56 Worked Examples: Aircraft, Rocket, & Spacecraft Performance

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

Worked Example #1

A propeller-driven airplane has a fuel flow {\overbigdot{W}_\! f} to the engines of the form

    \[ \frac{dW_f}{dt} = {\overbigdot{W}_\! f}} = c_{b} \left( A V_{\infty}^3 + \frac{B}{V_{\infty}} \right) \]

where {\overbigdot{W}_\! f} is the weight of fuel burned per unit time, V_{\infty} is the true airspeed, c_b is the brake specific fuel consumption. The values of A and B depend on the characteristics of the aircraft, the aircraft weight, and the density altitude at which it is flying, i.e.,

    \[ A = \frac{1}{2} \varrho_{\infty} S C_{D_{0}} \mbox{~~~~and~~~~} B = \frac{2 W^2}{\varrho_{\infty} S (\pi AR e)} \]

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

    \[ V_{\infty} = V_x = \left(\frac{B}{3A}\right)^{1/4} \]

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

    \[ V_{\infty} = V_y = \left(\frac{B}{A}\right)^{1/4} \]

Explain the operational significance of flying at V_x and V_y, 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

    \[ \frac{dW_f}{dt} = {{\overbigdot{W}_\! f}} = c_b\left( A V_{\infty}^3 + \frac{B}{V_{\infty}} \right) \]

the lowest fuel burn (hence maximum flight endurance) can be determined by finding when {\overbigdot{W}_\! f} is a minimum. Differentiating the fuel flow equation with respect to V_{\infty} gives

    \[ \frac{d{\overbigdot{W}_\! f}}{dV_{\infty}} = c_b \left( 3 A V_{\infty}^2 - \frac{B}{V_{\infty}^2} \right) \]

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

    \[ 3 A V_{\infty}^2 - \frac{B}{V_{\infty}^2} = 0 \]

and so

    \[ V_{\infty} = V_x = \left( \frac{B}{3A} \right)^{1/4} \]

The best range is obtained when the ratio {\overbigdot{W}_\! f}/V_{\infty} is a minimum. In this case

    \[ \frac{{\overbigdot{W}_\! f}}{V_{\infty}} = c_b \left( A V_{\infty}^2 + \frac{B}{V_{\infty}^2} \right) \]

so that

    \[ \frac{d({\overbigdot{W}_\! f}/V_{\infty})}{dV_{\infty}} = c_b \left( 2 A V_{\infty} - 2\frac{B}{V_{\infty}^3} \right) \]

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

    \[ 2 A V_{\infty} - \frac{2B}{V_{\infty}^3} = 0 \]

and so

    \[ V_{\infty} = V_y = \left( \frac{B}{A} \right)^{1/4} \]

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

Starting from the steady, level-flight flight assumption of L = W and T = D, and assuming the engine thrust specific fuel consumption TSFC (c_t) is constant, then show that for a jet-propelled airplane lying at a constant altitude that the fuel flow to the engines in terms of weight of fuel burned per unit time {\overbigdot{W}_\! f} can be expressed in the form

    \[ {{\overbigdot{W}_\! f}} = A V_{\infty}^2 + \frac{B}{V_{\infty}^2} \]

where you should evaluate A and B in terms of the the wing area, S, wing aspect ratio, A\!R, flight weight,  W, parasitic drag coefficient & C_{D_{0}}, Oswald’s efficiency factor, e, engine TSFC, c_t.

The standard formula gives the lift L on the airplane.

    \[ L = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_L \]

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

    \[ D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_D \]

where the drag coefficient C_D is given by

    \[ C_D = C_{D_{0}} + C_{D_{i}} = C_{D_{0}} + \frac{{C_L}^2}{\pi AR e} \]

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

    \[ C_L = \frac{2 L}{\varrho_{\infty} V_{\infty}^2 S} = \frac{2 W}{\varrho_{\infty} V_{\infty}^2 S} \]

Also, because the thrust T equals drag, then

    \[ T = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_{D_{0}} + \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \left( \frac{2 W}{\varrho_{\infty} V_{\infty}^2 S} \right)^2 \left( \frac{1}{\pi AR e} \right) \]

and after some rearrangement, then

    \[ T = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_{D_{0}} + \frac{2 W^2}{\varrho_{\infty} V_{\infty}^2 S (\pi AR e)} \]

