Magneto-Hydrodynamics (MHD) Pouya Pourkarim Niloofar Pourjafari Outline ο MHD equations ο Behavior of a fluid in the presence of electromagnetic fields ο Flux freezing: An MHD case study MHD equations ο ππ + π». ππ£ = 0 ππ‘ (continuity equation) ο ππ£ 1 π = −π»π + π½ × π΅ + πΉπ£ + ππ ππ‘ π (force equation) ο 1 ππ΅ π»×πΈ =− π ππ‘ (Faraday’s law) ο 4π π»×π΅ = J π (Ampere’s law (neglecting displacement current)) ο 1 π½ = π(πΈ + π£ × π΅) π (Ohm’s law) ο π= πΎππΎ (equation of state) Ideal Ohm’s law π½ = ππΈ E’ and J’ are measured in the rest frame of the medium π½=π½ 1 πΈ = πΈ + (π£ × π΅) π 1 π½ = π(πΈ + π£ × π΅) π 1 πΈ+ π£×π΅ =0 π For a medium moving with velocity v relative to the laboratory Behavior of a fluid in the presence of electromagnetic fields ο Electromagnetic effect ο Mechanical effect Electromagnetic effect Time dependence of a magnetic field: ππ΅ ππ‘ =π»× π£×π΅ + π2 π»2π΅ 4ππ Derivation: 1 π½ 1 π½ = π πΈ+ π£×π΅ →πΈ = − π£×π΅ π π π ππ΅ ππ‘ = −cπ» × πΈ = π» × π£ × π΅ − π»×π΅ = 4π π½ πΆ →π»×π½ = π π» 4π π π (π» × π½) × π»×π΅ = π 4π π» π». π΅ − π΅(π». π») ππ΅ π2 2 =π»× π£×π΅ + π» π΅ ππ‘ 4ππ ππ΅ π2 2 = π» π΅ ππ‘ 4ππ (Fluid at rest) 4πππΏ2 π= π2 This means that an initial configuration of magnetic field will decay away in a diffusion time. ππ΅ = π» × (π£ × π΅) ππ‘ πΈ×π΅ π€=π π΅2 (Infinite conductivity) Time dependence of a magnetic field: (Infinite conductivity) ππ΅ =π»× π£×π΅ ππ‘ π π = + π£. π» ππ‘ ππ‘ convective derivative: total time rate of change of a quantity moving instantaneously with the velocity v. ππ΅ ππ΅ = + π£. π» . π΅ ππ‘ ππ‘ π» × π£ × π΅ = π£ π». π΅ − π΅ π£. π» → π£. π» π΅ = −π» × (π£ × π΅) → ππ΅ ππ΅ = − π» × (π£ × π΅) ππ‘ ππ‘ π∅π΅ π = ππ‘ ππ‘ π΅. πππ = π π π» × π£ × π΅ . πππ = π π ππ΅ . πππ − ππ‘ π£ × π΅ . ππ π» × (π£ × π΅). πππ π Stokes theorem π ππ΅ . πππ = ππ‘ π» × π£ × π΅ . πππ = π π∅π΅ =0 ππ‘ π£ × π΅ . ππ π Magnetic flux through any loop moving with the local fluid velocity is constant in time. In other words, the lines of force are frozen into the fluid and are carried along with it. The component of v perpendicular to B can be identified as the velocity w of the lines of magnetic force. 1 πΈ + (π£ × π΅) = 0 π (Ohm’s ideal law) → π€ × π΅ = −ππΈ πΈ×π΅ →π€=π π΅2 πΆ =π΄×π →π = πΆ×π΄ π΄. π΄ Magnetic Reynolds number: A useful parameter to distinguish between situations in which diffusion of the field lines relative to the fluid occurs and those in which the lines of force are frozen in is the magnetic Reynolds number. π π = π» × (π£ × π΅) π2 2 π» π΅ 4ππ π π β« 1 Transport of the lines of force with the fluid dominates over diffusion π π βͺ 1 Diffusion dominates over transport of the lines of force with the fluid Mechanical effect ππ£ 1 π = −π»π + π½ × π΅ + πΉπ£ + ππ ππ‘ π (force equation) ππ£ 1 π = πΉ+ π½×π΅ ππ‘ π π½ = π(πΈ + 1 π£ π × π΅) ππ΅2 π½ × π΅ = σπΈ × π΅ − π£⊥ π ππ£ ππ΅2 π = πΉ − 2 (π£⊥ − π€) ππ‘ π The velocity of flow of the fluid perpendicular to B, decays from some initially arbitrary value in a time of the order of: ππ 2 π= ππ΅2 to a value: ππ 2 π£⊥ = π€ + πΉ ππ΅2 ⊥ 4π π½ π 1 1 (π½ × π΅) = π»×π΅ ×π΅ π 4π π»×π΅ = 1 π΅2 1 π½ × π΅ = −π» + π΅. π» π΅ π 8π 4π π΅2 ππ = 8π 1 π» π΅2 = π΅. π» π΅ + π΅ × π» × π΅ 2 ππ£ 1 π = −π»π − π»ππ + ππ + π΅. π» π΅ ππ‘ 4π π