Colorado School of Mines CHEN403
Setting up the Mathematical Model
Review of Heat & Material Balances
Topic Summary...................................................................................................................................................... 1
Introduction ............................................................................................................................................................ 2
Conservation Equations ..................................................................................................................................... 3
Use of Intrinsic Variables .............................................................................................................................. 4
Well-Mixed Systems ........................................................................................................................................ 4
Conservation of Total Mass:......................................................................................................................... 4
Component Balances....................................................................................................................................... 5
Energy balance:................................................................................................................................................. 5
Additional Relationships.................................................................................................................................... 6
Exceptions to Well-Mixed Process Assumptions...................................................................................... 8
Processes with Dead Time............................................................................................................................ 8
Flow Approximated by Combination of Well-Mixed Blocks............................................................ 8
Examples .................................................................................................................................................................. 9
Tank Liquid Level with Flow Through Valve......................................................................................... 9
State Variables:.................................................................................................................................................. 9
Total Mass Balance Leads to a Volume Balance:.................................................................................. 9
Other Tank Geometries .......................................................................................................................... 10
Right Circular Cone .................................................................................................................................. 10
Horizontal Cylinder ................................................................................................................................. 10
Tank flow — Change in Inlet Concentration....................................................................................... 11
Overall & Component Mass Balances: .............................................................................................. 12
Component Material Balance: ............................................................................................................. 12
Using Concentration................................................................................................................................ 12
Using Mass Fraction ................................................................................................................................ 13
Tank Flow — Chemical Reaction ............................................................................................................ 13
Tank Flow — Change in Input Stream’s Temperature ................................................................... 15
State Variables:.......................................................................................................................................... 15
Initial state: ................................................................................................................................................. 15
Transient solutions.................................................................................................................................. 16
What if the heat input is described by Newton’s law? ............................................................... 17
Example – Tank Flow Controlled by Valve.......................................................................................... 17
Bernoulli’s Law:......................................................................................................................................... 18
Total mass balance reduces to volume balance for constant density:................................. 19
Example: Chemical Reaction..................................................................................................................... 19
Topic Summary
•
Definition of well-mixed system. Combining well-mixed sub-processes to describe
overall process that is not well-mixed.
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
•
•
•
Dynamic equations from material & energy balance equations
Typical simplifications & modifications
Additional relationships
We will show that certain dynamic relations typically come most directly from certain
conservation equations. These typical relationships are shown in the following table.
Conservation Equation
Typical Dynamic Relation
Outlet volumetric flowrates vs. inlet rates
Liquid: Liquid level in system vs. time
Gas: Gas pressure in system vs. time
Outlet concentration vs. time
Outlet mole/mass fraction vs. time
Outlet temperature vs. time
Overall Material Balance
Component Balance
Thermal Energy Balance
Introduction
Transient behavior of a process:
• Start up.
• “Steady state” — random disturbances.
• Change of set points.
• Shut down.
Steps for math modeling:
• Develop the relationships/equations.
• Simplify.
• Solve:
⇒ Analytical — Laplace transforms
⇒ Numerial — Euler, Runge-Kutte methods
Needs of math model:
• Quantities whose values describe the nature of the process. These are the state
variables.
• Equations that use the quantities & describe how the variables change with
time.
For algebraic equations, where N is the number of variables & E is the number of
independent equations:
• E = N , deterministic system.
• E > N , under-determined system.
• E < N , over-determined system.
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December 21, 2008
Colorado School of Mines CHEN403
For differential conditions, also need boundary conditions. For transient problems, these
are normally initial conditions. We will mostly be working with lumped systems – i.e.,
there will be no spatial variation. Often termed as a well-mixed system.
Conservation Equations
We will use the basic principles of chemical engineering to guide us in our descriptions of
our dynamic processes: conservation of mass, energy, & momentum. So, the types of
conservation equations will be:
• The overall mass balance.
• Component/chemical species balance (including reaction rate terms).
• Thermal energy/heat balance.
• Momentum balance (though we won’t usually work with this in this class).
General form of the “stuff” balance equation:
 Rate of
 Rate  Rate   Rate of   Rate of 

