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6.055 / Art of approximation 72
8.3 Drag
Pendulum motion is not a horrible enough problem to show the full benefit of dimensional analysis. Instead try fluid mechanics – a subject notorious for its mathematical and physical complexity; Chandrasekhar’s books [10, 11] or the classic textbook of Lamb [12] show that the mathematics is not for the faint of heart.
Density ρfl Viscosity ν
ρobj
R
v
The next examples illustrate two extremes of fluid flow: oozing and turbu- lent. An example of oozing flow is ions transporting charge in seawater ( Section 8.3.6 ). An example of turbulent flow is a raindrop falling from the sky after condens- ing out of a cloud ( Section 8.3.7 ). To find the terminal velocity, solve the partial-differential equations of fluid mechanics for the incompressible flow of a Newtonian fluid: ∂ v ∂t +^ ( v · ∇ ) v^ =^ −^
ρ ∇ p^ +^ ν ∇
(^2) v , (3 eqns)
∇ · v = 0. (1 eqn) Here v is the fluid velocity, ρ is the fluid density, ν is the kinematic viscosity, and p is the pressure. The first equation is a vector shorthand for three equa- tions, so the full system is four equations. All the equations are partial-differential equations and three are nonlinear. Worse, they are coupled: Quantities appear in more than one equation. So we have to solve a system of coupled, nonlinear, partial-differential equations. This solution must satisfy boundary con- ditions imposed by the marble or raindrop. As the object moves, the boundary conditions change. So until you know how the object moves, you do not know the boundary condi- tions. Until you know the boundary conditions, you cannot find the motion of the fluid or of the object. This coupling between the boundary conditions and solution compounds the difficulty of the problem. It requires that you solve the equations and the boundary conditions together. If you ever get there, then you take the limit t → ∞ to find the terminal velocity. Sleep easy! I wrote out the Navier–Stokes equations only to scare you into using dimen- sional analysis and special-cases reasoning. The approximate approach is easier than solv- ing nonlinear partial-differential equations.
8.3.1 Naive dimensional analysis
To use dimensional analysis, follow the usual steps: Choose relevant variables, form di- mensionless groups from them, and solve for the terminal velocity. In choosing quantities, do not forget to include the variable for which you are solving, which here is v. To decide on the other quantities, split them into three categories (divide and conquer):
- characteristics of the fluid,
- characteristics of the object, and
- characteristics of whatever makes the object fall.
8 Special cases 73
The last category is the easiest to think about, so deal with it first. Gravity makes the object fall, so g is on the list.
Consider next the characteristics of the object. Its velocity, as the quantity for which we are solving, is already on the list. Its mass m affects the terminal velocity: A feather falls more slowly than a rock does. Its radius r probably affects the terminal velocity. Instead of listing r and m together, remix them and use r and ρobj. The two alternatives r and m or r and ρobj provide the same information as long as the object is uniform: You can compute ρobj from m and r and can compute m from ρobj and r.
Choose the preferable pair by looking ahead in the derivation. The relevant properties of the fluid include its density ρfl. If the list also includes ρobj, then the results might contain pleasing dimensionless ratios such as ρobj/ρfl (a dimensionless group!). The ratio ρobj/ρfl has a more obvious physical interpretation than a combination such as m/ρflr^3 , which, ex- cept for a dimensionless constant, is more obscurely the ratio of object and fluid densities. So choose ρobj and r over m and r.
Scaling arguments also favor the pair ρobj and r. In a scaling argument you imagine varying, say, a size. Size, like heat, is an extensive quantity: a quantity related to amount of stuff. When you vary the size, you want as few other variables as possible to change so that those changes do not obscure the effect of changing size. Therefore, whenever possible replace extensive quantities with intensive quantities like temperature or density. The pair m and r contains two extensive quantities, whereas the preferable pair ρobj and r contains only one extensive quantity.
Now consider properties of the fluid. Its density ρfl affects the terminal velocity. Perhaps its viscosity is also relevant. Viscosity measures the tendency of a fluid to reduce velocity differences in the flow. You can observe an analog of viscosity in traffic flow on a multilane highway. If one lane moves much faster than another, drivers switch from the slower to the faster lane, eventually slowing down the faster lane. Local decisions of the drivers reduce the velocity gradient. Similarly, molecular motion (in a gas) or collisions (in a fluid) transports speed (really, momentum) from fast- to slow-flowing regions. This transport reduces the velocity difference between the regions. Oozier (more viscous) fluids probably produce more drag than thin fluids do. So viscosity belongs on the list of relevant variables.
