sg2215 exam questions.pdf
Post on 21-Apr-2017
273 Views
Preview:
TRANSCRIPT
SG2215 Exam Questions 7. Describe the Fanno flow and how it can be calculated Fanno flow refers to an adiabatic flow through a constant area where friction is taken into consideration. The flow is assumed to be steady and one-‐dimensional, and no mass is added within the duct. It is an irreversible process due to viscous friction changing the flow properties along the duct. This is calculated by assuming a shear stress on the wall acting on the fluid with uniform properties over any cross section of the duct. 8. How does the total temperature vary over a stationary normal shock? Explain! Stationary shock wave is when the fluid travels at the same speed of the shock but in the opposite direction. The relative velocity between the two is zero. The total temperature doesn’t change across a stationary normal shock. The shock wave does no work and there is no heat addition, therefore total enthalpy and total temperature are constant. Governing equations
ℎ! +𝑢!!
2 = ℎ! +𝑢!!
2 𝐸𝑁𝐸𝑅𝐺𝑌
ℎ! = ℎ +𝑢!
2 , 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑠ℎ𝑜𝑐𝑘
𝐶𝑎𝑙𝑜𝑟𝑖𝑐𝑎𝑙𝑙𝑦 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠, 𝑐! =ℎ𝑇
𝑇!"𝑇!"
=𝑇!"𝑇!𝑇!𝑇!𝑇!𝑇!"
𝑐!𝑇!" = 𝑐!𝑇!" ℎ!" = ℎ!" 𝐹𝑜𝑟 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠,𝑇!" = 𝑇!"
9. What is the Mach Angle? The mach angle is the angle of the mach wave relative to the flow direction. This mach angle is dependent on the upstream Mach number and is independent of the gas used.
𝜇 = 𝑎𝑟𝑐𝑠𝑖𝑛1𝑀
10. Show how one determines the properties across an oblique shock. Show that the velocity component parallel to the shock does not change across the shock. Use the shock as a control volume. Therefore, has to satisfy continuity, momentum and energy equation. For steady flow, Continuity
𝜌𝑉!.!
∙ 𝑑𝑆 = 0 = −𝜌!𝑢!𝐴 + 𝜌!𝑢!𝐴 → 𝜌!𝑢! = 𝜌!𝑢!
Momentum Steady flow without body forces
(𝜌𝑉!.!
∙ 𝑑𝑆)𝑉 = − (𝑝)𝑑𝑆!.!
(𝜌𝑢!.!
)𝑉𝑑𝐴 = − 𝑝𝑑𝑆!.!
Tangential component
0 = −𝜌!𝑢!𝑤!𝐴 + 𝜌!𝑢!𝑤!𝐴 but, 𝜌!𝑢! = 𝜌!𝑢! from continuity therefore,
𝑤! = 𝑤! and tangential velocity is constant across oblique shockwave Normal component
−𝜌!𝑢!!𝐴 + 𝜌!𝑢!!𝐴 = −(𝑝! − 𝑝!)𝐴 → 𝑝! + 𝜌!𝑢!! = 𝑝! + 𝜌!𝑢!! Energy
ℎ! +𝑉!!
2 = ℎ! +𝑉!!
2
𝑉!! = 𝑢!! + 𝑤!! 𝑉!! = 𝑢!! + 𝑤!!
𝑤! = 𝑤! therefore,
ℎ! +𝑢!!
2 = ℎ! +𝑢!!
2 𝑀!! = 𝑀!𝑠𝑖𝑛𝛽
𝐹𝑜𝑟 𝑎 𝑐𝑎𝑙𝑜𝑟𝑖𝑐𝑎𝑙𝑙𝑦 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠 𝜌!𝜌!=
(𝛾 + 1)𝑀!!!
(𝛾 − 1)𝑀!!! + 2
𝑝!𝑝!= 1+
2𝛾𝛾 + 1 (𝑀!!
! − 1)
𝑀!!! =
𝑀!!! + [2/(𝛾 − 1)]
[2𝛾/(𝛾 − 1)]𝑀!!! − 1
𝑇!𝑇!=𝑝!𝑝! 𝜌!𝜌!
𝑀! =𝑀!!
