Question

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.

Answer 1

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

Step-by-step 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