HW3_Solution - 3.12 v f = specific volume of liquid phase v...

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Unformatted text preview: 3.12 v f = specific volume of liquid phase v g =specific volume of vapor phase T sat = saturation temperature p = vapor pressure T = temperature v = specific volume (a) water, = 0.5 m 3 /kg, p = 3 bar, Find T in C Table A-3 v f = 0.0010732 m 3 /kg; v g = 0.6058 m 3 /kg Since f < = 0.5 m 3 /kg < g (1) The state is in the two-phase, liquid-vapor region. Thus, T = T sat (3 bar) = 133.6 C (2) (b) Ammonia, p = 11 lbf/in 2 , T= - 20 F. Find in ft 3 /lb Table A-14E T sat (10 lbf/in 2 ) = -41.33 F, T sat (12 lbf/in 2 ) = -35.14 F -41.33 < T sat (11 lbf/in 2 ) < -35.14 F T = -20 F > T sat (11 lbf/in 2 ) Therefore it is in the region of superheated vapor (Table A-15E) (1) v (10 lbf/in 2 ) = 27.241 ft 3 /lb v (12 lbf/in 2 ) = 22.621 ft 3 /lb Interpolate v (11 lbf/in 2 ) = 24.931 ft 3 /lb (2) (c) Propane, p = 1 MPa, T = 85 C, Find v in m 3 /kg Table A-17 T sat (1 MPa = 10 bar) = 26.95 C Therefore it is in the region of superheated vapor. (1) Table A-18 v (80 C) = 0.05992 m 3 /kg v (90 C) = 0.06226 m 3 /kg Interpolate v (85 C) = 0.06109 m 3 /kg (2) 3.42 Water is the substance v f = specific volume of liquid phase v g =specific volume of vapor phase T sat = saturation temperature p = vapor pressure T = temperature v = specific volume u = specific internal energy x = mass fraction of vapor phase T c = critical temperature P c = critical pressure h = enthalpy (a) p = 3 bar, T = 240 C, find v in m 3 /kg and u in kJ/kg T = 240 C > T sat (3 bar) = 133.6 C (Table A-3) Therefore it is in the region of superheated vapor. (1) Table A-4 v = 0.7805 m3/kg, u = 2713 kJ/kg (2) (b) p = 3 bar, v = 0.5 m 3 /kg, find T and u v f = 0.0010732 m 3 /kg; v g = 0.6058 m 3 /kg Since f < = 0.5 m 3 /kg < g (Table A-3) It is in the two phase region (1) Mass fraction of vapor phase x = (v-v f )/(v g-v f ) = (0.5-0.0010732 m 3 /kg)/(0.6058-0.0010732 m 3 /kg)...
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HW3_Solution - 3.12 v f = specific volume of liquid phase v...

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