Fluid and its physical properties

A viscometer measures the dynamic viscosity $\mu$ of a fluid from a measured relation between shear stress and velocity gradient, using Newton's law $\tau=\mu\,\dfrac{du}{dy}$.

Rotating-cylinder (Couette) viscometer

The fluid fills the thin gap $t$ between a fixed outer cylinder and an inner cylinder (radius $R$, immersed height $h$) turned at angular speed $\omega$. The torque $T$ on the inner cylinder is measured.

   fixed outer cylinder
   |   |  fluid gap t |   |
   |   |<->|  inner   |   |
   |   |   | cylinder |   |
   |   |   | (omega)  |   |

For a thin gap the velocity profile is linear, so

$$\tau=\mu\frac{\omega R}{t},\qquad T=\tau\,(2\pi R h)\,R$$ $$\mu=\frac{T\,t}{2\pi\omega R^{3}h}$$

Capillary-tube (Ostwald) viscometer

The time $t_1$ for a fixed volume of the test liquid to flow through a capillary under its own head is compared with the time $t_2$ for a reference liquid (water) of known viscosity:

$$\frac{\mu_1}{\mu_2}=\frac{\rho_1 t_1}{\rho_2 t_2}$$

For laminar flow (Hagen-Poiseuille), $\mu=\dfrac{\pi\,\Delta p\,r^{4}}{8QL}$.

Falling-sphere viscometer

A small sphere of diameter $d$ and density $\rho_s$ falls at terminal velocity $v$ in the liquid (density $\rho$). Balancing weight, buoyancy and Stokes drag $3\pi\mu d v$:

$$\mu=\frac{d^{2}(\rho_s-\rho)\,g}{18\,v}$$

Other types

The Saybolt (efflux) viscometer gives kinematic viscosity from the time taken for 60 mL of oil to flow out through a standard orifice.

Data: $A=1\ \text{m}^2$, $W=350$ N, $V=1.5$ m/s, $t=1\ \text{mm}=0.001$ m, $\tan\theta=5/12$ so $\sin\theta=5/13$.

        /|
       / |  5
      /  |
     /___|
       12
  plate slides down, oil film t = 1 mm

At uniform velocity there is no acceleration, so the shear force from the oil balances the component of weight along the plane:

$$F=W\sin\theta=350\times\frac{5}{13}=134.62\ \text{N}$$

The velocity profile across the thin film is linear:

$$\tau=\mu\frac{V}{t}=\frac{F}{A}$$ $$\mu=\frac{F\,t}{A\,V}=\frac{134.62\times0.001}{1\times1.5}=0.0897\ \text{Pa·s}$$

Answer: $\mu\approx0.0897\ \text{N·s/m}^2$ (0.897 poise).

No, the water does not overflow. If the tube is shorter than the calculated rise, the meniscus simply becomes less curved so that the pulling force is still balanced by the weight of the (shorter) column.

For a tube of radius $r$ with contact angle $\theta$, equating the vertical surface-tension pull to the weight of the column gives

$$2\pi r\sigma\cos\theta=\rho g\,\pi r^{2}h\quad\Rightarrow\quad h=\frac{2\sigma\cos\theta}{\rho g r}$$

If the tube height is $h_t<h$, the water rises to the top and the meniscus adjusts to a new contact angle $\theta'$ (larger than $\theta$) such that

$$\cos\theta'=\frac{\rho g r\,h_t}{2\sigma}<\cos\theta$$

Equivalently, the radius of curvature of the meniscus becomes $R'=r/\cos\theta'$, which is larger than before. The column height stays $h_t$ and the water stays in the tube. Only if the tube were so short that $\cos\theta'$ would have to exceed the limit of a flat surface could the liquid reach the rim, and even then it is held by surface tension at the edge rather than flowing over.

