Exercises — Viscosity — dynamic μ, kinematic ν = μ - ρ; Newtonian vs non-Newtonian

(a) $\nu =\frac{\mu }{\rho }=\frac{0.90}{900}=1.0\times 10^{-3}{\text{m}}^2/\text{s}$. Why this operation: $\nu$ is defined as dynamic viscosity divided by density — that division converts a force-measure into a momentum-diffusion-measure (units m²/s). (b) The reasoning is wrong. Being Newtonian means $\mu$ is constant with shear rate, not that $\mu$ is large. Honey is very viscous and Newtonian; ketchup is less viscous but non-Newtonian. Magnitude of $\mu$ says nothing about Newtonian-ness.

Rearrange: $\mu =\frac{\tau }{\dot{\gamma }}$.

• $\tau$ is a stress = force/area = $\frac{\text{N}}{{\text{m}}^2}=\frac{{\text{kg⋅m/s}}^2}{{\text{m}}^2}=\frac{\text{kg}}{{\text{m⋅s}}^2}$.
• $\dot{\gamma }$ is a gradient = (m/s)/m = ${\text{s}}^{-1}$. So $\mu =\frac{\text{kg}/({\text{m⋅s}}^2)}{1/\text{s}}=\frac{\text{kg}}{{\text{m⋅s}}^2}\cdot \text{s}=\frac{\text{kg}}{\text{m⋅s}}$. ✓ (This equals Pa·s.)

$\dot{\gamma }=du/dy=0$ because there is no difference in speed between layers. Then $\tau =\mu \times 0=0$ Pa. Viscous stress needs a gradient, not a speed — the whole slab could move at any speed with zero internal friction.

Look at Figure 1 (above): the fluid speed rises in a straight line from $0$ at the fixed floor to $U$ at the moving lid, so the gradient (the tilt of the red arrows' tips) is constant everywhere. (a) $\dot{\gamma }=\frac{U}{h}=\frac{0.6}{0.0015}=400{\text{s}}^{-1}$. Why $U/h$: a straight-line profile has slope = (top speed − bottom speed)/(gap) = $U/h$. (b) $\tau =\mu \dot{\gamma }=0.65\times 400=260\text{Pa}$. Why: Newton's law of viscosity. (c) $F=\tau A=260\times 0.4=104\mathbf{N}$. Why: stress is force per area, so force = stress × area.

Build the forward chain first, one physical link at a time, then invert it:

1. Force spreads over area to make a stress: $\tau =F/A$. (Bigger plate → same force gives less stress.)
2. That stress is carried by the velocity gradient through Newton's law: $\tau =\mu \dot{\gamma }$.
3. For the straight Couette profile the gradient is $\dot{\gamma }=U/h$. Substitute 3 into 2 and set equal to 1: $\frac{F}{A}=\mu \frac{U}{h}$. Only $\mu$ is unknown, so solve for it: $\mu =\frac{Fh}{UA}=\frac{30\times 0.002}{0.4\times 0.25}=\frac{0.06}{0.1}=0.6\text{Pa⋅s}.$ Why the algebra is legal: every physical link is a genuine equality, so rearranging them is just isolating the one quantity we don't know — no approximation is hidden. Sanity check the units: $\frac{\text{N}\cdot \text{m}}{(\text{m/s})\cdot {\text{m}}^2}=\frac{\text{N⋅s}}{{\text{m}}^2}=\text{Pa⋅s}$ ✓.

${\nu }_{\text{gly}}=\frac{1.5}{1260}=1.19\times 10^{-3}{\text{m}}^2/\text{s}$. ${\nu }_{\text{water}}=\frac{1.0\times 10^{-3}}{1000}=1.0\times 10^{-6}{\text{m}}^2/\text{s}$. Ratio $\approx 1190$. Glycerin spreads momentum faster (larger $\nu$) — even though it feels far gooier — because $\nu$ measures diffusion of momentum, and glycerin's huge $\mu$ wins over its modestly larger $\rho$.

Figure 2 shows this paint as the magenta curve, bending below the straight Newtonian line — so its slope-through-origin drops as we shear harder. ${\mu }_{app}=\frac{K{\dot{\gamma }}^n}{\dot{\gamma }}=K{\dot{\gamma }}^{n-1}=3{\dot{\gamma }}^{-0.6}.$

• At $\dot{\gamma }=1$: ${\mu }_{app}=3\times 1^{-0.6}=3.0\text{Pa⋅s}$.
• At $\dot{\gamma }=100$: ${\mu }_{app}=3\times 100^{-0.6}=3\times 10^{-1.2}=3\times 0.0631=0.189\text{Pa⋅s}$. (Units check: ${\mu }_{app}=K{\dot{\gamma }}^{n-1}$ has units ${\text{Pa⋅s}}^{0.4}\cdot ({\text{s}}^{-1}{)}^{-0.6}={\text{Pa⋅s}}^{0.4}\cdot {\text{s}}^{0.6}=\text{Pa⋅s}$ ✓.) Interpretation: on the wall (near-zero shear) the paint is thick — $3\text{Pa⋅s}$ — so it doesn't drip. On the brush (high shear) it thins to $\sim 0.19\text{Pa⋅s}$ and spreads easily. That's exactly what shear-thinning ($n<1$) buys you.