Fluid mechanicians have defined two viscosities: dynamic viscosity η and kinematic vis- cosity ν. [Sadly, we could not use the mellifluous term fluid mechanics to signify a host of physicists agonizing over the equations of fluid mechanics; it would not distinguish the toilers from their toil.] The two viscosities are related by η = ρflν. Life in Moving Fluids [13, pp. 23–25] discusses the two types of viscosity in detail. For the analysis of drag force, you need to know only that viscous forces are proportional to viscosity. Which viscosity should we use? Dynamic viscosity hides ρfl inside the product νρfl; a ratio of ρobj and η then looks less dimensionless than it is because ρobj’s partner ρfl is buried inside η. Therefore the kine- matic viscosity ν usually gives the more insightful results. Summarizing the discussion, the table lists the variables by category.
8 Special cases 75
The figure shows the roadmap updated with this information.
higher p
buoyant force
lower p
Sphere
The remaining pieces are drag and buoyancy. Buoyancy is easier, so do it first (the principle of maximal laziness). It is an upward force that results because gravity affects the pressure in a fluid. The pressure increases according to p = p 0 + ρfl gh, where h is the depth and p 0 is the pressure at zero depth (which can be taken to be at any level in the fluid). The pressure difference between the top and bottom of the object, which are separated by a distance ∼ r, is ∆p ∼ ρfl gr. Pressure is force per area, and the pressure difference acts over an area A ∼ r^2. Therefore the buoyant force created by the pressure difference is
Fb ∼ A∆p ∼ ρflr^3 g.
terminal velocity
weight drag buoyancy
ρspr^3 g ρflr^3 g
As a check on this result, Archimedes’s principle says that the buoyant force is the ‘weight of fluid displaced’. This weight is mass ︷ ︸︸ ︷ ρfl^4 π 3 πr^3 ︸ ︷︷ ︸ volume
g.
Except for the factor of 4 π/ 3 , it matches the buoyant force so Archimedes’s principle con- firms our estimate for Fb. That result updates the roadmap. The main unexplored branch is the drag force, which we solve using dimensional analysis.
8.3.3 Dimensional analysis for the drag force
The weight and buoyancy were solvable without dimensional analysis, but we still need to use dimensional analysis to find the drag force. The purpose of breaking the problem into parts was to simplify this dimensional analysis relative to the brute-force approach in Section 8.3.1. Let’s see how the list of variables changes when computing the drag force rather than the terminal velocity. The drag force Fd has to join the list: not a promising beginning when trying to eliminate variables. Worse, the terminal velocity v remains on the list, even though we are no longer computing it, because the drag force depends on the velocity of the object. However, all is not lost. The drag force has no idea what is inside the sphere. Picture the fluid as a huge computer that implements the laws of fluid dynamics. From the viewpoint of this computer, the parameters v and r are the only relevant attributes of a moving sphere. What lies underneath the surface does not affect the fluid flow: Drag is only skin deep. The computer can determine the flow (if it has tremendous processing power) without knowing the sphere’s density ρobj, which means it vanishes from the list. Progress!
6.055 / Art of approximation 76
Var Dim What Fd MLT−^2 drag force ν L^2 T−^1 kinematic viscosity ρfl ML−^3 fluid density r L object radius v LT−^1 terminal velocity
Now consider the characteristics of the fluid. The fluid supercomputer still needs the density and viscosity of the fluid to determine how the pieces of fluid move in response to the object’s motion. So ρfl and ν remain on the list. What about gravity? It causes the object to fall, so it is responsible for the terminal velocity v. However, the fluid supercomputer does not care how the object acquired this velocity; it cares only what the velocity is. So g vanishes from the list. The updated tabled shows the new, shorter list.
terminal velocity
weight drag buoyancy
ρspr^3 g DA ρflr^3 g
group 1 group 2
The five variables in the list are composed of three basic dimensions. From the Buckingham Pi theorem ( Section 7.6 ), we expect two dimen- sionless groups. We find one group by dividing and conquering. The list already includes a velocity (the terminal velocity). If we can con- coct another quantity V with dimensions of velocity, then v/V is a di- mensionless group. The viscosity ν is almost a velocity. It contains one more power of length than velocity does. Dividing by r eliminates the extra length: V ≡ ν/r. A dimensionless group is then
G 1 ≡ v V =^
vr ν.
