Phys 239
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Quantitative Physics
Problem Set 4 Solutions
Winged Migration
Take a stab at estimating the power requirements for a migrating bird, using the basic relationships developed
in lecture. I will allow you to look up mass and wingspan of your bird. If metabolically converting stored fat at
25% efficiency, what fraction of the bird’s mass is lost per day in flight? If the answer is scary, try different
assumptions, because we know they can do long-haul migration! Let that fact inform your estimations. It
may be easiest to assume periods of gliding, followed by flap/climb, repeated all day.
I will choose the Arctic Tern, which is a beautifully efficient and graceful flyer. It has a wing span of 70 cm
and a mass around 100 g. It migrates between the Arctic and Antarctic regions! I will guess its speed to
be something like 15 m/s (might be a tad high—also tempted by 10 m/s). At this speed, the ∼ 15, 000 km
journey (pole-to-pole would be 20,000 km, but it satisfies itself with latitudes well short of 90◦ ) would take
about one million seconds, which would be about 12 days of continuous flying.
I want to estimate a glide slope. My initial guess is that a graceful Tern does well, likely achieving a glide
ratio of at least 20:1. Watching birds glide, the more efficient flyers seem better than the Cessna-type plane
with a 10:1 ratio, but probably not as amazing as the tightly-engineered sailplanes (gliders) that achieve
40:1. Halfway (logarithmically) is 20:1. I’m not sure it’s right, but let’s see where that takes us.
Stepping back to the Lecture 7 material, we had an expression ρAeff vvdown = mbird g, and we found that
Aeff ∼ 0.4L2 for the B747 and Aeff ∼ 0.12L2 for the Cessna, where L is the wingspan. We might suspect
that the low-speed Tern shares more in common with the low-speed Cessna, and adopt Aeff ∼ 0.1L2 . Also,
recall that vdown = vθ, so that we can solve for theta for our Tern to find θ around 0.07, using v = 15 m/s.
Since the glide slope is θ/2, this corresponds to a glide ratio around 25. We had to guess the Aeff scaling,
but at least had some models against which to compare. At least we seem roughly justified in adopting 20:1
for now.
We can pretend that once per second, the bird flaps to regain its lost altitude. In this case, it’s 0.75 m
(15 m of horizontal progress divided by 20:1 glide ratio). The associated potential energy is 0.75 J for the
0.1 kg mass. That means the bird must put out 0.75 W of mechanical power. At an efficiency of 25%, this
corresponds to a metabolic rate of 3 W, accumulating to 250 kJ per day, or 60 kcal. At 9 kcal/g in fat, we
need about 7 g per day, or 7% of body weight. After 12 days, this would amount to almost all the body
weight. So something is wrong: it can’t be this dire. Perhaps the Tern finds food along the way, but I’m
not sure this is viable (they spend months building up enough reserve for the migration). If we decrease the
speed, we reduce the power required but lengthen the journey by the same amount for a net wash. Therefore
I conclude the glide ratio must be better. I probably need to go up to 40:1, cutting the daily loss to 3.5%
(at 15 m/s) and a total loss of 40%. That’ll be one hungry bird!
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Frequent Flyer MPG
Estimate the fuel efficiency of flying commercially. Use the fact that the amount of force needed to sustain
lift is equal to the drag force, as developed in the lecture. Imagine you’re flying Southwest Airlines, which
means a Boeing 737 aircraft. In this case, don’t just look up the weight: work out estimates possibly from
various angles, similar to the lecture approach. Parallel somewhat the fuel efficiency for a vehicle done in
lecture. Put the result into familiar context (e.g., miles per gallon for Americans). The absolute number is
pretty horrible, but per passenger it’s not bad. How many people do you have to put in a typical car to get a
similar per-passenger fuel economy?
I have flown on Southwest more times than I would care to count. A fine airline, but the thrill is gone.
Anyway, the whole fleet consists of B737 aircraft, having approximately 25 rows of 6 seats each (150 people,
seated in three groups of “60,” but the C group doesn’t go all the way up to 60). Each row takes about
a meter, and the radius is about 1.5 m. This makes the volume approximately 200 m3 , considering some
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room beyond the passenger cabin in both directions. As in lecture, I imagine the plane floating with wing
tops still dry, displacing perhaps a quarter of the volume: 50 m3 → 50 tons of water. I can also use the tire
inflation trick, thinking that each truck has 4 tires 1 m in diameter, each with a ground footprint around
0.1 m2 (something like 0.3 m by 0.3 m). I imagine inflation around 45 psi (3 atm). If the pressure is much
higher, the firmer tire will have less area in contact with the ground. My sense of 0.3 m flat spot is based on
familiarity with tires and their shapes at 30–40 psi. Anyhow, the pressure 3 × 105 Pa times 0.8 m2 leads to
24 ton, which is about half the previous estimate. If I guess that passengers constitute 10% of the load, and
have 150 70 kg meat bags on board, I get 10 tons of people and 100 tons for the plane, or twice the earlier
estimate. This means that the 50 ton estimate is the geometric mean of my three, and I’ll proceed with this
number. Looking online, I see numbers between 60–70 ton.
