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3.5: Drag Using Conversion of Energy

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    24097
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    A box model will next help us estimate drag forces. Drag, one of the most difficult subjects in physics, is also one of the most important forces in everyday life. If it weren’t for drag, bicycling, flying, and driving would be a breeze. Because of drag, locomotion requires energy. Rigorously calculating a drag force requires solving the Navier–Stokes equations:

    \[(\mathbf{v} \cdot \nabla) \mathbf{v} + \frac{\delta \mathbf{v}}{\delta t} = -\frac{1}{\rho} \nabla p + v \nabla^{2} \mathbf{v}\]

    They are coupled, nonlinear, partial-differential equations. You could read many volumes describing the mathematics to solve these equations. Even then, solutions are known only in a few circumstances—for example, a sphere moving slowly in a viscous fluid or moving at any speed in a nonviscous fluid. However, a nonviscous fluid—what Feynman [14, Section II-40-2], quoting John von Neumann, rightly disparages as “dry water”—is particularly irrelevant to real life because viscosity is the cause of drag, so a zero-viscosity solution predicts zero drag! Using a box model and conservation of energy is a simple and insightful alternative.

    3.5.1 Box model for drag

    We will first estimate the energy lost to drag as an object moves through a fluid, as in Section 3.2.1. From the energy, we will find the drag force. To quantify the the problem, imagine pushing an object of cross-sectional area Acs at speed v for a distance d. The object sweeps out a tube of fluid. (The tube length d is arbitrary, but it will cancel out of the force.)

    clipboard_e5c9a91eedd303c22b2a80a031de2b49e.png

    How much energy is consumed by drag?

    Energy is consumed because the object gives kinetic energy to the fluid (say, water or air); viscosity, as we will model in Section 6.4.4, then turns this energy into heat. The kinetic energy depends on the mass of the fluid and on the speed it is given. The mass of fluid in the tube is \(\rho A_{cs} d\), where \(\rho\) is the fluid density. The speed imparted to the fluid is roughly the speed of the object, which is v. Therefore, the kinetic energy given to the fluid is roughly \(\rho A_{cs} v^{2} d\):

    \[E_{\textrm{kinetic}} \sim \underbrace{\rho A_{cs} d}_{mass} \times v^{2} = \rho A_{cs} v^{2}d.\]

    This calculation ignores the factor of one-half in the definition of kinetic energy. However, the other approximations, such as assuming that only the swept-out fluid is affected or that all the swept-out fluid gets speed v, are at least as inaccurate. For this rough calculation, there is little point in including the factor of one-half.

    This kinetic energy is roughly the energy converted into heat. Therefore, the energy lost to drag is roughly \(\rho A_{cs} v^{2} d\). The drag force is then given by

    \[\underbrace{\textrm{energy lost to drag}}_{\sim \rho A_{cs} v^{2} d} = \underbrace{\textrm{drag force}}_{F_{drag}} \times \underbrace{\textrm{distance}}_{d}.\]

    Now we can solve for the drag force:

    \[F_{drag} \sim \rho A_{cs} v^{2}.\]

    As expected, the arbitrary distance d has canceled out.

    3.5.2 Testing the analysis with a home experiment

    To test this analysis, try the following home experiment. Photocopy or print this page at 200 percent enlargement (a factor of 2 larger in width and height), cut out the template, and tape the two straight edges together to make a cone:

    clipboard_e8382e1a6f688a898ca9b003a5085b849.png

    We could use many other shapes. However, a cone is easy to construct, and also falls without swishing back and forth (as a sheet of paper would) or flipping over (as long as you drop it point down).

    clipboard_e78e7032643b0dafb9c298ddde0ba2d37.png

    We’ll test the analysis by predicting the cone’s terminal speed: that is, its steady speed while falling. When the cone is falling at this constant speed, its acceleration is zero, so the net force on it is, by Newton’s second law, also zero. Thus, the drag force Fdrag equals the cone’s weight mg (where m is the cone’s mass and g is the gravitational acceleration):

    \[\rho_{air} v^{2} A_{cs} \sim mg\]

