Summary of Aerodynamics A Formulas 1 1.1 1.1
Relations Relations betw between heigh height, t, pressure pressure,, densit density y and temperat temperature ure Defini Definiti tion onss
g = Gravitational acceleration at a certain altitude ( g0 = 9.81 81m/s m/s2 ) (m/s2 ) r = Earth radius (6378km (6378 km)) (m) hg = Height above the ground (Geometric height) ( m) ha = Height above the center of the earth ( ha = h g + r + r)) (m) h = Geopotential altitude (Geopotential height) ( m) p = p = Pressure (P (P a = N/m = N/m 2 ) ρ = Air density (kg/m (kg/m3 ) ν = = ρ1 = Specific volume (m ( m3 /kg) /kg ) T = T = Temperature (K ( K ) R = 287. 287.05 05J/ J/((kgK ) kgK ) = Gas constant P s = 1.01325 × 105 N/m2 = Pressure at sea level ρs = 1.225 225kg/m kg/m3 = Air density at sea level
288.15 15K = Temperature at sea level T s = 288. K = a = dT = Temperature gradient (a ( a = 0.0065 0065K/m K/m in in the troposphere (lowest part) of the earth atmodh sphere) (K/m (K/m))
1.2
Relation Relation betwe between en geopotent geopotential ial height height and and geometric geometric heigh heightt
Newton’s gravitational law implicates: g = g = g0
2
r ha
= g 0
r r + h + hg
2
The hydrostatic equation is: dp = dp =
−ρgdhg
However, g However, g is variable here for different heights. Since a variable gravitational acceleration is difficult to work with, the geopotential height h has been introduced such that: dp = dp =
−ρg0 dh
(1.2.1)
So this means that: dh = dh =
g r2 dhg = dhg g0 (r + h + hg )2
And integration gives the general relationship between geopotential height and geometric height: r h = hg (1.2.2) r + hg
1
Ahilan A, Noorul Islam College of Engg,
[email protected]
1.3
Relation Relationss between between pressur pressure, e, densit density y and heigh heightt
The famous equation of state is: p = p = ρRT ρRT
(1.3.1)
Dividing the hydrostatic equation (1.2.1) by the equation of state (1.3.1) gives as results: dp − ρgo dh go = = − dh p ρRT RT If we assume an isothermal environment (the temperature stays the same), then integration gives: p2
h
dp go =− p RT
p1
dh
h1
Solving this gives the following equation: g0 p2 = e ( RT )(h2 p1 −
h1 )
(1.3.2)
−
And combining this with the equation of state gives the following equation: g0 ρ2 p2 T R p2 = = = e ( RT )(h2 ρ1 p1 T R p1 −
1.4
h1 )
(1.3.3)
−
Relation Relationss between between pressur pressure, e, densit density y and temperat temperature ure
We now again divide the hydrostatic equation (1.2.1) by the equation of state (1.3.1), but this time we don’t assume an isothermal environment, but we substitute dh = dh = dT in it, to to get: a dp − ρgo dh go dT = =− p ρRT aR T Integration gives: p2 = p1
T 2 T 1
−
g0 aR
(1.4.1)
Which is a nice formula. But by using the equation of state, we can also derive the following: ρ2 p2 T 1 R p2 = = ρ1 p1 T 2 R p1
1
−
−
g0 aR
1
−
T 2 T 1
=
T 2 T 1
T 2 T 1
=
T 2 T 1
g0 +1) ( aR
−
(1.4.2)
All those relations can be written in a simpler way: p2 = p1
ρ2 ρ1
g0 g0 +aR
=
T 2 T 1
−
g0 aR
(1.4.3)
These relations are the standard atmospheric relations in gradient layers.
2
Ahilan A, Noorul Islam College of Engg,
[email protected]
2 2.1
Continuity equation and Bernoulli’s equation Definitions
m = ˙ Mass flow (kg/s) ρ = Air density (kg/m3 ) A = Area (m2 ) V = Speed (m/s) p = Pressure (P a = N/m 2 ) h = Height (m) S = Surface (m2 ) r = Radius of the curvature (m) v = Volume (m3 ) V a = Velocity difference (m/s) S d = Actuator disk surface area ( m2 ) T = Thrust (N ) P = Power (W = J/s) η = Efficiency (dimensionless)
2.2
Continuity equation
The continuity equation states that in a tube, the following must be true: m ˙ in = m ˙ out This equation can be rewritten to: ρ1 A1 V 1 = ρ 2 A2 V 2
(2.2.1)
Or for incompressible flows (where ρ 1 = ρ2 ): A1 V 1 = A 2 V 2
2.3
(2.2.2)
(2.3.1)
Bernoulli’s equation
The Euler equation, when gravity forces and viscosity are neglected, is: dp =
−ρV dV
This formula is also valid for compressible flows. Integration for 2 points along a streamline gives: ( p2 − p1 ) + ρ
1 2 1 2 V − V 1 = 0 2 2 2
3 Ahilan A, Noorul Islam College of Engg,
[email protected]
This can be easily transformed to the Bernoulli equation, which says that the total pressure is constant along a streamline: 1 1 p1 + ρV 1 2 = p 2 + ρV 2 2 2 2
(2.3.2)
However, if gravity forces are included, the following variant of the Bernoulli equation can be derived: 1 1 p1 + ρV 1 2 + ρgh1 = p 2 + ρV 2 2 + ρgh2 2 2
(2.3.3)
But do remember that the Bernoulli equation is only valid for inviscid (frictionless) incompressible flows.
