Answer:
m = 0.239kg/s
V₂ = 0.119m^3/s
v² = 5.92m/s
Explanation:
Given that,
Diameter = 16 cm
Enter pressure= 200 kPa
Enter temperature = 20°C
Velocity = 5 m/s
Exit pressure = 180 kPa
Exit temperature = 40°C
Mass flow rate is determine by the properties of the air at the initial state fromit initial velocity and the cross sectional area of the pipe
[tex]m = p_1V_1[/tex]
[tex]= \frac{P_1}{RT_1} A_1v_1[/tex]
[tex]= \frac{P_1}{RT_1} \frac{D^2}{4} \pi v_1\\= \frac{200}{0.287} \times \frac{0.16^2}{4} \times \pi \times 5\\= 0.239kg/s[/tex]
The final volume rate is determine from the mass flow rate and the final properties of the air
[tex]V_2 = \frac{m}{P_2} \\= \frac{RT_2m}{P_2} \\= \frac{0.287 \times 313 \times0.239}{180} \\= 0.119m^3/s[/tex]
Final velocity is determine from final volume flow rate and the cross sectin area
[tex]v^2 = \frac{V_2}{A} \\=\frac{4V_2}{D^2\pi } \\= \frac{4 \times 0.119}{0.16^2 \times \pi } \\= 5.92m/s[/tex]