=
−
+
−

 Accumulation   In   Out  Generation  Consumption 
Basic principles of ChE:
F1
Q1
F6
F2
Q3
F3
F5
Q2
F4
Summary of the balance equations:
Overall Mass Balance
dm
= ∑ mɺ i − ∑ mɺ j
dt i:inlet
j:outlet
Component Balance
dmA
= ∑ mɺ A ,i − ∑ mɺ A , j + ∑ RkV
dt
i :inlet
j:outlet
k :rxns
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Colorado School of Mines CHEN403
dNA
= ∑ Nɺ A ,i − ∑ Nɺ A , j + ∑ rkV
dt i:inlet
j:outlet
k :rxns
dE
= ∑ Eɺ i − ∑ Eɺ j + ∑ Qk − ∑ Ws ,m
dt i:inlet
j:outlet
k
m
Energy Balance
Use of Intrinsic Variables
It is useful to factor out intrinsic variables — properties that depend upon the state of the
system (pressure, temperature, phase condition) but not the magnitude. It is often
convenient to use the product of the mass density and volumetric flowrate instead of the
mass flowrate. Some of the useful relationships will be:
Mass
Moles
Energy
Cumulative Expression
m = ρV
ɶV = CtotalV
n=ρ
E = mEˆ = ρVEˆ
Rate Expression
mɺ = ρF
ɶF = Ctotal F
nɺ = ρ
ɺ ˆ = ρFEˆ
Eɺ = mE
H = mHˆ = ρVHˆ
= nHɶ = C VHɶ
ɺ ˆ = ρFHˆ
Hɺ = mH
ɺ ɶ = C FHɶ
= nH
Enthalpy
total
total
Well-Mixed Systems
The intrinsic properties of the system at any point will be the same. Added implication —
every outlet stream will possess the same intrinsic properties as the system itself. The
intrinsic properties of the inlets can still be independent.
Conservation of Total Mass:
dm
= ∑ mɺ i − ∑ mɺ j
dt i:inlet
j:outlet
d 
ρdV  = ∑ ρi Fi − ∑ ρ j F j
 i:inlet
dt  ∫∫∫
j:outlet
d ( ρV )
dt
=
∑ρF − ∑
i i
i :inlet
(use the intrinsic variable ρ )
ρF j
(well-mixed system)
j:outlet
where Fi is a volumetric flow rate (in or out). Note that all outlet streams will have the
same density ρ as the density within the system’s volume.
Conservation of total moles? NO! When there is a chemical reaction the total number of
moles may not necessarily be conserved.
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December 21, 2008
Colorado School of Mines CHEN403
Component Balances
The component balances can be expressed either in terms of mass (and mass fraction):
dmA
= ∑ mɺ A ,i − ∑ mɺ A , j + ∑  ∫∫∫ Rk dV 
A
dt
i:inlet
j:outlet
k :rxns
d 
ωA ρdV  = ∑ ωA ,i ρi Fi − ∑ ωA , j ρ j F j + ∑  ∫∫∫ Rk dV 
 i:inlet