2.4
Bernoulli’s equation of curved flow
When looking at an infinitely small part of air, there is a pressure p working on one side of it, and a pressure p + dp working on the other side. This pressure difference causes the flow to bend. If S is the surface and dr is the length of the part of air, then the resultant force on the part of air is: F = (( p + dp) − p) · S = dp · S =
dp dp · S · dr = v dr dr
Where v is the volume of the air. However, when performing a circular motion, the resultant force is: F =
mV 2 ρV 2 = v r r
Combining these data, we find: dp ρV 2 = dr r
(2.4.1)
And therefore the following formula remains constant along a curving flow: p2
p1
r2
dp =
r1
ρV 2 dr r
(2.4.2)
And this is the exact reason why concave shapes have a lower pressure in flows, and convex areas have a higher pressure.
2.5
Actuator disk thrust
Suppose we have a propeller, blowing air from left to right. Let’s call point 0 a point infinitely far to the left (undisturbed flow), point 1 just to the left of the propeller (infinitely close to it), point 2 identical to point 1, but then on the right, and point 3 identical to point 0, but also on the right. In every point n, the airflow has a pressure pn , a velocity V n and an area S n . Since point 1 and 2 are infinitely close to each other, the air velocity in both points is equal, so V 1 = V 2 . Since point 0 and point 3 are both in the undisturbed flow, the pressure in those points is equal, so p0 = p3 . Let’s define V a1 = V 1 − V 0 , V a2 = V 3 − V 2 and V at = V 3 − V 0 , so V at = V a1 +V a2 . From the Bernoulli equation, the following equations can be derived: 1 1 p0 + ρV 0 2 = p 1 + ρ(V 0 + V a1 )2 2 2 1 1 p2 + ρ(V 0 + V a1 )2 = p 0 + ρ(V 0 + V at )2 2 2 4
Ahilan A, Noorul Islam College of Engg,
[email protected]
Note that it is not allowed to use bernoulli between point 1 and point 2, since energy is added to the flow there. Combining these two equations, we find the pressure difference: 1 p2 − p1 = ρV at (V 0 + V at ) 2 If S d is the surface area of the actuator disk, then the thrust is: T = S d ( p2 − p1 ) Combining this equation with the previous one gives: 1 T = ρS d (V 0 + V at )V at 2 However, the thrust is also equal to the mass flow times the change in speed. The mass flow is ρS d (V 0 + V a1 ), and the change in speed is simply equal to V at . So: T = ρS d (V 0 + V a1 )V at
(2.5.1)
Combining this equation with the previous one, we find another important conclusion: V a1 = V a2 =
2.6
1 V a 2 t
(2.5.2)
Actuator disk efficiency
Now let’s look at the power and the efficiency of the actuator disk. The power input is equal to the change in kinetic energy of the air per unit time: P in =
1 ρS d (V 0 + V a1 )((V 0 + V a3 )2 − V 0 2 ) = ρS d (V 0 + V a1 )2 V at 2
The power that really is outputted, is simply equal to the force times the velocity of the airplane: P out = T V 0 = ρS d (V 0 + V a1 )V at V 0 Combing these two equations, we can find the propeller efficiency: η =
P out T V 0 ρS d (V 0 + V a1 )V at V 0 = = P in P in ρS d (V 0 + V a1 )2 V at
Simplifying this gives: η =
V 0 1 = V V 0 + V a1 1 + V a01
5
(2.6.1)
Ahilan A, Noorul Islam College of Engg,
[email protected]
3 3.1
Enthalpy, specific heat and isentropic flows Definitions
h = Enthalpy (J/kg) e = The specific internal energy in a gas ( J/kg) (Note: This is in some books given the name u) R = Gas constant (Value is: 287 J/(kgK )) (J/(kgK )) T = The temperature of a certain amount of gas ( K ) p = Pressure (P a = N/m 2 ) v = Specific volume (defined as
1 ρ
) (m3 /kg)
δq = The added heat into a certain amount of gas ( J ) c = Specific heat (J/(kgK )) cv = Specific heat at constant volume (For air: cv = 717J/(kgK )) (J/(kgK )) c p = Specific heat at constant pressure (For air: c p = 1004J/(kgK )) (J/(kgK )) γ = Ratio of specific heats (Value for normal air is: 1 .4) (dimensionless) ρ = Density (kg/m3 )
3.2
Enthalpy
The enthalpy h is defined as follows: h = e + RT = e + pv
(3.2.1)
To differentiate that, the chain rule must be applied, since neither p or v are constant. So: dh = de + pdv + vdp
3.3
(3.2.2)
Specific heat
Note: Almost
all of the following formulas apply in general, as long as the gas is a perfect gas.