terminal velocity
weight drag buoyancy
ρspr^3 g DA ρflr^3 g
group 1 group 2
vR ν
Our knowledge, including this group, is shown in the figure. This group is so important that it has a name, the Reynolds number , which is abbreviated Re. It is important because it is a dimensionless mea- sure of flow speed. The velocity, because it contains dimensions, can- not distinguish fast from slow flows. For example, 1000 m s−^1 is slow for a planet, whose speeds are typically tens of kilometers per second, but fast for a pedestrian. When you hear that a quantity is small, fast, large, expensive, or almost any adjective, your first reaction should be to ask, ‘compared to what?’ Such a comparison suggests dividing v by another velocity; then we get a dimensionless quantity that is propor- tional to v. The result of this division is the Reynolds number.
Low values of Re indicate slow, viscous flow (cold honey oozing out of a jar). High values indicate turbulent flow (a jet flying at 600 mph). The excellent Life in Moving Fluids [13] discusses many more dimensionless ratios that arise in fluid mechanics.
The Reynolds number looks lonely in the map. To give it company, find a second dimen- sionless group. The drag force is absent from the first group so it must live in the second; otherwise we cannot solve for the drag force.
Instead of dreaming up the dimensionless group in one lucky guess, we construct it in steps (divide-and-conquer reasoning). Examine the variables in the table, dimension by dimension. Only two (Fd and ρfl) contain mass, so both or neither appear in the group. Because Fd has to appear, ρfl must also appear. Each variable contains a first power of mass, so the group contains the ratio Fd/ρfl. A simple choice is
6.055 / Art of approximation 78
In highly dissipative flows, when energy is burned directly up by viscosity, the numerator is much larger than the denominator, so this ratio (which will turn out to measure drag) is much greater than 1. In highly streamlined flows (a jet wing), the the work done against drag is small because the fluid returns most of the imparted kinetic energy to the object. So in the ratio, the numerator will be small compared to the denominator. To solve for Fd, which is contained in G 2 , use the form G 2 = f (G 1 ), which becomes Fd ρflr^2 v^2 = f
( (^) vr ν
The drag force is then
Fd = ρflr^2 v^2 f
( (^) vr ν
The function f is a dimensionless function: Its argument is dimensionless and it returns a dimensionless number. It is also a universal function. The same f applies to spheres of any size, in a fluid of any viscosity or density! Although f depends on r, ρfl, ν, and v, it depends on them only through one combination, the Reynolds number. A function of one variable is easier to study than is a function of four variables: A good table of functions of one variable may require a page; that of a function of two variables a volume; that of a function of three variables a bookcase; and that of a function of four variables a library. —Harold Jeffreys [6, p. 82] Dimensional analysis cannot tell us the form of f. To learn its form, we specialize to two special cases:
- viscous, low-speed flow (Re 1 ), the subject of Section 8.3.4 ; and
- turbulent, high-speed flow (Re 1 ), the subject of Section 8.3..
8.3.4 Viscous limit
As an example of the low-speed limit, consider a marble falling in vegetable oil or glycerin. You may wonder how often marbles fall in oil, and why we bother with this example. The short answer to the first question is ‘not often’. However, the same physics that determines the fall of marbles in oil also determines, for example, the behavior of fog droplets in air, of bacteria swimming in water [14], or of oil drops in the Millikan oil-drop experiment. The marble problem not only illustrates the physical principles, but also we can check our results with a home experiment. In slow, viscous flows, the drag force comes directly from – surprise! – viscous forces. These forces are proportional to viscosity because viscosity is the constant of proportionality in the definition of the viscous force. Therefore
Fd ∝ ν.
8 Special cases 79
The viscosity appears exactly once in the drag result, repeated here:
Fd = ρflr^2 v^2 f
( (^) vr ν
To flip ν into the numerator and make Fd ∝ ν, the function f must have the form f (x) ∼ 1 /x. With this f (x) the result is
Fd ∼ ρflr^2 v^2 ν vr =^ ρflνv.
Dimensional analysis alone is insufficient to compute the missing magic dimensionless con- stant. A fluid mechanician must do a messy and difficult calculation. Her burden is light now that we have worked out the solution except for this one constant. The British mathe- matician Stokes, the first to derive its value, found that
Fd = 6 πρflνvr.
