Fluid Mechanics Concept Page - 8

Definition

Experimental verification of principle of floatation

Consider a wood of mass m$m$ with known density which is less than the density of water, uniform cross section A$A$ and length L$L$. Mark equally spaced points on the wood and label them to facilitate the measuring of length of wood immersed in water. When slowly immersed in the liquid, it floats at the surface of water. Measure the length immersed in the fluid l$l$.
Since the wood is at rest, it means that there is no net force acting on the wood. Hence, liquid is applying an upward force equal to the weight of the wood. This is the buoyant force (Fb$F_b$). 
Fb=mg=ρwoodLAwoodg$F_b=mg={\rho }_{wood}L{A}_{wood}g$

Calculate the weight of displaced liquid, Fd=ρwaterlAwoodg$F_d={\rho }_{water}l{A}_{wood}g$

Find the ratio, FbFd=ρwoodLρwaterl${\displaystyle \frac{F_b}{F_d}}={\displaystyle \frac{{\rho }_{wood}L}{{\rho }_{water}l}}$
The ratio comes out to be 1 hence proving principle of floatation.

Definition

Applications of Principle of Floatation

Following are some application of principle of floatation:

1. Floating of ships.
2. Submarines: Ballast tanks in submarine is filled with sea water so that it sinks.
3. Icebergs floating on water
4. Swimming
5. Hot air balloon

Result

Submerged volume

Let total volume of the body be Vb$V_b$, submerged volume be Vs$V_s$, density of  body be ρ$\rho$ and density of fluid be ρw${\rho }_w$.
Case 1, ρ<ρw: Vbρ=Vsρw$\rho <{\rho }_w:V_b\rho =V_s{\rho }_w$, Body floats
Case 2, ρ≥ρw:Vs=Vb$\rho \ge {\rho }_w:V_s=V_b$ i.e. Body sinks 

Definition

Experimental determination of density of immiscible solid

Consider a solid with density ρs>ρw${\rho }_s>{\rho }_w$ where ρw${\rho }_w$ is the density of water.
Pour a volume V$V$ of water in a volume measuring container and weigh it using a weighing balance. Let the reading of mass be M$M$.
Now, put the solid inside water. Measure the volume and mass of the new apparatus. Let the new readings be M2$M_2$ and V2$V_2$.
Now, density of solid is given by, ρs=M2−MV2−V${\rho }_s={\displaystyle \frac{M_2-M}{V_2-V}}$

Example

Measuring the density of a liquid

1. A measuring cylinder is put on a balance and its mass found(A).
2. The liquid is poured into the measuring cylinder and its volume measured(V).
3. The mass of the measuring cylinder and the liquid it contains is found(B).
4. The mass of the liquid is found by subtracting A from B.
5. The density of the liquid is calculated by dividing the mass of the liquid by its volume:
Density=B−AV$Density={\displaystyle \frac{B-A}{V}}$

Example

Measuring the density of a rectangular solid block

1. The mass of the solid block is found by placing the block on a balance and reading the scale.
2. The mass in grams is recorded and the volume is found by measuring the length, breadth and height of the solid block.
3. Then density is found by taking ratio of mass to volume of the block.

Definition

Relationship between volume submerged, density of fluid and density of object

The fraction submerged is the ratio of the volume submerged to the volume of the object,
Fraction submerged=VsubmergedVobj=ρobjρfl$={\displaystyle \frac{V_{submerged}}{V_{obj}}}={\displaystyle \frac{{\rho }_{obj}}{{\rho }_{fl}}}$
where, ρobj${\rho }_{obj}$ is the average density of the object and ρfl${\rho }_{fl}$ is the density of the fluid.

Example

Archimedes' principle combined with volume expansion

A solid floats in a liquid at 200${}^0$C with 75% of it immersed in a liquid. When the liquid is heated to 1000${}^0$C the same body floats with 80% of it immersed in the liquid. The coefficient of real expansion of the liquid is:For the body to float with 75% inside liquid, the buoyant force at that temperature must be equal to its weight,⟹Mg=0.75Vρ20∘g$⟹Mg=0.75V{\rho }_{{20}^{\circ }}g$
Similarly for body to float with 80% inside liquid, the buoyant force at that temperature must be equal to its weight,⟹Mg=0.80Vρ100∘g$⟹Mg=0.80V{\rho }_{{100}^{\circ }}g$
So, ρ100∘ρ20∘=0.750.80=11+γΔT${\displaystyle \frac{{\rho }_{{100}^{\circ }}}{{\rho }_{{20}^{\circ }}}}={\displaystyle \frac{0.75}{0.80}}={\displaystyle \frac{1}{1+\gamma \mathrm{\Delta }T}}$⟹γ=8.33×10−4/∘C$⟹\gamma =8.33\times {10}^{-4}{/}^{\circ }C$

Example

Problem on density

Example:
A weather forecasting plastic balloon of volume 15m3$15m^3$ contains hydrogen of density 0.09kgm−3$0.09kg{m}^{-3}$. Find the total mass of the balloon and relative density of the balloon if the mass of empty balloon is 7.15 kg.
Solution:The mass of a body is calculated from the formula, Mass=Density×Volume$Mass=Density\times Volume$.
The volume of the balloon is the volume of hydrogen that will be present in the balloon, that is, 15m3$15m^3$.  The density of hydrogen is 0.09kg/m3$0.09kg/m^3$.
Therefore, the mass of hydrogen in the balloon = 15 x 0.09 = 1.35 Kg. The mass of the balloon is given as 7.15 kg.
So, mass of hydrogen + mass of the balloon = 1.35 + 7.15 = 8.50 kg 
Density of air + balloon, ρ=mg+mbV=8.515=0.567kgm−3$\rho ={\displaystyle \frac{m_g+m_b}{V}}={\displaystyle \frac{8.5}{15}}=0.567kg{m}^{-3}$
Relative density RD=ρρwater=0.567$RD={\displaystyle \frac{\rho }{{\rho }_{water}}}=0.567$ 

Definition

Dynamic LIft

Dynamic Lift: A fluid flowing past the surface of a body exerts a force on it. Dynamic Lift is the component of this force that is perpendicular to the oncoming flow direction. It contrasts with the drag force, which is the component of the surface force parallel to the flow direction. If the fluid is air, the force is called an aerodynamic force. In water, it is called a hydrodynamic force.

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