Thermodynamics

cen84959_ch03.qxd 4/1/05 12:31 PM Page 136136 | Thermodynamics151.83T, °C75P = 500 kPasaturation table (Table A–4) and determine the u f and u g values at the giventemperature. At 125°C, we read u f 524.83 kJ/kg and u g 2534.3 kJ/kg.Next we compare the given u value to these u f and u g values, keeping inmind thatifu 6 u f we have compressed liquidifu f u u g we have saturated mixtureifu 7 u g we have superheated vaporIn our case the given u value is 1600, which falls between the u f and u g valuesat 125°C. Therefore, we have saturated liquid–vapor mixture. Then thepressure must be the saturation pressure at the given temperature:P P sat @ 125°C 232.23 kPa1Table A–42The quality is determined fromx u u f 1600 524.83 0.535u fg 2009.5The criteria above for determining whether we have compressed liquid,saturated mixture, or superheated vapor can also be used when enthalpy h orspecific volume v is given instead of internal energy u, or when pressure isgiven instead of temperature.(c) This is similar to case (b), except pressure is given instead of temperature.Following the argument given above, we read the u f and u g values at thespecified pressure. At 1 MPa, we have u f 761.39 kJ/kg and u g 2582.8kJ/kg. The specified u value is 2950 kJ/kg, which is greater than the u g valueat 1 MPa. Therefore, we have superheated vapor, and the temperature at thisstate is determined from the superheated vapor table by interpolation to beT 395.2°C1Table A–62We would leave the quality column blank in this case since quality has nomeaning for a superheated vapor.(d) In this case the temperature and pressure are given, but again we cannottell which table to use to determine the missing properties because we donot know whether we have saturated mixture, compressed liquid, or superheatedvapor. To determine the region we are in, we go to the saturationtable (Table A–5) and determine the saturation temperature value at thegiven pressure. At 500 kPa, we have T sat 151.83°C. We then compare thegiven T value to this T sat value, keeping in mind thatifT 6 T sat @ given P we have compressed liquidifT T sat @ given P we have saturated mixtureifT 7 T sat @ given P we have superheated vaporu = ~ u f @ 75°CFIGURE 3–44At a given P and T, a pure substancewill exist as a compressed liquid ifT T sat @ P .uIn our case, the given T value is 75°C, which is less than the T sat value atthe specified pressure. Therefore, we have compressed liquid (Fig. 3–44),and normally we would determine the internal energy value from the compressedliquid table. But in this case the given pressure is much lower thanthe lowest pressure value in the compressed liquid table (which is 5 MPa),and therefore we are justified to treat the compressed liquid as saturated liquidat the given temperature (not pressure):u u f @ 75°C 313.99 kJ>kg1Table A–42