Conservation of mass, momentum
and energy in fluid dynamics
Classical mechanics (Newtonian)
Consider a particle of mass m subject to some force F(t,r,v) which possible depends on time, position and velocity of the particle. The position of the particle in space at time t is described by vector r(t) with respect to selected frame of reference. The goal of classical mechanics is to determine this vector as a function of time. This is done by applying Newton’s second law and solving a set of ordinary differential equations for the components of the vector r(t) = [x(t), y(t), z(t)]. In vector form it reads
This is a set of ordinary differential equations (ODEs):
2 2
m d
dt r F
2 2
2 2
2 2
1 , , , , , ,
1 , , , , , ,
1 , , , , , ,
x
y
z
d x dx dy dz
F t x y z
dt m dt dt dt
d y dx dy dz
F t x y z
dt m dt dt dt
d z dx dy dz
F t x y z
dt m dt dt dt
Classical mechanics (Newtonian)
The equations of motion are supplemented with the initial conditions (for the position and velocity of the point mass at the initial time t = t0)
that is
0 0 0 0
( ) t , ( ) t ,
r r
0 0 0 0 0 0
0 0 0 0 0 0
( ) , ( ) , ( ) ,
( )
x, ( )
y, ( )
z. x t x y t y z t z
x t v y t v z t v
Classical mechanics (Newtonian)
Example (A simple gravity pendulum). Newton’s equation of motion (ma = F) for a small material particle attached to a rod with negligible mass is best expressed in the polar coordinate with (t)=angle measured with respect to a downward vertical direction. The linear acceleration along the path a = l hence
Wprowadzając w2=g/l otrzymujemy
Zamieniamy na układ wprowadzając:
sin .
ml
g
2
sin 0.
w
1 2
2
2 1
,
sin . y y
y w y
1
,
2.
y
y
A simple implementation in MATLAB and a plot presenting dependence of the angle on time t.
For readability the ange is given in the plot in degrees. The solution is periodic. For small amplitudes it is almast the sinus function. But in general it is dufferent function. Moreover, the period of the motion depend on the amplitude.
0 5 10 15 20 25 30 35 40
-15 -10 -5 0 5 10 15
czas
kąt
Classical mechanics (Newtonian)
Example: The one-body problem: determine the motion of a point mass (m) moving in the gravity force field produces by a stationary mass (M).
This is a set of the following equations in 3D:
In fact, this system can be reduced to 2D (planar) problem, because the trajectory (x(t),y(t),z(t)) lies in the plane defined by the initial position and velocity.
Let the mass M is placed in the center of a coordinate system XYZ. Then the equation of motion for the point mass m is
2
2 2
ˆ
d mM
m G
dt r r
r
2
2 2 2 2 3/ 2
2
2 2 2 2 3/ 2
2
2 2 2 2 3/ 2
( ) ,
( ) ,
( ) .
d x x
dt GM x y z
d y y
dt GM x y z
d z x
dt GM x y z
Classical mechanics (Newtonian)
Example (continuation)
Possible orbits, due to universal gravity, of a small mass around a single large mass.
2
2 2 2 2 3/ 2
2
2 2 2 2 3/ 2
2
2 2 2 2 3/ 2
( ) ,
( ) ,
( ) .
d x x
dt GM x y z
d y y
dt GM x y z
d z x
dt GM x y z
Classical mechanics (Newtonian)
Energy in classical mechanics appears as follows: when the force is conservative (rot F = 0), then it can be expressed as the gradient of a scalar function V(t,r)
This function is called the potential of the force or simply the potential energy. It is also denoted by Ep or Epot. Then the mechanical energy Emech is defined as the sum of the kinetic and potential energy
Important property of E is that this quantity is preserved: for any motion r(t) in the force field F it holds that
We say that the mechanical energy is conserved.
.
V F
1
22 .
mech kin pot
E E E m V
( ( )) .
E
mechr t const
Classical mechanics (Newtonian)
Example: Motion of a point mass near the Earth surface.
In the vicinity of the Earth surface, the graviation is almost constant. It dependancy on the hight may be neglected for typical experiments of every day (say up to sever kilometers).
Such homogeneous field gives rise to the following potential energy
Hence
The notion of such conserved quantity helps solving many problems. For example, what is the maximum height that a stone thrown vertically upward will attain if the initial velocity is v0?
