p0150 - · 10 4 N/m 2 . There-fore, since the...

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1.50. CHAPTER 1, PROBLEM 50 51 1.50 Chapter 1, Problem 50 Problem: Assuming temperature is 15.6 o C, compute the fractional change in density, ρ / ρ , for air and water caused by a change in pressure, p =0 . 25 atm. The initial pressure is p =1 atm. Solution: As noted in the text, in terms of compressibility, τ , the fractional change in density is given by ρ ρ = τ p The given pressure change is p = 0.25 atm = 25.25 kPa = 2.525
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Unformatted text preview: · 10 4 N/m 2 . There-fore, since the compressibility air is 10 − 5 m 2 /N and the compressibility of water is 4 . 65 · 10 − 10 m 2 /N, we have the following. Air : ∆ ρ ρ = p 10 − 5 m 2 / N Q p 2 . 525 · 10 4 N / m 2 Q = 0 . 2525 Water : ∆ ρ ρ = p 4 . 65 · 10 − 10 m 2 / N Q p 2 . 525 · 10 4 N / m 2 Q = 1 . 17 · 10 − 5...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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