sin(𝛽 − 𝜗)
Type equation here. 𝑁𝑜𝑡𝑒: 𝐶ℎ𝑎𝑛𝑔𝑒𝑠 𝑎𝑐𝑟𝑜𝑠𝑠 𝑛𝑜𝑟𝑚𝑎𝑙 𝑠ℎ𝑜𝑐𝑘 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑜𝑛𝑙𝑦 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑀𝑎𝑐ℎ 𝑛𝑢𝑚𝑏𝑒𝑟
𝐶ℎ𝑎𝑛𝑔𝑒𝑠 𝑎𝑐𝑟𝑜𝑠𝑠 𝑜𝑏𝑙𝑖𝑞𝑢𝑒 𝑠ℎ𝑜𝑐𝑘 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀! 𝑎𝑛𝑑 𝛽 𝑀! 𝑐𝑎𝑛′𝑡 𝑏𝑒 𝑓𝑜𝑢𝑛𝑑 𝑢𝑛𝑡𝑖𝑙 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑔𝑙𝑒 𝜃 𝑖𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑
𝐵𝑢𝑡 𝜃 𝑐𝑎𝑛 𝑏𝑒 𝑓𝑜𝑢𝑛𝑑 𝑎𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀 𝑎𝑛𝑑 𝛽 𝑠𝑒𝑒 𝑛𝑒𝑥𝑡 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛
11. Describe how the deflection angle, the shock-‐wave angle and the Mach number are related for oblique, attached shock waves. What happens for large deflection angles? Deflection angle, θ Shock wave angle, β Mach number, M
tan𝜃 = 2 cot𝛽𝑀!
! sin! 𝛽 − 1𝑀!
!(𝛾 + cos 2𝛽)+ 2
For a large deflection angle, the oblique shock wave is no longer attached to the corner and is replaced by a detached bow shock. For relatively low values of theta, there will be two solutions for beta. Therefore there are two possible shock waves. The shockwave with the higher beta is the strong shockwave, while the lower one is the weak shock wave. The weak shock wave usually almost always exists and so the smallest beta can usually be used. 12. What is a Prandtl-‐Meyer expansion and how are the conditions downstream the expansion calculated? A Prandtl-‐Meyer expansion is when supersonic flow turns around a convex corner. This process is also isentropic. Because this process is isentropic, equations can be used to determine the fluid properties.
𝑇!𝑇!=
1+ 𝛾 − 12 𝑀!!
1+ 𝛾 − 12 𝑀!!
𝑝!𝑝!=
1+ 𝛾 − 12 𝑀!!
1+ 𝛾 − 12 𝑀!!
!!!!
𝜌!𝜌!=
1+ 𝛾 − 12 𝑀!!
1+ 𝛾 − 12 𝑀!!
!!!!
The mach number after the turn M2 is related to the initial mach number M1 and the turn angle, θ, by
𝜃 = 𝜈(𝑀!)− 𝜈(𝑀!) v(M) is the Prandtl-‐Meyer function. This number is defined through an equation where you can then calculate v(M2). From v(M2), M2 can be determined and thus the properties of the fluid downstream can be calculated. 13. What is wave drag and how is it calculated for a parallelogram without angle of attack? Wave drag is a component of drag (forces opposing the direction of motion with respect to a surrounding fluid) that is caused by the presence of shock waves. Wave drag is calculated for a parallelogram without angle of attack through the thickness of the parallelogram, t, and the pressures along the parallelogram.
23. Under what conditions is the following equation for the disturbance velocity potential obtained?:
(𝟏−𝑴!𝟐)𝝏𝟐𝝓𝝏𝒙𝟐 +
𝝏𝟐𝝓𝝏𝒚𝟐 +
𝝏𝟐𝝓𝝏𝒛𝟐 = 𝟎
For steady flow, the potential flow theory can be used to model irrotational, isentropic flow. For either subsonic or supersonic (not transonic or hypersonic) flows, at small angles of attack and thin bodies, we can assume that the velocity potential is split into an undisturbed onflow velocity V∞ in the x-‐direction, and a small perturbation velocity
∇Φ = 𝑉!𝑥 + ∇𝜑 From this assumption, the linearized small-‐perturbation potential equation – an approximation to the full equation – can be used.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛
−2𝜙!𝜙!𝑎! 𝜙!" −
2𝜙!𝜙!𝑎! 𝜙!" −
2𝜙!𝜙!𝑎! 𝜙!" = 0
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑡ℎ𝑖𝑠 𝑏𝑦 𝑎! 𝑎𝑛𝑑 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑓𝑙𝑜𝑤
ℎ! +𝑉!!
2 = ℎ +𝑉!
2 = ℎ +(𝑉! + 𝑢′)! + 𝑣′! + 𝑤′!
2
𝑎! = 𝑎!! −𝛾 − 12 (2𝑢′𝑉! + 𝑢′! + 𝑣′! + 𝑤′!)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑡ℎ𝑖𝑠 𝑖𝑛𝑡𝑜 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛𝑠,
𝑢′𝑉!,𝑣′𝑉!,𝑤′𝑉!
≪ 1
𝑢′𝑉!
!
,𝑣′𝑉!
!
,𝑤′𝑉!
!
