Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 15, Problem 93RQ

(a)

To determine

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

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Total mass is 150000 lbm.

Wing area is 1800ft2.

Cruising speed is 550mi/h.

Altitude is 38000 ft.

Density of air is 0.0208lbm/ft3.

Density of air on the ground is 0.075lbm/ft3.

Calculation:

Refer figure 15-45 and obtain the maximum lift coefficient with and without flaps as follows:

  CLmax,1=3.48CLmax,2=1.52

During takeoff of the aircraft,

  W=FLW=12CLρAV2V=2WCLρA

Calculate the stall speed with flaps.

  Vmin,1=2WCLmax,1ρA=2(150000 lbf)3.48(0.075lbm/ft3)(1800 ft2)=143ft/s

Calculate the stall speed without flaps.

  Vmin,2=2WCLmax,2ρA=2(150000 lbf)1.52(0.075lbm/ft3)(1800 ft2)=217ft/s

Calculate the safe speed with flaps.

  V1=1.2Vmin,1=1.2(143ft/s)(1 mi/h1.4667ft/s)=117mph

Thus, the safe speed with flaps is 117mph.

Calculate the safe speed without flaps.

  V2=1.2Vmin,2=1.2(217ft/s)(1 mi/h1.4667ft/s)=178mph

Thus, the safe speed without flaps is 178mph.

(b)

To determine

The angle of attack.

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The lift coefficient is,

  CL=FL12ρAV2=150000 lbf12(0.0208lbm/ft3)(1800 ft2)(550×1.4667 ft/s)2(32.2lbmft/s21lbf)=0.40

From Figure 15-45, the angle of attack corresponding to the lift coefficient is 3.5°.

Thus, the angle of attack is 3.5°.

(c)

To determine

The power required to overcome drag.

(c)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

From Figure 15-45, the drag coefficient corresponding to the lift coefficient is 0.015.

The drag force is,

  FD=CDρAV22=0.015(0.0208lbm/ft3)(1800 ft2)(550×1.4667 ft/s)22(1 lbf32.2lbmft/s2)=5675 lbf

The power required to overcome the drag is,

  W˙drag=FDV=(5675lbf)(550mi/h×5280ft1mi×1h3600s)(1 kW737.56 lbfft/s)=6200kW

Thus, the power required to overcome the drag is 6200 kW.

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Chapter 15 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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