For a jet engine, the fuel burn rate {\overbigdot{W}_\! f} will be the product of the thrust and the thrust-specific fuel consumption, i.e.,

    \[ \frac{d W_f}{dt} = {\overbigdot{W}_\! f} = T c_t \]

and substituting for T gives

    \[ {\overbigdot{W}_\! f} = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S c_t C_{D_{0}} + \frac{2 c_t W^2}{\varrho_{\infty} V_{\infty}^2 S (\pi AR e)} \]

or

    \[ {\overbigdot{W}_\! f} = \left( \frac{1}{2} \varrho_{\infty} S c_t C_{D_{0}} \right) V_{\infty}^2 + \left( \frac{2 c_t W^2}{\varrho_{\infty} S (\pi AR e)} \right) \frac{1}{V_{\infty}^2} \]

which is of the form

    \[ {\overbigdot{W}_\! f} = A V_{\infty}^2 + \frac{B}{V_{\infty}^2} \]

where

    \[ A = \frac{1}{2} \varrho_{\infty} S c_t C_{D_{0}} \]

and

    \[ B = \frac{2 c_t W^2}{\varrho_{\infty} S (\pi AR e)} \]

Both constants are based on the stated assumptions of all the values involved for a constant weight W and altitude, i.e., \varrho_{\infty} = constant.

Worked Example #3

Using the fuel flow equation from the previous worked example, i.e.,

    \[ {{\overbigdot{W}_\! f}} = \frac{d W_f}{dt} = A V_{\infty}^2 + \frac{B}{V_{\infty}^2} \]

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

    \[ V_{\infty} = V_x = \left(\frac{B}{A}\right)^{1/4} \]

In this problem, we are asked to start from

    \[ \frac{d W_f}{dt} = {\overbigdot{W}_\! f} = A V_{\infty}^2 + \frac{B}{V_{\infty}^2} \]

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

    \[ \frac{d{\overbigdot{W}_\! f}}{dV_{\infty}} = 2 A V_{\infty} - \frac{2 B}{V_{\infty}^3} \]

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

    \[ 2 A V_{\infty} - \frac{2B}{V_{\infty}^3} = 0 \]

and so

    \[ V_{\infty} = V_x = \left( \frac{B}{A} \right)^{1/4} \]

where

    \[ A = \frac{1}{2} \varrho_{\infty} S c_t C_{D_{0}} \]

and

    \[ B = \frac{2 c_t W^2}{\varrho_{\infty} S (\pi AR e)} \]

So V_x 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^3. 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, b = 36 ft
  • Wing area, S = 174 ft^2
  • Non-lifting drag coefficient of aircraft, C_{D_{0}} = 0.02
  • Average propeller efficiency, \eta_p = 0.85
  • Oswald’s efficiency factor, e = 0.81
  • Engine BSFC, c_b = 0.45 lb hp^{-1} hr^{-1}

Calculate the following:
(a) The operating lift coefficient of the wing, C_L.
(b) The induced drag coefficient, C_{D_{i}} and the total drag coefficient, C_D.
(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 W_0, 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 W_f = 48 lb of fuel. Therefore, the weight to perform the analysis is

    \[ W = W_0 - \frac{W_f}{2} = 2,100 - \frac{48.0}{2} = 2, 076~\mbox{lb} \]

An airspeed of 120 kts is equivalent to 120 \times 1.688 = 202.54 ft/s = V_{\infty}.

