Complete the following table for H20. Use data from the steam tables. T, °C P, kPa 400 u, kJ/kg 1450 Phase description Superheated vapor v 220 190 2500 4000 Saturated vapor Saturated vapo ' Superheated vapor ' 3040 Complete the following table for Refrigerant-134a. Use data from the steam tables. P. psia 80 h, Btu/lbm 78 Phase description Saturated liquid v Saturated mixture Superheated vapor Saturated mixture Y Compressed liquid ▼ 15 10 0.6 70 180 Click to select) 110

## Answer

Q1 (1) From table "Saturated water: Pressure table", at P = 400kPa check the internal energy. u, = 604.31kJ/kg u = 2553.6 kJ/kg The internal energy given (=1450kJ/kg) is in-between u, and u, hence H,O is in saturated mixture. Obtain the saturation temperature from table "Saturated water: Pressure table", at P=400 kPa. T =143.63°C (2) Obtain the pressure and internal energy at T = 220°C from table “Saturated water temperature table". P= Psa P=2.318 MPa = 2318kPa u=2602.4kJ/kg Since the phase is saturated vapor u=u. (3) Obtain the pressure at T = 190°C from table "Saturated water temperature table". Pa = 1.2544 MPa Given pressure is greater than the saturation pressure therefore H,O is compressed liquid. Obtain the internal energy from "Saturated water temperature table" at T = 190°C. u=u u=806kJ/kg

From table “Saturated water: Pressure table”, at P=4000 kPa check the internal energy. un = 1082.31kJ/kg ug = 2602.3kJ/kg The internal energy given (= 3040 kJ/kg) is greater than u, hence H,O is in super- heated vapor phase. Obtain the temperature and internal energy, at P = 4000 kPa from table super-heated water. 3040-3011 T = 450+ - 3100.3-3017 (500 – 450) T = 466.23°C Thus, the results are tabulated below: T, °C Р. kPa 143.63 400 220 2318 190 2500 466.23 4000 u, kJ/kg 1450 2602.4 806 3040 Phase description satuarted mixture Saturated vapor compressed liquid superheated vapor Q2:

a) Given P = 80 psia h = 78 Btu/lbm Only pressure and enthalpy are given. We do not know which table to use to determine the missing properties because we have no clue as to whether we have saturated mixture, compressed liquid or superheated vapor, which can be determined from the enthalpy of saturated liquid (ht) and enthalpy of saturated vapor (ha) from the saturated refrigerant-134a- pressure table At P= 80 psia h = 33.394 Btu/lbm hig = 78.804 Btu/lbm ha = 112.20 Btu/bm Thus, the given value h = 78 Btu/bm is in between the values of hand h,. Therefore, we have saturated liquid-vapor mixture at pressure of 80 psia. The temperature is the saturation temperature at given pressure. Hence, T = Tsat @ 80 psia = 65.89°F The quality is determined from x= h-h, hogy 78-33.394 78.804 = 0.566 and Saturated mixture

b) Given Temperature, T = 15 °F Quality, x = 0.6 The quality, x = 0.6 implies that 60 percent of the mass is in the vapor state and the remaining 40 percent is in the liquid phase. Therefore, we have saturated liquid-vapor mixture at temperature of 15 °F. From the saturated refrigerant-134a-temperature table P= Psat @ 15 F = 29.759 psia and Saturated mixture and also he = 16.889 Btu/bm hig = 88.377 Btu/bm Thus, the average enthalpy of the mixture is h = h + xhg = 16.889 + (0.6)*(88.377) = 69.915 Btu/lbm

C) Given T = 10 °F P = 70 psia In this case, the temperature and pressure are given, but again we do not know whether we have saturated mixture, compressed liquid or superheated vapor. To determine the region we are in, we go the saturation refrigerant-134a-pressure table (Table A-12E) and determine the saturation temperature at the given pressure. Thus, for P = 70 psia, we have Tsat = 58.30 °F. We then compare the given T value to this Tsat value, keeping in mind that if T< Tsat @given p we have compressed liquid if T = Tsat @given p we have saturated mixture if T > Tsat @given p we have superheated vapor. In our case, the given T value is 10 °F, which is less than the Tsat value at the given pressure. Therefore, we have compressed liquid and normally we would determine the enthalpy value from the compressed liquid tables of refrigerant 134a. But in this case, we do not have compressed liquid tables for the refrigerant and therefore we are justified to treat the compressed liquid as saturated liquid at the given temperature (not pressure) and obtain the value of enthalpy from Table. h = ht @ 10 °F = 15.318 Btu/lbm As we have compressed liquid, there is no meaning for quality in the compressed liquid region.

d) Given P = 180 psia h = 129.46 Btu/lbm Only pressure and enthalpy are given. We do not know which table to use to determine the missing properties because we have no clue as to whether we have saturated mixture, compressed liquid or superheated vapor, which can be determined from the enthalpy of saturated liquid (ht) and enthalpy of saturated vapor (hq) from the saturated refrigerant-134a- pressure table (Table A-12E). At P= 180 psia h = 51.497 Btu/bm hg = 117.96 Btu/lbm Thus, the given value h = 124.96 Btu/bm is greater than the value of hg. Therefore, we have superheated vapor and the temperature is determined from the superheated vapor table at pressure, P=180 psia and h= 124.96 Btu/lbm Hence, T=160°F The quality has no meaning as we have Superh eated vapor

e) Given Temperature, T = 110 °C Quality, X = 1.0 The quality, x = 1.0 implies that 100 percent of the mass is in the vapor state. Therefore, we have satuarted vapor temperature of 110 °C. From the saturated refrigerant-134a-temperature table. P= Psat @ 110 °F = 161.16 psia and h = hg = 117.23 Btu/lbm

T, °F P , psia h, Btu/lbm 65.891 80 78 15 29.75969.915 10 70 15.318 160 180 129.46 110 T161.16 | 117.23 10.5661 0.6 (Click to select) (Click to select) 1.0 Phase description saturated mixture saturated mixture Compressed liquid Superheatd vapor Superheated vapor