A
dt  ∫∫∫
j:outlet
k :rxns
d ( ω A ρV )
dt
=
∑ω
A ,i
∑
ρi Fi −
i :inlet
ω A ρF j +
j:outlet
∑ [R dV ]
k
A
k :rxns
or in terms of moles (and molar concentration, though mole fractions could also be used):
dN A
= ∑ Nɺ A ,i − ∑ Nɺ A , j + ∑  ∫∫∫ rk dV 
A
dt i:inlet
j:outlet
k :rxns
d 
c A dV  = ∑ c A ,i Fi − ∑ c A , j F j + ∑  ∫∫∫ rk dV 
A
dt  ∫∫∫
i :inlet
j:outlet
k :rxns
d ( c AV )
dt
=
∑c
F−
A ,i i
i :inlet
∑
cA Fj +
j:outlet
∑ [r V ]
k
A
k :rxns
where Rk is the mass reaction rate expression (kg A per unit time per unit volume) and rk
is the molar reaction rate expression (moles A per unit time per unit volume). It is usually
easier to work with moles when there is a chemical reaction. Note that the reaction term
can be positive for generation & negative for consumption.
Energy balance:
dE
= ∑ Eɺ i − ∑ Eɺ j + ∑ Qk − ∑ Ws ,m
dt i:inlet
j:outlet
k
m
d (U + K + P )
dt
=
∑ w Hˆ
i
i
−
i :inlet
=
∑ ρ F Hˆ
i i
i :inlet
∑
j:outlet
i
−
w j Hˆ j + ∑ Qk − ∑ Ws ,m
∑
j:outlet
k
m
ρ j F j Hˆ j + ∑ Qk − ∑Ws ,m
k
m
where Hˆ i is the specific enthalpy (per unit mass). The energy balance can also be put on a
molar basis:
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
d (U + K + P )
dt
=
∑ Nɺ Hɶ
i
i
−
i :inlet
∑
j:outlet
Nɺ j Hɶ j + ∑ Qk − ∑Ws ,m
k
m
where Hɶi is the specific enthalpy per unit mole basis.
There are additional assumptions normally made to the energy balance:
d (U + K + P )
dt
=
∑ ρ F Hˆ
i i
i :inlet
i
−
∑
j:outlet
ρ j F j Hˆ j + ∑ Qk − ∑Ws ,m
k
m
dU
= ∑ ρi Fi Hˆ i − ∑ ρ j F j Hˆ j + ∑ Qk − ∑Ws ,m
dt i:inlet
j :outlet
k
m
(internal energy dominant term)
For liquid systems we will generally use the assumption that U ≈ H :
dH
= ∑ ρi Fi Hˆ i − ∑ ρ j F j Hˆ j + ∑ Qk − ∑Ws ,m
dt i:inlet
j :outlet
k
m
d 
ˆ =
ρHdV
ρi Fi Hˆ i − ∑ ρ j F j Hˆ j + ∑ Qk − ∑Ws ,m
 i:∑
dt  ∫∫∫
inlet
j:outlet
k
m
ˆ
d ρHV
= ∑ ρi Fi Hˆ i − ∑ ρF j Hˆ + ∑ Qk − ∑ Ws ,m
dt
i :inlet
j :outlet
k
m
(
)
(well-mixed).
For gas systems this is not necessarily the case. However, since we ultimately want a
dynamic expression for temperature this is not a significant problem.
d 
ˆ =
ρUdV
ρi Fi Hˆ i − ∑ ρ j F j Hˆ j + ∑ Qk − ∑ Ws ,m
 i:∑
dt  ∫∫∫
inlet
j:outlet
k
m
ˆ
d ρUV
= ∑ ρi Fi Hˆ i − ∑ ρF j Hˆ + ∑ Qk − ∑Ws ,m
dt
i :inlet
j:outlet
k
m
(
)
(well-mixed).
Additional Relationships
These are the basic equations, but now we also need relationships to the measured process
variables.
Relationship between mass & volume:
m = ρV = ρAh (for constant cross sectional area)
mɺ = ρF
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December 21, 2008
Colorado School of Mines CHEN403
Thermodynamic relationships:
T
 ∂Hɶ 
ɶ ⇒ Hɶ − Hɶ = Cɶ dT ≈ Cɶ (T − T )
C
≡


