Consider the following parameters:
- B (unit magnetic field)
- H (magnetization of the material in response to B)
- pm (magnetic dipole moment)
- 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"]
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 + µ∇2v + f
The equation of flow of the magnetic fluid in the ducts becomes
ρ{∂v/∂t + v(∇v)} = -∇p + η∇v + ρg + M∇B
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 + η∇v + ρg + M∇B
ρ = ρo(1 – βT)
M = Mo(1 – KΘ)
Here, v(∇v) is the convective acceleration, -∇p the pressure gradient, η∇v the Newtonian viscosity, and ρg the kinetic energy of the fluid.
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