1
2mech kin pot
2
E E E m mgh const
p
,
E V mgh
2 2
0 max
2
2 0
0 max max
1 1
0 0
2 2
1 .
2 2
mech kin pot
E E E const m mg m mg h
m mg h h
g
Classical mechanics (Newtonian)
Example: In the motion of a point mass (m) in the gravitational field of a big stationary planet/star (M) the potential energy of the mass m is
hence
In this particular case the mass m disappears as we can divide by it to obtain
Even if we do not know the solution (position r(t) and velocity v(t)) we do know that the above quantity must be constant with respect to time during the motion.
2
2
1
mech kin pot
2
E E E m G mM const
r
p
,
E G Mm
r
2
2
2G M const r
Classical mechanics (Newtonian)
If a constant force is applied to a particle and a displacement is r , the work is defined as
In general case when the force may vary in space the work is defined by the line integral
where AB is a path along which the mass is moved from point A to B.
Note that in general this definition implies that the work may depend on the path.
This is not true if the force is conservative. In the case of conservative force we have
W F r
AB
W d
F r
( ) ( )
AB
B A
W d E E
F r r r
Fluid mechanics (concept of a continuum)
Materials (solids, liquids and gases) are composed of molecules separated by empty space. But the continuum model as a mathematical concept assumes that material exists as a continuous entity. It means that the matter in the body is continuously distributed and fills the entire region of space it occupies. A continuum body can, for example, be infinitely sub-divided into smaller and smaller elements preserving properties of the bulk material. It also means that two points in such body may be close at arbitrary small distance. Due to these assumption we can consider physical quantities (density, pressure, velocity, forces etc.) as defined at every point as continuous functions of space position and that space derivatives can be also defined.
Example (density). A function r(x,y,z,t) is defined at every point (x,y,z) and is continuous. The meaning of this function
is as follows: (x,y,z)
W
is the mass contained in the region W.
( , ) ( , , , ) ( , )
W W
m W t r x y z t dxdydz r r t dV
Fluid mechanics
Let
be a region in two dimensional (2D) or three dimensional (3D) space filled with a fluid (gas or liquid). Our object is to describe the motion of such a fluid.Let r ∈
be a point and consider the particle of fluid moving through r at time t.In a standard Euclidean coordinates system in space, we write r = (x, y, z). This particle traverses a well-defined trajectory r(t)=(x(t),y(t),z(t). Let v(r, t) denote the velocity of the particle of fluid that is moving through r at time t. Thus, for each fixed time, v is a vector field on
, as in figure below. We call v the (spatial) velocity field of the fluid.We will use the notion of some region W which all particles (points) move according to the velocity field. The symbol Wt denots the set of points of the initial region W which were displaced to new locations after time t. Thus we have W0 = W. The obvious property of such defined set Wt is the its mass is conserved:
This formula is just a statement of the mass conservation law. It is also called the continuity equation or mass balance equation.
Let us notice that it can also be written as
Fluid mechanics (mass conservation)
0
0
mass of mass of In short: ( ) ( )
t
t
W W
m W m W
( )
0
t
t
t
W
W
m W dV const
d dV dt
r
r
The form of mass conservation law from the previous slide
may be converted to a more useful differential form. It can be done in two ways:
(1) from the above integral form by applying Reynold’s transport theorem; (2) By direct application of Gauss’ divergence theorem. The final form is
Fluid mechanics (mass conservation)
Some textbooks use the notation for the divergence operator: u = div u.
or 0
t t
W W
dV const d dV
r dt r
div ( ) 0 t
r r
v
Fluid mechanics (material derivative)
The material derivative of a scalar function f(t,r) in the region filled with a moving fluid with the velocity field v(t,r) is defined as
It takes into account the fact that the fluid is moving and that the positions of fluid particles change with time. Thus, if r(t) = [x(t), y(t), z(t)] describes the movement of a fluid particle (the path followed by the fluid particle), then the change of any quantity f along this path can be computed as follows
Hence we can write for the change of function f(t,r) along the path r(t):
: ( )
Df f
Dt t f
3 1
( , ( )) ( , ( ))
( , ( ), ( ), ( )) ( , ( ))
( ) ( ) ( )
( , ( )) ( , ( )) ( , ( )) ( )
i
i i
t t t t
dx
d f f
f t x t y t z t t t
dt t x dt
f dx t f dy t f dz t Df
t t t t t t f
x dt y dt z dt t Dt
r r
r
r r r
( , ( )) ( , ( )).
