Transcribed Text
2. Flowthrough tanks
The system illustrated below is a twochambered ethanol tank with a fixedrate inflow and a level
dependent outflow. Initially, the total volume in both chambers of the tank is 12.06 m³.
Q1
T
Q2
5
(a) You have measurements of the flow rates into (Q1) and out of (Q2 the tank, in terms of time. In
tegrate using trapezoidal rule to determine the volume of ethanol in the tank at 10 S intervals.
The relevant integral looks like
V = + 
t (s)
Q1 (m³/s)
Q2 (m³/s)
t (s)
Q1 (m³/s)
Q2 (m³/s)
0.0630
0.0854
40
0.0630
0.0828
10
0.0630
0.0850
50
0.0630
0.0820
20
0.0630
0.0842
60
0.0630
0.0814
30
0.0630
0.0836
5
(b) The data has been collected such that trapezoidal rule or Simpson's rule is appropriate. Gauss
quadrature has far better error behaviour, especially with many points per interval and exact data.
Explain the practical difficulties associated with collecting data suitable for Gauss quadrature.
5
(c) The perinterval formulas for trapezoidal rule and Simpson's 1/3 rule have a significant advantage
over the entireintegral formulas in a modern environment. Explain this advantage.
5
(d) Explain why Gauss quadrature tables are given over a normalized range of (1, +1).
3. Falling object in air
The velocity of an object falling vertically through air is known to be wellapproximated by a two
parameter model, at least through the initial acceleration:
U=gtkt³ =
where time is in seconds, and g & k are the two unknown parameters.
t (s)
U (m/s)
2
22
4
35
6
51
8
79
10
80
10
(a) Transform the system into standard form for least squares, including specifying the parameters X
and y, and any fit function(s), Q((x).
10 (b) Write out the matrix problem that must be solved, in terms of the fit function(s) Calculate all of
the required sums from the above data.
5
(c) Solve the matrix problem for the unknown coefficients (hints: you expect that the unknown g will
be close to the standard value of 9.81 m²/s, and the coefficient k is a reasonably round positive num
ber, something like 0.0150062)
5
(d) In this case, the root mean square error is relatively large: it's equal to 4.8 m/s. Explain why a large
RMS error does not necessarily mean the model is a poor fit. There is no need to calculate the RMS
error here, except perhaps as a check.
2. Flowthrough tanks
The system illustrated below is a twochambered ethanol tank with a fixedrate inflow and a level
dependent outflow. Initially, the total volume in both chambers of the tank is 12.06 m³.
Q1
Q2
t (s)
Q1 (m³/s)
Q2 (m³/s)
t (s)
Q1 (m³/s)
Q2 (m³/s)
0.0630
0.0854
40
0.0630
0.0828
10
0.0630
0.0850
50
0.0630
0.0820
20
0.0630
0.0842
60
0.0630
0.0814
30
0.0630
0.0836
(a) You have measurements of the flow rates into (Q1) and out of (Q2) the tank, in terms of time. In
tegrate using trapezoidal rule to determine the volume of ethanol in the tank at 10 S intervals.
The relevant integral looks like
=
The simplest approach here is to integrate based on a table of the actual integrand, which in this
case is Q1  Q2. The divisions in the table are increments of 10 S, so here h = 10 S. trapezoidal rule
over an interval is
I
and in this case each incremental integral represents an incremental change in volume.
The resulting table is as follows:
3. Falling object in air
solution3
The
velocity of an object falling vertically through air is known to be wellapproximated by a two
parameter model, at least through the initial acceleration:
U=gtkt³ 
where time is in seconds, and 8 & k are the two unknown parameters.
solution
Numerical Methods
5 of 8
3
t (s)
U (m/s)
2
22
4
35
6
51
8
79
10
80
(a) Transform the system into standard form for least squares, including specifying the parameters X
and y, and any fit function(s), Q((x).
No reorganization is required to apply least squares. The y parameter is best defined asy = U. The
X parameter is most simply defined as time, x = t. The model equation simply becomes
U=gtkt³
y=axazx3
The model has two unknown parameters, each multiplied by a known function, so that the coeffi
cients and fit functions are
a,=8 91(x)=: = x
a2=k 02(x)=x3 = 3
(b) Write out the matrix problem that must be solved, in terms of the fit function(s) Calculate all of
the required sums from the above data.
For two fit functions, the matrix equation has the form
N
20((x)102(x))
y,01(x)
a
as
N
(9(x))92(x))
[02((x)
i=1
i=1
solutio. ~
3
Numerical Methods
6 of 8
which simplifies to
4
xiyi
The most straightforward approach here is to tabulate the fit functions and their products:
x,
y,
Ó2 2(x,)
x2
x1
x,6
xy
x3y
2
22
2
8
4
16
64
44
176
4
35
4
64
16
256
4096
140
2240
6
51
6
216
36
1296
46656
306
11016
8
79
8
512
64
4096
262144
632
40448
10
80
10
1000
100
10000
106
800
80000
sum:
220
15664
1.31296x106
1922
133880
(c) Solve the matrix problem for the unknown coefficients (hints: you expect that the unknown g will
be close to the standard value of 9.81 m²/s, and the coefficient k is a reasonably round positive num
ber, something like 0.0150062)
The matrix problem is therefore
220
15664
a
1922
15664 1.31296x10
a
133880
which has the solutions
a = 9.80478
a2 = 0.0150058
or, in the original form,
8 = 9.80478
k =0.0150058
U = 9.80510.015t³ 
2. Density of water
At atmospheric pressure, the density of water has a maximum at a temperature of 4°C, and follows a
nearparabolic curve around that temperature. The density of water at 4°C is known to be
999.9750 kg/m³. The resulting model has the form
p=p(4°C)Kf2
where k is an unknown coefficient, and where t = T  4°C.
