13-21
( )
( ) ( )
( )
( )
( )
!
!
!
!
!
!
!
!
!
!
d
F
eE
e
C
C
d
F
E
k
d
F
E
k
Ao
A“#
$
%
&
‘
(
)
*
+*
$
%
&
‘
(
)
*
+*
$
$
$
%
&
‘
‘
‘
(
)
*
“
*=
“
1
1
1
( )
( ) ( )
( )
( )
( )
!
!
!
!
!
!
!
!
!
!
!
d
F
eeE
e
C
C
d
F
E
k
d
F
E
k
Ao
A“#
$
%
&
‘
(
)
*
*
*
$
%
&
‘
(
)
*
+*
$
$
$
%
&
‘
‘
‘
(
)
*
“
*=
“
1
1
1
by definition
( ) ( ) ( )
!!!
dFdE =
gives:
( )
( ) ( ) ( )( )[ ] ( )( )
!
!
!
!
!
!
Feee FLnd
F
dF
d
F
E
“==
#
=
#“
“
“
$
%
&
‘
(
)
“
“
1
1
1
1
changing the variables from λ to t in the RHS integral:
( )( ) ( ) ( )( )
( ) dt
tF
tFetE
Fe
C
Ckt
k
Ao
A!“#
#$
%
&
‘
(
)
#
#
=#
*
**
1
1
1
( ) ( ) dtetE
F
e
CC kt
k
AoA !“#
#
=
$
$
$
1
(1)
Exit concentration is when λ=0, F(0)=0 hence eqn (1) becomes:
( ) dtetECC kt
AoA !“#
=0
This is the same expression as for the exit concentration for the Segregation model.
P13-4 (a)
Mean Residence Time
By definition
!
“
=
0
1)( dttE
. The area of the semicircle representing the E(t) is given by
1
2
2
==
!”
A
and
!
“
2
=
. For constant volumetric flow
min8.0
2===
!
“
m
t
.
P13-4 (b)
Variance
( ) ( )
! !
” “
#=#=
0
2
0
2
2
2)(
$$%
dttEtdttEt
13-22
( ) ( ) ( ) ( )
[ ] ( )
! ! !
“
=++#=##=
0
2
0
0
4224
2
222
8
5
sin1cos2cos
$
%
$
%
$$$
dxxxxdtttdttEt
159.0
2
1
8
5242 ==!=
“
##
“
$
Using Polymath:
See Polymath program P13-4-b.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 1.596 1.596
sigma 0 0 0.1593161 0.1593161
tau 0.7980869 0.7980869 0.7980869 0.7980869
t1 1.5961738 1.5961738 1.5961738 1.5961738
E2 0 0 0.7980614 0.0166534
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(sigma)/d(t) = (t-tau)^2*E
Explicit equations as entered by the user
[1] tau = (2/3.14)^0.5
[2] t1 = 2*tau
[3] E2 = (t*(2*tau-t))^(1/2)
[4] E = if (t<t1) then (E2) else (0)
P13-4 (c)
Conversion predicted by the Segregation model
( )
!
“
=
0
)( dttEtXX
kt
etX !
!=1)(
13-23
( )
( )
!“““=“
#
##
2
0
2/1
2
2
1dtteX kt
See Polymath program P13-4-c.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 1.596 1.596
Xbar 0 0 0.4447565 0.4447565
tau 0.7980869 0.7980869 0.7980869 0.7980869
t1 1.5961738 1.5961738 1.5961738 1.5961738
E2 0 0 0.7980671 0.0166534
X 0 0 0.7210716 0.7210716
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] tau = (2/3.14)^0.5
[2] t1 = 2*tau
[3] E2 = (t*(2*tau-t))^(1/2)
[4] E = if (t<t1) then (E2) else (0)
[5] k = .8
[6] X = 1-exp(-k*t)
%5.44=X
P13-4 (d)
Conversion predicted by the Maximum Mixedness model
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
( )
XkCkCr AoAA !!=!=1
13-24
( ) ( )
( )X
F
E
Xk
d
dX
!
!
!
“
+““=1
1
( ) ( )
( )X
F
E
Xk
dz
dX
!
!