In honor of Stokes, this result is called Stokes drag.
Let’s sanity check the result. Large or fast marbles should feel a lot of drag, so r and v should be in the numerator. Viscous fluids should produce a lot of drag, so ν should be the numerator. The proposed drag force passes these tests. The correct location of the density
- in the numerator or denominator – is hard to judge.
You can make an educated judgment by studying the Navier–Stokes equations. In those equations, when v is ‘small’ (small compared to what?) then the ( v · ∇ ) v term, which con- tains two powers of v, becomes tiny compared to the viscous term ν ∇^2 v , which contains only one power of v. The second-order term arises from the inertia of the fluid, so this term’s being small says that the oozing marble does not experience inertial effects. So per- haps ρfl, which represents the inertia of the fluid, should not appear in the Stokes drag. On the other hand, viscous forces are proportional to the dynamic viscosity η = ρflν, so ρfl should appear even if inertia is unimportant. The Stokes drag passes this test. Using the dynamic instead of kinematic viscosity, the Stokes drag is
Fd = 6 πηvr,
often a convenient form because many tables list η rather than ν.
This factor of 6 π comes from doing honest calculations. Here, it comes from solving the Navier–Stokes equations. In this book we wish to teach you how not to suffer, so we do not solve such equations. We usually quote the factor from honest calculation to show you how accurate (or sloppy) the approximations are. The factor is often near unity, although not in this case where it is roughly 20! In fancy talk, it is usually ‘of order unity’. Such a number suits our neural hardware: It is easy to remember and to use. Knowing the approximate derivation and remembering this one number, you reconstruct the exact result without solving difficult equations.
Now use the Stokes drag to estimate the terminal velocity in the special case of low Reynolds number.
8 Special cases 81
v ∝ g
ρobj ρfl^. We retain the g in the proportionality for the following reason: The true solution returns if we replace g by an effective gravity g^0 where
g^0 ≡ g
ρfl ρobj
So, one way to incorporate the effect of the buoyant force is to solve the problem without buoyancy but with the reduced g. Check this replacement in two limiting cases: ρfl = 0 and ρfl = ρobj. When ρobj = ρfl gravity vanishes: People, whose density is close to the density of water, barely float in swimming pools. Then g^0 should be zero. When ρfl = 0 , buoyancy vanishes and gravity retains its full effect. So g^0 should equal g. The effective gravity definition satisfies both tests. Between these two limits, the effective g should vary linearly with ρfl because buoyancy and weight superpose linearly in their effect on the object. The effective g passes this test as well. Another test is to imagine ρfl > ρobj. Then the relation correctly predicts that g^0 is negative: helium balloons rise. This alternative to using buoyancy explicitly is often useful. If, for example you forget to include buoyancy (which happened in the first draft of this chapter), you can correct the results later by replacing g with the g^0. If we carry forward the constants of proportionality, starting with the magic 6 π in the Stokes drag and including the 4 π/ 3 that belongs in the weight, we find
v ∼
gr^2 ν
( (^) ρ obj ρfl^ −^1
8.3.6 Conductivity of seawater
σ =^1 ρ
ρ
R = VI block geometry
V = El I^ =^ qnvA
E l n^ v^ A Fq Fd
As an application of Stokes drag and a rare example of a realis- tic situation with low Reynolds numbers, let’s estimate the elec- trical conductivity of seawater. Solving this problem is hopeless without breaking it into pieces. Conductivity σ is the reciprocal of resistivity ρ. (Apologies for the convention that overloads the density symbol with yet another meaning.) Resistivity, as its name suggests, is related to resistance R. Why have both ρ and R? Re- sistance is a useful measure for a particular wire, but not for wires in general because it depends on the diameter and cross-sectional area of the wire. It is not an intensive quantity. Before examin- ing the relationship between resistivity and resistance, let’s finish sketching the solution tree, leaving ρ as depending on R plus geometry. We can find R by placing a voltage V across a block of seawater and measuring the current I; then R = V/I. To find V or I we need a physical model. First, why does seawater conduct at all? Con- duction requires the transport of charge, which is produced by an electric field. Seawater
6.055 / Art of approximation 82
is mostly water and table salt (NaCl). The ions that arise from dissolving salt can transport charge. The resulting current is
I = qnvA,
where A is the cross-sectional area of the block, q is the ion charge, n is the ion concentration, and v is its terminal speed.