24
(a) Use linear least squares method to fit the data below to the above oneparameter model, using all
six points (include the 4°C point, despite its incorporation in the model). Determine k.
T (°C)
p (kg/m³)
2
999.9429
4
999.9750
6
999.9430
8
999.8509
10
999.7021
12
999.4996
8
(b) The RMS error for the above fit is 0.003. Confirm this by calculating the RMS error directly.
4. Multiparameter ODE solution
Consider a fourparameter system: u(t), v(t), w(t), and x(t). The derivatives are given such that
where each of the derivatives depends on all of the other parameters as well as time. The system is to
be solved using RK2 method.
10
(a) Write out the values to be calculated to take two steps in RK2 method, with step size h, starting at
an initial point of (up, Vo Wo xo), where to = 0. For credit, you must indicate precisely the order in
which the values are to be calculated, as well as what is to be calculated.
10
(b) Explain the role of intermediate/fictitious/fake points in RungeKutta steps, especially for multi
parameter systems such as this one. Highlight such points in your solution to the first part of this
question, or alternatively, detail them here.
2. Density of water
At atmospheric pressure, the density of water has a maximum at a temperature of 4°C, and follows a
nearparabolic curve around that temperature. The density of water at 4°C is known to be
999.9750 kg/m³. The resulting model has the form
p=p(4°C)Kt2
where k is an unknown coefficient, and where t = T  4°C.
(a) Use linear least squares method to fit the data below to the above oneparameter model, using all
six points (include the 4°C point, despite its incorporation in the model). Determine k.
T (°C)
P (kg/m³)
2
999.9429
4
999.9750
6
999.9430
8
999.8509
10
999.7021
12
999.4996
The system must be reorganized to apply least squares, since the density at 4°C is known:
x=t=T4
p=p(1°C)Kt2
There is just one unknown parameter in the resulting model equation, so that
a1=k
y=kx²
Q1(x)==2 =
The equation for a singleparameter least squares equation is
a1 =
Therefore, from the values and sums tabulated below,
a
= Ex,1 2 4 = 42.4920 5680 = 0.00748099
so that k = 0.00748099 = 0.75.
Ti
2
4
2
2
Pi
Xi
Yi
xi
xi
xixi
anx,
(yi  a1x,²) 2
2
999.9429
2
0.0321
4
16
0.1284
0.029924
4.73522x106
4
999.9750
0.0000
0.000000
6
999.9430
2
0.0320
4
16
0.1280
0.029924
4.31001x106
8
999.8509,
4
0.1241
16
256
1.9856
0.119696
1.93972x10"5
10
999.7021
6
0.2729
36
1296
9.8244
0.269315
1.28487x105
12
999.4996
8
0.4754
64
4096
30.4256
0.478783
1.14454x10°5
sums:
5680
42.4920

5.27365x10
(b) The RMS error for the above fit is 0.003. Confirm this by calculating the RMS error directly.
The rootmeansquare error can be calculated either in terms of y or of p, but both approaches
give the same result here. As it is simpler, the following outlines the determination of RMS error
in y. The above table shows the model estimates for y (secondlast column) and the squares of the
differences between the actual values Y1 and the estimated values a,x,2 (last column). The RMS
error follows as
i=6
1/2
1
5
=
0.00296469 0 0.00
i=1
4. Multiparameter ODE solution
Consider a fourparameter system: u(t), v(t), w(t), and x(t). The derivatives are given such that
f,(u,v,w,x,t)=v'(
fw(w,w,w,x,t)=w'(
fr(u,v,w,x,t)=x'(t
where each of the derivatives depends on all of the other parameters as well as time. The system is to
be solved using RK2 method.
(a) Write out the values to be calculated to take two steps in RK2 method, with step size h, starting at
an initial point of (uo, Yo Wo xo), where to = 0. For credit, you must indicate precisely the order in
which the values are to be calculated, as well as what is to be calculated.
First, determine the k1 values for the first step:
k1u=hfu
KIW
The k2 values for the first step then depend on a fictitious (intermediate) point:
(uo + KIu,
so that
+kivs Wo KIN, xo +k1x h)
K2v = =hfy (uo + KIII VO +KIV Wo +kiw xo +kix h)
k2w=hfw Wo KIW Xo + kix ,h)
The (actual) point after the first step follows as
u = 24++1(kin+k2u)
V1 = vo+1(kivtk2v)
W1
4=h
The process is then repeated for the second step, in sequence, starting at the new point. The k1
values follow directly,
KIU
leading to a fictitious (intermediate) point,
which gives the k2 values,
k2v
know h)
K2x
and ultimately the (actual and final) point after the second step:
=11+1((kin+k2u)
=
ww++kim+k2w)
t2=2h
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