Modelling and scientific computing

From Process Model Formulation and Solution: 3E4
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Process modelling slides

Please download the lecture slides. 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)

Approximation and computer representation

Please download the lecture slides. 22 and 23 September (updated)

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))

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