≪≪ 1
𝑊𝑖𝑡ℎ 𝑡ℎ𝑖𝑠 𝑖𝑛 𝑚𝑖𝑛𝑑, 𝑡𝑒𝑟𝑚𝑠 𝑐𝑎𝑛 𝑐𝑎𝑛𝑐𝑒𝑙 𝑜𝑢𝑡 𝑎𝑠 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑡𝑜𝑜 𝑠𝑚𝑎𝑙𝑙 𝑒𝑥𝑐𝑒𝑝𝑡 𝑓𝑜𝑟 𝑀! 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 0− 0.8 𝑎𝑛𝑑 1.2− 5
𝑇ℎ𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑀!! (𝛾 + 1)
𝑢′𝑉!+⋯
∂u′∂x
𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 𝑐𝑜𝑚𝑝𝑎𝑟𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓
(1−𝑀!! )𝜕𝑢′𝜕𝑥
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑖𝑔𝑛𝑜𝑟𝑒 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑒𝑟 𝑡𝑒𝑟𝑚 𝑊𝑖𝑡ℎ 𝑡ℎ𝑒𝑠𝑒, 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑟𝑒𝑑𝑢𝑐𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑛𝑒𝑒𝑑𝑒𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛.
This is a linearized equation that can’t be associated with transonic flow (extreme sensitivity to geometry changes at sonic conditions) or hypersonic flow (strong shock waves close to geometric boundaries). 24. Derive an expression for the linearized pressure coefficient Definition of pressure coefficient for incompressible flow
𝑐! =𝑝 − 𝑝!12𝜌!𝑉!
!
When flow is compressible, more sense to use M∞ as normalizing parameter. 12𝜌!𝑉!
! =12𝛾𝑝!𝛾𝑝!
𝜌!𝑉!! =12 𝛾𝑝!𝑀!
!
𝑎!! = 𝛾𝑝!𝜌!
Therefore,
𝑐! =𝑝 − 𝑝!12 𝛾𝑝!𝑀!
!=
2𝛾𝑀!
! (𝑝𝑝!
− 1)
Energy Equation
ℎ +𝑉!
2 = ℎ! +𝑉!!
2 ℎ = 𝑐!𝑇 𝑓𝑜𝑟 𝑐𝑎𝑙𝑜𝑟𝑖𝑐𝑎𝑙𝑙𝑦 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑔𝑎𝑠
𝑇 +𝑉!
2𝐶!= 𝑇! +
𝑉!!
2𝐶!
𝑇 − 𝑇! =12𝐶!
(𝑉!! − 𝑉!)
𝑇𝑇!
= 1+1
2𝐶!𝑇!(𝑉!! − 𝑉!) = 1+
𝛾𝑅2𝐶!𝑎!!
(𝑉!! − 𝑉!)
𝛾𝑅𝐶!
=𝐶!𝐶!(𝐶! − 𝐶!)
𝐶!= 𝛾 − 1
𝑇𝑇!
= 1+𝛾 − 12𝑎!!
(𝑉!! − 𝑉!)
𝑇𝑇!
= 1−𝛾 − 12𝑎!!
(𝑉! − 𝑉!!)
𝑠𝑖𝑛𝑐𝑒,𝑉! = (𝑉! + 𝑢′)! + 𝑣′! + 𝑤′! 𝑇𝑇!
= 1−𝛾 − 12𝑎!!
(2𝑢′𝑉! + 𝑢′! + 𝑣′! + 𝑤′!)
𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝐹𝑙𝑜𝑤 𝑝𝑝!
=𝑇𝑇!
!!!!
𝑝𝑝!
= 1−𝛾 − 12𝑎!!
𝑀!! (2𝑢′𝑉!
+𝑢′! + 𝑣′! + 𝑤′!
𝑉!)
!!!!
𝑆𝑚𝑎𝑙𝑙 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛𝑠 𝑠𝑜 𝑖𝑡 𝑟𝑒𝑑𝑢𝑐𝑒𝑠 𝑡𝑜 𝑝𝑝!
= 1− (𝛾 − 1)𝑢!!𝑢𝑢!
!!!!
which is of the form, (1− 𝜀)! = 1+ 𝑛𝜀 +⋯ expanded using binomial theorem. Therefore,
𝑝𝑝!
= 1−𝛾2𝑀!
! (2𝑢′𝑉!
+𝑢′! + 𝑣′! + 𝑤′!
𝑉!)+⋯
𝐶! =2
𝛾𝑀!! 1−
𝛾2𝑀!
! (2𝑢′𝑉!
+𝑢′! + 𝑣′! + 𝑤′!
𝑉!)+⋯− 1
= −2𝑢′𝑉!
−𝑢′! + 𝑣′! + 𝑤′!
𝑉!+⋯ = −
2𝑢′𝑉!
25. Show how the Prandtl-‐Glauert rule is derived. What does it mean? Prandtl-‐Glauert rule allows for solving of compressible flow problems through the use of incompressible flow calculation methods. Valid for 2D subsonic flow over thin aerofoil. From Boundary Conditions
𝑑𝑓𝑑𝑥 =
𝑣′𝑉! + 𝑢′
= 𝑡𝑎𝑛𝜃
𝐹𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛𝑠,𝑢′ ≪ 𝑉! 𝑎𝑛𝑑 𝑡𝑎𝑛𝜃 = 𝜃 𝑑𝑓𝑑𝑥 =
𝑣′𝑉!
