Every efficiency on this page answers the same sentence — what you got, over what you paid for — and they differ only in what the question hands you.

Thermal efficiency: $\eta = \dfrac{W}{Q_h}$, where $W$ is the net work delivered and $Q_h$ the heat supplied, both energies in the same unit. Use it when the totals are already in hand.

Carnot efficiency: $\eta = 1 - \dfrac{T_c}{T_h}$, with $T_h$ the hot reservoir temperature and $T_c$ the cold one — and both must be absolute, in kelvin. This is not a machine, it is a ceiling: no engine between those temperatures does better, whatever it is built from. Celsius dropped into that ratio is the classic wound, and it always flatters the answer.

Rankine cycle: $\eta = \dfrac{(h_1 - h_2) - (h_4 - h_3)}{h_1 - h_4}$. Four state points, all kJ/kg, going round: $h_1$ turbine inlet, $h_2$ turbine exhaust, $h_3$ condensate at the pump inlet, $h_4$ feedwater leaving the pump. The first bracket is the turbine's work, the second is the pump's — a small debt, but net means net.

Otto: $\eta = 1 - \dfrac{1}{r^{\gamma-1}}$, with $r$ the compression ratio. Brayton: $\eta = 1 - \dfrac{1}{r_p^{(\gamma-1)/\gamma}}$, with $r_p$ the pressure ratio. $\gamma$ (gamma) is the heat capacity ratio, 1.4 for cold air. Both are dimensionless, both say the same thing — squeeze harder, gain efficiency — and the only difference is the exponent. Brayton's is divided by $\gamma$ as well, so using Otto's on a gas turbine is the named mistake here.

Carnot is the rail under all of it. Compute the ceiling first, and any cycle answer that climbs above it is arithmetic to go back and find, not a discovery.