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23 July, 01:35

Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s and leaves at 375°C and 400 kPa while losing heat at a rate of 27 kW. For an inlet area of 800 cm^2, determine the velocity and the volume flow rate of the steam at the nozzle exit. Use steam tables.

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  1. 23 July, 05:07
    0
    Using the equation of continuity:

    A

    1

    v

    1

    =

    A

    2

    v

    2

    0.08

    (

    10

    )

    =

    A

    2

    (

    225

    )

    A

    2

    =

    3.55

    *

    10

    -

    3

    m

    2

    Q

    2

    =

    A

    2

    v

    2

    Q

    2

    =

    3.55

    *

    10

    -

    3

    *

    225

    Q

    2

    =

    0.798

    m

    3

    /

    s

    Explanation:

    Steady Flow Energy Equation:

    The steady flow energy equation is a representation of the first law of thermodynamics. It is the conservation of energy law for an open system. A nozzle is an open system in the context of thermodynamics. It is used to produce a high velocity by reducing its pressure.

    The steady flow energy equation can be given by the following formula:

    h

    1

    +

    1

    2

    v

    2

    1

    +

    g

    z

    1

    +

    q

    =

    h

    2

    +

    1

    2

    v

    2

    2

    +

    g

    z

    2

    +

    w

    where 'h' is enthalpy, 'v' is velocity, 'z' is height, 'q' is the heat and 'w' is work.

    h

    =

    C

    p

    d

    T

    Answer and Explanation:

    Given:

    initial temp,

    T

    1

    =

    400

    0

    C

    initial Pressure,

    p

    1

    =

    800

    k

    P

    a

    Initial Velocity,

    v

    1

    =

    10

    m

    /

    s

    Final temp,

    T

    2

    =

    300

    0

    C

    Final Pressure,

    p

    2

    =

    200

    k

    P

    a

    Rate of heat loss, Q = 25 KW

    Inlet Area,

    A

    1

    =

    800

    c

    m

    2

    As per the steady flow energy equation:

    h

    1

    +

    1

    2

    v

    2

    1

    +

    g

    z

    1

    +

    q

    =

    h

    2

    +

    1

    2

    v

    2

    2

    +

    g

    z

    2

    +

    w

    Since, there is external work, w = 0. Also, consider there is a negligible change in KE.

    h

    1

    +

    1

    2

    v

    2

    1

    +

    q

    =

    h

    2

    +

    1

    2

    v

    2

    2

    h

    1

    -

    h

    2

    +

    1

    2

    v

    2

    1

    +

    q

    =

    1

    2

    v

    2

    2

    C

    p

    (

    T

    1

    -

    T

    2

    )

    +

    1

    2

    (

    10

    )

    2

    +

    25000

    =

    1

    2

    v

    2

    2

    2

    (

    400

    -

    300

    )

    +

    50

    +

    25000

    =

    1

    2

    v

    2

    2

    2

    (

    400

    -

    300

    )

    +

    50

    +

    25000

    =

    1

    2

    v

    2

    2

    25250

    =

    1

    2

    v

    2

    2

    v

    2



    225

    which is the answer.

    Using the equation of continuity:

    A

    1

    v

    1

    =

    A

    2

    v

    2

    0.08

    (

    10

    )

    =

    A

    2

    (

    225

    )

    A

    2

    =

    3.55

    *

    10

    -

    3

    m

    2

    Now, volume flow rate,

    Q

    2

    =

    A

    2

    v

    2

    Q

    2

    =

    3.55

    *

    10

    -

    3

    *

    225

    Q

    2

    =

    0.798

    m

    3

    /

    s
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