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

    \[ C_L = \frac{2W}{\varrho V^2 S} = \frac{2 \times 2,076}{0.00216 \times 202.54^2 \times 174.0} = 0.269 \]

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

    \[ AR = \frac{b^2}{S} = \frac{(36.0)^2}{174.0} = 7.45 \]

The induced drag coefficient C_{D_{i}} will be

    \[ C_{D_{i}} = \frac{{C_L}^2}{\pi AR e} = \frac{0.269^2}{\pi \times 7.45 \times 0.81} = 0.00382 \]

The total drag coefficient, C_D, is

    \[ C_D = C_{D_0} + C_{D_i} = C_{D_0} + \frac{{C_L}^2}{\pi AR e} = 0.02 + 0.00382 = 0.02382 \]

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

    \[ \frac{L}{D} = \frac{C_L}{C_D} = \frac{0.269}{0.02382} = 11.3 \]

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

    \[ \frac{W}{C_L/C_D} = \frac{2,076}{11.3} = 183.7~\mbox{lb} \]

The power required for flight P_{\rm req} will be

    \[ P_{\rm req} = \frac{ T \, V_{\infty} }{\eta_p} = \frac{183.7 \times 202.54}{0.85 \times 550} = 79.6~\mbox{hp} \]

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

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

    \[ \frac{d W_f}{dt} = {\overbigdot{W}_\! f} = P_{\rm req} \, c_b = 79.82 \times 0.45 = 35.8~\mbox{lb hr$^{-1}$} \]

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

    \[ \mbox{Time} = T_i = \frac{48.0}{35.92} = 1.34~\mbox{hrs} \]

so the potential approximate range of the aircraft is

    \[ \mbox{Range} = R = 1.34 \times 120 \approx 160~\mbox{nautical~miles} \]

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^3. 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 C_r = 9 ft and a tip chord C_t = 4 ft. The other characteristics of the aircraft are: Wing span, b = 48 ft; Non-lifting drag coefficient, C_{D_{0}} = 0.02; Oswald’s efficiency factor, e = 0.81; Engine TSFC = c_t = 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, C_L.
(b) The induced drag coefficient, C_{D_{i}} and the total drag coefficient, C_D.
(c) The lift-to-drag ratio of the aircraft at the conditions of flight.
(d) The thrust T 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

    \[ S = b\frac{(C_r + C_t)}{2} = 48.0 \left( \frac{9.0 + 4.0}{2} \right) = 312.0~\mbox{ft}^2 \]

The aspect ratio can be calculated using

    \[ AR = \frac{b^2}{S} = \frac{(48.0)^2}{312.0} = 7.38 \]

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

    \[ W = W_0 - \frac{W_f}{2} = 21,000 - \frac{850.0}{2} = 20,575~\mbox{lb} \]

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

    \[ C_L = \frac{2W}{\varrho V^2 S} = \frac{2 (20,575)}{0.0015 (422.0)^2 312.0} = 0.494 \]

(b) The total drag coefficient, C_D, is

    \[ C_D = C_{D_0} + C_{D_i} = C_{D_0} + \frac{{C_L}^2}{\pi AR \, e} = 0.02 + \frac{0.494^2}{\pi\times 7.38 \times 0.81} = 0.033 \]

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

    \[ \frac{L}{D} = \frac{C_L}{C_D} = \frac{0.494}{0.033} = 14.925 \]

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

    \[ \frac{W}{C_L/C_D} = \frac{20,575}{14.925} = 1378.56~\mbox{lb} \]

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

    \[ \mbox{Time} = T_i = \frac{W_f}{c_t \, T} = \frac{850.0}{(0.5/3600)1378.56} = 4,439~\mbox{seconds} = 1.23~\mbox{hours} \]

so the potential approximate range of the aircraft is

    \[ \mbox{Range} = R = 1.23 \times 250 = 308~\mbox{nautical~miles} \]

Worked Example #6

Consider an aircraft that has a wing with a lifting planform area S = 60 m^2, an aspect ratio A\!R = 12, and an Oswald’s efficiency factor e = 0.90. The wing has a non-lifting profile drag coefficient of 0.01. The remainder of the aircraft has a non-lifting drag coefficient of 0.03. All force coefficients are based on wing area S. The mass of the airplane is 16,000 kg. If the aircraft is flying at a density altitude of 10,000 ft and its true airspeed is 253 kts, then calculate: (a) The lift force produced by the wing; (b) The lift coefficient of the wing; (c) The drag force on the wing; (d) The lift-to-drag ratio of the wing; (e) The total drag force on the aircraft;  (f) The lift-to-drag ratio of the aircraft.

