or discussion. Demonstrates understanding of at least one of the crescent phase set times, but discussion regarding implausibility is missing, or somehow interprets results as plausible.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Diagram and discussion problematic. May instead discuss a gibbous phase.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Section 70158
Exam code: midterm01neVs
p: 23 students
r: 5 students
t: 3 students
v: 12 students
x: 2 students
y: 0 students
z: 0 students

Section 70160
Exam code: midterm01s4Tn
p: 20 students
r: 2 students
t: 4 students
v: 6 students
x: 4 students
y: 0 students
z: 0 students

A sample "p" response (from student 1776), eliminating both waxing crescent and waning crescent phases as possibilities:
A sample "p" response (from student 1123), instead starting with phases visible just before dawn, and eliminating each as possibilities:
A sample "x" response (from student 6392), concerned with the local topography:

Astronomy quiz archive: telescopes

Astronomy 210 Quiz 3, fall semester 2014
Cuesta College, San Luis Obispo, CA

Section 70158, version 1
Exam code: quiz03s8tE


Section 70158

0- 8.0 :	* [low = 4.0]
8.5-16.0 :	******
16.5-24.0 :	***********
24.5-32.0 :	************** [mean = 25.3 +/- 8.6]
32.5-40.0 :	********** [high = 40.0]

Section 70160, version 1
Exam code: quiz03n6n4


Section 70160

0- 8.0 :
8.5-16.0 :	**** [low = 8.5]
16.5-24.0 :	************
24.5-32.0 :	******** [mean = 26.0 +/- 7.8]
32.5-40.0 :	********** [high = 36.5]