Equation Solving with Python & SymPy

In engineering applications, the same equation will be solved over and over with different values or measurements as inputs. Anticipating this, we can either write one function for each variable which inputs all other variables, or take a much easier route using SymPy.

The following example is a simple implementation of Manning's formula:

\[ v = \frac{k}{n} {R_h}^{2/3} S^{1/2} \]

with \(k = 1.49\) since we most often work in English units in American hydrology. Note that the variable mannings_eqn is actually the above expression set equal to 0, so that we see a \(-v\) term.

The following code creates Sympy Symbols for each variable in the expression. In general, if we have \(n\) variables in the expression, we can pass in a dict with \(n-1\) key-value pairs (equations) with known values to solve for the \(n\)th variable. We can create dicts with keys, the set of knowns, and pass in any (independent) combination of \(n-1\) variables from our expression to get a number.

from sympy import symbols, solve

v, n, Rh, S = symbols('v, n, Rh, S')

knowns_1 = {n: 0.025, Rh: 10, S: 0.01}
knowns_2 = {v: 100, n: 0.01, S: 0.05}
knowns_3 = {n: 0.05, Rh: 5}

def solve_mannings(input_dict):
    mannings_eqn = -v + 1.49/n * Rh**(2/3) * S**(1/2)
    # Use dict flag to get {variable: value} output, not anonymous [value]
    solution = solve(mannings_eqn.subs(input_dict), dict=True)
    print(solution)

solve_mannings(knowns_1)
solve_mannings(knowns_2)
solve_mannings(knowns_3)

However, the third set of knowns gives us an underdetermined system since we only have two equations for our four-variable system. Accordingly, we will have as output one variable as a function of another, in this case \(S\):

[{v: 27.6638694483322}]
[{Rh: 5.19987727958683}]
[{v: 87.1357285987434*sqrt(S)}]