Exercise set

Aerospace Propulsion and Flight Performance Exercises

Worked aerospace exercises for Mach, climb, OEI climb, nozzle flow, cruise fuel, takeoff margin, Breguet range, loiter and validation.

Branch: Aerospace Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.2.6 · reviewed

These exercises practise aerospace propulsion and flight-performance calculations as engineering evidence. They connect aerodynamic state, thrust required, excess power, one-engine-inoperative climb, sustained maneuver performance, cruise fuel, fuel or propellant use, mission reserve, and validation uncertainty.

The goal is not only to produce a number. The goal is to decide whether the number is consistent with the flight condition, propulsion boundary, mission rule, and evidence available for review.

Assume simplified screening models unless an exercise states otherwise. Real aircraft and spacecraft performance work requires atmosphere models, engine decks, propeller or fan maps, aerodynamic polars, mass properties, operational rules, flight-test corrections, uncertainty budgets, and configuration control.

How to Use These Exercises

For each exercise, define:

the flight or mission condition: altitude, temperature, density, speed, Mach number, mass, and configuration;
the propulsion boundary: available thrust, installed thrust, shaft power, fuel flow, electric input power, or ideal exhaust velocity;
the aerodynamic boundary: reference area, coefficient source, drag build-up, and whether the value is clean, configured, measured, or predicted;
the mission rule: climb requirement, range target, reserve policy, delta-v need, duty cycle, or acceptance threshold;
the validation evidence required before using the calculation for a design decision.

The common mistake is mixing boundaries. Static thrust, net installed thrust, shaft power, electric input power, effective propulsive power, and mission fuel are not interchangeable.

Release Evidence Notes

Aerospace performance calculations should be tied to the flight or mission state. For each result, record altitude, temperature, density, speed basis, Mach number, mass or weight, configuration, propulsion setting, reference area, aerodynamic data source, mission segment, reserve rule and validation dataset. A performance number without that state cannot support clearance or mission acceptance.

Propulsion evidence should preserve the boundary. Static thrust, installed thrust, shaft power, electric input power, propulsive power, fuel flow, propellant mass flow and ideal exhaust velocity describe different interfaces. The release record should state which boundary is measured or modeled, which losses are included and which installation or environmental correction applies.

Flight-performance evidence should include the limiting case. Climb rate, climb gradient, service ceiling, takeoff distance, sustained-turn specific excess power, hot-and-high operation and reserve margins depend on weight, configuration, atmosphere, runway or trajectory condition, engine setting, uncertainty guard and operating rule. A nominal pass should not clear a segment when the guarded result fails.

Mission evidence should separate ideal equations from operational rules. Breguet range, Breguet loiter endurance, delta-v, fuel reserve, electric propulsion duty cycle and thermal limits should state unusable fuel or residual propellant, routing, wind, contingency maneuvers, pointing, power, thermal state and recovery or hold requirements. Range, endurance and delta-v should not be treated as interchangeable margins.

The practical release question is whether aerodynamic state, propulsion boundary, mission rule, uncertainty and validation evidence all support the same decision. If one element is missing, the result is a sizing screen or request for more evidence rather than a release-ready performance claim.

Engineering Boundary Notes

These exercises use simplified aerospace performance and propulsion screens. They do not replace certified aircraft performance data, engine deck analysis, propeller or fan maps, aerodynamic polar validation, flight-test correction, mission analysis software, operating procedures or safety review. A calculated performance margin applies only to the stated aircraft or spacecraft configuration, atmosphere, speed basis, mass state, propulsion boundary and mission rule.

Separate aerodynamic, propulsion and mission boundaries before release. Equivalent airspeed, true airspeed, Mach number, installed thrust, shaft power, fuel flow, propellant mass, electric input power, range reserve and delta-v are not interchangeable. Release evidence should state which boundary owns the decision and which conversion or loss model connects it to the next boundary.

Common Release Mistakes

using static or sea-level thrust as installed in-flight thrust without correction;
mixing indicated, equivalent and true airspeed when evaluating lift, drag, climb or range;
accepting a nominal climb, takeoff or range result without weight, temperature, wind and reserve guards;
treating Breguet range, loiter endurance, fuel reserve and spacecraft delta-v as the same margin;
using a clean aerodynamic polar for a configured or damaged aircraft state;
releasing a performance change without matching model, test, operational rule and uncertainty evidence.

Scenario Map

Scenario	Main calculation	Engineering decision
Flight condition	Mach number, dynamic pressure, lift and drag	Check whether the aerodynamic state is consistent with weight and configuration.
Aircraft performance	excess thrust, climb rate, climb gradient, one-engine-inoperative climb, service ceiling, sustained turn Ps and takeoff distance	Decide whether a flight segment or maneuver has enough operational margin.
Mission range, endurance and reserve	Breguet range, Breguet endurance, route fuel, headwind, fuel reserve and landing fuel	Separate ideal range or loiter time from dispatch or mission acceptance.
Propulsion boundary	propulsive efficiency, nozzle mass flow, pressure thrust and useful power	Decide whether exhaust velocity, mass flow and pressure mismatch are matched to the mission speed.
Space propulsion	rocket delta-v, propellant reserve, electric thrust and duty cycle	Verify that total impulse, power, pointing and thermal limits close together.
Validation evidence	model discrepancy, combined uncertainty and guard band	Decide whether a performance result can be released or needs more evidence.

Validation Package Checklist

aircraft or spacecraft configuration, mass state, atmosphere, speed basis and operating rule are documented;
aerodynamic data source, reference area, drag build-up and configuration state are controlled;
propulsion boundary, installation corrections, fuel or propellant basis and power limits are explicit;
climb, takeoff, range, reserve, duty-cycle or delta-v requirement is linked to the governing mission rule;
hot-day, one-engine-inoperative, wind, runway, thermal and uncertainty guards are included where relevant;
model, simulation, test, flight data and operational evidence are not mixed without transfer justification;
final release decision states accept, derate, retest, revise model, change mission rule, increase reserve or hold.

Exercise 1: Mach Number, Dynamic Pressure, Lift, and Drag

An aircraft is analysed at a flight condition with:

T=250\ \text{K}

\gamma=1.4

R=287\ \text{J/(kg K)}

V=230\ \text{m/s}

\rho=0.50\ \text{kg/m}^3

Reference wing area is:

S=16\ \text{m}^2

At this condition the aerodynamic data give:

C_L=0.55

C_D=0.035

Compute speed of sound, Mach number, dynamic pressure, lift, and drag.

