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# drag coefficient units

From Wikipedia

Coefficient

In mathematics, a coefficient is a multiplicative factor in some term of an expression (or of a series); it is usually a number, but in any case does not involve any variables of the expression. For instance in

7x^2-3xy+1.5+y

the first three terms respectively have the coefficients 7, âˆ’3, and 1.5 (in the third term there are no variables, so the coefficient is the term itself; it is called the constant term or constant coefficient of this expression). The final term does not have any explicitly written coefficient, but is usually considered to have coefficient 1, since multiplying by that factor would not change the term. Often coefficients are numbers as in this example, although they could be parameters of the problem, as a, b, and c in

ax^2+bx+c

when it is understood that these are not considered as variables.

Thus a polynomial in one variable x can be written as

a_k x^k + \cdots + a_1 x^1 + a_0,

for some integer k, where ak, ... a1, a0 are coefficients; to allow this kind of expression in all cases one must allow introducing terms with 0 as coefficient. For the largest i with (if any), ai is called the leading coefficient of the polynomial. So for example the leading coefficient of the polynomial

\, 4x^5 + x^3 + 2x^2

is 4.

Specific coefficients arise in mathematical identities, such as the binomial theorem which involves binomial coefficients; these particular coefficients are tabulated in Pascal's triangle.

## Linear algebra

In linear algebra, the leading coefficient of a row in a matrix is the first nonzero entry in that row. So, for example, given

M = \begin{pmatrix} 1 & 2 & 0 & 6 \\ 0 & 2 & 9 & 4 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 \end{pmatrix}.

The leading coefficient of the first row is 1; 2 is the leading coefficient of the second row; 4 is the leading coefficient of the third row, and the last row does not have a leading coefficient.

Though coefficients are frequently viewed as constants in elementary algebra, they can be variables more generally. For example, the coordinates (x_1, x_2, ..., x_n) of a vectorv in a vector space with basis \lbrace e_1, e_2, ..., e_n \rbrace , are the coefficients of the basis vectors in the expression

v = x_1 e_1 + x_2 e_2 + ... + x_n e_n .

## Examples of physical coefficients

1. Coefficient of Thermal Expansion(thermodynamics) (dimensionless) - Relates the change in temperature to the change in a material's dimensions.
2. Partition Coefficient(KD) (chemistry) - The ratio of concentrations of a compound in two phases of a mixture of two immiscible solvents at equilibrium.
3. Hall coefficient(electrical physics) - Relates a magnetic field applied to an element to the voltage created, the amount of current and the element thickness. It is a characteristic of the material from which the conductor is made.
4. Lift coefficient(CLor CZ) (Aerodynamics) (dimensionless) - Relates the lift generated by an airfoil with the dynamic pressure of the fluid flow around the airfoil, and the planform area of the airfoil.
5. Ballistic coefficient(BC) (Aerodynamics) (units of kg/m2) - A measure of a body's ability to overcome air resistance in flight. BC is a function of mass, diameter, and drag coefficient.
6. Transmission Coefficient(quantum mechanics) (dimensionless) - Represents the probability flux of a transmitted wave relative to that of an incident wave. It is often used to describe the probability of a particle tunnelling through a barrier.
7. Damping Factora.k.a. viscous damping coefficient (Physical Engineering) (units of newton-seconds per meter) - relates a damping force with the velocity of the object whose motion is being

## Chemistry

A coefficient is a number placed in front of a term in a chemical equation to indicate how many molecules (or atoms) take part in the reaction. For example, in the formula 2H_2 + O_2 \rarr 2H_2O, the number 2's in front of H_2 and H_2O are stoichiometric coefficients.

Drag coefficient - Wikipedia, the free encyclopedia

It is used in the drag equation, where a lower drag coefficient indicates ...

Lift-induced drag

In aerodynamics, lift-induced drag, induced drag, vortex drag, or sometimes drag due to lift, is a drag force that occurs whenever a moving object redirects the airflow coming at it. This drag force occurs in airplanes due to wings or a lifting body redirecting air to cause lift and also in cars with airfoil wings that redirect air to cause a downforce. With other parameters remaining the same, induced drag increases as the angle of attack increases.

## Source of induced drag

Lift is produced by the changing direction of the flow around a wing. The change of direction results in a change of velocity (even if there is no speed change, just as seen in uniform circular motion), which is an acceleration. To change the direction of the flow therefore requires that a force be applied to the fluid; lift is simply the reaction force of the fluid acting on the wing.

When producing lift, air below the wing is generally at a higher pressure than atmospheric pressure, while air above the wing is generally at a lower than atmospheric pressure. On a wing of finite span, this pressure difference causes air to flow from the lower surface wing root, around the wingtip, towards the upper surface wing root. This spanwise flow of air combines with chordwise flowing air, causing a change in speed and direction, which twists the airflow and produces vortices along the wing trailing edge. The vortices created are unstable, and they quickly combine to produce wingtip vortices. The resulting vortices change the speed and direction of the airflow behind the trailing edge, deflecting it downwards, and thus inducing downwash behind the wing.

Wingtip vortices modify the airflow around a wing. Compared to a wing of infinite span, vortices reduce the effectiveness of the wing to generate lift, thus requiring a higher angle of attack to compensate, which tilts the total aerodynamic force rearwards. The angular deflection is small and has little effect on the lift. However, there is an increase in the drag equal to the product of the lift force and the angle through which it is deflected. Since the deflection is itself a function of the lift, the additional drag is proportional to the square of the lift.

The total aerodynamic force is usually thought of as two components, lift and drag. By definition, the component of force parallel to the oncoming flow is called drag; and the component perpendicular to the oncoming flow is called lift. At practical angles of attack the lift greatly exceeds the drag. Unlike parasitic drag on an object (which is proportional to the square of the airspeed), for a given lift, induced drag on an airfoil is inversely proportional to the square of the airspeed. In straight and level flight of an aircraft, lift varies only slowly because it is approximately equal to the weight of the aircraft. Consequently in straight and level flight, the induced drag is inversely proportional to the square of the airspeed. At the speed for minimum drag, induced drag is equal to parasitic drag.

## Reducing induced drag

Theoretically a wing of infinite span and constant airfoil section would produce no induced drag. The characteristics of such a wing can be measured on a section of wing spanning the width of a wind tunnel, since the walls block spanwise flow and create what is effectively two-dimensional flow.

A rectangular wing produces much more severe wingtip vortices than a tapered or elliptical wing, therefore many modern wings are tapered. However, an elliptical planform is more efficient as the induced downwash (and therefore the effective angle of attack) is constant across the whole of the wingspan. Few aircraft have this planform because of manufacturing complications â€” the most famous examples being the World War IISpitfire and Thunderbolt. Tapered wings with straight leading and trailing edges can approximate to elliptical lift distribution. Typically, straight wings produce between 5â€“15% more induced drag than an elliptical wing.

Similarly, a high aspect ratio wing will produce less induced drag than a wing of low aspect ratio because the size of the wing vortices will be much reduced on a longer, thinner wing. Induced drag can therefore be said to be inversely proportional to aspect ratio. The lift distribution may also be modified by the use of washout, a spanwise twist of the wing to reduce the incidence towards the wingtips, and by changing the airfoil section near the wingtips. This allows more lift to be generated at the wing root and less towards the wingtip, which causes a reduction in the strength of the wingtip vortices.

Some early aircraft had fins mounted on the tips of the tailplane which served as endplates. More recent aircraft have wingtip mounted winglets or wing fences to oppose the formation of vortices. Wingtip mounted fuel tanks may also provide some benefit, by preventing the spanwise flow of air around the wingtip.

## Calculation of Induced drag

For a wing with an elliptical lift distribution, induced drag is calculated as follows:

D_i = \frac{1}{2} \rho V^2 S C_{Di} = \frac{1}{2} \rho_0 V_e^2 S C_{Di}

where

C_{Di} = \frac{C_L^2}{ \pi e AR} and
C_L = \frac{L}{ \frac{1}{2} \rho_0 V_e^2 S}

Thus

C_{Di} = \frac{L^2}{\frac{1}{4} \rho_0^2 V_e^4 S^2 \pi e AR}

Hence

D_i = \frac{L^2}{\frac{1}{2} \rho_0 V_e^2 S \pi e AR}

Where:

AR \, is the aspect ratio,
C_{Di} \, is the induced drag coefficient (see Lifting-line theory),
C_L \, is the lift coefficient,
D_i \, is the induced drag,
e \, is the Oswald efficiency number by which the induced drag exceeds th

Coefficient of performance

The coefficient of performance or COP (sometimes CP), of a heat pump is the ratio of the change in heat at the "output" (the heat reservoir of interest) to the supplied work.

## Equation

The equation is:

COP = \frac{Q_{H}}{ W}

where

• Q_{H} \ is the heat supplied to the hot reservoir
• W \ is the work consumed by the heat pump.

The COP for heating and cooling are thus different, because the heat reservoir of interest is different. When one is interested in how well a machine cools, the COP is the ratio of the heat removed from the cold reservoir to input work. However, for heating, the COP is the ratio of the heat removed from the cold reservoir plus the heat added to the hot reservoir by the input work to input work:

COP_{heating}=\frac{ W}
COP_{cooling}=\frac{ W}

where

• Q_{C} \ is the heat supplied to the cold reservoir.

## Derivation

According to the first law of thermodynamics, in a reversible system we can show that Q_{hot}=Q_{cold}+W and W=Q_{hot}-Q_{cold}, where Q_{hot} is the heat given off by the hot heat reservoir and Q_{cold} is the heat taken in by the cold heat reservoir.
Therefore, by substituting for W,

COP_{heating}=\frac{Q_{hot}}{Q_{hot}-Q_{cold}}

For a heat pump operating at maximum theoretical efficiency (i.e. Carnot efficiency), it can be shown that \frac{Q_{hot}}{T_{hot}}=\frac{Q_{cold}}{T_{cold}} and Q_{cold}=\frac{Q_{hot}T_{cold}}{T_{hot}}, where T_{hot} and T_{cold} are the absolute temperatures of the hot and cold heat reservoirs respectively.

At maximum theoretical efficiency,

COP_{heating}=\frac{T_{hot}}{T_{hot}-T_{cold}}

Which is equal to the inverse of the ideal Carnot cycle efficiency because a heat pump is a heat engine operating in reverse. Similarly,

COP_{cooling}=\frac{Q_{cold}}{Q_{hot}-Q_{cold}} =\frac{T_{cold}}{T_{hot}-T_{cold}}

It can also be shown that COP_{cooling}=COP_{heating}-1 . Note that these equations must use the absolute temperature (the Kelvin or Rankine scale.)

COP_{heating} applies to heat pumps and COP_{cooling} applies to air conditioners or refrigerators. For heat engines, see Efficiency. Values for actual systems will always be less than these theoretical maximums. In Europe, ground source heat pump units are standard tested at {T_{hot}} is 35 Celsius (95 Fahrenheit) and {T_{cold}} is 0 Celsius (32 Fahrenheit). According to above formula, the maximum achievable COP would be 7.8. Test results of the best systems are around 4.5. When measuring installed units over a whole season and one also counts the energy needed to pump water through the piping systems, then seasonal COP's are around 3.5 or less. This indicates room for improvement.

## Improving COP

As the formula shows, to improve the COP of a heat pump system, one needs to reduce the temperature gap T_{hot} minus T_{cold} at which the system works. For a heating system this would mean two things. One is to reduce output temperature to around 30 Celsius (86 Fahrenheit) which requires piped floor- or wall- or ceiling heating, or oversized water to air heaters. The other is to increase input temperature (by using an oversized ground source). For an air cooler, COP could be improved by using ground water as an input instead of air, and by reducing temperature drop on output side through increasing air flow. For both systems, also increasing the size of pipes and air canals would help to reduce noise and the energy consumption of pumps (and ventilators).

