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Propulsion Systems, Lecture 1
Introduction
Classifications of propulsion systems:
1) Airbreathing: reciprocating, turbojet, turbofan, turboprop, turboshaft, ramjet, scramjet,
pulsejet, pulse detonation engine
2) Non-airbreathing: chemical rockets (gas, liquid, solid, hybrid), electric, nuclear, solar, laser,
biological
Overall goal of these systems:
1) Take mass from surroundings and throw it backwards
a
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g
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g
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2) However, the text provides derivations/formulas for either SI or English
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174 ft*lbm/(lbf*s2): ∑ 𝐹⃗ = 𝑔
6)

Don’t get confused when you see gc in book
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Equations of motion for a fixed wing aircraft

x,z: aircraft coordinate
x – through the fuselage centerline
x1 – direction of flight
x’, z’: inertial coordinate, earth frame
ϒ – angle of climb
α – angle of attack
T – thrust, along axis of aircraft coordinate
D – drag, along axis of direction of flight
mg – mass of aircraft, along inertial coordinate
L – lift, perpendicular to direction of flight
⃗⃗

𝑑(𝑚𝑉 )
EOM (Newton’s 2nd law): ∑ 𝐹⃗ = 𝑑𝑡

𝑚𝑔⃗ + 𝐹⃗𝑎𝑒𝑟𝑜 + 𝐹⃗𝑡ℎ𝑟𝑢𝑠𝑡 =

⃗⃗ )
𝑑(𝑚𝑉
𝑑𝑡

Assumptions: dm/dt = 0, excellent for aircraft but not for rockets; aircraft is rigid (excellent)
In wind coordinates (x1, z1)
x1: -mg sinϒ + T cosα – D = m udotX1 (aircraft acceleration in x1)
z1: mg cosϒ - T sinα – L = m udotZ1 (aircraft acceleration in z1)
assume small α: sinα ≈ α and cosα ≈ 1
and for steady flight (no acceleration) and let W = mg
-W sinϒ + T – D = 0

W cosϒ - Tα – L = 0
Compare magnitude of W and T for some examples
Cessna 182 (common 4 seater): W = 2800 lbs, T = 400 lbs; W/T = 7
Boeing 737: W = 105000 lbs, T = 24000 lbs; W/T = 4
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F15: W = 44600 lbs (fuel full) and 28000 lbs (fuel empty), T = 46900 lbs; W/T = 0
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95
In commercial cases L >> Tα
T – D = Wsinϒ
L = Wcosϒ
Special case (steady level flight ϒ = 0)
T=D
L=W
Note during climb L < W since W is somewhat supported by T
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Must consider take off, single engine operation, evasive
maneuvering, etc
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Momentum theorem
⃗⃗ = momentum of mass m (vector quantity)
Elementary mechanics: 𝑚𝑉
⃗⃗⃗ = ∑ 𝑚𝑖 𝑉
⃗⃗𝑖 – momentum of a system of masses mi
𝑀
Fluid dynamics: ρV = momentum density of a fluid with mass density ρ
⃗⃗⃗𝑠𝑦𝑠 = ∫
⃗⃗ 𝑑𝑉 - momentum of fluid system with fixed identity
𝑀
𝜌𝑉
𝑠𝑦𝑠(𝑡)
(dV = differential volume)
In Newtonian mechanics, momentum theorem is:
For fluid mechanics:

⃗⃗⃗𝑠𝑦𝑠
𝑑𝑀
𝑑𝑡

⃗⃗⃗
𝑑𝑀
𝑑𝑡

= ∑ 𝐹⃗𝑖

= 𝐹⃗ , 𝐹⃗ is net force on all mass within system

Let’s define a system which moves with a fluid

𝑑
⃗⃗ 𝑑𝑉 = 𝐹⃗
∫
𝜌𝑉
𝑑𝑡 𝑠𝑦𝑠(𝑡)
Using the Reynolds Transport Theorem (RTT) for any quantity B = mb (B - extensive, b - intensive)
𝑑𝐵𝑠𝑦𝑠
𝑑𝑡

=

𝜕𝐵𝐶𝑉
𝜕𝑡

+ 𝐵̇𝑜𝑢𝑡 − 𝐵̇𝑖𝑛 at time t when CV and system are coincident

General form:
𝑑𝐵𝑠𝑦𝑠
𝜕
⃗⃗ ∙ 𝑛̂)𝑑𝑆
= ∫ 𝜌𝑏𝑑𝑉 + ∫ 𝜌𝑏(𝑉
𝑑𝑡
𝜕𝑡 𝐶𝑉
𝐶𝑆
Moving stationary CV CS enclosing stationary CV
dS is a differential surface element and 𝑛̂ a unit outward normal vector
Apply RTT to mass (Bsys = msys*1, b = 1)
𝑑𝑚𝑠𝑦𝑠
𝑑𝑡