Solution

Speed of sound:

a=\sqrt{\gamma R T}

a=\sqrt{1.4(287)(250)}=316.9\ \text{m/s}

Mach number:

\displaystyle M=\frac{V}{a}=\frac{230}{316.9}=0.726

Dynamic pressure:

\displaystyle q=\frac{1}{2}\rho V^2

\displaystyle q=\frac{1}{2}(0.50)(230)^2=13225\ \text{Pa}

Lift:

L=qSC_L

L=13225(16)(0.55)=116380\ \text{N}

Therefore:

L=116\ \text{kN}

Drag:

D=qSC_D

D=13225(16)(0.035)=7406\ \text{N}

Therefore:

D=7.41\ \text{kN}

Engineering Comment

The aircraft is in the subsonic regime but close enough to transonic operation that compressibility and airframe-specific drag rise should not be ignored. If aircraft weight at this condition is $110\ \text{kN}$ , the calculated lift is about $6\%$ above weight, so either the aircraft is climbing, manoeuvring, or the lift coefficient would be trimmed slightly lower for steady level flight.

Plausibility Check

A dynamic pressure near $13\ \text{kPa}$ at $230\ \text{m/s}$ and reduced density is plausible. A drag near $7.4\ \text{kN}$ is also consistent with a small aircraft using a clean drag coefficient of $0.035$ .

Exercise 2: Excess Power and Rate of Climb

Use the drag value from Exercise 1:

D=7.41\ \text{kN}

The aircraft weight is:

W=110\ \text{kN}

Available installed thrust at the same condition is:

T_a=10.0\ \text{kN}

True airspeed is:

V=230\ \text{m/s}

Estimate excess power and approximate rate of climb.

Solution

Excess thrust:

T_a-D=10.0-7.41=2.59\ \text{kN}

Convert to newtons:

T_a-D=2590\ \text{N}

Excess power:

P_{ex}=(T_a-D)V

P_{ex}=2590(230)=595700\ \text{W}

Therefore:

P_{ex}=596\ \text{kW}

For a small climb angle, rate of climb can be estimated from:

\displaystyle ROC=\frac{P_{ex}}{W}

Use weight in newtons:

\displaystyle ROC=\frac{595700}{110000}=5.42\ \text{m/s}

The climb gradient estimate is:

\displaystyle \frac{T_a-D}{W}=\frac{2590}{110000}=0.0235

or about:

2.35\%

Engineering Comment

The aircraft has positive climb capability, but the margin is not large. A few kilonewtons of installation loss, icing drag, bleed extraction, degraded engine performance, or higher actual weight could materially change the climb decision.

Plausibility Check

A $596\ \text{kW}$ excess-power estimate producing about $5.4\ \text{m/s}$ climb for a $110\ \text{kN}$ aircraft is consistent because power divided by weight has units of velocity.

Exercise 3: Breguet Range with Reserve Derating

A simplified cruise estimate uses the Breguet range form:

\displaystyle R=\frac{V}{c}\frac{L}{D}\ln\left(\frac{W_i}{W_f}\right)

where $c$ is a weight-specific fuel consumption parameter in $\text{s}^{-1}$ for this simplified form.

Use:

V=230\ \text{m/s}

c=1.7\times10^{-4}\ \text{s}^{-1}

\displaystyle \frac{L}{D}=15

\displaystyle \frac{W_i}{W_f}=1.18

Estimate ideal cruise range. Then apply a $15\%$ derating for reserves, routing, wind, and operational allowances.

Solution

Compute the logarithmic fuel-weight term:

\ln(1.18)=0.1655

Compute:

\displaystyle R=\frac{230}{1.7\times10^{-4}}(15)(0.1655)

R=3.36\times10^6\ \text{m}

Therefore:

R=3360\ \text{km}

Apply the derating:

R_{usable}=0.85(3360)=2856\ \text{km}

So the usable planning range is approximately:

R_{usable}=2860\ \text{km}

Engineering Comment

The logarithm shows why range does not increase linearly with fuel fraction. Adding fuel also increases initial weight, and the range benefit depends on aerodynamic efficiency, speed, propulsion efficiency, and mission rules.

Plausibility Check

The ideal range is several thousand kilometers, not tens of thousands, because the chosen $c$ value is in $10^{-4}\ \text{s}^{-1}$ rather than $10^{-5}\ \text{s}^{-1}$ . A unit mistake in $c$ would create an unrealistic answer.

Exercise 4: Rocket Equation Delta-V

A spacecraft propulsion stage has:

I_{sp}=320\ \text{s}

Initial mass:

m_0=5000\ \text{kg}

Final mass after the burn:

m_f=2600\ \text{kg}

Use:

g_0=9.80665\ \text{m/s}^2

Estimate ideal delta-v. Then subtract $180\ \text{m/s}$ for gravity, steering, and finite-burn losses. Compare the result with a mission need of $1750\ \text{m/s}$ .

Solution

Rocket equation:

\displaystyle \Delta v=I_{sp}g_0\ln\left(\frac{m_0}{m_f}\right)

Mass ratio:

\displaystyle \frac{m_0}{m_f}=\frac{5000}{2600}=1.923

Logarithmic term:

\ln(1.923)=0.6539

Ideal delta-v:

\Delta v=320(9.80665)(0.6539)

\Delta v=2052\ \text{m/s}

After losses:

\Delta v_{usable}=2052-180=1872\ \text{m/s}

Margin against the mission need:

M=1872-1750=122\ \text{m/s}

Percentage margin relative to the mission need:

\displaystyle \frac{122}{1750}=0.0697

or:

7.0\%

Engineering Comment

The stage closes the simplified delta-v requirement, but not with a large margin. Actual acceptance would also check residual propellant, mixture-ratio uncertainty, trapped propellant, thrust-vector losses, engine performance tolerance, navigation dispersion, and contingency maneuvers.

Plausibility Check

An $I_{sp}$ near $320\ \text{s}$ and mass ratio near $1.9$ commonly produce delta-v on the order of $2\ \text{km/s}$ . The result has the right scale.

Exercise 5: Electric Propulsion Thrust and Propellant Flow

An electric thruster operates at:

P_{in}=2.0\ \text{kW}

Thruster efficiency is:

\eta=0.60

Specific impulse is:

I_{sp}=1600\ \text{s}

Assume the useful jet power relation:

\displaystyle P_{jet}=\frac{1}{2}\dot{m}v_e^2

and:

T=\dot{m}v_e

Estimate exhaust velocity, thrust, and propellant mass flow. Then estimate ideal delta-v for a $600\ \text{kg}$ spacecraft after $30$ days of continuous thrust.

Solution

Exhaust velocity:

v_e=I_{sp}g_0

v_e=1600(9.80665)=15691\ \text{m/s}

Useful jet power:

P_{jet}=\eta P_{in}=0.60(2000)=1200\ \text{W}

Combining jet power and thrust:

\displaystyle T=\frac{2P_{jet}}{v_e}=\frac{2\eta P_{in}}{v_e}

Substitute:

\displaystyle T=\frac{2(0.60)(2000)}{15691}=0.153\ \text{N}

Mass flow:

\displaystyle \dot{m}=\frac{T}{v_e}

\displaystyle \dot{m}=\frac{0.153}{15691}=9.75\times10^{-6}\ \text{kg/s}

Convert to kilograms per day:

\dot{m}_{day}=9.75\times10^{-6}(86400)=0.842\ \text{kg/day}

Acceleration for a $600\ \text{kg}$ spacecraft:

\displaystyle a=\frac{T}{m}=\frac{0.153}{600}=2.55\times10^{-4}\ \text{m/s}^2

Time for $30$ days:

t=30(86400)=2592000\ \text{s}

Ideal delta-v:

\Delta v=at

\Delta v=(2.55\times10^{-4})(2592000)=661\ \text{m/s}

Engineering Comment

The thrust is very small, but the propellant use is also low. Electric propulsion is therefore useful for long-duration orbit raising, station keeping, and deep-space trajectory shaping, not for impulsive maneuvers that require high acceleration.

Plausibility Check

A few kilowatts producing a fraction of a newton is realistic for high-exhaust-velocity electric propulsion. The ideal $661\ \text{m/s}$ assumes continuous pointing, power availability, thermal acceptability, and no duty-cycle interruption.

Exercise 6: Fuel Reserve Decision

A mission planning review uses:

m_{trip}=1800\ \text{kg}

Cruise fuel flow for diversion and holding:

\dot{m}_f=650\ \text{kg/h}

Required diversion time:

t_{div}=0.65\ \text{h}

Required holding time:

t_{hold}=0.45\ \text{h}

Contingency fuel is $5\%$ of trip fuel. Unusable fuel allowance is:

m_{unusable}=60\ \text{kg}

Planned landing fuel is:

m_{landing}=790\ \text{kg}

Compute required reserve fuel and decide whether the plan is acceptable.

Solution

Diversion plus holding time:

t_{res}=0.65+0.45=1.10\ \text{h}

Fuel for diversion and holding:

m_{div+hold}=650(1.10)=715\ \text{kg}

Contingency fuel:

m_{cont}=0.05(1800)=90\ \text{kg}

Total required reserve:

m_{reserve}=715+90+60=865\ \text{kg}

Compare with planned landing fuel:

790<865

Shortfall:

865-790=75\ \text{kg}

The plan is short by:

75\ \text{kg}

Engineering Comment

The aircraft may meet cruise range but still fail the mission rule because reserve fuel is a separate operational requirement. Corrective actions could include reducing payload, adding fuel, changing routing, selecting a closer alternate, changing cruise speed, or using a more conservative performance model.

Plausibility Check

The reserve is dominated by diversion and holding fuel. A $75\ \text{kg}$ shortfall is small relative to trip fuel but large enough to reject a dispatch or mission plan if the reserve rule is mandatory.

Exercise 7: Thrust Model Validation Against Flight Evidence

An engine deck predicts installed net thrust at a measured flight condition:

T_{model}=9.80\ \text{kN}

Flight-test reconstruction from acceleration, weight, and drag estimates gives:

T_{flight}=9.20\ \text{kN}

The one-sigma uncertainty of the flight-test reconstruction is:

u_{flight}=0.35\ \text{kN}

The one-sigma uncertainty of the engine model at this condition is:

u_{model}=0.25\ \text{kN}

Assume the uncertainty sources are independent. Compute the discrepancy, combined uncertainty, and normalized discrepancy. Decide whether the model should be accepted if the review criterion is:

|z|<2

Solution

Discrepancy:

\Delta T=T_{model}-T_{flight}

\Delta T=9.80-9.20=0.60\ \text{kN}

Combined one-sigma uncertainty:

u_c=\sqrt{u_{flight}^2+u_{model}^2}

u_c=\sqrt{(0.35)^2+(0.25)^2}=0.430\ \text{kN}

Normalized discrepancy:

\displaystyle z=\frac{\Delta T}{u_c}

\displaystyle z=\frac{0.60}{0.430}=1.40

Since:

1.40<2

the single-point comparison passes the stated criterion.

Engineering Comment

Passing one point does not prove the engine deck is valid over the whole envelope. The model is biased high at this point. If repeated points at nearby Mach number, altitude, temperature, or throttle setting show the same sign, the review should treat the difference as a possible systematic bias rather than random scatter.

Plausibility Check

A $0.60\ \text{kN}$ difference on a roughly $10\ \text{kN}$ thrust value is about $6\%$ . That is large enough to matter for climb margin, even if it is not enough by itself to fail the stated uncertainty criterion.

Exercise 8: Climb-Gradient Release with an Uncertainty Guard Band

A flight-test point is being used to release a climb-gradient model. At the review condition:

T_a=18.5\ \text{kN}

D=15.1\ \text{kN}

W=130\ \text{kN}

The combined one-sigma uncertainty in excess thrust is:

u_{T-D}=0.45\ \text{kN}

The required climb gradient is $2.4\%$ . Compute nominal climb gradient, a two-sigma guarded climb gradient, and the release decision.

