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

i don’t know what the answer is i wish i

magnified images possible images can be real = projected onto a screen