d Df
f t t t t
dt r Dt r
Fluid mechanics (ideal fluid)
For any continuum, forces acting on a piece of material are of two types:
• External, or body, forces such as gravity, intertial (pseudoforce) , a magnetic field, which exert a force per unit volume on the continuum
• forces of stress, whereby the piece of material is acted on by forces across its surface by the rest of the continuum.
An ideal fluid has the following property:
For any motion of the fluid there is a scalar function p(t,r) called the pressure such that if S is a surface in the fluid with a normal vector n, the force of stress exerted across the surface S per unit area at r S at time t is p(t,∈ r)n, i.e.
force across per unit area = ( , ) . S p t r n
Fluid mechanics (ideal fluid – momentum equation)
Intuitively, the absence of tangential forces implies that there is no way for rotation to start in a fluid, nor, if it is there at the beginning, to stop. This may be too much simplification and we can detect physical trouble for ideal fluids because of the abundance of rotation in real fluids (near the oars of a rowboat, in tornadoes, etc.).
If W is a region in the fluid at a particular instant of time t, the total force exerted on the fluid inside W by means of stress on its boundary is
If e is any fixed vector in space, then the divergence theorem gives
hence
W
W
p dV S
={force on W through the surface S} ( , ) .
W
W
p t dA
S r n
div( ) ( )
W
W W W W
p dA p dV p dV p dV
e S e n e e e
Fluid mechanics (ideal fluid – momentum equation)
If b(t,r) denotes the given body force per unit mass, then the total body force is
Thus the total force acting on the fluid domain W in the case of ideal fluid is
Hence
,
={body force on W } .
b W
W
r dV
F b
,
( ) ( )
b W W
W W W
p dV r dV p r dV
F + S b b
( )
W
W
p r dV
F b
force per unit volume b p r
Fluid mechanics (ideal fluid)
Now we want to apply Newton’s second law for the fluid element W force = mass × acceleration
Acceleration: let r(t) is the motion of the fluid particle. From the very definition of the velocity we get
The acceleration a=d2r/dt2 thus
But, by the chain rule one can show
( ) ( , ( ))
d t t t
dt r v r
2
( ) d
2( ) d ( , ( ))
t t t t
dt dt
r v
a r
( ) d ( , ( )) ( )
t t t
dt t
v v v v
a r
Fluid mechanics (ideal fluid – Euler’s equations)
Let us summarize:
From the above relations we get
The above equation is called Euler’s equation for ideal fluid. In fact it is the momentum balance for the ideal fluid.
t p
r r
v v v b
force = mass acceleration force per unit volume
acceleration
p t
r
v v v
b
a
Fluid mechanics (NavierStokes equations)
Now we consider a more general fluid than the ideal one. Real fluids possess the phenomenon of viscosity. The microscopic basis of the effect of viscosity is connected with the transfer of momentum between layers of moving fluid.
If the forces are only normal to S, there will be no transfer of momentum between the fluid volumes denoted by B1 and B2. However, in reality faster molecules from above (B2) will diffuse across S and impart momentum to the fluid (B1), and, likewise, slower molecules from below S will diffuse across S to slow down the fluid above S. This is how the viscosity arises.
In the previous slide the viscosity was explained in the qualitative way. Here is how we can measure it.
The dynamic viscosity (sometimes ) of a fluid describes its resistance to shearing flows. In the picture (the speed of the top plate is parallel to the x axis.
We see that an external force F is required in order to keep the top plate moving at constant speed v.
Quantifying the dynamic viscosity
F A A A
y
F
A y
Instead of assuming that (ideal fluid)
Fluid mechanics (NavierStokes equations)
where n is the normal vector to S, we now assume that (viscous fluid)
where is the stress tensor which describes the shear forces (tensions) in the fluid. It is called the Cauchy stress tensor. In a fixed coordinate system the tensor stress is just a matrix 3 x 3. Hence the expression n should be understood as the matrix multiplication (n is written here as a column).
