Question 5 of 13: Viscosity and Tank Equilibrium Level — Gasoline vs. Lubricating Oil

Question 5: Viscosity and Tank Equilibrium Level — Gasoline vs. Lubricating Oil (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) μ_gasoline, μ_oil at 20°C; (b) equilibrium level, gasoline; (c) equilibrium level, oil; (d) comment.

Approach. (a) read absolute viscosity directly off the attached chart at 20°C; (b)/(c) at equilibrium, inflow equals orifice outflow, so solve the standard orifice equation Q = C_dA_o√(2gh) for h; (d) compare the two required levels against the tank's actual 500 mm height.

1. Part (a) — Read viscosities from the Absolute Viscosity chart at 20°C. Follow the vertical 20°C line up to the "Gasoline, s=0.716" curve and the "SAE 30 Eastern lubricating oil" curve: $$\mu_{gasoline} \approx 2.9\times10^{-4}\ \text{N}\cdot\text{s/m}^2, \qquad \mu_{oil} \approx 3.8\times10^{-1}\ \text{N}\cdot\text{s/m}^2$$ Oil is roughly 1300 times more viscous than gasoline at the same temperature.
2. Part (b) — Orifice area and gasoline equilibrium level. Solving the orifice equation $Q_{in}=C_dA_o\sqrt{2gh}$ for h, with $A_o=\frac{\pi}{4}(0.010)^2=7.854\times10^{-5}\ \text{m}^2$: $$h = \frac{1}{2g}\left(\frac{Q_{in}}{C_dA_o}\right)^2 = \frac{1}{19.62}\left(\frac{2.0\times10^{-4}}{0.90(7.854\times10^{-5})}\right)^2 = \frac{1}{19.62}(2.829)^2 = \boxed{0.408\ \text{m} = 408\ \text{mm}}$$ This is comfortably inside the 500 mm tank.
3. Part (c) — Oil equilibrium level (same formula, $C_d=0.75$): $$h = \frac{1}{19.62}\left(\frac{2.0\times10^{-4}}{0.75(7.854\times10^{-5})}\right)^2 = \frac{1}{19.62}(3.395)^2 = \boxed{0.588\ \text{m} = 588\ \text{mm}}$$
4. Part (d) — Compare and check the Reynolds numbers through the orifice. The orifice exit velocities are $V_{gas}=2.83$ m/s and $V_{oil}=3.40$ m/s, giving $$Re_{gas} = \frac{\rho_{gas}V_{gas}d_o}{\mu_{gas}} = \frac{716(2.83)(0.010)}{2.9\times10^{-4}} \approx 7.0\times10^4\ \text{(turbulent, high }C_d\text{)}$$ $$Re_{oil} = \frac{\rho_{oil}V_{oil}d_o}{\mu_{oil}} = \frac{891(3.40)(0.010)}{3.8\times10^{-1}} \approx 80\ \text{(deep laminar/viscous regime, low }C_d\text{)}$$ The oil's much lower orifice Reynolds number is exactly why its discharge coefficient is lower (0.75 vs. 0.90) — viscous losses dominate, so a higher head is needed to push the same flow through. Since the required oil level (588 mm) EXCEEDS the tank's actual height (500 mm), the oil tank never reaches a true equilibrium: it fills to the rim and overflows continuously, whereas the gasoline tank settles cleanly at 408 mm.