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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.

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Answer:

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) + [tex]\frac{v2^2-v1^2}{2} + \Delta p e[/tex]  ]  

so that will be

-Q - 0 = [tex]\frac{A1\times V1}{v1}[/tex]  [ (h2-h1) + [tex]\frac{v2^2-v1^2}{2} + 0[/tex]  ]  

put here value

- 26 × 10³ = [tex]\frac{800\times 10^{-4}\times 10}{0.38429} [ (3222.2-3267.7)10^3 + \frac{v2^2-10^2}{2} + 0 ][/tex]

v = 257.14 m/s

and mass is

m = [tex]\frac{A1\times V1}{v1}[/tex] = [tex]\frac{A2\times V2}{v2}[/tex]  

so

A2 V2 = [tex]\frac{v2}{v1} ( A1 \times V1 )[/tex]  

A2 V2 = [tex]\frac{0.74321}{0.38439} ( 800 \times 10^{-4}\times 10 )[/tex]  

A2 V2 = 1.547 m³/s

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