From this formulation we see that shear force term is linear with respect to n. It is not obvious! It can be proved under a general assumption (continuous dependence on n) and using balance of momentum. This result is called Cauchy’s theorem.
force on per unit area S r n p ( , ) , t
( )
force on per unit area ( , ) ( , ) ,
normal forces possible shear tangential forces
S p t t
r n r n
As in the case of the ideal fluid we can formulate for viscous fluid Newton’s second law for the moving fluid particle Wt
Fluid mechanics (NavierStokes equations)
where dV indicates the volume integral and dA is the surface integral. The internal forces acting through surface are now composed of two terms: normal (represented by scalar function p – the pressure) and shear/tangential (represented by matrix function – the stress tensor). It can be also proved that the stress tensor is symmetric.
internal forces external (body) forces
( )
t t t
W W W
d dV p dA dV
dt r r
v b
n n
To obtain a more specific equation for the motion of the fluid we have to analyze in more detail the stress tensor . The following assumptions are physically sound:
1. depends linearly on the velocity gradients v.
2. is invariant under rotations in space.
3. is symmetric (as was pointed out earlier this can be deduce as a consequence of balance of angular momentum.
The velocity gradient is just a matrix of first order space derivatives of all components of the velocity v=[vx, vy, vz]
Fluid mechanics (NavierStokes equations)
v v v
v v v
v v v
x x x
y y y
z z z
x y z
x y z
x y z
v
Under the assumptions 1,2,3 (previous slide) it can be shown that the stress tensor has the following form
Fluid mechanics (NavierStokes equations)
Where and are non-negative parameters characterazing the viscous fluid and
(div ) + 2
v I D
v 1 v v 1 v v
2 2
v v v
v v
1 1
2 2
v v v v v
1 1
2 2
( ) ( )
D = 1 ( ) ( ) ( )
2
( ) ( )
x x y x z
y y y
x z
x z y z z
x y x z x
T
y x y z y
z x z y z
v v
If we now employ the transport theorem and the divergence theorem to the balance of momentum with the stress tensor = (div v)I + 2D after some algebra we will obtain the famous Navier-Stokes equations (NS):
Fluid mechanics (NavierStokes equations)
This form is for a general fluid – at that stage we do not assume incompressibility, so the unknowns are: v(t,r), r(t,r), p(t,r). Thus we have five unknown functions, but the NS system gives only three equations. One more equation is obvious – it is the mass balance equation
The fifth equation (for compressible fluid) comes usually in the form of a constitutive relation between pressure, density and temperature:
( ) p ( ) (div )
r t
v v v v v
div ( ) 0 t
r r
v
( , ). p T
r r
Assumption: the viscous fluid is incompressible, i.e. r = const. The mass balanace equation gives now
Fluid mechanics
(incompressible NavierStokes equations)
and the term ( + )(div v) vanishes from the NS system. Thus we obtain the NavierStokes equations for incompressible fluid (usually written with the kinematic viscosity = /r >0):
( )
div = 0 t p
r
v v v v
v
div ( ) 0 0 div ( ) 0
0div = 0.
t
r r r
r
v v v
We will consider only two basic boundary conditions. Boundary is denoted by .
Fluid mechanics boundary conditions
1)
The expression means that the normal component of the velocity on the boundary is zero. It is rather obvious statement that the wall is impenetrable.
The tangential component may be non-zero (fluid can slip along the wall. This condition is consistent with the inviscid fluid and is used with Euler’sequation.
2)
This condition says that the velocity on the boundary is completely zero. The tangential component must vanish, hence this is called non-slip boundary condition. It is consistent with viscous fluid ( > 0) and is used in the Navier- Stokes equations.
0 on
v n
0 on
v
Fluid dynamics introduction:
the Reynolds number
The physical nature of a fluid flow may be characterized by several dimensionless numbers. One of them is the Reynolds number (Re). It primarily measures the effect of viscosity on the flow.
