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Numerical Analysis and Scientific Computing: Workshop in Honour of BulentKarasozen’s 67th Birthday
Ankara, Turkey, November 24, 2017
Numerical Methods in Fluid Flow and Heat TransferApplication:Boundary Integral Solution of MHD Pipe Flow
Munevver Tezer-Sezgin and Canan Bozkaya
Department of MathematicsMiddle East Technical University
Ankara, [email protected]
M. Tezer (METU) IAM2017 November 24, 2017 1 / 43
Outline
1 Numerical methods in fluid flow and heat transfer
The physical problem and mathematical modelingSolution approachesSome challenging numerical computations
Global climate modelingHeart simulationVehicle aerodynamicsTurbomachinery analysisComputational fluid dynamics (CFD)
Components of a Numerical SolutionNumerical grid (grid generation)Solution of discretized equations
M. Tezer (METU) IAM2017 November 24, 2017 2 / 43
The physical problem and mathematical modeling
The mathematical formulation of the problem is the reduction of the physicalproblem to a set of either algebraic or differential equations subject to certainassumptions.
The process of modeling of phsical systems in the real world should generally followthe path illustrated schematically in the chart below:
M. Tezer (METU) IAM2017 November 24, 2017 3 / 43
Physical System+
side conditions(Problems in reality)
−→1
Known physical laws+
assumptions
↓ 2
Efficient NumericalMethod
←−3
Mathematical ModelGoverning equations
(usually PDEs in FluidDynamics)
+boundary and/or initial
conditions
−→3
Analyticalsolution (if itexists and canbe obtained)
↓ 4
Discretization−→
5Systems of Algebraicequations or ODEs
−→6
Solver usinghigh
performancecomputer
↓ 7
Numericalsolution
M. Tezer (METU) IAM2017 November 24, 2017 4 / 43
Solution approaches
Three approaches or methods are used to solve a problem in fluid mechanics and heattransfer:
1 Experimental methods:Requires most realistic experiment, e.g. In a laboratory a flowing river is setup andits surface is modeled as a moving boundary problem. There are scaling problems,measurement difficulties, operating costs.
2 Theoretical (analytical) methods:Gives closed form solution as a formula or as a function. But, it is usually restrictedto simple geometry, constant coefficients and usually restricted to linear problemswith some assumptions.
3 Numerical (computational) methods (CFD):
No restriction to linearityComplicated physics can be treated as variable coefficientsTime evolution of the flow can be depictedLarge problem parameters as Re, Ha and Ra can be tackled.
Disadvantages in CFDTruncation errorsBoundary condition problemsComputer costsConvergence, accuracy and stability problems.
M. Tezer (METU) IAM2017 November 24, 2017 5 / 43
Some challenging numerical computations (H. Gunes)
Science:
Global climate modelingAstrophysical modelingBiology: Genome analysis, protein folding, drug design, heart simulation, blood flow inconstricted arteries
Engineering:Automotive designVehicle aerodynamicsCrash simulationElectronic device designEarthquake and structural modeling
Business:
Financial and economic modelingWeb services, search engines
Defense:Nuclear weaponsCryptography
M. Tezer (METU) IAM2017 November 24, 2017 6 / 43
Global climate modeling
Problem is to compute:
f (latitude, longitude,elevation, time)→ temperature,pressure, humidity, windvelocity
Approach:
Discretize the domain, e.g. ameasurement point every 1 km.
Devise an algorithm to predictweather at time t+ 1 if knownat time t.
Computation:
Solve the Navier-stokesequations in 3D-modeling thecompressible fluid flow inatmosphere.
Also, the energy equation mustbe added.
Sea Surface temperature from aneddy resolving ocean model. Source:http://www.csm.ornl.gov/chammp/chammp.html
M. Tezer (METU) IAM2017 November 24, 2017 7 / 43
Heart simulation
Problem is to compute bloodflow in the heart
Heart is modeled as anelastic structure in anincompressible fluid.
The model can be used todesign heart valves.
The model helps tounderstand effects ofdisease (leaky valves).
The model can be used tolook at the behavior of theheart during a heart attack.
