# lecture 0
80% in person exam

# lecture 1

## rockets
\+ momentum out is all thrust
\- you have to carry the oxidiser with you

## air breathing
\+ don't have to carry oxidiser
\- there is also momentum in too

# lecture 2

## momentum conservation

### theory
using control volumes, resultant forces can be found
![[momentum conservation]]
$\dot M_{x, out} - \dot M_{x, in} = \sum F_x$
$\dot M_2 V_2 - \dot M_1 V_1 = A_1 P_1 - A_2 P_2 + F$

This can be used to analyse engines

### applications

#### rocket engine

$V_{in} = 0$
$V_{out} = V_j$
$A_2$ is the area of the rocket exhaust
$\dot m_p$ is the mass flow rate of the fuel

$\dot M_{x, out} - \dot M_{x, in} = \sum F_x$
$\dot M_{x, in} = 0$
$\dot m_p V_j = p_1A_j - p_2A_j + F$

$P_1$ is the atmospheric pressure
$P_2$ is the pressure of the exhaust

$F = \dot m_p V_j + A_j(P_j-P_a)$
this shows that the force is strongly dependant on the altitude of the rocket

#### gas turbine engine

$V_{in} = V_o$, the speed of the aircraft
$V_{out} = V_j$
$\dot m_{in} = \dot {m}_a$, the mass flow rate of air
$\dot m_{out} = \dot m_a + \dot m_f$
the pressure at the front is just atmospheric pressure, but the pressure at the back is atmospheric pressure + pressure from the jet. To make things simpler, the area considered is just the area of the jet

$V_j(\dot m_a + \dot m_j) - V_o \dot m_a = A_j(p_a - p_j) + F$

$p_j \approx p_a$ **assumption**
the fuel to air ratio is $f =\frac{\dot m_f}{\dot m_a}$

$F = \dot m_a \left[(1+f)V_j - V_o \right]$

for many operations, $f \approx 0$ so $F = \dot m_a \left[V_j - V_o \right]$

## breguet range equation

assume steady level flight, so $L=W$, $T=D$ and $V_0 = \text{const}$

$W=L=D\frac{L}{D} = T\frac{L}{D}$

fuel consumption:
$\frac{dW}{dt} = -\dot m_f g$
$TSFC = \frac{\dot m_f}{F}$, thrust specific fuel consumption
rearrangements and insertions
$\frac{dW}{dt} = -TSFC \cdot F \cdot g = -g\cdot TSFC \cdot \frac{W}{\frac{L}{D}}$
$\frac{dW}{W} = -g\frac{TSFC}{\frac{L}{D}} dt$
$ds = V_o dt$
$\frac{dW}{W} = -\frac{g}{V_o}\frac{TSFC}{\frac{L}{D}} ds$
integrate things
$\int^{w_2} _{w_1} \frac{dW}{W} = -\frac{g}{V_o} \frac{TSFC}{\frac{L}{D}}\int_0^s ds$
$s=\frac{L}{D} \frac{V_o}{g\cdot TSFC} \ln \frac{w_1}{w_2}$

#### example: b777
Empty mass: $145,150kg$
fuel capacity: $145, 538kg$
cruise speed: $248m\ s^{-1}$
range: $15,843km$
L/D: $25$

estimate:
	fuel mass flow rate
	TSFC
	Thrust

$TSFC = \frac{L}{D} \frac{V}{g\cdot s} \ln \frac{w_1}{w_2}$
$TSFC = 25 \frac{248}{9.81 \cdot 15843 \times 10^3} \ln 2 = 2.77\times 10^{-5}$

$\dot m_f = \frac{m_f}{\Delta t} = \frac{m_f}{\frac{s}{v}} = 2.282.28 kg\ s^{-1}$

$F=\frac{\dot m_f}{TSFC}$ = 82.3 kN



# efficiency

overall efficiency  = propulsive efficiency $\times$ thermal efficiency
$\eta_o = \eta_p \times \eta_{th}$

## propulsive efficiency
$\eta_p = \frac{\dot W_{aircraft}}{\dot W_{jet}}$
$\dot W_{aircraft} = F\times V_o$
$F=\dot m_a [(1+f) V_f - V_o]$

$\dot W_{jet} = \frac{d}{dt} (E_{k, out} - E_{k, in})$
$\frac{1}{2} (\dot m_a + \dot m_f) V_j^2 - \frac{1}{2} \dot m_a V_o^2$

things from before:
$\frac{1}{2} \dot m_a[(1+f)V_j^2-V_o^2]$

$\eta_p = \frac{V_o [(1+f) V_f - V_o]}{\frac{1}{2}[(1+f)V_j^2-V_o^2]}$
assume $f<<1$

$\eta_p = \frac{2V_o[V_j - V_o]}{(V_j^2 - V_o^2)} \approx \frac{2V_o}{V_j + V_o} = \frac{2}{\frac{V_j}{V_o}+1}$
want to make $V_j$ as close as possible to $V_o$ for maximum efficiency
![[Pasted image 20230203141306.png]]
Because $F=\dot m_a (V_j - V_o)$, big thrust means low efficiency and high efficiency means high thrust.
little known life hack is increases mass flow rate of air while keeping $V_j - V_o$ good

## thermal efficiency
$\eta_{th} = \frac{\dot W}{\dot Q_{in}}$
work transfer is known, it is $\dot W_{jet}$ from above

$LCV$ is the lower calorific value
$\dot Q_{in} = \dot m_f \times LCV$

$\eta_{th} = \frac{\frac{1}{2} \dot m_a (V_j^2 - V_a^2)}{\dot m_f \cdot LCV} = \frac{V_j^2 - V_o}{2\times f \times LCV}$

from thermofluids, best way to increase efficiency is to increase maximum temperature. This is done by increasing pressure ratio
![[Brayton cycle]]

## overall efficiency
$\eta_o  = \frac{(V_j - V_o)V_o}{f\times LCV}$
$\eta_o = \frac{V_o}{TSFC \times LCV}$
$\eta_o = \eta_p \times \eta_{th}$

## Altitude tables
The atmosphere is complicated, so just use one standard atmosphere


# problem class 1

## 1.2
### a
$TSFC = \frac{\dot m_f}{F} \to TSFC\times F = \dot m_f$
$\dot m_f = 2.38 \dot{kg}$

$\dot m_a = V\times A_1 \times \rho = 666\dot{kg}$
$f = \frac{\dot m_f}{\dot m_a} = 3.57\times 10^{-3}$

### b
$F=\dot m_a \left[(1+f) (V_j - V_0)\right] \to \frac{F}{\dot m_a(1+f)}+V_0 = V_j$
$V_j  = 409$

### c
$\eta_p = \frac{\dot W_{aircraft}}{\dot W_{jet}}$
$\dot W_{aircraft} = F \times V_0 = 28\times 10^6$
$\dot W_{jet} = \frac{1}{2} (\dot m_a + \dot m_f) V^2_j  - \frac{1}{2}\dot m_a V_0^2 = 43\times 10^6$