Also the heat pump itself can be improved a lot. The two most simple ways to improve heat pump units, is to double the size of the internal heat exchangers relative to the power of the compressor, and to reduce the system's internal temperature gap over the compressor. This last measure however, makes such heat pumps unsuitable to produce output above roughly 40 Celsius (104 Fahrenheit) which means that a separate machine is needed for producing hot tap water.

One reason that heat pump manufacturers often show little interest into making such improvements, is that they often also earn a lot of money in producing boilers, turbines and transformers for electricity plants. More efficient heat pumps would mean that less growth can be achieved in the industrial boilers and turbines sector.

For the same reason, there is little progress in engineering small heat pumps, driven on oil or piped gas. Such systems can also produce electricity, and, especially when combined with seasonally storing excess heat underground, they would be even more energy efficient then electricity driven heat pumps, because they cut away the heat wastage of central electricity production, by using exhaust and cool water heat for house warming. However, they would also vastly reduce the need for centrally produced electricity.

## Example

A geothermal heat pump operating at COP_{heating} 3.5 provides 3.5 units of heat for each unit of energy consumed (i.e. 1&nbsp;kWh consumed would provide 3.5&nbsp;kWh of output heat). The output heat comes from both the heat source and 1&nbsp;kWh of input energy, so the heat-source is cooled by 2.5&nbsp;kWh, not 3.5&nbsp;kWh.

A heat pump of COP_{heating} 3.5, such as in the example above, could be less expensive to use than even the most efficient gas furnace.

A heat pump cooler operating at COP_{cooling} 2.0 removes 2 units of heat for each unit of energy consumed (e.g. an air conditioner consuming 1&nbsp;kWh would remove 2&nbsp;kWh of heat from a building's air).

Given the same energy source and operating conditions, a higher COP heat pump will consume less purchased energy than one with a lower COP. The overall environmental impact of a heating or air conditioning installation depends on the source of energy used as well as the COP of the equipment. The operating cost to the consumer depenends on the cost of energy as well as the COP or efficiency of the unit. Some areas provide two or more sources of energy, for example, natural gas and electricity. A high COP of a heat pump may not entirely overcome a relatively high cost for electicity compared with the same heating value from natural gas.

For example, the 2009 US average price per therm (100,000 BTU) of electricity was $3.38 while the average price per therm of natural gas was$1.16. Using these prices, a heat pump with a COP of 3.5 in moderate climate would cost $0.97 to provide one therm of heat, while a high efficiency gas furnace with 95% efficiency would cost$1.22 to provide one therm of heat. With these average prices, the heat pump costs 20% less to provide the same amount of heat. At 0 fahrenheit (-18 Celsius) COP is much lower. Then, the same system costs as much to operate as an efficient gas heater. The yearly savings will depend on the actual cost of electricity and natural gas, which can both vary widely.

However, a COP may help make a determination of system choice based on carbon contribution. Although a heat pump may cost more to operate than a conventional natural gas or electric heater, depending on the source of electricity generation in one's area, it may contribute less net carbon dioxide to the environment than burning natural gas or heating fuel. If locally no green electricity is available, then carbon wise the best option would be to drive a heat pump on piped gas or oil, to store excess heat in the ground source for use in winter, while using the same machine also

Question:Alright, so I know what the equation to determine drag is and I have figured out the drag coefficients of various parachutes using American units of measure (ft, lbs etc.) but my teacher wants me to do it using metric values. Anyway, I have attempted to do it, but my answers were waaay off. Just wondering if anyone can help me? An example: air pressure - 1.2kg/m3, weight - 0.15kg, velocity - 2.63m/s, reference area - 3.7m2 I used my kg values for lbs, m/s for ft/s, m2 for ft2 etc. Just wondering if that is incorrect? Sorry, I think you misunderstood what I was asking - my fault. Instead of say slugs/ft3 I used the equivalent kg/m3 value, instead of having the velocity in ft/s I have the equivalent velocity in m/s etc. What I am getting at is, my equation isn't working