Solution

Nominal excess thrust:

T_a-D=18.5-15.1=3.4\ \text{kN}

Nominal climb gradient:

\displaystyle \gamma_{nom}\approx\frac{T_a-D}{W}=\frac{3.4}{130}=0.0262

Therefore:

\gamma_{nom}=2.62\%

Two-sigma guarded excess thrust:

T_{guard}=3.4-2(0.45)=2.50\ \text{kN}

Guarded climb gradient:

\displaystyle \gamma_{guard}=\frac{2.50}{130}=0.0192

Therefore:

\gamma_{guard}=1.92\%

The nominal result exceeds the $2.4\%$ requirement, but the two-sigma guarded result does not. The model should not be released as a robust pass at this condition.

Engineering Comment

This is exactly why flight-performance reviews should not stop at nominal excess thrust. A small climb-margin pass can disappear when installation loss, drag reconstruction uncertainty, engine-deck uncertainty, weight error or atmospheric correction is treated explicitly.

Plausibility Check

The nominal margin is only $0.22$ percentage points above the requirement. A two-sigma guard band of $0.90\ \text{kN}$ on a $130\ \text{kN}$ aircraft changes climb gradient by about $0.69$ percentage points, enough to reverse the decision.

Exercise 9: Service-Ceiling Screen from Rate of Climb

At a high-altitude condition, an aircraft has:

W=96\ \text{kN}

V=170\ \text{m/s}

T_a=6.4\ \text{kN}

D=5.8\ \text{kN}

A preliminary service-ceiling rule requires at least:

ROC_{min}=0.50\ \text{m/s}

Estimate excess power, rate of climb, and margin against the rule.

Solution

Excess thrust:

T_a-D=6.4-5.8=0.60\ \text{kN}=600\ \text{N}

Excess power:

P_{ex}=(T_a-D)V

P_{ex}=600(170)=102000\ \text{W}

Therefore:

P_{ex}=102\ \text{kW}

Rate of climb:

\displaystyle ROC=\frac{P_{ex}}{W}

\displaystyle ROC=\frac{102000}{96000}=1.06\ \text{m/s}

Margin:

M=1.06-0.50=0.56\ \text{m/s}

The condition passes the simplified service-ceiling screen with a $0.56\ \text{m/s}$ rate-of-climb margin.

Engineering Comment

The pass is useful but still preliminary. Near service ceiling, small errors in drag, thrust lapse, temperature, bleed extraction, propeller or fan efficiency, and aircraft weight can strongly affect the result. A certified or operational ceiling should be based on a validated performance model, not a single point.

Plausibility Check

Only $0.60\ \text{kN}$ of excess thrust is available, so the rate of climb should be small. Dividing about $102\ \text{kW}$ by about $96\ \text{kN}$ gives a value close to $1\ \text{m/s}$ , which matches the computed result.

Exercise 10: Takeoff-Distance Margin at Hot-and-High Conditions

A preliminary takeoff estimate starts from a reference balanced field length:

s_{ref}=1200\ \text{m}

at reference mass:

m_{ref}=6000\ \text{kg}

The planned takeoff mass is:

m=6300\ \text{kg}

The density ratio at the runway is:

\sigma=0.88

A headwind correction factor is:

f_w=0.95

Use the screening relation:

\displaystyle s=s_{ref}\left(\frac{m}{m_{ref}}\right)^2\frac{1}{\sigma}f_w

Runway available is:

s_{avail}=1450\ \text{m}

Compute required distance and runway margin.

Solution

Mass ratio:

\displaystyle \frac{m}{m_{ref}}=\frac{6300}{6000}=1.05

Required distance:

\displaystyle s=1200(1.05)^2\frac{1}{0.88}(0.95)

s=1428\ \text{m}

Runway margin:

M=s_{avail}-s=1450-1428=22\ \text{m}

Relative margin:

\displaystyle \frac{22}{1450}=0.015

or:

1.5\%

Engineering Comment

The calculation nominally fits the runway, but the margin is too thin for a serious release decision. A real review would include accelerate-stop and accelerate-go rules, obstacle clearance, runway slope, surface condition, wind variability, engine deterioration, anti-ice or bleed configuration, pilot technique and regulatory safety factors.

Plausibility Check

The hot-and-high density correction increases distance by about $14\%$ , while the headwind correction reduces it by $5\%$ . A final value just under the $1450\ \text{m}$ available runway is therefore plausible but operationally fragile.

Exercise 11: Remaining Delta-V Reserve After Executed Maneuvers

After planned maneuvers, a spacecraft has current mass:

m_0=950\ \text{kg}

Usable remaining propellant is:

m_p=28\ \text{kg}

The propulsion system has:

I_{sp}=225\ \text{s}

A contingency maneuver requires:

\Delta v_{req}=60\ \text{m/s}

Mission rules add a $15\%$ execution uncertainty allowance to the contingency requirement. Compute remaining ideal delta-v and the guarded reserve decision.

Solution

Final mass after using the remaining propellant:

m_f=950-28=922\ \text{kg}

Remaining ideal delta-v:

\displaystyle \Delta v=I_{sp}g_0\ln\left(\frac{m_0}{m_f}\right)

\displaystyle \Delta v=225(9.80665)\ln\left(\frac{950}{922}\right)

\Delta v=66.0\ \text{m/s}

Guarded requirement:

\Delta v_{guard}=1.15(60)=69.0\ \text{m/s}

Guarded margin:

M=66.0-69.0=-3.0\ \text{m/s}

The remaining propellant satisfies the unguarded $60\ \text{m/s}$ contingency maneuver, but it fails the guarded reserve rule by about $3.0\ \text{m/s}$ .

Engineering Comment

Remaining delta-v should not be treated as a single optimistic number. Navigation dispersions, finite-burn losses, residual propellant, pressure and temperature effects, minimum impulse bit, attitude constraints and valve performance can consume a reserve that appears adequate in an ideal rocket-equation calculation.

Plausibility Check

The remaining propellant is only about $3\%$ of spacecraft mass. With $I_{sp}=225\ \text{s}$ , a remaining delta-v near $60$ to $70\ \text{m/s}$ is the right order of magnitude.

Exercise 12: Electric-Propulsion Duty Cycle and Thermal Limit

Use the electric-propulsion result from Exercise 5:

T=0.153\ \text{N}

\dot{m}=9.75\times10^{-6}\ \text{kg/s}

Spacecraft mass is:

m=600\ \text{kg}

The maneuver is planned for $30$ calendar days, but thermal limits allow thrusting only $70\%$ of the time. Pointing and communications constraints remove another $8\%$ of the calendar time. Estimate effective thrusting time, delivered ideal delta-v, propellant consumed, and margin against a $480\ \text{m/s}$ maneuver target.

