A1 — Time to drain a tank through a pipe: 23-Chem-B1 Transport Phenomena, May 2018, Q1

Question 1: A1 — Time to drain a tank through a pipe (25 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.

Quantity	Symbol	Value
Tank diameter	$D_t$	$16\ \text{ft}$
Pipe length, inside diameter	$L,\ D$	$42\ \text{ft},\ 4\ \text{in}=0.3333\ \text{ft}$
Initial head (tank surface above reservoir)	$z_1$	$26\ \text{ft}$
Volume to discharge	$\mathcal{V}$	$1500\ \text{ft}^3$
Darcy friction factor	$f$	$0.008\ (f_{\text{Fanning}}=0.002)$
Entrance / exit loss coefficients	$K_{en},K_{ex}$	$0.5,\ 1.0$

Find. The time $t$ required for the falling tank level to pass $1500\ \text{ft}^3$ of water through the pipe into the reservoir.

Fig. A1: Water drains from the tank through a single pipe into a large reservoir. The instantaneous head $z$ (difference of the two free surfaces) drives the pipe flow; as the tank empties, $z$ falls and the flow slows — a quasi-steady draining problem.

Approach. At each instant apply the steady mechanical-energy balance between the two free surfaces to get the pipe velocity as a function of the current head $z$, then equate the tank’s volume-loss rate to the pipe throughput and integrate that ordinary differential equation over the required volume.

Quasi-steady head–velocity relation. Between the tank surface and the reservoir surface (both at atmospheric pressure, both with negligible surface velocity), the energy balance charges the head $z$ against friction plus minor losses: $$z=\left(f\frac{L}{D}+K_{en}+K_{ex}\right)\frac{V^2}{2g}.$$ The bracket collects all resistances as an equivalent loss coefficient.
Evaluate the loss coefficient. $$f\frac{L}{D}+K_{en}+K_{ex}=0.008\cdot\frac{42}{0.3333}+0.5+1.0=1.008+0.5+1.0=2.508.$$ Solving the balance for the velocity, $$V=\sqrt{\frac{2g\,z}{2.508}}=\sqrt{25.68\,z}\quad(\text{ft/s, }z\text{ in ft}),$$ using $g=32.2\ \text{ft/s}^2$. At the start ($z=26$ ft) $V=25.8\ \text{ft/s}$; at the end ($z=18.5$ ft) $V=21.8\ \text{ft/s}$.
Relate the falling level to the head. The reservoir is large, so its surface is essentially fixed and $z$ falls only because the tank level drops. Discharging $1500\ \text{ft}^3$ lowers the tank a distance $$\Delta z=\frac{\mathcal V}{A_t}=\frac{1500}{\frac{\pi}{4}(16)^2}=\frac{1500}{201.06}=7.46\ \text{ft},$$ so the head runs from $z_1=26$ ft to $\boxed{z_2=26-7.46=18.54\ \text{ft}.}$
Volume balance on the tank. The rate the tank loses liquid equals the pipe throughput: $$-A_t\,\frac{dz}{dt}=A_p V=A_p\sqrt{K\,z},\qquad K=\frac{2g}{2.508}=25.68.$$ Separating variables, $$dt=-\frac{A_t}{A_p\sqrt{K}}\,\frac{dz}{\sqrt z}.$$
Integrate over the drawdown. With $A_t/A_p=\dfrac{201.06}{0.08727}=2304$ and $\sqrt K=5.067$, $$t=\frac{A_t}{A_p\sqrt K}\int_{z_2}^{z_1}\frac{dz}{\sqrt z}=\frac{2304}{5.067}\Big[2\big(\sqrt{z_1}-\sqrt{z_2}\big)\Big].$$
Evaluate. $2(\sqrt{26}-\sqrt{18.54})=2(5.099-4.306)=1.586$, hence $$t=\frac{2304}{5.067}\times1.586\ \Longrightarrow\ \boxed{t\approx721\ \text{s}\approx12.0\ \text{min}.}$$

Quantity	Result
Equivalent loss coefficient	$fL/D+K_{en}+K_{ex}=2.508$
Level drop for 1500 ft³	$\Delta z=7.46\ \text{ft}$ ($z:26\to18.54$ ft)
Pipe velocity (start → end)	$25.8\to21.8\ \text{ft/s}$
Discharge time	$t\approx721\ \text{s}\approx12.0\ \text{min}$

Check — constant f, fixed reservoir, submerged discharge

The Darcy $f=0.008$ is taken constant (given); at these velocities the flow is strongly turbulent so $f$ is nearly Reynolds-independent, and the tank draws down only 29 %, so the assumption is sound. The reservoir surface is assumed stationary (large reservoir) and the pipe discharges submerged, giving the exit-loss coefficient $K_{ex}=1.0$ (full velocity head dissipated); a free jet to atmosphere would instead retain the exit velocity head — the same numeric bracket here.

Concept & Theory

Topic. Quasi-steady draining of a tank through a pipe with friction and minor (entrance/exit) losses.

Key relations. Mechanical-energy balance $z=\big(fL/D+\sum K\big)V^2/2g$; Darcy–Weisbach friction $h_f=f\frac{L}{D}\frac{V^2}{2g}$; tank continuity $-A_t\,dz/dt=A_pV$; integrated draw-down time $t=\dfrac{2A_t}{A_p\sqrt K}\big(\sqrt{z_1}-\sqrt{z_2}\big)$.

Why this works. The pipe transit time is far shorter than the draining time, so at every instant the flow is effectively steady at the current head (“quasi-steady”). Because $V\propto\sqrt z$, the falling level obeys a separable first-order ODE whose $\int z^{-1/2}dz$ gives the characteristic $\sqrt{z_1}-\sqrt{z_2}$ draining law.

Common pitfall. Mixing Darcy and Fanning factors ($f_{Darcy}=4f_{Fanning}$) — the paper gives both to avoid this; using $0.002$ in $fL/D$ would understate friction fourfold. Also forgetting that $z$ (not the absolute tank depth) is the driving head, and omitting the entrance ($0.5$) and exit ($1.0$) coefficients.

Source. Welty et al., Momentum, Heat and Mass Transfer, Ch. 13 (pipe friction and minor losses); White, Fluid Mechanics, §3.7 / Ch. 6 (tank-draining energy balance).

Question 1 of 6: A1 — Time to drain a tank through a pipe

Question 1: A1 — Time to drain a tank through a pipe (25 marks)