(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.,

    \[ L = W = M \, g = 16,000 \times 9.81 = 156,960~\mbox{N} \]

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

    \[ C_L = \frac{L}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} \]

To find C_L, 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

    \[ \varrho_{\infty} = 0.9046~\mbox{kg/m$^3$} \]

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

    \[ V_{\infty} = 130.154~\mbox{m/s} \]

Inserting the numbers gives

    \[ C_L = \frac{L}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} = \frac{156,960}{\frac{1}{2} \times 0.9046 \times 130.154^2 \times 60.0} = 0.341 \]

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

    \[ D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \, C_D \]

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

    \[ C_{D_{\rm wing}} = 0.01 + \frac{{C_L}^2}{\pi \, A\!R \, e} \]

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

    \[ C_{D_{\rm wing}} = 0.01+ \frac{0.341^2}{\pi \times 12 \times 0.9} = 0.01 + 0.00343 = 0.01343 \]

Therefore, the drag force on the wing is

    \[ D_{\rm wing} = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \, C_D = 0.5 \times 0.906 \times 130.154^2 \times 60.0 \times 0.01343 = 6,186.32~\mbox{N} \]

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

    \[ \frac{L}{D_{\rm wing}} = \frac{156,960}{6,186.32} = 25.37 \]

(e) The total drag force on the aircraft is

    \[ D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \, C_D \]

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

    \[ C_D = 0.03 + C_{D_{\rm wing}} \]

where we are given that for the remainder of the aircraft then C_{D_{0}} = 0.03 (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

    \[ C_D = 0.03 + 0.01343 = 0.04343 \]

Therefore, the drag force on the aircraft is

    \[ D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S \, C_D = 0.5 \times 0.906 \times 130.154^2 \times 60.0 \times 0.04343 = 20,005.3~\mbox{N} \]

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

    \[ \frac{L}{D} = \frac{156,960}{20,005.4} = 7.85 \]

 

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 b 35.8 ft
Wing area S 174 ft^2
Wing aspect ratio A\!R 7.37
Gross takeoff weight W_{\rm GTOW} 2,950 lb
Fuel capacity (tankage) V_f 65 U.S. gals AVGAS
Engine rated power P_{\rm bhp} 230 hp @ MSL ISA
Engine BSFC c_b 0.45 lb hp^{-1} hr^{-1}
Parasitic drag coefficient C_{D_{0}} 0.025
Oswald’s efficiency factor e 0.8
Average propeller efficiency \eta_p 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 L = W =W_{\rm \small GTOW} and so

    \[ L = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_L =W_{\rm \small GTOW} = W \]

Rearranging for the lift coefficient gives

    \[ C_L = \frac{W}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} \]

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

    \[ C_D = C_{D_{0}} + \frac{{C_L}^2}{\pi A\!R e} \]

and so the total drag is

    \[ D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S C_D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S\left(C_{D_{0}} + \frac{{C_L}^2}{\pi A\!R e}\right) \]

For level flight then the brake power required is

    \[ P_{\rm req} = \frac{D V_{\infty}}{\eta_p} \]

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 \varrho_{\infty} = 0.002048.

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

    \[ \frac{dW_f}{dt} = {\overbigdot{W}_\! f} = c_b P_{\rm req} \]

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 C_L/C_D and C_L^{3/2}/C_D. 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 W_0 = 2,950-60 = 2,890 lb and W_1 = 2,890 – 270 = 2,620 lb.

The estimated maximum endurance can be found using

    \[ E = \frac{ \eta_p}{c} \left(\frac{C_L^{3/2}}{C_D}\right) \sqrt{2 \varrho_{\infty} S} \left( \frac{1}{\sqrt{W_0 - W_f}} - \frac{1}{\sqrt{W_0} } \right) \]

From the graph below then the maximum value of C_L^{3/2}/C_D is 12.79 at 81 mph or 118.8 ft/s.