Solution

Effective duty fraction:

f_{duty}=0.70(1-0.08)=0.644

Calendar time:

t_{cal}=30(86400)=2592000\ \text{s}

Effective thrusting time:

t_{eff}=0.644(2592000)=1669248\ \text{s}

Acceleration:

\displaystyle a=\frac{T}{m}=\frac{0.153}{600}=2.55\times10^{-4}\ \text{m/s}^2

Delivered ideal delta-v:

\Delta v=a t_{eff}

\Delta v=(2.55\times10^{-4})(1669248)=426\ \text{m/s}

Propellant consumed:

m_p=\dot{m}t_{eff}

m_p=(9.75\times10^{-6})(1669248)=16.3\ \text{kg}

Maneuver-target margin:

M=426-480=-54\ \text{m/s}

The calendar schedule does not close the maneuver target once duty cycle and pointing constraints are included.

Engineering Comment

Electric propulsion is often limited by operations, not just thruster physics. A trajectory plan should connect thrust level, power availability, thermal rejection, pointing duty cycle, communications windows, eclipse periods, contamination constraints and propellant bookkeeping.

Plausibility Check

The ideal continuous-thrust delta-v from Exercise 5 was about $661\ \text{m/s}$ . Multiplying by the effective duty fraction $0.644$ gives about $426\ \text{m/s}$ , so the reduced result is internally consistent.

Exercise 13: Propulsive Efficiency and Jet Velocity Mismatch

A cruise propulsion concept produces the required net thrust:

T=8.0\ \text{kN}

at flight speed:

V_0=230\ \text{m/s}

The effective jet velocity is:

V_e=410\ \text{m/s}

Use the ideal propulsive-efficiency screen:

\displaystyle \eta_p=\frac{2V_0}{V_e+V_0}

and the momentum relation:

T=\dot{m}(V_e-V_0)

Estimate propulsive efficiency, air mass flow, useful propulsive power, and ideal power added to the flow. Then repeat for a redesigned fan concept with:

V_{e,1}=330\ \text{m/s}

at the same thrust and flight speed.

Solution

Base propulsive efficiency:

\displaystyle \eta_p=\frac{2(230)}{410+230}=\frac{460}{640}=0.719

Therefore:

\eta_p=71.9\%

Base air mass flow:

\displaystyle \dot{m}=\frac{8000}{410-230}=\frac{8000}{180}=44.4\ \text{kg/s}

Useful propulsive power:

P_{useful}=TV_0

P_{useful}=8000(230)=1.84\times10^6\ \text{W}

So:

P_{useful}=1.84\ \text{MW}

Ideal power added to the flow:

\displaystyle P_{flow}=\frac{P_{useful}}{\eta_p}=\frac{1.84}{0.719}=2.56\ \text{MW}

Power not converted to useful propulsive power:

P_{loss}=2.56-1.84=0.72\ \text{MW}

For the redesigned fan:

\displaystyle \eta_{p,1}=\frac{2(230)}{330+230}=\frac{460}{560}=0.821

Therefore:

\eta_{p,1}=82.1\%

Redesigned air mass flow:

\displaystyle \dot{m}_1=\frac{8000}{330-230}=80.0\ \text{kg/s}

Redesigned ideal flow power:

\displaystyle P_{flow,1}=\frac{1.84}{0.821}=2.24\ \text{MW}

Flow-power reduction:

2.56-2.24=0.32\ \text{MW}

Relative reduction:

\displaystyle \frac{0.32}{2.56}\times100=12.5\%

Engineering Comment

The redesigned fan improves propulsive efficiency because the jet velocity is closer to flight speed. That does not make it automatically better. Producing the same thrust with a smaller velocity increment requires higher mass flow, which can mean a larger fan, larger inlet, higher nacelle drag, more weight, installation constraints, noise changes, ground-clearance issues, or different off-design behavior.

The calculation is a mission-speed matching screen. It should be connected to engine maps, inlet recovery, fan pressure ratio, nacelle drag, bypass ratio, structural integration, acoustic limits, and fuel-flow evidence before it becomes a design decision.

Plausibility Check

For the base concept, the jet is $180\ \text{m/s}$ faster than the aircraft, so only $44.4\ \text{kg/s}$ of air is needed for $8$ kN of thrust, but propulsive efficiency is about $72\%$ . The redesigned concept uses a smaller $100\ \text{m/s}$ velocity increment, so it needs $80.0\ \text{kg/s}$ of air but improves efficiency to about $82\%$ . That tradeoff is the expected direction.

Exercise 14: Sustained-Turn Specific Excess Power Gate

An aircraft is reviewed at:

V=140\ \text{m/s}

with weight:

W=70\ \text{kN}

wing area:

S=20\ \text{m}^2

and density:

\rho=0.90\ \text{kg/m}^3

The drag polar at the reviewed configuration is:

C_D=C_{D0}+kC_L^2

with:

C_{D0}=0.025

and:

k=0.055

Installed thrust available is:

T_A=18.0\ \text{kN}

The maneuver command is a sustained level turn at:

n=2.5

The release rule requires guarded specific excess power:

P_{s,guard}\ge 4.0\ \text{m/s}

after allowing a $15\%$ thrust shortfall. Estimate dynamic pressure, turn lift coefficient, drag, nominal specific excess power, sustained turn rate, guarded specific excess power, and the maximum load factor that would meet the guarded $P_s$ rule at the same speed.

