Fluid Mechanics Test 1

At base P_B = 750 mm of Hg

At top, P_A = 600 mm of Hg

∴ Pressure difference = 150 mm of Hg = 150 × 10^-3 × 13.6 × 103 × 9.81

(ΔP) = hρg

\( \Rightarrow \;h\; = \;\frac{{{\rm{\Delta }}P}}{{\rho g}}\; = \;\frac{{150 × 13.6 × {{10}^3} × {{10}^{ - 3}} × 9.81}}{{1 × 9.81}}\; = \;2040\;m\)

⇒ Height of mountain = 2040 m

CONCEPT

• Surface Tension: It is defined as force per unit length in the plane of the liquid surface at right angles to either side of an imaginary line drawn to the surface.
  • It is the tendency of the liquid surface to shrink into a minimum surface area.
  • Mathematically, \(Surface\;Tension = \frac{{Force}}{{Lenght}}\)
  • The SI Unit of surface tension is N m^-1
• Surface Energy: In order to increase the surface area, work has to be done on the surface of the liquid. This extra work stored in the form of potential energy and known as surface energy.
  • W = S ΔA
  • W = work done, S = surface tension, ΔA = change in area,
  • The SI unit of surface energy is Joule. 

CALCULATION

Radius of one small drop r = 0.1 mm = 10^-4 m

Surface area of the water drop = 4πr² = 4π(10^-4)² = 4π × 10^-8 sq. m

Surface energy of the water drop = surface tension x area = 0.072 × 4π × 10^-8 J

Total surface energy of 125 drops = 125 × 0.072 × 4π × 10^-8 J = 11.3 × 10^-7 J

Let R be the radius of the big drop then,

Volume of the big water drop = 125 x volume of the small water drop

\(\frac{4}{3}\;\pi {R^3} = 125 \times \frac{4}{3}\;\pi {r^3}\;\)

R³ = 125 r³

R = 5r = 5 × 10^-4 m

Surface area of the big water drop \(= 4\pi {R^2} = 4\pi {\left( {5 \times {{10}^{ - 4\;}}} \right)^2}\)

Surface energy of big water drop \(= 0.072 \times 4\pi {\left( {5 \times {{10}^{ - 4\;}}} \right)^2} = 2.26 \times {10^{ - 7}}\;J\)

Concept:

Weight of any liquid, W = mg {m is the mass of liquid)

The specific weight of any liquid, S = weight per unit volume i.e. S = W/V

For a compressible fluid, its specific weight is not constant it will change due to change in its volume.

Mass density of liquid, ρ = mass/volume or m/V

The compressibility of liquid means the change in volume due to the change in pressure.

At the bottom surface, the pressure will be more as compared to the top surface, therefore there will be a change in the volume of liquid and hence, the specific weight will change.

The bulk modulus of elasticity of a liquid, K is given as:

\(K = \frac{{dp}}{{\frac{{dV}}{V}}}\)

dp is the magnitude of the change in pressure

dV is the magnitude of the change in volume

V is the original Volume

Calculation:

For top surface, p₁ = 0

Given that 3 liters of that liquid at the surface weights 23.7 N.

Specific weight at top surface, S₁ = 23.7× 1000/3 = 7900 N/m³ or 7.9 kN/m³

At certain depth, p₂ = 1000 N/m².

Now

dp = 1000-0 = 1000 N/m²

dV = V₁ – V₂ = 0.003 – V₂

\(K = \frac{{dp}}{{\frac{{dV}}{V}}}\)

\(3840 = \frac{{1000}}{{\frac{{0.003 - {V_2}}}{{0.003}}}}\)

⇒ V₂ = 2.22 × 10^-3 m³

Specific weight at certain depth, S₂ = W/V₂ = 23.7/2.22 × 10^-3 = 10675.68 N/m³ or 10.68 kN/m³

Change in specific Weight, ΔS = S₂ – S₁ = 10.68 – 7.9 = 2.78 kN/m³