Notice that the BSFC value must be converted into appropriate engineering units so c_b = 0.45/550/3600 = 2.27 \times 10^{-7} in units of (lb) (lb ft s^{-1})^{-1} (s)^{-1}. Substituting the appropriate values gives

    \begin{eqnarray*} E & = & \frac{0.8 \times 12.79}{2.27 \times 10^{-7}} \sqrt{2 \times 0.002048 \times 174} \left( \frac{1}{\sqrt{2,620}} - \frac{1}{\sqrt{2,890} } \right) \\ & = & 35,582 \mbox{s} = 593~\mbox{mins} = 9.9 \mbox{hrs} \nonumber \end{eqnarray*}

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

    \[ R = \frac{\eta_p}{c} \left( \frac{C_L}{C_D}\right) \ln \left( \frac{W_0}{W_0-W_f}\right) \]

We see that the maximum value of C_L/C_D is 13.6 at 106 mph or 155 ft/s.
Substituting the appropriate values gives

    \begin{eqnarray*} R & = & \frac{0.8 \times 13.6}{2.27 \times 10^{-7}} \ln \left( \frac{2,890}{2,620} \right) \\ & = & 3.73\times 10^6 \mbox{ft} = 707 \mbox{miles} \nonumber \end{eqnarray*}

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 b 53.3 ft
Wing area S 318 ft^2
Wing aspect ratio A\!R 8.93
Gross takeoff weight W_{\rm GTOW} 19,815 lb
Fuel capacity (tankage) V_f 1,119 U.S. gals JET-A
Engine rated thrust (\times 2) T_A 3,650 lb @ MSL ISA
Engine TSFC c_t 0.6 lb lb^{-1} hr^{-1}
Parasitic drag coefficient C_{D_{0}} 0.02
Oswald’s efficiency factor e 0.81
We can assume for the following calculations that the airplane’s weight is the gross takeoff weight, so L = W =W_{\rm \small GTOW}, although this assumption will tend to overpredict the fuel burn. Alternatively, one can use the W_{\rm GTOW} less half the total fuel weight, which is usually considered the preferred approach.The drag coefficient for the airplane is

    \[ C_D = C_{D_{0}} + \frac{{C_L}^2}{\pi A\!R e} \]

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

    \[ T_{\rm req} = D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 S\left(C_{D_{0}} + \frac{{C_L}^2}{\pi A\!R e}\right) \]

where

    \[ C_L = \frac{W}{\frac{1}{2} \varrho_{\infty} V_{\infty}^2 S} \]

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 \varrho_{\infty} = 0.0014963.

Because this is a twin-engine airplane each engine would produce half of this required thrust.Assuming a constant TSFC gives the total fuel flow rate as

    \[ \frac{dW_f}{dt} = {\overbigdot{W}_\! f} = c_t T_{\rm req} \]

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.

We can estimate the endurance and range of this jet airplane using the appropriate Breguet formulas. Notice that we should always be careful to use the appropriate formulas for a jet airplane and not confuse them with the propeller airplane.
The fuel capacity of this jet airplane is 1,119 gals of JET-A, and at 6.8 lb per gallon, the maximum fuel weight would be 7,609.2 lb, which is about 38% of the gross weight of the airplane. This is why a more accurate estimate of endurance and range would be done using W_{\rm GTOW} less half the total fuel weight. But we will continue here with W =W_{\rm \small GTOW}. Not all of this fuel would be usable, and allowances are needed for previously discussed reasons. We can proceed by assuming that W_0 = 19,815 -1,000 (the 1,000 lb being the allowance) = 18,815 lb and W_1 = 18,815 – 5,600 = 13,215 lb.The endurance is evaluated by using

    \[ E = \frac{1}{c_t} \left( \frac{C_L}{C_D} \right) \ln \left(\frac{W_0}{W_0 - W_f} \right) \]

The best C_L/C_D 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 C_L and C_D, the needed ratios can easily be calculated too.

Substituting the numbers and remembering to convert c_t into units of s^{-1} by dividing by 3,600 gives

    \[ E = \left(\frac{1}{0.6/3,600} \right) \left( 16.85 \right) \ln \left(\frac{18,815}{13,215} \right) = 9.9 \mbox{~hrs} \nonumber \]

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

    \[ R = \frac{2}{c_t} \sqrt{ \frac{2}{\varrho_{\infty} S}} \left(\frac{C_L^{1/2}}{C_D} \right) \left( W_0^{1/2} - W_1^{1/2} \right) \]

The best C_L^{1/2}/C_D ratio of 23.4 occurs at an airspeed of 274 kts or 462 ft/s. Substituting the actual numbers gives

    \begin{eqnarray*} R & = & \frac{2 \times 3,600}{0.6} \sqrt{ \frac{2}{0.0014963 \times 318}} \left( 23.4 \right) \left(18,815^{1/2} - 13,215^{1/2} \right) \nonumber \\ & = & 2,422 \mbox{~statute~miles} \end{eqnarray*}

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.