Flying at speeds similar to the B747 example in class, I will assume a similar glide ratio, or equivalently
downwash angle, θ. So whether we want to use the P = 12 mgvθ approach or do something similar to the
bird example, we’ll achieve equivalent results. For amusement, we’ll do the bird approach. At a speed of
240 m/s, a 20:1 ratio translates to a descent rate of 12 θv = 12 m/s (of course in actual glide, the plane may
be going only half this speed). 12 m/s times the weight means a power expenditure of 6 MW, or a force of
25 kN. Incidentally, at actual (lower) best-glide speed, the power required to maintain lift is greater than at
the cruise optimum (when power for lift equals power to fight drag), but drag is less. So I assume things
roughly add up the same in both cases. (Effectively, θ is less at high speeds, if v 2 θ ∼constant.)
A gallon of gasoline is about 4 L, with a mass of 3 kg. At 10 kcal/g, we get 120 MJ per gallon. Jet engine
efficiency is likely around 25% (better than car average of 15–20%), so we utilize 30 MJ per gallon, which
lasts us 5 seconds at 6 MW power output. In this time, the plane travels 1.2 km, or about 0.7 miles. So we
have 0.7 MPG, or over 300 L per 100 km traveled, in European parlance.
Seems abysmal, but on a per-passenger basis, we get about 100 MPG per passenger. Now it seems pretty
decent. If a car is typically 30–35 MPG, three people in the car (or two in a Prius) achieves the same fuel
economy. When the all the seats are full in a car, it does better than the airplane. Conversely, if seats are
empty on the airplane (often true), it does worse.
Wikipedia (Fuel economy in aircraft) puts a B737 fuel burn at about 3.2 kg/km, which is pretty much a gallon
per kilometer, right around what we calculated based on glide ratio (or equivalently, θ). The Wikipedia page
also has tables putting medium and long-haul aircraft in the range 70–10 MPG/passenger, typically. I swear
I didn’t work backwards (or revise accordingly). Really.
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Turbulent Pipe Flow
In class, we covered flow in a pipe at low-enough Reynolds number (<2000) to stay laminar (Poiseuille flow).
Now we will explore what happens in the turbulent regime. So let’s take a larger pipe of the sort you’d find
in a house. And let’s require a flow typical of what a sink faucet delivers.
a. First, establish your numbers and figure out the pressure differential (along a typical length of pipe in a
home) if Poiseuille Flow obtains. What is the characteristic Reynolds number of your flow, and would
you expect this result to be valid?
b. Next, use hobo derivatives to evaluate each term in the Navier-Stokes equation (all but gravity for
an assumed horizontal pipe) in the context of turbulent flow. Imagine swirls in the flow so that at
any given point the time derivative term and convective term are non-zero. Using this approach,
compute the pressure differential now. What term(s) dominate(s)? The art of the hobo is to establish
meaningful scales for changes in (in this case) velocity, time, and distance, possibly as a function of
direction/components.
I’ll use a pipe diameter of 1 cm for a household pipe, which corresponds to the smaller of two common
choices in houses (1/2-inch pipe vs. 3/4-inch pipe). We’ll ask for a flow rate of 1.5 gallons per minute, or
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F = 0.1 L/s (10−4 m3 /s). Our run will be 10 m. The average velocity is F/A ≈ 1.3 m/s. The Reynolds
number is about 6500, so we should have transcended the laminar flow regime.
Part a: In Poiseuille flow, we have that
F = 10−4 m3 /s =
1
10−8
παR4
,
= 0.4α 163
8ρν
10 · 10−6
so that α ≈ 400 Pa/m for a total pressure differential across the pipe of 4000 Pa. This corresponds to a
reservoir height h = 4000/ρg
≈ 0.4 m. The free-stream velocity from equating kinetic energy to reservoir
√
potential energy is v = 2gh ≈ 2.8 m/s.
Part b: We will assume that we have swirls filling the pipe, so that the scale of the swirls matches the radius
of the pipe. Even though we picture the swirl as a circular loop, the actual velocities will not be entirely
circular, as we still have net velocity/flow down the pipe. We will have to guess at the scale of turbulent
velocity superimposed on the flow velocity—maybe something like one tenth? We are after the pressure
gradient along the pipe, which we label as the z-direction. So looking at the Navier-Stokes equation:
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∂u
+ (u · ∇)u = ν∇2 u − ∇p + g,
∂t
ρ
(1)
we ignore the external force (last term), and are only concerned with the z-component of the pressure
gradient. This means we want to estimate the uz components of the other three terms. We imagine the swirl
velocity superimposed on the average velocity to be at a tenth the main flow velocity scale.