= 𝜃
𝑣′ =𝜕𝜃𝜕𝑦
𝜕𝜃𝜕𝑦 = 𝑉!
𝑑𝑓𝑑𝑥
𝑇ℎ𝑖𝑠 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡 𝑤𝑖𝑡ℎ 𝑙𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑒𝑑 𝑡ℎ𝑒𝑜𝑟𝑦
𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑒𝑑 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞. 𝑠𝑢𝑏𝑠𝑜𝑛𝑖𝑐
𝛽!𝜙!! + 𝜙!! = 0 𝛽 = 1−𝑀!
!
𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝜉 = 𝑥 𝜂 = 𝛽𝑦
𝜙(𝜉, 𝜂) = 𝛽𝜙(𝑥,𝑦) 𝑁𝑜𝑡𝑒
𝜕𝜉𝜕𝑥 = 1
𝜕𝜉𝜕𝑦 = 0
𝜕𝜂𝜕𝑥 = 0
𝜕𝜂𝜕𝑦 = 𝛽
𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 𝑜𝑓 𝜙(𝑥,𝑦) 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 𝑜𝑓 𝜙(𝜉, 𝜂)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑒𝑠 𝑖𝑛𝑡𝑜 𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑒𝑑 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞.
𝛽!(1𝛽𝜙!!)+ 𝛽𝜙!! = 0
𝜙!! + 𝜙!! = 0 𝑇ℎ𝑖𝑠 𝑖𝑠 𝐿𝑎𝑝𝑙𝑎𝑐𝑒′𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛,𝑤ℎ𝑖𝑐ℎ 𝑔𝑜𝑣𝑒𝑟𝑛𝑠 𝑖𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒 𝑓𝑙𝑜𝑤 𝜙 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑖𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒 𝑓𝑙𝑜𝑤 𝑖𝑛 (𝜉, 𝜂),𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑟𝑒𝑙𝑎𝑡𝑒𝑑
𝑡𝑜 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒 𝑓𝑙𝑜𝑤 𝜙 𝑖𝑛 (𝑥,𝑦)
𝑆ℎ𝑎𝑝𝑒 𝑜𝑓 𝑎𝑖𝑟𝑓𝑜𝑖𝑙
𝑉!𝑑𝑓𝑑𝑥 =
𝜕𝜙𝜕𝑦 =
1𝛽𝜕𝜙𝜕𝑦 =
𝜕𝜙𝜕𝜂
𝑉!𝑑𝑞𝑑𝜉 =
𝜕𝜙𝜕𝜂
𝑑𝑓𝑑𝑥 =
𝑑𝑞𝑑𝜉
𝑇ℎ𝑖𝑠 𝑖𝑠 𝑖𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑡 𝑎𝑠 𝑖𝑡 𝑑𝑒𝑚𝑜𝑛𝑠𝑡𝑟𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑠ℎ𝑎𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑖𝑟𝑓𝑜𝑖𝑙 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑖𝑛 𝑏𝑜𝑡ℎ 𝑠𝑝𝑎𝑐𝑒𝑠
𝐶! = −2𝑢′𝑉!
= −2𝑉!𝜕𝜙𝜕𝑥 = −
2𝑉!1𝛽𝜕𝜙𝜕𝑥 = −
2𝑉!1𝛽𝜕𝜙𝜕𝜉
𝐶! =1𝛽 (−
2𝑢𝑉!)
𝐶!" = −2𝑢𝑉!
𝐶! =𝐶!"1−𝑀!
!
26. Show, using the linearized theory, how the position of a streamline above an airfoil changes as the Mach number (<1) is increased? 27. Show how the pressure coefficient for linearized supersonic flow is derived. How do shock waves influence the solutions? For the derivation you can assume that the expression cp=-‐2u/u∞
𝐿𝑖𝑛𝑒𝑎𝑟𝑖𝑧𝑒𝑑 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛 − 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 2𝐷 𝑓𝑙𝑜𝑤 𝐹𝑜𝑟 𝑠𝑢𝑝𝑒𝑟𝑠𝑜𝑛𝑖𝑐 𝑓𝑙𝑜𝑤 𝜆!𝜙!! − 𝜙!! = 0
𝑤ℎ𝑒𝑟𝑒 𝜆 = 𝑀!! − 1
𝐶𝑙𝑎𝑠𝑠𝑖𝑐𝑎𝑙 𝑤𝑎𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑎𝑐𝑜𝑢𝑠𝑡𝑖𝑐 𝑡ℎ𝑒𝑜𝑟𝑦 𝜙 = 𝑓(𝑥 − 𝜆𝑦)+ 𝑔(𝑥 + 𝜆𝑦)
𝑙𝑒𝑡 𝑔 = 0 𝜙 = 𝑓(𝑥 − 𝜆𝑦)
𝑢′ =𝜕𝜙𝜕𝑥 = 𝑓′
𝑣′ =𝜕𝜙𝜕𝑦 = −𝜆𝑓′
𝑢′ = −𝑣′𝜆
𝐵𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛: 𝑑𝑦𝑑𝑥 =
𝑣′𝑉! + 𝑢′
= 𝑡𝑎𝑛𝜃
𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛𝑠 𝑢′ ≪ 𝑉! 𝑎𝑛𝑑 𝑡𝑎𝑛𝜃 ≈ 𝜃 𝑣′ = 𝑉!𝜃
→ 𝑢′ = −𝑉!𝜃𝜆
𝐶! = −2𝑢′𝑉!
