SOLUTIONS MANUAL TO ACCOMPANY
INTRODUCTION TO FLIGHT
8th Edition
By
John D. Anderson, Jr.
Chapter 2
2.1
= p/RT = (1.2)(1.01105 )/(287)(300)
= 1.41 kg/m2
v = 1/ = 1/1.41= 0.71 m3/kg
...
SOLUTIONS MANUAL TO ACCOMPANY
INTRODUCTION TO FLIGHT
8th Edition
By
John D. Anderson, Jr.
Chapter 2
2.1
= p/RT = (1.2)(1.01105 )/(287)(300)
= 1.41 kg/m2
v = 1/ = 1/1.41= 0.71 m3/kg
2.2 Mean kinetic energy of each atom
= 3 3
23
20
k T =
2 2
(1.38 10 ) (500) = 1.035 10 J
One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has
1 (6.02 1026 ) = 1.505 1026
4
atoms.
Total internal energy = (energy per atom)(number of atoms)
= (1.035´ 10- 20)(1.505 ´ 1026) = 1.558 ´ 106 J
2.3
= p =
2116
= 0.00237 slug
RT (1716)(460 + 59) ft3
Volume of the room = (20)(15)(8) = 2400 ft3
Total mass in the room = (2400)(0.00237) = 5.688slug Weight = (5.688)(32.2) = 183lb
2.4
= p RT
= 2116
(1716)(460 -
= 0.00274 slug
10) ft3
Since the volume of the room is the same, we can simply compare densities between the two problems.
= 0.00274 -
% change =
0.00237 = 0.00037 slug
ft3
= 0.00037 ´ (100) = 15.6% increase
0.00237
2.5 First, calculate the density from the known mass and volume,
= 1500/ 900 = 1.67 lbm /ft3
In consistent units,
= 1.67/32.2 = 0.052 slug/ft3.
Also, T = 70 F = 70 + 460 = 530 R.
Hence,
p = RT
= (0.52)(1716)(530)
p = 47, 290 lb/ft2
or
p = 47, 290 / 2116 = 22.3 atm
2.6
p = RT
l np = l np + l nR + l nT
Differentiating with respect to time,
1 dp = 1 d + 1 dT
p dt dt T dt
or,
dp = p d + p dT
dt dt T dt
or,
dp = RT d + R dT dt dt dt
(1)
At the instant there is 1000 lbm of air in the tank, the density is
= 1000 / 900 = 1.11lb /ft3
= 1.11/32.2 = 0.0345slug/ft3
Also, in consistent units, is given that
T = 50 + 460 = 510 R
and that
dT = 1F/min = 1R/min = 0.016R/sec dt
From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we
............................................Continued..........................................
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