Determine the work input and entropy generation during the compression of steam from 100 kPa to 1 MPa in an adiabatic compressor if the inle

Question

Determine the work input and entropy generation during the compression of steam from 100 kPa to 1 MPa in an adiabatic compressor if the inlet state is saturated vapor and the isentropic efficiency is 85 percent.

in progress 0
Thiên Ân 4 years 2021-08-20T00:12:10+00:00 1 Answers 47 views 0

Answers ( )

  1. RuslanHeatt
    0
    2021-08-20T00:13:26+00:00
    Reply

    Answer:

    The value of work input to the compressor W_{in} = 1505.7 \frac{KJ}{kg}

    The value of entropy generation inside the compressor

    S_{gen} = - 1.064 \frac{KJ}{kg K}

    Explanation:

    Initial pressure P_{1} = 100 k pa

    Final pressure P_{2} = 1 M pa = 1000 k pa

    At initial pressure P_{1} = 100 k pa the steam is saturated vapour. From the steam tables the value of volume at this state is v = 1.673 \frac{m^{3} }{kg}

    Thus the work input to the compressor W_{in} = v dP

    ⇒ W_{in} = v  ( P_{2} – P_{1} )

    ⇒ W_{in} = 1.673 ( 1000 – 100 )

    ⇒ W_{in} = 1505.7 \frac{KJ}{kg}

    The isentropic efficiency is  = 85 %

    So the work input to the compressor W_{in} = \frac{1505.7}{0.85}

    ⇒ W_{in} =  1771.42 \frac{KJ}{kg}

    This is the value of work input to the compressor.

    S_{gen} = - R \ln \frac{P_{2} }{P_{1} }

    S_{gen} = - 0.462 \ln \frac{1000}{100}

    S_{gen} = - 1.064 \frac{KJ}{kg K}

    This is the value of entropy generation inside the compressor.

Leave an answer

Browse

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