p
ref
p
ref
∫ p
 ∂T P
Tref
T
Hˆ − Hˆ ref =
∫ Cˆ dT ≈ Cˆ (T − T )
p
p
ref
Tref
U = H + PV ≈ H
U = H + RT
(for liquid systems)
(for ideal gas systems)
Equations of state relating density to pressure, temperature, & composition:
ρ = ρ (T , P , x ) (the equation of state). Some simple equations of state:
PM
P
=
∑ x i Mi
RT RT i
ω
xM
1
1
=∑ i = ∑ i i
ρ i ρi M i ρi
ρ=
(ideal gas)
(ideal volume of mixing/additive volumes)
Energy relationships:
1
1
K = mv 2 = ρVv 2
2
2
P = mg ( h − h0 ) = ρVg ( h − h0 ) (gravitational potential energy. May be other forms)
Relationship between heat transfer rate & temperature driving force:
Q = UA (T − Ta )
(Newton’s law of heating/cooling)
(
Q = A1 ε1T14 − α12T24
)
(Kirschoff’s law of radiant heat exchange)
Relationship between flow through a valve & pressure driving force:
F ∝ ∆p ⇒ F = Ap ∆p = C v ∆h
(non-linear valve flow expression)
Chemical reaction rate relationships:
rA = rA (T , x , P ) (the reaction rate function). For example:
rA = k (T ) c A = k0e − E / RT c A
(1st order reaction)
rA = k (T ) c 2A = k0e − E / RT c 2A or rA = k0e − E / RT c AcB
(2nd order reactions)
where k (T ) = k0e − E / RT is the Arrenhius rate expression.
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
Exceptions to Well-Mixed Process Assumptions
Processes with Dead Time
One exception to the well-mixed assumption is when there is some piece to the process
that introduces a significant time delay between when something happens and when this
might be measured. One can picture the situation as being similar to plug flow through a
pipe — the material does not change in the pipe, but there is a time difference between
when it enters and when it reappears at the other end. For example, if we are interested in
the temperature at the exit of a tank, To , but the thermocouple is in a pipe a distance L
away, then there will be a delay before the temperature can be measured. This time delay,
t o , can be estimated from:
to =
LA
distance
L
=
= c
F0
velocity F0 / Ac
where: Ac is the cross sectional area of the pipe.
Fo is the volumetric flow rate.
If there is no heat loss in the pipe then the relationships between the outlet temperature of
the tank at that temperature measured, Tm , is:
Tm ( t ) = To ( t − t o ) .
Flow Approximated by Combination of Well-Mixed Blocks
Sometimes the flow patterns within a process do not produce a well-mixed system. We may
still be able to approximate the overall process as a combination of well-mixed subprocesses. One obvious process that is not well-mixed is an annular heat exchanger with
counter-current flow. Though the fluids might be well-mixed across any face perpendicular
to the flow there will be a temperature gradient along the direction of the flow. However,
the overall process could be approximated by two series of well-mixed sub-processes that
flow from one to another, transferring heat across the boundaries of sub-processes.
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
Examples
Tank Liquid Level with Flow Through Valve
A
F0, ?0
h
?
F1, ?1
Look at how the level in a tank changes with changes of flows in and out of a tank.
State Variables:
•
•
Densities
Flow rates
What are we assuming if we say that the density out is the same as the density in the
tank?
What are we assuming if we say that the density is not changing?
Total Mass Balance Leads to a Volume Balance:
Total mass balance:
d ( ρV )
= ∑ ρi Fi − ∑ ρF j
dt
i :inlet
j:outlet
d ( ρAh)
dt
= ρ0F0 − ρF1
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
Apply other considerations:
• Constant cross-sectional area: A = constant
• Constant density: ρ = ρ1 = ρ0
Total mass balance becomes a volume balance leading to a differential equation for the
level in the tank with respect to time:
A
dh
= F0 − F1
dt
Other Tank Geometries
The above geometry assumes that the tank is an upright cylinder. There are other common
geometries:
Right Circular Cone
r
tan θ =
1
1
V = πr 2h = πh3 tan2 θ
3
3
dV
∴
= π tan2 θ h2
dh
(
θ
h
)
Horizontal Cylinder
a
s = arc length
b
θ
r
Mathematical Models
r
h
1
A =  sr − a ( r − b ) 
2
s = rθ
θ r −b
cos =
2
r
a
2
= r 2 − (r − b)
2
2
r −b
2
∴ A = r 2 cos −1 
 − (r − b) r − (r − b)
 r 
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December 21, 2008
Colorado School of Mines CHEN403
Still need the volume of the cylinder with respect to level:

2
−1  r − h 
2
2
 Lr cos  r  − L ( r − h) r − ( r − h)



V =
πr 2 L − Lr 2 cos −1  h − r  + L ( h − r ) r 2 − ( h − r )2



 r 
if
h≤r
if
h>r
The amazing part is that the derivative is the same for both halves:
dV
= 2L h ( 2r − h)
dh
Sphere. Using similar definitions are for the horizontal cylinder:
1 2

