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Solution Manual For Introduction to Flight, 8th Edition by Anderson
- Exam (elaborations) • 9 pages • 2021
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Chapter 2
2.1 ρ= p RT/ = (1.2)(1.01 10 )/(287)(300)× 5 ρ=1.41kg/m2
v=1/ρ=1/1.41= 0.71m /kg3
3 kT= 3 (1.38× 10−23)(500)=1.035× 10−20J
2.2 Mean kinetic energy of each atom =
2 2
One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has (6.02 ×1026)=1.505 ×1026 atoms.
Totalinternalenergy = (energyperatom)(numberof atoms)
= (1.035´10-20)(1.505´1026) = 1.558 ´106J
RTp (1716)( + 0.00237 slug3
2.3 ρ= = =
59) ft
Volumeof theroom = (20)(15)(8...
Exam (elaborations)
Solution Manual For Introduction to Flight, 8th Edition by Anderson
Last document update:
ago
Chapter 2 2.1 ρ= p RT/ = (1.2)(1.01 10 )/(287)(300)× 5 ρ=1.41kg/m2 v=1/ρ=1/1.41= 0.71m /kg3 3 kT= 3 (1.38× 10−23)(500)=1.035× 10−20J 2.2 Mean kinetic energy of each atom = 2 2 One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has (6.02 ×1026)=1.505 ×1026 atoms. Totalinternalenergy = (energyperatom)(numberof atoms) = (1.035´10-20)(1.505´1026) = 1.558 ´106J RTp (1716)( + 0.00237 slug3 2.3 ρ= = = 59) ft Volumeof theroom = (20)(15)(8...
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