Modelling and scientific computing

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[edit] Process modelling slides

Class date: 13 September 2010 (slides 1 to 8)
15 September 2010 (slides 9 to 15)
16 September 2010 (slides 16 to 19)
20 September 2010 (slides 20 to the end)
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[edit] Approximation and computer representation

Class date: 22 and 23 September (updated)
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calculating relative error working with integers
import numpy as np
y = 13.0
n = 3                           # number of significant figures
rel_error = 0.5 * 10 ** (2-n)   # relative error calculation
x = y / 2.0
x_prev = 0.0
iter = 0
while abs(x - x_prev)/x > rel_error:
    x_prev = x
    x = (x + y/x) / 2.0
    print(abs(x - x_prev)/x)
    iter += 1
 
print('Used %d iterations to calculate sqrt(%f) = %.20f; '
      'true value = %.20f\n ' % (iter, y, x, np.sqrt(y)))
import numpy as np
 
print(np.int16(32767))
print(np.int16(32767+1))
print(np.int16(32767+2))
 
# Smallest and largest 16-bit integer
print(np.iinfo(np.int16).min, np.iinfo(np.int16).max)
 
# Smallest and largest 32-bit integer
print(np.iinfo(np.int32).min, np.iinfo(np.int32).max)
working with floats special numbers
import numpy as np
 
help(np.finfo)          # Read what the np.finfo function does
 
f = np.float32          # single precision, 32-bit float, 4 bytes
f = np.float64          # double precision, 64-bit float, 8 bytes
 
print('machine precision = eps = %.10g' % np.finfo(f).eps)
print('number of exponent bits = %.10g' % np.finfo(f).iexp)
print('number of significand bits = %.10g' % np.finfo(f).nmant)
print('smallest floating point value = %.10g' % np.finfo(f).min)
print('largest floating point value = %.10g' % np.finfo(f).max)
 
# Approximate number of decimal digits to which this kind
# of float is precise.
print('decimal precision = %.10g' % np.finfo(f).precision)
import numpy as np
 
# Infinities
print(np.inf, -np.inf)
print(np.float(1E400))   # inf, number exceeds maximum value that
                         # is possible with a 64-bit float: overflow
print(np.inf * -4.0)     # -inf
print(np.divide(2.4, 0.0)) # inf
 
# NaN's
a = np.float(-2.3)
print(np.sqrt(a))  # nan
print(np.log(a))   # nan
 
# Negative zeros
a = np.float(0.0)
b = np.float(-4.0)
c = a/b
print(c)        # -0.0
print(c * c)    #  0.0
 
eps = np.finfo(np.float).eps
e = eps/3.0  # create a number smaller than machine precision
# Question: why can we create a number smaller than eps?
print(e)
 
# Interesting property: non-commutative operations can occur
# when working with values smaller than eps.  Why?
# The print out here should "True", but it prints "False"
print((1.0 + (e + e)) == (1.0 + e + e))

[edit] Practice questions

  1. From the Hangos and Cameron reference, (available here] - accessible from McMaster computers only)
    • Work through example 2.4.1 on page 33
    • Exercise A 2.1 and A 2.2 on page 37
    • Exercise A 2.4: which controlling mechanisms would you consider?
  2. Homework problem, similar to the case presented on slide 18, except
    • Use two inlet streams \(\sf F_1\) and \(\sf F_2\), and assume they are volumetric flow rates
    • An irreversible reaction occurs, \(\sf A + 3B \stackrel{r}{\rightarrow} 2C\)
    • The reaction rate for A = \(\sf -r_A = -kC_\text{A} C_\text{B}^3\)
    1. Derive the time-varying component mass balance for species B.
      • \( V\frac{dC_B}{dt} = F^{\rm in}_1 C^{\rm in}_{\sf B,1} + F^{\rm in}_2 C^{\rm in}_{\sf B,2} - F^{\rm out} C_{\sf B} + 0 - 3 kC_{\sf A} C_{\sf B}^3 V \)
    2. What is the steady state value of \(\sf C_B\)? Can it be calculated without knowing the steady state value of \(\sf C_A\)?
      • \( F^{\rm in}_1 C^{\rm in}_{\sf B,1} + F^{\rm in}_2 C^{\rm in}_{\sf B,2} - F^{\rm out} \overline{C}_{\sf B} - 3 k \overline{C}_{\sf A} \overline{C}^3_{\sf B} V \) - we require the steady state value of \(C_{\sf A}\), denoted as \(\overline{C}_{\sf A}\), to calculate \(\overline{C}_{\sf B}\).

More exercises available here

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