THEORY.md

Theory

This document derives the equations used by main.py and the gasturb package to compute the design point of a single-shaft jet engine with bleed air, turbine cooling and a bypass stream. Everything is one-dimensional, steady and adiabatic towards the environment; the only heat input is the combustor.

1. Station numbering

The stations follow SAE ARP 755A; numbers 6 and 8 are reserved for an afterburner section and are skipped here.

Station	Location
0	free stream
1	captured stream tube, still at free-stream conditions
2	compressor face, behind the inlet cone
3	compressor exit; the bypass stream splits off here
4	combustor exit = turbine inlet
5	turbine exit
7	core nozzle throat
9	core nozzle exit

Three secondary streams are taken by mass fraction:

• Bleed air (mu_bleed), extracted halfway through the compression and dumped overboard — it costs work and ram drag and returns nothing.
• Cooling air (mu_cool), extracted halfway through the compression and mixed back into the hot gas ahead of the turbine.
• Bypass air (mu_bypass), split off at the compressor exit and expanded to ambient pressure through its own nozzle.

2. Working fluid

The gas model (gasturb/gas.py) follows Walsh & Fletcher, Gas Turbine Performance, chapter 3. Air and combustion products are semi-perfect gases: they obey the ideal gas law, but the specific heat depends on temperature. For dry air

$$c_p(T) = \sum_{i=0}^{8} A_i , (T/1000)^i \qquad \text{(F3.23)}$$

and for the products of kerosene combustion at fuel–air ratio $\mathrm{FAR} = \dot m_f / \dot m_\mathrm{air}$

$$c_p(T,\mathrm{FAR}) = c_{p,\mathrm{air}}(T) + \frac{\mathrm{FAR}}{1+\mathrm{FAR}} \sum_{i=0}^{7} B_i ,(T/1000)^i \qquad \text{(F3.24)}$$

The weight $\mathrm{FAR}/(1+\mathrm{FAR})$ is the mass fraction of stoichiometric combustion products in the mixture — one kilogram of mixture contains $1/(1+\mathrm{FAR})$ kilograms of air and $\mathrm{FAR}/(1+\mathrm{FAR})$ kilograms of burned fuel.

Enthalpy and entropy follow by integration,

$$h(T) = \int c_p , dT, \qquad s(T,p) = \int \frac{c_p}{T} , dT - R \ln \frac{p}{p_\mathrm{ref}},$$

which keeps the polynomial form (F3.26/F3.27); the $1/T$ integration turns the constant term of $c_p$ into a logarithm, which is carried separately. The integration constants are chosen so that $h$ and $s$ match the NASA CEA reference values for air at 298.15 K, which makes the model directly comparable with equilibrium calculations.

The gas constant of the products (F3.29) barely differs from that of air, because the molar masses of CO₂ and H₂O straddle that of the O₂ they replace; at FAR = 0.02 the correction is about $2\cdot10^{-4}$ J/(kg K).

Two inversions are needed throughout the cycle and are solved by Newton's method:

• $T(h)$ — temperature from enthalpy, using $dh/dT = c_p$;
• isentropic temperature from a pressure ratio, $s^0(T_2) = s^0(T_1) + R\ln(p_2/p_1)$, using $ds/dT = c_p/T$,

and the inverse of the second, the isentropic pressure for a given temperature change, is explicit:

$$p_2 = p_1 \exp!\left(\frac{s^0(T_2)-s^0(T_1)}{R}\right).$$

3. Atmosphere

gasturb/atmosphere.py implements the U.S. Standard Atmosphere 1976. The geometric altitude $z$ is first converted to geopotential altitude

$$H = \frac{r_E , z}{r_E + z},$$

which absorbs the altitude dependence of gravity ($r_E$ = 6 356 766 m). The model then marches through layers of prescribed temperature gradient $\lambda$. Within a layer, hydrostatic equilibrium $dp = -\rho g_0,dH$ together with the ideal gas law integrates to

$$\frac{p}{p_b} = \left(\frac{T}{T_b}\right)^{-g_0/(R\lambda)} \quad (\lambda \ne 0), \qquad \frac{p}{p_b} = \exp!\left(-\frac{g_0 (H-H_b)}{R,T}\right) \quad (\lambda = 0),$$

where the subscript $b$ marks the base of the layer. At 40 000 ft the flight condition sits in the isothermal lower stratosphere: $T_0$ = 216.65 K, $p_0$ = 0.188 bar.

