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Outline Material and Energy Balances CHEN 2120 Class Meeting #26 March 21st, 2007 – Reference states “Energy balances on open systems” Specific properties • Specific property: an intensive quantity obtained by dividing an extensive property (or its flow rate) by the total amount (or flow rate) of the process material. • Extensive properties: mass, volume, kinetic energy, potential energy, internal energy • Intensive properties: specific mass, specific volume, specific kinetic energy, etc…. • The symbol ^ denotes specific properties. Open systems inlet • Specific properties and enthalpy • Flow work and shaft work • The steady-state open-system energy balance • Tables of thermodynamic data • Steam tables Specific properties • Thus, when a specific property is multiplied by mass or moles, the total extensive property can be determined • Example: ˆ U ( J / kg ) = m(kg )U ( J / kg ) Flow work and shaft work outlet • Material can flow into and out of an open system • Doesn’t necessarily have to be at steady state (but we’ll do mostly steady state) • W = Ws + Wfl • Ws = shaft work, or rate of work done by the process fluid on a moving part within the system (e.g., pump or piston) • Wfl = flow work, or rate of work done by the fluid at the system outlet minus the rate of work done on the fluid at the system inlet 1 Flow work inlet Enthalpy outlet • Wfl,in = - Pin·Vin (negative since work done on system) • Wfl,out = Pout·Vout • Wfl = PoutVout – PinVin • Enthalpy: H = U + PV • “Designed” to take into account ‘PV’ work for open systems • Example: If the specific internal energy of a gas at 300 K and 1 atm is 4200 J/mol, and the specific molar volume at the same temperature and pressure is 25.34 L/mol, the specific enthalpy of the gas is: Ĥ Ĥ Enthalpy • • • • ∆U + ∆Ek + ∆Ep = Q – W But W = Ws + Wfl And Wfl = PoutVout – PinVin Therefore, Uout – Uin + Ek,out – Ek,in + Ep,out – Ep,in = Q – Ws – (PoutVout – PinVin) Re-arrange: (Uout+PoutVout) – (Uin+PinVin) + Ek,out–Ek,in+Ep,out–Ep,in= Q–Ws Æ ∆H + ∆Ek + ∆Ep = Q - Ws • Specific Enthalpy: Ĥ = 4200 J/mol + (1 atm)(25.34 L/mol)(8.314 J/molK)/(0.08206 L-atm/K-mol) = 6770 J/mol Changes in enthalpy • Mainly when: 1) Heating or cooling a solid, liquid, or gas 2) Phase changes (evaporation, condensation, freezing, melting) ∆H& + ∆E& k + ∆E& p = Q& − W& s ∆Hˆ + ∆Eˆ k + ∆Eˆ p = Qˆ − Wˆ s Reference states Reference states • It is not possible to know the absolute value of U or H for a process material, but you can determine the change in U (∆U) or in H (∆H) corresponding to a specified change of state (temp., pressure, phase) Mt. Everest (29,029 ft) Denali (20,320 ft) Bear Peak (8,250 ft) 2 Reference states Mt. Everest (29,029 ft) Denali (20,320 ft) Bear Peak (8,250 ft) Now you try… (Clicker Prob. 26.1) How tall is Bear Peak? (depends upon ref!) How much enthalpy does stream X have? (depends upon ref!) How much taller is Mt. Everest than Denali (does not depend upon ref, but you do need a ref. state) What is the change in enthalpy going from state A to B? (does not depend upon ref, but you do need a ref. state) Mariana Trench (-35,000 ft) Now you try… (Clicker Prob. 26.1) State T(K) P(bar) V’(L/mol) U’(kJ/mol) Liquid 300 0.310 0.0516 0.000 Vapor 300 0.310 79.94 28.24 Vapor 340 1.33 20.92 29.62 • State T(K) P(bar) V’(L/mol) U’(kJ/mol) Liquid 300 0.310 0.0516 0.000 Vapor 300 0.310 79.94 28.24 Vapor 340 1.33 20.92 29.62 • What reference state was used to generate the listed specific internal energies? A) Liquid at 300 K B) Vapor at 300 K C) Vapor at 340 K Steam Tables Calculate ∆U’(kJ/mol) for a process in which bromine vapor at 300 K is condensed at constant pressure. Then calculate ∆H’(kJ/mol) for the same process. Example • Determine the vapor pressure, specific internal energy, and specific enthalpy of saturated steam at 133.5°C. Example • Determine the vapor pressure, specific internal energy, and specific enthalpy of saturated steam at 133.5°C. • From Table B.6: P = 3.0 bar, U’ = 2543.0 kJ/mol, H’ = 2724.7 kJ/mol 3 Now you try… (Clicker Prob. 26.2) inlet outlet • Steam enters an isobaric heat exchanger at a mass flow rate of 0.08 kg/sec as a superheated vapor at 350°C and 1.0 bar and exits the heat exchanger as a superheated vapor at 150°C. How much heat (kJ/s) is transferred from the steam to the heat exchanger? A) 2.4 kJ/s B) 45.2 kJ/s C) 32.0 kJ/s D) 10.4 kJ/s Now you try… (Clicker Prob. 26.2) inlet outlet • H’in = 3176 kJ/kg • H’out = 2776 kJ/kg • Q = ∆H = (0.08 kg/s)(3176 – 2776)kJ/kg = 32 kJ/s (or 32 kW) (Ans. C) 4