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3-24

3-61 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 @ 100°C = 0.001043 m 3 /kg (0.72%error)
u ≅ u f @ 100°C = 419.06 kJ/kg h ≅ h f @ 100°C = 419.17 kJ/kg (1.02% error) (2.61% error)

From compressed liquid table (Table A-7),

v = 0.001036 m 3 /kg P = 15 MPa  u = 414.85 kJ/kg T = 100°C   h = 430.39 kJ/kg
The percent errors involved in the saturated liquid approximation are listed above in parentheses.

3-62 EES Problem 3-61 is reconsidered. Using EES, the indicated propertiesof compressed liquid are to be determined, and they are to be compared to those obtained using the saturated liquid approximation. Analysis The problem is solved using EES, and the solution is given below.

Fluid$='Steam_IAPWS' T = 100 [C] P = 15000 [kPa] v = VOLUME(Fluid$,T=T,P=P) u = INTENERGY(Fluid$,T=T,P=P) h = ENTHALPY(Fluid$,T=T,P=P) v_app = VOLUME(Fluid$,T=T,x=0) u_app =INTENERGY(Fluid$,T=T,x=0) h_app_1 = ENTHALPY(Fluid$,T=T,x=0) h_app_2 = ENTHALPY(Fluid$,T=T,x=0)+v_app*(P-pressure(Fluid$,T=T,x=0)) SOLUTION Fluid$='Steam_IAPWS' h=430.4 [kJ/kg] h_app_1=419.2 [kJ/kg] h_app_2=434.7 [kJ/kg] P=15000 [kPa] T=100 [C] u=414.9 [kJ/kg] u_app=419.1 [kJ/kg] v=0.001036 [m^3/kg] v_app=0.001043 [m^3/kg]

3-25

3-63E A rigid tank contains saturated liquid-vapor mixture of R-134a. Thequality and total mass of the refrigerant are to be determined. Analysis At 50 psia, vf = 0.01252 ft3/lbm and vg = 0.94791 ft3/lbm (Table A-12E). The volume occupied by the liquid and the vapor phases are

V f = 3 ft 3 and V g = 12 ft 3
Thus the mass of each phase is
mf = mg =

Vf vf Vg vg

= =

3 ft 3 0.01252 ft 3 /lbm 12 ft 3 0.94791 ft 3 /lbm

R-134a 15 ft3 50 psia

= 239.63 lbm =12.66 lbm

Then the total mass and the quality of the refrigerant are mt = mf + mg = 239.63 + 12.66 = 252.29 lbm
x= mg mt = 12.66 lbm = 0.05018 252.29 lbm

3-64 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,
T = Tsat@1 MPa = 179.88°C

(Table A-5)
H2O 300°C 1 MPa T
1

(c) The quality at the final state is specified to be x2 = 0.5. The specific volumes at the initial and the final states are
P1 = 1.0 MPa  3  v = 0.25799 m /kg T1 =300 o C  1

(Table A-6)

P2 = 1.0 MPa x2 = 0.5

  v 2 = v f + x2v fg  = 0.001127 + 0.5 × (0.19436 − 0.001127) = 0.09775 m3/kg

2

Thus,
∆V = m(v 2 − v 1 ) = (0.8 kg)(0.09775 − 0.25799)m 3 /kg = −0.1282 m 3

v

3-26

3-65 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be determined. Analysis This is a constantvolume process (v = V /m = constant), and the initial specific volume is equal to the final specific volume that is

v 1 = v 2 = v g @150°C = 0.39248 m 3 /kg
since the vapor starts condensing at 150°C. Then from Table A-6,
T1 = 250°C   P = 0.60 MPa 3 v1 = 0.39248 m /kg  1

(Table A-4)
H2O T1= 250°C P1 = ?

T °C
250 1

150 2

v

3-66 Water is boiled in a pan by supplying electricalheat. The local atmospheric pressure is to be estimated. Assumptions 75 percent of electricity consumed by the heater is transferred to the water. Analysis The amount of heat transfer to the water during this period is
Q = fEelect time = (0.75)(2 kJ/s)(30 × 60 s) = 2700 kJ

The enthalpy of vaporization is determined from
h fg = Q 2700 kJ = = 2269 kJ/kg m boil 1.19 kg

Using the data by a...
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