Mechanical Properties of Fluids • Topic 2 of 3

Buoyancy, Viscosity & Bernoulli's Principle

When a body is immersed in a fluid, the pressure on its lower surface is greater than on its upper surface (because pressure increases with depth). The net upward force is called buoyant force (upthrust).

Archimedes' principle states that a body wholly or partly immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces:

  • $F_B=V_{disp}\,\rho_{fluid}\,g$, where $V_{disp}$ is the volume of fluid displaced.
  • Law of flotation: a body floats when its weight equals the buoyant force, i.e. it displaces a weight of fluid equal to its own weight. A body floats if its average density is less than that of the fluid; it sinks if it is more.

Viscosity is the internal friction between adjacent layers of a fluid moving with different velocities. The viscous force opposing relative motion is given by Newton's law of viscosity, $F=-\eta A\frac{dv}{dx}$, where $\eta$ is the coefficient of viscosity (SI unit $\text{Pa}\cdot\text{s}$) and $\frac{dv}{dx}$ is the velocity gradient. Honey is more viscous than water.

Stokes' law gives the viscous drag on a small sphere of radius $r$ moving with speed $v$ through a fluid: $F=6\pi\eta r v$.

Terminal velocity is the constant maximum speed a body attains when the net force on it becomes zero (weight = buoyancy + viscous drag). For a sphere falling through a fluid:

  • $v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}$, where $\rho$ is the body's density and $\sigma$ the fluid's density.
  • Notice $v_t\propto r^2$ — a larger raindrop falls faster than a small one.

Equation of continuity. For the steady flow of an incompressible fluid, the mass flowing per second is constant, so the volume flow rate is the same at every cross-section:

  • $A_1v_1=A_2v_2$, i.e. $Av=\text{constant}$.
  • Where the pipe is narrow the fluid moves fast, and where it is wide it moves slowly — which is why a river speeds up where it narrows.

Bernoulli's principle. For the streamline (non-turbulent), incompressible, non-viscous flow of a fluid, the total energy per unit volume — pressure energy, kinetic energy and potential energy — stays constant along a streamline:

  • $P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}$.
  • A key consequence: where the speed is high, the pressure is low. This explains aeroplane lift (faster air over the curved wing top means lower pressure above), the lift of a spinning ball (Magnus effect), the action of an atomiser/sprayer, and the venturimeter used to measure flow rate.
Streamline flow through a constricted pipe (continuity and Bernoulli)A1, v1 (slow)A2, v2 (fast)A1 v1 = A2 v2narrow = fast = low P
Solved Examples
1
Worked Example
A body of volume $2\times10^{-3}\ \text{m}^3$ is fully immersed in water. Find the buoyant force on it. (Take $\rho_{water}=1000\ \text{kg/m}^3$, $g=9.8\ \text{m/s}^2$.)
Solution
  1. Step 1: Use $F_B=V\rho g$.
  2. Step 2: Substitute: $F_B=2\times10^{-3}\times1000\times9.8$.
  3. Step 3: Compute: $F_B=19.6\ \text{N}$.

Answer: $F_B=19.6\ \text{N}$ (upward).

2
Worked Example
An ice cube of density $900\ \text{kg/m}^3$ floats in water ($1000\ \text{kg/m}^3$). What fraction of its volume lies below the surface?
Solution
  1. Step 1: For flotation, weight $=$ buoyancy: $V\rho_{ice}g=V_{sub}\rho_{water}g$.
  2. Step 2: So $\frac{V_{sub}}{V}=\frac{\rho_{ice}}{\rho_{water}}=\frac{900}{1000}$.
  3. Step 3: Compute: $\frac{V_{sub}}{V}=0.9$.

Answer: 90% of the cube is submerged.

3
Worked Example
A spherical drop of radius $1\times10^{-4}\ \text{m}$ and density $1000\ \text{kg/m}^3$ falls through air (density negligible) of viscosity $1.8\times10^{-5}\ \text{Pa}\cdot\text{s}$. Find its terminal velocity. (Take $g=9.8\ \text{m/s}^2$.)
Solution
  1. Step 1: Use $v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}$ with $\sigma\approx0$.
  2. Step 2: Numerator $=2\times(10^{-4})^2\times1000\times9.8=2\times10^{-8}\times9800=1.96\times10^{-4}$.
  3. Step 3: Divide by $9\eta=9\times1.8\times10^{-5}=1.62\times10^{-4}$: $v_t=\frac{1.96\times10^{-4}}{1.62\times10^{-4}}\approx1.21\ \text{m/s}$.

Answer: $v_t\approx1.21\ \text{m/s}$.

4
Worked Example
Water flows through a pipe whose cross-section narrows from $4\ \text{cm}^2$ to $1\ \text{cm}^2$. If the speed in the wide section is $2\ \text{m/s}$, find the speed in the narrow section.
Solution
  1. Step 1: Use the equation of continuity $A_1v_1=A_2v_2$.
  2. Step 2: So $v_2=\frac{A_1v_1}{A_2}=\frac{4\times2}{1}$.
  3. Step 3: Compute: $v_2=8\ \text{m/s}$.

Answer: $v_2=8\ \text{m/s}$.

5
Worked Example
Water flows horizontally through a pipe. At a wide point the pressure is $2\times10^5\ \text{Pa}$ and speed $3\ \text{m/s}$; at a narrow point the speed is $5\ \text{m/s}$. Find the pressure at the narrow point. (Take $\rho=1000\ \text{kg/m}^3$.)
Solution
  1. Step 1: For horizontal flow ($h$ same), Bernoulli gives $P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$.
  2. Step 2: $P_2=P_1+\frac{1}{2}\rho(v_1^2-v_2^2)=2\times10^5+\frac{1}{2}\times1000\times(9-25)$.
  3. Step 3: $P_2=2\times10^5+500\times(-16)=2\times10^5-8000=1.92\times10^5\ \text{Pa}$.

Answer: $P_2=1.92\times10^5\ \text{Pa}$ (lower, as the speed is higher).

6
Worked Example
A solid weighs 50 N in air and 40 N when fully immersed in water. Find the volume of the solid. (Take $\rho_{water}=1000\ \text{kg/m}^3$, $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: Loss of weight = buoyant force $=50-40=10\ \text{N}$.
  2. Step 2: $F_B=V\rho g$, so $V=\frac{F_B}{\rho g}=\frac{10}{1000\times10}$.
  3. Step 3: Compute: $V=10^{-3}\ \text{m}^3=1000\ \text{cm}^3$.

Answer: $V=10^{-3}\ \text{m}^3$ (1 litre).

Key Points

  • Archimedes' principle: buoyant force $F_B=V_{disp}\,\rho_{fluid}\,g$ equals the weight of fluid displaced; a body floats if its density is less than the fluid's.
  • Viscosity is internal fluid friction; Stokes' law gives drag on a sphere $F=6\pi\eta r v$.
  • Terminal velocity: $v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}$, attained when net force is zero; $v_t\propto r^2$.
  • Equation of continuity: $A_1v_1=A_2v_2$ — fluid speeds up where the pipe narrows.
  • Bernoulli's equation: $P+\frac{1}{2}\rho v^2+\rho gh=\text{const}$; high speed means low pressure (aeroplane lift, atomiser, venturimeter).
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Archimedes' principle states that the buoyant force equals the:
Explanation: $F_B=V_{disp}\rho g$, the weight of the displaced fluid.
Q2.The terminal velocity of a sphere falling through a viscous fluid is proportional to:
Explanation: $v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}\propto r^2$.
Q3.The equation of continuity $A_1v_1=A_2v_2$ is a statement of the conservation of:
Explanation: For an incompressible fluid in steady flow, the volume flowing per second $Av$ is constant — conservation of mass.
Q4.According to Bernoulli's principle, where the speed of a flowing fluid is high, the pressure is:
Explanation: $P+\frac{1}{2}\rho v^2+\rho gh$ is constant, so a larger $v$ means a smaller $P$.
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