πh (3r − h)
if h ≤ r

3
V =
 2 πr 3 − 1 π ( 2r − h )2 ( r + h) if h > r
 3
3
and, again, the derivative is the same for both halves:
dV
= πh ( 2r − h)
dh
Tank flow — Change in Inlet Concentration
F0, ?0, CA,0, CB,0
V, ?,
C A,
CB
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F1, ?1, CA,1, CB,1
December 21, 2008
Colorado School of Mines CHEN403
Look at how the concentration in a tank changes with changes of concentration into the
tank.
Overall & Component Mass Balances:
Total mass balance:
d ( ρV )
= ∑ ρi Fi − ∑ ρF j
dt
i :inlet
j:outlet
dρ
V
= ρ0F0 − ρF1
dt
What have we assumed here? Constant volume overflow (& well-mixed).
If we assume constant density:
0 = ρF0 − ρF1 ⇒ F1 = F0 .
Component Material Balance:
We can deal with multi-component mixtures with the concept of concentrations or
mole/mass fractions.
• Concentration gives the amount of the component per unit volume. This is
multiplied by the volumetric flow rate to get the flux of the component.
• Mole/mass fraction gives the fraction of the total amount corresponding to the
component. This must be multiplied by the overall molar/mass density and the
volumetric flow rate to get the flux of the component.
Using Concentration
Total mole balance using concentration:
d ( NA )
= ∑ C A ,i Fi − ∑ C A F j
dt
i :inlet
j:outlet
(no chemical reaction & well-mixed)
d ( C AV )
= C A ,0F0 − C A F1
dt
dC
V A = C A ,0F0 − C A F1
dt
(constant volume)
What have we assumed here? Well-mixed, no chemical reaction, & constant volume
overflow.
If we assume constant density:
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
V
dC A
dC A F0
= F0 ( C A ,0 − C A ) ⇒
= ( C A ,0 − C A )
dt
dt
V
Using Mass Fraction
Total mass balance using mass fraction ( ωA ):
d ( mA )
dt
=
∑ mɺ
A ,i
−
i :inlet
d ( ω A ρV )
dt
=
∑ω
i :inlet
∑
mɺ A , j
(no chemical reaction & well-mixed)
j:outlet
A ,i
ρi Fi −
∑
ω A ρF j
j:outlet
d ( ωAρ )
= ∑ ωA ,i ρi Fi − ∑ ω AρF j
dt
i :inlet
j:outlet
dωA
= ∑ ω A ,i Fi − ∑ ωA F j
V
dt i:inlet
j:outlet
V
(constant volume)
(constant density)
Tank Flow — Chemical Reaction
Let’s assume we have an isothermal, constant volume CSTR with a chemical reaction:
k1
A 
→B
The inlet stream has no B in it. The molar balances will be:
dC A
= F0C A0 − F0C A − k1C AV
dt
dC
V B = −F0C B + k1C AV
dt
V
To determine CB ( t ) we could first solve for C A ( t ) from the first equation & then plug it
into the 2nd equation.
However, if the reaction is actually
k1
A ←
→B
k2
then the molar balances will be:
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
dC A
= F0C A0 − F0C A − k1C AV + k2CBV
dt
dC
V B = −F0C B + k1C AV − k2CBV
dt
V
Now the equations are coupled. Some additional manipulation must be done to separate
the CB ( t ) terms from the C A ( t ) terms & visa versa. From the 2nd equation:
CA =
1 dC B F0 + k2V
+
CB
k1 dt
k1V
and (assuming F0 constant):
dC A 1 d 2CB F0 + k2V dC B
=
+
dt k1 dt 2
k1V
dt
Substituting these expressions into the 1st equation gives an expression in just C B , not C A :
 1 d 2C B F0 + k2V dCB 
 1 dCB F0 + k2V 
+
+
V
C B  + k2C BV
 = F0C A0 − ( F0 + k1V ) 
2
k1V
dt 
k1V
 k1 dt
 k1 dt

2
F + k V dC B ( F0 + k1V )( F0 + k2V )
V d C B F0 + k2V dCB
+
= F0C A0 − 0 1
+
CB + k2C BV
2
k1 dt
k1
dt
k1
dt
k1V

V d 2C B  2F0 + k1V + k2V  dCB  ( F0 + k1V )( F0 + k2V )
+
−
+ k2V  C B = q0C A0

2
k1 dt
k1
k1V

 dt 

Notice the system of 2 1st order ODEs has been replaced by 1 2nd order ODE & one of the
original 1st order ODEs (for C A ( t ) ).
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
Tank Flow — Change in Input Stream’s Temperature
F0, ?0, T0, Cp0
V 1,
?,
T,
Cp
Q1
h1
F1, ?1, T1, Cp1
Look at a variable inlet temperature T0 ( t ) . Start from steady state condition.
State Variables:
•
•
•
Densities
Temperatures
Flow rates
Initial state:
Steady State total mass balance:
d ( ρV )
= ∑ ρi Fi − ∑ ρF j
dt
i :inlet
j:outlet
0 = ρ0F0 − ρ1F1
(well-mixed)
(steady state)
Note that this also implies that ρ0 F0 = ρ1F1 ≡ w0 for any ρ(T ) .
Steady state energy balance:
(
ˆ
d ρHV
dt
)=
∑ ρ F Hˆ
i i
i :inlet
i
−
∑
ρF j Hˆ + Q − Ws
0 = ρ0F0 Hˆ 0 − ρ1 F1 Hˆ 1 + Q1
0 = mɺ 0 Cˆ p0 (T0 − Tref ) + Hˆ ref 0  − mɺ 0 Cˆ p1 (T1 − Tref ) + Hˆ ref 1  + Q1