ππ£ ππ‘ = −π» π + ππ + ππ + 1 4π π΅. π» π΅ π = −π»π π + ππ + ππ = ππππ π‘πππ‘ Apart from gravitational effects, any change in mechanical pressure must be balanced by an opposite change in magnetic pressure. If the fluid is to be confined within a certain region so that p falls rapidly to zero outside that region, the magnetic pressure must rise equally rapidly in order to confine the fluid. MHD Flow between Boundaries with Crossed Electric and Magnetic Fields Illustrating the behavioral differences between freezing force lines and diffusion through them, and the E × B drift. οΌ Ignore gravity οΌ No pressure gradient in x direction οΌ Incompressible fluid οΌ Viscous fluid οΌ Conducting fluid οΌ Nonconducting boundary surfaces at z = 0 and z = a οΌ Planes moving with different speeds π1 and π2 in x direction οΌ Uniform magnetic field π΅0 in z direction οΌ System is infinite in x – y plane Figure 1. Flow of viscous conducting fluid in a magnetic field between two plane surfaces moving with different velocities. Any electric fields? Faraday′s Law ππ΅ =0 π» × πΈ = 0 → πΈ = −π»π ππ‘ (π΅ × π£) × π΅ πΈ×π΅ π€= =π π΅2 π΅2 → A constant electric field is present → Electric field is in the y direction Any other magnetic fields? Moving fluid tend to carry field lines → An x component π΅π₯ (π§) of magnetic induction in addition to the z component π΅0 οΌ Incompressible fluid → Continuity equation becomes π» ⋅ v = 0 The steady state form of force equation becomes: π»π = 1 π½ × π΅ + ππ» 2 π£ π The only non vanishing component of J is π½π¦ (π§): 1 π½π¦ (π§) = π πΈ0 − π΅0 π£(π§) π (v(z) is the x component of velocity) 2 The force equation separate in 3 dimensions: ππ = ππ΅0 πΈ − π΅0 π£ + π π π£ 0 2 ππ₯ π π ππ =0 ππ¦ ππ ππ΅π₯ π΅0 =− πΈ0 − π£ ππ§ π π ππ§ οΌ No pressure gradient in x direction: The first equation can be written as π2π£ π − ππ§ 2 π 2 π π£=− π 2 ππΈ0 π΅0 ππ΅02 π2 Where M is the Hartmann number and is defined as: π = ππ 2 1 2 Hartmann number can be thought as the ratio of magnetic viscosity to normal viscosity. The solution to the equation using the boundary conditions π£ 0 = π1 and π£ π = π2 is: π£ π§ = π1 π ππβ sinh π π π−π§ π + π2 ππ§ π ππβ sinh π π + ππΈ0 π΅0 1− π ππβ π π−π§ π ππ§ +π ππβ π sinh π ο First limiting case: π΅0 → 0 (π → 0) π§ π£ π§ = π1 + (π2 − π1 ) (Laminar Flow) π ο Second limiting case: π β« 1: (Magnetic viscosity dominates and flow mostly follows E × B drift) For π β« π§ we obtain ππΈ0 ππΈ0 −ππ§ π£ π§ β + (π1 − )π π΅0 π΅0 Figure 2. Velocity profile of the two limits π While π π§ = π1 exactly at the surface, there is a rapid transition in a distance of order (a/M) to the E × B drift value. Near z = a, we simply switch π1 with π2 and z with a-z Figure 3. Transport of lines of magnetic induction The magnetic field π΅π₯ (π§) can be determined by the equation ππ΅π₯ 4π 4ππ 1 = π½ = (πΈ0 − π΅0 π£) ππ§ π π¦ π π Assuming that π΅π₯ vanishes at z = a and z = 0 and substituting π with its calculated formula we have π΅π₯ z = π΅0 4πππ2 π2 π2 −π1 2π π π ππ§ πππ β 2 −πππ β 2 − π π π π ππβ 2 π π Figure 4. Axial component of magnetic induction Which in the two limiting cases of π β« 1 and π βͺ 1 reduces to π΅π₯ z β π π π΅0 π§ π§ 1− , π π 1 1 − (π −ππ§ π πππ π βͺ 1 π + π −π π−π§ π ), πππ π β« 1