Transcript Provedení, principy činnosti a základy výpočtu pro výměníky tepla

EXM6
Experimental methods E181101
Material testing
constitutive
equations
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010
Constitutive equations
EXM6
Constitutive equations represent description of material properties
Kinematics (deformation) – stress (dynamic response to deformation)
kinematics is described by deformation of a body In case of solids
kinematics describes relative motion (rate af deformation) In case of
fluids
Deformations and internal stresses are expressed as tensors in 3D case.
Example: stress tensor describes distribution of internal stresses at an arbitrary cross section
y
y
s
s
xy
s
xz
y
s
yy
s
xx
s
x
yz
s
yx
x
s
zz
z
z
zy
s
x
zx
s ij
z
Index of plane
(cross section)
index of force component
(force acting upon the cross section i)
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Stress in solids/fluids
s ij
Tensor of stresses is formally the same in solids and fluids (in both
cases this tensor expresses forces acting to an arbitrary oriented
plane at a point x,y,z) , however physical nature of these forces is
different.
Solids – intermolecular forces (of electrical nature)
Fluids – stresses are caused by diffusional transfer of molecules
(momentum flux) between layers of fluid with different velocities
Total stress

= pressure

+
dynamic stress

s   p  
Unit tensor
Viscous stresses affected by fluid
flow. Stress is in fact momentum flux
due to molecular diffusion
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Viscous Fluids (kinematics)
Delvaux
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Viscous Fluids (kinematics)


1 
Rate of deformation   (u  (u )T )
2
u
j
Gradient of velocity is  u 
i j
tensor with components
xi
(in words: rate of deformation is
symmetric part of gradient of velocity)
Special kinematics: Simple shear flow ux(y) (only one nonzero component of velocity, dependent
on only one variable). Example: parallel plates, one is fixed, the second moving with velocity U
U=ux(H)
y
x
The only nonzero component of deformation rate tensor in case of simple shear flow
1
1 u x 1
 xy  ( xu y   yu x ) 
 
2
2 y 2
 is called shear rate
EXM6
Viscous Fluids (rheology)
Constitutive equations expressed for special case of simple shear flows
Newtonian fluid
  
Model with one parameter – dynamic viscosity [Pa.s]
Power law fluid
  K n
Model has two parameters K-consistency, n-power law index.
EXM6
Viscous Fluids (rheology)
General formulation for fully three-dimensional velocity field

  2 ( II )

Viscosity  is constant in Newtonian fluids and depends upon second invariant of
deformation[1] rate (more specifically upon three scalar invariants I, II, III of this
deformation rate tensor, but usually only the second invariant II is considered because the
first invariant I is zero for incompressible liquids). General definition of second invariant II
  3 3
II   :     ij  ji
i 1 j 1
(double dot product, give scalar value as a result)
The second invariant of rate of deformation tensor can be expressed easily in
simple shear flows
1 u
1
II  22xy  ( x ) 2   2
2 y
2
Power law fluid
[1]
  K ( 2 II ) n 1
Invariant is a scalar value evaluated from 9 components of a tensor, and this value is independent of the change (e.g. rotation) of coordinate system
(mention the fact that the rotation changes all 9 components of tensor! but invariant remains). Therefore invariant is an objective characteristics of
tensor, describing for example measure of deformation rate. It can be proved that the tensor of second order has 3 independent invariants.
EXM6
Viscous Fluids (rheology)
More complicated constitutive equations exist for fluids exhibiting
yield stress (fluid flows only if stress exceeds a threshold),
thixotropic fluids (viscosity depends upon the whole deformation history)
viscoelastic fluids (exhibiting recovery of strains and relaxation of stresses).
Oscillating rheometer: sinusoidaly applied stress and measured strain (not rate of strain!)
Hookean solid-stress is in phase with
strain (phase shift =0)
Viscoelastic material – phase shift
0<<90
Viscous liquid- zero stress corresponds
to zero strain rate (maximum ) =900
Examples of
Newtonian fluids are water, air, oils.
Power law, and viscoelastic fluids are polymer melts, foods.
Thixotropic fluids are paints and plasters.
Yield stress exhibit for example tooth paste, ketchup, youghurt.
EXM6
Rheograms (shear rate-shear stress)
n=1 Newtonian fluid, n<1 pseudoplastic fluids (n is power law index)
1
 [ Pa ] Shear stress
0.81
0.71
0.61
n=0.5
0.1
n=0.5
0.51
0.41
n=0.8
n=1.5
n=0.8
n=1.5
0.31
n=1
n=1
0.21
Shear rate
0.01
0.01
0.1
 [1/ s]
1
0.11
0.01
0.01
 [1/ s]
0.11
0.21
0.31
0.41
0.51
0.61
EXM6
DMA Dynamic Material Analysis and Oscilograms
   0 ( ) cos(t )
   0 ( ) cos(t   )
0
cos 
0