Mathematical Models
(well-mixed)
j:outlet
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(steady state)
December 21, 2008
Colorado School of Mines CHEN403
0 = mɺ 0Cˆ p (T0 − T1 ) + Q1
T1 = T0 +
(constant heat capacity & reference enthalpy)
Q1
mɺ 0Cˆ p
Transient solutions
After change, total mass balance:
d ( ρ1V )
dt
= ρ0F0 − ρ1F1 = mɺ 0 − mɺ 1
(well-mixed)
What does this assume? Well mixed.
Energy balance:
(
d ρ1VHˆ 1
) = ρ F Hˆ
0 0
0
− ρ1 F1 Hˆ 1 + Q1
dt
d ( ρ1V )
dHˆ
Hˆ 1
+ ρ1V 1 = ρ0F0 Hˆ 0 − ρ1F1 Hˆ 1 + Q1
dt
dt
d ( ρ1V )
dHˆ dT
Hˆ 1
+ ρ1V 1 1 = ρ0F0 Hˆ 0 − ρ1F1 Hˆ 1 + Q1
dt
dT1 dt
d ( ρ1V )
dT
Hˆ 1
+ ρ1VCˆ p 1 = ρ0F0 Hˆ 0 − ρ1F1 Hˆ 1 + Q1
dt
dt
d ( ρ1V )
dT
Hˆ 1
+ ρ1VCˆ p 1 = mɺ 0 Hˆ 0 − mɺ 1 Hˆ 1 + Q1
dt
dt
dT
Hˆ 1 ( mɺ 0 − mɺ 1 ) + ρ1VCˆ p 1 = mɺ 0 Hˆ 0 − mɺ 1 Hˆ 1 + Q1
dt
dT
ρ1VCˆ p 1 = mɺ 0 Hˆ 0 − Hˆ 1 + Q1
dt
(
)
(well-mixed)
(chain rule)
(chain rule)
(definition of heat capacity)
(derivative from mass balance)
(mathematical manipulation)
Note that this expression does not depend upon assumptions of constant volume or
constant density!
dT
ρ1VCˆ p 1 = mɺ 0Cˆ p (T0 − T1 ) + Q1
dt
(constant Cˆ p & reference state)
Can do some additional math. We could normalize the form of the ODE so that the
coefficient on the time derivative is 1; in this class, however, we will normally want the
coefficient on the variable without the derivative to be 1:
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
ρ1V dT1
Q
= T0 − T1 + 1
mɺ 0 dt
mɺ 0Cˆ p
ρ1VCˆ p dT1
Q
+ T1 = T0 + 1
mɺ 0Cˆ p dt
mɺ 0Cˆ p
Notice that the term ρ1V / mɺ 0 has units of time. This can be thought of as a characteristic
time constant for the system.
Note — Even though the derivation of the ODE does not depend upon whether the system
has constant volume and/or density, the integration with time will depend upon this!
What if the heat input is described by Newton’s law?
Let’s assume Q1 = UA (Ts − T ) . Then the ODE becomes:
dT
ρ1VCˆ p 1 = mɺ 0Cˆ p (T0 − T1 ) + UA (Ts − T )
dt
dT
ρ1VCˆ p 1 = mɺ 0Cˆ pT0 + UATs − mɺ 0Cˆ p + UA T1
dt
ρ1VCˆ p dT1
mɺ 0Cˆ p
UA
T0 +
Ts .
+ T1 =
ˆ
ˆ
ˆ
mɺ 0C p + UA dt
mɺ 0C p + UA
mɺ 0C p + UA
(
)
The form of the solution is the same, but the characteristic time constant is now
ρ1VCˆ p / mɺ 0Cˆ p + UA & is dependent upon heating parameters.
(
)
Example – Tank Flow Controlled by Valve
F0, ?
h
A, ?
F 1, ?
Look at flow in valve at outlet of tank.
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
Bernoulli’s Law:
∆p 1 2
+ ∆v + g∆h = w f
ρ 2
What do the terms represent?
Let’s apply Bernoulli’s law from the surface in the tank to the entrance of the pipe. We
will neglect entrance effects & other friction:
pe − ps 1 2 2
+ ve − v s + g ( 0 − h1 ) = 0
ρ
2
(
)
Assume the surface is open to the atmosphere, so ps = patm . Because the pipe has a much
smaller cross-sectional area than the tank ve >> v s and ve2 − v s2 ≈ ve2 so:
pe − patm 1 2
+ ve − gh1 = 0 .
ρ
2
( )
If the kinetic energy effect is small compared to the potential energy effect then:
pe − patm 1 2
p − patm
+ ve − gh1 ≈ e
− gh1 = 0 ⇒ pe − patm = ρgh1 .
2
ρ
ρ
( )
It is normally assumed that at the valve outlet the kinetic energy effect is what creates
the pressure drop, so:
2( pe − pv )
pv − pe 1 2 2
p − pe 1 2
+ vv − ve ≈ v
+ vv = 0 ⇒ vv2 =
ρ
ρ
ρ
2
2
(
)
( )
If pv ≈ patm , then:
vv2 =
2( pe − patm )
ρ
= 2 gh1 ⇒ vv = 2 gh1
The flow rate through a valve is Fv = Av vv which means that:
Fv = Av 2 gh1 = C v h1 where C v = Av 2 g .
Opening and closing the valve will change Av and hence C v .
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
Total mass balance reduces to volume balance for constant density:
d ( ρV )
dV
= F0 − F1
dt
= ρF0 − ρF1 ⇒
dt
dh
A = F0 − C v h
dt
This is a non-linear ODE. Sometimes we can get an analytical solution for this particular
ODE. For example, for a step change in flow such that Fin ( t ) → F0 from F0* :
∫
h
A
t
dh = ∫ dt
0
F0 − C v h
y
A
2
∫y0 F0 − Cv y d ( y ) = t
y
2 Ay
∫y0 F0 − Cv y dy = t
h0
y
 y F

