Uncertainty Calculator

Help you get rid of Physical Chemistry Experiment in CCME!

Package Requirements

pip install sympy

or

conda install -c conda-forge sympy

Define equation: string = expression

string is the name of what you want to calculate.

expression is a formula written in Python format.

eg.

equation = [
    x.strip() for x in
    r'W = (-Q_V*m-Q_N-Q_M)/Dt-rho*V*C'
    .split('=')
]

Define variable: ('symbol = mu +- sigma', string)

symbol is a temporary abbreviation of each variable, which should be consistent with expression, and will not exist in output.

mu is the expectation of each variable.

sigma is the standard deviation of each variable.

string is the name of each variable.

eg.

variable = [
    ('Q_V = -26.414 +- 0', r'Q_V'),
    ('m = 0.9547 +- 0.0004/sqrt(3)', r'm'),
    ('Q_N = -0.323 +- 3.243*0.0004/sqrt(3)', r'Q_\ce{Ni}'),
    ('Q_M = -0.01 +- 0.0002*16.736/sqrt(3)', r'Q_\text{cotton}'),
    ('rho = 0.99865 +- 0', r'\rho_\ce{H2O}'),
    ('V = 3000 +- 0.01', r'V'),
    ('C = 4.1824e-3 +- 0', r'C_\ce{H2O}'),
    ('Dt = 1.770 +- 0.009', r'\Delta T'),
]

Set digits of results: {'mu': integer, 'sigma': integer}

eg.

result_digit = {
    'mu': 4,
    'sigma': 2,
}

Set units of results: result_unit = string

string is a unit written in $\LaTeX$ format, which should be set to 1 if what you want to calculate is dimensionless.

eg.

result_unit = r'\text{kJ}/{}^\circ\text{C}'

Separate equation or not: separate = 0 or 1

eg.

separate = 1

Insert numbers or not: insert = 0 or 1

eg.

insert = 1

Include equation number or not: include_equation_number = 0 or 1

eg.

include_equation_number = 1

Output

$\LaTeX$ code of calculation details, including normal calculation, each partial derivative, total uncertainty combination, and the final result.

eg.

\begin{equation}
W=- C_\ce{H2O} V \rho_\ce{H2O} + \frac{- Q_\text{cotton} - Q_\ce{Ni} - Q_V m}{\Delta T}=- \left(0.0041824\right) \times \left(3000\right) \times \left(0.99865\right) + \frac{- \left(-0.01\right) - \left(-0.323\right) - \left(-26.414\right) \times \left(0.9547\right)}{\left(1.77\right)}=1.905\ \text{kJ}/{}^\circ\text{C}
\end{equation}

\begin{equation}
\begin{aligned}
\frac{\partial W }{\partial m }&=- \frac{Q_V}{\Delta T}=- \frac{\left(-26.414\right)}{\left(1.77\right)}=15.0\\
\frac{\partial W }{\partial Q_\ce{Ni} }&=- \frac{1}{\Delta T}=- \frac{1}{\left(1.77\right)}=-0.57\\
\frac{\partial W }{\partial Q_\text{cotton} }&=- \frac{1}{\Delta T}=- \frac{1}{\left(1.77\right)}=-0.57\\
\frac{\partial W }{\partial V }&=- C_\ce{H2O} \rho_\ce{H2O}=- \left(0.0041824\right) \times \left(0.99865\right)=-0.0042\\
\frac{\partial W }{\partial \Delta T }&=\frac{Q_\text{cotton} + Q_\ce{Ni} + Q_V m}{\Delta T^{2}}=\frac{\left(-0.01\right) + \left(-0.323\right) + \left(-26.414\right) \times \left(0.9547\right)}{\left(1.77\right)^{2}}=-8.2\\
\end{aligned}
\end{equation}

\begin{equation}
\begin{aligned}
\sigma_{W}&=\sqrt{\left(\frac{\partial W }{\partial m } \sigma_{m}\right)^2+\left(\frac{\partial W }{\partial Q_\ce{Ni} } \sigma_{Q_\ce{Ni}}\right)^2+\left(\frac{\partial W }{\partial Q_\text{cotton} } \sigma_{Q_\text{cotton}}\right)^2+\left(\frac{\partial W }{\partial V } \sigma_{V}\right)^2+\left(\frac{\partial W }{\partial \Delta T } \sigma_{\Delta T}\right)^2}\\
&=\sqrt{\left(15.0 \times 0.00023\right)^2+\left(-0.57 \times 0.00075\right)^2+\left(-0.57 \times 0.0019\right)^2+\left(-0.0042 \times 0.01\right)^2+\left(-8.2 \times 0.009\right)^2}\\
&=\sqrt{\left(0.0034\right)^2+\left(-0.00042\right)^2+\left(-0.0011\right)^2+\left(-0.000042\right)^2+\left(-0.073\right)^2}\\
&=0.073\ \text{kJ}/{}^\circ\text{C}
\end{aligned}
\end{equation}

\begin{equation}
W=\left (1.905 \pm 0.073 \right )\ \text{kJ}/{}^\circ\text{C}
\end{equation}

Acknowledgements

Sympy: https://github.com/sympy/sympy

$\LaTeX$公式编辑器: https://latexlive.com