Fluid Mechanics Concept Page - 14

Example

Viscous Force on a plate moving with linear velocity

Example: A flat plate of area 0.1 m2${}^2$ is placed on a horizontal surface and
is separated from it by an oil film 10−5${}^{-5}$ m thick. If the coefficient of viscosity of oil is 1.5 kg m−1s−1$m^{-1}{s}^{-1}$, what is the force required to cause the plate to slide on the surface at a constant speed of 1 mm/s ?

Solution:
(Viscous force)
ηoil=1.5kgm−1s−1${\eta }_{oil}=1.5kg{m}^{-1}{s}^{-1}$
thickness of oil film =10−5m${10}^{-5}m$
So,
dvdh=10−310−5${\displaystyle \frac{dv}{dh}}={\displaystyle \frac{{10}^{-3}}{{10}^{-5}}}$---------(1)
102s−1${10}^2{s}^{-1}$
Now, From on the plate =Aηdvdh=0.1×1.5×102$A\eta {\displaystyle \frac{dv}{dh}}=0.1\times 1.5\times {10}^2$[from(1)]
=15N$=15N$

Example

Find viscous force on a surface for a given velocity profile

Example:
A spherical solid ball of volume V$V$ is made of a material of density ρ1${\rho }_1$. It is falling through a liquid of density ρ2<ρ1${\rho }_2<{\rho }_1$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v$v$, i.e. Fviscous=−kv2(k>0)${\mathrm{F}}_{\mathrm{v}\mathrm{i}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{u}\mathrm{s}}=-\mathrm{k}{v}^2(\mathrm{k}>0)$. Find the terminal speed of the ball.

Solution:
At terminal velocity, the weight of the ball is balance by the viscous force and the buoyant force upwards. Thus, Vρ1g=kv2t+Vρ2g$V{\rho }_{1}g=k{v}_t^2+V{\rho }_{2}g$⟹vt=Vg(ρ1−ρ2)k−−−−−−−−−−√$⟹v_t=\sqrt{{\displaystyle \frac{Vg({\rho }_1-{\rho }_2)}{k}}}$

Definition

Assumptions involved in Stoke’s Law:

1) This law is valid for spherical particles only.
2) The value of Reynolds number should be less than 1.      Where,
     Reynolds number, Re=ρVDμ$Re={\displaystyle \frac{\rho VD}{\mu }}$             
     Here
     ρ$\rho$ is density of fluid             
      V is the flow velocity relative to the object             
      D is the diameter of the spherical object          
      And μ$\mu$ is the viscosity of fluid

Definition

Stoke's Law

When any object rises or falls through a fluid it will experience a viscous drag (frictional force) due to the fluid.  This object can be skydiver falling through air, a stone falling through water or a bubble rising through water. 
Statement of Stoke's law: Stokes law states that the force of viscosity on a small sphere moving through a viscous fluid is given by:
F=6πμrv$F=6\pi \mu rv$
Where,
F is the frictional force  acting on the interface between the fluid and the particle. 
μ$\mu$ is the dynamic viscosity
R is the radius of the spherical object
V is the flow velocity relative to the object

Example

Problem based on terminal velocity

Question: The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20oC${20}^{o}C$ is 6.5cms−1$6.5cm{s}^{-1}$. Compute the viscosity of the oil at 20oC${20}^{o}C$.
Density of oil is 1.5×103Kgm−3$1.5\times {10}^{3}Kg{m}^{-3}$, density of copper is 8.9×103Kgm−3$8.9\times {10}^{3}Kg{m}^{-3}$.
Answer: we know that the terminal velocity of a body is 
 Vt=2r2(ρb−ρl)g9μ$V_t={\displaystyle \frac{2r^2({\rho }_b-{\rho }_l)g}{9\mu }}$
Hence μ=2r2(ρb−ρl)g9vt$\mu ={\displaystyle \frac{2r^2({\rho }_b-{\rho }_l)g}{9v_t}}$
Putting the values in the formula we get,
μ=29×(2×10−3)m2×9.8ms−26.5×10−2ms−1×7.4×103Kgm−3$\mu ={\displaystyle \frac{2}{9}}\times {\displaystyle \frac{(2\times {10}^{-3})m^2\times 9.8m{s}^{-2}}{6.5\times {10}^{-2}m{s}^{-1}}}\times 7.4\times {10}^{3}Kg{m}^{-3}$
    = 9.9×10−1Kgm−1s−1$9.9\times {10}^{-1}Kg{m}^{-1}{s}^{-1}$

