Final answers for each question is:

1. Around 12 kJ

2. Around 6 kJ

3. Around -1kJ

use 0.018 kg for the mass of material.

(a) The initial internal energy, volume, and entropy can be found at the superheated steam table. The dead state quantities cannot be found, but values for saturated liquid at 1 bar and 99.6 C are available (hypothetical state 2). Consider a process from the dead state to state 2 that reversibly heats water from 223 K and 1 bar to saturated liquid at 1 bar. For such a process, the changes in energy and entropy can be calculated assuming that volume change is negligible over the temperature range. Using the hypothetical state 2, the quantities in the exergy equation will be calculated as, e.g. internal energy, (U1-U2+U2-Uo = U1-U0). The exergy should be calculated in units of kJ. (b) Interpolate the internal energy and entropy in the superheated steam table to the point double the initial volume at the same pressure. Then the exergy can be calculated. (c) Same as (b) but interpolate with constant temperature instead of pressure.

Further notes:

a) should treat this as a closed system and use internal energies instead of enthalpies. Use a constant value heat capacity from any appendix that seems fit to calculate the internal energy and entropy differences between State 2 and the Dead State (for entropy this should turn out to be 1.0115 kJ/kg-K. Now you can plug into the exergy equation with the internal energy, pressure x specific volume, and temperature x entropy terms all present. The Po and To values to use will be the Dead State values of 1 bar (100 kPa) and 293 K. Each term will be written out in terms of the three states in the problem, e.g. (u1-u2+u2-uo). The exergy you get should be ~12 kJ.

b) You will be using the three-term exergy equation as in (a) but now you only need to use the initial and final values for internal energy, specific volume, and entropy. The final state of an isobaric heating until volume doubles can be found in the superheated steam table by interpolating to the point of double the initial volume at the same pressure. This state is 23% of the way from 600 C to 700 C at 1 MPa (internal energy interpolated is 3,338 kJ/kg). This exergy value should end up being about half of the answer in (a).

c) The final state of an isothermal expansion until volume doubles can be found using the superheated steam table by interpolating to the point of double the initial volume at the same temperature. This state is 82% of the way from 600 kPa to 500 kPa (entropy interpolated will be 7.04 kJ/kg-K). This exergy value should be about -1 kJ.

1mol of steam is initially at 10bar and 200C. The surroundings are at 20C and 1 bar. (a) Calculate the exergy of the system. (b) Calculate the change in exergy for a process where the steam is heated at constant pressure until the volume doubles. (c) Calculate the change in exergy for a process where the steam isothermally expands until its volume doubles