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

Question

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.

in progress 0
Bơ 5 years 2021-09-04T15:48:32+00:00 1 Answers 764 views 0

Answers ( )

  1. mit
    0
    2021-09-04T15:50:00+00:00
    Reply

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )