Bài tập chương 10 Hệ Thống Lực Đẩy Máy Bay

Assuming the stream of air to be the fully developed slipstream behind an ideal actuator disc, and ignoring mixing between the jet and the surrounding air, estimate the fan diameter and

Bài tập chương 10

Hệ Thống Lực Đẩy Máy Bay 1

10.1 If an aircraft of wing area S and drag coefficient C D is flying at speed V in air

of density ρ, and if its single airscrew, of disc area A, produces a thrust equal to the aircraft drag, show that the speed in the slipstream V S is, on the basis of Froude’s momentum theory

S

V VC 1

A

Solution The Drag:

2 D

1

D V SC 2

 

The Thrust:

T AV V V A V V A V V



We have T = D

A V V V SC

A V V C SV

S

V V C 1

A S

V V C 1

A

   

    

10.2 A cooling fan is required to produce a stream of air 0.5 m in diameter with a speed of 3 m/s when operating in a region of otherwise stationary air of standard density Assuming the stream of air to be the fully developed slipstream behind an ideal actuator disc, and ignoring mixing between the jet and the surrounding air, estimate the fan diameter and power input required

(Answer: 0.707-m diameter; 3.24 W)

Solution The Fan speed

V V V V 3 1.5 m / s

     

The Fan diameter:

2 2

e

S e 0

m SV S V const

D D

D D 0.5 0.707 m

V 1.5

   







The power input required:

2

P TV S P P V V S V 1.225 3 1.5 3.246 W

 

           

10.3 Repeat Example 10.2 in the text for the case where two airscrews absorb

equal power, and find

(1) the thrust of the second airscrew as a percentage of the thrust of the first; (2) the efficiency of the second;

(3) the efficiency of the combination

(Answer: 84%; 75.5%; 82.75%)

Solution

0.9 a

1 a    9

Thus

V V 1 a V and V V 1 2a V

The Power of the front airscrew is:

P TV   S V V  V V

The second airscrew is working entirely in the slipstream of the first, so the speed

of the approaching flow is V S, The power is:

  2    2 

P  S V ' V ' V '   S V ' V ' V 

Now, by continuity,

S V S V '

 

Also, the powers from the two airscrews are equal, so

2 2

2 2

S S

S V V V S V ' V ' V S V V ' V ' V

V V V V ' V ' V

V ' V

V V V V ' V

2 2V V V V ' V

2 V V V V ' V

161

V ' V 1.41V

9



        

(1) the thrust of the second airscrew as a percentage of the thrust of the first:

2

11 1.41

V ' V ' V ' V

84.5%

11

9



(2) the efficiency of the second

Then, if the rate of mass flow through the discs is m  , Thrust of rear airscrew:

 S   S S

11

T m V ' V ' m V ' V m 1.41V V 0.188mV

9

       

The useful power given by the second airscrew is TV, not TVs; therefore

Useful power from 2nd airscrew  0.188mV  2

Kinetic energy added per second by the second airscrew, which is the power supplied by (and to) the second disc, is

2

P m V ' V m (1.41V) V 0.247mV

        

Thus the efficiency of the rear components is:

2 2

0.188mV

76.11%

0.247mV

   



(3) The efficiency of the combination

Power input to front airscrew TV

0.9



Power input to rear airscrew TV

0.7611



Total power input TV TV 2.425TV

0.7611 0.9

Total useful power output  2TV

Efficiency of combination 2TV 82.47%

2.425TV

10.4 Calculate the flight speed at which the airscrew in Example 10.3 produces a

thrust of 7500 N Also calculate the power absorbed at the same rotational speed (Answer: 93 m/s; 840 kW)

Solution

Từ các công thức:

 Vận tốc: V JnD 

Q

Q k n D  

 Công suất lực đẩy: P Q  

 Lực đẩy: T P

V



ta tính được các giá trị V, Q, P, T ứng với các giá trị J, k Q,  tương ứng (1, 2, 3 và 4)

Sử dụng phương pháp nội suy ứng với lực đẩy T=7500N, ta tính được các giá trị

V,  tương ứng (xem bảng)

1 1.06 0.041 0.76 75.083 6863.811 682837.736 9094.398

2 1.19 0.04 0.80 84.292 6696.401 701245.4285 8319.273

3 1.34 0.0378 0.84 94.917 6328.099 695810.7765 7330.754

4 1.44 0.0355 0.86 102.000 5943.056 669031.9667 6559.137

Vậy vận tốc dòng: V 93.098 m/s 

Năng lượng hấp thụ:

P T *V /    7500*93.098 / 0.83 841.246 kW 

10.5 At 1.5-m radius, the thrust and torque gradings on each blade of a three bladed airscrew revolve at 1200 rpm at a flight speed of 90 m/s TAS at an altitude where   0.725 are 300 Nm  1 and 1800 Nm m  1 , respectively

If the blade angle is 28 degrees, find the blade section absolute incidence Ignore compressibility

(Answer: 1 48' o )

Solution The thrust gradings on each blade of a three bladed airscrew is 300 Nm  1, thus

 

       

2

2

dT

4 rV a 1 a dr

3 300 4 0.725 1.225 1.5 90 a 1 a

a 0.0066

  

 

The torque gradings on each blade of a three bladed airscrew is 1800 Nm m  1, thus

     

3

dQ

4 r Vb 1 a 4 r Vb 1 a 2 n dr

1800 3 4 1.5 0.725 1.225 90 1 0.0066 2 20 b

b 0.0126

         

 

The blade section absolute incidence:

 

 

o

V 1 a 90 1 0.0066

r 1 b 2 20 1.5 1 0.0126 25,95

28 25,95 2 3'

    

  

       

10.6 At 1.25-m radius on a three-bladed airscrew, the airfoil section has the following characteristics:

L

Solidity 0.1 ;    29 7 ' ; =4 7 ' ; C   0.49 ; L/D=50

Allowing for both axial and rotational interference, find the local efficiency of the element

(Answer: 0.885)

Solution

o

D 1 tan =1.15

L 50

29 7 ' 4 7 ' 25

    

      

   0 

Geometric pitch = 2 rtan     2 1.25 tan 29 7 '  4.37m

L

L

q C sin 0.49sin 25 1.15 0.216

t C cos 0.49cos 25 1.15 0.440

      

      

 

   

 

   

o

0.1 0.216

0.0141

1 b 2sin 2 2sin 2 25

0.1 0.440

0.0187

1 a 4sin 2 4sin 2 25





o

V V 1 b 1 0.0187

tan tan 25 0.451

r 2 nr 1 a 1 0.0141

   

The local efficiency of the element:

l

V t 0.440 0.451 91.87%

r q 0.216

    



10.7 The thrust and torque gradings at 1.22-m radius on each blade of a two-bladed airscrew are 2120 N/m and 778 Nm/m, respectively Find the speed of rotation (in 1

rads  ) of the airstream immediately behind the disc at 1.22-m radius (Answer: 735 rads  1)

10.8 A four-bladed airscrew is required to propel an aircraft at 125 m/s at sea level, with a rotational speed of 1200 rpm The blade element at 1.25-m radius has

an absolute incidence of 6 degrees, and the thrust grading is 2800 N/m per blade Assuming a reasonable value for the sectional lift-curve slope, calculate the blade chord at 1.25-m radius Neglect rotational interference, sectional drag, and compressibility

(Answer: 240 mm)

Solution The thrust grading is 2800 N/m per blade:

 

     

2

2

dT

4 V a 1 a dr

2800 4 4 1.225 125 a 1 a

a 0.036

  

 

Neglect rotational interference:

 

 

 

 

   

      o

V 1 a V 1 a 125 1 0.036

r 1 b 2 nr 1 b 2 20 1.25 1 0 39.5

     

 

Neglect sectional drag: D o

L

     

We have:

   

o L

V V 1 a 125 1 0.036 203.6 m / s

sin sin 39.5

t C cos 0.6cos 39.5 0.463







     



The blade chord at 1.25-m radius:

     

2 R

2

dT 1

c V t

dr 2

2800 c 0.5 1.225 203.6 0.463

c 0.238m 23.8cm

 

10.9 A three-bladed airscrew is driven at 1560 rpm at a flight speed of 110 m/s at sea level At 1.25-m radius, the local efficiency is estimated to be 87%, while the lift/drag ratio of the blade section is 57.3 Calculate the local thrust grading, ignoring rotational interference

(Answer: 9000 N/m per blade)

propeller in terms of disc area, air density, and axial velocities of the air a long way ahead, and in the plane, of the propeller disc A helicopter has an engine developing 600 kW and a rotor of 16-m diameter with a disc loading of 170 N/m 2 When ascending vertically with constant speed at low altitude, the product of lift and axial velocity of the air through the rotor disc is 53% of the power available Estimate the velocity of ascent

(Answer: 110 m/min)