Respuesta :
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