If a swirl scale is set by the pipe size, and moving at uavg /10, the associated timescale for a swirl motion
to execute is time τ ∼ 10πR/uavg (the π is to complete half-a-cycle, constituting the relevant time scale for
change). So if we concentrate on one place in the pipe, we may see velocities deviating from uavg by ±uavg /10
on a time scale of τ . This puts the first term (time derivative) at the scale uavg /10τ = u2avg /100πR.
For the convective term, we want
ux
∂uz
∂uz
∂uz
+ uy
+ uz
.
∂x
∂y
∂z
The non-axial components of velocity, ux and uy , will reach up to the swirl velocity scale of uavg /10, but
may typically be about half this, while uz will hover around uavg . The derivatives are all of uz (since only
looking at z-component of pressure gradient), and essentially asking how fast the z-component of velocity
changes as a function of distance. The magnitude of change of uz will be the swirl scale of uavg /10. The
relevant dimension for the cross-dimensions (x and y) will be approximately R, while the scale for changes
in the z-direction may be about 10R, since the eddies are sort-of stretched out along the z-direction owing
to average flow. In the end, each term contributes similarly:
u2avg
uavg uavg
uavg uavg
uavg
+
+ uavg
∼
.
20 10R
20 10R
100R
50R
Now the viscous piece. The curvature of the z-component of the velocity field, as characterized by ν∇2 uz ,
operates on the scale of the radius (curve dimension), so we get νuavg /R2 , or possibly less, if uavg does not
change by its full scale.
Evaluating these pieces for our uavg ≈ 1.3 m/s, R = 0.005 m, and ν = 10−6 m2 /s, we find the time derivative
piece is approximately 1 m/s2 , the convective piece at around 7 m/s2 , and the viscous piece at a paltry
0.003 m/s2 . This is another way to say that viscosity is not at all important in this regime.
The fields we describe are fluctuating, and we have found the amplitude of this fluctuation. The typical scale
will be the standard deviation, and the different terms may combine in quadrature. The convective term
dominates, and is about 5 m/s2 in a RMS sense.
We therefore expect a pressure gradient in the neighborhood of ρ · 5 m/s2 ∼ 5 × 103 Pa/m. The pressure
differential across the 10 m pipe would therefore be about 5 × 104 Pa, or about half-an-atmosphere (about
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7 psi, and household water is usually pressurized at about 60 psi). This result is about ten times that derived
if Poiseuille flow applied.
How well does this work? Passably, perhaps. The Darcy-Weisbach equation gives the pressure gradient as
α = 14 fD ρu2avg /R, where fD ≈ 0.03 for the regime we face here. Putting this into the same form, we would
find a 120 in the denominator rather than 50 as in the convective term result. So off by a bit more than a
factor of two. Of course, the uavg /10 scale for swirl velocity was a crude guess, and this comparison suggests
that uavg /15 would have nailed it better. I will admit that when I first approached this problem I used a
swirl velocity equal to the bulk flow velocity and got results I knew to be preposterous (would have said that
a vertical pipe of this size would barely flow even under gravity’s pull). So I adopted uavg /10 (realizing that
snapshots of velocity vectors would be dominated by the z-component, and that swirls would look like swirls
only when moving along with fluid) and was careful about actual time scales and along-the-pipe dimensions
of the swirl, etc. Turbulence is hard to even model well numerically, as a complex 3-D phenomenon. The
hobo approach does not account for signed cancellations, etc. that a real flow exhibits. Whatever. It was
an interesting exercise that I hope was still illuminating in exploring a technique.
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Raindrop Radius
Use the fact that water has a surface tension of γ = 0.07 N/m to compute the largest a falling raindrop (in
terminal velocity) may be without the ram pressure from air ripping it apart.
The energy scale associated with the surface of a drop of water is expressed as the surface tension times the
surface area: E ≈ γAsurf . The energy density (pressure) associated with surface tension can be approximated
as P ≈ E/V . For a sphere, then, P ≈ 3γ/R. Setting this to the ram pressure from drag (drag force over
projected area), we get P ≈ 3γ/R ≈ mg/Aproj , where we have assumed terminal velocity such that the
drag force equals the weight of the drop (thuspno need to mess with v 2 drag formula). Inserting spherical
geometry, we find 3γ/R ≈ 43 ρgR, so that R ≈ 9γ/4ρg, evaluating to 4 mm. Not an outlandish result, but
it strikes me as being on the large side.
In practice, several things may modify this result. First, instabilities and dynamics may drive the breakup
sooner: we have assumed an orderly sphere for our geometry, which is certainly unfitting for a breaking-up
droplet. Also, we have essentially distributed the surface tension energy around the entire spherical area,
while said instabilities concentrate the action into smaller zones. Still, we got an answer that is not entirely
preposterous: I could believe the largest drops approach this size.
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