Mechanical Properties of Fluids: Full Notes

9.1 Pressure in a Fluid and Pascal's Law

A solid can push back against a force sideways; a fluid cannot. At rest, a fluid can only push perpendicular to any surface it touches, and that push, force per unit area, is pressure:

P = \dfrac{F}{A}

Pressure is a scalar (it has no direction of its own; the force it produces on a surface is always normal to that surface, whichever way the surface happens to face) and its SI unit, newton per square metre, is given the name pascal (Pa).

9.1.1 Pressure increases with depth

Consider a thin horizontal slab of fluid of thickness $dh$ and cross-section $A$ , sitting at depth $h$ below the surface. It is held up against its own weight by the difference between the (larger) pressure pushing up on its bottom face and the (smaller) pressure pushing down on its top face:

(P+dP)A - PA = \rho A\,dh\,g \;\Rightarrow\; dP = \rho g\,dh

Integrating from the open surface (pressure $P_0$ ) down to depth $h$ :

P = P_0 + \rho g h

Depth, not shape or total volume, is all that matters, which is why a narrow deep well and a wide shallow pond can have the same pressure at the bottom, and why a dam is built thicker at its base than at its top.

9.1.2 Pascal's Law and the hydraulic lift

Because fluids can’t sustain a shear, any pressure applied at one point of an enclosed fluid is transmitted undiminished to every other point and to the walls of the container. This is Pascal’s law, and it is what lets a hydraulic lift multiply force: a small force $F_1$ on a narrow piston of area $A_1$ creates a pressure $P = F_1/A_1$ that, transmitted undiminished, pushes on a wide piston of area $A_2$ with a much larger force:

Hydraulic lift

\textcolor{#e08a1e}{F_2} = F_1\dfrac{A_2}{A_1}

Nothing is gained for free: the wide piston moves up only a fraction $A_1/A_2$ as far as the narrow piston moves down, so the work done, force times distance, comes out equal on both sides, exactly as energy conservation demands.

Worked example

Force multiplication in a car lift

A garage lift has a narrow piston of radius 5 cm and a wide piston of radius 20 cm. A mechanic pushes on the narrow piston with 400 N. What load can the wide piston support, and what pressure does the mechanic’s push create in the oil?

The pressure created at the narrow piston:

P = \dfrac{F_1}{A_1} = \dfrac{400}{\pi (0.05)^2} = \dfrac{400}{7.854\times10^{-3}} \approx 5.09\times10^{4}\text{ Pa}

This same pressure acts on the wide piston, giving the supportable load:

F_2 = P A_2 = 5.09\times10^{4}\times\pi(0.20)^2 = 5.09\times10^{4}\times0.1257 \approx \textcolor{#e08a1e}{6400\text{ N}}

a sixteen-fold force gain, exactly $(A_2/A_1) = (20/5)^2 = 16$ , the square of the radius ratio, enough to lift a small car.

9.2 Streamline Flow and the Equation of Continuity

In streamline (laminar) flow, every fluid particle passing a given point follows the exact path traced by the particles before it, a streamline, and streamlines never cross (if two did, a particle arriving at the crossing point would have two different velocities at once, which is meaningless). Push the speed past a critical value and this orderly picture breaks down into turbulent flow: chaotic, swirling, with no fixed map of paths at all.

Now take an ideal fluid, incompressible and in steady flow, moving through a pipe whose cross-section changes from $A_1$ to a narrower $A_2$ . Mass cannot pile up anywhere inside a steady flow, so whatever mass enters a stretch of pipe per second must also leave it per second. The volume crossing area $A$ in time $dt$ is $A\cdot v\,dt$ , so equal mass flow (at constant density) means:

Equation of continuity

\textcolor{#e08a1e}{A_1 v_1} = A_2 v_2

A narrower cross-section does not restrict how much fluid gets through per second, the volume flow rate $Av$ stays fixed along the whole pipe; it simply forces the same volume through in the same time by speeding the fluid up. This is exactly what the Venturi-tube simulation shows: visibly, the flow particles bunch together and accelerate the moment the pipe narrows.