Data: $\mu=0.7$ Pa·s, $A=0.25\times0.25=0.0625\ \text{m}^2$, $V=0.15$ m/s. Total gap $=25$ mm; the plate is 6 mm from one wall, so its distance from the other wall is $25-6=19$ mm (plate thickness neglected).

 wall |<-- 6 mm -->[plate]<---- 19 mm ---->| wall

With linear velocity profiles on both sides, the shear stresses are

$$\tau_1=\mu\frac{V}{y_1}=0.7\times\frac{0.15}{0.006}=17.5\ \text{N/m}^2$$ $$\tau_2=\mu\frac{V}{y_2}=0.7\times\frac{0.15}{0.019}=5.526\ \text{N/m}^2$$

Forces on the two faces (both oppose the motion):

$$F_1=\tau_1A=17.5\times0.0625=1.094\ \text{N},\qquad F_2=\tau_2A=0.345\ \text{N}$$ $$F=F_1+F_2=1.094+0.345=1.439\ \text{N}$$

Answer: the force exerted by the liquid on the plate is about 1.44 N, opposing the motion.

A Newtonian fluid obeys $\tau=\mu\,\dfrac{du}{dy}$, so the ratio $\tau/(du/dy)$ is constant and $\tau$ against $du/dy$ is a straight line through the origin.

Apparent viscosity $\mu_a=\tau\big/\dfrac{du}{dy}$ for each reading:

The ratio is not constant: it falls steadily from 0.0178 to 0.0047 Pa·s as the shear rate rises, so the plot of $\tau$ against $du/dy$ is a curve, not a straight line.

A log-log fit of the data gives the power-law (Ostwald-de Waele) form

$$\tau=K\left(\frac{du}{dy}\right)^{n},\qquad n\approx0.74,\quad K\approx0.0196\ \text{Pa·s}^{n}$$

Since $n<1$ and the apparent viscosity decreases with shear rate, the liquid is non-Newtonian, of the pseudoplastic (shear-thinning) type.

Reading of the figure: the cone has a full apex angle of $60^\circ$ (half-angle $\alpha=30^\circ$). The wetted frustum extends from 40 mm to 100 mm measured along the axis from the apex (40 mm to the small end, and 60 mm height of frustum). Gap measured normal to the surface $t=0.2$ mm.

Data: $\mu=2$ poise $=0.2$ Pa·s, $N=100$ rpm, so $\omega=\dfrac{2\pi\times100}{60}=10.472$ rad/s.

        apex
        /\
       /  \   z1 = 40 mm
      /----\
     /      \  z2 - z1 = 60 mm
    /________\
    gap t = 0.2 mm, 60 deg apex

Take an element at axial distance $z$ from the apex:

• radius $r=z\tan\alpha$
• slant length $ds=\dfrac{dz}{\cos\alpha}$
• shear stress $\tau=\mu\dfrac{\omega r}{t}$

$$dT=\tau\,(2\pi r\,ds)\,r=\frac{2\pi\mu\omega}{t}\,\frac{r^{3}}{\cos\alpha}\,dz$$ $$T=\frac{2\pi\mu\omega\tan^{3}\alpha}{t\cos\alpha}\int_{z_1}^{z_2}z^{3}dz=\frac{\pi\mu\omega\tan^{3}\alpha}{2\,t\cos\alpha}\left(z_2^{4}-z_1^{4}\right)$$

Substituting ($\tan30^\circ=0.5774$, $\cos30^\circ=0.8660$, $z_1=0.04$ m, $z_2=0.10$ m):

$$T=\frac{\pi(0.2)(10.472)(0.5774)^{3}}{2(0.0002)(0.8660)}\left(0.10^{4}-0.04^{4}\right)=0.356\ \text{N·m}$$

Answer: $T\approx0.356\ \text{N·m}$. (If the figure dimensions differ, use the same formula with the correct $z_1,z_2$.)

Capillarity is the rise or fall of a liquid in a narrow tube relative to the surrounding free surface, caused by surface tension and the contact angle between the liquid and the tube wall.

• If cohesion is weaker than adhesion (water-glass, $\theta<90^\circ$), the liquid wets the wall and rises.
• If cohesion is stronger (mercury-glass, $\theta>90^\circ$), the liquid is depressed.

