Fluid Statics

(b) 2 : 1

By Jurin's law, $h = \dfrac{2T\cos\theta}{r\rho g}$, so for the same liquid $h \propto \dfrac1r$:

$$\frac{h_A}{h_B} = \frac{r_B}{r_A} = \frac{4}{2} = \frac{2}{1}$$

(c) greater than $90^\circ$

When the angle of contact is obtuse (for example mercury on glass, about $140^\circ$), cohesion is stronger than adhesion. The liquid forms a convex surface and does not wet the solid. An acute angle ($< 90^\circ$) means the liquid wets the surface.

(a) Surface tension

Surface tension is the property of a liquid by which its free surface behaves like a stretched membrane. It is measured as the force per unit length acting perpendicular to an imaginary line drawn on the liquid surface, tangential to the surface:

$$T = \frac{F}{l}$$

SI unit: N m$^{-1}$.

(b) Bernoulli's theorem

For the steady, streamline flow of an ideal (incompressible and non-viscous) liquid, the total energy per unit volume, i.e. the sum of pressure energy, kinetic energy and potential energy, is the same at every point along a streamline:

$$P + \tfrac12\rho v^2 + \rho gh = \text{constant}$$

(c) Numerical

Given:

• Coefficient of viscosity of castor oil, $\eta = 2.42$ N s m$^{-2}$
• Density of oil, $\sigma = 940$ kg m$^{-3}$
• Radius of steel ball, $r = 2$ mm $= 2\times10^{-3}$ m
• Density of steel, $\rho = 7800$ kg m$^{-3}$
• $g = 10$ m s$^{-2}$

At terminal velocity, weight = upthrust + viscous drag, which gives

$$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$$

Substituting:

$$v_t = \frac{2\times(2\times10^{-3})^2\times(7800 - 940)\times10}{9\times2.42} = \frac{2\times4\times10^{-6}\times6860\times10}{21.78} = \frac{0.5488}{21.78}$$ $$\boxed{v_t \approx 2.52\times10^{-2}\ \text{m s}^{-1}} \quad (\approx 2.5\ \text{cm s}^{-1})$$

(d) 3 m/s

By the equation of continuity, $A_1v_1 = A_2v_2$. Area $\propto d^2$, so doubling the diameter makes the area 4 times:

$$v_2 = v_1\frac{A_1}{A_2} = \frac{12}{4} = 3\ \text{m s}^{-1}$$

(a) Stokes' law

When a small sphere of radius $r$ moves with velocity $v$ through a fluid of coefficient of viscosity $\eta$, the viscous drag on it is

$$F = 6\pi\eta r v$$

(b) Terminal velocity of a sphere falling through a viscous liquid

1. A sphere of radius $r$ and density $\rho$ falls through a liquid of density $\sigma$ and coefficient of viscosity $\eta$. Three forces act on it:
   • weight (down): $W = \tfrac43\pi r^3\rho g$
   • upthrust (up): $U = \tfrac43\pi r^3\sigma g$
   • viscous drag (up), by Stokes' law: $F = 6\pi\eta r v$
2. At first $v$ is small, so $W > U + F$ and the sphere accelerates. As $v$ increases, $F$ increases, until the net force becomes zero. From then on the sphere falls with a constant velocity, the terminal velocity $v_t$.
3. At terminal velocity:

$$W = U + F \;\Rightarrow\; \tfrac43\pi r^3\rho g = \tfrac43\pi r^3\sigma g + 6\pi\eta r v_t$$

4. Rearranging:

$$6\pi\eta r v_t = \tfrac43\pi r^3(\rho - \sigma)g$$

5. Hence

$$\boxed{v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}}$$

Measuring it: a steel ball is dropped into a tall jar of glycerin with two marks on it, the upper mark well below the surface so that the ball has already reached terminal velocity. The time $t$ taken to fall the distance $d$ between the marks is measured with a stopwatch, and $v_t = d/t$. This agrees with the formula above (and the formula can then be used to find $\eta$).

(c) Two raindrops coalesce

Given: terminal velocity of each small drop, $v_1 = 10$ cm s$^{-1}$. Let each small drop have radius $r$, and the big drop radius $R$.

1. Volume is conserved:

$$\tfrac43\pi R^3 = 2\times\tfrac43\pi r^3 \;\Rightarrow\; R = 2^{1/3}r$$

2. Terminal velocity $v_t \propto r^2$ (same liquid, same air), so

$$\frac{v_2}{v_1} = \frac{R^2}{r^2} = 2^{2/3} \approx 1.587$$

3. Hence

$$v_2 = 1.587\times10 \approx \boxed{15.9\ \text{cm s}^{-1}}$$

a) Forces on the ball (free body diagram)

                 ^  U = (4/3) π r^3 σ g   (upthrust)
                 |
                 |  ^  F = 6 π η r v      (viscous drag)
                 |  |
               .-+--+-.
              (  ball  )   radius r, density ρ
               '--+---'
                  |
                  v  W = (4/3) π r^3 ρ g  (weight)

• Weight acting downward: $W = \tfrac43\pi r^3\rho g$
• Upthrust (buoyant force) of glycerin acting upward: $U = \tfrac43\pi r^3\sigma g$
• Viscous drag acting upward (opposite to motion): $F = 6\pi\eta r v$

b) Velocity–time graph

   v
   ^
v_t|- - - - - - - - - - - - - - - - - - - -
   |               .  *  *  *  *  *  *  *
   |         *
   |     *
   |   *
   |  *
   | *
   |*
   +----------------------------------------> t
   0

The velocity rises quickly at first. The rise becomes slower as the drag grows, and the velocity finally becomes constant at the terminal velocity $v_t$.

c) Expression for terminal velocity

1. A sphere of radius $r$ and density $\rho$ falls through a liquid of density $\sigma$ and coefficient of viscosity $\eta$. Three forces act on it:
   • weight (down): $W = \tfrac43\pi r^3\rho g$
   • upthrust (up): $U = \tfrac43\pi r^3\sigma g$
   • viscous drag (up), by Stokes' law: $F = 6\pi\eta r v$
2. At first $v$ is small, so $W > U + F$ and the sphere accelerates. As $v$ increases, $F$ increases, until the net force becomes zero. From then on the sphere falls with a constant velocity, the terminal velocity $v_t$.
3. At terminal velocity:

$$W = U + F \;\Rightarrow\; \tfrac43\pi r^3\rho g = \tfrac43\pi r^3\sigma g + 6\pi\eta r v_t$$

4. Rearranging:

$$6\pi\eta r v_t = \tfrac43\pi r^3(\rho - \sigma)g$$

5. Cancelling $\pi r$:

$$\boxed{v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}}$$

So the terminal velocity is proportional to $r^2$ and to $(\rho - \sigma)$, and inversely proportional to $\eta$.

a) Relation between surface tension and surface energy

    ______________________ P         P'
   |                      |  - - - - :
   |      soap film       |  ----> F :
   |   (two surfaces)     |  l        :
   |______________________| Q - - - - Q'
   U-shaped wire frame     |<-- x -->|

1. Take a U-shaped wire frame with a straight wire PQ of length $l$ that can slide on it, and form a soap film on the frame.
2. The film has two free surfaces, so the surface tension $T$ pulls PQ inward with a force

$$F = 2Tl$$

3. Pull PQ outward slowly through a small distance $x$ (at constant temperature). Work done against surface tension:

$$W = Fx = 2Tlx$$

4. Increase in surface area (two surfaces):

$$\Delta A = 2lx$$

5. This work is stored as surface energy. Surface energy per unit area is

$$E = \frac{W}{\Delta A} = \frac{2Tlx}{2lx}$$ $$\boxed{E = T}$$

So surface tension is numerically equal to the surface energy per unit area (and the energy needed to increase the area by $\Delta A$ is $W = T\,\Delta A$).

b) i) Relation between $v_2$ and $v_1$

(We neglect the small upthrust of the air compared with the weight of the drop.)

1. Let each small drop have radius $r$ and the big drop radius $R$. Volume is conserved:

$$\tfrac43\pi R^3 = N\cdot\tfrac43\pi r^3 \;\Rightarrow\; R = N^{1/3}r$$

2. Terminal velocity: $v = \dfrac{2r^2\rho g}{9\eta}$, so $v \propto r^2$.
3. Therefore

$$\frac{v_2}{v_1} = \frac{R^2}{r^2} = N^{2/3} \;\Rightarrow\; \boxed{v_2 = N^{2/3}v_1}$$

b) ii) For $N = 2$, $v_1 = 0.4$ m s$^{-1}$

$$v_2 = 2^{2/3}\times0.4 = 1.587\times0.4 \approx \boxed{0.635\ \text{m s}^{-1}}$$

(B) 0.24 m/s

For the combined drop, $R = 2^{1/3}r$, and terminal velocity $v \propto r^2$, so

$$v' = 2^{2/3}\times0.15 = 1.587\times0.15 \approx 0.24\ \text{m s}^{-1}$$

a) Laminar and turbulent flow

• Laminar (streamline) flow: the liquid moves in smooth, parallel layers that do not mix. The velocity at any given point stays constant in magnitude and direction with time. It happens at low speeds.
• Turbulent flow: the motion of the liquid is irregular, with eddies and whirls. The velocity at a point changes randomly with time and the layers mix. It happens when the speed exceeds the critical velocity.

b) Numerical

Given:

• At point 1: $P_1 = 4\times10^4$ N m$^{-2}$, $v_1 = 2$ m s$^{-1}$, $A_1 = 200$ cm$^2$
• At point 2: $A_2 = 50$ cm$^2$, $P_2 = ?$
• Density of water, $\rho = 1000$ kg m$^{-3}$
• The pipe is horizontal, so $h_1 = h_2$.

Step 1: speed at point 2 (equation of continuity).

$$A_1v_1 = A_2v_2 \;\Rightarrow\; v_2 = \frac{A_1v_1}{A_2} = \frac{200\times2}{50} = 8\ \text{m s}^{-1}$$

(The area ratio is the same in cm$^2$ or m$^2$.)

Step 2: pressure at point 2 (Bernoulli's equation for a horizontal pipe).

$$P_1 + \tfrac12\rho v_1^2 = P_2 + \tfrac12\rho v_2^2$$ $$P_2 = P_1 - \tfrac12\rho(v_2^2 - v_1^2) = 4\times10^4 - \tfrac12\times1000\times(64 - 4)$$ $$P_2 = 4\times10^4 - 3\times10^4$$ $$\boxed{P_2 = 1.0\times10^4\ \text{N m}^{-2}}$$

The pressure is lower where the pipe is narrower and the water flows faster.

(c) Graph C

The cork starts from rest at the bottom. Its upthrust is larger than its weight, so it accelerates upward. As its speed increases, the viscous drag increases, the net force falls to zero, and the cork then rises with a constant terminal velocity until it reaches the surface. Graph C (velocity increasing, then constant) shows this behaviour. (Strictly, the rising part is a smooth curve that flattens out, not a straight line.)

a) Stokes' law and coefficient of viscosity

Stokes' law: the viscous drag on a small sphere of radius $r$ moving with velocity $v$ through a fluid of coefficient of viscosity $\eta$ is

$$F = 6\pi\eta r v$$

Expression for $\eta$ (falling-ball method):

1. A small steel ball of radius $r$ and density $\rho$ is dropped into a tall jar of the given liquid of density $\sigma$. Forces on it: weight $\tfrac43\pi r^3\rho g$ (down), upthrust $\tfrac43\pi r^3\sigma g$ (up), and viscous drag $6\pi\eta r v$ (up).
2. The drag increases with speed until the net force is zero. The ball then falls with a constant terminal velocity $v$:

$$6\pi\eta r v + \tfrac43\pi r^3\sigma g = \tfrac43\pi r^3\rho g$$

3. So

$$6\pi\eta r v = \tfrac43\pi r^3(\rho - \sigma)g$$ $$\boxed{\eta = \frac{2r^2(\rho - \sigma)g}{9v}}$$

4. In practice, $v$ is found by timing the ball over a measured distance $d$ between two marks on the jar (after it has reached terminal velocity): $v = d/t$. Putting $r$, $\rho$, $\sigma$, $g$ and $v$ into the formula gives $\eta$.

b) Equation of continuity and applications

For steady flow of an incompressible liquid through a tube of varying cross-section, the volume of liquid passing any section per second is the same:

$$A_1v_1 = A_2v_2 \qquad \text{or} \qquad Av = \text{constant}$$

So the liquid flows faster where the tube is narrow and slower where it is wide.

Applications:

1. When we partly close the mouth of a garden hose (or fit a nozzle), the area decreases, so water comes out faster and goes farther.
2. A river flows slowly where it is wide and deep, and fast where it is narrow and shallow. (Also: water falling from a tap speeds up, so the stream becomes narrower as it falls.)

(b) Energy

Bernoulli's principle follows from the work-energy theorem applied to a flowing ideal fluid: $P + \tfrac12\rho v^2 + \rho g h = \text{constant}$ expresses conservation of energy per unit volume.

(a)(i) Coefficient of viscosity

It is the tangential viscous force per unit area needed to maintain a unit velocity gradient between two layers of a fluid: $\eta = \dfrac{F}{A\,(dv/dx)}$. SI unit: Pa s.

(a)(ii) Finding $\eta$ using Stokes' law

Drop a small steel ball of radius $r$ and density $\rho$ in the liquid of density $\sigma$ in a tall jar. After falling a short distance it reaches terminal velocity $v$, when the net force is zero:

$$\text{weight} = \text{upthrust} + \text{viscous force}$$ $$\tfrac43\pi r^3\rho g = \tfrac43\pi r^3\sigma g + 6\pi\eta r v$$ $$\eta = \frac{2r^2(\rho-\sigma)g}{9v}$$

Measure $v$ by timing the ball over a measured distance between two marks, and calculate $\eta$.

(b) Pressure change

Given $A_1 = 25\ \text{cm}^2$, $v_1 = 3$ m/s, $A_2 = 15\ \text{cm}^2$.

By the equation of continuity:

$$v_2 = \frac{A_1v_1}{A_2} = \frac{25\times3}{15} = 5\ \text{m/s}$$

By Bernoulli's equation for a horizontal pipe:

$$P_1 - P_2 = \tfrac12\rho\,(v_2^2 - v_1^2) = \tfrac12(1000)(25-9) = 8000\ \text{Pa}$$

The pressure falls by $8\times10^3$ Pa on going into the narrower pipe.

(c) $90^\circ$

The capillary rise is $h = \dfrac{2T\cos\theta}{r\rho g}$. There is no rise or fall only when $h = 0$, i.e. $\cos\theta = 0$, so $\theta = 90^\circ$. For $\theta<90^\circ$ the liquid rises, and for $\theta>90^\circ$ it falls.

(a) For water the adhesive force between water and glass is greater than the cohesive force among water molecules, so the angle of contact is acute and the surface is concave. For mercury the cohesive force is greater than the adhesive force, so the angle of contact is obtuse and the surface is convex.

(b) The viscosity of air (a gas) increases with temperature. Viscosity in gases comes from the transfer of momentum between layers by molecules moving across them. At higher temperature the molecules move faster, so more momentum is transferred and the viscous drag increases. (In liquids it is the opposite.)

(c) For steady flow of an ideal liquid in a horizontal pipe, continuity gives $A_1v_1 = A_2v_2$:

$v_2 = v_1\left(\frac{0.1}{0.04}\right)^2 = 6.25\,v_1$

Bernoulli's equation gives $\Delta P = \tfrac12\rho(v_2^2 - v_1^2)$:

$10 = \tfrac12(1.25\times10^3)(39.06 - 1)v_1^2 = 2.379\times10^{4}\,v_1^2 \;\Rightarrow\; v_1 = 0.0205\ \text{m/s}$

The rate of flow is

$Q = A_1v_1 = \pi(0.1)^2(0.0205) \approx 6.4\times10^{-4}\ \text{m}^3/\text{s}$

(a) Capillarity is the rise or fall of a liquid in a tube of very fine bore (a capillary tube) dipped in it.

(b) The adhesive force between water and glass is greater than the cohesive force among water molecules, so water wets the glass, the angle of contact is acute and the water rises. For mercury the cohesive force is greater than the adhesive force, so the angle of contact is obtuse and the mercury is depressed.

(c) The height of rise is $h = \dfrac{2T\cos\theta}{r\rho g}$, so $h\propto 1/r$:

$\frac{h_2}{h_1} = \frac{r_1}{r_2} = 3 \;\Rightarrow\; h_2 = 3\times 2 = 6\ \text{cm}$

(a) Bernoulli's principle

For the steady, streamline flow of an ideal (incompressible, non-viscous) liquid, the sum of pressure energy, kinetic energy and potential energy per unit volume is constant at every point along a streamline:

$$P + \tfrac12\rho v^2 + \rho gh = \text{constant}$$

(b) Numerical

Given:

• Density, $\rho = 1200$ kg m$^{-3}$
• At A: $A_A = 1.0$ cm$^2$ $= 1.0\times10^{-4}$ m$^2$, $v_A = 10$ cm s$^{-1}$ $= 0.10$ m s$^{-1}$
• At B: $A_B = 20$ mm$^2$ $= 20\times10^{-6} = 2.0\times10^{-5}$ m$^2$
• A and B are at the same height.

(i) Speed at B

By the equation of continuity:

$$A_Av_A = A_Bv_B \;\Rightarrow\; v_B = \frac{A_Av_A}{A_B} = \frac{1.0\times10^{-4}\times0.10}{2.0\times10^{-5}}$$ $$\boxed{v_B = 0.50\ \text{m s}^{-1}}$$

(ii) Pressure difference between A and B

By Bernoulli's equation with $h_A = h_B$:

$$P_A + \tfrac12\rho v_A^2 = P_B + \tfrac12\rho v_B^2$$ $$P_A - P_B = \tfrac12\rho(v_B^2 - v_A^2) = \tfrac12\times1200\times\left[(0.50)^2 - (0.10)^2\right] = 600\times0.24$$ $$\boxed{P_A - P_B = 144\ \text{Pa}}$$

The pressure at A (wide part) is higher than at B (narrow part) by 144 N m$^{-2}$.

(b) goes down

By Bernoulli's principle, the fast-moving air under the pan has a lower pressure than the still air above it. The greater pressure above pushes the pan downward.

(c) $6h$

From $h = \dfrac{2T\cos\theta}{r\rho g}$, $h \propto \dfrac1g$. On the Moon $g_m = g/6$, so the liquid rises to $6h$ (provided the tube is long enough).

(i) Surface tension

Surface tension is the property of a liquid by which its free surface behaves like a stretched elastic membrane. It is the force per unit length acting perpendicular to an imaginary line drawn on the liquid surface, tangential to the surface: $T = F/l$. SI unit: N m$^{-1}$.

(ii) Relation between surface tension and surface energy

    ______________________ P         P'
   |                      |  - - - - :
   |      soap film       |  ----> F :
   |   (two surfaces)     |  l        :
   |______________________| Q - - - - Q'
   U-shaped wire frame     |<-- x -->|

1. A soap film is formed on a U-shaped wire frame closed by a sliding wire PQ of length $l$.
2. The film has two surfaces, so the inward pull of surface tension on PQ is

$$F = 2Tl$$

3. If PQ is pulled out slowly through a distance $x$, the work done is

$$W = Fx = 2Tlx$$

4. The increase in surface area (both sides) is

$$\Delta A = 2lx$$

5. This work is stored as surface energy. So the surface energy per unit area is

$$E = \frac{W}{\Delta A} = \frac{2Tlx}{2lx} \;\Rightarrow\; \boxed{E = T}$$

Surface tension is numerically equal to the surface energy per unit area of the liquid surface.

(iii) Terminal velocity of the combined drop

1. Volume is conserved: $\tfrac43\pi R^3 = 2\times\tfrac43\pi r^3$, so $R = 2^{1/3}r \approx 1.26\,r$.
2. Terminal velocity $v = \dfrac{2r^2(\rho - \sigma)g}{9\eta}$, so $v \propto r^2$.
3. Therefore

$$\frac{v_{\text{new}}}{v} = \left(\frac{R}{r}\right)^2 = 2^{2/3} \approx 1.59$$

The new drop falls faster: its terminal velocity is about 1.59 times the original. Its weight grows as $r^3$, while the viscous drag grows only as $r$, so a larger speed is needed before the drag balances the weight.

(i) Bernoulli's principle

For the steady, streamline flow of an ideal (incompressible, non-viscous) liquid, the total energy per unit volume (pressure energy + kinetic energy + potential energy) is constant at every point along a streamline:

$$P + \tfrac12\rho v^2 + \rho gh = \text{constant}$$

(ii) Derivation of Bernoulli's equation

                               (2) narrow and higher:
                                   A2, P2, v2, h2
                          ____________
            _____________/   ---->    |
  (1)      /    flow                  |
  A1, P1, |  ---->       _____________|
  v1, h1  |_____________/
  ------------------------------------------- reference level

1. Consider an ideal liquid of density $\rho$ in steady flow through a tube. At section 1: area $A_1$, pressure $P_1$, speed $v_1$, height $h_1$. At section 2: $A_2$, $P_2$, $v_2$, $h_2$.
2. In a short time $\Delta t$, the same mass $\Delta m$ (volume $\Delta V$) enters at 1 and leaves at 2:

$$\Delta V = A_1v_1\Delta t = A_2v_2\Delta t, \qquad \Delta m = \rho\Delta V$$

3. Net work done on the liquid by the pressure forces:

$$W = P_1A_1v_1\Delta t - P_2A_2v_2\Delta t = (P_1 - P_2)\Delta V$$

4. Gain in kinetic energy:

$$\Delta K = \tfrac12\Delta m(v_2^2 - v_1^2) = \tfrac12\rho\Delta V(v_2^2 - v_1^2)$$

Gain in potential energy:

$$\Delta U = \Delta m\,g(h_2 - h_1) = \rho\Delta V g(h_2 - h_1)$$

5. By the work–energy theorem, $W = \Delta K + \Delta U$:

$$(P_1 - P_2)\Delta V = \tfrac12\rho\Delta V(v_2^2 - v_1^2) + \rho\Delta V g(h_2 - h_1)$$

6. Dividing by $\Delta V$ and collecting terms:

$$\boxed{P_1 + \tfrac12\rho v_1^2 + \rho gh_1 = P_2 + \tfrac12\rho v_2^2 + \rho gh_2}$$

i.e. $P + \tfrac12\rho v^2 + \rho gh = \text{constant}$.

(iii) Covering the hose opening with a thumb

• By the equation of continuity, $Av = \text{constant}$. Partly covering the opening with the thumb reduces the area $A$ of the outlet, so the water comes out with a much larger speed $v$.
• In Bernoulli's terms, the pressure energy of the water inside the hose is converted into more kinetic energy at the narrow outlet.
• The jet behaves like a projectile, and its horizontal range increases with the exit speed (for a jet aimed at angle $\theta$, $R = v^2\sin2\theta/g$; for a jet held horizontally at height $h$, $R = v\sqrt{2h/g}$). So the water reaches a much greater distance.