Note: For all aircraft there is a maximum useful load that can be carried, which is the sum of the payload and the fuel load. More payload (e.g., passengers) may, and usually does, allow for less fuel load.

Worked Example #9

Consider a small single-stage rocket with the following design characteristics: propellant mass = 7,200 kg; structural mass = 800 kg; payload mass = 50 kg. The specific impulse for this rocket is 275 s. The rocket blasts off from Earth and climbs vertically. The burning the fuel occurs at a steady rate and the burnout time is 60 s. The aerodynamic drag can be neglected but the effects of gravity should be included.  (a) What will be the value of the burnout velocity? (b) Find the thrust generated by the rocket. (c) What is the initial acceleration of the rocket? (d) What will be the speed and acceleration of the rocket at 30 seconds into the flight?

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

    \[ V_{\rm eq} = I_{\rm sp} \, g = 275 \times 9.81 = 2,698~\mbox{m/s} \]

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

    \[ M_b = M_S + M_L = 800 + 50 = 850 \mbox{kg} \]

The initial mass of the rocket M_0 has all the unburned fuel so

    \[ M_0 = M_S + M_L + M_p = 800 + 50 + 7,200 = 8,050~\mbox{kg} \]

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

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b = 2698 \times \ln(8050/850) - 9.81 \times 60 = 5,477~\mbox{m/s} \]

(b) The thrust generated by the rocket will be

    \[ T = \overbigdot{m}_P \, V_{\rm eq} = \frac{M_P}{t_b} V_{\rm eq} = 7,200/60 \times 2,698 = 323.76~\mbox{kN} \]

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

    \[ a = \frac{T - M_0 \, g_0}{M_0} = \frac{323,760 - 8,050 \times 9.81}{8,050} = 30.41~\mbox{m/s$^2$ } \]

so about 3g.

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

    \[ M_{30} = M_0 - \overbigdot{m}_P \, t = 8,050 - 7,200 \times 30 /60 = 4,450~\mbox{kg} \]

and so the acceleration at 30 secs will be

    \[ a_{30} = \frac{T - M_{30} \, g}{M_{30}} = \frac{ 323,760 - 4,450 \times 9.81}{4,450} = 62.94~\mbox{m/s$^2$} \]

so about 6g.

    \[ \Delta V_{30} = V_{\rm eq} \ln \left( \frac{M_0}{M_{30}} \right) -g \, t_b = 2698 \times \ln(8050/4450) - 9.81 \times 30 = 1,305~\mbox{m/s} \]

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

Worked Example #10

A single-stage rocket has a total mass of 1.14 \times 10^5 kg and a burnout mass of 1.11 \times 10^4 kg, including engines, structural shell, and payload. The rocket blasts off from Earth and climbs vertically, exhausting all of its propellant in 2.2 minutes. The burning of propellant occurs a steady rate, and the specific impulse of the propulsion system is 240 seconds. (a) If air resistance and gravity are neglected, what will be the velocity of the rocket at burnout conditions? (b) What thrust does the rocket engine develop at liftoff? (c) What is the initial acceleration of the rocket if gravity is not neglected? (d) What is the acceleration of the rocket at 60 seconds into the flight?

(a) We have that specific impulse I_{\rm sp} is 240 seconds and the initial mass M_0 is 1.04 \times 10^5 kg and the burnout mass M_b is 1.11\times 10^4 kg. Therefore, the propellant mass M_P will be

    \[ M_P = M_0 - M_b = 1.14 \times 10^5 - 1.11 \times 10^4 = 1.029 \times 10^5~\mbox{kg} \]

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

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \]

The equivalent exhaust velocity V_{\rm eq} is given in terms of the specific impulse, i.e.,

    \[ V_{\rm eq} = I_{\rm sp} g_0 = 240.0 \times 9.81 = 2,354.4~\mbox{m/s} \]

