# theory
a gas turbine engine only needs a compressor, heat addition (usually through combustion), and a turbine to power the compressor. This is a pure turbojet.
After this, an additional low pressure turbine can be added to power other things, like a fan for a turbofan, propellors for ships or turboprops, or a generator for power generation

## the brayton cycle
this consists of isentropic compression, isobaric heat addition, isentropic expansion, and isobaric heat rejection. 
the thermal efficiency is then $\eta_{th} = \frac{w_{x, net}}{q_{in}}$
in aircraft, thrust is produced rather than shaft work, and it is easier to think of chemical energy of the fuel rather than the heat added. 
brayton cycle analysis is useful, as it says increasing the pressure ratio is how to increase the efficiency, but it is flawed.

# analysis of a real engine
## improvements
turbomachinery isn't isentropic
the combustor has a pressure loss
$c_p$ and $\gamma$ vary with temperature, and $c_p$ is different after combustion
adding fuel increases mass flow rate
turbine exhaust is released to the atmosphere so it isn't a cycle

**for aircraft specifically:**
the inlet conditions vary with speed and mach number
there is no net work from the spool
the back pressure downstream may be different to inefficiencies in the nozzle

# example of a turbojet in flight
a turbojet is flying at Mach 2 at 31000ft
![[images/Pasted image 20230407103713.png]]
![[images/Drawing 2023-04-07 10.37.31.excalidraw]]
1 is the inlet, 2 is the inlet to the fan, 3 is the inlet to the combustor, 4 is the inlet to the turbine, 5 is the inlet to the nozzle and 6 is the exhaust

## inlet conditions
the altitude is known, so the databook can be used to find the static temperature and pressure
$T_1 = T_{atm} = 226.73K$ and $p_1 = p_{atm} = 27.8kPa$
because the engine is moving at mach 2, the perfect gas relations for compressible flow must be used, $\frac{T_0}{T} = 1+\frac{\gamma +1}{2}Ma^2$ and $\frac{p_0}{p} = \left( 1+\frac{\gamma +1}{2} Ma^2 \right)^\frac{\gamma}{\gamma - 1}$
from these, $T_{01} = 408.1K$ and $p_{01} = 224.6kPa$
by finding the speed of sound at the **static** temperature, the velocity of the engine can be found as $603.7ms^{-1}$

## $1\to 2$
this part is adiabatic and isentropic. This means that $q_{1,2} = 0$ there is no net work either, so the stagnation enthalpy remains constant
$T_{02} = T_{01}  =408.1K$ $p_{02} = p_{01} = 224.6kPa$

## $2\to 3$
the pressure ratio is given as 30, meaning that $p_{03} = 30 p_{02} = 6738kPa$
the isentropic heat can be found with $T_{03s} = T_{02} (\frac{p_{03}}{p_{02}})^\frac{\gamma - 1}{\gamma} = 1078.45K$. **THIS ISN'T GIVEN IN THE DATA BOOK**
the compressor isn't isentropic, meaning that there will be additional heat. $\eta_C = \frac{h_{03s} - h_{02}}{h_{03} - h_{02}}$. Because $\Delta h = c_p\Delta T$, the enthalpies can be replaced with the calculated temperatures. 
$0.9 = \frac{1078.45 - 408.1}{T_{02} - 408.1}$ gives $T_{02} = 1152.9K$

using the steady flow energy equation:
$\dot q - w_{x, 23} = c_p(T_3 - T_2)$
heat addition is 0 so specific work is dependant on the temperature change 
$-w_{x, 23} = 1000(1152.9 - 408.1) = 744800 J kg^{-1}$
this is the power needed to drive the compressor, so it is obvious how a lower isentropic efficiency means a less efficient compressor

## $3\to 4$
the burner has a pressure loss of 4%, so $p_{04} = 0.96p_{03} = 6468.48$
### finding fuel-to-air ratio $f$
use steady flow energy equation
there is no work done, and no heat is added. The energy is instead added as chemical potential energy

$\dot Q - \dot W_x = \dot m_{air} c_{p, air} (T_{ref} - T_{03}) - \dot m_f LCV$