It needs real-time clinicalwork (electrical response ofthe heart model, details ofmuscles, lungs, circulatorysystems).
Heart simulation calculations
It involves solving Navier-Stokesequations for incompressible fluid.Energy equation is also coupled fordetermining the temperature of theblood.
Electrical model of the heart, muscles,lungs and circulatory systems. Ref:
Chris Johnson, Andrew McCullochM. Tezer (METU) IAM2017 November 24, 2017 8 / 43
Vehicle aerodynamics
Flow around a moving truck in a wind tunnel
Modeling the truck and the blowing air around it.
The road also has to move at the air speed, the modeling is a difficult task.
The drag coefficient, lift coefficient, moment coefficient are going to be obtainedfrom the solution of the Navier-Stokes equations for compressible fluid in terms ofstreamlines.
M. Tezer (METU) IAM2017 November 24, 2017 9 / 43
Turbomachinery analysis
Flow in an inclined duct fan (need to consider rotating fluid)
Navier-Stokes equations in spherical coordinates will be solved for rotational flow. Then,absolute and angular velocities need to be computed.
M. Tezer (METU) IAM2017 November 24, 2017 10 / 43
The Heat Equation
The heat equation governs the temperature distribution in an object.In 3D conservation of energy
c(x, y, z)ρ(x, y, z)∂u
∂t= −∇.φ+Q(x, y, z, t)
and Fourier’s law gives heat flux φ as
φ(x, y, z, t) = −K0(x, y, z)∇u
whereu(x, y, z, t) : Temperaturec(x, y, z, t) : Specific heatρ(x, y, z, t) : Density of the materialQ(x, y, z, t) : Heat sourceK0(x, y, z, t) : Thermal conductivity of the materialφ(x, y, z, t) : Heat flux (amount of thermal energy that flows in or out of the object)
φ > 0 → heat is addedφ < 0 → heat is removed
M. Tezer (METU) IAM2017 November 24, 2017 11 / 43
The Heat Equation (Paul’s Online Math Notes)
Then,
c(x, y, z)ρ(x, y, z)∂u
∂t= ∇.(K0(x, y, z)∇u) +Q(x, y, z, t) .
This form of the heat equation is not actually can be solved. However, we can obtainFourier series solution of the following simplified equation in which c, ρ,K0 are constants:
∂u
∂t= k∇2u+
Q
cρ, k =
K0
cρ.
Related boundary conditions are
Initial condition : u(x, y, z, 0) = f(x, y, z)Prescribed condition : u(x, y, z, 0) = T (x, y, z) on the boundaryHeat flux condition : −K0∇u.~n = φ(t)
Newton’s law of cooling : −K0∇u.~n = H(u− uB)(Robins condition) (when the object is in a moving fluid)
H > 0 experimentally determineduB is the temperature of the fluid on the object boundary
M. Tezer (METU) IAM2017 November 24, 2017 12 / 43
The Navier-Stokes Equations
The traditional form of Navier-Stokes equations
du
dt= −(u.∇)u− 1
ρ∇p+ ν∇2u+ f
u : velocity of the fluidρ : density of the fluidp : pressureν : kinematic viscosityf : external force (any other force acting on the fluid)−(u.∇)u : convection term which is the divergence on the velocity of the fluid− 1ρ∇p : determines how the fluid particles move as pressure changes
(the tendency to move away from areas of higher pressure)ν∇2u : determine what a fluid particle does and what its neighbors do.
( high viscous fluid induces nearby particles to move, in contrast,less viscous fluid induces a lower effect on its neighbors)
du
dt: time rate of change of the fluid velocity
Navier-Stokes equations can be viewed as an application of Newton’s second law
F = ma .
M. Tezer (METU) IAM2017 November 24, 2017 13 / 43
The Navier-Stokes Equations
Non-dimensional form of Navier-Stokes equations:
du
dt+ (u.∇)u = −∇p+
1
Re∇2u+ f ,
where Re =LU
ν.
High viscosity (small Re), Creeping flow︸ ︷︷ ︸low velocity
:1
Re∇2u+ f = ∇p+
du
dt
Low viscosity (large Re), Ideal fluid︸ ︷︷ ︸high velocity
:du
dt+ (u.∇)u = −∇p+ f .
