Viscosity of oil from laminar pipe flow

Data. $D = 0.06$ m ($R = 0.03$ m), s.g. 0.8 so $\rho = 800$ kg/m³, $\Delta p = 200$ kPa over $L = 5$ km. Weight collected: 1962 N in 4 min = 240 s.

Discharge

Weight flow rate $= 1962/240 = 8.175$ N/s, so

$$Q = \frac{W}{\rho g} = \frac{8.175}{800(9.81)} = 0.001042\ \text{m}^3/\text{s} = 1.042\ \text{l/s}$$

Dynamic viscosity (laminar flow, Hagen-Poiseuille)

$$\Delta p = \frac{128\mu L Q}{\pi D^4} \;\Rightarrow\; \mu = \frac{\pi D^4\,\Delta p}{128 L Q} = \frac{\pi(0.06)^4(200\,000)}{128(5000)(0.001042)} = 0.01221\ \text{Pa s}$$

Verification that the flow is laminar

$$V = \frac{Q}{\pi D^2/4} = 0.3684\ \text{m/s},\qquad Re = \frac{\rho VD}{\mu} = \frac{800(0.3684)(0.06)}{0.01221} = 1448 < 2000$$

So the flow is laminar and the assumption is correct.

Velocity at 20 mm from the wall

Radius at that point: $r = R - 0.020 = 0.010$ m. For laminar flow

$$u = \frac{1}{4\mu}\frac{\Delta p}{L}\left(R^2 - r^2\right) = \frac{1}{4(0.01221)}\left(\frac{200\,000}{5000}\right)\left(0.03^2 - 0.01^2\right) = 0.655\ \text{m/s}$$

(The centre-line velocity is 0.737 m/s, twice the mean velocity.)

Answer: $\mu = 0.0122$ Pa s; $Re \approx 1448$ (laminar); $u$ at 20 mm from the wall $= 0.655$ m/s.