2 A  − − 02 ln F0 − C v y  = t
 Cv Cv
 y0
 y − y0 F0
F − Cv y 
2 A −
− 2 ln 0
 =t
Cv
Cv
F0 − C v y0 

 h − h0 F
F −C h 
 =t
2A −
− 02 ln 0 v
Cv
Cv
F0 − C v h0 

If the inlet flow is more complicated or if this equation is part of a larger set of equations,
there is no guarantee that an analytical solution exists.
Example: Chemical Reaction
F0
F1
Tcoil
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
In the main section we started to analyze a CSTR with a first order reaction A → B . To
start setting up the equations lets only make a couple assumptions:
•
•
•
•
•
Well-mixed system within the reactor.
There is a constant liquid volume within the reactor.
A heat transfer fluid is used within the coils to control the temperature. A phase
change occurs to provide the heating or cooling (e.g., steam condensation for
heating or refrigerant boiling for cooling). This will keep the temperature uniform
throughout the coil.
Pure A is fed to the reactor.
The reaction has elementary fist order kinetics:
r = kC A = k0e − E / RT C A
Total mass balance:
d ( ρ1V )
dt
V
= ρ0F0 − ρ1F1
dρ1
= ρ0F0 − ρ1F1 = mɺ 0 − mɺ 1
dt
(Constant volume)
Mole balance on A:
dN A d ( C A1V )
=
= C A0F0 − C A1F1 − rV
dt
dt
V
dC A1
= C A0F0 − C A1F1 − k0e − E / RT1 C A1V
dt
(Kinetic expression)
Mole balance on B:
dNB d ( C B1V )
=
= −C B1F1 + rV
dt
dt
V
dC B1
= −CB1F1 + k0e − E / RT1 C A1V
dt
(Kinetic expression)
Energy balance:
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
(
d ρ1VHˆ 1
0 0
dt
V
) = ρ F Hˆ
(
d ρ1 Hˆ 1
dt
0
) = ρ F Hˆ
0 0
− ρ1F1 Hˆ 1 + UA (Tcoil − T1 )
0
− ρ1F1Hˆ 1 + UA (Tcoil − T1 )
(Constant volume)
ρ1V
dHˆ 1 ˆ dρ1
+ H1V
= ρ0F0 Hˆ 0 − ρ1F1 Hˆ 1 + UA (Tcoil − T1 )
dt
dt
ρ1V
dHˆ 1 ˆ
+ H1 ( ρ0F0 − ρ1F1 ) = ρ0F0Hˆ 0 − ρ1F1Hˆ 1 + UA (Tcoil − T1 )
dt
ρ1V
dHˆ 1
+ ρ0F0 Hˆ 1 = ρ0F0 Hˆ 0 + UA (Tcoil − T1 )
dt
ρ1V
dHˆ 1
= ρ0F0 Hˆ 0 − Hˆ 1 + UA (Tcoil − T1 )
dt
ρ1VCˆ p1
(
)
(
)
dT1
= ρ0F0 Hˆ 0 − Hˆ 1 + UA (Tcoil − T1 )
dt
(Chain rule)
(Mass balance)
(Cancel like terms)
(A little algebra)
(Chain rule & Cˆ p definition)
This energy balance equation is general and has the same form whether there is a heat of
reaction or not. So, where is the heat of reaction? (Or heats of mixing, or temperature
dependent heat capacities, or composition dependent heat capacities for that matter.) It
was noted before that the heat of reaction is embedded in the difference in the enthalpies,
Hˆ 0 − Hˆ 1 ; this is all well and good to say, but it doesn’t really help in the practical matter of
setting up the energy balance equation to relate all of the relevant temperatures.
To simplify the math, let’s make two other assumptions:
• The heat capacity is constant with respect to temperature (though not necessarily
with respect to composition).
• The enthalpies mix ideally (i.e., no heat of mixing effects).
With these assumptions the enthalpies can be expressed as:
Hˆ 0 = Cˆ p0 (T0 − Tref ) + Hˆ ref 0
Hˆ 1 = Cˆ p1 (T1 − Tref ) + Hˆ ref 1
and the energy balance is:
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
dT
ρ1VCˆ p1 1 = ρ0F0 Cˆ p0 (T0 − Tref ) + Hˆ ref 0 − Cˆ p1 (T1 − Tref ) − Hˆ ref 1  + UA (Tcoil − T1 )