M. Tezer (METU) IAM2017 November 24, 2017 14 / 43
Effect of convection term Re(u.∇)u
Re large Re small
When the river converges in thenarrowing part, the overall veloc-ity of the flow increases.
If the river diverges, the particlesspread out, and the overall speedof the flow decreases.
(Steven Dobek, Fluid Dynamics and Navier-Stokes equations, (2012). )
M. Tezer (METU) IAM2017 November 24, 2017 15 / 43
Computational fluid dynamics (CFD)
CFD obtains approximate solutions to complex problems numerically.
It needs to use a discretization method which approximates the differential equationsby a system for algebraic equations, which can then be solved on a computer.
Accuracy of numerical solutions ⇔ Quality of discretization
M. Tezer (METU) IAM2017 November 24, 2017 16 / 43
Components of a Numerical Solution
1 Mathematical modelSet of PDEs or integro-differential equations and the corresponding boundaryconditions
2 Discretization methodsFinite Difference Method (FDM)Finite Volume Method (FVM)Finite Element Method (FEM)Spectral Element Methods (DQM, RBF,...)Boundary Element Method (BEM)
Common character in each discretization method:
PDEs (continuous problem) ⇐⇒ Discrete equations (Discrete problem, a set oflinear or nonlinear algebraic equations or ODEs)
M. Tezer (METU) IAM2017 November 24, 2017 17 / 43
Numerical grid (grid generation)
Numerical grid or mesh is the discrete locations at which the variables (solution) areto be calculated.
Grid generation depends on the numerical method.
FDM: Approximates the derivatives at the grid points (derivative approximations areobtained from Taylor series expansion of the solution). FDM discretization is done byusing rectangles.
FEM: A weighted residual statement is needed for the DEs and boundary conditions.Discretization is performed using triangles or rectangles.
BEM: A boundary integral equation must be obtained with the correspondingfundamental solution of DE. Discretization is only on the boundary of the problem.
DQM, RBF can use nonuniform grid points.
M. Tezer (METU) IAM2017 November 24, 2017 18 / 43
Solution of discretized equations
Discretization =⇒ A large system of linear on nonlinear algebraic equations
Linear equations =⇒ Algebraic equation solvers
Non-linear equa-tions
=⇒ Iteration schemes are used
Unsteady flows: Methods are based on marching in time.
Steady flows: A pseudo-time marching or equivalent iteration schemes are used.
For iterative procedures a convergence criteria needs to be used as:
Absolute convergence: |a− a ∗ | < ε (tolerance)
Relative convergence:∣∣∣a− a∗
a
∣∣∣ < ε
M. Tezer (METU) IAM2017 November 24, 2017 19 / 43
Application: Boundary integral solution of MHD
pipe flow
M. Tezer (METU) IAM2017 November 24, 2017 20 / 43
Outline
1 Application: Boundary integral solution of MHD pipe flow
The Physical Problem (MHD flow)Mathematical ModelSolution Procedure
Inner and Outer ProblemsLinear System of Equations
Numerical ResultsConclusionBibliography
M. Tezer (METU) IAM2017 November 24, 2017 21 / 43
The Physical Problem
Physical laws are:
Conservation of mass+
Conservation of momentum(defining fluid flowincluding Lorentz force
) +Electromagnetic process (Ampere’s and Ohm’s Laws)(relating electric and magnetic fields to the fluid flow)
Electricallyconducting
fluid, viscous,incompressible
B0
Magnetic Field
Electricallyconducting
medium
Pressure
gradient∂p∂z
z
x
y
Figure: MHD flow in a circular pipe
M. Tezer (METU) IAM2017 November 24, 2017 22 / 43
The Physical Problem
MHD flow trough pipes has many practical applications in the design of
cooling systems with liquid metals for nuclear reactors
MHD generators
electromagnetic pumps
MHD flow-meters measuring blood pressure
biomedical instruments using magnetic sources.