For a given problem, let x
cand u
cbe a characteristic length and characteristic velocity, respectively. Then, the characteristic time is t
c=u
c/x
c. When we introduce the following dimensionless quantities
u’= u/u
c, x’=x/x
c, t’=t/t
c,
and express the NS equations in these variable we obtain:
where
( ) 1 ( , ),
p Re t
t
u u u u f x
Re x u
c cx u
c c, / . v v
r r
Fluid dynamics introduction:
the Reynolds number
The flows with the same Reynolds number in regions of the same shape related by scaling are called similar. This notion is very helpful in designing of experimental models. For example, it allows to perform the initial tests on a scaled-down (smaller) version of the model. The results will be relevant to the actual (full-scale) flow as long as the Reynolds number for the flow in our experiment matches that of the actual flow.
For flow in a pipe of diameter d, experimental observations show that for
"fully developed" flow, laminar flow occurs when Re < 2300 and turbulent flow occurs when Re > 2900
( ) 1 ( , ),
Re
Re
c c c c, / .
p t
t
x u x u v v
r r
u u u u f x
Fluid dynamics example of 2D flow
a=2[m]
b=0.4[m]
0.2a
center: (x,y)=(a/2,b/2) 3b radius: r=b/4
Fluid dynamics example of 2D flow
Fluid mechanics (energy conservation)
For any moving fluid region of volume Wt we can define its energy as
where r is the density, v is the velocity field, and e is the density of internal energy. We identify here
Now we are interested in how the quantity Wt changes with time when the fluid particle moves in the system.
the kinetic energy of the moving volume Wt the internal energy of the moving volume Wt
EWt
2
2 ,
t
t
W
W
E e dV
r
2
t 2
kin W
E r dV
t
int W
E
re dV Fluid mechanics (energy)
If we want to move the time derivative under the symbol of integral we must be cautious because the region of integration Wt also depend on time. It is not stationary!
The rate of change with time can be rewritten
where we applied the transport theorem (Reynolds)
valid for any quantity f(t,r)=f(t,x,y,z) that may depend on time and location. And moving region Wt. Here, Df/Dt is defined earlier the material derivative.
Reynold’s transport theorem
2 2
2 2
t t
W W
d D
e dV e dV
dt Dt
r r
,
t t
W W
d Df
f dV dV
dt r r Dt
Fluid mechanics (energy conservation)
The previous slide described a defintion of energy of a moving element and how its change can be written with derivative under the integral. But these were merely mathematical manipulations of purely kinematical nature.
Now we have to add some constitutive relations to bind the energy of the moving element Wt with other processes present in the system. This leads to the following general form of energy balance in the integral form:
where:
rate of work by (external) mass forces rate of work by (internal) pressure forces
flux of the ”heat” energy
energy gain/loss due the internal sources/sinks (e.g. reactions)
2
( ) ,
t
2
t t t tQ q
W W W W W
d e dV dV dA dA R dV
dt
r r r
F p J n
( )
t
t
t
t
W
W
Q W
q W
dV dA
dA R dV
r
r
F
J
pn
Fluid mechanics (energy conservation)
The integral form:
can be transformed to differential if we apply some formulas of vector analysis.
where the basic tool – Gauss’ divergence theorem was used:
Here u stand for any vector field in space (3D or 2D).
2
( ) ,
t
2
t t t tQ q
W W W W W
d e dV dV dA dA R dV
dt
r r r
F p J n
div( ) ,
( ) div ,
t t
t t
W W
Q Q
W W
dA dV
dA dV
J
J
p
n
div .
t t
W W
dV dA
u
u nFluid mechanics (energy conservation)
The previous slide contained the expression
where a major assumption concerning the nature of the internal stress p was made.
First, p=p(t,r,n) which means that it not only depend on time and location – but also on the normal to the surface!
Second, it can be proved under some broad assumptions that it is a linear function of n, hence:
where is called the stress tensor. It is symmetric tensor that can be expressed as a matrix
div( ) ,
t t
W W
dA dV
p
p n
xx xy xz
xy yy yz
xz yz zz
p p p
p p p
p p p
Fluid mechanics (energy conservation)
Now we write all integrals on the right hand side as volume integrals
and use the transport theorem
Because this integral equality holds for any domain Wt we obtain following general form of energy conservation equation
2
div( ) div ,
t
2
t t t tQ q
W W W W W
d e dV dV dV dV R dV
dt
r r r
F
J
,
V V
d Df
f dV dV
dt
r
r
Dt2
div( ) div .
t
2
t t t tQ q
W W W W W
D e dV dV dV dV R dV
Dt
r r r
F
J
2
div( ) div .
