# 56 Worked Examples: Aircraft, Rocket, & Spacecraft Performance

These worked examples have been fielded as homework problems or exam questions.

Worked Example #1

where is the weight of fuel burned per unit time, is the true airspeed, is the brake specific fuel consumption. The values of and depend on the characteristics of the aircraft, the aircraft weight, and the density altitude at which it is flying, i.e.,

Based on this fuel flow equation, show that the airspeed for the best endurance for the airplane will be obtained when the true airspeed is

and the corresponding airspeed for the best range for the airplane will be obtained when the true airspeed is

Explain the operational significance of flying at and , and give an example of an aircraft flight profile (or a part of one) when such airspeeds might be specifically used.

The first term in the fuel flow equation is the contribution from the profile/parasitic (non-lifting) part of the drag, which grows with the cube of the airspeed. The second part is the contribution from the induced drag, which is inversely proportional to airspeed. Starting from

the lowest fuel burn (hence maximum flight endurance) can be determined by finding when is a minimum. Differentiating the fuel flow equation with respect to gives

which is zero for a minimum, i.e.,

and so

The best range is obtained when the ratio is a minimum. In this case

so that

which is zero for a minimum, i.e.,

and so

There are a variety of missions where flight at the speed for the best range and best endurance are essential. For example, a long-range ferry mission of the aircraft (say, over water) might need the aircraft to be flown at or near its airspeed for the best range, even though this airspeed is typically lower than the best cruise speed of the aircraft. Likewise, flight at an airspeed for best endurance might be needed for a reconnaissance mission where the aircraft might have to fly at or near a particular location for an extended period.

Worked Example #2

where you should evaluate and in terms of the the wing area, , wing aspect ratio, , flight weight, , parasitic drag coefficient & , Oswald’s efficiency factor, , engine TSFC, .

The standard formula gives the lift on the airplane.

which is equal to weight . The drag on the airplane is

where the drag coefficient is given by

i.e., the sum of non-lifting and lifting (induced) parts. is the wing’s aspect ratio, and is Oswald’s efficiency factor. Using the equation for lift (as well as the assumption that ) then, the lift coefficient is

Also, because the thrust equals drag, then

and after some rearrangement, then

For a jet engine, the fuel burn rate will be the product of the thrust and the thrust-specific fuel consumption, i.e.,

and substituting for gives

or

which is of the form

where

and

Both constants are based on the stated assumptions of all the values involved for a constant weight and altitude, i.e., = constant.

Worked Example #3

then show that the airspeed for the best endurance for a jet airplane will be obtained when

In this problem, we are asked to start from

The lowest fuel burn (hence maximum flight endurance) can be determined by differentiating the fuel flow equation with respect to , i.e.,

which is zero for a minimum, i.e.,

and so

where

and

So would be the airspeed to fly for minimum fuel flow. Notice that this speed depends on the weight of the aircraft and the altitude at which it is flying.

Worked Example #4

An ERAU Cessna 172 Skyhawk aircraft is flying along at an initial estimated in-flight weight of 2,100 lb and an airspeed of 120 knots at 2,000 ft where the air density is 0.00216 slugs/ft. The student pilot is flying solo on a long cross-country flight. When flying over Jacksonville airport on the way back to ERAU (125 nautical miles left to fly), the pilot notices from the fuel gauge that only 8 gallons of usable AVGAS fuel remain in the tanks. The engineering characteristics of the aircraft are given below:

- Wing span, = 36 ft
- Wing area, = 174 ft
- Non-lifting drag coefficient of aircraft, = 0.02
- Average propeller efficiency, = 0.85
- Oswald’s efficiency factor, = 0.81
- Engine BSFC, = 0.45 lb hp hr

Calculate the following:

(a) The operating lift coefficient of the wing, .

(b) The induced drag coefficient, and the total drag coefficient, .

(c) The lift-to-drag ratio of the aircraft.

(d) The propeller thrust and engine power (in hp) required for the aircraft to fly.

(e) The fuel flow in gallons per minute. Will the pilot make it back to ERAU on the remaining fuel?

(f) Assume that the aircraft’s weight for this analysis is the initial in-flight weight less half the remaining fuel weight. Assume also that AVGAS fuel weighs 6.0 lb per gallon.