M. Tezer (METU) IAM2017 November 24, 2017 23 / 43
Mathematical Model
Governing equations
Physical equations are: Continuity and Navier-Stokes equations + Maxwell’s equations+ Equation of Ohm’s Law
Continuity equation: divV = 0
Navier-Stokes equations: (V.∇)V = −gradp+1
Re∇2V +RhJ×B
Maxwell’s equations: curlE = 0, divB = 0, curlB = J, divE = 0
Ohm’s Law: J = Rm(E + V ×B)
Ω
Ω′
Γ
B0
y
x
Figure: Cross-section of the pipe in 2D
(V.∇)B =1
Rm∇2B + (B.∇)V
V = (0, 0, V (x, y)) : velocityB = (0, B0, B(x, y)) : induced magnetic fieldE = 0 : electric field (absent)J = Rm(V ×B) : electric currentJ×B = Rm(V ×B×B) : Lorentz force
M. Tezer (METU) IAM2017 November 24, 2017 24 / 43
Mathematical Model
For fully developed flow (pipe is long enough to attain fully developed flow,∂V
∂z= 0)
problem is 2-dimensional in the section of the pipe.
Dimensionless equations (Dragos, 1975; Carabineanu et.al., 1995, 2006) are:z-components of Navier-Stokes and Maxwell’s equations (coupled in Ω as)
∇2V +ReRh∂BΩ
∂y= −1
in Ω
∇2BΩ +Rm1∂V
∂y= 0
(1)
∇2BΩ′ = 0 in Ω′ (2)
Boundary conditions (also coupled on Γ)
V = 0BΩ = BΩ′ on Γ
1
Rm1
∂BΩ
∂n=
1
Rm2
∂BΩ′
∂n′.
(3)
Rewrite equations (1) by taking B1 =ReRh
MBΩ and B2 = BΩ′ .
M. Tezer (METU) IAM2017 November 24, 2017 25 / 43
Mathematical Model
∇2V +M∂B1
∂y= −1
in Ω
∇2B1 +M∂V
∂y= 0
(4)
∇2B2 = 0 in Ω′ (5)
On the pipe boundary
V = 0
B1 =M
Rm1B2 on Γ
1
M
∂B1
∂n= − 1
Rm2
∂B2
∂n
(6)
where Hartmann, Reynolds, magnetic Reynolds and magnetic pressure numbers:
M =√ReRhRm1, Re =
L0V0
ν, Rm1 = σfµfL0V0,
Rm2 = σexµexL0Vex, Rh =B2
0
ρfµfV 20
.
M. Tezer (METU) IAM2017 November 24, 2017 26 / 43
Mathematical Model
Coupled equations (4) are transformed with
u1 = V +B1 +1
My, u2 = V −B1 −
1
My (7)
to two decoupled diffusion-advection equations:
Inner problem:
∇2u1 +M∂u1
∂y= 0
in Ω
∇2u2 −M∂u2
∂y= 0
(8)
Outer Problem:∇2B2 = 0 in Ω′ . (9)
The coupled boundary conditions in terms of new variables become
u1 = −u2,∂u2
∂n=∂u1
∂n+ 2
M
Rm2
∂B2
∂n− 2
M
∂y
∂n
B1 =M
Rm1B2,
1
M
∂B1
∂n= − 1
Rm2
∂B2
∂n, u1 =
M
Rm1B2 +
y
M
(10)
M. Tezer (METU) IAM2017 November 24, 2017 27 / 43
Solution Procedure
Inner Problem: Assuming u1 and u2 are assigned on the boundary Γ
Diffusion-advection equations
∇2u1 +M∂u1
∂y= 0
in Ω(11)
∇2u2 −M∂u2
∂y= 0 . (12)
Fundamental solutions for the equations (11) and (12) are given as
g1(x− ξ, y − η) =1
2πe
M2rxK0(
M
2r)
g2(x− ξ, y − η) =1
2πe−
M2ryK0(
M
2r)
(13)
where r = (rx, ry).
