Thermodynamics exercises
========================

.. include:: /_generated/nav/thermodynamics.rst

Three problems built on the :doc:`thermodynamics gallery </api/gallery/thermodynamics/index>`.
Try each one before opening its solution. The solutions run when the
documentation is built, and the test suite checks every number they print.
The tutorial :doc:`/tutorials/power_plant_to_cooling_tower` follows a
steam cycle through its condenser.

1. Why diesels are more efficient
---------------------------------

At the same compression ratio the ideal Otto cycle is more efficient than
the ideal Diesel cycle. Show it for a compression ratio of 18 and a cutoff
ratio of 2. Yet diesel engines are the more efficient in practice. Explain
why by comparing a diesel at 18 with a petrol engine limited by knock to a
compression ratio of 10.

*Builds on:* :doc:`/api/gallery/thermodynamics/index` (the Otto and Diesel comparison).

.. dropdown:: Solution

   .. plot::
      :context: reset

      import numpy as np

      from engineeringkit.thermodynamics import diesel_efficiency, otto_efficiency

      print(f"r = 18: Otto {otto_efficiency(18):.3f}, Diesel (cutoff 2) {diesel_efficiency(18, 2):.3f}")
      print(f"petrol engine at r = 10: Otto {otto_efficiency(10):.3f}")
      # r = 18: Otto 0.685, Diesel (cutoff 2) 0.632
      # petrol engine at r = 10: Otto 0.602
      assert otto_efficiency(18) > diesel_efficiency(18, 2) > otto_efficiency(10)
      assert np.isclose(otto_efficiency(10), 1 - 10 ** (1 - 1.4))

   A diesel compresses air alone, so it cannot knock, and its higher
   compression ratio more than makes up for adding heat at constant
   pressure.

2. The best pressure ratio of a gas turbine
-------------------------------------------

A gas turbine draws air at 300 K, and its turbine inlet is limited to
1400 K. In the ideal Brayton cycle (cold air, :math:`\gamma = 1.4`,
:math:`c_p = 1005` J/(kg K)) the efficiency rises with pressure ratio,
but the net work per kilogram of air has a maximum. Find the pressure ratio
of maximum work, and check it against
:math:`r_p = (T_3/T_1)^{\gamma/(2(\gamma-1))}`.

*Builds on:* :doc:`/api/gallery/thermodynamics/index`.

.. dropdown:: Solution

   .. plot::
      :context: reset

      import matplotlib.pyplot as plt
      import numpy as np

      from engineeringkit.thermodynamics import brayton_efficiency, isentropic_temperature_ratio

      T1, T3, cp, gamma = 300.0, 1400.0, 1005.0, 1.4
      rp = np.linspace(2, 60, 5801)
      tau = isentropic_temperature_ratio(rp, gamma)              # T2/T1 = T3/T4
      work = cp * (T3 - T3 / tau) - cp * (T1 * tau - T1)
      best = rp[np.argmax(work)]
      print(f"maximum net work {work.max() / 1e3:.1f} kJ/kg at r_p = {best:.1f} (formula {(T3 / T1) ** (gamma / (2 * (gamma - 1))):.2f})")
      efficiency = np.array([brayton_efficiency(r) for r in rp])
      print(f"efficiency there {brayton_efficiency(best):.3f}")
      # maximum net work 405.9 kJ/kg at r_p = 14.8 (formula 14.82)
      # efficiency there 0.537
      assert np.allclose(efficiency, 1 - 1 / tau)
      assert abs(best - (T3 / T1) ** (gamma / (2 * (gamma - 1)))) < 0.02

      fig, ax = plt.subplots(figsize=(5, 3.2))
      ax.plot(rp, work / 1e3)
      ax.set(xlabel="pressure ratio", ylabel="net work [kJ/kg]")
      ax2 = ax.twinx()
      ax2.plot(rp, efficiency, "C1")
      ax2.set_ylabel("efficiency", color="C1")
      fig.tight_layout()

   At maximum work the compressor outlet and turbine outlet temperatures
   are equal, :math:`T_2 = T_4 = \sqrt{T_1 T_3}`. Aircraft engines, where
   weight matters, run near this ratio; combined-cycle plants accept less
   work per kilogram for more efficiency, or recover the exhaust heat.

3. A heat pump against the second law
-------------------------------------

A house needs 10 kW of heat at 20 °C when it is 0 °C outside. A heat pump
achieves 45% of the Carnot COP. Find its COP, its electrical input, and
the minimum work the second law allows, from the exergy of the delivered
heat. Compare with an electric heater.

*Builds on:* :doc:`/api/gallery/thermodynamics/index`.

.. dropdown:: Solution

   .. plot::
      :context: reset

      import numpy as np

      from engineeringkit.thermodynamics import carnot_cop_heat_pump, exergy_heat, heat_pump_cop

      Q, T_house, T_out = 10e3, 293.15, 273.15
      cop = 0.45 * carnot_cop_heat_pump(T_house, T_out)
      W = Q / cop
      W_min = exergy_heat(Q, T_house, T_out)
      print(f"Carnot COP {carnot_cop_heat_pump(T_house, T_out):.2f}, real COP {cop:.2f}, electrical input {W / 1e3:.2f} kW")
      print(f"minimum work {W_min:.0f} W: second-law efficiency {W_min / W:.0%} (electric heater {W_min / Q:.1%})")
      # Carnot COP 14.66, real COP 6.60, electrical input 1.52 kW
      # minimum work 682 W: second-law efficiency 45% (electric heater 6.8%)
      assert np.isclose(heat_pump_cop(Q, W), cop) and np.isclose(W_min, Q / carnot_cop_heat_pump(T_house, T_out))

   The heat pump's second-law efficiency equals its fraction of Carnot,
   45%. The electric heater turns 10 kW of work into heat worth only
   682 W of work: by the second law it is less than 7% efficient, though
   by the first law it is perfect.