GATE CE 2022 Set 2 Fluid Mechanics and Hydraulic Machines - Drag and Lift Question 1 English Explanation 1

$$\rho$$ = density of water

$$\rho_b$$ = density of particle

Diameter of P, $D_1$ = 0.5 mm

Diameter of Q, $D_2$ = 1 mm

The ratio of the drag force on particle Q to the drag force on particle P is :

$ \frac{F_{D2}}{F_{D1}} = \frac{Drag\,force\,on\,particle\,Q}{Drag\,force\,on\,particle\,P} = ? $

To solve this, let's consider the basic case of Stoke's law for a particle of diameter $d$ moving through a fluid.

GATE CE 2022 Set 2 Fluid Mechanics and Hydraulic Machines - Drag and Lift Question 1 English Explanation 2

$F_B$ = Buoyancy force

$F_D$ = Drag force

$V_s$ = Settling velocity

$W$ = weight of particle

$W_g = F_B + F_D$

For a particle moving in fluid :

$ mg - F_B = 3\pi \mu D V_s $

Considering the volume of a sphere :

$ (\rho_b - \rho) V_{sphere} = 3\pi \mu D V_s $

The volume of the sphere is :

$ V_{sphere} = \frac{4}{3} \pi r^3 $

The settling velocity is :

$ V_s = \frac{(\rho_b - \rho) D^2 g}{18 \mu} $

Now, the ratio of drag forces is :

$ \frac{F_{D2}}{F_{D1}} = \frac{3\pi \mu D_2 V_{s2}}{3\pi \mu D_1 V_{s1}} = \frac{D_2}{D_1} \times \frac{D_2^2}{D_1^2} $

Therefore :

$ \frac{F_{D2}}{F_{D1}} = \left( \frac{D_2}{D_1} \right)^3 = \left( \frac{1}{0.5} \right)^3 $

Thus :

$ \frac{F_{D2}}{F_{D1}} = 8 $

The drag force on particle Q is 8 times the drag force on particle P.

Drag and Lift · Fluid Mechanics and Hydraulic Machines · GATE CE

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