“
““=1
1
See Polymath program P13-4-d.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 1.596 1.596
x 0 0 0.4445289 0.4445289
F 1 -5.053E-04 1 -5.053E-04
k 0.8 0.8 0.8 0.8
lam 1.596 0 1.596 0
E1 0.0166534 0 0.7980666 0
E 0.0166534 0 0.7980666 0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] k = .8
[2] lam = 1.596-z
[3] tau = (2/3.14)^0.5
[4] E1 = (tau^2-(lam-tau)^2)^0.5
[5] E = if (lam<=2*tau) then (E1) else (0)
%5.44=X
as for the Segregation Model, but we knew this because for first order
reactions Xseg = XMM
P13-5 (a)
The cumulative distribution function F(t) is given:
The real reactor can be modelled as two parallel PFRs:
The relative
( ) ( )
!
“
#$+$=21 4
3
4
1
)(
%&%&
tttE
Mean Residence Time
min25)75.0min*20()1min*10(
1
0
=+== !tdFtm
or
( ) ( ) min25
4
3
4
1
4
3
4
1
)( 2121
0
=+=
!
“
#
$
%
&‘+‘== ((
)
***+*+
tttdtttEtm
P13-5 (b)
Variance
( ) ( ) ( ) ( ) ( ) ( )
2
2
2
2
121
2
0
2
2
min75
4
3
4
1
4
3
4
1
)(
=
=!+!=
“
#
$
%
&
‘!+!!=!=((
)
mmm ttdtttttdttEt
***+*+*,
P13-5 (c)
For a PFR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /mol⋅min–1, τ = 25
min and CAo = 1.25 mol/dm3
PFR
PFR
13-26
758.0
1=
+
=
Ao
Ao
Ck
Ck
X
!
!
For a CSTR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /mol⋅min-1, τ = 25
min and CAo = 1.25 mol/dm3
( ) Ao
Ck
X
X
!
=
“2
1
→
572.0=X
For two parallel PFRs, τ1 = 10 min and τ2 = 30 min, Fa01 = 1/4Fa0 and Fa02 = 3/4Fa0 , second
order, liquid phase, irreversible reaction with k = 0.1 dm3 /mol⋅min-1 and CAo = 1.25 mol/dm3
3
1
1
1/556.0
1
dmmolC
Ck
Ck
CC Ao
Ao
Ao
AoA =
+
!=
“
“
3
2
2
2/263.0
1
dmmoC
Ck
Ck
CC Ao
Ao
Ao
AoA =
+
!=
“
“
731.0
4
3
4
1
21
=
!!
=
Ao
AAAo
vC
vCvCvC
X
P13-5 (d)
1-Conversion predicted by the Segregation Model
( ) ( ) ( )
731.0
14
3
14
1
4
3
4
1
1
)(
2
2
1
1
0
21
0
=
+
+
+
=
!
“
#
$
%
&‘+‘
+
== ((
))
*
*
*
*
*+*+
Ao
Ao
Ao
Ao
Ao
Ao
kC
kC
kC
kC
dttt
tkC
tkC
dttEtXX
2-Conversion predicted by the Maximum Mixedness model
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
( )222 1XkCkCr AoAA !!=!=
( ) ( )
( )X
F
E
XkC
d
dX
Ao
!
!
!
“
+““=1
12
We need to change the variable such the integration proceeds forward:
( ) ( )
( )X
zTF
zTE
XkC
dz
dX
Ao !!
!
!!=1
12
See Polymath program P13-5-d.pol
13-27
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 40 40
x 0 0 0.7125177 0.7061611
F 0.9999 -1.081E-04 0.9999 -1.081E-04
cao 1.25 1.25 1.25 1.25
k 0.1 0.1 0.1 0.1
lam 40 0 40 0
ca 1.25 0.3614311 1.25 0.3672986
t2 30 30 30 30
E3 0 0 0 0
E1 1.25 1.25 1.25 1.25
E 0 0 1.25 0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] cao = 1.25
[2] k = .1
[3] lam = 40-z
[4] ca = cao*(1-x)
[5] t1 = 10
[6] t2 = 30
[7] E3 = 0
[10] E1 = 0.25/(t1*2*(1-0.99))
[11] E = if ((lam>=0.99*t1)and(lam<1.01*t1)) then (E1) else( if ((lam>=0.99*t2)and(lam<1.01*t2)) then
(E2) else (E3))
[12] EF = E/(1-F)
P13-5 (e)
Adiabatic Reaction E=10000cal/mol and
XT 500325 !=
Introducing the enthalpy balance:
XT 500325 !=
and the constitutive equation for
))/1325/1(*314.8/45000(
325
T
ekk !