=2𝜃𝜆
𝑜𝑟 𝐶! =2𝜃
𝑀!! − 1
28. What is meant by the critical Mach number for an airfoil?
The critical Mach number for an airfoil is the lowest Mach number at which the airflow over some point of the aircraft reaches the speed of sound. A thicker airfoil would have a lower critical Mach number because a thicker wing accelerates the airflow to a faster speed than a thinner airfoil. OR Critical Mach number is the free stream Mach number at which sonic flow is first encountered on the airfoil. This would happen at the minimum pressure point on the airfoil. 29. Consider subsonic flow with the Mach number M∞ over a wavy wall described by yw=hcos(2*pi*x/l), where yw is the ordinate of the wall, h is the amplitude and l is the wave length. The amplitude h can be assumed to be small as compared to the wavelength l. Using linearized theory, derive the surface pressure coefficient (cpw).
(1−𝑀!!)𝜙!! + 𝜙!! = 0
𝜙(𝑥,𝑦) = 𝐹(𝑥).𝐺(𝑥)
(1−𝑀!! )𝑑!𝐹𝑑𝑥! 𝐺 + 𝐹
𝑑!𝐺𝑑𝑦! = 0
1𝐹𝑑!𝐹𝑑𝑥! = −
1(1−𝑀!
! )1𝐺𝑑!𝐺𝑑𝑦!
LHS is only function of x, RHS is only function of y
1𝐹𝑑!𝐹𝑑𝑥! = 𝑐𝑜𝑛𝑠𝑡 = −𝑘! = −
1(1−𝑀!
! )1𝐺𝑑!𝐺𝑑𝑦!
𝑑!𝐹𝑑𝑥! + 𝐹𝑘
! = 0 → 𝐹 = 𝐵! sin(𝑘𝑥)+ 𝐵! cos(𝑘𝑥) 𝑑!𝐺𝑑𝑦! − 𝑘
!(1−𝑀!!)𝐺 = 0
𝐺 = 𝐴!𝑒!! !!!!!! + 𝐴!𝑒
!! !!!!!! second term zero as disturbance should go to zero as y approaches infinity
𝜙 = 𝐹 ∙ 𝐺 = 𝑒!! !!!!!(𝐶! sin 𝑘𝑥 + 𝐶! cos 𝑘𝑥) Cn=AnBn and C2=AnB2 Boundary conditions at the wall
𝑑𝑦!𝑑𝑥 tan𝜃 =
𝑣(𝑥,𝑦!)𝑢! + 𝑢(𝑥,𝑦!)
≈𝑣𝑢!
𝑣(𝑙,𝑦!) = 𝑣(𝑥, 0)+ 𝑦!𝑑𝑢𝑑𝑦 |!!! +⋯ ≈ 𝑣(𝑥, 0)
𝑑𝑦!𝑑𝑥 =
𝑣(𝑥, 0)𝑢!
𝑦! = ℎ cos2𝜋𝑥𝑙 →
𝑑𝑦!𝑑𝑥 = −ℎ
2𝜋𝑙 sin
2𝜋𝑥𝑙
𝑣 = 𝜙! = −𝑘 1−𝑀!! 𝑒!! !!!!! !(𝐶! sin 𝑘𝑥 + 𝐶! cos 𝑘𝑥)
2𝜋ℎ𝑙 sin
2𝜋𝑥𝑙 = −
𝑘𝑢!
1−𝑀!! (𝐶! sin 𝑘𝑥 + 𝐶! cos 𝑘𝑥)
C2 = 0, k = 2*pi/l 2𝜋ℎ𝑙 = −
𝑘𝑢!
1−𝑀!! ∙ 𝐶!
𝐶! =2𝜋ℎ𝑙
𝑢!2𝜋/𝑙
11−𝑀!
!=
ℎ𝑢!1−𝑀!
!
𝜙 =ℎ𝑢!1−𝑀!
!𝑒!!! !!!!!
!! sin
2𝜋𝑥𝑙
𝑢 = 𝜙! =2𝜋𝑢!1−𝑀!
!
ℎ𝑙 𝑒
!!! !!!!!!! cos
2𝜋𝑥𝑙
𝑣 = 𝜙! = −2𝜋𝑢!ℎ𝑙 𝑒
!!! !!!!!!! sin
2𝜋𝑥𝑙
𝐶! =−2𝑢𝑢!
