1. ChemE 260 Internal Energy, Enthalpy & The NIST Webbook April 5, 2005 Dr. William Baratuci Senior Lecturer Chemical Engineering Department University of Washington TCD 3: A & B CB 2: 9 – 11, Supplement

2. Internal Energy & Enthalpy Internal Energy –Non-nuclear energy stored within molecules –Sum of the vibrational, translational and rotational kinetic energies –U = strong fxn of T and a weak fxn of P –U  sharply as T  but U  slightly as P . –Ideal Gas, Incompressible Liquids, Solids U = fxn(T) only U  fxn(P) Enthalpy –H = U + P VdH = dH + d(PV)  H =  U +  (PV) –H = strong fxn(T) –H = moderate fxn(P) –Ideal Gas: H  fxn(P)S Baratuci ChemE 260 April 5, 2005

3. NIST Webbook Baratuci ChemE 260 April 5, 2005

4. Reference State We cannot determine an absolute U or H in the way we can determine an absolute T. We must choose a reference state and assign = 0 or = 0 at that state. Calculate all other values of and relative to the reference state. You cannot use thermodynamic data from different sources that are based on different reference states without correcting for the difference in reference state !! Baratuci ChemE 260 April 5, 2005

5. Generating a Saturated Temperature Table Un-check this box ! Baratuci ChemE 260 April 5, 2005

6. Saturated Liquid Properties Baratuci ChemE 260 April 5, 2005

7. Saturated Vapor Properties Baratuci ChemE 260 April 5, 2005

8. Extra Info from the NIST Webbook Baratuci ChemE 260 April 5, 2005

9. Example #1 Determine the T sat, and of saturated liquid ammonia at 300 kPa. (Default ref. state) Ans.: T sat = -9.2243 o C U = 300.25 kJ/kg H = 300.71 kJ/kg Baratuci ChemE 260 April 5, 2005

10. Example #2 Determine the, and of butane at 14.696 psia and 77 o F in units of Btu, lb m and ft 3. (Default ref. state) Ans.: V = 6.5394 ft 3 /lb m U = 251.92Btu/lb m H = 269.71 Btu/lb m Baratuci ChemE 260 April 5, 2005

11. Example #3 Determine the, and of a saturated mixture of R-123 at –40 o C and x = 0.30. (Default ref. state, kJ, mole, m 3 ) Ans.: P sat = 3.5752 Kpa U sat liq = 24.660 kJ/mol, U sat vap = 52.800 kJ/mol, U x=0.03 = 33.102 kJ/mol H sat liq = 24.660 kJ/mol, H sat vap = 54.731 kJ/mol, H x=0.03 = 33.681 kJ/mol V sat liq = 9.4405 x 10 -5 m 3 /mol, V sat vap = 0.54014 m 3 /mol, V x=0.03 = 0.16211 m 3 /mol Baratuci ChemE 260 April 5, 2005

12. Next Class Heat Capacities –How much does the temperature of 1 mole or kg of a substance change when 1 J is added ? Phase Changes –Latent heats of vaporization, fusion and sublimation Hypothetical Process Paths –HPP’s make it easier to calculate how much a property changes during any real process. Baratuci ChemE 260 April 5, 2005