Gas Power Cycles (Otto, Diesel, Brayton)

The engine in your car, the turbine under a jet's wing, and the power plant that peaks during summer afternoons all share a common family tree: the gas power cycles. Unlike vapor cycles, the working fluid stays gaseous throughout — no phase change. This makes them lighter and faster-responding, at the cost of a much larger back-work penalty.

In this lesson, we'll idealize three cornerstone cycles — Otto (spark ignition), Diesel (compression ignition), and Brayton (gas turbines) — and analyze them with Python.

Air-Standard Assumptions

Real combustion engines are messy. To make them tractable, we apply the air-standard assumptions:

• The working fluid is air, behaving as an ideal gas.
• All processes are internally reversible.
• Combustion is replaced by heat addition from an external source.
• Exhaust is replaced by heat rejection that restores the initial state.

When we additionally assume constant specific heats at room temperature ($c_p = 1.005$, $c_v = 0.718$ kJ/kg·K, $k = 1.4$), it's called the cold-air-standard analysis.

The Otto Cycle (Spark Ignition)

The Otto cycle models gasoline engines. Its four processes:

The defining parameter is the compression ratio:

$$r = \frac{V_1}{V_2} = \frac{V_{max}}{V_{min}}$$

Using the isentropic relations and constant-volume heat exchanges, the thermal efficiency simplifies beautifully:

$$\eta_{Otto} = 1 - \frac{1}{r^{k-1}}$$

Efficiency depends only on compression ratio and $k$. Higher $r$ is better — but knocking limits real gasoline engines to $r \approx 8$–11.

The Diesel Cycle (Compression Ignition)

The Diesel cycle differs in one key process: heat is added at constant pressure (fuel injected into hot compressed air), not constant volume.

Besides $r$, the Diesel cycle introduces the cutoff ratio:

$$r_c = \frac{V_3}{V_2}$$

The efficiency is:

$$\eta_{Diesel} = 1 - \frac{1}{r^{k-1}} \left[ \frac{r_c^{k} - 1}{k (r_c - 1)} \right]$$

The bracketed term is always greater than 1, so for the same $r$, a Diesel cycle is less efficient than Otto. But Diesel engines run at much higher $r$ (15–22), so in practice they win.

The Brayton Cycle (Gas Turbines)

The Brayton cycle powers jet engines and gas-turbine power plants. It's an open or closed loop of:

The key parameter is the pressure ratio $r_p = P_2/P_1$. The efficiency is:

$$\eta_{Brayton} = 1 - \frac{1}{r_p^{(k-1)/k}}$$

Back-Work Ratio

In gas turbines, the compressor consumes a large fraction of the turbine's output:

$$\text{BWR} = \frac{w_{compressor}}{w_{turbine}}$$

BWR is often 40–80% — enormous compared to the ~1% of vapor cycles. This is why gas turbines need very efficient compressors.

Regeneration

When turbine exhaust (state 4) is hotter than compressor exit (state 2), a regenerator preheats the combustor air with exhaust heat, cutting fuel use. Effectiveness $\epsilon$ sets the recovered heat. Regeneration helps only at low pressure ratios (where $T_4 > T_2$).

Python: Otto Efficiency vs Compression Ratio

import numpy as np
import matplotlib.pyplot as plt

k = 1.4
r = np.linspace(4, 22, 200)
eta_otto = 1 - 1 / r ** (k - 1)

plt.figure(figsize=(9, 5))
plt.plot(r, eta_otto * 100, 'b-', lw=2)
plt.axvline(10, color='g', ls='--', label='Typical gasoline limit (~10)')
plt.xlabel('Compression ratio r')
plt.ylabel('Otto thermal efficiency (%)')
plt.title('Otto Cycle Efficiency vs Compression Ratio')
plt.legend()
plt.grid(True, alpha=0.3)
plt.show()

for ri in [8, 10, 12, 18]:
    print(f"r = {ri:2d} -> eta = {(1 - 1/ri**(k-1))*100:.1f}%")

r =  8 -> eta = 56.5%
r = 10 -> eta = 60.2%
r = 12 -> eta = 63.0%
r = 18 -> eta = 68.5%

Note these air-standard numbers overstate real engines (~25–35%) because of finite combustion, heat loss, and friction.

Python: Diesel Cycle State Analysis

A full state-by-state Diesel analysis ($r = 18$, cutoff $r_c = 2$, intake at 300 K, 100 kPa).

import numpy as np

# Air properties
k = 1.4
cp = 1.005   # kJ/kg.K
cv = 0.718   # kJ/kg.K
R  = 0.287   # kJ/kg.K

# Given
T1, P1 = 300.0, 100.0   # K, kPa
r  = 18.0               # compression ratio
rc = 2.0                # cutoff ratio

