Advance Chemical Engineering Thermodynamics

1. Advance Chemical Engineering Thermodynamics By Dr.DangSaebea

2. Heat changes at constant pressure and volume: the Enthalpy • For n moles of a homogeneous fluid contained in a closed system the energy balance is: The work for a mechanically reversible, here written: (1) (2) These two equations combine: (3)

3. Constant-Volume Process • If the process occurs at constant total volume, the work is zero. Moreover, for closed systems the last term of Eq. (3) is also zero, because n and V are both constant. Thus, (4) Integration yields: (5) Thus the heat transferred in a mechanically reversible, constant-volume, closed-system process equals the internal-energy change of the system.

4. Constant-Pressure Process (6) For a constant-pressure change of state: (7) (8) Thus, the mathematical (and only) definition of enthalpy is: (9) The differential form is: (10) It becomes an equation for a finite change in the system: (11)

5. Example 5 Calculate ΔU and ΔH for 1 kg of water when it is vaporized at the constant temperature of 100°C and the constant pressure of 101.33 kPa. The specific volumes of liquid and vapor water at these conditions are 0.00104 and 1.673 m3 kg-1. For this change, heat in the amount of 2,256.9 kJ is added to the water.

6. HEAT CAPACITY Heat Capacity at Constant Pressure The constant-pressure heat capacity is defined as: (1) (2) (3) For a mechanically reversible, constant-P process (4)

7. HEAT CAPACITY Heat Capacity at Constant Volume The constant-volume heat capacity of a substance is defined as: (1) (2) Integration yields: (3) (4)

8. Example 6 Air at 1 bar and 298.15 K (25°C) is compressed to 5 bar and 298.15 K by two different mechanically reversible processes: (a) Cooling at constant pressure followed by heating at constant volume. (b) Heating at constant volume followed by cooling at constant pressure. Calculate the heat and work requirements and ΔU and ΔH of the air for each path. The following heat capacities for air may be assumed independent of temperature: Cv = 20.78 and Cp = 29.10 J mol-1 K-1 Assume also for air that PV / T is a constant, regardless of the changes it undergoes. At 298.15 K and 1 bar the molar volume of air is 0.02479 m3 mol-1.

9. MASS AND ENERGY BALANCES FOR OPEN SYSTEMS • Importantly, mass and molar flowrates relate to velocity: • Open systems are characterized by flowing streams; there are four common measures of flow: • Mass flowrate, • Molar flowrate, • Volumetric flowrate, q • Velocity, u Importantly, mass and molar flow rates relate to velocity:

10. Mass Balance for Open Systems • The rate of change of mass within the control volume, equals the net rate of flow of mass into the control volume. The mass balance is expressed mathematically by: where the second term for the control volume of Figure is steady state Schematic representation of a control volume

11. The General Energy Balance • The rate of energy accumulation within the control volume includes this quantity in addition to the heat transfer rate and work rate: a total energy of each steam • Work term • Shaft work =Ws • Flow work = (+ or -) PV

12. Example 7 An evacuated tank is filled with gas from a constant-pressure line. What is the relation between the enthalpy of the gas in the entrance line and the internal energy of the gas in the tank? Neglect heat transfer between the gas and the tank.

13. Example 8 Air at 1 bar and 25°C enters a compressor at low velocity, discharges at 3 bar, and enters a nozzle in which it expands to a final velocity of 600 m s-1 at the initial conditions of pressure and temperature. If the work of compression is 240 kJ per kilogram of air, how much heat must be removed during compression?

14. The End