# lecture 1
a gas turbine engine has a diffuser at the inlet. This causes pressure and temperature to rise, and the rise is dependant on the airspeed $T_2 = T_1 + \frac{V_o^2}{2C_p} = T_1 (1+\frac{\gamma-1}{2}Ma^2)$
This assumes that $T_2 = T*$

when flying at high mach numbers, the high inlet temperature can basically be at the limit of materials. This means not much fuel can be burned, and therefore not much work can be done. This makes gas turbines ineffective without an afterburner. When increasing mach number more, you can't burn fuel without exceeding material properties.

The way around this is to get rid of the compressor and let the diffuser slow down the air. This also means you can get rid of the turbine because work doesn't need to be extracted for the compressor

## supersonic diffuser
in a subsonic diffuser, the area is expanded to slow down the air

in a supersonic diffuser, oblique shocks (ones that leave the flow supersonic) are used to slow down the air. There is finally a normal shock which brings it to be subsonic. Oblique shocks are used to minimise entropy
After the flow is subsonic, it is then expanded as normal

## supersonic nozzle
a converging-diverging nozzle has to be used to accelerate flow faster than the freestream velocity

## applications

gas turbines typically operate with $500K < T* < 1000K$
ramjets are good from about $2.5 < Ma < 4$
at high speeds, the stagnation temperature is just too high. The pressure losses are also high, reducing efficiency

an ideal ramjet has a constant pressure the whole way through, which is the stagnation pressure.

# lecture 2 part 1

## calculating thrust of a ramjet
$F = \dot m_a [(1+f)V_j - V_o] + A_j(p_j - p_a)$
for a ramjet, $f$ cannot be ignored as they consume a lot of fuel

specific thrust is just thrust per mass flow rate of air. the $A_j$ term can typically be ignored

## components
### diffuser for an ideal ramjet
this is isentropic

the diffuser is  typically adiabatic so the stagnation temperature is constant

$T_{0, x}$ is the stagnation temperature

$q=0$
$T_{0,1} = T_{0, 2}$
$T_2 = T_{0,1} = T_1 (1+\frac{\gamma -1}{2} Ma_1^2)$
this gives the temperature at the outlet of the diffuser

calculating the pressure from the temperature:
$p_{0, 2}= p_{0, 1} = p_1 \left(\frac{T_{0, 1}}{T_1}\right)^{\frac{\gamma}{\gamma -1}}$

### combustor for an ideal ramjet
this is isobaric

![[Drawing 2023-02-24 13.12.37.excalidraw]]
$\dot Q = (\dot m_a + \dot m_f)h_{0, 3} - \dot m_a h_{0, 2}$
$\dot Q = \dot m_f \cdot LCV$
$\dot Q = (\dot m_a + \dot m_f) c_p T_3 - \dot  m_a c_p T_2$
$f \cdot LCV = (1+f) c_pT_3 - c_pT_2$

$T_3$ is often the highest temperature

### nozzle for an ideal ramjet
this is isentropic
$T_3 = T_{0, 3} = T_{0, 4}$. The fuel has no kinetic energy so the stagnation temperature is constant
$T_{0, 4}  =T_4(1+\frac{\gamma - 1}{2} Ma_4^2) = T_4 + \frac{V_4^2}{2c_p}$
$T_4 = T_3 \left(\frac{p_4}{p_3}\right) ^\frac{\gamma -1}{\gamma}$
ideally, $p_4$ is atmospheric pressure

## example
![[Pasted image 20230224132654.png]]
$Ma_1 = 3$
$T_3 = 2500K$

$\frac{F}{\dot m_a} =  [(1+f)V_j - V_o]$

find $V_o, f,V_j$
$V_o = Ma_1 \sqrt{\gamma RT_1} = 951 ms^{-1}$

$f = \frac{T_3 - T_2}{\frac{LCV}{c_p} - T_3}$ (from [[propulsion/wk 4#combustor for an ideal ramjet]])
know everything other than $T_2$

