RATE OF CLIMB, TIME TO CLIMB, AND CEILINGS

The rate of climb was given previously by Equation 7.13. For a steady climb this equation becomes

dh V(T-D) dt W

It can be expanded and expressed, as noted previously, as

dh Pa – Pr dt W

where Pa and Pr are the available power and required power, respectively.

Pa = TV Pr = DV

For a gas turbine engine, T is known as a function of altitude and velocity, so that Equation 7.15 is the obvious form to use in this case to determine the rate of climb, R/C. In the case of a propeller-driven airplane,

Pa Г/F shaft

where Pshaft is the shaft power delivered to the propeller and 17 is the propeller efficiency.

As a function of V, the required power is determined by

Pr = DV

Let us apply Equation 7.15 to the 747-100 example using the rated climb thrust given in Figure 6.37. Flaps and gear during the climb are assumed to be retracted, so that / reduces to lOO. ft2 (9.29 m2). As an example, take a speed of 200 m/s at an altitude of 6000 m. The acoustic velocity of this altitude equals 318 m/s. Hence the operating Mach number equals 0.629. From Figure 6.37.

T = 311 kN (4 engines)

The mass density at 6096 m equals 0.662 kg/m3. V and p result in a dynamic pressure of

q = 13,240 Pa

With a weight of 3260 kN and a wing area of 511 m2, the lift coefficient equals 0.482. Hence Cd, =0.0152. Therefore,

D = q(f+C»S)

= 226 kN

Equation 7.15 gives, for the rate of climb,

nl„ 200(311-226) R/C 3260

= 5.2 m/s

In this way, the curves presented in Figure 7.12 were obtained. Here, the rate of climb is plotted as a function of true airspeed at four different altitudes. At the highest altitude and air speeds, the curve is probably optimistic, since the drag rise beyond the drag-divergence Mach number is not considered.

At a given altitude, there exists an optimum airspeed for maximum rate of climb. In this case, the airspeed for the best rate of climb is seen to correspond to a nearly constant indicated air-speed of 296 kt.

The speed for the best climb angle is less than the speed for the best R/C. The angle of climb, 0C, in radians, is given by

Figure 7.12 Calculated rate of climb for Boeing 747-100 as a function of true airspeed for several altitudes.

For any velocity, this angle is represented by the angle between the abscissa and a straight line from the origin to the R/C curve on Figure 7.12. As shown for the sea level curve, this angle is a maximum for the straight line that is tangent to the curve. Thus, to clear an obstacle ahead, a pilot should fly slightly slower than the speed that gives the maximum rate of climb.

The optimum airspeed for maximum rate of climb can be obtained approximately from Equation 7.15 for a turbojet or turbofan.

Figure 7.13 Calculated maximum rate of climb for Boeing 747-100 as a function Of altitude.

The maximum rates of climb obtained from Figure 7.12 are plotted as a function of altitude in Figure 7.13. The absolute ceiling’ is defined as the altitude for which the R/C equals zero. In this case, this corresponds to an altitude of 9.20 km.

Consider now the rate-of-climb calculation for a propeller-driven air­plane. We will use as an example the Piper Cherokee Arrow that is similar to the Cherokee 180 except that the former has a retractable landing gear, a 200-bhp (149-kW) piston engine, and a constant speed propeller. The engine is rated at 2700 rpm but, for a continuous climb in accordance with recom­mended practice, an rpm of only 2500 will be used. At standard sea level, the engine develops 185 bhp (138 kW) at this rpm. For higher altitudes, the engine power is estimated on the basis of Figure 6.2. The following values are known or have been estimated for this airplane.

The propeller performance curves for this airplane were presented in Figures 6.16 and 6.17. These curves, together with the engine power, are used to estimate the available power. As an example, consider an altitude of

10,0 ft (305 m). The engine power at this altitude equals 130 bhp. At a speed of, say, 140 fps, the advance ratio will be

J = VlnD

= 140(0.1467)/(2500/60)/6.17 = 0.389

The power coefficient equals

Lp pn3D5

= 130(550)/(0.00176X41.7)3(6.17) = 0.063

This power coefficient and advance ratio lead to a blade angle of 2 Г from Figure 6.16. Using this blade angle, together with J, results in a propeller efficiency, 17, of 0.70 from Figure 6.17. Thus the available power at this speed
and altitude equals 91 thp where thp stands for “thrust horsepower.” The equivalent term in the SI system would be tkW, for “thrust kilowatts.”

The power required for the Arrow is calculated from

Pr = DV

which can be expressed in the form

p _ Pfvз і 2(Wlb)2 1 r 2 пре V

The first term on the right side is the parasite power and the second term the induced power. The shape of this relationship will be discussed in more detail later. For now let us simply evaluate Equation 7.21 at the altitude of 10,000 ft and a speed of 140 fps. The obvious substitutions result in a required power of 65.3 hp. The rate of climb can now be calculated from Equation 7.16. However, in so doing, the excess power must be expressed in foot-pounds per second.

(91 – 65.3)550
2650

= 5.34 fps

It is current practice in the American aviation industry to express the R/C in feet per minute, so that the above becomes 320 fpm (1.63 m/s).

In this manner, the rate of climb can be calculated over a range of speeds for several altitudes for the Arrow. The maximum rates of climb thus determined are presented in Figure 7.14, where the R/C is seen to decrease almost linearly with altitude. The altitude for which the R/C equals 100 fpm (0.5 m/s) is shown on Figure 7.14 and is called the service ceiling. In this example, the calculated service ceiling and sea level rate of climb are close to the corresponding values quoted by the manufacturer.

The power-required and power-available curves calculated for this example at sea level are presented in Figure 7.15. Similar to the drag curve in Figure 4.52, the power-required curve has a minimum value at some speed. Below this speed, it actually requires more power to fly slower. This part of the curve is referred to as the backside of the power curve.

‘ The speed for minimum required power can be found by setting to zero

Figure 7.15 Sea level power-required and power-available curves for the Cherokee Arrow.

the derivative of equation 7.21 with respect to V. This leads to

(7.22)

The preceding speed is, of course, not necessarily the speed for maxi­mum rate of climb, since the available power varies with airspeed. In the case of the Arrow, Figure 7.15 shows the optimum speed to be greater than the value given by Equation 7.22. Nevertheless, one might expect the two speeds to be related. Equation 7.22 shows the minimum required power to occur at a constant indicated airspeed. This suggests that the maximum rate of climb will also occur at a constant indicated airspeed. This was found to be true for the 747-100 and also holds for piston engine airplanes.