2
Q qD e R
Dt
r r r
F
JThe heat flux Jq in direction Ox. A domain is in the form of a parallelepiped with edges parallel to the coordinate axes.
cross-sectional are perpendicular to the flux (m2) density of the body (kg/m3)
specific heat of the material (J/kgm3)
Basic definitions and properties:
Change of internal energy (heat) with the change of temperature T Mass of a body with density r and volume V
The mount of energy which flows through the surface of unit area in unit time in the direction perpendicular to the surface
The amount of energy which flows through area A in time t in a direction perpendicular to A
volumetric heat sources (J/m3s)]
The amount of energy produced/consumed in the volume of V in time t
Heat transport equation – elementary derivation
w
A
c S r
Q mc
w T m r V J
qJ A t
q
S V t
x x+x
Q
tot= m c
wT
J
q(x,t) J
q(x+ x,t)
Q
src= S(x,t) V
Balance of energy in the domain in time t:
Qtotal = (the flux through boundary) + (sources)
m cw T = Jq(x,t) A t Jq(x+x,t) A t + S(x,t) V t r A x cw T = Jq(x,t) A t Jq(x+x,t) A t + S(x,t) A x t
Take the limit x 0, t 0 Divide both sides by (A x t):
( , ) ( , )
( , )
q q
w
J x x t J x t
c T S x t
t x
r
( , )
q w
T J
c S x t
t x
r
Heat transport – boundar conditions
1) Dirichlet boundary condition: temperature is given/controlled on the boundary or part of it
2) System or part of it is closed to flux transfer (thermal isolation)
3) Newton’s law of cooling (the transfer of heat through boundary is proportional to the differnce between temperatures at both sides, of the boundary
Here h is the heat transfer coefficient (W/m2K).
on (or part of the boundary) T T
env
0 on (or part of the boundary)
J
q
n
( ) on (or part of the boundary)
q
h T T
env J
n
Poisson’s Equation
This partial differential equation is encounterd in many application. For exmaple, it can describe a stedy-state (stationary) distribution of temperature, electric potential or gravitation potential energy. The mathematical from of Poisson’s equation in a Cartesian coordinate system is
where f is a given function defined in a domain and u is an unknown function (that is a function that is sought). The minus sign in the equation is a matter of tradition and convenience. Most applications us the form with minus sign, althought the equation without it also may be encountered.
The Laplacian , also written as 2, in a Cartesian system is defined as
In pure mathematics the number of dimensions (n) can by any positive integer, but in applications physical reality dictates n=3. When a physical system some possesses some symmetry it may be reduced to n=2 or n=1. In the case of n=1 we have in fact an ordinary differential equation:
in n
u f
R
2 2
2 2
1
.
n
u u
u x x
2 2 2 2 2 2 2
2 2 2 2 2 2 2
3 2 1
, , .
n n n
u u u u u u d u
u u u
x y z x y x dx
Poisson’s Equation – boundary condition
In a typical situation we use two basic form of boundary condition: the Dirichlet boundary condition or the Neumann boundary condition.
The Dirichlet boundary condition requires that an unknown function must have some specified (given) values on the boundary points. The Neumann boundary condition (in a strict sense) requires that the normal derivative of the unknown function is given on the boundary points. In a more general sense, Neumann type boundary conditions are any boundary conditions that involve a normal derivative of au unknown functions on the boundary.
a domain where the equation is defined
- the boundaro of a domain .
D – a part of the boundary where the Dirichlet boundary condition holds.
N – a part of the boundary where the Neumann boundary condition holds.
( , )
u f x y
( , ) ( , ) ( , ) D
u x y g x y x y
( , ) ( , ) ( , ) N u x y h x y x y
n
Poisson’s Equation – boundary condition
Example. Consider a stationary distribution of temperature in a two dimensional slab of homogeneous and isotropic material. Two sides are kept at a given temperature (40°C and 90°) and two others are isloted.
A mathematical formulation is as follows:
Problem Solution
2 2
2 2 0 ( , ) (0, ) (0, )
( ,0) 40 , (0, ) 90 , ( , ) 0, ( , ) 0
T T
x y a b
x y
T T
T x C T y C a y x b
x y
2 2
2 2 0
T T
x y
( ,0) 40 T x (0, ) 90
T y
( , ) 0 T x b
y
( , ) 0 T a y
x
Chemical kinetics
(introduction to formal descriptions of homogeneous reactions)
Chemical reactions can be roughly divided in two broad categories: homogeneous and heterogeneous.