Solution

Dynamic pressure:

\displaystyle q=\frac{1}{2}\rho V^2

\displaystyle q=\frac{1}{2}(0.90)(140)^2=8820\ \text{Pa}

In a level turn:

L=nW

so:

\displaystyle C_L=\frac{nW}{qS}

Use $W=70000\ \text{N}$ :

\displaystyle C_L=\frac{2.5(70000)}{(8820)(20)}=0.992

Drag coefficient:

C_D=0.025+0.055(0.992)^2=0.0791

Drag:

D=qSC_D

D=(8820)(20)(0.0791)=13950\ \text{N}=14.0\ \text{kN}

Nominal excess thrust:

T_A-D=18.0-14.0=4.05\ \text{kN}

Specific excess power:

\displaystyle P_s=\frac{(T_A-D)V}{W}

\displaystyle P_s=\frac{(4050)(140)}{70000}=8.1\ \text{m/s}

Sustained turn rate for a level turn is:

\displaystyle \omega=\frac{g\sqrt{n^2-1}}{V}

\displaystyle \omega=\frac{9.81\sqrt{2.5^2-1}}{140}=0.160\ \text{rad/s}

Convert to degrees per second:

\displaystyle \omega=0.160\left(\frac{180}{\pi}\right)=9.2^\circ/\text{s}

Guarded thrust after a $15\%$ shortfall:

T_{guard}=0.85(18.0)=15.3\ \text{kN}

Guarded excess thrust:

T_{guard}-D=15.3-14.0=1.35\ \text{kN}

Guarded specific excess power:

\displaystyle P_{s,guard}=\frac{(1350)(140)}{70000}=2.7\ \text{m/s}

The guarded condition fails the release rule:

2.7<4.0\ \text{m/s}

To meet the guarded $P_s$ rule, allowable drag is:

\displaystyle D_{allow}=T_{guard}-\frac{WP_{s,req}}{V}

\displaystyle D_{allow}=15300-\frac{70000(4.0)}{140}=13300\ \text{N}

Allowable drag coefficient:

\displaystyle C_{D,allow}=\frac{13300}{(8820)(20)}=0.0754

Allowable lift coefficient from the polar:

\displaystyle C_{L,allow}=\sqrt{\frac{C_{D,allow}-C_{D0}}{k}}

\displaystyle C_{L,allow}=\sqrt{\frac{0.0754-0.025}{0.055}}=0.957

Maximum load factor at the same speed:

\displaystyle n_{allow}=\frac{C_{L,allow}qS}{W}

\displaystyle n_{allow}=\frac{(0.957)(8820)(20)}{70000}=2.41

The maneuver should be limited to about:

n\le 2.4

unless additional thrust margin or lower drag is validated.

Engineering Comment

Specific excess power makes the maneuver tradeoff visible. The nominal turn looks comfortable, but the guarded thrust case does not preserve the required climb or acceleration energy margin. A release decision should connect the drag polar, thrust deck, load factor, airspeed, configuration, engine setting, temperature correction and flight-test evidence before clearing the maneuver.

Plausibility Check

At $2.5g$ , the required lift coefficient is close to $1.0$ , so induced drag is a large part of the drag polar. Nominal thrust leaves $8.1\ \text{m/s}$ of specific excess power, but a $15\%$ thrust guard consumes most of the excess thrust and lowers $P_s$ to $2.7\ \text{m/s}$ . Reducing allowable load factor to about $2.4g$ is therefore a plausible operational correction.

Exercise 15: Breguet Loiter Endurance and Holding Margin

A jet aircraft enters a loiter segment with weight:

W_i=92.0\ \text{kN}

The flight plan protects reserve and landing fuel by requiring the loiter segment to end no lower than:

W_f=88.5\ \text{kN}

At the planned loiter speed and altitude, the performance model gives:

\displaystyle \frac{L}{D}=14.2

and the weight-specific fuel consumption parameter is:

c=1.55\times10^{-4}\ \text{s}^{-1}

Use the jet Breguet endurance relation:

\displaystyle E=\frac{1}{c}\frac{L}{D}\ln\left(\frac{W_i}{W_f}\right)

Estimate available loiter endurance. Then check a planning requirement of $45\ \text{min}$ holding plus a $10\ \text{min}$ operational guard. Finally, estimate how much additional fuel weight would be needed if the required loiter time increased to $65\ \text{min}$ at the same $L/D$ and $c$ .

Solution

Weight ratio:

\displaystyle \frac{W_i}{W_f}=\frac{92.0}{88.5}=1.0395

Logarithmic fuel-weight term:

\ln(1.0395)=0.0388

Available endurance:

\displaystyle E=\frac{1}{1.55\times10^{-4}}(14.2)(0.0388)

E=3554\ \text{s}

Convert to minutes:

\displaystyle E=\frac{3554}{60}=59.2\ \text{min}

Required protected holding time:

t_{req}=45+10=55\ \text{min}

Time margin:

M_t=59.2-55=4.2\ \text{min}

The planned loiter fuel boundary passes the $55\ \text{min}$ protected holding requirement, but only with a small margin.

For a $65\ \text{min}$ requirement:

E_{65}=65(60)=3900\ \text{s}

Solve the endurance relation for the required weight ratio:

\displaystyle \ln\left(\frac{W_i}{W_{f,65}}\right)=\frac{cE_{65}}{L/D}

\displaystyle \ln\left(\frac{W_i}{W_{f,65}}\right)=\frac{(1.55\times10^{-4})(3900)}{14.2}=0.0426

Therefore:

\displaystyle \frac{W_i}{W_{f,65}}=e^{0.0426}=1.0435

Required final weight after a $65\ \text{min}$ loiter:

\displaystyle W_{f,65}=\frac{92.0}{1.0435}=88.16\ \text{kN}

The current protected final weight is $88.5\ \text{kN}$ , so the additional fuel weight needed is:

88.5-88.16=0.334\ \text{kN}

Convert this fuel weight to mass:

\displaystyle m_{fuel,extra}=\frac{0.334\times1000}{9.81}=34.1\ \text{kg}

Engineering Comment

Endurance is not the same decision as range. A still-air Breguet range check can pass while holding time is inadequate for weather, traffic, alternate sequencing, or mission loiter. This result should be tied to the actual loiter speed schedule, altitude, engine setting, anti-ice or bleed configuration, reserve policy, unusable fuel, and whether $L/D$ and $c$ are validated for the loiter condition rather than borrowed from cruise.

Plausibility Check

The available fuel-weight change is only $3.5\ \text{kN}$ on a $92.0\ \text{kN}$ aircraft, so an endurance near one hour is plausible rather than many hours. Extending the requirement from $55$ to $65\ \text{min}$ consumes a small additional protected fuel mass because the calculation is close to the original boundary and the logarithmic weight-ratio term changes modestly.