E ' '  0 sin 
0
E' 
storage modulus
loss modulus
polyoxymethylene
Viscous properties E’’
Elastic properties E’
EXM6
Viscoelastic effects
Weissenberg effect (material climbing up on the rotating rod)
Barus effect (die swell)
Kaye effect
EXM6
Viscoelastic models
Oldroyd B model
Extra stress S
Deformation rate
Upper convective derivative
EXM6
Rheometers
Rheometry (identification of constitutive models).
-Rotational
rheometers use
different configurations of cylinders, plates,Rotating
cylinder
and cones. Rheograms are evaluated from
measured torque (stress) and frequency of
rotation (shear rate).
-Capillary rheometers evaluate
rheological equations from experimentally
determined relationship between flowrate
and pressure drop. Theory of capillary
viscometers, Rabinowitch equation, Bagley
correction.
Plate-plate, or
cone-plate
EXM6
Capillary rheometer
1–glass cylinder, 2-metallic piston, 3-pressure transducer Kulite, 4tested liquid, 5-plastic holder of needle, 6-needle, 7-calibrated
resistor (electric current needle-tank), 8-calibrated resistor (current
flowing in tank), 9-AC source (3-30V), 10-SS source for pressure
transducer (10V), 11-A/D converter, 12-procesor, 13-metallic head,
14-push bar, 15-scale of volume
EXM6
Capillary rheometer
Example: Relationship between flowrate and pressure drop for power law
fluid
p dp
3n  1 n V n

 2K (
) 3n 1
L dx
n
R
Consistency variables
R p
w 
2L
4V
 3
R
Model parameters K,n are evaluated from diagram of consistency variables
3n  1 n n
w  K(
) 
4n
EXM6
Capillary rheometer
Ptotal  Pres  Pe  Pcap
EXM6
Elastic solids
Lempická
EXM6
Elastic solids
Constitutive equations are usually designed in a different way for different
materials: one class is represented by metals, crystals,… where arrays of atoms
held together by interatomic forces (elastic stretches can be of only few
percents). The second class are polymeric materials (biomaterials) characterized
by complicated 3D networks of long-chain macromolecules with freely rotating
links – interlocking is only at few places (cross-links). In this case the stretches
can be much greater (of the order of tens or hundreds percents) and their
behavior is highly nonlinear.
"Dogbone" sample
EXM6
Elastic solids Deformation tensor
transforms a vector of a material segment from reference configuration to loaded configuration. Special
case - thick wall cylinder
z
loaded configuration
x( , z , r )
t
R
reference configuration
r
X (, Z , R)  x( , z , r )
r
X (, Z , R )
t r and z are principal stretches (stretches in the principal directions). There are always three principal
direction characterized by the fact that a material segment is not rotated, but only extended (by the stretch
ratio). In this specific case and when the pipe is loaded only by inner pressure and by axial force, there is no
twist and the principal directions are identical with directions of axis of cylindrical coordinate system. In this
case the deformation gradient has simple diagonal form
  t
F  0
0

0
z
0
0  r / R 0
0 
 

0  0 l/L
0 
r   0
0 h / H 
EXM6
Elastic solids Cauchy Green tensor
Disadvantage of deformation gradient F - it includes a rigid body rotation. And this rotation cannot effect
the stress state (rotation is not a deformation). The rotation is excluded in the right Cauchy Green tensor
C defined as
  
