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Example 1 A closed, rigid container of volume 0.5 m3 is placed on a hot plate. Initially the container holds a two-phase mixture of saturated liquid water and saturated water vapor at P1= 1 bar with a quality of 0.5. After heating, the pressure in the container is P2=1.5 bar. Indicate the initial and final states on a T-v diagram, and determine: a) the temperature, in °C, at each state. b) the mass of vapor present at each state, in kg. c) if heating continues, determine the pressure, in bar, when the container holds only saturated vapor. Solution: Assumptions: 1. Water in the container is a closed system. 2. States 1, 2, and 3 are equilibrium states. 3. The volume of container remains constant. Two independent properties are required to fix state 1 and 2. At the initial state, the pressure and quality are known. Thus state 1 is known, as mentioned in the problem. The specific volume at state 1 is found using the given quality:

v1  v f 1  x1 v g1  v f 1 

From Table A - 5 at P  1 bar  100 kPa v1  0.001043  0.5 (1.694  0.001043)  0.8475 m 3 / kg At state 2, the pressure is known. Volume and mass remain constant during the heating process within the container, so v2=v1. For P2= 0.15 MPa, Table A-5 gives vf2= 0.001053 and vg2=1.1593 m3/kg. Since vf2 < v2 < vg2 State 2 must be in the two-phase region as well. Since state 1 and 2 are in the two-phase liquidvapor region, the temperatures correspond to the saturation temperatures for the given. Table A5: T1 = 99.63 °C and T2 = 111.4 °C To find the mass of water vapor present, we first find the total mass, m.

V 0.5m 3   0.59kg v 0.8475m 3 / kg m g1  x1 m  0.50.59kg   0.295kg

m

T

3

P2 = 1.5 bar

2 P1 = 1 bar 1

v

The mass of vapor at state 2 is found similarly using quality x2. From Table A-5, for P2 = 1.5 bar, we have:

x2  x2  mg 2

v vf2 vg 2  v f 2 0.8475  0.001053  0.731 1.159  0.001053  0.731 0.59kg   0.431 kg

Example2 Determine the specific volume of R-134a at 1 MPa and 50°C, using (a) ideal gas equation (b) the generalized compressibility chart. Compare the values obtained with the actual value of 0.02171 m3/kg. Solution: From Table A-1, for R-134a, R = 0.0815 kPa.m3/(kg.K), Pcr = 4.067 MPa, and Tcr = 374.3 K (a) Ideal gas equation of state

v





0.0815 kPa.m 3 / kg.K  323 K  RT   0.02632 m 3 / kg 1000 kPa P

Comparing with the tabulated value, using ideal gas equation one would get an error of (0.026320.02171)/0.02171=0.212 or 21.2%. (b) To determine the correction factor Z,

PR 

P 1MPa   0.246 Pcr 4.067 MPa

TR 

T 323K   0.863 Tcr 374.3K

From Fig. A-28, Z= 0.84. Thus, v = Z videal = 0.84 (0.02632 m3/kg) =0.02211 m3/kg

Example 3 Determine the enthalpy of 1.5 kg of water contained in a volume of 1.2 m3 at 200 kPa. Recall we need two independent, intensive properties to specify the state of a simple substance. Pressure P is one intensive property and specific volume is another. Therefore, we calculate the specific volume.

v

Volume 12 . m3 m3   0.8 mass 15 . kg kg

Using Table A-5 at P = 200 kPa, vf = 0.001061 m3/kg,

vg = 0.8858 m3/kg

Is v  v f ? No Is v f  v  v g ? Yes Is v g  v ? No We see that the state is in the two-phase or saturation region. So we must find the quality x first.

v  v f  x (v g  v f ) 𝑥=

𝑣 − 𝑣𝑓 𝑣𝑔 − 𝑣𝑓

0.8 − .0001061 0.8858 − 0.001061 =0.903 Then,

h  h f  x h fg  504.7  (0.903)(2201.6)  2492.7

kJ kg

Example 4 Determine the internal energy of refrigerant-134a at a temperature of 0C and a quality of 60%. Using Table A-11, for T = 0C, uf = 51.63 kJ/kg

ug =230.16 kJ/kg

u  u f  x (u g  u f )  51.63  (0.6)(230.16  51.63)  158.75

kJ kg

Example 5 Calculate the specific volume of nitrogen at 300 K and 8.0 MPa and compare the result with the value given in a nitrogen table as v = 0.011133 m3/kg. From Table A.1 for nitrogen Tcr = 126.2 K, Pcr = 3.39 MPa R = 0.2968 kJ/kg-K

T 300 K   2.38 Tcr 126.2 K P 8.0 MPa PR    2.36 Pcr 3.39 MPa

TR 

Since T > 2Tcr and P < 10Pcr, we use the ideal gas equation of state

Pv  RT v

RT  P

0.2968

m3  0.01113 kg

kJ (300 K ) 3 m MPa kg  K 8.0 MPa 103 kJ

Example 6 Determine the internal energy of refrigerant – 134a at a temperature of 0°C and a quality of 60%. From table A-5: Uf = 51.63 kJ/kg Ug = 230.16 kJ/kg The internal energy of R 134a at a given condition: U = Uf + x(Ug – Uf) = 51.63 + (0.6)(230.16 – 51.63) = 158.75 kJ/kg Example 7 An ideal gas is contained in a closed assembly with an initial pressure and temperature of 220 kPa and 70°C respectively. If the volume of the system is increased 1.5 times and the temperature drops to 15°C, determine the final pressure of the gas. State 1: P1 = 220kPa T1 = 70 + 273K = 343K State 2: T2 = 15 +273 = 288K V2 = 1.5V1 From ideal gas law: 𝑃1𝑉1 𝑇1

P2 =

= 𝑉1

1.5𝑉1

𝑃2𝑉2 𝑇2 288

(343) ÷ (220𝑥103 )

= 123.15 kPa

Example 8 A closed assembly contains 2 kg of air at an initial pressure and temperature of 140 kPa and 210°C respectively. If the volume of the system is doubled and temperature drops to 37°C, determine the final pressure of the air. Air can be modelled as an ideal gas. State 1: P1 = 140 kPa T1 = 210 + 273K = 483K State 2: T2 = 37 + 273 = 310K V2 = 2V1 From ideal gas law:

𝑃1𝑉1 𝑃2𝑉2 = 𝑇1 𝑇2 P2 =

𝑉1 2𝑉1

310

(483) ÷ (140𝑥103 )

= 44.93 kPa Example 9 An automobile tire with a volume of 0.6 m^3 is inflated to a gage pressure of 200kPa calculate the mass of air in the tire if the temperature is 20°C. State 1: P = 200 + 100 kPa T = 20 + 273K = 293K From ideal gas law: 𝑚= =

𝑃𝑉 𝑅𝑇 𝑁 (0.6𝑚2 ) 𝑚3 𝑁𝑚 287 (293𝐾) 𝑘𝑔.𝐾

300𝑥103

= 2.14 kg

Example 10 A rigid tank contains 50kg of saturated liquid water at 90°C. Determine the pressure in tank and the volume of the tank. m = 50kg

T = 90°C

From table A-4, P = 70.183 kPa v= 0.001036 m^3/kg V = mv = (50)(0.001036) = 0.0518 m^3 Example 11 A piston – cylinder device contains 0.06m^3 of saturated water vapor at 350kPa pressure. Determine the temperature and mass of vapor inside the cylinder. P = 350kPa

V = 0.06m^3

From table A-5, T = 138.86°C V = 0.22422 m^3/kg V = mv m = 0.06 / 0.52422 = 0.114 kg