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.

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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explanation:

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