Balancing the vertical pull of surface tension against the weight of the column in a tube of diameter $d$:

$$h=\frac{4\sigma\cos\theta}{\rho g d}$$

The rise is larger for smaller tubes. Examples: rise of water in soil pores and plant stems, wick action, and errors in piezometer and manometer readings when the tube is very narrow.

Data: gap $=2.2$ cm, plate thickness $=0.2$ cm, so the clearance on each side is $\dfrac{2.2-0.2}{2}=1.0$ cm $=0.01$ m. $A=1.5\times1.5=2.25\ \text{m}^2$, $V=0.15$ m/s, $\mu=1.75$ N·s/m², $W=40$ N, $\rho=0.9\times1000=900\ \text{kg/m}^3$.

  wall |  1 cm | plate | 1 cm |  wall
       |  oil  | 0.2cm | oil  |
                  ^ lifted at 0.15 m/s

Viscous drag on both faces (linear profile):

$$F_v=2\mu A\frac{V}{t}=2\times1.75\times2.25\times\frac{0.15}{0.01}=118.13\ \text{N}$$

Buoyancy (plate volume $=1.5\times1.5\times0.002=0.0045\ \text{m}^3$):

$$F_b=\rho g V_p=900\times9.81\times0.0045=39.73\ \text{N}$$

Force balance for upward motion at constant velocity:

$$F+F_b=W+F_v\ \Rightarrow\ F=40-39.73+118.13=118.39\ \text{N}$$

Answer: $F\approx118.4$ N upward (about 118 N; if buoyancy is ignored, $F=158.1$ N).

Data: shaft radius $R=8$ cm $=0.08$ m, bearing radius $8.025$ cm, so the clearance $t=0.025$ cm $=0.25$ mm. $L=0.30$ m, $N=200$ rpm, $\mu=2.5$ poise $=0.25$ Pa·s.

   bearing  |<-t->| shaft
   (fixed)  |     | R = 8 cm, 200 rpm

Angular speed and surface velocity:

$$\omega=\frac{2\pi\times200}{60}=20.944\ \text{rad/s},\qquad V=\omega R=1.6755\ \text{m/s}$$

Shear stress (linear profile across the thin film):

$$\tau=\mu\frac{V}{t}=0.25\times\frac{1.6755}{0.00025}=1675.5\ \text{N/m}^2$$

Shear force over the shaft surface $A=2\pi RL=2\pi(0.08)(0.30)=0.1508\ \text{m}^2$:

$$F=\tau A=1675.5\times0.1508=252.7\ \text{N}$$

Torque:

$$T=F\,R=252.7\times0.08=20.21\ \text{N·m}$$

Answer: $T\approx20.2\ \text{N·m}$.

Control volume

A control volume is a fixed region of space, of chosen shape and size, through whose boundary (the control surface) fluid may flow in and out. Instead of following a particular mass (system approach), we study what enters and leaves the region. It is the basis of the Eulerian approach, used to write the continuity, momentum and energy equations for flow through pipes, nozzles, turbines and bends.

Continuum

Fluids are made of molecules with empty spaces between them. The continuum concept ignores this and treats the fluid as continuously distributed matter, so properties such as density, pressure and velocity are smooth functions of position and time and can be differentiated. It is valid when the smallest length of interest is much larger than the mean free path of the molecules (Knudsen number $\ll1$). It fails for rarefied gases at very high altitude.

Viscosity

Viscosity is the property of a fluid by which it resists relative motion between its layers. By Newton's law of viscosity, the shear stress is proportional to the velocity gradient:

$$\tau=\mu\frac{du}{dy}$$

where $\mu$ is the dynamic viscosity (unit Pa·s = N·s/m²; 1 poise = 0.1 Pa·s). The kinematic viscosity is $\nu=\mu/\rho$ (m²/s; 1 stokes = $10^{-4}\ \text{m}^2/\text{s}$). Viscosity of liquids decreases and that of gases increases with temperature.

Surface tension

Surface tension $\sigma$ is the tensile force per unit length acting along the free surface of a liquid (N/m), due to cohesion between molecules.

