Bài tập chương 10 Hệ Thống Lực Đẩy Máy Bay

Hệ Thống Lực Đẩy Máy Bay Bài tập chương 10 Hệ Thống Lực Đẩy Máy Bay Phan Hoàng Đức 1610786 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 VS is, on the basis of Froude’s momentum theory VS = VC D + S A Solution The Drag: D = ρV 2SC D The Thrust: T = ρAV0 ( VS − V ) = ρA VS + V ( VS − V ) = ρA ( VS2 − V ) 2 We have T = D 1 ρA ( VS2 − V ) = ρV 2SCD 2 2 ⇔ A ( VS − V ) = CDSV S   ⇔ VS2 = V  CD + 1÷ A   ⇔ VS = V CD S +1 A 10.2 A cooling fan is required to produce a stream of air 0.5 m in diameter with a speed of 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 Phan Hoàng Đức 1610786 Hệ Thống Lực Đẩy Máy Bay (Answer: 0.707-m diameter; 3.24 W) Solution The Fan speed V0 = 1 ( VS + V ) = VS = × = 1.5 m / s 2 The Fan diameter: &= ρSV0 = ρSe VS ( ρ = const ) m πDe2 πD V0 = VS 4 VS ⇔ D = De = 0.5 = 0.707 m V0 1.5 ⇔ The power input required: 1 π× 0.707  1  P = TV0 = S ( P2 − P1 ) V0 = VS2S ữV0 = 1.225 ì 32 ì ÷( 1.5 ) = 3.246 W 2 2   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 1 = 0.9 ⇒ a = 1+ a Thus V0 = V ( + a ) = 10 11 V and VS = V ( + 2a ) = V 9 The Power of the front airscrew is: P = TV0 = ρS1V0 ( VS − V ) V0 Phan Hoàng Đức 1610786 Hệ Thống Lực Đẩy Máy Bay The second airscrew is working entirely in the slipstream of the first, so the speed of the approaching flow is VS , The power is: P = ρS2 ( V0 ' ) ( VS '− V ') = ρS2 ( V0 ') ( VS '− VS ) Now, by continuity, ρS1V0 = ρS2 V0 ' Also, the powers from the two airscrews are equal, so ρS1V02 ( VS − V ) = ρS2 ( V0 ' ) ( VS '− VS ) = ρS1V0 V0 ' ( VS '− VS ) ⇔ V0 ( VS − V ) = V0 ' ( VS '− VS ) VS '+ VS ( VS '− VS ) ⇔ 2V0 ( VS − V ) = VS '2 − VS ⇔ V0 ( VS − V ) =  10  11   11  ⇔  V ÷ V − V ÷ = VS '2 −  V ÷    9  ⇔ VS ' = 161 V ≈ 1.41V (1) the thrust of the second airscrew as a percentage of the thrust of the first: 11 1.41 − T2 VS '− V ' VS '− VS = 84.5% = = = 11 T1 VS − V VS − V −1 (2) the efficiency of the second Then, if the rate of mass flow through the discs is m&, Thrust of rear airscrew: 11 &( VS '− V ' ) = m &( VS '− VS ) = m &1.41V − V ÷ = 0.188mV & T=m   The useful power given by the second airscrew is TV, not TVs; therefore & Useful power from 2nd airscrew = 0.188mV Kinetic energy added per second by the second airscrew, which is the power supplied by (and to) the second disc, is  dE 1  11 &( VS '2 − VS2 ) = m & (1.41V) −  V ÷ ÷ = 0.247mV & =P= m  ÷ dt 2  9   Phan Hoàng Đức 1610786 Hệ Thống Lực Đẩy Máy Bay Thus the efficiency of the rear components is: η= & 0.188mV = 76.11% & 0.247mV (3) The efficiency of the combination Power input to front airscrew = Power input to rear airscrew = TV 0.9 TV 0.7611 TV TV + = 2.425TV 0.7611 0.9 Total power input = 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 Các đại lượng đề cho Lực đẩy T Khối lượng riêng ρ Đường kính chong chóng D Tốc độ quay n Từ công thức: + Vận tốc: V = JnD + Moment xoắn Q: Q = k Qρn D5 + Công suất lực đẩy: P = Q × η + Lực đẩy: T = Phan Hoàng Đức P V 1610786 Giá trị Đơn vị 7500 N 0.848925 kg/m3 3.4 m 20.83333 rps Hệ Thống Lực Đẩy Máy Bay ta tính giá trị V, Q, P, T ứng với giá trị J, k Q , η tương ứng (1, 2, 4) Sử dụng phương pháp nội suy ứng với lực đẩy T=7500N, ta tính giá trị V, η tương ứng (xem bảng) J kQ η V (m/s) Q (N.m) P (W) T (N) 1.0 0.041 0.76 75.083 6863.811 682837.736 9094.398 1.1 0.04 0.80 84.292 6696.40 701245.428 8319.273 1.3 0.0378 0.84 94.917 6328.09 695810.776 7330.754 1.4 0.0355 