4. Free stream and inlet (stations 0–2)

With the flight Mach number the velocity is $u_0 = \mathrm{Ma}_0 , c(T_0)$, where $c = \sqrt{\gamma R T}$. The total state follows from bringing the flow to rest adiabatically and reversibly:

$$h_{t0} = h(T_0) + \tfrac12 u_0^2, \qquad T_{t0} = T(h_{t0}), \qquad p_{t0} = p_0 \exp!\left(\frac{s^0(T_{t0})-s^0(T_0)}{R}\right).$$

At Ma 0.85 this gives $T_{t0} \approx 248$ K and $p_{t0}/p_0 \approx 1.60$ — useful sanity anchors for the printout.

The engine swallows the stream tube that fills the compressor face, $A_1 = \pi D_o^2/4$, so the mass flow is $\dot m_1 = \rho_0 u_0 A_1$. Between stations 1 and 2 the inlet cone blocks the centre of the duct, $A_2 = A_1 - \pi D_i^2/4$. The contraction is solved as an isentropic duct: the static state at 2 satisfies conservation of mass, entropy and total enthalpy,

$$\rho_2 u_2 A_2 = \dot m_1, \qquad s(T_2,p_2) = s(T_1,p_1), \qquad h(T_2) + \tfrac12 u_2^2 = h_t,$$

solved with a damped Newton method (Gas.compress). A subsonic flow accelerates through a contraction, so Ma₂ is slightly above Ma₀. The inlet is treated as adiabatic and lossless, hence the total state of the free stream is preserved: $T_{t2} = T_{t0}$, $p_{t2} = p_{t0}$.

5. Compressor (station 3)

The compressor is specified by its overall pressure ratio and isentropic efficiency. On total quantities,

$$p_{t3} = \mathrm{OPR} \cdot p_{t2}, \qquad w = \frac{h_t(T_{t3s}) - h_{t2}}{\eta_c},$$

where $T_{t3s}$ is the isentropic exit temperature for the same pressure ratio. The efficiency divides the ideal work because a real compressor must invest more work to reach the same pressure. The real exit state is $h_{t3} = h_{t2} + w$.

Bleed and cooling extraction. Both secondary flows are drawn off where half of the specific work has been done, i.e. at the total enthalpy $h_{t3c} = h_{t2} + w/2$. The full flow $\dot m_1$ absorbs the first half of the work, only the remainder $\dot m_3 = \dot m_1(1-\mu_\mathrm{cool}-\mu_\mathrm{bleed})$ continues through the second half, so the shaft must provide

$$P_\mathrm{comp} = \left(\dot m_1 + \dot m_3\right) \tfrac12 w .$$

The static state at the exit area $A_3 = A_2/\mathrm{AR}$ is recovered from the total state and continuity (Gas.static_from_total): starting from rest, the iteration

$$u \leftarrow \frac{\dot m}{\rho(T,p),A}, \qquad T \leftarrow T!\left(h_t - \tfrac12 u^2\right), \qquad p \leftarrow p_t \exp!\left(\frac{s^0(T)-s^0(T_t)}{R}\right)$$

converges to the subsonic branch.

Bypass split. The bypass stream leaves the core at station 3. Both streams share the same static state, so the duct areas divide in proportion to the mass flows: $A_{3a}/A_3 = \dot m_\mathrm{core}/\dot m_3$.

6. Combustor (station 4)

Only the core stream is burned. The fuel–air ratio defines the fuel flow $\dot m_f = \mathrm{FAR} \cdot \dot m_\mathrm{core}$, and the energy balance over the burner, per kilogram of mixture, reads

$$h_{t4} = \frac{h_{t3} + \eta_b , H_u , \mathrm{FAR}}{1+\mathrm{FAR}} .$$

The combustion efficiency $\eta_b$ acts only on the heat release: unburned fuel still passes through the turbine as mass, it just contributes no heat. From $h_{t4}$ the turbine inlet temperature follows with the product-gas polynomials, $T_{t4} = T(h_{t4}, \mathrm{FAR})$.

Pressure loss. The total pressure drops for two reasons: friction and mixing in the liner (the cold loss), and the fundamental loss of adding heat to a moving gas (the hot or Rayleigh loss). Both scale with the dynamic head of the incoming flow, which is captured by the non-dimensional flow parameter

$$X = \frac{\dot m_\mathrm{core}\sqrt{R,T_{t3}}}{p_{t3},A_4}, \qquad X^2 \approx \gamma,\mathrm{Ma}^2 ,$$

so that

$$\frac{\Delta p_t}{p_{t3}} = X^2\left(k_\mathrm{aer} + k_\mathrm{th}\left(\frac{T_{t4}}{T_{t3}} - 1\right)\right).$$

The defaults are anchored in the literature: $k_\mathrm{aer} = 10$ corresponds to a pressure-loss factor $\Delta p_t / q_\mathrm{ref} \approx 20$, typical of annular combustors (Lefebvre), and $k_\mathrm{th} = 0.5$ reproduces the low-Mach Rayleigh result $\Delta p_t/p_t \approx \tfrac{\gamma}{2}\mathrm{Ma}^2 (T_{t4}/T_{t3}-1)$. With the present geometry the burner inlet Mach number is about 0.08, giving a total loss of roughly 10 %.

7. Turbine (station 5)

Cooling-air mixing. Ahead of the turbine the cooling air rejoins the hot gas. The mixing is modelled at constant total pressure; the mixed enthalpy is the mass-weighted average

$$h_{t4m} = \frac{\dot m_4 h_{t4} + \dot m_\mathrm{cool}, h_{t3c}}{\dot m_5}, \qquad \dot m_5 = \dot m_4 + \dot m_\mathrm{cool},$$

and the fuel–air ratio is diluted accordingly, $\mathrm{FAR}_5 = \dot m_f / (\dot m_5 - \dot m_f)$.

Shaft balance. On a single shaft the turbine must deliver exactly the compressor power plus the mechanical losses:

$$P_\mathrm{shaft} = \frac{P_\mathrm{comp}}{\eta_\mathrm{shaft}}, \qquad h_{t5} = h_{t4m} - \frac{P_\mathrm{shaft}}{\dot m_5}.$$

This fixes the real exit enthalpy — energy conservation leaves no freedom here. What the isentropic efficiency determines is the pressure this work costs: an ideal turbine would extract the same work with a smaller expansion ratio. The ideal end state

$$h_{t5s} = h_{t4m} - \frac{h_{t4m} - h_{t5}}{\eta_t} ;\le; h_{t5}$$

lies on the isentrope through the inlet state, so the exit pressure follows from the entropy relation evaluated at $T_{t5s} = T(h_{t5s})$:

$$p_{t5} = p_{t4} \exp!\left(\frac{s^0(T_{t5s}) - s^0(T_{t4m})}{R}\right).$$

Because $h_{t5s} < h_{t5}$, the real machine expands further in pressure than the ideal one for the same work — the entropy production is paid in pressure, not in energy.

8. Nozzles (stations 7 and 9)

Choking. A convergent–divergent nozzle only works as such if the pressure ratio exceeds the critical value,

$$\frac{p_{t5}}{p_0} > \left(1 + \frac{\gamma-1}{2}\right)^{\gamma/(\gamma-1)} \approx 1.85,$$

which the deck verifies (here $p_{t5}/p_0 \approx 3$).