    The terminal speed thus reveals the drag force. (Even though the drag force equals the weight, the left side is only an approximation to the drag force, so we connect the left and right sides with a single approximation sign ~.) The terminal speed vterm is then

    \[v_{term} \sim \sqrt{\frac{mg}{A_{cs}\rho_{air}}.}\]

    The mass of the cone is

    \[m = A_{paper} = \underbrace{\textrm{areal density of paper}}_{\sigma_{paper}}.\]

    Here, Apaper is the area of the cone template; and the areal density \(\sigma_{paper}\), named in analogy to the regular (volume) density, is the mass per area of paper. Although areal density seems like a strange quantity to define, it is used worldwide to describe the “weight” of different papers.

    The quotient \(m/A_{cs}\) contains the ratio \(A_{paper}/A_{cs}\). Rather than estimating both areas and finding their ratio, let’s estimate the ratio directly.

    How does the cross-sectional area Acs compare to the area of the paper?

    clipboard_eb01522820b162c93a18cdbe5678d1202.png

    How does the cross-sectional area Acs compare to the area of the paper?

    Because the cone’s circumference is three-quarters of the circumference of the full circle, its cross-sectional radius is three-quarters of the radius r of the template circle. Therefore,

    \[A_{cs} = \pi (\frac{3}{4}r)^{2}\]

    Because the template is three-quarters of a full circle,

    \[A_{paper} = \frac{3}{4} \pi r^{2}.\]

    The paper area has one factor of three-quarters, whereas the cross-sectional area has two factors of three-quarters, so \(A_{paper}/A_{cs} = 4/3\). Now vterm simplifies as follows:

    \[v_{term} \sim (\frac{\overbrace{A_{paper}\sigma_{paper}}^{m} \times g}{A_{cs} \rho_{air}})^{1/2} = (\frac{\frac{4}{3} \sigma_{paper} g}{\rho_{air}})^{1/2}.\]

    The only unfamiliar number is the areal density \(\sigma_{paper}\), the mass per area of paper. Fortunately, areal density is used commercially, so most reams of printer paper state their areal density: typically, 80 grams per square meter.

    Is this \(\sigma_{paper}\) consistent with the estimates for a dollar bill in Section 1.1?

    There we estimated that the thickness t of a dollar bill, or of paper in general, is approximately 0.01 centimeters. The regular (volume) density \(\rho\) would then be 0.8 grams per cubic centimeter:

    \[\rho_{paper} = \frac{\sigma_{paper}}{t} \approx \frac{80 g m^{-2}}{10^{-2} cm} \times \frac{1m^{2}}{10^{4} cm^{2}} = 0.8 \frac{g}{cm^{3}}.\]

    This density, slightly below the density of water, is a good guess for the density of paper, which originates as wood (which barely floats on water). Therefore, our estimate in Section 1.1 is consistent with the proposed areal density of 80 grams per square meter.

    After putting in the constants, the cone’s terminal speed is predicted to be roughly 0.9 meters per second:

    \[v_{term} \sim (\frac{4}{3} \times \frac{\overbrace{8 \times 10^{-2} \textrm{kg m}^{-2}}^{\sigma_{paper}} \times \overbrace{10 \textrm{m s}^{-2}}^{g}}{\underbrace{1.2 \textrm{kg m}^{-3}}_{\rho_{air}}})^{1/2} \sim 0.9 \textrm{m s}^{-1}.\]

    To test the prediction and, with it, the analysis justifying it, I held the cone slightly above my head, from about 2 meters high. After I let the cone go, it fell for almost exactly 2 seconds before it hit the ground—for a speed of roughly 1 meter per second, very close to the prediction. Box models and conservation triumph again!

    3.5.3 Cycling

    In introducing the analysis of drag, I said that drag is one of the most important physical effects in everyday life. Our analysis of drag will now help us understand the physics of a fantastically efficient form of locomotion—cycling (for its efficiency, see Problem 3.34).

    What is the world-record cycling speed?

    The first task is to define the kind of world record. Let’s analyze cycling on level ground using a regular bicycle, even though faster speeds are possible riding downhill or on special bicycles. In bicycling, energy goes into rolling resistance, friction in the chain and gears, and air drag. The importance of drag rises rapidly with speed, due to the factor of v2 in the drag force, so at high-enough speeds drag is the dominant consumer of energy.