Example

Example of Terminal Velocity

Here are some of the real-life examples of terminal velocity.
1) When a raindrop falls from the sky, it accelerates initially due to gravity. After some time, the raindrop achieves a constant terminal velocity.
2) Skydiver dives from the sky. As he opens the parachute, the drag force on him suddenly increases. The skydiver ultimately achieves a constant terminal velocity.

Formula

Derivation of terminal velocity

The condition for a body to acquire terminal speed is:
Fg=FD+FB$F_g=F_D+F_B$

Let us use the following notations.
ρb${\rho }_b$: density of the body
ρl${\rho }_l$ : density of the fluid
Vb$V_b$ : volume of the body
g$g$: acceleration due to gravity
μ$\mu$: viscosity of the fluid
r$r$: radius of the sphere
v$v$ : velocity of the body
vt$v_t$: terminal velocity of the body
Hence the above equation can be written as 
ρbVbg=ρlVbg+6πμrv${\rho }_b{V}_{b}g={\rho }_l{V}_{b}g+6\pi \mu rv$
Putting the value of Vb=4πr33$V_b={\displaystyle \frac{4\pi r^3}{3}}$ and rearranging we can get the expression for terminal velocity as
Vt=2(ρb−ρl)r2g9μ$V_t={\displaystyle \frac{2({\rho }_b-{\rho }_l)r^{2}g}{9\mu }}$

Definition

Understanding terminal velocity

Suppose a spherical object is dropped inside a container containing a certain fluid at time t=0 as shown in the figure(i).
Let us study the subsequent motion of the body.

1. Initially when the body is dropped, there are two forces acting on the body: a) force due to gravity Fg$F_g$ and b) buoyancy force FB$F_B$ as shown in the image (i). The drag force is zero as the velocity of the body is zero. Since the density of the body is more than density of the fluid, the net force will be downwards. Hence the body will accelerate downwards. 
2. As the body acquires some velocity, a drag force starts acting on the fluid upwards as shown in the image (ii). But still the net downward force would be greater as shown in the figure (ii).
3. As body further moves down the velocity of the body increases and hence drag force will also increase. (Since drag force is directly proportional to velocity)
4. A moment comes when the net upward force will be equal to net downward force as shown in the image(iii). Then the body moves with a constant speed as net acceleration on the body is zero.
5. The speed thus acquired by the body is called terminal velocity.

Definition

Terminal Velocity

When an object falls through a fluid, it attains a constant velocity through its subsequent motion. This happens because the net force on the body due to gravity and fluid becomes zero.

This constant velocity is termed as terminal velocity.

Example

Problem based on Stoke's law

Question: Assume that a spherical body is flowing through water. The velocity of the body at particular instant is 2 m/s. Find the drag force on the body due to the fluid. Assume that Stokes law is valid. (Given: viscosity of water = 0.001 Kgm−1s−1$Kg{m}^{-1}{s}^{-1}$, radius of spherical object = 2 mm$2mm$)
Ans: We know that the drag force on the body will be 
F=6πμrV$F=6\pi \mu rV$
putting the values in the expression,
F=6×3.14×0.001Kgms×2×10−3×2ms$F=6\times 3.14\times 0.001{\displaystyle \frac{Kg}{ms}}\times 2\times {10}^{-3}\times 2{\displaystyle \frac{m}{s}}$
hence F=75.36×10−6N

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