$T_2 = T_{0,1} = T_1 (1+\frac{\gamma -1}{2} Ma_1^2)$([[propulsion/wk 4#diffuser for an ideal ramjet]])
$T_2 = 700K$

$f = 0.0458$

$T_{0, 4} =T_4 + \frac{V_4^2}{2c_p}$
$V_j = V_4 = \sqrt{ 2c_p (T_3 - T_4)}$ ([[propulsion/wk 4#nozzle for an ideal ramjet]])

need to find $T_4$
$T_4 = T_3 \left(\frac{p_4}{p_3}\right) ^\frac{\gamma -1}{\gamma}$
need to find $p_3$
$p_3 = p_2 = p_{0, 2} = 367kPa$

$T_4 = 893 K$
$V_j = 1797 ms^{-1}$

$\frac{F}{\dot m_a} = (1+f)V_j - V_o = 929ms^{-1}$


# lecture 2 part 2

## losses in real ramjets
ideally:
![[Drawing 2023-02-24 13.58.28.excalidraw]]

in reality, none of these components have no losses
ideal: $p_{0, 1} = p_{0, 2} = p_{0, 3} = p_{0, 4}$
real: $p_{0, 1} \neq p_{0, 2} \neq p_{0, 3} \neq p_{0, 4}$

this stagnation pressure loss is characterised by $\Gamma$, the stagnatino pressure ratio. the closer to 1, the better
$\Gamma_d =  \frac{p_{0, 2}}{p_{0, 1}}$
$\Gamma_c =  \frac{p_{0, 3}}{p_{0, 2}}$
$\Gamma_n =  \frac{p_{0, 4}}{p_{0, 3}}$
multiplying them gives the loss across the whole ramjet

$V_4 = \sqrt{sc_p(T_3 - T_4)}$
$T_4 = T_3 \left( \frac{p_4}{p_{0, 4}} \right)^\frac{\gamma -1}{\gamma}$
$p_{0, 4} = \Gamma_d \Gamma_c \Gamma_n$
$V_4 =  \sqrt{2c_p T_3 (1-\left( \frac{p_4}{p_{0, 4}}\right)}^\frac{\gamma -1 }{\gamma}$
most of the losses are due to viscous effects

$\dot Q = \eta_c \times LCV \times \dot m_f$
combustion efficiency represents heat flow from the combustor, and the potential generation of other molecules that take up more energy


![[Drawing 2023-02-24 14.18.14.excalidraw]]
a non-ideal compressor will have to do more work to reach the same pressure, whcih will lead to loss and entropy. A non-ideal nozzle doesn't do work, so just will reach a lower temperature


## equations

### in the diffuser:
$T_2 = T_1 (1+\frac{\gamma-1}{2} Ma^2)$
$p_{0, 2} = p_{0, 1} \Gamma _d$
$p_2 = \Gamma_d \left(\frac{T_2}{T_1}\right)^\frac{\gamma -1}{\gamma} p_2

### in the combustor:
$\dot Q = (\dot m_a + \dot m_f) c_p T_3 - \dot m_a c_p T_2 = \dot m_f \times LCV \times \eta_c$
$f \times \eta_c \times LCV = (1+f) c_p T_3 - c_p T_2$

$p_3 = \Gamma_c p_2 = \Gamma _c \Gamma_d p_{0, 1}$

### in the nozzle
$V_4 = \sqrt{2 c_p(T_3 - T_4)}$
$p_{0, 4} = \Gamma_n p_{0, 3} = \Gamma_n \Gamma_c \Gamma_d p_{0, 1}$
$T_4 = T_3 \left( \frac{p_1}{p_{0, 4}}\right) ^ \frac{\gamma - 1}{\gamma}$

## going faster
high mach numbers lead to high temperatures, reducing combustion efficiency and melting everything ever

this can be worked around by slowing down the air but not making it subsonic, so it doesn't get too hot. This also reduces the pressure losses
This leads to supersonic combustion, which is problematic
this leads to a scramjet
![[Pasted image 20230224143524.png]]