Solution Manual for Introduction to Flight 8th Edition by Anderson

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SOLUTIONS MANUAL TO ACCOMPANY

INTRODUCTION TO FLIGHT

8^th Edition

By

John D. Anderson, Jr.

Chapter 2

2.1     

2.2      Mean kinetic energy of each atom

            One kg-mole, which has a mass of 4 kg, has 6.02 × 10²⁶ atoms. Hence 1 kg has  atoms.

           

2.3     

           

2.4     

            Since the volume of the room is the same, we can simply compare densities between the two problems.

           

           

2.5      First, calculate the density from the known mass and volume,

            In consistent units,  Also, T = 70 F = 70 + 460 = 530 R.

            Hence,

                    

                    

            or     

2.6     

           

            Differentiating with respect to time,

           

            or,    

            or,                                                                                                                         (1)

            At the instant there is 1000 lb_m of air in the tank, the density is

                    

                    

            Also, in consistent units, is given that

                     T = 50 + 460 = 510 R

            and that

                    

            From the given pumping rate, and the fact that the volume of the tank is 900 ft³, we also have

                    

                     

            Thus, from equation (1) above,

           

2.7      In consistent units,

                    

            Thus,

                    

2.8     

2.9     

           

           

            Magnitude of the resultant aerodynamic force =

           

2.10  

            Minimum velocity occurs when sin q = 0, i.e., when q = 0° and 180°.

                      V_min = 0 at q = 0° and 180°, i.e., at its most forward and rearward points.

            Maximum velocity occurs when sin q = 1, i.e., when q = 90°. Hence,

                    

            i.e., the entire rim of the sphere in a plane perpendicular to the freestream direction.

2.11   The mass of air displaced is

                    

            The weight of this air is

                    

            This is the lifting force on the balloon due to the outside air. However, the helium inside the balloon has weight, acting in the downward direction. The weight of the helium is less than that of air by the ratio of the molecular weights

                    

            Hence, the maximum weight that can be lifted by the balloon is

                     0.168 - 0.0233 = 0.145 lb.

2.12   Let p₃, r₃, and T₃ denote the conditions at the beginning of combustion, and p₄, r₄, and T₄ denote conditions at the end of combustion. Since the volume is constant, and the mass of the gas is constant, then p₄ = r₃ = 11.3 kg/m³. Thus, from the equation of state,

                    

            or,

                    

2.13   The area of the piston face, where the diameter is 9 cm = 0.09 m, is

                    

            (a)     The pressure of the gas mixture at the beginning of combustion is

                    

                     The force on the piston is

                    

                     Since 4.45 N = l lbf,

                    

            (b)    

                     The force on the piston is

                    

                    

2.14   Let p₃ and T₃ denote conditions at the inlet to the combustor, and T₄ denote the temperature at the exit. Note:

            (a)    

            (b)    

2.15   1 mile = 5280 ft, and 1 hour = 3600 sec.

            So:

                               

            A very useful conversion to remember is that

                                   

            also,                            1 ft = 0.3048 m

                                   

            Thus                           

2.16  

           

2.17   On the front face

                                   

            On the back face

                                   

            The net force on the plate is

                          

            From Appendix C,

                                   

            So,

                                   

            This force acts in the same direction as the flow (i.e., it is aerodynamic drag.)

2.18 

            In SI units:

                                   

            In terms of kilogram force,

                                   

2.19  

           

2.20  

2.21   From Fig. 2.16,

            length of fuselage = 33 ft, 4.125 inches = 33.34 ft

                                   

            wing span = 40 ft, 11.726 inches = 40.98 ft

                                   

2.22   (a)     From App. C 1 ft. = 0.3048 m.

                     Thus,

                                354,200 ft = (354,000)(0.3048) = 107,960 m = 107.96 km

            (b)     From Example 2.6: 60 mph = 26.82 m/sec

                     Thus,

                                   

2.23

                                   

                     From Newton’s 2^nd Law

                                               

                                   

2.24

                                   

2.25   From Appendix C, one pound of force equals 4.448 N. Thus, the thrust of the Rolls-Royce Trent engine in pounds is

                                   

2.26

         (a)    

         (b)