Helium gas contained in a piston cylinder assembly undergoes an isentropic polytropic process from the given initial state with P1 = 0.02 ba

Question

Helium gas contained in a piston cylinder assembly undergoes an isentropic polytropic process from the given initial state with P1 = 0.02 bar, T1 = 200 K to the final state with P2 = 0.14 bar. Determine the work and heat transfer (per unit mass) involved in this process.

in progress 0
Ben Gia 5 years 2021-08-22T23:39:03+00:00 1 Answers 10 views 0

Answers ( )

  1. thienthi
    0
    2021-08-22T23:40:16+00:00
    Reply

    Answer:

    Heat transfer during the process = 0

    Work done during the process = – 371.87 KJ

    Explanation:

    Initial pressure P_{1} = 0.02 bar

    Initial temperature T_{1} = 200 K

    Final pressure P_{2} = 0.14 bar

    Gas constant for helium R = 2.077 \frac{KJ}{kg k}

    This is an isentropic polytropic process so temperature – pressure relationship is given by the following formula,

    \frac{T_{2} }{T_{1} } = [\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }

    Put all the values in above formula we get,

    ⇒ \frac{T_{2} }{200} = [\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }

    ⇒  \frac{T_{2} }{200} = 1.74

    ⇒ T_{2} = 348.72 K

    This is the final temperature of helium.

    For isentropic polytropic process heat transfer to the system is zero.

    ⇒ ΔQ = 0

    Work done W = m × ( T_{1} – T_{2} ) × \frac{R}{\gamma - 1}

    ⇒ W = 1 × ( 200 – 348.72 ) × \frac{2.077}{1.4 - 1}

    ⇒ W = 371.87 KJ

    This is the work done in this process. here negative sign shows that work is done on the gas in the compression of gas.

Leave an answer

Browse

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