9.3 Bernoulli's Principle

Continuity alone says a narrower pipe means faster flow; it says nothing about pressure. That connection comes from applying the work-energy theorem to a fluid element as it moves along a streamline from a wide, low section (area $A_1$ , speed $v_1$ , height $h_1$ , pressure $P_1$ ) to a narrow, higher section ( $A_2,\,v_2,\,h_2,\,P_2$ ).

The pressure behind the element does positive work pushing it forward, $P_1A_1\,dx_1$ , while the pressure ahead does negative work resisting it, $-P_2A_2\,dx_2$ . Since $A_1\,dx_1 = A_2\,dx_2 = dV$ is the same small volume moved at each end (continuity again), the net work done by pressure per unit volume is $(P_1-P_2)$ , and this must account for the resulting change in the element’s kinetic and gravitational potential energy per unit volume:

P_1 - P_2 = \left(\tfrac12\rho v_2^2 - \tfrac12\rho v_1^2\right) + \left(\rho g h_2 - \rho g h_1\right)

Rearranged so each side refers to a single point, this is Bernoulli’s equation:

Bernoulli's equation

\textcolor{#e08a1e}{P} + \dfrac12\rho v^2 + \rho g h = \text{constant along a streamline}

valid for steady, incompressible, non-viscous flow. For a horizontal pipe, $h_1=h_2$ , and the height terms cancel, leaving the qualitative result worth remembering above all else: where a horizontal flow is faster, its pressure is lower, and vice versa.

9.4 Applications of Bernoulli's Principle

9.4.1 The Venturi meter

A Venturi meter is exactly the horizontal narrow-throated pipe of the simulation, built specifically to measure flow speed by measuring a pressure drop. Combine continuity (§9.2) with horizontal Bernoulli (§9.3): from $A_1v_1=A_2v_2$ , the throat speed is $v_2 = v_1(A_1/A_2)$ , and substituting into Bernoulli with $h_1=h_2$ gives the pressure drop directly in terms of the inlet speed and the area (or radius) ratio:

Venturi pressure drop

\textcolor{#e08a1e}{P_1 - P_2} = \tfrac12\rho\left(v_2^2 - v_1^2\right) = \tfrac12\rho\,v_1^2\left[\left(\dfrac{A_1}{A_2}\right)^{2}-1\right]

which is precisely $\Delta P = \tfrac12\rho(v_2^2-v_1^2)$ as read off the simulation’s manometer columns: the throat column is always the shortest, because that is exactly where $v$ is largest and $P$ smallest. Measuring this pressure drop (the column-height difference) and knowing the area ratio lets you back out $v_1$ , and hence the flow rate $A_1v_1$ , without inserting anything into the pipe that would disturb the flow.

9.4.2 Atomisers and lift

The same principle drives a perfume atomiser or a carburettor: air blown rapidly across the top of a narrow tube lowers the pressure there below atmospheric, and atmospheric pressure then pushes liquid up the tube to be swept away as a fine spray. Aircraft wing lift is popularly explained the same way, faster air over the curved top meaning lower pressure there, but the usual justification for why the top air is faster (the “equal transit time” argument, that air splitting at the leading edge must rejoin at the trailing edge) is simply false; Bernoulli’s principle remains valid, but the full story of lift needs the wing’s effect on the surrounding airflow (circulation), not just this one popular argument.

Worked example

Flow speed from a Venturi pressure drop

Water ( $\rho=1000\text{ kg/m}^3$ ) flows through a horizontal Venturi meter whose throat area is half the inlet area. A manometer reads a pressure drop of 3000 Pa between inlet and throat. Find the inlet speed.

With $A_2 = A_1/2$ , continuity gives $v_2 = 2v_1$ . Substituting into the pressure-drop formula:

3000 = \tfrac12(1000)\left[(2v_1)^2 - v_1^2\right] = 500\times3v_1^2 = 1500\,v_1^2

v_1^2 = 2.0 \;\Rightarrow\; v_1 \approx \textcolor{#e08a1e}{1.41\text{ m/s}}

and correspondingly $v_2 \approx 2.83\text{ m/s}$ at the throat.