    Therefore, let’s simplify the analysis by assuming that drag is the only consumer of energy. At the maximum cycling speed, the power consumed by drag equals the maximum power that the rider can supply. The problem therefore divides into two estimates: the power consumed by drag (PDrag) and the power that an athlete can supply (Pathlete).

    Power is force times velocity:

    \[\textrm{power} = \frac{\textrm{energy}}{\textrm{time}} = \frac{\textrm{force} \times \textrm{distance}}{\textrm{time}} = \textrm{force} \times \textrm{velocity}\]

    Therefore,

    \[P_{drag} = F_{drag}v_{max} sim \rho v^{3} A_{cs}.\]

    Setting \(P_{drag} = P_{athlete}\) allows us to solve for the maximum speed:

    \[v_{max} \sim (\frac{P_{athlete}}{\rho_{air}A_{cs}})^{1/3},\]

    where Acs is the cyclist’s cross-sectional area. In Section 1.7.2, we estimated Pathlete as 300 watts. To estimate the cross-sectional area, divide it into a width and a height. The width is a body width—say, 0.4 meters. A racing cyclist crouches, so the height is roughly 1 meter rather than a full 2 meters. Then Acs is roughly 0.4 square meters. Plugging in the numbers gives

    \[v_{max} \sim (\frac{300W}{1kg m^{-3}} \times 0.4 m^{2})^{1/3}.\]

    That formula, with its mix of watts, meters, and seconds, looks suspicious. Are the units correct?

    Let’s translate a watt stepwise into meters, kilograms, and seconds, using the definitions of a watt, joule, and newton:

    \[W \equiv \frac{J}{s}, \: \: J \equiv NM, \: \: N \equiv \frac{kg m }{s^{2}}.\]

    The three definitions are represented in the next divide-and-conquer tree, one definition at each nonleaf node. Propagating the leaves toward the root gives us the following expression forthe watt in terms of meters, kilograms, and seconds (the fundamental units in the SI system):

    \[W \equiv \frac{kg m^{2}}{s^{3}}.\]

    The units in vmax become

    \[(\frac{\overbrace{\cancel{\textrm{kg m}^{2}} \textrm{ s}^{-3}}^{W}}{\cancel{\textrm{kg }} \textrm{m}^{-3} \times \cancel{m^{2}}})^{1/3} = (\frac{s^{-3}}{m^{-3}})^{1/3}.\]

    The kilograms cancel, as do the square meters. The cube root then contains only meters cubed over seconds cubed; therefore, the units for vmax are meters per second.

    clipboard_e03cdbcc1a92859c6f8c08b5f09da5049.png

    Let’s estimate how many meters per second. Don’t let the cube root frighten you into using a calculator. We can do the arithmetic mentally, if we massage (adjust) the numbers slightly. If only the power were 400 watts (or instead the area were 0.3 square meters)! Instead of wishing, make it so—and don’t worry about the loss of accuracy: Because we have neglected the drag coefficient, our speed will be approximate anyway. Then the cube root becomes easy an calculation:

    \[v_{max} \sim (\frac{\cancel{300} 400W}{1 kg m^{-3} \times 0.4 m^{2}})^{1/3} = (1000)^{1/3} m s^{-1} = 10 m s^{-1}. \]

    In more familiar units, the record speed is 22 miles per hour or 36 kilometers per hour. As a comparison, the world 1-hour record—cycling as far as possible in 1 hour—is 49.7 kilometers or 30.9 miles, set in 2005 by Ondřej Sosenka. Our prediction, based on the conservation analysis of drag, is roughly 70 percent of the actual value.

    How can such an estimate be considered useful?

    High accuracy often requires analyzing and tracking many physical effects. The calculations and bookkeeping can easily obscure the most important effect and its core idea, costing us insight and understanding. Therefore, almost everywhere in this book, the goal is an estimate within a factor of 2 or 3. That level of agreement is usually enough to convince us that our model contains the situation’s essential features.