Exercise 16: One-Engine-Inoperative Climb-Gradient Guard

A twin-engine aircraft is reviewed for a takeoff climb segment after one engine has failed. At the reviewed speed and configuration, aircraft weight is:

W=120\ \text{kN}

The operating engine can provide installed net thrust:

T_{OEI}=21.0\ \text{kN}

Configured airframe drag at the segment speed is:

D_{config}=16.2\ \text{kN}

Windmilling and failed-engine installation drag add:

D_{wm}=1.4\ \text{kN}

The required climb gradient is:

\gamma_{req}=2.4\%

For release, apply a guard by subtracting:

\Delta T=0.7\ \text{kN}

from available thrust and adding:

\Delta D=0.4\ \text{kN}

to configured drag.

Estimate nominal and guarded climb gradient, the nominal live-engine thrust margin to the rule, and the guarded release decision.

Solution

Nominal excess thrust:

T_{ex,nom}=T_{OEI}-D_{config}-D_{wm}

T_{ex,nom}=21.0-16.2-1.4=3.4\ \text{kN}

Nominal climb gradient:

\displaystyle \gamma_{nom}=\frac{T_{ex,nom}}{W}

\displaystyle \gamma_{nom}=\frac{3.4}{120}=0.0283

or:

\gamma_{nom}=2.83\%

Nominal margin to the rule:

2.83\%-2.40\%=0.43\ \text{percentage points}

Required excess thrust for the rule:

T_{ex,req}=\gamma_{req}W

T_{ex,req}=0.024(120)=2.88\ \text{kN}

Nominal live-engine thrust required:

T_{OEI,req}=D_{config}+D_{wm}+T_{ex,req}

T_{OEI,req}=16.2+1.4+2.88=20.48\ \text{kN}

Nominal live-engine thrust margin:

21.0-20.48=0.52\ \text{kN}

Now apply the guard:

T_{OEI,g}=21.0-0.7=20.3\ \text{kN}

D_{config,g}=16.2+0.4=16.6\ \text{kN}

Guarded excess thrust:

T_{ex,g}=20.3-16.6-1.4=2.3\ \text{kN}

Guarded climb gradient:

\displaystyle \gamma_g=\frac{2.3}{120}=0.0192

or:

\gamma_g=1.92\%

Guarded margin:

1.92\%-2.40\%=-0.48\ \text{percentage points}

The nominal calculation passes, but the guarded OEI climb-gradient release fails. The segment should not be released without reducing weight, improving thrust margin, reducing configured drag, changing the procedure, or validating a smaller guard.

Engineering Comment

One-engine-inoperative climb is a boundary-matching problem. The calculation must use installed live-engine thrust at the actual temperature, altitude, bleed and anti-ice state; drag for the exact flap, gear, speed and failed-engine configuration; and the weight used for the departure segment. Using all-engine thrust, clean drag, or an unguarded engine deck can turn a failed climb segment into a false pass.

Plausibility Check

The nominal excess thrust is only $3.4\ \text{kN}$ on a $120\ \text{kN}$ aircraft, so a climb gradient below $3\%$ is expected. The nominal thrust margin to the rule is just $0.52\ \text{kN}$ ; subtracting $0.7\ \text{kN}$ of thrust and adding $0.4\ \text{kN}$ of drag should consume more than that margin, which explains the guarded failure.

Exercise 17: Choked-Nozzle Mass Flow and Pressure Thrust

A small turbojet nozzle is reviewed at a steady test point. The nozzle total conditions upstream of the throat are:

p_t=420\ \text{kPa}

T_t=840\ \text{K}

Use:

\gamma=1.4

R=287\ \text{J/(kg K)}

The throat area is:

A_t=0.018\ \text{m}^2

The estimated exit velocity is:

V_e=590\ \text{m/s}

The aircraft flight speed is:

V_0=180\ \text{m/s}

The nozzle exit pressure and ambient pressure are:

p_e=85\ \text{kPa}

p_0=70\ \text{kPa}

The exit area is:

A_e=0.030\ \text{m}^2

The installed-thrust release target for this boundary is:

T_{req}=4.60\ \text{kN}

Check whether the throat can choke, estimate ideal choked mass flow, calculate momentum thrust, pressure thrust and total nozzle thrust. Then apply a 6 percent mass-flow loss for throat roughness or total-pressure measurement uncertainty and decide whether the release target still passes.

Solution

Critical pressure ratio for choking is:

\displaystyle \left(\frac{p^*}{p_t}\right)=\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}}

\displaystyle \left(\frac{p^*}{p_t}\right)=\left(\frac{2}{2.4}\right)^{3.5}=0.528

The ambient-to-total pressure ratio is:

\displaystyle \frac{p_0}{p_t}=\frac{70}{420}=0.167

Since (0.167<0.528), the throat can choke in this simplified screen.

Choked ideal mass flow is:

\displaystyle \dot{m}=A_t p_t\sqrt{\frac{\gamma}{RT_t}}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}

\displaystyle \dot{m}=0.018(420000)\sqrt{\frac{1.4}{287(840)}}\left(\frac{2}{2.4}\right)^3=10.5\ \text{kg/s}

Momentum thrust is:

T_m=\dot{m}(V_e-V_0)

T_m=10.5(590-180)=4323\ \text{N}

Pressure thrust is:

T_p=(p_e-p_0)A_e

T_p=(85000-70000)(0.030)=450\ \text{N}

Total simplified nozzle thrust:

T=T_m+T_p=4323+450=4773\ \text{N}=4.77\ \text{kN}

Nominal margin to the release target:

M_T=4.77-4.60=0.17\ \text{kN}

The nominal nozzle calculation passes, but narrowly.

Guarded mass flow with a 6 percent loss is:

\dot{m}_g=0.94(10.5)=9.91\ \text{kg/s}

Guarded thrust:

T_g=9.91(590-180)+450=4513\ \text{N}=4.51\ \text{kN}

Guarded margin:

M_{T,g}=4.51-4.60=-0.09\ \text{kN}

The guarded release fails. The nozzle should not be accepted at this boundary until the throat area, total-pressure measurement, exit pressure, loss model or installed-thrust requirement is resolved.

Engineering Comment

Nozzle thrust is a boundary calculation. The mass flow belongs to total conditions and throat area, while the pressure thrust belongs to exit pressure, ambient pressure and exit area. A static-test result, a sea-level thrust quote and an installed in-flight nozzle result can differ materially. Release evidence should state whether the nozzle is choked, which total-pressure station was used, whether throat area includes thermal growth and manufacturing tolerance, how losses were applied, and whether the pressure mismatch is underexpanded, overexpanded or inside the validated engine-deck range.

