The Workbook

1. \(F = \eta A\dfrac{dv}{dy}\): the viscous force is proportional to the area and to the velocity gradient.

   The oil is at rest at the fixed surface and moves at \(0.5\) m/s at the plate: \(\dfrac{dv}{dy} = \dfrac{0.5}{10^{-3}} = 500\) s\(^{-1}\).

2. \(F = 0.2 \times 0.1 \times 500 = 10\) N.

\(\eta = \dfrac{F}{A\,(dv/dy)}\): unit \(\dfrac{\text{N}}{\text{m}^2\,\text{s}^{-1}} = \) N s/m\(^2 = \) Pa s.

Dimensions: \(\dfrac{\mathrm{MLT^{-2}}}{\mathrm{L^2 T^{-1}}} = \mathrm{ML^{-1}T^{-1}}\).

Water: \(0.01\) poise \(= 0.01 \times 0.1 = 10^{-3}\) Pa s.

1. Fastest on the axis, at rest at the wall. The layer touching the wall sticks to it. Each layer is dragged back by the slower layer outside it, so the speed grows towards the centre.

2. The gradient is the slope of the profile, which is steepest at the wall and zero on the axis. The viscous stress \(\eta\,dv/dy\) is therefore largest at the wall.

A liquid becomes less viscous when heated. Its viscosity comes from the attraction between neighbouring molecules, and faster molecules break free of each other more easily.

A gas becomes more viscous when heated. Its viscosity comes from molecules crossing between layers and carrying momentum, and hotter molecules cross more often.

With the same \(A\) and the same gradient, \(F \propto \eta\). Four times the force means \(\eta_B = 4\eta_A\).

Stokes' law: \(F = 6\pi\eta r v\) for a small sphere in slow streamline motion.

\(F = 6 \times 3.14 \times 0.1 \times 10^{-3} \times 0.1 = 1.9 \times 10^{-4}\) N.

1. Weight \(W = \dfrac{4}{3}\pi r^3\rho g\) downward. Buoyancy \(F_B = \dfrac{4}{3}\pi r^3\sigma g\) upward. Viscous drag \(F_v = 6\pi\eta r v\) upward, growing with the speed.

2. At the terminal velocity the acceleration is zero: \(W = F_B + F_v\).

   \(\dfrac{4}{3}\pi r^3(\rho - \sigma)g = 6\pi\eta r v_t\), so \(v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}\).

\(v_t = \dfrac{2 \times (3 \times 10^{-3})^2 \times (7500 - 1500) \times 10}{9 \times 1.2} = \dfrac{2 \times 9 \times 10^{-6} \times 6 \times 10^{4}}{10.8} = \dfrac{1.08}{10.8} = 0.1\) m/s.

\(v_t = \dfrac{2 \times (3 \times 10^{-5})^2 \times 1000 \times 10}{9 \times 2 \times 10^{-5}} = \dfrac{1.8 \times 10^{-5}}{1.8 \times 10^{-4}} = 0.1\) m/s.

Cloud droplets are even smaller, and \(v_t \propto r^2\). They drift down at a few millimetres per second, slower than the air currents that lift them.

1. Volume is conserved: \(8 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi R^3\), so \(R^3 = 8r^3\) and \(R = 2r\).

2. \(v_t \propto r^2\). Doubling the radius makes the terminal velocity \(4v_0\).

1. At the start the ball is at rest, so there is no viscous force. Net force \(= W - F_B\), so \(a = g\left(1 - \dfrac{\sigma}{\rho}\right) = 10 \times \dfrac{6000}{7500} = 8\) m/s\(^2\).

2. The graph levels off at \(0.1\) m/s, the same as the terminal velocity of Question 8.

3. As the ball speeds up the drag \(6\pi\eta r v\) grows. The net force, and so the acceleration, shrinks. When the drag equals \(W - F_B\) the acceleration is zero and the speed stays constant.

With \(\rho \approx 0\): \(v_t = \dfrac{2r^2(0 - \sigma)g}{9\eta} = -\dfrac{2 \times 10^{-6} \times 900 \times 10}{0.9} = -0.02\) m/s. The minus sign means upward: \(2\) cm/s.

The viscous force always opposes the motion, so on a rising bubble it acts downward, together with the (tiny) weight, against the buoyancy.

1. \(R_e = \dfrac{1000 \times 0.1 \times 0.02}{10^{-3}} = 2000\). This is at the upper limit of streamline flow; the flow is just streamline.

2. \(R_e \propto v\): \(v = 0.1 \times \dfrac{3000}{2000} = 0.15\) m/s.

\(Q = \dfrac{\pi P r^4}{8\eta l} = \dfrac{\pi \times 800 \times (10^{-3})^4}{8 \times 10^{-3} \times 1} = \dfrac{\pi \times 8 \times 10^{-10}}{8 \times 10^{-3}} = \pi \times 10^{-7} \approx 3.1 \times 10^{-7}\) m\(^3\)/s, about \(0.3\) mL/s.

1. \(Q \propto r^4\): \(2^4 = 16\) times.

2. \(Q \propto \dfrac{1}{l}\): half.

3. \(Q \propto P\): twice.

From \(Q = \dfrac{\pi P r^4}{8\eta l}\), \(P = \dfrac{8\eta l}{\pi r^4}Q\), so \(R_f = \dfrac{8\eta l}{\pi r^4}\), like an electrical resistance with \(P\) as the voltage and \(Q\) as the current.

In series the same \(Q\) passes through both tubes and the pressure differences add. \(R_f \propto \dfrac{1}{r^4}\), so the narrow tube has \(16\) times the resistance of the wide one.

Fraction across the narrow tube \(= \dfrac{16}{16 + 1} = \dfrac{16}{17}\). Almost all the pressure is used up in the narrow tube.

Each tube carries the same \(Q\) as before, so the total flow is twice that of one tube.

Combined resistance \(= \dfrac{P}{2Q} = \dfrac{R_f}{2}\), half that of one tube, as for two equal resistors in parallel.

1. \(Q \propto r^4\): the flow falls to \(\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}\).

2. \(Q \propto P\): the pressure must rise \(16\) times to restore the flow. This is why a narrowed artery raises blood pressure.

Left side: \([Q] = \) m\(^3\)/s.

Right side: \(\dfrac{[P][r^4]}{[\eta][l]} = \dfrac{\text{Pa} \times \text{m}^4}{\text{Pa s} \times \text{m}} = \) m\(^3\)/s. The two sides agree.

1. \(P = h\rho g = 0.10 \times 1000 \times 10 = 1000\) Pa.

2. \(r^4 = (5 \times 10^{-4})^4 = 6.25 \times 10^{-14}\) m\(^4\).

   \(Q = \dfrac{3.14 \times 1000 \times 6.25 \times 10^{-14}}{8 \times 10^{-3} \times 0.2} = \dfrac{1.96 \times 10^{-10}}{1.6 \times 10^{-3}} = 1.2 \times 10^{-7}\) m\(^3\)/s \(= 0.12\) mL/s.

3. \(t = \dfrac{12}{0.12} = 100\) s.

Poiseuille's law assumes streamline flow. A long narrow tube and a small pressure difference keep the speed, and so the Reynolds number, small. The tube must also be long so that the flow settles into its parabolic profile.

At a very large pressure difference the flow becomes turbulent. The eddies waste energy, and the flow rate is less than the formula predicts.