Answer:
[tex]n=2.9\times 10^9[/tex]
[tex]A=1.88\times 10^{-8}\ m^2[/tex]
Explanation:
Given that
Q= 5 L/min
1 L = 10⁻³ m³/s
1 min = 60 s
Q=0.083 x 10⁻³ m³/s
d= 6 μm
v= 1 mm/s
So the discharge flow through one tube
q = A v
[tex]A=\dfrac{\pi}{4}d^2[/tex]
[tex]A=\dfrac{\pi}{4}\times (6\times 10^{-6})^2\ m^2[/tex]
A=2.8 x 10⁻¹¹ m²
v= 1 x 10⁻³ m/s
q= 2.8 x 10⁻¹⁴ m³/s
Lets take total number of tube is n
Q= n q
n=Q/q
[tex]n=\dfrac{0.083\times 10^{-3} }{ 2.8\times 10^{-14}}[/tex]
[tex]n=2.9\times 10^9[/tex]
Surface area A
A= π d L
[tex]A=\pi \times 6\times 10^{-6}\times 10^{-3}\ m^2[/tex]
[tex]A=1.88\times 10^{-8}\ m^2[/tex]