Throat. At the throat the flow is exactly sonic. The static temperature $T_7$ satisfies the implicit energy equation

$$h(T_7) + \tfrac12 c(T_7)^2 = h_{t5},$$

solved by fixed-point iteration; pressure follows from the isentrope and the throat area from continuity, $A_7 = \dot m_5 / (\rho_7 u_7)$.

Exit. The divergent section expands the flow isentropically to ambient pressure ($p_9 = p_0$, an adapted nozzle, which maximizes thrust). The exit temperature comes from the isentrope, the velocity from the energy equation:

$$T_9 = T_s(T_{t5}, p_{t5} \to p_0), \qquad u_9 = \sqrt{2\left(h_{t5} - h(T_9)\right)} .$$

Bypass nozzle. The bypass stream carries the full compressor exit state $(T_{t3}, p_{t3})$ and expands through its own adapted nozzle by the same two equations. Because it was compressed through the full pressure ratio but never heated, it leaves cold and fast — an unusual design compared to a turbofan, where the bypass air only passes the fan, but a thermodynamically consistent one.

9. Thrust bookkeeping

The net thrust is the momentum theorem applied to a control volume around the whole engine:

$$F = \underbrace{\dot m_5 u_9 + A_9 (p_9 - p_0)}_{\text{core, gross}} + \underbrace{\dot m_\mathrm{byp} u_{9b}}_{\text{bypass, gross}} - \underbrace{\dot m_1 u_0}_{\text{ram drag}} .$$

Conventions used here:

• Both nozzles are adapted, so the pressure terms vanish; the core term is kept in the code for clarity.
• The ram drag is charged on the full captured flow $\dot m_1$, including the bleed air. The bleed stream is dumped overboard with no thrust credit — its momentum is simply lost.
• The fuel adds exhaust momentum ($\dot m_5$ contains $\dot m_f$) but no ram drag, since it is carried on board.

Two figures of merit close the summary: the specific thrust $F/\dot m_1$ and the thrust specific fuel consumption $\mathrm{TSFC} = \dot m_f / F$.

Efficiency chain. The fuel power is converted to thrust power in two stages, and each stage has its own efficiency (Walsh & Fletcher). The thermal efficiency measures how much of the fuel heat ends up as kinetic energy of the jets, evaluated in the aircraft-fixed frame:

$$\eta_\mathrm{th} = \frac{\Delta \dot E_\mathrm{kin}}{\dot m_f H_u}, \qquad \Delta \dot E_\mathrm{kin} = \tfrac12\left(\dot m_5 u_9^2 + \dot m_\mathrm{byp} u_{9b}^2\right) - \tfrac12 \dot m_1 u_0^2 .$$

The propulsive efficiency measures how much of that kinetic energy becomes thrust power — the remainder is left behind as swirl-free wake energy:

$$\eta_p = \frac{F,u_0}{\Delta \dot E_\mathrm{kin}} .$$

Their product is the overall efficiency, which links fuel flow directly to thrust power:

$$\eta_0 = \eta_\mathrm{th},\eta_p = \frac{F,u_0}{\dot m_f H_u} .$$

10. References

• P. P. Walsh, P. Fletcher, Gas Turbine Performance, 2nd ed., Blackwell, 2004. — Gas property polynomials (F3.23–F3.29), cycle conventions.
• A. H. Lefebvre, D. R. Ballal, Gas Turbine Combustion, 3rd ed., CRC Press, 2010. — Combustor pressure-loss factors.
• U.S. Standard Atmosphere 1976, NOAA/NASA/USAF, Washington D.C., 1976.
• S. Gordon, B. J. McBride, Computer Program for Calculation of Complex Chemical Equilibrium Compositions and Applications (NASA CEA), NASA RP-1311, 1994. — Reference state of air.
• SAE ARP 755A, Gas Turbine Engine Performance Station Identification and Nomenclature. — Station numbering.