C  FT  F
2


  t
C  0

 0
0
z 2
0
0 

0 

r 2 
4



 t
C2   0

 0
0
z 4
0
0 

0 

r 4 
Deformed state can be expressed in terms of Cauchy Green tensor. Each tensor of the second order
can be characterized by three scalars independent of coordinate system (mention the fact that if you
change coordinate system all matrices F,C will be changed). The first two invariants (they characterize
“magnitudes” of tensor) are defined as


I c  trC  t2  2z  2r
2
1 2
II c  ( I c  trC )  t2 2z  2r 2z  2r 2z
2
Material of blood vessel walls can be considered incompressible, therefore the volume of a loaded part
is the same as the volume in the unloaded reference state. Ratio of volumes can be expressed in terms
Vloaded
of stretches
Vreference
 t  r  z  1
(unit cube is transformed to the brick having sides t r z)
Therefore only two stretches are independent and invariants of C-tensor can be expressed only in terms of
these two independent (and easily measurable) stretches
I c  t2  2z 
1
t2 2z
II c  t2 2z 
1
t2

1
2z
EXM6
Elastic solids Mooney Rivlin model
Using invariants it is possible to suggest several different models defining energy of deformation W
(energy related to unit volume – this energy has unit of stress, Pa)
J
m
N
m3
N
m3

m2
 Pa
Example: Mooney Rivlin model of hyperelastic material
W  c1 ( I c  3)  c2 ( II c  3).
W (t ,  z )  c1 (t2  2z 
1

2
t
2
z
 3)  c 2 (t2 2z 
1

2
t

1

2
z
 3)
Remark: for an unloaded sample are all stretches 1 (t =r = z=1) and Ic=3, IIc=3, therefore deformation
energy is zero (as it should be).
Stresses are partial derivatives of deformation energy W with respect stretches (please believe it wihout proof)
W (t , z )
1
1
2
2 2
s t  t
 2(c1 (t  2 2 )  c2 (t z  2 ))
t
t z
t
W (t , z )
1
1
s z  z
 2(c1 (2z  2 2 )  c2 (t22z  2 ))
z
t z
z
st
sz
These equations represent constitutive equation, model calculating stresses for arbitrary stretches and for
given coefficients c1, c2. At unloaded state with unit stretches, the stresses are zero (they represent only
elastic stresses and an arbitrary isotropic hydrostatic pressure can be added giving total stresses).
EXM6
Evaluation of stretches and stresses
Only two stretches is to be evaluated from measured outer radius after and before loading ro, Ro, from
initial wall thickness H, and lengths of sample l after and L before loading.
2ro  h
2ro  H /( z t )
r
t  

R 2 Ro  H
2 Ro  H
l
z 
L
This is quadratic equation
ro  ro2  H / z (2 Ro  H )
t 
2 Ro  H
Therefore it is sufficient to determine Ro,H,L before measurement and only outer radius ro and length l
after loading, so that the kinematics of deformation will be fully described.
Corresponding stresses can be derived from balance of forces acting upon annular and transversal cross
section of pipe
s texp 
s
exp
z

r 1
r 1
pr p(ro  ri ) p(2ro  h)


 p( o  )  p( o t z  )
h
2h
2h
h 2
H
2
s texp
2

Gz
H (2 Ro  H )
G-force
st
r0
h
sz
p
EXM6
Elastic solids instruments
Uniaxial testers
Sample in form af a rod, stripe, clamped at ends and
stretched
Static test
Creep test
Relaxation test
EXM6
Elastic solids instruments
Biaxial testers
Sample in form of a plate, clamped at 4 sides to
actuators and stretched
Anisotropy
Homogeneous inflation
EXM6
Elastic solids instruments
Inflation tests
Tubular samples inflated by inner
overpressure.
CCD cameras of
correlation system
Q-450
Pressure
transducer
Internal pressure load
Axial load
Torsion
Pressurized
sample (latex tube)
Laser scanner
Axial loading
(weight)
Confocal probe