Answer:
a. The energy of the air at the initial and the final states is 0kJ and 0.171kJ respectively
b. 0.171kJ
c. 0.143
Explanation:
a.
Because there are same conditions of the state of air at the surroundings and at the Initial stage, the energy of air at the Initial stage is 0kJ.
Calculating energy at the final state;
We start by calculating the specific volume of air in the environment and at the final state.
U2 = At the final state, it is given by
RT2/P2
U1= At the Initial state, it is given by
RT1/P1
Where R = The gas constant of air is 0.287 kPa.m3/kg
T2 = 150 + 273 = 423K
T1 = 25 + 273 = 298K
P2 = 600KPa
P1 = 100KPa
U2 = 0.287 * 423/600
U2 = 0.202335m³/kg
U1 = 0.287 * 298/100
U1 = 0.85526m³/kg
Then we Calculate the mass of air using ideal gas relation
PV = mRT
m = P1V/RT1 where V = 2*10^-3kg
m = 100 * 2 * 10^-3/(0.287 * 298)
m = 0.00234kg
Then we calculate the entropy difference, ∆s. Which is given by
cp2 * ln(T2/T1) - R * ln(P2/P1)
Where cp2 = cycle constant pressure = 1.005
∆s = 1.005 * ln (423/298) - 0.287 * ln(600/100)
∆s = -0.1622kJ/kg
Energy at the final state =
m(E2 - E1 + Po(U2 - U1) -T0 * ∆s)
E2 and E1 are gotten from energy table as 302.88 and 212.64 respectively
Energy at the final state
= 0.00234(302.88 - 212.64 + 100(0.202335 - 0.85526) - 298 * -0.1622)
Energy at the final state = 0.171kJ
b.
Minimum Work = ∆Energy
Minimum Work = Energy at the final state - Energy at the initial state
Minimum Work = 0.171 - 0
Minimum Work done = 0.171kJ
c. The second-law efficiency of this process is calculated by ratio of meaningful and useful work
= 0.171/1.2
= 0.143
