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Thursday, 10 March 2011

Physical Analysis Of An ADR: An exploration

Adiabatic demagnetization

Consider the following parameters:

  1. B (unit magnetic field)

  2. H (magnetization of the material in response to B)

  3. pm (magnetic dipole moment)

  4. µ and µo


pm represents the maximum torque experienced in a unit magnetic field.

τ = pm x B

B and M are associated by the following relationship:

B = µH = µo(H + M)

The units of M are N m T-1 m-3 (force-meter per degrees Celsius per unit volume).

Assuming the given paramagnetic sample is isotropic and that pm and B are in the same direction, we can eliminate the need to represent them as vectors and work with them as scalars alone.

The total work done on a paramagnetic sample in increasing its magnetization from M to M + dM in a magnetic field B is given by:

W = B.dM

If energy is transferred to the material without allowing for a change in volume:

dU = T.dS + B.dM

Now, we are in a position to define the state functions H, A and G.

H = U – BM

A = U – TS

G = H – TS = A – BM

Differentiating,

dH = T.dS – M.dB

dA = -S.dT + B.dM

dG = -S.dT – M.dB

Therefore, adiabatic decompression can be represented as (∂T/ ∂P)S, and adiabatic demagnetization as (∂T/∂B)S.

f (S, T, P) = 0

Therefore: dP = (∂P/ ∂S)T.dS + (∂P/ ∂T)S.dT

In order to parameterize T in terms of S, we can safely assume that the curve defined by the above function moves through a point where dP = 0 and where the abscissa is expressible in terms of the ordinate. Thus,

dT = (∂T/∂S)P.dS

0 = (∂P/∂S)T.dS + (∂P/∂T)S.( ∂T/∂S)P.dS

Dividing by dS and rearranging,

(∂P/∂S)T = - (∂P/∂T)S.(∂T/∂S)P

Therefore: (∂S/∂T)P.( ∂T/∂P)S.( ∂P/∂S)T = -1

In order to deduce an expression for (∂T/∂P)S, we assume that the process is reversible and isobaric.


[caption id="" align="alignright" width="224" caption="P-v diagram of an isobaric process"]pv[/caption]


For a reversible process: dQ = T.dS

For an isobaric process: dQ = Cp.dT

Therefore: (∂S/∂T)P = Cp/T

Introducing a Maxwell’s relation,

(∂S/∂P)T = - (∂V/∂T)P

Cp/T.( ∂T/∂P)S.( ∂V/∂T)P = 1

(∂T/∂P)S = T/Cp.(∂V/∂T)P

If the gas is an ideal gas: PV = RT

Therefore,

(∂V/∂T)P = R/P = V/T

(∂T/∂P)S = V/Cp

Since S, T and B are involved in a cyclic relationship: (∂S/∂T)B.( ∂T/∂B)S.( ∂B/∂S)T = -1

For a reversible process: dQ = T.dS

In a constant magnetic field: dQ = CB.dT

Therefore: (∂S/∂T)B = CB/T

Introducing another of Maxwell’s relations: (∂S/∂B)T = (∂M/∂T)B

Therefore: (∂T/∂B)S = -T/CB.(∂M/∂T)B

For a paramagnetic material,

(∂M/∂T)B = -aB/T2 = -M/T

Therefore: (∂T/∂B)S = M/CB

Magnetorheology

If B is the magnetic field and E, the electric field, then they are defined as

B = x A

E = -φ - A/∂t, where

A is the vector field, φ the electric potential and , the curl vector.

Therefore,

∇.B = ∇ . (x A) = 0 (Gauss’ law: magnetic field is non-divergent)

x E = x (-∇φ - A/t) = - ∂/∂t(x A)

By applying the above equations Maxwell’s equations for potential formulation for an electromagnetic field and performing a gauge fixing (i.e. accounting for unphysical degrees of freedom in the system),

. A + (1/c2)(∂φ/∂t) = 0

2φ – (1/c2)( ∂2φ/∂t2) = ρ/Єo

2A – (1/c2)(∂2A/∂t2) = - µoJ, where

ρ is the charge density and J, the current density.

Therefore, the magnetic field of the solenoid using vector potential formulations is

x (/µ x A) = J

Applying the Navier-Stokes equation,

ρ(∂v/∂t + v . ∇v) = -p + µ2v + f

The equation of flow of the magnetic fluid in the ducts becomes

ρ{∂v/∂t + v(v)} = -p + ηv + ρg + MB

Thus, the potential parts of magnetic and gravitational forces have been included in the pressure-gradient in the Navier-Stokes equation. Here, ρ is the density of the fluid.

By approximating a linear-dependency between the temperature-gradient and the density and the magnetic moment, we get

ρ = ρo(1 – βT)

M = Mo(1 – KΘ)

Therefore, we have a simple set of three equations with which to model fluid-flow in the ADR ducting during operation.

ρ{∂v/∂t + v(v)} = -p + ηv + ρg + MB

ρ = ρo(1 – βT)

M = Mo(1 – KΘ)

Here, v(v) is the convective acceleration, -p the pressure gradient, ηv the Newtonian viscosity, and ρg the kinetic energy of the fluid.

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