Question 1.10: A Newtonian fluid having viscosity, μ = 0.4 Ns/m², fills the......
A Newtonian fluid having viscosity, μ = 0.4 Ns/m², fills the gap between two parallel infinite plates which are 25 mm apart (Figure 1.10). Calculate the force required to move a thin plate of 0.2 m² at a constant velocity at 1 m/s which has been placed (a) at middle of two plates (b) at 10 mm apart from one plate.
Given,
Viscosity of fluid, μ= 0.4 Ns/m²
Distance of parallel plates = 25 mm = 0.025 m
Surface area of thin plate = 0.2 m²
When thin plate will move in the fluid, a friction (viscous) drag force will act in opposite direction of both the faces of thin plate. So, let force on top surface be F_1 and force on the bottom surface = F_2 .
(a) From Newton’s law of viscosity,
F_1=\tau_1 \times A=\mu \frac{d u}{d y} A
Now, assuming the linear velocity distribution,
F_1=0.4 \times \frac{1}{12.5 \times 10^{-3}} \times 0.2=6.4 N
Similarly, F_2=6.4 N
Total force required to move the plate at constant velocity,
F=F_1+F_2=6.4+6.4=12.8 N
(b)
\begin{aligned} & F_1=\tau_1 \times A=\mu \frac{d u}{d y_1} A=0.4 \times \frac{1}{10} \times 0.2=8 N \\ & F_2=\mu \frac{d u}{d y_2} A=0.4 \times \frac{1}{15} \times 0.2=5.33 N \end{aligned}
Total force required will be
\begin{aligned} & F=F_1+F_2 \\ & F=8+5.33=13.33 N \end{aligned}