M. Tezer (METU) IAM2017 November 24, 2017 28 / 43
Integral equations
Now multiplying (11)-(12) by their fundamental solutions and applying DivergenceTheorem two times
cpu1(ξ, η) =
∫Γ
(g1∂u1
∂n− u1
∂g1
∂n
)dΓ +M
∫Γ
g1u1nydΓ
Similarly,
cpu2(ξ, η) =
∫Γ
(g2∂u2
∂n− u2
∂g2
∂n
)dΓ−M
∫Γ
g2u2nydΓ
where P = (ξ, η) and
∫Ω
δ(x− ξ, y − η)u(x, y)dΩ = cpu(ξ, η).
cP =α(P )
2π
and α(p) is the internal angle.
P
Γ
(x, y)
P = (ξ, η)
α(P ) = π ⇒ cP = 12
on Γα(P ) = 2π ⇒ cP = 1 on Ω/Γ
M. Tezer (METU) IAM2017 November 24, 2017 29 / 43
Integral equations
When
P = (ξ, η) ∈ Γ, u1 = −u2,∂u2
∂n=∂u1
∂n+
2M
Rm2
∂B2
∂n− 2
M
∂y
∂n
1
2u1(ξ, η) =
∫Γ
(g1∂u1
∂n− u1
∂g1
∂n
)dΓ +M
∫Γ
g1u1nydΓ (14)
−1
2u1(ξ, η) =
∫Γ
(g2
(∂u1
∂n+
2M
Rm2
∂B2
∂n− 2
M
∂y
∂n
)− (−u1)
∂g2
∂n
)dΓ
−M∫
Γ(−u1)g2nydΓ
(15)
These are two integral equations for the unknowns u1,∂u1
∂nand
∂B2
∂non Γ. We need
another equation for∂B2
∂non Γ.
M. Tezer (METU) IAM2017 November 24, 2017 30 / 43
Outer Problem
The Laplace equation ∇2B2 = 0 in Ω′ must be transformed to an integral equationdefined on Γ.
This is achieved by considering the outer problem with Neumann boundaryconditions on the pipe wall.
The problem requires the solvability condition∫ ′Ω
∇2B2dΩ′ =
∫Γ
∂B2
∂n′dΓ = 0 .
Consider the exterior of the circle with radius R
Laplace equation becomes
1
r′∂
∂r′
(r′∂B2
∂r′
)+
1
r′2∂B2
∂φ2= 0
r′ is the radial distance
r′ =√x2 + y2
for a point Q = (x, y) outside or onthe boundary Γ.
Ω
Ω′
Γ
x
(ξ, η)
Rr
φ
r′
ψ
ψr′
∇2B2 = 0Q = (x, y)
Q = (x, y)
Here, r′ is different from the radial r distance used before for inner problem.M. Tezer (METU) IAM2017 November 24, 2017 31 / 43
Solution of Outer Problem (Neumann)
Laplace equation will be solved for B2 in Ω′ as ifnormal derivative of B2 is known on Γ.
∇2B2 = 0 in Ω′
∂B2
∂r′= given = f(φ) at r = R
where∂
∂n=
∂
∂r′and
∂
∂n′= − ∂
∂n.
Γ
R
Ω
Ω′
~n
~n′
Existence of solution of this exterior problem is from Divergence theorem (Polyanin(2002)) ∫
Ω′∇2B2︸ ︷︷ ︸
0
dΩ′ =
∫Γ
∂B2
∂n′dΓ = −
∫Γ
f(φ)dΓ = −∫ 2π
0
f(φ)dφ
implies that ∫ 2π
0
f(φ)dφ = 0
which is the solvability condition for Neumann problem.
M. Tezer (METU) IAM2017 November 24, 2017 32 / 43
Solution of Outer Problem (Neumann)
The solution for the outer (exterior) Neumann problem is given as (Carrier and Pearson(1976), Polyanin (2002))
B2(r′, φ) = − R
2π
∫ 2π
0
f(ψ) ln
(r′
2 − 2Rr′ cos (φ− ψ) +R2
r′2
)dψ + C (16)
(solution exists up to an additive constant).On the boundary Γ, r′ = R and
B2(R,φ) = − R
2π
∫ 2π
0
∂B2
∂r′(R,ψ) ln
(2R2 − 2R2 cos (φ− ψ)
R2
)dψ + C (17)
B2(R,φ) = − R
2π
∫ 2π
0
∂B2
∂r′(R,ψ) ln [2(1− cos (φ− ψ))]dψ + C .