On Figure 2 this is the orange curve, bending above the straight line — the mirror image of the paint. ${\mu }_{app}=K{\dot{\gamma }}^{n-1}=0.8{\dot{\gamma }}^{0.5}.$

• At $\dot{\gamma }=1$: ${\mu }_{app}=0.8\times 1^{0.5}=0.8\text{Pa⋅s}$.
• At $\dot{\gamma }=100$: ${\mu }_{app}=0.8\times 100^{0.5}=0.8\times 10=8.0\text{Pa⋅s}$. Interpretation: the faster you shear ($n>1$), the thicker it gets. Step slowly and you sink (low shear, thin); stamp fast and it jams solid under your foot (high shear, ~10× thicker) — so you can run across it. This is the full opposite end of the power-law family from the paint in L3.1.

(a) Below the yield stress there is no flow: since $150<200$, $\dot{\gamma }=0$ — the paste stays put in the tube. This is why toothpaste holds its shape on the brush. (b) Above yield, invert: $\dot{\gamma }=\frac{\tau -{\tau }_0}{{\mu }_p}=\frac{350-200}{5}=30{\text{s}}^{-1}$. Why subtract ${\tau }_0$ first: only the stress in excess of the yield value drives flow; the first $200$ Pa is "spent" unlocking the jammed structure before any sliding begins.

(a) ${\nu }_{\text{air}}=\frac{1.8\times 10^{-5}}{1.2}=1.5\times 10^{-5}{\text{m}}^2/\text{s}$. ${\nu }_{\text{water}}=1.0\times 10^{-6}{\text{m}}^2/\text{s}$. Air is 15× larger. Air's tiny $\mu$ is beaten by its even tinier $\rho$. (b) $\ell =\sqrt{\nu t}$.

• Air: $\sqrt{1.5\times 10^{-5}\times 10}=\sqrt{1.5\times 10^{-4}}\approx 0.0122\text{m}\approx 1.2\text{cm}$.
• Water: $\sqrt{1.0\times 10^{-6}\times 10}=\sqrt{10^{-5}}\approx 0.00316\text{m}\approx 3.2\text{mm}$. Momentum "reaches out" farther in air per unit time — Momentum Diffusion governed by $\nu$, not by how gooey the fluid feels.

First $\nu =\mu /\rho =10^{-3}/1000=10^{-6}{\text{m}}^2/\text{s}$. $Re=\frac{UD}{\nu }=\frac{2\times 0.05}{10^{-6}}=\frac{0.1}{10^{-6}}=1.0\times 10^5.$ Since $10^5\gg 2300$, the flow is turbulent. See Reynolds Number: $Re$ is the ratio of inertial push to viscous smoothing; here inertia dominates $100,000$-to-1, so viscosity can't keep the flow orderly.

$R^4=(2\times 10^{-3}{)}^4=1.6\times 10^{-11}{\text{m}}^4$. $Q=\frac{\pi \times 1.6\times 10^{-11}\times 8000}{8\times 0.1\times 1}=\frac{\pi \times 1.28\times 10^{-7}}{0.8}=\pi \times 1.6\times 10^{-7}\approx 5.03\times 10^{-7}{\text{m}}^3/\text{s}.$ If $R$ doubles: $Q\propto R^4$, so $Q$ grows by $2^4=16\times$. The fourth-power on radius is the punchline — a small pipe-widening massively boosts flow, which is why arteries and pipes are so radius-sensitive.

Set up the three vertical forces on the sinking sphere (volume $V=\frac{4}{3}\pi a^3$):

• Weight (down): $W={\rho }_{s}Vg={\rho }_s\frac{4}{3}\pi a^{3}g$.
• Buoyancy (up): $B={\rho }_{f}Vg={\rho }_f\frac{4}{3}\pi a^{3}g$ (weight of displaced glycerin).
• Stokes drag (up, opposes sinking): $F_d=6\pi \mu av$.