Chemical reactions
Homogeneous reactions
Heterogeneous reactions
Homogeneous reactions take place in the same phase in the whole volume. These reactions usually take place in a liquid or gaseous state. An example is the reaction between oxygen and nitrogen gases (without any catalyst)
N2 + O2 2NO
Heterogeneous reactions usually take place at the boundaries of two different phases.
In most cases one of the phase is in a solid state.
In a description of the chemical reaction we may be interested in how quickly it proceeds and how the concentration of reagents change with time.
In an homogeneous reaction
A + B C
Is elementary, the we can assume the rate („speed”) of this reaction is given by v = k[A][B]
Here we used a standard notation used in chemistry: [A] denotes the molar concentration per volume, i.e.
[A] = n/V = (number of moles of A)/(volume)
The SI unit of molar concentration is mole/m3 but in laboratory practice the unit mole/dm3 is used.
The rate of reaction v means in this example that
[ ] v [ ][ ]
d C k A B
dt
The equation for [C] is an ordinary differential equation. But, we have to eliminate the functions [A] and [B ] to close the system.
There are simple relations between [A], [B], and [C] coming from the fact that they are related by the transformation A + B C. If for simplicity we assume that at the beginning there was no product C in the system ([C]0 =0), then
[A] = [A]0 – [C], [B] = [B]0 – [C].
where [A]0 = the initial (t=0) concentration for A, [B]0 = the initial (t=0) concentration for B.
Thus we have ODE with initial condition problem
where parameters [A]0, [B]0, and k are given.
0 0
0
[ ] ([ ] [ ])([ ] [ ]), [ ] 0,
d C k A C B C
dt C
Synthesis of hydrogen bromide
It is a complex reaction with the balance formula
The kinetics of this reaction (how the concentration of the product HBr changes with time) is given by the following ODE
The kinetic constants k1 and k2 depend on temperature. In standard conditions k2≈0,1.
Sometimes written in equivalent form
2 2
H Br 2HBr
3/ 2
2 2
1
2 2
[H ][Br ] [HBr]
[Br ] [HBr].
d k
dt k
1/ 2
2 2
1
2
2
[H ][Br ] [HBr]
[HBr] .
1 [Br ]
d k
dt k
Synthesis of hydrogen bromide (contd.)
Let us denote y(t) = [HBr]. Taking the overall reaction
We have the molar balance equations
Inserting these realtions int the equation fo d[HBr]/dt we obtain
2 2
H Br 2HBr
3/ 2
2 0 2 0
1
2 0 2
([H ] 0.5 )([Br ] 0.5 ) [Br ] ( 0.5) .
y y
dy k
dt k y
1 1
2 2 0 2 2 0 2
1 1
2 2 0 2 2 0 2
[H ] [H ] [HBr] [H ] ,
[Br ] [Br ] [HBr] [Br ] . y
y
Synthesis of hydrogen bromide (contd.)
The simulation of the reaction (solving the ODE from the previous slide) Plots present the concentration (mol/dm
3) vs time (s).
True kinetics of HBr synthesis
If the kinetics was as if for elementary reaction
3/ 2
2 2
1
2 2
[H ][Br ] [HBr]
[Br ] [HBr].
d k
dt k
1 2 2
[HBr]
[H ][Br ].
d k
dt
Lotka-Volterra model (predator-prey system)
In 1920 Alfred Lotka published a scheme of theoretical chemical reactions where the first and the second stage were autocatalytic:
where k1, k2, and k3 are rate constants. This is an overall reaction of converting reagent A to product B: A B, and X1, X2 are intermediates. Such reaction mechanism leads to the following system of ODEs for the concentrations of both intermediates:
1 2
2 3
[ ] [ ][ ] [ ][ ],
[ ] [ ][ ] [ ].
d X k A X k X Y
dt
d Y k X Y k Y dt
1
2
3
2 (autocatalytic production of X) 2 (autocatalytic production of Y)
(decay of Y)
k k k
A X X
X Y Y
Y B