Since B2 = (u1 −1
My)Rm1
M, we have
u1(R,φ)− 1
MR sinφ = − MR
2πRm1
∫ 2π
0
∂B2
∂n(R,ψ) ln [2(1− cos (φ− ψ))]dψ + C (18)
M. Tezer (METU) IAM2017 November 24, 2017 33 / 43
Now adding this equation to inner problem integral equations (14)-(15), we have threeintegral equations
1
2u1(ξ, η) =
∫Γ
(g1∂u1
∂n− u1
∂g1
∂n
)dΓ +M
∫Γu1g1nydΓ
1
2u1(ξ, η) = −
∫Γ
(g2
(∂u1
∂n+
2M
Rm2
∂B2
∂n−
2
M
∂y
∂n
)+ u1
∂g2
∂n
)dΓ−M
∫Γu1g2nydΓ
(19)
u1(R,φ) = − MR
2πRm1
∫ 2π
0
∂B2
∂n(R,ψ) ln [2(1− cos (φ− ψ))]dψ + C +
1
MR sinφ
for the unknowns u1,∂u1
∂nand
∂B2
∂non Γ.
The solvability condition∫ 2π
0
∂B2
∂r′(R,φ)dφ = 0
is going to be inserted to Eqn. (19).
Ω
Ω′
Γ1
N
Q
Q(x, y)
r
R
r′
(ξ, η)
Γj
y
xφψ
(ξ, η) is the fixed source point, and Q = (x, y) is thefield point ranging on Γ or Ω′.
M. Tezer (METU) IAM2017 November 24, 2017 34 / 43
Linear System of Equations
Inner Dirichlet-Outer Neumann Problem:
The integral
∫Γ
will be subdivided into N boundary subintegrals
∫Γj
where (ξ, η) and
(x, y) are ranging from 1 to N on Γ as midpoints of these subintervals Γj . On each Γj
unknowns u1,∂u1
∂n,∂B2
∂nare assumed to be constant. Then, the systems to be solved
for N collocating points on Γ will take the form
[H1]u1+ [G1]
∂u1
∂n
= 0
[H2]u1+ [G2]
∂u1
∂n
+ [K2]
∂B2
∂n
=
2
M[G2]
∂y
∂n
(20)
[I]u1+ [K3]
∂B2
∂n
=
R
Msinφ
in which all the matrices are of the size N ×N .Solvability condition is also discretized as∫ 2π
0
∂B2
∂n(R,φ)dφ =
N∑j=1
∂B2
∂n|Γj (φj+1 − φj) = 0 (21)
and added to the third equation.M. Tezer (METU) IAM2017 November 24, 2017 35 / 43
Linear System of Equations
The entries of the matrices are
H1ij =
1
2+
∫Γj
(−Mg1ny +
∂g1
∂n
)dΓ, i, j = 1, · · · , N
H2ij = −1
2+
∫Γj
(−Mg2ny −
∂g2
∂n
)dΓ, Γj subintervals on Γ.
G1ij = −
∫Γj
g1dΓ, G2ij = −
∫Γj
g2dΓ, K2ij = −
∫Γj
2M
Rm2g2dΓ
K3ij =
∫Γj
MR
2πRm1ln[2(1− cos (φ− ψ))]dψ .
M. Tezer (METU) IAM2017 November 24, 2017 36 / 43
Numerical Results
Inner Dirichlet-Outer Neumann Problem
V
B
Flow and induced magnetic fields are visualized interms of equivelocity and current lines forRm1 = Rm2 = 1, Re = 1, Rh = 10.
Flow attains its maximum value at the center ofthe pipe and reduces to V (x, y) = 0 on the pipewall.
Inner and outer induced magnetic fields continueon the pipe wall obeying to the boundary
condition B1 =M
Rm1B2.
M. Tezer (METU) IAM2017 November 24, 2017 37 / 43
Inner Dirichlet-Outer Neumann Problem
Effect of Rm1(= 10, 50, 100) on the velocity and induced magnetic field whenRe = Rm2 = 1 and Rh = 10 (Outer Neumann problem).