We are told to assume that the aircraft’s weight for this analysis is the initial in-flight weight, say , less half the remaining fuel weight. We have only 8 gallons of AVGAS fuel, and we are told to assume AVGAS weighs 6.0 lb per gallon, so = 48 lb of fuel. Therefore, the weight to perform the analysis is

An airspeed of 120 kts is equivalent to 120 1.688 = 202.54 ft/s = .

(a) The operating lift coefficient of the wing, , is

(b) To find the induced drag coefficient then, we need the aspect ratio of the wing, , i.e.,

The induced drag coefficient will be

The total drag coefficient, , is

(c) The lift-to-drag ratio of the aircraft at the given conditions of flight is

(d) The thrust from the propeller required for the aircraft to fly is

The power required for flight will be

where the factor 550 has been used to convert the base units to horsepower (hp).

(e) The fuel flow now follows directly, i.e.,

so the aircraft is using about 6 gallons per hour.

(f) The time to burn off all of the relatively small amount of available fuel is

so the potential approximate range of the aircraft is

In conclusion, at least in theory the aircraft probably has enough fuel to get back to ERAU. But, in practice, the FAA requires a minimum fuel reserve of 30 minutes of flight time (visual rules) and 45 minutes of flight time (instrument rules), so legally the student pilot is probably going to have to land and pick up more fuel before completing the final leg back to ERAU.

Worked Example #5

Consider a small jet-powered aircraft with an initial in-flight weight of 21,000 lb that is flying in unaccelerated level flight at an airspeed of 250 knots where the air density is 0.0015 slugs/ft. The aircraft has a weight of 850 lb of usable fuel remaining in its tanks. The wing span of the aircraft is 48.0 ft, and the wing panels have a trapezoidal shape with a root chord = 9 ft and a tip chord = 4 ft. The other characteristics of the aircraft are: Wing span, = 48 ft; Non-lifting drag coefficient, = 0.02; Oswald’s efficiency factor, = 0.81; Engine TSFC = = 0.5 lb lb^{-1} hr^{-1}. Assume for the following analysis that the weight of the aircraft from the burning of fuel is the initial weight minus half the remaining fuel weight. Calculate the following:

(a) The operating lift coefficient of the wing, .

(b) The induced drag coefficient, and the total drag coefficient, .

(c) The lift-to-drag ratio of the aircraft at the conditions of flight.

(d) The thrust required from the engines for the aircraft to fly.

(e) The approximate maximum potential remaining flight range of the aircraft.

(f)The approximate flight time it will take to reach the remaining flight range.

We need some initial information, including the wing area and aspect ratio as well as the weight at which to run the calculations. Calculating the area of the wing gives

The aspect ratio can be calculated using

We are told to use the weight of the aircraft at a point which is its initial weight minus half the remaining fuel weight, so

(a) An airspeed of 250 kts is equal to 422 ft/s. The operating lift coefficient of the wing, , is

(b) The total drag coefficient, , is

(c) The lift-to-drag ratio of the aircraft at the given conditions of flight is

(d) The thrust required for the aircraft to fly is

(e) The time to burn off all of the relatively small amount of available fuel is

so the potential approximate range of the aircraft is

Worked Example #6

(a) With the assumption that the aircraft is flying along in steady unaccelerated flight, the lift force produced by the wing will equal the weight of the aircraft, i.e.,

(b) The lift coefficient of the wing is given by

To find , therefore, we need the density of the air (in this case at 10,000~ft) and the true airspeed in units of m/s. The density can be found from the ISA (assume standard temperature) so

and converting from nautical miles per hour (kts) to m/s (the conversion factor is on the formula sheet) gives

Inserting the numbers gives

(c) The drag force on the wing will be given by

where is the drag coefficient of the wing, which will be given by

where the second part is the induced drag (i.e., drag due to lift). Inserting the numbers gives

Therefore, the drag force on the wing is

(d) Now the lift and drag on the wing are known then the lift-to-drag ratio of the wing is

(e) The total drag force on the aircraft is

where is the net drag coefficient of the aircraft, which will be given by

where we are given that for the remainder of the aircraft then (the non-lifting part) and the second part will be from the wing (which has already been calculated). Notice that all drag coefficients are defined using wing area as a reference. For the entire aircraft then

Therefore, the drag force on the aircraft is

(f) Now the lift and drag are known then the lift-to-drag ratio of the entire aircraft is

Worked Example #7

It is desired to estimate the endurance and range of general aviation airplane similar to a Cessna 182 Skylane, the parameters describing this propeller-driven airplane being given in the table below.

