CHAPTER3_SECTION2 - Chapter 3 Properties of Pure Substances...

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Chapter 3 Properties of Pure Substances 3-59 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables, and also by using the saturated liquid approximation, and the results are to be compared. Analysis Compressed liquid can be approximated as saturated liquid at the given temperature. Then from Table A-4, T = 100°C v v f C f C f C m kg error u u kJ kg error h h kJ kg error @ @ @ . / ( . ) . / ( . ) . / ( . ) 100 3 100 100 0001044 076% 41894 101% 41904 2 61% From compressed liquid table (Table A-7), kJ/kg 430.28 kJ/kg 414.74 /kg m 0.0010361 C 100 MPa 15 3 h u v T P The percent errors involved in the saturated liquid approximation are listed above in parentheses. 3-60 Problem 3-59 is reconsidered. Using EES, the indicated properties of compressed liquid are to be determined, and they are to be compared to those obtained using the saturated liquid approximation. T = 100 "[C]" P = 15000 "[kPa]" v = VOLUME(Steam,T=T,P=P) "[m^3/kg]" u = INTENERGY(Steam,T=T,P=P) "[kJ/kg]" h = ENTHALPY(Steam,T=T,P=P) "[kJ/kg]" v_app = VOLUME(Steam,T=T,x=0) "[m^3/kg]" u_app = INTENERGY(Steam,T=T,x=0) "[kJ/kg]" h_app_1 = ENTHALPY(Steam,T=T,x=0) "[kJ/kg]" h_app_2 = ENTHALPY(Steam,T=T,x=0)+v_app*(P-pressure(Steam,T=T,x=0)) "[kJ/kg]" SOLUTION Variables in Main h=430.3 [kJ/kg] h_app_1=419.1 [kJ/kg] h_app_2=434.6 [kJ/kg] P=15000 [kPa] T=100 [C] u=414.7 [kJ/kg] u_app=419 [kJ/kg] v=0.001036 [m^3/kg] v_app=0.001043 [m^3/kg] 3-15
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Chapter 3 Properties of Pure Substances 3-61E A rigid tank contains saturated liquid-vapor mixture of R-134a. The quality and total mass of the refrigerant are to be determined. Analysis At 30 psia, v f = 0.01209 ft 3 /lbm and v g = 1.5408 ft 3 /lbm. The volume occupied by the liquid and the vapor phases are V V f g 1.5 ft and 13.5 ft 3 3 Thus the mass of each phase is m V m V f f f g g g v v 1.5 ft 0.01209 ft / lbm 124.1lbm 13.5 ft 1.5408 ft / lbm 8.76 lbm 3 3 3 3 Then the total mass and the quality of the refrigerant are m m m x m m t f g g t 124.1 8.76 8.76 132.86 132.86 lbm . 0 0659 3-16 R-134a 15 ft 3 30 psia
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Chapter 3 Properties of Pure Substances 3-62 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final temperature and the volume change are to be determined, and the process should be shown on a T-v diagram. Analysis (b) At the final state the cylinder contains saturated liquid-vapor mixture, and thus the final temperature must be the saturation temperature at the final pressure, C 179.91 MPa [email protected] T T (c) The quality at the final state is specified to be x 2 = 0.5. The specific volumes at the initial and the final states are /kg m 0.2579 C 300 MPa 1.0 3 1 1 1 v T P /kg m .0978 0 ) 001127 . 0 19444 . 0 ( 5 . 0 001127 . 0 5 . 0 MPa 1.0 3 2 2 2 2 fg f v x v v x P Thus, V m kg m kg  ( ) ( . )( . . ) / v v 2 1 3 08 0 0978 0 2579 0.128m 3 3-63 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be determined.
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.

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CHAPTER3_SECTION2 - Chapter 3 Properties of Pure Substances...

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