Question 12 of 13: Draining-Time Comparison — Upright vs Inverted Conical Tanks

Question 12: Draining-Time Comparison — Upright vs Inverted Conical Tanks (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.

Both tanks obey Torricelli's law at the discharge hole, dV/dt = −CdA0√(2gy), where y is the instantaneous water-surface height above the hole; the two tanks differ only in how their cross-sectional area A(y) varies with y, which changes how the remaining volume relates to the (shrinking) driving head.

1. Cross-sectional area as a function of height above the hole, for each tank (radius R, full height H). Tank A (apex at the hole, radius grows linearly with height): $A_A(y) = \pi R^2(y/H)^2$. Tank B (apex at the top, radius shrinks linearly with height above the hole): $A_B(y) = \pi R^2(1-y/H)^2$.
2. Draining-time integral. Since $A(y)\,dy = -C_dA_0\sqrt{2gy}\,dt$, the total time to drain from y=H to y=0 is $$t = \frac{1}{C_dA_0\sqrt{2g}}\int_0^H \frac{A(y)}{\sqrt y}\,dy$$
3. Evaluate the integral for each tank (k=πR²/H² common to both). $$t_A \propto \int_0^H \frac{y^2}{\sqrt y}\,dy = \int_0^H y^{3/2}\,dy = \tfrac{2}{5}H^{5/2}$$ $$t_B \propto \int_0^H \frac{(H-y)^2}{\sqrt y}\,dy = H^{5/2}\,B(3,\tfrac12) = \tfrac{16}{15}H^{5/2}$$ (the second integral evaluated via the Beta function, $B(3,\tfrac12)=\Gamma(3)\Gamma(\tfrac12)/\Gamma(\tfrac72)=16/15$).
4. Compare. $$\frac{t_B}{t_A} = \frac{16/15}{2/5} = \boxed{\frac{8}{3} \approx 2.67}$$ Tank B takes 8/3 times as long as Tank A, so Tank A (upright, apex-down) empties faster — it drains in only 3/8 of Tank B's time.