To understand, and be able to rederive this formula, first check its dimensions. Current is charge per time. Is the right side also charge per time? Yes: q takes care of the charge; and vA has dimensions of L^3 T−^1 so nvA, which has dimensions of T−^1 , takes care of the time.
As a second check, watch a cross-section of the block for a time ∆t. How much charge flows in that time? The charges move at speed v, so all charges in block of width v∆t and area A cross the cross-section. This block has volume vA∆t. The ion concentration is n, so the block contains nvA∆t charges. If each ion has charge q, then the total charge on the ions is Q = qnvA∆t. It took a time ∆t for this charge to flow, so the current is I = Q/∆t = qnvA. The terminal speed v depends on the applied force Fq and on the drag force Fd, just as for the falling marble but with an electrical force instead of a gravitational force. The result of this subdividing is the preceding map.
A
l
Now let’s find expressions for the unknown nodes. Only three remain: ρ, v, and n. The figure illustrates the relation between ρ and R:
ρ = RA l
To find v we follow the same procedure as for the marble. The applied force is Fq = qE, where q is the ion charge and E is the electric field. The electric field produced by the voltage V is E = V/l, where l is the length of the block, so
Fq = qV l
an expression in terms only of known quantities. The drag is Stokes drag. Equating this drag to the applied force gives the terminal velocity v in terms of known quantities:
v ∼ qV 6 πηlr
where r is the radius of the ion.
Only the number density n remains unknown. We estimate it after getting a symbolic result for σ, which you can do by climbing up the solution tree. First, find the current in terms of the terminal velocity:
I = qnvA ∼ q^2 nAV 6 πηlr
Use the current to find the resistance:
R ∼ V I
6 πηlr q^2 nA
6.055 / Art of approximation 84
η = ρwaterν ∼ 103 kg m−^3 × 10 −^6 m^2 s−^1 = 10 −^3 kg m−^1 s−^1.
Here I switched to SI (mks) units. Although most calculations are easier in cgs units – also known as God’s units – than they are in SI units, the one exception is electromagnetism, which is represented by the e^2 in the conductivity. Electromagnetism is conceptually eas- ier in cgs units – which needs no ghastly μ 0 or 4 π 0 , for example – than it is in SI units. However, the cgs unit of charge, the electrostatic unit, is unfamiliar. So, for numerical cal- culations, use SI units.
water
water
water
water
The final quantity required is the ion radius. A positive ion (sodium) attracts an oxygen end of a water molecule; a negative ion (chloride) attracts the hydrogen end of a water molecule. Either way, the ion, being charged, is surrounded by one or maybe more layers of water molecules. As it moves, it drags some of this baggage with it. So rather than use the bare ion radius you should use a larger radius to include this shell. But how thick is the shell? As an educated guess, assume that the shell includes one layer of water molecules, each with a radius of 1. 5 Å. So for the ion plus shell, r ∼ 2 Å.
With these numbers, the conductivity becomes:
σ ∼
e^2 ︷ ︸︸ ︷ (1. 6 · 10 −^19 C)^2 ×
n ︷ ︸︸^ ︷ 7 · 1026 m−^3 ︸︷︷︸^6 ×^3 6 π
× 10 −^3 kg m−^1 s−^1 ︸ ︷︷ ︸ η
× (^2) ︸ · 10 ︷︷−^10 m︸ r
You can do the computation mentally: Take out the big part, apply the principle of maximal laziness, and divide and conquer by first counting the powers of ten (shown in red) and then worrying about the small factors. Then divide and conquer again by counting the top and bottom contributions separately. The top contributes -12 powers of ten: − 38 from e^2 and
- 26 from n. The bottom contributes -13 powers of ten: − 3 from η and − 10 from r. The division produces one power of ten.
Now account for the remaining small factors:
- 62 × 7 6 × 3 × 2.
Slightly overestimate the answer by pretending that the 1. 62 on top cancels the 3 on the bottom. Slightly underestimate the answer – and maybe compensate for the overestimate
- by pretending that the 7 on top cancels the 6 on the bottom. After these lies, only 1 / 2 remains. Multiplying it by the sole power of ten gives
σ ∼ 5 Ω−^1 m−^1.
Using a calculator to do the arithmetic gives 4. 977... Ω−^1 m−^1 , which is extremely close to the result from mental calculation.