The burnout velocity is

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) = 2,354.4 \ln \left( \frac{114,000}{11,100} \right) = 5,484~\mbox{m/s} = 5.48~\mbox{km/s} \]

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 M varies over time as

    \[ M = M_0 - M_P \left( \frac{t}{t_b} \right) = M_0 - \left( \frac{M_P}{t_b} \right) t = M_0 - \overbigdot{m}_P \, t \]

where t_b is the burnout time and \overbigdot{m}_P is the rate of burning propellant. The thrust produced T will be

    \[ T = \overbigdot{m}_P V_{\rm eq} = \left( \frac{M_P}{t_b} \right) V_{\rm eq} = \left( \frac{1.029 \times 10^5}{2.2 \times 60} \right) 2,354.4 = 1.835~\mbox{MN} \]

(c) The initial acceleration will be

    \[ a = \frac{T - M_0 g}{M_0} = \frac{1.835 \times 10^6 - 1.14 \times 10^5 \times 9.81}{1.14 \times 10^5} = 6.29~\mbox{m/s$^{2}$} \]

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

    \[ M_{60} = M_0 - \left( \frac{M_P}{t_b}\right) t = 1.14 \times 10^5 - \left( \frac{1.029 \times 10^5}{2.2 \times 60} \right) 60.0 = 67,227.3~\mbox{kg} \]

so the acceleration will be

    \[ a = \frac{T - M_{60} g}{M_{60}} = \frac{1.835\times 10^6 - (67,227.3) (9.81)}{67,227.3} = 17.49~\mbox{m/s$^{2}$} \]

Worked Example #11

Use the rocket equation to determine the burnout velocity and the maximum achievable height of a rocket, assuming that it was launched vertically. Neglect the aerodynamic drag forces. Solve for the burnout velocity and maximum altitude if the burnout time is 60 seconds. The specific impulse is 250 seconds, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg.

We are given that specific impulse I_{\rm sp} is 250 seconds, the initial mass M_0 is 12,700 kg, and the propellant mass M_P is 8,610 kg. The rocket equation gives us change in the velocity of the vehicle \Delta V, i.e.,

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \]

where M_0 is the initial mass of the vehicle and M_b is the final or burnout mass. If gravity is included (but no aerodynamic drag) then

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b. \]

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

The equivalent exhaust velocity V_{\rm eq} is given in terms of the specific impulse, i.e.,

    \[ V_{\rm eq} = I_{\rm sp} g = 250 \times 9.81 = 2,452.5~\mbox{m/s} \]

The burnout mass M_b is given by

    \[ M_b = M_0 - M_P = 12,700 - 8,610 = 4,090~\mbox{kg} \]

So now we have the burnout velocity, which is

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) = 2,452.5 \ln \left( \frac{12,700}{4,090} \right) \]

so

    \[ \Delta V = 2,452 \times 1.133 = 2,778.16~\mbox{m/s} = 2.77~\mbox{km/s} \]

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

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b \]

so \Delta V at the burnout is reduced to

    \[ \Delta V = 2,778.16 - 9.81 \times 60 = 2,778.16 - 588.6 = 2,189.56~\mbox{m/s} = 2.18~\mbox{km/s} \]

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

    \[ M = M_0 - M_P \frac{t}{t_b} = M_0 - \left( M_0 - M_b \right) \frac{t}{t_b} \]

The velocity then is

    \[ V = V_{\rm eq} \ln \left( \frac{M_0}{M} \right) - g t \]

The height achieved at the burnout time is then

    \[ H_b = \int_0^{t_b} V dt \]

which after some mathematics gives

    \[ H_b = V_{\rm eq} t_b \left( \frac{ \ln \left( \displaystyle{\frac{M_0}{M_b}} \right) }{\displaystyle{\frac{M_0}{M_b} - 1} } \right) - \frac{1}{2} g t_b^2 \]

Inserting the values gives

    \[ H_b = 2,452.5 \times 60 \left( \frac{ \ln \left( \displaystyle{\frac{12,700}{4,090}} \right) }{\displaystyle{\frac{12,700}{4,090} - 1} } \right) - \frac{1}{2} \times 9.81 \times 60^2 \]

so

    \[ H_b = 2,452.5 \times 60 \left( \frac{1.133}{2.105} \right) - \frac{1}{2} \times 9.81 \times 60^2 = 61,544.5~\mbox{m} = 61.54~\mbox{km} \]

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

A rocket vehicle must provide a speed of 6,000 m/s to a payload of 12,000 kg. The structural coefficient of the vehicle is 0.06. The propellants used in the engines have a specific impulse of 325 secs. What must be the initial mass and propellant mass be to meet these requirements?