Wing span | 35.8 ft | |

Wing area | 174 ft | |

Wing aspect ratio | 7.37 | |

Gross takeoff weight | 2,950 lb | |

Fuel capacity (tankage) | 65 U.S. gals AVGAS | |

Engine rated power | 230 hp @ MSL ISA | |

Engine BSFC | 0.45 lb hp hr | |

Parasitic drag coefficient | 0.025 | |

Oswald’s efficiency factor | 0.8 | |

Average propeller efficiency | 0.8 |

From the information given, many of the performance characteristics of the airplane can be established including the power requirements for flight. It can be assumed that the weight of the airplane is the gross takeoff weight, so and so

Rearranging for the lift coefficient gives

Also, the drag coefficient for the airplane (profile drag plus induced drag) is

and so the total drag is

For level flight then the brake power required is

remembering to factor in the propeller efficiency; the power curve is shown in the figure below for a pressure altitude of 5,000 ft where = 0.002048.

Finally, assuming a constant BSFC gives the fuel flow rate as

which is shown graphically in the figure below.

The airplane’s endurance and range can be estimated using the appropriate Breguet formulas. This airplane’s fuel capacity is 65 gallons of AVGAS and at 6.01 lb per gallon the maximum fuel weight that could be carried would be 390.65 lb. This is about 13% of the gross weight, so the assumption of constant weight over the flight time is reasonable for the evaluation of the maximum values of and . While these values and their corresponding airspeeds they will change somewhat with weight and altitude, another representative weight that can be used for analysis is the GTOW less half the total fuel weight.

Not all of this fuel would be usable, however, because at least some quantity fuel would be needed for startup, taxi, takeoff and climb and well as for the descent and landing plus an allowance of 30 minutes of VFR flight or 45 minutes of IFR flight must be included to comply with FAA regulations. If an average fuel burn allowance of 120 lb is assumed, which seems reasonable for takeoff and landing, this leaves 270 lb of fuel for actual flight use. So, for the endurance and range estimates in this case then = 2,950-60 = 2,890 lb and = 2,890 – 270 = 2,620 lb.

The estimated maximum endurance can be found using

From the graph below then the maximum value of is 12.79 at 81 mph or 118.8 ft/s.

Notice that the BSFC value must be converted into appropriate engineering units so = 0.45/550/3600 = 2.27 in units of (lb) (lb ft s) (s). Substituting the appropriate values gives

So, if we could loiter the airplane at or around an airspeed of 81 mph would could theoretically stay in the air for nearly 10 hrs.

The estimated maximum range can be found using

We see that the maximum value of is 13.6 at 106 mph or 155 ft/s.

Substituting the appropriate values gives

which would only be obtained if this airplane was flown at or around 106 mph. Of course, this is really rather slow compared to the top speed of the airplane, which is around 170 mph, but at that airspeed the range would be reduced by nearly half.

Remember that these latter results are only estimates of the maximum range and endurance, but they are typically within 10% of the actual values that can be demonstrated in flight.

Worked Example #8

It is desired to estimate the endurance and range of a jet-powered airplane in the form of a small jet similar to Cessna Citation. The parameters describing this airplane a listed in the table below.

Wing span | 53.3 ft | |

Wing area | 318 ft | |

Wing aspect ratio | 8.93 | |

Gross takeoff weight | 19,815 lb | |

Fuel capacity (tankage) | 1,119 U.S. gals JET-A | |

Engine rated thrust ( 2) | 3,650 lb @ MSL ISA | |

Engine TSFC | 0.6 lb lb hr | |

Parasitic drag coefficient | 0.02 | |

Oswald’s efficiency factor | 0.81 |

assuming no transonic wave drag. For level flight then the total thrust required is

where

Substituting the known values for our exemplar jet airplane gives the thrust required curve shown below for a pressure altitude of 15,000 ft where = 0.0014963.

which is shown in the figure below. Notice that the fuel flow mimics the thrust requirements, and for flight at higher airspeeds the fuel flow increases rapidly, reflecting the fact that for any aircraft there is a high fuel cost involved in flying fast.