# State 2: isentropic compression
T2 = T1 * r ** (k - 1)
P2 = P1 * r ** k

# State 3: constant-pressure heat addition, V3/V2 = rc
T3 = T2 * rc
P3 = P2

# State 4: isentropic expansion back to V1
# V4/V3 = V1/V3 = r/rc  -> T4 = T3 * (V3/V4)^(k-1)
T4 = T3 * (rc / r) ** (k - 1)
P4 = P1 * (T4 / T1)     # constant volume from 4->1

q_in  = cp * (T3 - T2)
q_out = cv * (T4 - T1)
w_net = q_in - q_out
eta   = w_net / q_in

print(f"State 1: T={T1:6.1f} K, P={P1:7.1f} kPa")
print(f"State 2: T={T2:6.1f} K, P={P2:7.1f} kPa")
print(f"State 3: T={T3:6.1f} K, P={P3:7.1f} kPa")
print(f"State 4: T={T4:6.1f} K, P={P4:7.1f} kPa")
print("-" * 38)
print(f"q_in  = {q_in:6.1f} kJ/kg")
print(f"q_out = {q_out:6.1f} kJ/kg")
print(f"w_net = {w_net:6.1f} kJ/kg")
print(f"eta   = {eta*100:5.1f} %")

State 1: T= 300.0 K, P=  100.0 kPa
State 2: T= 953.3 K, P= 5720.0 kPa
State 3: T=1906.6 K, P= 5720.0 kPa
State 4: T= 791.7 K, P=  263.9 kPa
--------------------------------------
q_in  =  958.1 kJ/kg
q_out =  353.0 kJ/kg
w_net =  605.1 kJ/kg
eta   =  63.2 %

Python: Brayton Net Work & Optimal Pressure Ratio

For a Brayton cycle with fixed inlet temperature $T_1$ and maximum temperature $T_3$, net work peaks at a specific pressure ratio. The optimum occurs at:

$$r_{p,opt} = \left( \frac{T_3}{T_1} \right)^{\frac{k}{2(k-1)}}$$

import numpy as np
import matplotlib.pyplot as plt

k  = 1.4
cp = 1.005   # kJ/kg.K
T1 = 300.0   # K  (compressor inlet)
T3 = 1300.0  # K  (turbine inlet, metallurgical limit)

rp = np.linspace(2, 40, 300)
# Isentropic temperature ratios
T2 = T1 * rp ** ((k - 1) / k)
T4 = T3 / rp ** ((k - 1) / k)

w_comp = cp * (T2 - T1)
w_turb = cp * (T3 - T4)
w_net  = w_turb - w_comp
q_in   = cp * (T3 - T2)
eta    = 1 - 1 / rp ** ((k - 1) / k)
bwr    = w_comp / w_turb

# Optimal pressure ratio for max net work
rp_opt = (T3 / T1) ** (k / (2 * (k - 1)))

fig, ax1 = plt.subplots(figsize=(9, 5))
ax1.plot(rp, w_net, 'b-', lw=2, label='Net work')
ax1.axvline(rp_opt, color='k', ls='--', label=f'rp_opt = {rp_opt:.1f}')
ax1.set_xlabel('Pressure ratio rp')
ax1.set_ylabel('Net work (kJ/kg)', color='b')
ax2 = ax1.twinx()
ax2.plot(rp, eta * 100, 'r--', lw=2, label='Efficiency')
ax2.set_ylabel('Thermal efficiency (%)', color='r')
ax1.legend(loc='upper left')
plt.title('Brayton Cycle: Net Work and Efficiency vs Pressure Ratio')
ax1.grid(True, alpha=0.3)
plt.show()

i = np.argmax(w_net)
print(f"Optimal pressure ratio (max work) = {rp_opt:.2f}")
print(f"At rp_opt: net work = {w_net[i]:.1f} kJ/kg, eta = {eta[i]*100:.1f}%, BWR = {bwr[i]*100:.1f}%")
print(f"At rp=20 : eta = {(1 - 1/20**((k-1)/k))*100:.1f}%")

Optimal pressure ratio (max work) = 11.31
At rp_opt: net work = 339.2 kJ/kg, eta = 50.7%, BWR = 50.0%
At rp=20 : eta = 57.5%

Notice the tension: net work peaks near $r_p \approx 11$, but efficiency keeps rising with $r_p$. Designers choose a pressure ratio that balances power output against fuel economy — and the back-work ratio of 50% shows why compressor efficiency is critical.

Cycle Comparison

For the same compression ratio: $\eta_{Otto} > \eta_{Diesel}$. For the same maximum pressure and heat input (the fairer comparison for real engines): $\eta_{Diesel} > \eta_{Otto}$.

Common Pitfalls

• Using Celsius in the cycle formulas. All isentropic relations need absolute temperatures (kelvin).
• Thinking Otto efficiency depends on heat input. It depends only on $r$ and $k$ — heat input changes the work, not the efficiency.
• Ignoring the back-work ratio in Brayton. A 50%+ BWR means small compressor inefficiencies devastate net output.
• Confusing "max work" with "max efficiency". In Brayton they occur at different pressure ratios — pick based on the design goal.
• Over-trusting air-standard numbers. Real efficiencies are far lower due to combustion losses, friction, and heat transfer.

Key Takeaways

• Air-standard assumptions turn messy combustion engines into solvable ideal-gas cycles.
• Otto efficiency $1 - 1/r^{k-1}$ rises with compression ratio but is capped by knock.
• Diesel adds the cutoff ratio; for equal $r$ it's less efficient than Otto, but real diesels run at much higher $r$.
• Brayton efficiency rises with pressure ratio, while net work peaks at $r_{p,opt} = (T_3/T_1)^{k/[2(k-1)]}$; its large back-work ratio demands efficient compressors.
• Regeneration improves Brayton efficiency only at low pressure ratios where exhaust is hotter than compressor exit.

You've now covered the major power cycles — the thermodynamic engines that move the modern world.