Going deeper12 min read4 / 9

Turbojet cycle analysis

This is the course’s engine room: the complete ideal cycle analysis of a turbojet, in the τ-notation that professionals use and that the demo’s solver implements line for line. Master this one derivation and every other engine family is a variation — the turbofan adds one term, the ramjet deletes one.

The four τ’s

Everything is non-dimensionalised as a total-temperature ratio across a component. For flight at Mach $M_0$ with ambient temperature $T_0$ :

\tau_r = 1 + \frac{\gamma-1}{2}M_0^2, \qquad \tau_c = \pi_c^{(\gamma-1)/\gamma}, \qquad \tau_\lambda = \frac{T_{t4}}{T_0}

(1)

$\tau_r$: ram heating from slowing the freestream (station 0 → 2) [–]
$\tau_c$: compressor temperature rise, from its pressure ratio πc [–]
$\tau_\lambda$: the cycle temperature ratio — the hottest point over the coldest [–]
$\tau_t$: turbine temperature drop (≤ 1), found from the work balance below [–]

Multiplying τ’s chains components: the burner-entry temperature is $T_{t3} = T_0\,\tau_r \tau_c$ , and the demo’s Tt3 readout is literally that product.

Step 1 — the turbine must feed the compressor

The turbine’s only job in a turbojet is to drive the compressor through the shaft. Ideal work balance, per kg of air, with $w = c_p \Delta T_t$ :

c_p\,(T_{t4} - T_{t5}) = c_p\,(T_{t3} - T_{t2})

(2)

Divide by $c_p T_0$ and write in ratios ( $T_{t4} = T_0\tau_\lambda$ , $T_{t3}-T_{t2} = T_0\,\tau_r(\tau_c - 1)$ ):

\tau_t = 1 - \frac{\tau_r}{\tau_\lambda}\,(\tau_c - 1)

(3)

This is the cycle’s central constraint. If you demand more compression ( $\tau_c\uparrow$ ) or fly faster ( $\tau_r\uparrow$ ) without raising $\tau_\lambda$ , the turbine must drop further — until $\tau_t$ hits the point where nothing is left for thrust. The demo’s feasible flag trips exactly when this bookkeeping fails.

Step 2 — the nozzle cashes in what remains

After the turbine, the gas still holds total temperature $T_{t5} = T_0\,\tau_\lambda \tau_t$ at high pressure. The ideal nozzle expands it isentropically all the way back to ambient pressure, converting the excess into velocity. Energy conservation (with total-to-static bookkeeping across the whole cycle) gives the classic result:

\frac{V_9^2}{2} = c_p\,T_{t5}\left(1 - \frac{1}{\tau_r\,\tau_c\,\tau_t}\right)

(4)

— the expansion ratio in the bracket is the total-temperature ratio the gas can fall through, which for the ideal cycle (all pressure ratios preserved) is the product of every temperature ratio on the way in and through.

Step 3 — assemble the outputs

The three numbers on every demo readout follow directly:

\frac{F}{\dot m} = V_9 - V_0 \qquad\text{(specific thrust)}

(5)

f = \frac{c_p\,T_0\,(\tau_\lambda - \tau_r\tau_c)}{h_{PR}} \qquad\text{(fuel/air ratio: heat to climb from } T_{t3}\text{ to } T_{t4})

(6)

\text{TSFC} = \frac{\dot m_f}{F} = \frac{f}{F/\dot m} \qquad\text{(fuel per newton-second)}

(7)

$h_{PR}$: fuel heating value — 43 MJ/kg for jet fuel in the demo [J/kg]
$F/\dot m$: thrust per unit air flow; numerically a velocity [N·s/kg]
$\text{TSFC}$: demo units: grams of fuel per kilonewton-second [g/(kN·s)]

What the solution teaches

There is a best πc for thrust. More compression raises thermal efficiency but starves the nozzle (equation 3 again). Specific thrust peaks at a finite pressure ratio — the demo’s Mach-sweep chart shows the whole ridge.
Speed eats the turbojet from both ends. As $M_0$ rises, $V_0$ grows while ram heating shrinks the temperature margin $\tau_\lambda - \tau_r\tau_c$ , so $V_9 - V_0$ pinches from both sides. Past M ≈ 2–3 the pure turbojet fades — precisely where the ramjet takes over.
Its exhaust is fast, so its ηₚ is poor at subsonic speed — the momentum article’s trade, and the reason the next article invents the turbofan.