𝜕

⃗⃗ ∙ 𝑛̂)𝑑𝑆 = 0 (Conservation Of Mass -COM)
= 𝜕𝑡 ∫𝐶𝑉 𝜌𝑑𝑉 + ∫𝐶𝑆 𝜌(𝑉

CV and CS fixed in space with 1-D inlets and outlets
𝑑𝑚𝐶𝑉
= ∑𝜌𝑖 𝐴𝑖 𝑉𝑖 |𝑖𝑛𝑙𝑒𝑡𝑠 − ∑𝜌𝑖 𝐴𝑖 𝑉𝑖 |𝑜𝑢𝑡𝑙𝑒𝑡𝑠
𝑑𝑡

⃗⃗ 𝑑𝑉, b = V)
Apply RRT to momentum (Bsys = mV = ∫𝑠𝑦𝑠 𝜌𝑉
𝑑
𝜕
⃗⃗ 𝑑𝑉 = ∫ 𝜌𝑉
⃗⃗ 𝑑𝑉 + ∫ 𝜌𝑉
⃗⃗ (𝑉
⃗⃗ ∙ 𝑛̂)𝑑𝑆
∫ 𝜌𝑉
𝑑𝑡 𝑠𝑦𝑠
𝜕𝑡 𝐶𝑉
𝐶𝑆
𝜕
⃗⃗⃗𝐶𝑉 + ∫ 𝜌𝑉
⃗⃗ (𝑉
⃗⃗ ∙ 𝑛̂)𝑑𝑆 - CV and CS fixed in space (Momentum Theorem – MT)
𝐹⃗ = 𝜕𝑡 𝑀
𝐶𝑆

Application of conservation of mass and the momentum theorem to jet engine

Assumptions: 1) steady, 2) uniform P and V upstream, 3) uniform exit, 4) forces only act along x
COM:

𝜕
∫ 𝜌𝑑𝑉
𝜕𝑡 𝐶𝑉

⃗⃗ ∙ 𝑛̂)𝑑𝑆 = 0  0 (𝑠𝑡𝑒𝑎𝑑𝑦) + ∫ 𝜌(𝑉
⃗⃗ ∙ 𝑛̂)𝑑𝑆 = 0  𝜌0 𝑉0 𝐴0 − 𝜌𝑒 𝑉𝑒 𝐴𝑒 = 0
+ ∫𝐶𝑆 𝜌(𝑉
𝐶𝑆
𝜕

⃗⃗⃗𝐶𝑉 (0, 𝑠𝑡𝑒𝑎𝑑𝑦) + ∫ 𝜌𝑉
⃗⃗ (𝑉
⃗⃗ ∙ 𝑛̂)𝑑𝑆
MT: 𝐹⃗ (𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 + 𝑇) = 𝜕𝑡 𝑀
𝐶𝑆
⃗⃗ (𝑉
⃗⃗ ∙ 𝑛̂)𝑑𝑆
𝐹𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙,𝑥 − ∫ 𝑃𝑛̂𝑑𝑆 = ∫ 𝜌𝑉
𝐶𝑆

𝐶𝑆

𝐹𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙,𝑥 + 𝑃0 𝐴𝑒 − 𝑃𝑒 𝐴𝑒 = ∫

𝑒𝑥𝑖𝑡

𝜌𝑈𝑒2 𝑑𝑆 − ∫

𝑖𝑛𝑙𝑒𝑡

𝜌𝑈02 𝑑𝑆

𝑚𝑒 ̇ = ∫𝑒𝑥𝑖𝑡 𝜌𝑈𝑒 𝑑𝑆 (mass flow rate at exit)
𝑚0 ̇ = ∫𝑖𝑛𝑙𝑒𝑡 𝜌𝑈0 𝑑𝑆 (mass flow rate at inlet)

𝑇 + 𝑃0 𝐴𝑒 − 𝑃𝑒 𝐴𝑒 = 𝑚̇𝑒 𝑈𝑒 − 𝑚̇0 𝑈0
𝑇 = 𝑚̇𝑒 𝑈𝑒 − 𝑚̇0 𝑈0 + 𝐴𝑒 (𝑃𝑒 − 𝑃0 ) - general result, valid for all steady propulsion systems
*“ram effect” always results in a loss of thrust for increasing U0 (flight speed)
If we would like to separate air and fuel flow rates: 𝑚̇𝑒 = 𝑚̇0 + 𝑚̇𝑓
𝑇 = 𝑚̇0 (𝑈𝑒 − 𝑈0 ) + 𝑚̇𝑓 𝑈𝑒 + 𝐴𝑒 (𝑃𝑒 − 𝑃0 )



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