The best ratio of 16.85 for this airplane occurs at an airspeed of 208 kts or 351. ft/s according to the results given below. Because we have formulas for calculating both and , the needed ratios can easily be calculated too.

Substituting the numbers and remembering to convert into units of s by dividing by 3,600 gives

where the final result has been converted back from seconds to hours.The range is evaluated by using

The best ratio of 23.4 occurs at an airspeed of 274 kts or 462 ft/s. Substituting the actual numbers gives

which seems fairly reasonable for this class of jet-powered airplane. Of course, all of these results would depend on the payload, which may need to be traded for fuel.

Worked Example #9

(a) The equivalent velocity from the rocket motor will be

The burnout mass has no fuel left so this mass will just be the sum of the structural mass and payload mass, i.e.,

The initial mass of the rocket has all the unburned fuel so

With the gravity term included, then the burnout velocity of the rocket will be

(b) The thrust generated by the rocket will be

(c) With the consideration of gravity, the initial acceleration will be

so about 3g.

(d) After 30 secs from liftoff the the mass of the rocket will be

and so the acceleration at 30 secs will be

so about 6g.

Of course, the inclusion of aerodynamic drag on the rocket would reduce this value slightly.

Worked Example #10

(a) We have that specific impulse is 240 seconds and the initial mass is 1.04 kg and the burnout mass is 1.11 kg. Therefore, the propellant mass will be

The rocket equation (we are told to ignore the gravity term in this equation) gives us change in the velocity of the vehicle , i.e.,

The equivalent exhaust velocity is given in terms of the specific impulse, i.e.,

The burnout velocity is

if we do not include the gravity term, as stated.

(b) We need to find the net mass flow rate to the engines. Assuming the the rate of propellant consumption is constant, then the mass of the rocket varies over time as

where is the burnout time and is the rate of burning propellant. The thrust produced will be

(c) The initial acceleration will be

(d) The mass of the rocket at 60 seconds into the flight will be

so the acceleration will be

Worked Example #11

We are given that specific impulse is 250 seconds, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg. The rocket equation gives us change in the velocity of the vehicle , i.e.,

where is the initial mass of the vehicle and is the final or burnout mass. If gravity is included (but no aerodynamic drag) then

where is the burnout time, which is given as 60 seconds in this case.

The equivalent exhaust velocity is given in terms of the specific impulse, i.e.,

The burnout mass is given by

So now we have the burnout velocity, which is

so

if we do not include the gravity term. With the gravity term included then

so at the burnout is reduced to

Assuming the propellant consumption rate is constant, then the mass of the rocket varies over time as

The velocity then is

The height achieved at the burnout time is then

which after some mathematics gives

Inserting the values gives

so

The final additional coasting height of the rocket can then be determined by equating the kinetic energy of the rocket at its burnout time with its change in potential energy between that point and the maximum obtained height.

Worked Example #12

The equivalent exhaust velocity

so

and

Therefore, the initial mass is

and the burnout mass is

so the propellant mass is

Worked Example #13

We have that specific impulse is 250~seconds, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg. The rocket equation gives us change in the velocity of the vehicle , i.e.,

where is the initial mass of the vehicle and is the final or burnout mass. If gravity is included (but no aerodynamic drag) then

where is the burnout time, which is given as 60 seconds in this case.

The equivalent exhaust velocity is given in terms of the specific impulse, i.e.,

The burnout mass is given by

So now we have the burnout velocity, which is

so

if we do not include the gravity term. With the gravity term included then

so at the burnout is reduced to

Assuming the propellant consumption rate is constant, then the mass of the rocket varies over time as

The velocity is

The height achieved at the burnout time is then

which after some mathematics gives

Inserting the values gives

so

The final additional coasting height of the rocket can then be determined by equating the kinetic energy of the rocket at its burnout time with its change in potential energy between that point and the maximum obtained height.