“=
in the MM model.
13-28
P13-5 (f)
Conversion Predicted by an ideal laminar flow reactor
For a LFR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /mol⋅min-1, τ = 25
min and CAo = 1.25 mol/dm3
We apply the Segregation model, using Polymath:
Ao
Ao
ktC
ktC
X
+
=1
and
!
“
#
$
<
=%min5.12min)2/(625
min5.120
)( 13 tfort
tfor
tE
See Polymath program P13-5-e.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 300 300
xbar 0 0 0.7077852 0.7077852
cao 1.25 1.25 1.25 1.25
k 0.1 0.1 0.1 0.1
tau 25 25 25 25
E1 3.125E+06 1.157E-05 3.125E+06 1.157E-05
E 0 0 0.0991813 1.157E-05
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[1] cao = 1.25
[2] k = .1
[3] tau = 25
[4] E1 = tau^2/2/(t^3+0.0001)
[5] t1 = tau/2
[6] E = if (t<t1) then (0) else (E1)
[7] x = k*cao*t/(1+k*cao*t)
We can compare with the exact analytical formula due to Denbigh.
709.0)/21ln(
2
1=
!
“
#
$
%
&+
‘
(
)
*
+
,
–=Da
Da
DaX
with Da= kCAoτ
13-29
P13-6 (a)
Mean Residence Time
By definition
!
“
=
0
1)( dttE
. The area of the triangle representing the E(t) is given by
2
2.02 1== t
( )
!
!
!
“
!
!
!
#
$
%%&&
<
=
otherwise
tttiftt
t
ttif
t
t
tE
0
22
1
)( 111
2
1
1
2
1
( ) ( )
! ! !! =+“==
#1
0
1
2
1
1
1
2
11
2
1
2
2
1
2
0
2
t t
t
t
t
mtdt
t
t
dt
t
t
dt
t
t
dtttEt
.
P13-6 (b)
Variance
( ) ( )
! !
” “
#=#=
0
2
0
2
2
2)( mm tdttEtdttEtt
$
( ) 2
0
2
6
7
m
tdttEt =
!
“
and
2
2
222 min167.4
6
25
66
7===!=m
mm
t
tt
“
See Polymath program P13-6-b.pol
13-30
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 10 10
sigma 0 0 4.1666667 4.1666667
tau 5 5 5 5
t1 10 10 10 10
E1 0 0 0.4 0.4
Differential equations as entered by the user
[1] d(sigma)/d(t) = (t-tau)^2*E
Explicit equations as entered by the user
[1] tau = 5
[2] t1 = 2*tau
[3] E1 = t/tau^2
[4] E2 = -(t-2*tau)/tau^2
[5] E = if (t<tau) then (E1) else (if(t<=t1)then(E2)else(0))
5.0
1=
+
=
Ao
Ao
Ck
Ck
X
!
!
For a CSTR, second order, liquid phase, irreversible reaction with kCAo =0.2 min-1,τ=5 min
( ) Ao
Ck
X
X
!
=
“2
1
→
382.0=X
P13-6 (d)
1-Segregation model
See Polymath program P13-6-d-1.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 15 15
Xbar 0 0 0.4767547 0.4767547
k1 0.2 0.2 0.2 0.2
X 0 0 0.75 0.75
tau 5 5 5 5
E1 0 0 0.6 0.6
E2 0.4 -0.2 0.4 -0.2
13-31
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] k1 = .2
[2] X = k1*t/(1+k1*t)
[3] tau = 5
[4] t1 = 2*tau
[5] E1 = t/tau^2
[7] E = if (t<tau) then (E1) else(if(t<=t1)then(E2)else(0))
%7.47=X
kCAo =k’
( )
( ) !
!
“
#
$
$
%
&
‘‘
‘
+‘=X
zTF
zTE
C
r
dz
dX
Ao
A
1
( )222 1XkCkCr AoAA !!=!=
13-32
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 20 20
x 0 0 0.6642538 0.4669205
F 1 -7.513E-06 1 -7.513E-06
k 0.2 0.2 0.2 0.2
lam 20 0 20 0
t1 10 10 10 10
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)^2+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] k = 0.2
[2] lam = 20-z
[3] tau = 5
[4] E1 = lam/tau^2
[5] t1 = 2*tau
[7] E = if (lam<tau) then (E1) else(if (lam<=t1) then(E2)else (0))
%7.46=X
For a LFR, 2nd order, liq. phase, irreversible reaction kCAo =0.2 min-1,τ=5 min.
