AE481 · Lesson 7

Mission Analysis

Integrate the aircraft constraint equation through each mission segment to estimate fuel use.

Fuel used is the time integral of the constraint master equation

Adjust a representative mission segment. The equation and history update together.

Integrate fuel flow to find fuel used t₁∫t₀ c_tTW_TOdt = t₁∫t₀ −Ẇ_fW_TOdt = ΔW_fW_TO

Time Integral of Constraint Master Equation ΔW_fW_TO = c_t q C_DoW_TO/S + k₁ nβq² W_TO/S + k₂ nβq Δt + c_tβ ΔhV + ΔVg

The demonstration assumes constant aerodynamic and atmospheric inputs during one segment. Later panels can chain segments into a complete mission.

Altitude affects thrust available and thrust required

Endurance and range come at different points on the drag curve

For a turbojet, maximum endurance occurs at minimum thrust required. Maximum range occurs where a line from the origin is tangent to the thrust-required curve.

Cruise Legs

A steady, level cruise leg has no altitude or speed change, so fuel is used only to overcome aerodynamic drag.

Master equation reduction

ΔW_f/W_TO=drag burn × t_cruise+altitude-energy term+speed-energy term

Steady, level cruise: Δh = 0, ΔV = 0, and n = 1.

1 · Cruise time

For a constant-speed leg, convert the required cruise distance into time.

t_{cruise} = \frac{R_{cruise}}{V}

2 · Remove the energy-change terms

Set $Δ h = 0$ , $Δ V = 0$ , and $n = 1$ . The master equation retains only the drag-related fuel burn.

\frac{Δ W_{f}}{W_{TO}} = c_{t} q [\frac{C_{D o}}{W_{TO} / S} + k_{1} {(\frac{β}{q})}^{2} (\frac{W_{TO}}{S}) + k_{2} (\frac{β}{q})] t_{cruise}

Turns

SAME AS CRUISE EXCEPT: n > 1

Master equation reduction

ΔW_f/W_TO=drag burn × t_turn+altitude-energy term+speed-energy term

Level, constant-speed turn: Δh = 0 and ΔV = 0.

1 · Turn time

t_{turn} = \frac{Δ Ψ}{ω} = \frac{Δ Ψ V}{g \sqrt{n^{2} - 1}}

2 · Simplify the master equation

For a level, constant-speed turn, $Δ h = 0$ and $Δ V = 0$ . Only the drag term remains.

\frac{Δ W_{f}}{W_{TO}} = c_{t} q [\frac{C_{D o}}{W_{TO} / S} + k_{1} {(\frac{n β}{q})}^{2} (\frac{W_{TO}}{S}) + k_{2} (\frac{n β}{q})] t_{turn}

Take-Off

Master equation reduction

ΔW_f/W_TO=fuel flow × t_TO+altitude-energy term

Ground roll: Δh = 0; acceleration determines the time to reach V_TO.

1 · Ground acceleration

Use the conservative case $L = 0$ .

\frac{d V}{d t} = g (\frac{T - D}{W} - μ)

2 · Time to take off

Δ V = V_{TO} = 1.2 \sqrt{\frac{2 W_{TO}}{ρ S C_{L \max}}} t_{TO} \approx \frac{V_{TO}}{d V / d t}

\frac{W_{f TO}}{W_{TO}} = α \frac{T_{SL}}{W_{TO}} c_{t} t_{TO}

WTO/S

TSL/WTO

CLα, per deg

Climbs

Master equation reduction

ΔW_f/W_TO=drag burn+altitude-energy term+speed-energy term

Constant-speed climb: dV/dt = 0.

1 · Constant-speed simplification

Set $\frac{d V}{d t} = 0$ and solve the master equation for rate of climb.

\frac{d h}{d t} = V (\frac{α T_{SL}}{β W_{TO}} - \frac{q}{β} \frac{C_{D o}}{W_{TO} / S} - \frac{n^{2} β}{q} k_{1} \frac{W_{TO}}{S})

2 · Climb time

t_{climb} = \frac{h_{final} - h_{initial}}{d h / d t}

\frac{W_{f climb}}{W_{TO}} = α \frac{T_{SL}}{W_{TO}} c_{t} t_{climb}

Acceleration

Master equation reduction

ΔW_f/W_TO=drag burn+altitude-energy term+speed-energy term

Constant-altitude acceleration: dh/dt = 0.

1 · Constant-altitude simplification

Set $Δ h = 0$ . The speed-change energy term remains.

\frac{d V}{d t} = g (\frac{T}{W} - \frac{D}{W})

2 · Acceleration time

t_{accel} = \frac{V_{1} - V_{0}}{d V / d t}

\frac{Δ W_{f}}{W_{TO}} = c_{t} q [\frac{C_{D o}}{W_{TO} / S} + k_{1} {(\frac{n β}{q})}^{2} (\frac{W_{TO}}{S}) + k_{2} (\frac{n β}{q})] t_{accel} + c_{t} β \frac{1}{g} (V - V_{0})

Thrust Specific Fuel Consumption (TSFC)

A simple mission-analysis model relates thrust-specific fuel consumption to the local standard-day temperature ratio.

c_{t} = TSFC = C \sqrt{θ}

Definitions and assumptions

C is the engine’s sea-level-static TSFC.

θ is the local-to-sea-level standard temperature ratio.

TSFC is often treated as weakly dependent on speed and throttle for preliminary mission analysis.
A representative c_t may be selected for each mission segment.
More detailed engine models can replace this approximation when better data are available.

Mach-dependent example

Mattingly’s more detailed models include Mach number and power setting. The following low-bypass turbofan with afterburner example is identified in Section 3.3.2 of MHD.

c_{t} = TSFC = (0.9 + 0.30 M) \sqrt{θ}

Military power

c_{t} = TSFC = (1.6 + 0.27 M) \sqrt{θ}

Maximum power

Mission Fuel Calculations

Combine the fuel used by each mission leg to update the remaining aircraft weight fraction, β.

β_{accel} = β_{takeoff} - \frac{W_{f, takeoff}}{W_{TO}}

β_{climb} = β_{accel} - \frac{W_{f, accel}}{W_{TO}}

β_{cruise} = β_{climb} - \frac{W_{f, climb}}{W_{TO}}

β_{loiter} = β_{cruise} - \frac{W_{f, cruise}}{W_{TO}}

Total mission fuel fraction

\frac{W_{f, mission}}{W_{TO}} = \frac{W_{f, takeoff}}{W_{TO}} + \frac{W_{f, accel}}{W_{TO}} + \frac{W_{f, climb}}{W_{TO}} + \frac{W_{f, cruise}}{W_{TO}} + \frac{W_{f, loiter}}{W_{TO}}