A piston-cylinder device initially contains 2 L of air at 100 kPa and 25°C. Air is now compressed to a final state of 600 kPa and 150°C. The useful work input is 1.2 kJ. Assuming the surroundings are at 100 kPa and 25°C, determine (a) the exergy of the air at the initial and the final states, (b) the minimum work that must be supplied to accomplish this compression process, and (c) the second-law efficiency of this process.

Question

Physics

Cher

21 May, 11:27

A piston-cylinder device initially contains 2 L of air at 100 kPa and 25°C. Air is now compressed to a final state of 600 kPa and 150°C. The useful work input is 1.2 kJ. Assuming the surroundings are at 100 kPa and 25°C, determine (a) the exergy of the air at the initial and the final states, (b) the minimum work that must be supplied to accomplish this compression process, and (c) the second-law efficiency of this process.

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Answers (1)

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Hezekiah Crawford · Answer 1

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