We apply the segregation model, using Polymath:
Ao
Ao
ktC
ktC
X
+
=1
and
!
“
#
$
<
=%min5.2min)2/(25
min5.20
)( 13 tfort
tfor
tE
See Polymath program P13-6-e.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 300 300
xbar 0 0 0.4506243 0.4506243
kcao 0.2 0.2 0.2 0.2
tau 5 5 5 5
E1 1.25E+05 4.63E-07 1.25E+05 4.63E-07
E 0 0 0.0549822 4.63E-07
13-33
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[1] kcao = 0.2
[2] tau = 5
[3] E1 = tau^2/2/(t^3+0.0001)
[4] t1 = tau/2
[5] E = if (t<t1) then (0) else (E1)
[6] x = kcao*t/(1+kcao*t)
We can compare with the exact analytical formula due to Denbigh.
451.0)/21ln(
2
1=
!
“
#
$
%
&+
‘
(
)
*
+
,
–=Da
Da
DaX
with Da= kCAoτ
XLFR
XMM
Xseg
XPFR
XCSTR
0.451
0.467
0.477
0.5
0.382
P13-7
Irreversible Liquid phase, half order, Segregation model.
Mean conversion
( ) ( )
!“== 0
1.0dttEtXX
(1)
Assume a Gaussian distribution for E(t):
( ) ( ) ( ) !
“
#
$
%
&
‘
(
(=
!
“
#
$
%
&(
(=2
2
2
2
32
5
exp
.23
1
.2
exp
2
1tt
tE
)
*
+
)*
( ) 2/1
2/1
1X
C
C
k
dt
dX
Ao
Ao !=
and CAo=1 mol/dm3
The only unknown k1 is estimated solving with a trial and error method Eq(1).
Using POLYMATH: k1 = 0.0205 mol1/2/dm3/2 .s
13-34
P13-8 (a)
The E(t) is a square pulse
Third order liquid-phase reaction: rA = kCA
3 with CAo = 2mol/dm3 and k = 0.3 dm6/mol2/min (
Isothermal Operation)
X(t)=1“1
2kCAo
2t+1
( )
X =X(t)E(t)dt =
0
“
#1$1
2kCAo
2t+1
%
&
‘
‘
(
)
*
*
dt=
1
2
t$2kCAo
2t+1
kCAo
2=
1
2
#
0.53
See Polymath program P13-8-a-1.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2 2
Xbar 0 0 0.5296583 0.5296583
k 0.3 0.3 0.3 0.3
Cao 2 2 2 2
t1 1 1 1 1
X 0 0 0.5847726 0.5847726
13-35
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[2] Cao = 2
[3] t1 = 1
[4] E2 = 1
[5] E = if (t>=t1) then (E2) else (0)
[6] X = 1-1/(1+2*k*Cao^2*t)^(1/2)
2-Maximum Mixedness Model
( )
( ) !
!
“
#
$
$
%
&
‘‘
‘
+‘=X
zTF
zTE
C
r
dz
dX
Ao
A
1
( )333 1XkCkCr AoAA !!=!=
( ) ( )
( )X
zTF
zTE
XkC
dz
dX
Ao !!
!
!!=1
13
2
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 2 2
x 0 0 0.5215389 0.5215389
F 0.9999 -9.999E-05 0.9999 -9.999E-05
cao 2 2 2 2
k 0.3 0.3 0.3 0.3
lam 2 0 2 0
ra -2.4 -2.4 -0.2628762 -0.2628762
EF 1.0E+04 0 1.0E+04 0
13-36
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] cao = 2
[2] k = .3
[3] lam = 2-z
[4] ca = cao*(1-x)
[5] E1 = 1
[6] ra = -k*ca^3
[7] t2 = 2
[8] t1 = 1
[9] E = if ((lam>=t1)and(lam<=t2)) then (E1) else(0)
0
2
=
!
d
.
P13-8 (b)
Introducing in the Segregated Model and in the MM Model :
))/1300/1(*20000(
300
T
ekk !