= −4𝜋
1−𝑀!!
ℎ𝑙 𝑒
!!! !!!!!!! cos
2𝜋𝑥𝑙
𝐶!(𝑦 = 0) = −4𝜋
1−𝑀!!
ℎ𝑙 cos
2𝜋𝑥𝑙
30. Consider the same wall as in the previous problem. Derive the surface pressure coefficient for the case of supersonic flow over the wavy wall.
(𝑀!! − 1)𝜙!! − 𝜙!! = 0
𝜙(𝑥,𝑦) = 𝑓(𝑥 − 𝑀!! − 1𝑦)+ 𝑔(𝑥 + 𝑀!
! − 1𝑦) No disturbances from infinity Boundary Conditions
𝑑𝑦!𝑑𝑥 = tan𝜃 =
𝑣𝑢! + 𝑢
→𝑑𝑦!𝑑𝑥 =
𝑣(𝑥, 0)𝑢!
=𝜙!(𝑥, 0)𝑢!
𝜕𝜙𝜕𝑦 = 𝜙! = − 𝑀!
! − 1𝑓′
𝜙 = 𝑓(𝜉) 𝑤𝑖𝑡ℎ 𝜉 = 𝑥 − 𝑀!! − 1𝑦
→ 𝜙! =𝜕𝜉𝜕𝑦𝑑𝑓𝑑𝜉
−2𝜋ℎ𝑙 sin
2𝜋𝑥𝑙 =
1𝑢!
(− 𝑀!! − 1)𝑓′(𝑥)|!!!
→ 𝑓′(𝑥) = 2𝜋ℎ𝑙
𝑢!𝑀!! − 1
sin(2𝜋𝑥𝑙 )
→ 𝑓(𝑥) = −𝑙2𝜋
2𝜋ℎ𝑙
𝑢!𝑀!! − 1
cos(2𝜋𝑥𝑙 ) = −
ℎ𝑢!𝑀!! − 1
cos 2𝜋𝑥𝑙
𝜙(𝑥,𝑦) = −ℎ𝑢!𝑀!! − 1
cos(2𝜋𝑙 (𝑥 − 𝑀!
! − 1𝑦))
𝐶! = −2𝑢𝑢!
= −4𝜋ℎ𝑙
1𝑀!! − 1
sin(2𝜋𝑙 (𝑥 − 𝑀!
! − 1𝑦))
𝐶!(𝑦 = 0) = −4𝜋ℎ𝑙
1𝑀!! − 1
sin(2𝜋𝑥𝑙 )
31. What is the difference between transonic flow over an airfoil as compared to subsonic and supersonic flow respectively? With transonic flow, the shockwave can appear at any point along the airfoil. For lower mach numbers M≈0.79, the shockwave appears just downstream of the leading edge. This shockwave then gets terminated by a nearly normal shockwave as the measured pressure coefficient drops to a value below the critical pressure coefficient behind the shock. The flow over the bottom of the airfoil also doesn’t have shockwaves and is completely subsonic. As mach number increases, the shockwave moves further along the airfoil and shockwaves may start to appear on the bottom side of the airfoil – but further downstream of the airfoil. Because of the shockwave being terminated along the airfoil, the shock wave/boundary layer interaction appears. The pressure increases almost discontinuously across the shock wave. This represents an extremely large adverse pressure gradient (flow moving from lower pressure to higher pressure). Boundary layers separate from the surface in regions of adverse pressure gradients. When the shock wave impinges on the surface, the boundary layer encounters an extremely large adverse pressure gradient and it will almost always separate. Along with the total pressure losses caused by shockwaves, the
shock-‐induced separated flows create large drag forces. This is the drag-‐divergence phenomenon that is always associated with flight in transonic regime.
32. What is meant by the area-‐rule for transonic flow? The objective of the area rule is to reduce drag in the transonic regime. The area rule is a simple statement that the cross-‐sectional area of the body should have a smooth variation with longitudinal distance along the body – no rapid or discontinuous changes in the cross-‐sectional area distribution. For example, on a conventional wing-‐body, where the wings are, will be a sudden increase in cross-‐sectional area. The area rule says that to compensate for this, the point where the wings attach to the body, the body cross-‐sectional area at that point should
be reduced, to produce a bottle shape and reduce the overall cross-‐sectional area of the body.
Supercritical airfoil is shaped somewhat flat on the top surface in order to reduce the local Mach number inside the supersonic region below what it would be for a conventional airfoil. Therefore, the shockwave strength is lower, boundary layer effects are less severe and the free-‐stream Mach number can be higher before drag-‐divergence phenomenon sets in. 33. Show how the transonic similarity equation is derived. What is the transonic similarity parameter? How does the pressure coefficient vary with the thickness of the profile in the transonic flow regime?
𝐼𝑛𝑣𝑖𝑠𝑐𝑖𝑑 𝑡𝑟𝑎𝑛𝑠𝑜𝑛𝑖𝑐 𝑓𝑙𝑜𝑤𝑠 𝑔𝑜𝑣𝑒𝑟𝑛𝑒𝑑 𝑏𝑦 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑟 𝐸𝑢𝑙𝑒𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
𝐶𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦: 𝜕𝜌𝜕𝑡 + ∇ ∙ (𝜌𝑽) = 0
𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚: 𝜌𝐷𝑽𝐷𝑡 = −∇𝑝
𝐸𝑛𝑒𝑟𝑔𝑦: 𝜌𝐷ℎ!𝐷𝑡 =
𝜕𝑝𝜕𝑡
𝑇ℎ𝑒𝑠𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑤𝑒 𝑎𝑠𝑠𝑢𝑚𝑒 𝑖𝑛𝑣𝑖𝑠𝑐𝑖𝑑,𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑓𝑙𝑜𝑤 𝑤𝑖𝑡ℎ 𝑛𝑜 𝑏𝑜𝑑𝑦 𝑓𝑜𝑟𝑐𝑒𝑠 𝐸𝑛𝑡𝑟𝑜𝑝𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑖𝑛 𝑡𝑟𝑎𝑛𝑠𝑜𝑛𝑖𝑐 𝑓𝑙𝑜𝑤𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑠ℎ𝑜𝑐𝑘𝑤𝑎𝑣𝑒𝑠 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡ℎ𝑒𝑠𝑒 𝑓𝑙𝑜𝑤𝑠 𝑎𝑟𝑒 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑠 𝑑𝑒𝑚𝑜𝑛𝑠𝑡𝑟𝑎𝑡𝑒𝑑 𝑏𝑦 𝐶𝑟𝑜𝑐𝑐𝑜′𝑠
𝑡ℎ𝑒𝑜𝑟𝑒𝑚.
𝐻𝑜𝑤𝑒𝑣𝑒𝑟,𝑎𝑟𝑒 𝑠ℎ𝑜𝑐𝑘 𝑤𝑎𝑣𝑒𝑠 𝑤𝑒𝑎𝑘 𝑒𝑛𝑜𝑢𝑔ℎ 𝑡𝑜 𝑛𝑒𝑔𝑙𝑒𝑐𝑡 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑙𝑜𝑤? 𝐶ℎ𝑒𝑐𝑘 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑠ℎ𝑜𝑐𝑘 𝑤𝑎𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝑊𝑖𝑙𝑙 𝑟𝑒𝑎𝑐ℎ 𝑠! − 𝑠!𝑅 ≈
2𝛾3(𝛾 + 1)! (𝑀!
! − 1)!
𝑆ℎ𝑜𝑤𝑠 𝑡ℎ𝑎𝑡 𝑖𝑓 𝑀!! − 1 ≪ 1, 𝑡ℎ𝑒𝑛 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑠ℎ𝑜𝑐𝑘 𝑖𝑠 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙
𝐹𝑜𝑟 𝑡𝑟𝑎𝑛𝑠𝑜𝑛𝑖𝑐 𝑓𝑙𝑜𝑤 𝑤𝑖𝑡ℎ 𝑙𝑜𝑤𝑒𝑟 𝑀𝑎𝑐ℎ 𝑛𝑢𝑚𝑏𝑒𝑟𝑠,𝑤𝑒 𝑐𝑎𝑛 𝑎𝑠𝑠𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑒𝑠𝑠𝑒𝑛𝑡𝑖𝑎𝑙𝑙𝑦 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑎𝑛𝑑 𝑖𝑟𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙.𝐻𝑜𝑤𝑒𝑣𝑒𝑟, 𝑓𝑜𝑟
𝑀𝑎𝑐ℎ 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑖𝑛 𝑡ℎ𝑒 ℎ𝑖𝑔ℎ 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑎𝑛𝑠𝑜𝑛𝑖𝑐 𝑟𝑎𝑛𝑔𝑒, 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑚𝑎𝑦 𝑏𝑒 𝑡𝑜𝑜 𝑙𝑎𝑟𝑔𝑒 𝑡𝑜 𝑖𝑔𝑛𝑜𝑟𝑒.
