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3 December, 00:39

A piston-cylinder device containing carbon dioxide gas undergoes an isobaric process from 15 psia and 80°F to 170°F. Determine the work and the heat transfer associated with this process, in Btu/lbm. The properties of CO2 at room temperature are R = 0.04513 Btu/lbm·R and cv = 0.158 Btu/lbm·R.

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Answers (1)
  1. 3 December, 01:17
    0
    See explanation

    Explanation:

    Given:

    Initial pressure,

    p

    1

    =

    15

    psia

    Initial temperature,

    T

    1

    =

    80



    F

    Final temperature,

    T

    2

    =

    200



    F

    Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.

    R

    =

    0.04513

    Btu/lbm. R

    C

    v

    =

    0.158

    Btu/lbm. R

    Find the work done during the isobaric process.

    w

    1

    -

    2

    =

    p

    (

    v

    2

    -

    v

    1

    )

    =

    R

    (

    T

    2

    -

    T

    1

    )

    =

    0.04513

    (

    200

    -

    80

    )

    w

    1

    -

    2

    =

    5.4156

    Btu/lbm

    Find the change in internal energy during process.

    Δ

    u

    1

    -

    2

    =

    C

    v

    (

    T

    2

    -

    T

    1

    )

    =

    0.158

    (

    200

    -

    80

    )

    =

    18.96

    Btu/lbm

    Find the heat transfer during the process using the first law of thermodynamics.

    q

    1

    -

    2

    =

    w

    1

    -

    2

    +

    Δ

    u

    1

    -

    2

    =

    5.4156

    +

    18.96

    q

    1

    -

    2

    =

    24.38

    Btu/lbm
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