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Interpolation - Using Properties Tables 1 Identify...

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Using Properties Tables 2) Determine known properties (e.g., T & v, T & x, P & v, etc.) (1.5 - 0.001043)/(1.6941 - 0.001043) = 0.8854 = 88.54% Quality [88.54% of the mass P (kPa) u (kJ/kg) h (kJ/kg) s (kJ/kg*K) 99.61 100.000 1.500000 0.001043 1.694100 2266.20 417.40 2505.60 2416.19 417.51 2675.00 6.664600 1.302800 Phase Source Liq./vap. Mix Table A-5, pg. 916 Calculated values 1) Create Table of properties needed P (kPa) u (kJ/kg) h (kJ/kg) s (kJ/kg*K) Phase Source 300.00 1000.000 0.257990 2793.70 3051.60 7.124600 SH Table A-6, pg. 921 325.00 1000.000 0.270245 2834.70 3104.90 7.213750 SH 350.00 1000.000 0.282500 2875.70 3158.20 7.302900 SH Table A-6, pg. 921 2) Determine slope according to variable given (T in this example} (325-300)/(350-300) = 0.50 3) Calculate each variable according to variable = lower value + slope*(upper value - lower value) For example, v = .25799 + .5*(.28250 - .25799) = 0.270245 u = 2793.7 + .5*(2875.7 - 2793.7) = 2834.70 h = 3051.6 + .5*(3158.2 - 3051.6) = 3104.90 s = 7.1246 + .5*(7.3029 - 7.1246) = 7.213750 1) Identify substance (e.g., H 2 O) 3) If T given, use Table A-4, otherwise if P is given, use Table A-5 (for H 2 O) 4) Determine if property is less than sat. liquid variable (e.g., v < v f ); if so, substance is at a compressed liquid state. 5) Determine if property is greater than sat. liquid variable (e.g., v > v g ); if so, substance is at a superheated vapor state. 6) If property is between these states (e.g., v f < v < v g ), then it is a liquid/vapor mixture and given T or P yield P SAT or T SAT . You will need to have x to define state (1) Liquid-vapor example (What are H 2 O properties at 100 kPa and v = 1.5 m 3 /kg?) For P = 100 kPa, v f = 0.001043 m 3 /kg and v g = 1.6941 m 3

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