“=
See Polymath program P13-8-b-1.pol and P13-8-b-2.pol
300K
310K
320K
330K
340K
350K
Xseg
0.530
0.82
0.933
0.974
0.989
0.995
XMM
0.521
0.806
0.924
0.97
0.987
0.994
Introducing the enthalpy balance:
( ) XX
C
H
TT
Pii
RX
50
40000
305
0+=
!“
+= #
$
See Polymath program P13-8-c.pol
13-37
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 5 5
Xbar 0 0 0.8949502 0.8949502
ko 0.8948702 0.8948702 0.8948702 0.8948702
t1 1 1 1 1
E2 1 1 1 1
t2 2 2 2 2
Cao 2 2 2 2
E 0 0 1 0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] ko = 0.3*exp(20000*(1/300-1/305))
[2] t1 = 1
[3] E2 = 1
[4] t2 = 2
[5] Cao = 2
[6] E = if((t>=t1) and (t<=t2)) then (E2) else (0)
[7] k = ko*exp(20000*(1/300-1/310))
[8] To = 305
P13-9 (a)
3rd order, k=175 dm6/(mol2 min), CBo=0.0313 dm3/min
(LFR, PFR, CSTR with τ=100s)
PFR
Design equation
2
BAAoo CCk
dV
dX
Cv =
( )
XCC AoA !=1
( )
XCC BoB !=1
13-38
( )
!=
“
X
Bo v
V
kC
X
dX
0
2
3
1
( ) 12
1
12
2+=
!v
V
kC
XBo
Using the quadratic solution
832.0168.01
1
1
2
=!=
!=
V
X
The conversion for a PFR X=83.2%
CSTR
Design equation
( ) VrCCv AAAoo !=!
2
BAA CCkr =!
( )
XCC AoA !=1
( )
XCC BoB !=1
X
C
CC
Ao
AAo =
!
( ) v
V
XkCX Bo
32 1!=
The conversion for a CSTR X=66.2 %
LFR (completely segregated)
E(t)=0 for t<τ/2
=τ2 /2t3 for t>= τ/2
( )3
21XkC
dt
dX
Ao !=
Where
( ) tkC
tX
Bo
2
21
1
1
+
!=
See Polymath program P13-9-a.pol
13-39
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 1.0E-05 1.0E-05 100 100
xbar 0 0 0.1827616 0.1827616
cbo 0.0313 0.0313 0.0313 0.0313
k 175 175 175 175
tau 1.67 1.67 1.67 1.67
E1 1.394E+15 1.403E-06 1.394E+15 1.403E-06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[1] cbo = 0.0313
[2] k = 175
[3] tau = 1.67
[4] E1 = tau^2/(2*t^3)
[5] x = 1-(1/(1+2*k*cbo^2*t))^0.5
[6] E = if(t>=tau/2) then (E1) else (0)
“
P13-9 (b)
(Segregation Model and Maximum Mixedness Model applying RTD of Example 13-1)
Segregation model
Vr
dt
dN
A
A!=!
Batch reactor
( )
XNN AoA !=1
VCkC
dt
dX
NBAAo
2
=
( )
XCC AoA !=1
( )
XCC BoB !=1
( )
!=
“
X
Bo tkC
X
dX
0
2
3
1
Similarly
( ) !
!
“
#
$
$
%
&
+
‘=tkC
tX
Bo
2
21
1
1
( ) ( )
!
=
0
dttEtXX
and E(t) from the given data, fitted using Polymath
See Polymath program P13-9-b-regression.pol
POLYMATH Results
Polynomial Regression Report
Model: C02 = a1*C01 + a2*C01^2 + a3*C01^3 + a4*C01^4
Variable Value 95% confidence
a1 0.0889237 0.0424295
a2 -0.0157181 0.0163712
a3 7.926E-04 0.0019617
a4 -8.63E-06 7.288E-05
Regression not including free parameter
Number of observations = 13
Statistics
R^2 = 0.8653673
R^2adj = 0.8204897
Rmsd = 0.0065707
Variance = 8.107E-04
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 14 14
xbar 0 0 0.4106313 0.4106313
F 0 0 1.1137842 1.1137842
cbo 0.0313 0.0313 0.0313 0.0313
k 175 175 175 175
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = E*x
[2] d(F)/d(t) = E
Explicit equations as entered by the user
[1] cbo = 0.0313
[2] k = 175
[3] x = 1-(1/(1+2*k*cbo^2*t))^0.5
[4] E = 0.0899237*t-0.0157181*t^2+0.000792*t^3-0.00000863*t^4