Rm1 = 10 (M = 10) Rm1 = 50 (M ≈ 22.36)Rm1 = 100 (M ≈ 31.62)
B
V
With an increase in Rm1
fluid becomes stagnant atthe center of the pipe.
Velocity magnitudedecreases as Rm1, thus Mincreases, which is thewell-known retarding effectof magnetic field onvelocity.
Boundary layers are formednear the interior pipe walltending to be thinner in theparts parallel to the appliedmagnetic field.
As Rm1 increases, the inner induced magnetic field decreases in magnitude, which is theflattening tendency phenomena, and behaves as current in a pipe with insulated wall.
The current lines of both inner and outer induced magnetic field smoothly connect on thepipe wall.
M. Tezer (METU) IAM2017 November 24, 2017 38 / 43
Inner Dirichlet-Outer Neumann Problem
Effect of Re(= 10, 50, 100) on the velocity and induced magnetic field whenRm1 = Rh = 10 and Rm2 = 1 (Outer Neumann problem).
Re = 10 (M ≈ 31.62) Re = 50 (M ≈ 70.71) Re = 100 (M = 100)
B
V
Magnitudes of thevelocity and the inducedmagnetic fields drop withan increase in Re. This isan expected flatteningtendency of MHD flowwhen either B0 (intensityof applied magnetic field)or σ (electricalconductivity of the fluid)is increased. These are inthe definition of Rm andRh, respectively.
Continuation of induced magnetic fields on the pipe wall is more pronounced when Re islarge.
Velocity magnitude drops as Re increases.
As Re increases velocity forms boundary layers in terms of two eddies along the sidesparallel to applied magnetic field.
M. Tezer (METU) IAM2017 November 24, 2017 39 / 43
Inner Dirichlet-Outer Neumann Problem
Effect of Rh(= 5, 10, 20) on the velocity and induced magnetic field when Re = 1,Rm1 = 10 and Rm2 = 1 (Outer Neumann problem).
Rh = 5 (M ≈ 7.07) Rh = 10 (M = 10) Rh = 20 (M ≈ 14.14)
B
V
Magnitudes of the velocity and the induced magnetic fields drop with an increase in Rh.
Velocity magnitude drops as Rh increases.
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Conclusion
The MHD problem is formulated as an outer Neumann or Dirichlet problem coupledwith inner Dirichlet problem in a circular region.
Inner advection-diffusion equations are transferred to boundary integral equations byusing fundamental solutions and Divergence theorem. Solutions exist and are unique.
In the external Neumann problem, the solution is given with Poisson’ s integral andon the boundary is coupled to the solutions of inner problem in terms of normalderivative of external induced magnetic field. Solution exists up to an additiveconstant in Neumann problem.
Solution of both inner and outer problems are obtained numerically by solving3N × 3N system using collocating points on the boundary.
Then, the solution of the whole problem is extended to the interior and exterior ofthe circular region.
Flattening tendency of induced magnetic field is observed as Hartmann numberM =
√RmRhRe is increased which is the well-known behavior of MHD flow.
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Bibliography
G.F. Carrier and C.E. Pearson, Partial Differential Equations: Theory andTechnique, Academic Press, 1976.
D.A. Polyanin, Handbook of Linear PDE for Engineers and Scientists, Chapman andHall/CRC, 2002.
A. Carabineanu, A. Dinu, I. Oprea, The application of BEM to MHD duct flow,ZAMP , 46 (1995), 971-981.
L. Dragos, Magnetofluid Dynamics, Abacus Press, 1975.M. Tezer-Sezgin and C. Bozkaya. Boundary element method solution ofmagnetohydrodynamic flow in a rectangular duct with conducting walls parallel toapplied magnetic field. Computational Mechanics, 41(6):769775, 2007.
M. Tezer-Sezgin and S. Han Aydın. BEM solution of MHD flow in a pipe coupledwith magnetic induction of exterior region. Computing, 95(1):751770, 2013.
Hasan Gunes. Numerical Methods in Fluid Flow and Heat Transfer.http://atlas.cc.itu.edu.tr/ guneshasa.
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Thanks for your attendance.
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