    Here, our predicted speed is only 30 percent lower than the actual value, so our model of the energy cost of cycling must be broadly correct. Its main error arises from the factor of one-half that we ignored when estimating the drag force—as you can check by doing Problem 3.33.

    3.5.4 Fuel efficiency of automobiles

    Bicycles, in many places, are overshadowed by cars. From the analysis of drag, we can estimate the fuel consumption of a car (at highway speeds). Most of the world measures fuel consumption in liters of fuel per 100 kilometers of driving. The United States uses the reciprocal quantity, fuel efficiency—distance per volume of fuel—measured in miles per US gallon. To develop unit flexibility, we’ll do the calculation using both systems.

    For a bicycle, we compared powers: the power consumed by drag with the power supplied by an athlete. For a car, we are interested in the fuel consumption, which is related to the energy contained in the fuel. Therefore, we need to compare energies. For cars traveling at highway speeds, most of the energy is consumed fighting drag. Therefore, the energy consumed by drag equals the energy supplied by the fuel.

    Driving a distance d, which will be 100 kilometers, consumes an energy

    \[E_{drag} \sim \rho_{air} v^{2} A_{cs} d.\]

    The fuel provides an energy

    \[E_{fuel} \sim \underbrace{\textrm{energy density}}_{\varepsilon_{fuel}} \times \underbrace{\textrm{fuel mass}}_{\rho_{fuel}V_{fuel}} = \varepsilon_{fuel} \rho_{fuel} V_{fuel}.\]

    Because \(E_{fuel} sim E_{drag}\), the volume of fuel required is given by

    \[V_{fuel} \sim \frac{E_{drag}}{\rho_{fuel}\varepsilon_{fuel}} \sim \underbrace{\frac{\rho_{air}}{\rho_{fuel}} \frac{v^{2} A_{cs}}{\varepsilon_{fuel}}}_{A_{consumption}} d.\]

    Because the left-hand side, Vfuel, is a volume, the complicated factor in front of the travel distance d must be an area. Let’s make an abstraction by naming this area. Because it is proportional to fuel consumption, a self-documenting name is Aconsumption. Now let’s estimate the quantities in it.

    1. Density ratio \(\rho_{air}/rho_{fuel}\). The density of gasoline is similar to the density of water, so the density ratio is roughly 10−3.

    2. Speed v. A highway speed is roughly 100 kilometers per hour (60 miles per hour) or 30 meters per second. (A useful approximation for Americans is that 1 meter per second is roughly 2 miles per hour.)

    3. Energy density \(\varepsilon_{fuel}\). We estimated this quantity Section 2.1 as roughly 10 kilocalories per gram or 40 megajoules per kilogram.

    4. Cross-sectional area Acs. A car’s cross section is about 2 meters across by 1.5 meters high, so Acs ∼ 3 square meters

    clipboard_e71b2e76f55f2240c30ae1563df6ad2bf.png

    With these values,

    \[A_{consumption} \sim 10^{-3} \times \frac{\overbrace{10^{3} m^{2} s^{-2}}^{v^{2}} \times \overbrace{3 m^{2}}^{A_{cs}}}{\underbrace{4 \times 10^{7} \textrm{J kg} ^{-1}}_{\varepsilon_{fuel}}} \approx 8 \times 10^{-8} m^{2}.\]

    To find the fuel consumption, which is the volume of fuel per 100 kilometers of driving, simply multiply Aconsumption by d = 100 kilometers or 105 meters, and then convert to liters to get 8 liters per 100 kilometers:

    \[V_{fuel} \approx \underbrace{8 \times 10^{-8} \cancel{m^{2}}}_{A_{consumption}} \times \underbrace{10^{5} \cancel{m}}_{d} \times \frac{10^{3}l}{1 \cancel{m^{3}}} = 8 l.\]

    For the fuel efficiency, we use Aconsumption in the form \(d= V_{fuel}/A_{consumption}\) to find the distance traveled on 1 gallon of fuel, converting the gallon to cubic meters:

    \[d \sim \frac{\overbrace{1 \cancel{\textrm{ gallon}}}^{V_{fuel}}}{\underbrace{8 \times 10^{-8} \cancel{m^{2}}_{A_{consumption}}} \times \frac{4 \cancel{l}}{1 \cancel{\textrm{ gallon}}} \times \frac{10^{-3} m^{\cancel{3}}}{1 \cancel{l}} = 5 \times 10^{4} m.\]

    The struck-through exponent of 3 in the \(m^{\cancel{3}}\) indicates that the cubic meters became linear meters, as a result of cancellation with the m2 in the Aconsumption. The resulting distance is 50 kilometers or 30 miles. The predicted fuel efficiency is thus roughly 30 miles per gallon.