8 Special cases 85
The estimated resistivity is
ρ ∼ σ−^1 ∼ 0. 2 Ω m = 20 Ω cm,
where we converted to the conventional although not fully SI units of Ω cm. A typical experimental value for seawater at T = 15 ◦C is 23. 3 Ω cm (from [15, p. 14- 15]), absurdly close to the estimate! Probably the most significant error is the radius of the ion-plus-water combination that is doing the charge transport. Perhaps r should be greater than 2 Å, especially for a sodium ion, which is smaller than chloride; it therefore has a higher electric field at its surface and grabs water molecules more strongly than chloride does. In spite of such uncertainties, the continuum approximation produced more accurate results than it ought to. At the length scale of a sodium ion, water looks like a collection of spongy boulders more than it looks like a continuum. Yet Stokes drag worked. It works because the important length scale is not the size of water molecules, but rather their mean free path between collisions. Molecules in a liquid are packed to the point of contact, so the mean free path is much shorter than a molecular (or even ionic) radius, especially compared to an ion with its shell of water. The moral of this example, besides illustrating Stokes drag, is to have courage. Approxi- mate first and ask questions later. Maybe the approximations are correct for reasons that you do not suspect when you start solving a problem. If you agonize over each approxima- tion, you will never start a calculation, and then you will not find out that many approxi- mations would have been fine.. .if only you had had the courage to make them.
8.3.7 Turbulent limit
We now compute drag in the other flow extreme: high-speed, or turbulent, flow. The exam- ple will be to compute the terminal speed of a raindrop. These results apply to most flows. For example, when a child rises from a chair, the airflow around her is high-speed flow, as you can check by computing the Reynolds number. Say that the child is 0. 2 m wide, and that she rises with velocity 0. 5 m s−^1. Then
Re ∼ vr νair^ ∼^
- 5 m s−^1 × 0. 2 m 2 · 10 −^5 m^2 s−^1
Here viscosity of air is closer to
νair ≈ 1. 5 · 10 −^5 m^2 s−^1 ,
than to 2 · 10 −^5 m^2 s−^1 , but 2 · 10 −^5 m^2 s−^1 easily combines with the 0. 2 m in the numerator to allows us to do the calculation mentally. Using either value for the viscosity, the Reynolds number is much larger than unity, so the flow is turbulent. Larger objects, such as planes, trains, and automobiles, create turbulence even when they travel even more slowly than the child. In short, most fluid flow around us is turbulent flow.
8 Special cases 87
therefore prevents flow at the front from being radically different from the flow at the back, and thereby squelches any turbulence. In the other limit, when τV τv or Re 1 – momentum diffusion is outraced by fluid flow, so the fluid is free to shred itself into a turbulent mess. Once the viscosity is low enough to allow turbulence, its value does not affect the drag, which is why we can ignore it for Re 1. Here Re 1 means ‘large enough so that turbulence sets in’, which happens around Re ∼ 1000. A more complete story, which we discuss as part of boundary layers in Section 9.4 , slightly corrects this approximation. However, it is close enough for our purposes here.
terminal velocity
weight drag buoyancy
ρspr^3 g DA ρflr^3 g
group
Fd ρflr^2 v^2
Here the important point is that the viscosity vanishes from the analy- sis and so does group 1. Once it disappears, the dimensionless group that remains is
G 2 = Fd ρflr^2 v^2
Because it is the only group, the solution is
G 2 = dimensionless constant,
or
Fd ∼ ρflr^2 v^2.
h
A
fluid
Vol. = Ah m = ρflAh
This drag is for a sphere. What about other shapes, which are characterized by more parameters than a sphere is? So that the drag force generalizes to more com- plex shapes, we express it using the cross-sectional area of the object. Here A = πr^2 , so
Fd ∼ ρflAv^2.
This conventional choice has a physical basis. As an object moves, the mass of fluid that it displaces is proportional to its cross-sectional area:
mfl = ρflAh.
The fluid is given a speed comparable to v, so the fluid’s kinetic energy is
EK ∼ 1 2 mflv^2 ∼ 1 2 ρflAhv^2.
If all this kinetic energy is dissipated by drag, then the drag force is EK/h or
Fd ∼
2 ρflAv
In this form with the factor of 1 / 2 , the constant of proportionality is the drag coefficient cd.