The equivalent exhaust velocity

    \[ V_{\rm eq} = I_{\rm sp} g = 3,188.25~\mbox{m/s} \]

so

    \[ R = \exp \left( \frac{\Delta V}{V_{\rm eq}} \right) = 6.57 \]

and

    \[ \lambda = \frac{1 - \epsilon R}{R - 1} = 0.109 \]

Therefore, the initial mass is

    \[ M_0 = \frac{1+\lambda}{\lambda} M_L = 122,211~\mbox{kg} \]

and the burnout mass is

    \[ M_b = \frac{M_0}{R} = 18,613~\mbox{kg} \]

so the propellant mass is

    \[ M_p = M_0 - M_b = 103.598~\mbox{kg} \]

Worked Example #13

Use the rocket equation to determine the burnout velocity and the maximum achievable height of a rocket if its burnout time is 60 seconds. The specific impulse is 250 seconds, the initial mass is 12,700 kg,  and the propellant mass is 8,610 kg. Assume the rocket is launched vertically.

We have that specific impulse I_{\rm sp} is 250~seconds, the initial mass M_0 is 12,700 kg,  and the propellant mass M_p is 8,610 kg. The rocket equation gives us change in the velocity of the vehicle \Delta V, i.e.,

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \]

where M_0 is the initial mass of the vehicle and M_b is the final or burnout mass. If gravity is included (but no aerodynamic drag) then

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b. \]

where t_0 is the burnout time, which is given as 60 seconds in this case.

The equivalent exhaust velocity V_{\rm eq} is given in terms of the specific impulse, i.e.,

    \[ V_{\rm eq} = I_{\rm sp} g = 250 \times 9.81 = 2,452.5~\mbox{m/s} \]

The burnout mass M_b is given by

    \[ M_b = M_0 - M_P = 12,700 - 8,610 = 4,090~\mbox{kg} \]

So now we have the burnout velocity, which is

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) = 2,452.5 \ln \left( \frac{12,700}{4,090} \right) \]

so

    \[ \Delta V = 2,452 \times 1.133 = 2,778.16~\mbox{m/s} = 2.77~\mbox{km/s} \]

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

    \[ \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b \]

so \Delta V at the burnout is reduced to

    \[ \Delta V = 2,778.16 - 9.81 \times 60 = 2,778.16 - 588.6 = 2,189.56~\mbox{m/s} = 2.18~\mbox{km/s} \]

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

    \[ M = M_0 - M_P \frac{t}{t_b} = M_0 - \left( M_0 - M_b \right) \frac{t}{t_b} \]

The velocity is

    \[ V = V_{\rm eq} \ln \left( \frac{M_0}{M} \right) - g t \]

The height achieved at the burnout time is then

    \[ H_b = \int_0^{t_b} V dt \]

which after some mathematics gives

    \[ H_b = V_{\rm eq} t_b \left( \frac{ \ln \left( \displaystyle{\frac{M_0}{M_b}} \right) }{\displaystyle{\frac{M_0}{M_b} - 1} } \right) - \frac{1}{2} g t_b^2 \]

Inserting the values gives

    \[ H_b = 2,452.5 \times 60 \left( \frac{ \ln \left( \displaystyle{\frac{12,700}{4,090}} \right) }{\displaystyle{\frac{12,700}{4,090} - 1} } \right) - \frac{1}{2} \times 9.81 \times 60^2 \]

so

    \[ H_b = 2,452.5 \times 60 \left( \frac{1.133}{2.105} \right) - \frac{1}{2} \times 9.81 \times 60^2 = 61,544.5~\mbox{m} = 61.54~\mbox{km} \]

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.

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Introduction to Aerospace Flight Vehicles by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.