Activity factor

The activity factor is a measure of the power-absorbing capacity of the airscrew, which, for optimum performance, must be accurately matched to the power pro­duced by the engine.

Подпись:Consider an airscrew of diameter D rotating at n with zero forward speed, and consider in particular an element of the blade at a radius of r, the chord of the element being c. The airscrew will, in general, produce a thrust and therefore there will be a finite speed of flow through the disc. Let this inflow be ignored, however. Then the motion and forces on the element are as shown in Fig. 9.3. 1

Activity factor

2.1ГГП

 

Activity factor

Fig. 9.3

and therefore the torque associated with the element is

SQ = 2ir1pCj}nl(cri)Sr

Подпись: tip

It is further assumed that Co is constant for all blade sections. This will not normally be true, since much of the blade will be stalled. However, within the accuracy required by the concept of activity factor, this assumption is acceptable. Then the total torque required to drive an airscrew with В blades is

Подпись: tip

Thus the power absorbed by the airscrew under static conditions is approximately

Подпись: 0.5.D

In a practical airscrew the blade roots are usually shielded by a spinner, and the lower limit of the integral is, by convention, changed from zero (the root) to 0.1 D. Thus

Activity factor

Defining the activity factor (AF) as

Activity factor

leads to

Further work on the topic of airscrew coefficients is most conveniently done by means of examples.

Подпись: J kQ Подпись: 1.06 0.0410 0.76 Подпись: 1.19 0.0400 0.80 Подпись: 1.34 0.0378 0.84 Подпись: 1.44 0.0355 0.86

Example 9.3 An airscrew of 3.4 m diameter has the following characteristics:

Calculate the forward speed at which it will absorb 750 kW at 1250 rpm at 3660 m (a = 0.693) and the thrust under these conditions. Compare the efficiency of the airscrew with that of the ideal actuator disc of the same area, giving the same thrust under the same conditions.

Power = 2m Q

Therefore

„ 750000×60 „„Л1кТ

torque g= 27rxl250 = 5730 Nm

Подпись:= 20.83 rps n1 = 435 (rps)2

Activity factor

Therefore

Plotting the given values of kg and 77 against J shows that, for kg = 0.0368, J = 1.39 and 77 = 0.848. Now J = V/nD, and therefore

V = JnD = 1.39 x 20.83 x 3.4 = 98.4ms_I

Since the efficiency is 0.848 (or 84.8%), the thrust power is

750 x 0.848 = 635 kW

Therefore the thrust is

Подпись: SpeedПодпись: 98.4

Activity factor

Power 635000 T = г = = 6460 N

whence

a = 0.0417

Activity factor

Thus the ideal efficiency is

Thus the efficiency of the practical airscrew is (0.848/0.958) of that of the ideal actuator disc. Therefore the relative efficiency of the practical airscrew is 0.885, or 88.5%.

Подпись: Speed (rpm) Power (kW)

Подпись: 1800 1072 Подпись: 1900 1113 Подпись: 2000 1156 Подпись: 2100 1189

Example 9.4 An aeroplane is powered by a single engine with speed-power characteristic:

The fixed-pitch airscrew of 3.05m diameter has the following characteristics:

J

0.40

0.42

0.44

0.46

0.48

0.50

кт

0.118

0.115

0.112

0.109

0.106

0.103

kQ

0.0157

0.0154

0.0150

0.0145

0.0139

0.0132

and is directly coupled to the engine crankshaft. What will be the airscrew thrust and efficiency during the initial climb at sea level, when the aircraft speed is 45ms~’?

Preliminary calculations required are:

Q = kepn2D5 = 324.2 kQn2

after using the appropriate values for p and D.

J = V/nD = 14.75/n

The power required to drive the airscrew, Pr, is

Pr = 2 miQ

With these expressions, the following table may be calculated:

rpm

1800

1900

2000

2100

Fa(kW)

1072

1113

1156

1189

n (rps)

30.00

31.67

33.33

35.00

/72(rps)2

900

1003

1115

1225

J

0.492

0.465

0.442

0.421

kQ

0.013 52

0.01436

0.01494

0.015 38

G(Nm)

3950

4675

5405

6100

P r(kW)

745

930

1132

1340

In this table, P3 is the brake power available from the engine, as given in the data, whereas the values of kQ for the calculated values of J are read from a graph.

A graph is now plotted of P3 and PT against rpm, the intersection of the two curves giving the equilibrium condition. This is found to be at a rotational speed of 2010 rpm, i. e. n = 33.5 rps. For this value of n, J = 0.440 giving kT = 0.112 and kQ = 0.0150. Then

T = 0.112 x 1.226 x (33.5)2 x (3.05)4 = 13 330N

Activity factor Подпись: 0.523 or 52.3%

and

As a check on the correctness and accuracy of this result, note that

thrust power = TV = 13 300 x 45 = 599 кW

At 2010 rpm the engine produces 1158kW (from engine data), and therefore the efficiency is 599 x 100/1158 = 51.6%, which is in satisfactory agreement with the earlier result.

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