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5 January, 19:47

g 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 25.5 kW. For an inlet area of 800 cm2, determine the velocity and the volume flow rate of the steam at the nozzle exit. Use steam tables.

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  1. 5 January, 19:51
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    The answer for velocity = 1.55 m³/s and volume flow rate 259 m/s

    Explanation:

    To answer this answer, we have been told to consider the steam table.

    Now, referencing the steam table, it is noted that the specific volume and enthalpies are obtained from A - 6 for the given temperature and pressure in the question.

    Thus,

    The volume flow rate at the outlet is obtained from the equality of mass flow rate:

    Therefore, M₁ = M₂

    Hence, V₁ / α₁ = V₂ / α₂

    A₁V₁ / α₁ = V₂ / α₂

    Now, if we make V₂ the subject of the formula, we have:

    V₂ = (α₂ / α₁) A₁V₁

    = 0.74321 / 0.38429 x 0.08. 10 m³/s

    = 1.55 m³/s

    Hence, the velocity at the outlet is determined from the energy balance:

    m (h₁ + V₁²/2) = Q + m (h₂ + V₂²/2)

    V₂ = √ 2 (h₁ - h₂) + V₁² - 2α₁Q / A₁V₁

    If we again refer back to the steam table, we have:

    = √ 2 (3267.7 - 3222.2) X 10³ + 10² - 2 X 0.38429 X 25 X 10³ / 10 X 0.08 m/s

    = 259 m/s

    So the answer for velocity = 1.55 m³/s and volume flow rate 259 m/s
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