Liquid droplet of diameter $d$: the pressure force over the cut section balances the surface-tension force round its circumference:

$$\Delta p\,\frac{\pi d^{2}}{4}=\sigma\,\pi d\ \Rightarrow\ \Delta p=\frac{4\sigma}{d}$$

For a soap bubble (two surfaces), $\Delta p=\dfrac{8\sigma}{d}$.

Capillarity

In a vertical tube of diameter $d$ dipped in a liquid with contact angle $\theta$, the upward component of the surface-tension force round the circumference supports the weight of the raised column of height $h$:

$$\sigma\cos\theta\,(\pi d)=\rho g\,\frac{\pi d^{2}}{4}\,h$$ $$h=\frac{4\sigma\cos\theta}{\rho g d}$$

For water in glass $\theta\approx0^\circ$ (rise); for mercury $\theta\approx130^\circ$ so $h$ is negative (depression).

Data: inner radius $R=0.075$ m, outer radius $0.0755$ m, so the gap $t=0.5$ mm $=0.0005$ m. Height $h=0.25$ m, $T=12$ N·m, $N=100$ rpm, so $\omega=\dfrac{2\pi\times100}{60}=10.472$ rad/s.

Surface velocity of the inner cylinder: $V=\omega R=10.472\times0.075=0.7854$ m/s.

For a thin gap the profile is linear: $\tau=\mu\dfrac{V}{t}$. The torque on the inner cylinder (bottom neglected):

$$T=\tau\,(2\pi Rh)\,R=\mu\frac{\omega R}{t}\,2\pi R^{2}h$$ $$\mu=\frac{T\,t}{2\pi\omega R^{3}h}=\frac{12\times0.0005}{2\pi(10.472)(0.075)^{3}(0.25)}$$ $$\mu=0.8646\ \text{Pa·s}$$

Answer: $\mu\approx0.865\ \text{N·s/m}^2$ (8.65 poise).

Data: $V_0=0.5\ \text{m}^3$, test pressure $\Delta p=3000\ \text{bar}=300\ \text{MPa}$, $K=2000$ MPa, the empty vessel expands by 0.6 %.

Step 1: volume of the vessel at 3000 bar (expanded by 0.6 %):

$$V_v=0.5\times1.006=0.503\ \text{m}^3$$

Step 2: compression of water. From $K=-\dfrac{\Delta p}{\Delta V/V}$:

$$\frac{\Delta V}{V}=\frac{\Delta p}{K}=\frac{300}{2000}=0.15$$

So the water occupies 85 % of its original (atmospheric) volume at 3000 bar. The water must fill the vessel (0.503 m³) at that pressure:

$$V_w(1-0.15)=0.503\ \Rightarrow\ V_w=\frac{0.503}{0.85}=0.5918\ \text{m}^3$$

Step 3: mass of water (density of water at atmospheric pressure 1000 kg/m³):

$$m=\rho V_w=1000\times0.5918=591.8\ \text{kg}$$

Answer: about 592 kg of water must be pumped in (about 92 kg more than the 500 kg that would fill the vessel at atmospheric pressure).

Data: $d_1=1.5$ mm, $d_2=2$ mm, $\sigma=0.075$ N/m, $\theta=0^\circ$, difference of menisci $\Delta h=2$ mm $=0.002$ m.

Capillary rise in each limb: $h=\dfrac{4\sigma\cos\theta}{\rho g d}$. The narrower limb rises higher, so

$$\Delta h=h_1-h_2=\frac{4\sigma}{\rho g}\left(\frac{1}{d_1}-\frac{1}{d_2}\right)$$ $$\rho=\frac{4\sigma}{g\,\Delta h}\left(\frac{1}{d_1}-\frac{1}{d_2}\right)=\frac{4(0.075)}{9.81(0.002)}\left(\frac{1}{0.0015}-\frac{1}{0.002}\right)$$ $$\rho=\frac{0.3}{0.01962}\times166.67=2548\ \text{kg/m}^3$$

Answer: $\rho\approx2548\ \text{kg/m}^3$.