MECH 213 - 9.11.07

# MECH 213 - 9.11.07 - p R = 0.9MPa/4.059MPa = 0.2217 • T R...

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MECH 213 – September 11, 2007 Method of Generalized Compressibility o pv = RT v ideal = RT/p v actual z = v actual /v ideal (measured) compressibility factor o iff you know z(p,T) pv = zRT v actual = zRT/p o T R ≡ T/T critical p R = p/p critical z = f(T R p R ) o As z approaches 1, the gas becomes more ideal o PAGE 141 shows this diagram  Substance o Look up critical Temperature and pressure (Table A-1) o Given (T, p) Use: T R ≡ T/T critical ; pR = p/p critical Use table on page 141 o Given (T,v) find p Given (p,v) find T Psudo-reduced volume (v R ) v R = v actual /(RT critical /p critical ) Problem 90; Chapter 3 o R-134a Table A-1 Find R 0.08149 kJ/kg-K Find T critical 374.2 K Find p critical 4.059 MPa pv = RT v ideal = [(0.08149)(70 + 273)] / (0.9*1000) = m 3 /kg 0.0311 m 3 /kg pv = zRT

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Unformatted text preview: p R = 0.9MPa/4.059MPa = 0.2217 • T R = (70 + 273)/(374.2)K = 0.9166 • z = 0.9 o z = v actual /v ideal therefore v actual = 0.9*v ideal o v actual = 0.028 m 3 /kg VALUE USING TABLES: 0.027413 • %Error (a) = (0.0311 – 0.027413)/(0.027413) = 13.5% • %Error (b) = (0.028 – 0.027413)/(0.027413) = 2.15% 1.0 z 1.0 p R At low p, all gases are ideal At high T, all gases are approximately ideal At high enough p ideal Z T R p R Problem 119; Chapter 3 o One lbm water; volume = 2.649 ft 3 ; Temp = 400F; cooled to 100F o Constant pressure process o Water real pure substance o T 1 = 400F v 1 = 2649 ft 3 /lbm v f = 0.0186 v g = 1.864 o T 2 = 100F p 2 = p 1 Superheated table p v...
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MECH 213 - 9.11.07 - p R = 0.9MPa/4.059MPa = 0.2217 • T R...

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