Answer:
Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × [tex]10^6[/tex] Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 [tex]\times 10^{-3}[/tex] )²× 9 [tex]\times 10^{-3}[/tex] × sin18 × cos18
Qd = 94.305 × [tex]10^{-6}[/tex] m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × [tex]10^{6}[/tex] × π × 85 [tex]\times 10^{-3}[/tex] × ( 9 [tex]\times 10^{-3}[/tex] )³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × [tex]10^{-6}[/tex] m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × [tex]10^{-6}[/tex] - 85.2 × [tex]10^{-6}[/tex]
Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s