Answer : The pressure difference required is, [tex]1.792\times 10^5Pa[/tex]
Explanation : Given,
Length of tube = 2.0 m
Diameter of tube = 1.0 mm = 1.0 × 10⁻³
Average speed of water = 4.0 m/s
Viscosity of water at 40°C = 0.7 × 10⁻³
The expression used for pressure difference is:
[tex]\Delta P=8\pi \eta \frac{L\nu_{avg}}{A}[/tex]
where,
[tex]\Delta P[/tex] = pressure difference
[tex]\eta[/tex] = viscosity of water = 0.7 × 10⁻³
L = length of tube = 2.0 m
A = area of tube = [tex]\pi r^2=\pi (\frac{Diameter}{2})^2=3.14\times (\frac{1.0\times 10^{-3}m}{2})^2=7.85\times 10^{-7}m^2[/tex]
[tex]\nu_{avg}[/tex] = average speed of water = 4.0 m/s
Now put all the given values in the above expression, we get:
[tex]\Delta P=8\times 3.14\times 0.7\times 10^{-3}\frac{2.0\times 4.0}{7.85\times 10^{-7}}[/tex]
[tex]\Delta P=179200Pa=1.792\times 10^5Pa[/tex]
Thus, the pressure difference required is, [tex]1.792\times 10^5Pa[/tex]