Answers:It is very incorrect. Simply changing the unit you write after a number while not changing the number itself is the same as saying "1 foot = 1 meter", which is clearly wrong. What you must do is convert the numbers as well as the units. Google offers a handy way to do this. Simply type in the quantity you know, then the word "in", then the units you want it converted into. For example, typing "0.15 pounds in kilograms" should get you an answer of "0.0680388555 kilograms". Google also recognizes some abbreviations, so "0.15 pounds in kg" is also acceptable. Here's how to convert the area: "2.63 ft^2 in m^2" You should get an answer of "0.244334995 m^2". After you have converted each quantity into metric units, then you can substitute them back into the equation. Make sure to keep writing the units in each step as you solve it, allowing the units to cancel. That way, if you forgot to change a value and it has inconsistent units, the units won't cancel and you'll know you did something wrong.

Question:Just a quick question about this, Is it always equal to 1.328/(Re^1/2) for Laminar Flow? Thanks The question should be, when you have a flate plate parallel to the flow, is the frictional drag coefficient equal to 1.328/Re^1/2 while the pressure coefficient is equal to 0 and when the plate is perpendicular to the flow, is the frictional coefficient = 0 while the pressure coefficient =1.9?

Answers:Don't we just love the limitless Reynolds number....???? Here's a spot on wikipedia that has drag coefficients for your plate ...and some other stuff bricks one of the limits in the lower table looks awfully like 1.3 to me...might check it out to see here's the link http://en.wikipedia.org/wiki/Drag_coefficient there ya go from the E....

Question:In aerodynamics you have this formula: Fd=1/2AP(v^2)Cd Fd is drag force in N A is the objects surface in m^2 P is the air compactness in kg/m^3 v is the speed in m/s Cd is the drag coefficient of the object Now how do you get that coefficient out of an object, in this case a balloon? Or at least does somebody knows its coefficient? thank you very much!

Answers:Yuko has taken some trouble in response to your question but it seems to me he doesn't properly address it. He tells you how to manipulate the calculation, but I feel you know that anyway. Also a disc, as he proposes, has much worse aerodynamics than a sphere so his proposed substitution is well off the mark. I don't quite see the relevance of his bit about the buoyancy. The only way to get this coefficient is to measure the drag on the balloon in an airstream of known velocity; for that, you normally need a wind tunnel! A balloon is approximately a sphere and flow past a sphere goes through a number of transitions with increasing velocity. At very low velocity, a stable pair of vortices is formed on the downwind side. As velocity increases, the vortices become unstable and are alternately shed downstream. As velocity is further increased, the boundary layer transitions to chaotic turbulent flow with vortices of many different scales being shed in a turbulent wake from the body. Each of these flow regimes produces a different amount of drag on the sphere. You would therefore need to know a wind speed range for which you want the coefficient. The coefficient for a sphere varies between 0.07 to 0.5, rising through that range with wind speed at the above transitions. I doubt if a balloon would withstand the turbulences of the chaotic high end very long so probably something like 0.1 is a reasonable compromise number if you can't do any measurements. Please spare a moment to award one of your answers "Best Answer". That's partly why we do this.

Question:In aerodynamics you have this formula: Fd=1/2AP(v^2)Cd Fd is drag force in N A is the objects surface in m^2 P is the air compactness in kg/m^3 Cd is the drag coefficient of the object Now how do you get that coefficient out of an object, in this case a balloon? Or at least does somebody knows its coefficient? thank you very much!

Answers:An empirical relation Cd = 2.7 105 R-1 is given between drag coefficient, Ca, and Reynolds number R, for a captive, nominally spherical, rubber balloon, for the range log10 R = 5.1 to 5.8. It is suggested that the drag increases as the shape of the balloon progressively deviates from the spherical. A calibration of a Meteorological Office radiosonde windmill as a cup anemometer is also included .