6.055 / Art of approximation 88
Object cd Sphere 0. 5 Cylinder 1. 0 Flat plate 2. 0 Car 0. 4
Like its close cousin f from the dimensionless drag force, the drag coefficient is a dimensionless measure of the drag force. It depends on the shape of the object – on how streamlined it is. The table lists cd for various shapes (at high Reynolds number). The drag coefficient, being proportional to the function f (Re) in the general solution, also depends on the Reynolds number. How- ever, using the reasoning that the flow at high Reynolds number is indepen- dent of viscosity, the drag coefficient should also be independent of Reynolds number. Using the drag coefficient instead of f (which implies using cross-sectional area instead of r^2 ), the turbulent drag force becomes
Fd =
2 cdρflv
2 A.
So we have an expression for the turbulent drag force. The weight and buoyant forces are the same as in the viscous limit. So we just need to redo the analysis of the viscous limit but with the new drag force. Because the weight and buoyant forces contain r^3 , we return to using r^2 instead of A in the drag force. With these results, the terminal velocity v is given by
ρflr^2 v^2 ︸︷︷︸ Fd
∼ g(ρobj − ρfl)r^3 ︸ ︷︷ ︸ Fg−Fb
so
v ∼
gr
( (^) ρ obj ρfl^ −^1
Pause to sanity check this result: Are the right variables upstairs and downstairs? We consider each variable in turn.
- ρfl: The terminal velocity is smaller in a denser fluid (try running in a swimming pool), so ρfl should be in the denominator.
- g: Imagine a person falling on a planet that has a gravitational force stronger than that of the earth. Gravity partially determines atmospheric pressure and density. Holding the atmospheric density constant while increasing gravity might be impossible in real life, but we can do it easily in a thought experiment. The drag force then does not depend on g, so gravity increases the terminal speed without opposition from the drag force: g should be upstairs.
- ρobj: Imagine a raindrop made of (very) heavy water. Relative to a standard raindrop, the gravitational force increases while the drag force remains constant, as shown using
the fluid-is-a-computer argument in ??sec:drag-force-DA. So ρobj should be upstairs.
- r: To determine where the radius lives requires a more subtle argument. Increasing r increases both the gravitational and drag forces. The gravitational force increases as r^3
6.055 / Art of approximation 90
8.3.8 Combining solutions from the two limits
You know know the drag force in two extreme cases, viscous and turbulent drag. The results are repeated here:
Fd =
6 πρflνvr (viscous), 1 2 cdρflAv (^2) (turbulent).
Let’s compare and combine them by making the viscous form look like the turbulent form. Compared to the turbulent form, the viscous form lacks one power of r and one power of v but has an extra power of ν. A combination of variables with a similar property is the Reynolds number rv/ν. So multiply the viscous drag by a useful form of unity:
Fd =
( (^) rv/ν Re
1
× 6 πρflvρflνr ︸ ︷︷ ︸ Fd
Re 6 πρflv^2 r^2 (viscous).
This form, except for the 6 π and the r^2 , resembles the turbulent drag Fortunately A = πr^2 so
Fd =
Re ρflv
(^2) A (viscous),
With
cd = 12 Re (viscous),
the turbulent drag and this rewritten viscous drag for a sphere have the same form:
Fd = 12 ρflAv^2 ×
Re (Re 1 ),
- 5 (Re 1 ).
12 Re
- 5
0.01 1 100 104
1
10
100
1000
Re
At high Reynolds number the drag coefficient remains cd constant. For a sphere, that constant is cd ∼ 1 / 2. If the low-Reynolds-number approximation for cd is valid at sufficiently high Reynolds numbers, then cd would cross 1 / 2 near Re ∼ 24 , where presumably the high-Reynolds- number approximation takes over. The crossing point is a reasonable estimate for the transition between low- and high-speed flow. Experiment or massive simulation are the only ways to get a more accurate result. Exper- imental data place the crossover near Re ∼ 5 , at which point cd ∼ 2. Why can’t you calculate this value analyt- ically? If a dimensionless variable, such as the Reynolds number, is close to unity, calcu- lations become difficult. Approximations that depend on a quantity being either huge or tiny are no longer valid. When all terms in an equation are roughly of the same magnitude, you cannot get rid of any term without making large errors. To get results in these situa- tions, you have to do honest work: You must do experiments or solve the Navier–Stokes equations numerically.
