Question

Two kg water at 120°C with a quality of 25% has its temperature raised 20°C in a constant volume process. According to the thermodynamic tables in your textbook, what is the new specific entropy (in kJ/(kg?K))? -another qustion-:

Consider the same ammonia heat engine as problems 6-5 and 6-6, which has the following reversible steps:

1 ? 2 Isothermal heating of saturated liquid @ 60°C to superheated vapor

2 ? 3 Adiabatic expansion to saturated vapor at -20°C

3 ? 4 Isothermal cooling to a 2 phase fluid

4 ? 1 Adiabatic compression back to state 1

Find W (in kJ/kg). Use the correct sign convention to indicate the direction of work.

Answer 1

Ans 1

Initial state at 120 °C

Saturated steam

Specific volume of fluid vf = 0.001060 m3/kg

Specific volume of gas vg = 0.89186 m3/kg

Specific volume at state 1

v = vf + x (vg - vf)

= 0.001060 + 0.25 (0.89186 - 0.001060)

= 0.22376 m3/kg

For constant volume process

v1 = v2 = 0.22376 m3/kg

Quality of steam at 120 + 20 = 140°C

x = ( v2 - vf) / (vg - vf)

= (0.22376 - 0.00108) / (0.50885 - 0.00108)

= 0.4385

Specific entropy

S2 = Sf + x (Sg - Sf)

= 1.739289 + 0.4385 (6.92927 - 1.739289)

= 4.01509 kJ/kg-K