Plausibility Check

The pressure term is only (450\ \text{N}), about 9 percent of total thrust, so the result is mostly controlled by mass flow and velocity change. A 6 percent mass-flow loss reduces the momentum term by about (259\ \text{N}), which is larger than the nominal (173\ \text{N}) margin. The guarded fail is therefore expected and not a rounding artifact.

Exercise 18: Cruise Fuel Headwind and Route-Release Gate

A route-release review checks whether a cruise segment preserves protected landing fuel. The planned remaining cruise distance is:

D=780\ \text{km}

The planned true airspeed is:

V_{TAS}=740\ \text{km/h}

The forecast headwind component is:

V_{hw}=85\ \text{km/h}

The installed cruise fuel-flow evidence gives:

\dot{m}_f=710\ \text{kg/h}

Fuel available at the start of the reviewed cruise segment is:

m_{start}=1280\ \text{kg}

Protected landing fuel, including reserve and unusable fuel, is:

m_{prot}=430\ \text{kg}

Apply a fuel forecast guard:

g_f=5\%

Calculate ground speed, cruise time, guarded cruise fuel, predicted landing fuel and protected-fuel margin. Then repeat for a direct routing option:

D_{direct}=720\ \text{km}

Solution

Ground speed:

V_g=V_{TAS}-V_{hw}

V_g=740-85=655\ \text{km/h}

Cruise time for the planned route:

\displaystyle t=\frac{D}{V_g}=\frac{780}{655}=1.191\ \text{h}

Unguarded cruise fuel:

m_{cruise}=\dot{m}_ft

m_{cruise}=710(1.191)=845.6\ \text{kg}

Guarded cruise fuel:

m_{cruise,g}=1.05(845.6)=887.9\ \text{kg}

Predicted protected landing fuel:

m_{landing}=1280-887.9=392.1\ \text{kg}

Protected-fuel margin:

M_f=m_{landing}-m_{prot}

M_f=392.1-430=-37.9\ \text{kg}

The planned route fails the protected-fuel gate under the stated wind and guard assumptions.

For the direct routing:

\displaystyle t_{direct}=\frac{720}{655}=1.099\ \text{h}

m_{direct}=710(1.099)=780.3\ \text{kg}

m_{direct,g}=1.05(780.3)=819.3\ \text{kg}

Predicted protected landing fuel for the direct route:

m_{landing,direct}=1280-819.3=460.7\ \text{kg}

Protected-fuel margin:

M_{f,direct}=460.7-430=30.7\ \text{kg}

The direct route passes this simplified release gate, but with a modest margin.

Engineering Comment

This is not a Breguet range problem; it is an operational route-fuel gate. The calculation depends on ground speed, forecast wind, installed fuel-flow evidence, protected landing-fuel rule and the mission boundary at which fuel is counted. A release record should state whether the fuel flow includes anti-ice, bleed, generator load, engine deterioration, altitude, Mach schedule, step climb and contingency allowances.

Plausibility Check

The headwind reduces ground speed by more than 11 percent, so a 780 km segment takes about 1.19 hours rather than about 1.05 hours at still-air TAS. At roughly 710 kg/h, that difference is large enough to erase a small fuel margin. Shortening the route by 60 km saves about 65 to 70 kg of guarded fuel, which explains why the direct route changes the decision.

Review Checklist

Before accepting an aerospace propulsion or flight-performance calculation, check:

whether the result is aerodynamic, installed-propulsion, mission-planning, spacecraft-propulsion, or validation evidence;
whether thrust is static, gross, net, installed, shaft-power-derived, electric-input-derived, or reconstructed from flight data;
whether nozzle mass flow, total conditions, throat area, pressure thrust and loss factors share the same propulsion boundary;
whether speed is true airspeed, equivalent airspeed, calibrated airspeed, Mach number, ground speed, or trajectory velocity;
whether weight, mass, fuel, propellant, reserve and unusable quantities are measured at the same mission boundary;
whether propulsive efficiency, exhaust velocity and mass flow are matched to the mission speed and installation boundary;
whether climb, ceiling and takeoff margins include uncertainty and environmental corrections;
whether one-engine-inoperative climb uses live-engine thrust, windmilling drag, exact configuration, speed schedule, weight and guard bands;
whether sustained maneuver performance uses the correct load factor, drag polar, thrust guard and specific excess power rule;
whether range calculations separate ideal physics from routing, reserves, wind and operating rules;
whether cruise route fuel uses ground speed, headwind, installed fuel-flow evidence, protected landing fuel and forecast guard consistently;
whether endurance or loiter calculations are checked as time-margin problems, not inferred from range alone;
whether rocket-equation results include finite-burn losses, residuals, execution error and contingency maneuvers;
whether electric-propulsion results include power, pointing, thermal, duty-cycle and operations constraints;
whether model validation checks repeated points, not only one favorable comparison;
whether the final decision is a sizing screen, dispatch rule, mission acceptance, model release or request for more evidence.

Common Mistakes

Mixing static thrust, installed thrust, shaft power, electric input power and propulsive power in one performance margin.
Treating choked-nozzle mass flow as a fixed number while changing total pressure, throat area, loss factor or pressure-thrust boundary.
Comparing true airspeed, equivalent airspeed, Mach number, ground speed or trajectory speed without preserving the required speed basis.
Treating a climb, ceiling or takeoff result as release-ready before applying atmosphere, configuration, weight and uncertainty guards.
Using all-engine thrust or clean drag in a one-engine-inoperative climb-gradient check.
Using an ideal Breguet range result as proof of dispatch range while ignoring reserve, routing, wind, unusable fuel and operating rules.
Using true airspeed as if it were ground speed when calculating cruise segment fuel under headwind.
Inferring loiter or holding endurance from range instead of checking the loiter speed, altitude, fuel-flow parameter and protected final weight.
Claiming sustained-turn capability from lift or thrust alone while specific excess power, drag polar and load-factor guard fail.
Using rocket-equation delta-v without residual propellant, finite-burn losses, execution errors and contingency maneuvers.
Releasing electric-propulsion duty cycles without power, pointing, thermal, operations and degradation constraints.
Validating a thrust or performance model from one favorable data point instead of repeated points over the intended envelope.
Reporting a performance number without the flight condition, propulsion boundary, mission rule, uncertainty and evidence consequence.

REF