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 26 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.

velocity and the volume flow rate of the steam is 257.14 m/s and 1.547 m³/s

Explanation:

given data

velocity = 10 m/s

temperature t1 = 400°C

pressure p1 = 800 kPa

temperature t21 = 375°C

pressure p2 = 400 kPa

rate of heat lose = 26 kW

inlet area = 800 cm²

solution

we use here steam tables A 6 ( super heated steam )

use for 400°C  and 800 kPa

v1 = 38429 m³/kg

h1 = 3267.7 KJ/kg

and use for 375°C   and 400 kPa

v2 = 0.74321  m³/kg

h2 = 3222.2 KJ/kg  

so here now

steady state energy equation that is express as

Q - w = mass [ (h2-h1) +  ]  

so that will be

-Q - 0 =  [ (h2-h1) +  ]  

put here value

- 26 × 10³ =

v = 257.14 m/s

and mass is

m = =  

so

A2 V2 =  

A2 V2 =  

A2 V2 = 1.547 m³/s

1) +60.0 J

The work done when lifting an object is equal to the change in gravitational potential energy of the object:

where m is the mass of the object, g is the gravitational acceleration, is the variation of height of the object.

In this problem, m = 3.4 kg, g = 9.8 m/s^2 and , so the work done is:

And the work done is positive, since the object has gained potential energy.

2) 0 J

In this case, there is no variation of the heigth of the object:

Therefore, the variation of potential energy is zero:

And so, the work done is zero as well:

3) -60.0 J

In this case, the variation of height of the object is negative, since the object has been lowered to the ground:

So, the object has lost potential energy:

And so the work done is negative: