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# solution1 - 650:460 Aerodynamics Homework Assignment No 1...

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Unformatted text preview: 650:460 Aerodynamics Homework Assignment No. 1 Prof. Doyle Knight Tel: 732 445 4464, Email: [email protected] Office hours: Tues and Thur, 4:30 pm - 6:00 pm and by appointment Fall 2009 1 Problem 1.6 Nitrogen is often used in wind tunnels as the test gas substitute for air. Compare the value of the kinematic viscosity ν = μ/ρ for nitrogen at a temperature of 350 deg F and 150 psia with that for air at the same conditions. Solution Converting to consistent English units, p = 2 . 16 · 10 4 lbf/ft 2 and T = 809 . 67 deg R. The gas constant for air is R = 1716 ft 2 /s 2 · deg R. The density of air is obtained from the ideal gas equation ρ = p RT = 2 . 16 · 10 4 1716 · 809 . 67 = 1 . 5546 · 10- 2 slug/ft 3 Note that lbf = slug · ft/s 2 2 Problem 1.6 The dynamic molecular viscosity for air is μ = 2 . 27 · 10- 8 T 1 . 5 T + 198 . 6 = 2 . 27 · 10- 8 (809 . 67) 1 . 5 (809 . 67 + 198 . 6) = 5 . 187 · 10- 7 lbf · s/ft 2 Thus, the kinematic viscosity for air is ν = 5 . 187 · 10- 7 1 . 5546 · 10- 2 = 3 . 337 · 10- 5 ft 2 /s 3 Problem 1.6 The gas constant for nitrogen is R = 1774 ft 2 /s 2 · deg R. The density of nitrogen is obtained from the ideal gas equation ρ = p RT = 2 . 16 · 10 4 1774 · 809 . 67 = 1 . 504 · 10- 2 slug/ft 3 The dynamic molecular viscosity for nitrogen is μ = 2 . 16 · 10- 8 T 1 . 5 T + 183 . 6 = 2 . 16 · 10- 8 (809 . 67) 1 . 5 (809 . 67 + 183 . 6) = 5 . 010 · 10- 7 lbf · s/ft 2 Thus, the kinematic viscosity for nitrogen is ν = 5 . 010 · 10- 7 1 . 504 · 10- 2 = 3 . 331 · 10- 5 ft 2 /s 4 Problem 1.8 A perfect gas undergoes a process whereby the pressure is doubled and its density is decreased to three-quarters of its original value. If the initial temperature is 200 deg F, what is the final temperature in deg F ? in deg C ? Solution Converting the temperature to absolute units, the initial temperature is T 1 = 659 . 67 deg R. From the perfect gas equation p 1 = ρ 1 RT 1 and p 2 = ρ 2 RT 2 Thus, p 2 p 1 = ρ 2 T 2 ρ 1 T 1 and T 2 T 1 = p 2 p 1 ρ 1 ρ 2 = 2 . 75 = 2 . 667 Thus, T 2 = (2 . 667)(659 . 67) = 1759 deg R and T 2 = 1299 deg F=704.1 deg C 5 Problem 1.24 A U-tube manometer is used to measure the pressure at the...
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solution1 - 650:460 Aerodynamics Homework Assignment No 1...

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