𝐼𝑓 𝑡𝑟𝑎𝑛𝑠𝑜𝑛𝑖𝑐 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑖𝑟𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑐ℎ𝑎𝑛𝑔𝑒𝑠) 𝑡ℎ𝑒𝑛 𝑎 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝜙 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
𝑽 = ∇𝜙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑐𝑎𝑛 𝑡ℎ𝑒𝑛 𝑏𝑒 𝑢𝑠𝑒𝑑.𝐴𝑠 𝑤𝑒 𝑎𝑠𝑠𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑖𝑟𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐. 𝐼𝑓 𝑤𝑒 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑓𝑙𝑜𝑤 𝑜𝑣𝑒𝑟 𝑎 𝑠𝑙𝑒𝑛𝑑𝑒𝑟 𝑏𝑜𝑑𝑦
𝑎𝑛𝑑 𝑠𝑚𝑎𝑙𝑙 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑎𝑡𝑡𝑎𝑐𝑘,𝑤𝑒 𝑎𝑠𝑠𝑢𝑚𝑒 𝑠𝑚𝑎𝑙𝑙 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛𝑠. 𝑇ℎ𝑖𝑠 𝑙𝑒𝑎𝑑𝑠 𝑡𝑜 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝜙 = 𝑉!𝑥 + 𝜙
𝑊𝑖𝑡ℎ 𝑡ℎ𝑖𝑠,𝑤𝑒 𝑟𝑒𝑎𝑐ℎ 𝑡𝑟𝑎𝑛𝑠𝑜𝑛𝑖𝑐 𝑠𝑚𝑎𝑙𝑙 − 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
(1−𝑀!!)𝜕!𝜙𝜕𝑥! +
𝜕!𝜙𝜕𝑦! +
𝜕!𝜙𝜕𝑧! = 𝑀!
![(𝛾 + 1)𝜙!𝑉!]𝜙!!
𝑁𝑜𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙𝑖𝑧𝑒 𝑡ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑠𝑙𝑒𝑛𝑑𝑒𝑟𝑛𝑒𝑠𝑠 𝑟𝑎𝑡𝑖𝑜 𝜏 = 𝑏/𝑐
𝑏 𝑖𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑐 𝑖𝑠 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑜𝑑𝑦 𝑠𝑙𝑒𝑛𝑑𝑒𝑟𝑛𝑒𝑠𝑠 𝑟𝑎𝑡𝑖𝑜 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑠𝑚𝑎𝑙𝑙 𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛𝑠
𝑊𝑖𝑡ℎ 𝑛𝑜𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙𝑖𝑧𝑖𝑛𝑔,𝑤𝑒 𝑜𝑏𝑡𝑎𝑖𝑛 (1−𝑀!
! )𝜏!/! −𝑀!
! (𝛾 + 1)𝜙! 𝜙!! + 𝜙!! + 𝜙!! = 0
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑤𝑖𝑡ℎ 𝑙𝑖𝑛𝑒 𝑎𝑏𝑜𝑣𝑒 𝑎𝑟𝑒 𝑛𝑜𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝐷𝑒𝑓𝑖𝑛𝑒 𝑡𝑟𝑎𝑛𝑠𝑜𝑛𝑖𝑐 𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝐾 𝑎𝑠
𝐾 =(1−𝑀!
! )𝜏!/!
𝐾 −𝑀!! (𝛾 + 1)𝜙! 𝜙!! + 𝜙!! + 𝜙!! = 0
𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑀𝑎𝑐ℎ 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑟𝑒 𝑛𝑒𝑎𝑟 𝑢𝑛𝑖𝑡𝑦 𝑓𝑜𝑟 𝑡𝑟𝑎𝑛𝑠𝑜𝑛𝑖𝑐 𝑓𝑙𝑜𝑤 𝑅𝑒𝑝𝑙𝑎𝑐𝑒 𝑀! 𝑤𝑖𝑡ℎ 𝑢𝑛𝑖𝑡𝑦, 𝑜𝑏𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝐾 − (𝛾 + 1)𝜙! 𝜙!! + 𝜙!! + 𝜙!! = 0
𝑊𝑖𝑡ℎ 𝑡𝑤𝑜 𝑓𝑙𝑜𝑤𝑠 𝑎𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑀! 𝑜𝑣𝑒𝑟 𝑎 𝑏𝑜𝑑𝑦 𝑤𝑖𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝜏
𝑏𝑢𝑡 𝑠𝑎𝑚𝑒 𝐾, 𝑡ℎ𝑒𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑏𝑜𝑡ℎ 𝑓𝑙𝑜𝑤𝑠 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝐴𝑙𝑠𝑜, 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
𝐶!𝜏!/! = −2𝜙! = 𝑓(𝐾, 𝑥,𝑦, 𝑧) 𝑎𝑛𝑑 𝑡ℎ𝑒𝑦 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒.
𝑆𝑡𝑒𝑝𝑠
1.𝐸𝑢𝑙𝑒𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 − 𝑒𝑥𝑎𝑐𝑡 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 2.𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙) − 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑤𝑒 𝑎𝑠𝑠𝑢𝑚𝑒 𝑠ℎ𝑜𝑐𝑘 𝑤𝑎𝑣𝑒 𝑖𝑠 𝑤𝑒𝑎𝑘 𝑒𝑛𝑜𝑢𝑔ℎ 𝑡𝑜 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑓𝑙𝑜𝑤 𝑎𝑠
𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑎𝑛𝑑 𝑖𝑟𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 3. 𝑆𝑚𝑎𝑙𝑙 − 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 − 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛
top related