example added

examples/hydrogen_production_thermodynamics.md

@@ -0,0 +1,47 @@
+# Thermodynamics of Hydrogen Production
+
+The minimal energy required to produce hydrogen from liquid water is given by the
+Higher Heating Value (HHV). The HHV is the sum of the difference
+between the enthalpies of products and educts (LHV: Lower Heating Value) and
+the Heat of Evaporation for water.
+
+```python
+import gaspype as gp
+#import numpy as np
+#import matplotlib.pyplot as plt
+
+lhv = gp.fluid({'H2': 1, 'O2': 1/2, 'H2O': -1}).get_H(25 + 273.15)
+dh_v = 43990  # J/mol  (heat of evaporation for water @ 25 °C)
+
+hhv = lhv + dh_v
+print(f'LHV: {lhv/1e3:.1f} kJ/mol')
+print(f'HHV: {hhv/1e3:.1f} kJ/mol')
+```
+
+Thermodynamics also defines which part of the energy must be
+provided as work (e.g., electric power) and which part can be supplied
+as heat. This depends on temperature and pressure. For generating 1 bar
+of hydrogen the temperature dependency can be calculated as follows:
+
+```python
+import numpy as np
+import matplotlib.pyplot as plt
+
+t = np.linspace(0, 2000, 128)  # 0 to 2000 °C
+p = 1e5  # Pa (=1 bar)
+
+g_products = gp.fluid({'H2': 1, 'O2': 1/2, 'H2O': 0}).get_G(t + 273.15, p)
+g_educts =   gp.fluid({'H2': 0, 'O2': 0,   'H2O': 1}).get_G(t + 273.15, p)
+
+work = g_products - g_educts  # J/mol
+heat = lhv - work  # J/mol
+
+fig, ax = plt.subplots(figsize=(6, 4), dpi=120)
+ax.set_xlabel("Temperature / °C")
+ax.set_ylabel("Energy / kWh/kg")
+k = 1e-3 / 3600 / 0.002  # Conversion factor from J/mol to kWh/kg for hydrogen
+ax.stackplot(t, k * work, k * heat, k * dh_v * np.ones_like(t))
+```
+
+Green is the heat of evaporation, orange the additional heat provided at
+the given temperature and blue the work.