Chapter 6, Problem 6.2E

find the changes in h  as appropriate. The initial state pressure is p1 = 0.5 MPa. the final state is 2.   a.  constant volume :                   v1 = 0.3 m3/kg,            p2 = 0.3 MPa; b. constant entropy :                    s1 = 6.3 kJ/kg K,        p2 = 0.15 MPa; c. constant volume :                    h1 = 2500 kJ/kg,          p2 = 0.2 MPa; d. constant enthalpy :                 s1 = 6.4 kJ/kg K,        p2 = 0.2 MPa;

A closed system undergoes a process in which work is done on the system and heat transfer occurs only at a location on the boundary where the temperature is Tb. Considering an adiabatic an internally reversible process, the system entropy change is O positive negative equal to zero O is not predictable from the info given.

For any irreversible process the net entropy change is Zero positive O negative O infinite O unity O

5. Steam undergoes a temperature heat addition process initially at 200 Kpa and 60% quality to 200 C. Determine the following: A. heat addition B. work done C. change in entropy 6. Steam undergoes a constant volume heat addition process initially at 200 KPa and 60% quality to 200 C. Determine the following: A. heat addition B. work done C. change in entropy

Problem: Identify the Enthalpy of State 1 to 4 (h1, h2, h3 and h4)Note: Point 2 and 4 are saturated.Process 1-2 IsometricProcess 2-3 IsothermalProcess 3-4 IsometricProcess 4-1 Isothermalwhere: at point1; Entropy = 9 KJ/Kg-K and Pressure = 0.64 MPaat point3 X=0.2

B5

Define the entropy of a pure substance is determined from the tables (like other properties).

The turbine of a vapor power plant its schematics is shown in Figure Q6 is evaluated for entropy generation. The details of the steam at the inlet and exit from the turbine are Given in Table Q6. The mass flow rate and the power produced by the steam turbine is also given in the Table. 1 Wev 2 Figure Q6: Schematics of the steam turbine Consider Case # 2 from the Table: Table Q6: State points of the steam turbine along with corresponding Case Numbers. Turbine exit Power velocity Output Turbine inlet Turbine inlet Turbine inlet Mass flowrate at Case Turbine exit Pressure Temperature Velocity the turbine inlet Number state (bar) (oC) (m/s) (kg/s) (m/s) (kW) Saturated 1 40 500 150 2.0 vapor at 60 1900 65°C Saturated 2 100 520 160 2.1 vapor at 70 1950 70°C Saturated 3 60 600 170 2.2 vapor at 80 2000 75°C Saturated 4 80 560 140 2.3 vapor at 90 2050 80°C Saturated 60 540 150 2.4 vapor at 100 2100 0.20 bar Saturated 6 80 520 160 2.5 vapor at 80 2150 0.30 bar Saturated 7 100 600 140 2.6 vapor…

Please show the complete solution (Formula, Given and Conversion of Units) The gain of entropy during isothermal non- flow process of 5 lb of air at 60°F is 0.462 Btu/°R. Find the V1/V2.