Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 11, Problem 101EP

A commercial airplane has a total mass of 150.000 Ibm and a wing planform area of 1700 ft2. The plane has a cruising speed of 625 mi1 and a cruising altitude of 38.000 ft where the air density is 0.0208 Ibm/fl3. The plane has double-slotted flaps for use during takeoff and landing. but it cruises with all flaps retracted. Assuming the lift and drag characteristics of the wings can be approximated by NACA 23012, detennine (a) the minimum safe speed for takeoff and landing with and without extending the flaps. (b) the anzle of attack to cruise steadily at the cruising altitude, and (c) the power that needs to be supplied to provide enough thrust to overcome drag. Take the air density on the ground to be 0.075 Ibm/ft3.

Expert Solution
Check Mark
To determine

(a)

The minimum safe speed for takeoff and landing with the flaps.

The minimum safe speed for takeoff and landing without the flaps.

Answer to Problem 101EP

The minimum safe speed for takeoff and landing with the flaps is 267.91ft/s.

The minimum safe speed for takeoff and landing without the flaps is 177.06ft/s.

Explanation of Solution

Given information:

The total mass of the airplane is 150000lbm, the wing platform area is 1700ft2, the cursing speed of the plane is 625mi/h, the cursing altitude is 38000ft, the density of the air at the altitude of 38000ft is 0.0208lbm/ft3 and the density of air at ground is 0.075lbm/ft3.

Write the expression for the initial velocity of the aircraft.

  V1=( 2W ρ a C L1 A)........ (I)

Here, the weight of the aircraft is W, the density of the air is ρa, the frontal area is A, and the lift coefficient at initial state is CL1.

Write the expression for the final velocity of the aircraft.

  V2=( 2W ρ a C L2 A)........ (II)

Here, the lift coefficient at final state is CL2.

Write the expression for the weight of the aircraft.

  W=mg........ (III)

Here, the mass of the airplane is m and the acceleration due to gravity is g.

Write the expression for the safe velocity with flaps.

  Vs1=1.2V1........ (IV)

Write the expression for the safe velocity without flaps.

  Vs2=1.2V2........ (V)

Calculation:

Substitute 150000lbm for m and 32.2ft/s2 for g in Equation (III).

  W=150000lbm(32.2ft/ s 2)=4830000lbmft/s2

Refer to Figure 15-45 titled "Effect of flaps on the lift and drag coefficient of an airfoil" to obtain maximum lift coefficient with flaps as 1.52 and without flaps as 3.48.

Substitute 1.52 for CL1, 4830000lbmft/s2 for W, 0.075lbm/ft3 for ρa and 1700ft2 for A in Equation (I).

  V1=( 2( 4830000 lbmft/ s 2 ) 1.52( 0.075 lbm/ ft 3 )( 1700 ft 2 ) )=( ( 9660000 ft 2 / s 2 ) ( 193.8 ) )=( 49845.20 ft 2 / s 2 )=223.26ft/s

Substitute 3.48 for CL2, 4830000lbmft/s2 for W, 0.075lbm/ft3 for ρa and 1700ft2 for A in Equation (II).

  V2=( 2( 4830000 lbmft/ s 2 ) 3.48( 0.075 lbm/ ft 3 )( 1700 ft 2 ) )=( ( 9660000 ft 2 / s 2 ) ( 443.7 ) )=( 21771.46 ft 2 / s 2 )=147.55ft/s

Substitute 223.26ft/s for V1 in Equation (IV).

  Vs1=1.2(223.26ft/s)=267.91ft/s

Substitute 147.55ft/s for V2 in Equation (V).

  Vs2=1.2(147.55ft/s)=177.06ft/s

Conclusion:

The minimum safe speed for takeoff and landing with the flaps is 267.91ft/s.

The minimum safe speed for takeoff and landing without the flaps is 177.06ft/s.

Expert Solution
Check Mark
To determine

(b)

The angle of attack to cruise steadily at the cruising altitude.

Answer to Problem 101EP

The angle of attack to cruise steadily at the cruising altitude is α=3.3°.

Explanation of Solution

Write the expression for the coefficient of the lift force.

  CL=FL12ρaV2A........ (VI)

Here, the lift force is FL, the velocity of the airplane is V, and the density of air at the given altitude is ρ

Calculation:

The lift force is equals to weight of the airplane that is FL=4830000lbmft/s2.

Substitute 4830000lbmft/s2 for FL, 0.0208lbm/ft3 for ρ, 1700ft2 for A and 625mi/h for V in Equation (VI).

  CL=4830000lbmft/ s 212( 0.0208 lbm/ ft 3 ) ( 625 mi/h )2( 1700 ft 2 )=4830000lbmft/ s 20.5( 0.0208 lbm/ ft 3 ) ( 625 mi/h ( 1.466 ft/s 1 mi/h ) )2( 1700 ft 2 )=483000014842608.63=0.325

Refer to Figure 15-45 "Effect of flaps on the lift and drag coefficients of an airfoil" to obtain the angle of attack of a clean flap as α=3.3°.

Conclusion:

The angle of attack to cruise steadily at the cruising altitude is α=3.3°.

Expert Solution
Check Mark
To determine

(c)

The power to be supplied to present enough thrust to overcome the drag.

Answer to Problem 101EP

The power to be supplied the enough thrust to overcome the drag is 8589.38kW.

Explanation of Solution

Write the expression for the drag force.

  FD=CD12ρaV2A........ (VII)

Here, the drag force coefficient is FD.

Write the expression for the drag force and the velocity of the airplane.

  P=FD×V........ (VIII)

Calculation:

Refer to Figure 15-45 "Effect of flaps on the lift and drag coefficients of an airfoil" to obtain the drag coefficient as 0.015 at the lift coefficient CD=0.325.

Substitute 0.015 for CD, 0.0208lbm/ft3 for ρ, 1700ft2 for A and 625mi/h for V in Equation (VII).

  FD=(0.015)12(0.0208lbm/ ft 3)(625 mi/h)2(1700 ft2)=(0.0075)(0.0208lbm/ ft 3)(625 mi/h( 1.466 ft/s 1 mi/h ))2(1700 ft2)=222639.12lbmft/s2( 1lbf 32.2 lbmft/ s 2 )=6914.25lbf

Substitute 6914.25lbf for FD and 625mi/h for V in Equation (VIII).

  p=(6914.25lbf)(625mi/h)=(6914.25lbf)(625mi/h( 1.466 ft/s 1 mi/h ))=6335189.51lbfft/s( 1kW 737.56 lbfft/s )=8589.38kW

Conclusion:

The power to be supplied the enough thrust to overcome the drag is 8589.38kW.

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Chapter 11 Solutions

Fluid Mechanics: Fundamentals and Applications

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