    This prediction is very close to the official values. For example, for new midsize American cars (in 2013), fuel efficiencies of nonelectric vehicles range from 16 to 43 miles per gallon, with a mean and median of 30 miles per gallon (7.8 liters per 100 kilometers)

    The fuel-efficiency and fuel-consumption predictions are far more accurate than we deserve, given the many approximations! For example, we ignored all energy losses except for drag. We also used a very rough drag force \(\rho_{air}v^{2}A_{cs}\), derived from a reasonable but crude conservation argument. Yet, like Pippi Longstocking, we came out right anyway.

    What went right?

    The analysis neglects two important factors, so such accuracy is possible only if these factors cancel. The first factor is the dimensionless constant hidden in the single approximation sign of the drag force:

    \[F_{drag} \sim \rho_{air} A_{cs}v^{2}.\]

    Including the dimensionless prefactor (shown in gray), the drag force is

    \[F_{drag} = \frac{1}{2}c_{d} \rho_{air}A_{cs} v^{2},\]

    where cd is the drag coefficient (introduced in Section 3.2.1). The factor of one-half comes from the one-half in the definition of kinetic energy. The drag coefficient is the remaining adjustment, and its origin is the subject of Section 5.3.2. For now, we need to know only that, for a typical car, cd ≈ 1/2. Therefore, the dimensionless prefactor hidden in the single approximation sign is approximately 1/4.

    Based on this more accurate drag force, will cars use more or less than 8 liters of fuel per 100 kilometers?

    Including the cd/2 reduces the drag force and the fuel consumption by a factor of 4. Therefore, cars would travel 120 miles on 1 gallon of fuel or would consume only 2 liters per 100 kilometers. This more careful prediction is far too optimistic—and far worse than the original, simpler estimate.

    What other effect did we neglect?

    The engine efficiency—a typical combustion engine, whether gasoline or human, is only about 25 percent efficient: An engine extracts only one-quarter of the combustion energy in the fuel; the remaining three-quarters turns into heat without doing mechanical work. Including this factor increases our estimate of the fuel consumption by a factor of 4.

    The engine efficiency and the more accurate drag force together give the following estimate of the fuel consumption, with the new effect in gray:

    \[V_{fuel} \approx \frac{\frac{1}{2}c_{d}}{0.25} \times \frac{\rho_{air}v^{2}A_{cs}}{\rho_{fuel}\varepsilon_{fuel}}d.\]

    The 0.25 in the denominator, from the engine efficiency, cancels the \(\frac{1}{2}c_{d}\) in the numerator. That is why our carefree estimate, which neglected both factors, was so accurate. The moral, which I intend only half jokingly: Neglect many factors, so that the errors can cancel one another out.

    Exercise \(\PageIndex{1}\): Adjusting the cycling record

    Our estimate of the world 1-hour record as roughly 35 kilometers (Section 3.5.3) ignored the drag coefficient. For a bicyclist, \(c_{d} \approx 1\). Will including the drag coefficient improve or worsen the prediction in comparison with the actual world record (roughly 50 kilometers)? Answer that question before making the new prediction! What is the revised prediction?

    Exercise \(\PageIndex{2}\): Bicyclist fuel efficiency

    What is the fuel consumption and efficiency of a bicyclist powered by peanut butter? Express your estimate as an efficiency (miles per gallon of peanut butter) and a consumption (liters of peanut butter per 100 kilometers). How does a bicycle compare with a car?


    This page titled 3.5: Drag Using Conversion of Energy is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Sanjoy Mahajan (MIT OpenCourseWare) .

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