9.5 Viscosity, Stokes' Law and Terminal Velocity

Real fluids resist relative sliding between their own layers, an internal friction called viscosity. Between two layers a distance $dz$ apart with a velocity difference $dv$ , the tangential force needed per unit area is proportional to the velocity gradient $dv/dz$ :

\dfrac{F}{A} = \eta\,\dfrac{dv}{dz}

with $\eta$ the coefficient of viscosity (SI unit: Pa·s). Unlike most liquids, a gas’s viscosity increases with temperature, while a liquid’s decreases, which is why warming honey makes it flow more easily.

9.5.1 Stokes' law and terminal velocity

A small sphere of radius $r$ moving at speed $v$ through a viscous fluid experiences a drag force given by Stokes’ law:

Stokes' law

\textcolor{#e08a1e}{F} = 6\pi\eta r v

Drop a sphere into a viscous fluid and three forces act on it: weight $mg$ downward, buoyancy $F_b$ upward, and viscous drag upward, growing with speed. The sphere accelerates only until drag plus buoyancy exactly balance weight, after which it falls at a constant terminal velocity $v_t$ . Writing $\rho$ for the sphere’s density and $\sigma$ for the fluid’s:

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

Solving for $v_t$ :

Terminal velocity

\textcolor{#e08a1e}{v_t} = \dfrac{2r^2(\rho-\sigma)g}{9\eta}

Terminal velocity grows with the square of the radius, which is why fine mist settles through air far more slowly than large raindrops, and why, in a centrifuge or a measuring cylinder of oil, bigger particles of the same material always reach the bottom first.

Worked example

Terminal velocity of a steel ball in oil

A steel ball of radius 1.0 mm and density 7800 kg/m³ is dropped into oil of density 900 kg/m³ and viscosity 0.20 Pa·s. Find its terminal velocity. ( $g=9.8\text{ m/s}^2$ )

v_t = \dfrac{2(1.0\times10^{-3})^2(7800-900)(9.8)}{9\times0.20} = \dfrac{2\times10^{-6}\times6900\times9.8}{1.8}

v_t = \dfrac{0.1352}{1.8} \approx \textcolor{#e08a1e}{0.075\text{ m/s}}

about 7.5 cm/s, reached almost instantly for a ball this small, which is exactly the sort of measurement used in Millikan-style experiments to find an unknown viscosity.

9.6 Reynolds Number

Whether a given flow stays laminar or turns turbulent is predicted, roughly, by a single dimensionless combination, the Reynolds number:

Re = \dfrac{\rho v D}{\eta}

with $D$ a characteristic length (a pipe’s diameter, for instance). As a rough rule of thumb, flow tends to stay laminar for $Re$ below about 1000–2000 and becomes turbulent above it; $Re$ is best read as comparing the inertial forces driving chaotic motion against the viscous forces that damp it out, large $Re$ meaning inertia wins.

9.7 Surface Tension and Capillarity

A molecule inside a liquid is pulled equally in all directions by its neighbours, but one at the surface has neighbours only below and beside it, giving it a net inward pull. The surface behaves like a stretched elastic skin, quantified as surface tension $S$ : either the energy needed to create unit extra surface area, or equivalently the force per unit length along any line drawn on that surface:

S = \dfrac{F}{l} = \dfrac{\text{Energy}}{\text{Area}}

9.7.1 Excess pressure inside a drop and a bubble

A curved surface under tension must have higher pressure on its concave (inner) side to stay in equilibrium, exactly as a stretched balloon skin pushes inward. For a liquid drop of radius $r$ (one surface):

\Delta P = \dfrac{2S}{r}

A soap bubble has two surfaces, inner and outer, both under tension, doubling the effect:

\Delta P = \dfrac{4S}{r}

which is why a smaller soap bubble needs a higher internal pressure to sustain the same tension, and, counterintuitively, why blowing air into two bubbles of unequal size joined by a tube makes the smaller one shrink and the larger one grow.

9.7.2 Capillary rise

Dip a fine open tube of radius $r$ into water and the water climbs up inside it, pulled by the curved meniscus the surface tension forms where the liquid meets the glass. Balancing the upward pull of surface tension around the tube’s circumference against the weight of the risen column gives the height of capillary rise:

Capillary rise

\textcolor{#e08a1e}{h} = \dfrac{2S\cos\theta}{\rho g r}

with $\theta$ the angle of contact. Water wets clean glass ( $\theta$ small, so $\cos\theta>0$ ) and rises; mercury does not wet glass ( $\theta>90^\circ$ , so $\cos\theta<0$ ) and is instead depressed below the surrounding level. Narrower tubes pull liquid higher, exactly inversely with $r$ , which is how capillary action moves water upward through the fine vessels of soil and plant stems.

Worked example

Capillary rise in a fine glass tube

Water ( $S = 0.072\text{ N/m}$ , contact angle $0^\circ$ ) rises in a glass capillary of radius 0.15 mm. Find the height of rise. ( $\rho=1000\text{ kg/m}^3$ , $g=9.8\text{ m/s}^2$ )

h = \dfrac{2(0.072)(\cos0^\circ)}{(1000)(9.8)(0.15\times10^{-3})} = \dfrac{0.144}{1.47}

h \approx \textcolor{#e08a1e}{0.098\text{ m} \approx 9.8\text{ cm}}

— From the NCERT exercises

A couple of the chapter’s own exercise questions, with original worked solutions.

NCERT Exercise 9.11

A plane is in level flight at constant speed, and its total wings area is 2.5 m². What is the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface, if the plane is in level flight at 720 km/h and the plane's mass is 50,000 kg? Take air density to be 1.2 kg/m³.

Solution

Level flight means the pressure difference across the wings supports the plane’s entire weight:

(P_l - P_u)\,A = mg \;\Rightarrow\; P_l - P_u = \dfrac{50000\times9.8}{2.5} = 1.96\times10^{5}\text{ Pa}

Bernoulli’s equation at the same height for the lower and upper surfaces gives:

P_l - P_u = \tfrac12\rho\left(v_u^2 - v_l^2\right)

With $v_l = 720\text{ km/h} = 200\text{ m/s}$ and writing $v_u = v_l(1+\epsilon)$ for a small fractional increase $\epsilon$ , so that $v_u^2 - v_l^2 \approx 2v_l^2\epsilon$ :

1.96\times10^{5} = \tfrac12(1.2)\left(2\times200^2\times\epsilon\right) = 4.8\times10^{4}\,\epsilon

\epsilon = \dfrac{1.96\times10^{5}}{4.8\times10^{4}} \approx \textcolor{#e08a1e}{4.08\text{, i.e.\ about a 4-fold increase}}

Such a large fractional change means the small- $\epsilon$ approximation is really only a rough guide here, but it correctly shows the upper-surface speed must be several times the lower-surface speed to support this particular aircraft.

NCERT Exercise 9.20

A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small hinged door of area 20 cm². The tank is filled with water in one compartment and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. Compute the force necessary to keep the door closed.

Solution

The door feels a net outward force from the pressure difference between the two liquids at its depth. Both liquids are filled to the same height, and the door sits at the bottom, so using $P=\rho g h$ at $h=4.0\text{ m}$ for each side:

P_{acid} - P_{water} = (\rho_{acid}-\rho_{water})\,g h = (1700-1000)(9.8)(4.0)

= 700\times9.8\times4.0 = 27440\text{ Pa}

With door area $A = 20\text{ cm}^2 = 20\times10^{-4}\text{ m}^2$ , the net force needed to hold it shut:

F = \Delta P \times A = 27440\times20\times10^{-4} \approx \textcolor{#e08a1e}{54.9\text{ N}}