At terminal velocity the sphere stops accelerating, so the ups balance the down: $W=B+F_d\Rightarrow \frac{4}{3}\pi a^{3}g({\rho }_s-{\rho }_f)=6\pi \mu a{v}_t.$ Solve for $v_t$ — divide both sides by $6\pi \mu a$: $v_t=\frac{\frac{4}{3}\pi a^{3}g({\rho }_s-{\rho }_f)}{6\pi \mu a}=\frac{4\pi a^{3}g({\rho }_s-{\rho }_f)}{3\cdot 6\pi \mu a}=\frac{2a^2({\rho }_s-{\rho }_f)g}{9\mu }.$ (The $\pi$ cancels, one power of $a$ cancels leaving $a^2$, and $\frac{4}{18}=\frac{2}{9}$.) Evaluate: $a^2=10^{-6}{\text{m}}^2$, ${\rho }_s-{\rho }_f=6540{\text{kg/m}}^3$: $v_t=\frac{2\times 10^{-6}\times 6540\times 9.8}{9\times 1.5}=\frac{0.128184}{13.5}\approx 9.5\times 10^{-3}\text{m/s}(\approx 9.5\text{mm/s}).$ Slow, because glycerin's large $\mu$ sits in the denominator. The $a^2$ shows small spheres fall dramatically slower.

Figure 3 (above) shows the kinked profile: two straight segments with different slopes because the two films have different $\mu$ but must transmit the same $\tau$. Each film obeys Newton's law with its own gradient. Film 1 spans speed $0\to u_i$ over height $h_1$; film 2 spans $u_i\to U$ over $h_2$: $\tau ={\mu }_1\frac{u_i}{h_1}={\mu }_2\frac{U-u_i}{h_2}.$ The velocity drop across each film is $\Delta u=\tau h/\mu$, and the two drops must add up to the full $U$: $U=\tau \left(\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}\right)\Rightarrow \tau =\frac{U}{\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}}.$ Numbers: $\frac{h_1}{{\mu }_1}=\frac{0.001}{0.2}=5\times 10^{-3}$, $\frac{h_2}{{\mu }_2}=\frac{0.001}{0.5}=2\times 10^{-3}$, sum $=7\times 10^{-3}$. $\tau =\frac{0.3}{7\times 10^{-3}}=\frac{0.3}{0.007}\approx 42.857\text{Pa}.$ Interface speed, from film 1: $u_i=\frac{\tau h_1}{{\mu }_1}=\frac{42.857\times 0.001}{0.2}=\frac{0.042857}{0.2}=0.2143\text{m/s}$. Check with film 2: its drop is $U-u_i=0.3-0.2143=0.0857\text{m/s}$, and $\frac{\tau h_2}{{\mu }_2}=\frac{42.857\times 0.001}{0.5}=0.0857$ ✓. The films add like resistors in series ($h/\mu$ = resistance); the more viscous film 2 has the gentler slope (it needs less gradient to carry the same stress).

Same $\tau ={\mu }_{eff}\frac{U}{h}$ as the real stack, so ${\mu }_{eff}=\frac{\tau h}{U}=\frac{42.857\times 0.002}{0.3}=\frac{0.085714}{0.3}=0.2857\text{Pa⋅s}.$ The naive average is $(0.2+0.5)/2=0.35$ — not equal. Series films combine as a thickness-weighted harmonic mean: $\frac{h}{{\mu }_{eff}}=\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}\Rightarrow {\mu }_{eff}=\frac{h_1+h_2}{\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}}=\frac{0.002}{7\times 10^{-3}}=\mathrm{0.2857.}$ The low-viscosity film dominates (it's the "easy" path), pulling the blend below the arithmetic mean — exactly like resistors in series where the smallest slope carries the most deformation.

From $Re=\frac{UD}{\nu }$, solve for $\nu$: $\nu =\frac{UD}{Re}=\frac{1\times 0.02}{500}=4.0\times 10^{-5}{\text{m}}^2/\text{s}.$ Then $\mu =\rho \nu =1200\times 4.0\times 10^{-5}=4.8\times 10^{-2}\text{Pa⋅s}$ ($=48$ cP, a light oil). Reasoning: low $Re$ means viscosity must dominate inertia, so you raise $\nu$ (thicker or lighter fluid). This is the inverse of L4.1 — there we computed $Re$ from a fluid; here we reverse-engineer the fluid from a target $Re$.

L1 answers: $1.0\times 10^{-3}{\text{m}}^2/\text{s}$; unit derivation; $\tau =0$. L2 answers: $\dot{\gamma }=400,\tau =260,F=104\text{N}$; $\mu =0.6$; ${\nu }_{gly}=1.19\times 10^{-3},{\nu }_w=10^{-6}$. L3 answers: ${\mu }_{app}=3.0$ then $0.189$ (thinning); $0.8$ then $8.0$ (thickening); no flow then $30{\text{s}}^{-1}$; ${\nu }_{air}=1.5\times 10^{-5}$, spreads $1.2\text{cm}$ vs water $3.2\text{mm}$. L4 answers: $Re=10^5$ turbulent; $Q\approx 5.03\times 10^{-7}$; $v_t\approx 9.5\text{mm/s}$. L5 answers: $\tau \approx 42.857,u_i\approx 0.2143$; ${\mu }_{eff}=0.2857$; $\nu =4\times 10^{-5},\mu =0.048$.