0.86 102.000 5943.05 669031.966 6559.137 0.83 93.098 Nội Suy 7500 Vậy vận tốc dòng: V ≈ 93.098 m/s Năng lượng hấp thụ: P = T * V / η = 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 σ = 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: 1o 48' ) Solution The thrust gradings on each blade of a three bladed airscrew is 300 Nm −1 , thus Phan Hoàng Đức 1610786 Hệ Thống Lực Đẩy Máy Bay dT = 4πρrV a ( + a ) dr ⇔ × 300 = 4π ( 0.725 × 1.225 ) ( 1.5 ) ( 90 ) × a ( + a ) ⇔ a = 0.0066 The torque gradings on each blade of a three bladed airscrew is 1800 Nm m −1 , thus dQ = 4πr 3ρVb ( + a ) Ω = 4πr 3ρVb ( + a ) ( 2πn ) dr ⇔ 1800 × = 4π ( 1.5 ) ( 0.725 ×1.225) ( 90 ) ( + 0.0066 ) ( 2π× 20 ) b ⇔ b = 0.0126 The blade section absolute incidence: tan φ = V (1+ a) 90 ( + 0.0066 ) = = 0.487 Ωr ( − b ) ( 2π× 20 ) ( 1.5 ) ( − 0.0126 ) ⇔ φ = 25,95o ⇒ α = θ − φ = 280 − 25,95o = 2o 3' 10.6 At 1.25-m radius on a three-bladed airscrew, the airfoil section has the following characteristics: Solidity = 0.1 ; θ = 29o ' ; α =4o ' ; C L = 0.49 ; L/D=50 Allowing for both axial and rotational interference, find the local efficiency of the element (Answer: 0.885) Solution D = ⇒ γ =1.15o L 50 φ = θ − α = 29o '− 4o ' = 25o tan γ = Geometric pitch = 2πrtanθ = 2π ( 1.25 ) tan ( 29 07 ' ) = 4.37m q = C L sin ( φ + γ ) = 0.49sin ( 25o + 1.15o ) = 0.216 t = C L cos ( φ + γ ) = 0.49cos ( 25o + 1.15o ) = 0.440 Phan Hoàng Đức 1610786 Hệ Thống Lực Đẩy Máy Bay ( 0.1) ( 0.216 ) = 0.0141 b σq = = − b 2sin ( 2φ ) 2sin ( × 25o ) ( 0.1) ( 0.440 ) = 0.0187 a σt = = + a 4sin ( 2φ ) 4sin ( × 25o ) V V 1− b − 0.0187 = = tan φ = tan 25o = 0.451 Ωr 2πnr + a + 0.0141 The local efficiency of the element: ηl = V t 0.440 = 0.451× = 91.87% Ωr q 0.216 10.7 The thrust and torque gradings at 1.22-m radius on each blade of a twobladed airscrew are 2120 N/m and 778 Nm/m, respectively Find the speed of rotation (in rads −1 ) 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 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: dT = 4πρV a ( + a ) dr ⇔ 2800 × = 4π ( 1.225 ) ( 1252 ) a ( + a ) ⇒ a = 0.036 Neglect rotational interference: tan φ = V (1+ a) V (1+ a) ( 125 ) ( + 0.036 ) = 0.824 = = Ωr ( − b ) 2πnr ( − b ) 2π ( 20 ) ( 1.25 ) ( − ) φ = 39.5o Phan Hoàng Đức 1610786 Hệ Thống Lực Đẩy Máy Bay Neglect sectional drag: tan γ = D = ⇒ γ = 0o L We have: 1   VR = V ( + a ) sin φ = 125 ( + 0.036 ) sin 39.5o = 203.6 ( m / s ) ( )   t = C cos φ + γ = 0.6 cos 39.5o = 0.463 ( ) ( ) L  The blade chord at 1.25-m radius: dT = c ρVR2 t dr ⇔ 2800 = c ( 0.5 ) ( 1.225 ) ( 203.62 ) 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) 10.10 Using simple momentum theory, develop an expression for the thrust of a 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 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) Phan Hoàng Đức 1610786 ... − V ) =  10  11   11  ⇔  V ÷ V − V ÷ = VS ''2 −  V ÷    9  ⇔ VS '' = 16 1 V ≈ 1. 41V (1) the thrust of the second airscrew as a percentage of the thrust of the first: 11 1. 41 − T2 VS... 1. 15o ) = 0. 216 t = C L cos ( φ + γ ) = 0.49cos ( 25o + 1. 15o ) = 0.440 Phan Hoàng Đức 16 10786 H? ?? Thống Lực Đẩy Máy Bay ( 0 .1) ( 0. 216 ) = 0. 014 1 b σq = = − b 2sin ( 2φ ) 2sin ( × 25o ) ( 0 .1) ... = = = 11 T1 VS − V VS − V ? ?1 (2) the efficiency of the second Then, if the rate of mass flow through the discs is m&, Thrust of rear airscrew: 11 &( VS ''− V '' ) = m &( VS ''− VS ) = m &? ?1. 41V −