dt
dT
ρ1VCˆ p1 1 = ρ0F0 Cˆ p0 (T0 − Tref ) − Cˆ p1 (T1 − Tref )  + ρ0 F0  Hˆ ref 0 − Hˆ ref 1  + UA (Tcoil − T1 ) .


dt
The heat of reaction is still embedded in the term relating the specific enthalpy values at
the reference conditions of temperature ( Tref ) and composition. Notice that assuming the
heat capacity is not composition dependent does not affect this reference state term, it only
simplifies the first term relating the net flow of enthalpy to the system; if we assume no
composition dependency then Cˆ p0 = Cˆ p1 ≡ Cˆ p and:
dT
ρ1VCˆ p 1 = ρ0F0 Cˆ p (T0 − Tref ) − Cˆ p (T1 − Tref )  + ρ0F0  Hˆ ref 0 − Hˆ ref 1  + UA (Tcoil − T1 )


dt
dT
ρ1VCˆ p 1 = ρ0F0Cˆ p (T0 − T1 ) + ρ0F0  Hˆ ref 0 − Hˆ ref 1  + UA (Tcoil − T1 ) .
dt
Couple things to note about the reference state term that has the heat of reaction
embedded in it:
• The units on the term are energy/time, such as cal/min. In this particular
formulation the bracketed term is energy/mass & the leading term is mass/time;
the two terms could just as easily be split in molar units.
• The heat of a reaction is usually calculated by determining the differences between
the heats of formation of the reactants and the products. Heat of formation is simply
a reference enthalpy – there is little difference between what we’ve done in thermo
class & what we want to do here.
• The change in the reference state enthalpy term only comes from that portion of the
stream that reacts. We can express the term reference state enthalpy term as:
ρ0F0  Hˆ ref 0 − Hˆ ref 1  = rV ( −∆Hɶrxn ) .
Using this expression in the energy balance equation:
dT
ρ1VCˆ p1 1 = ρ0F0 Cˆ p0 (T0 − Tref ) − Cˆ p1 (T1 − Tref )  + rV ( −∆Hɶrxn ) + UA (Tcoil − T1 )


dt
For the specific rate expression considered here:
ρ1VCˆ p1
dT1
= ρ0F0 Cˆ p0 (T0 − Tref ) − Cˆ p1 (T1 − Tref )  + k0e − E / RT1 C A1V ( −∆Hɶrxn ) + UA (Tcoil − T1 )


dt
Mathematical Models
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December 21, 2008
Colorado School of Mines CHEN403
or if the heat capacity is not composition dependent:
dT
ρ1VCˆ p 1 = ρ0F0Cˆ p (T0 − T1 ) + k0e − E / RT1 C A1V ( −∆Hɶrxn ) + UA (Tcoil − T1 ) .
dt
Mathematical Models
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December 21, 2008