From basic thermodynamics, it can be derived that: δq = de + pdv
(3.3.1)
When adding an amount of heat to 1 kg of a gas, the temperature of the gas increases. The relation is (per definition): c =
δq dT
(3.3.2)
The addition of heat can be done in multiple ways. If the volume is kept constant (thus dv = 0), the following formulas apply: cv =
6
δq dT
(3.3.3)
Ahilan A, Noorul Islam College of Engg,
[email protected]
cv dT = δq = de + pdv = de e = c v T
(3.3.4)
But if the pressure is kept constant (thus dp = 0), it is slightly more difficult. For that we will use the enthalpy. The following formulas now apply: c p =
δq dT
(3.3.5)
(3.3.6)
c p dT = δq = de + pdv = dh − vdp = dh h = c p T
Between the two specific heat constants are interesting relationships. Their difference can be derived: c p − cv =
c p T − cv T h − e RT = = = R T T T
And their ratio is (per definition): γ =
3.4
c p cv
(3.3.7)
Specific heat ratio in isentropic processes
An isentropic process is both an adiabatic process (δq = 0) as a reversible process (no friction forces). Suppose there is an isentropic process present. It can in that case be shown that: dp c p dv dv =− = −γ p cv v v Integrating that equation and working out the results would give us: ln
p 2 v 2 = −γ ln p1 v1
And since v = ρ1 , it can be derived that: p1 = p2
ρ1 ρ2
γ
Using the equation of state ( pρ = RT ) it can also be derived that: p1 = p2
T 1 T 2
γ γ −1
And by combining these two formulas, one can also derive the following formula: p1 = p2
γ
ρ1 ρ2
=
T 1 T 2
γ γ −1
(3.4.1)
These important relations are called the isentropic flow relations, and are only relevant to compressible flow.
7
Ahilan A, Noorul Islam College of Engg,
[email protected]
3.5
Energy equation
We still assume an isentropic process (so δq = 0). Therefore δq = dh − vdp = 0, and since the Euler equation says that dp = −ρV dV we know that dh + vρV dV = 0. However, v = ρ1 so also dh + V dV = 0. Integrating, and working out the results of it, would give: 1 1 h1 + V 1 2 = h 2 + V 2 2 2 2
(3.5.1)
And since h = c p T , also the following equation is true: 1 1 c p T 1 + V 1 2 = c p T 2 + V 2 2 2 2
(3.5.2)
This equation is called the energy equation.
8 Ahilan A, Noorul Islam College of Engg,
[email protected]
4 4.1
Applications of the Mach number Definitions
ρ = Air density (kg/m3 ) A = Area (m2 ) a = Speed of sound (m/s) p = Pressure (P a = N/m 2 ) R = Gas constant (Value is: 287 J/(kgK )) (J/(kgK )) T = The temperature of a certain amount of gas ( K ) V = Velocity (m/s) M = Mach number (dimensionless) γ = Ratio of specific heats (Value for normal air is: 1 .4) (dimensionless) cv = Specific heat at constant volume (For air: cv = 717J/(kgK )) (J/(kgK )) c p = Specific heat at constant pressure (For air: c p = 1004J/(kgK )) (J/(kgK )) V cal = Calibrated flight velocity (m/s)
4.2
Speed of sound
The speed of sound is an important thing in aerodynamics. A formula to calculate it can be derived. Therefore the continuity equation is used on a sound wave traveling through a tube with constant area: ρAa = (ρ + dρ)A(a + da) ⇒ a =
−ρ da dρ
Assuming an isentropic flow, it is allowed to fill in the speed of sound in the Euler equation, which gives dp = −ρ a d a. Combining this with the previous equation gives: a =
dp dρ
(4.2.1)
This is not a very easy formula to work with though. But by using the isentropic flow relations, it can be derived that dp = γ pρ , and therefore: dρ a =
p γ ρ
(4.2.2)
(4.2.3)
Using the equation of state, this can be transformed to: a =
γRT
So the speed of sound only depends on the temperature. But when the speed of sound is known, also the Mach number can be calculated: M =
V a
(4.2.4)
There are three important regions of Mach numbers: 9 Ahilan A, Noorul Islam College of Engg,
[email protected]
1. M 1: Supersonic flow These regions can be split up into more detailed, but less important regions, as can be seen in table 4.1. Mach number: Name:
M
