Question

What pressure difference is required between the ends of a 2.0-m-long, 1.0-mm-diameter horizontal tube for 40 C water to flow through it at an average speed of 4.0 m/s?

Answer 1

Answer : The pressure difference required is, $1.792\times 10^5Pa$

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:

$\Delta P=8\pi \eta \frac{L\nu_{avg}}{A}$

where,

$\Delta P$ = pressure difference

$\eta$ = viscosity of water = 0.7 × 10⁻³

L = length of tube = 2.0 m

A = area of tube = $\pi r^2=\pi (\frac{Diameter}{2})^2=3.14\times (\frac{1.0\times 10^{-3}m}{2})^2=7.85\times 10^{-7}m^2$

$\nu_{avg}$ = average speed of water = 4.0 m/s

Now put all the given values in the above expression, we get:

$\Delta P=8\times 3.14\times 0.7\times 10^{-3}\frac{2.0\times 4.0}{7.85\times 10^{-7}}$

$\Delta P=179200Pa=1.792\times 10^5Pa$

Thus, the pressure difference required is, $1.792\times 10^5Pa$