Elevated-Tank Discharge — Required Head: 23-Chem-A2 Unit Operations and Separation Processes, May 2013, Q1

Question A1: Elevated-Tank Discharge — Required Head (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. Water at 180 °F; target volumetric discharge $Q=100$ US gal/min; schedule-40 commercial steel throughout.

Quantity	Value
Discharge rate $Q$	100 US gal/min $=0.2228\ \mathrm{ft^3/s}$
4-in Sch-40 inside diameter / area	4.026 in / $0.0884\ \mathrm{ft^2}$
2-in Sch-40 inside diameter / area	2.067 in / $0.02330\ \mathrm{ft^2}$
Water $\rho$ / $\mu$ @ 180 °F	$60.55\ \mathrm{lb/ft^3}$ / $2.32\times10^{-4}\ \mathrm{lb/(ft\,s)}$
Roughness $\varepsilon$ (commercial steel)	$1.5\times10^{-4}$ ft
Lengths: 4-in / 2-in	20 ft / $125+10+50=185$ ft

Find. the vertical height $H$ of the free surface (point 1) above the open discharge (point 2) needed to drive 100 gal/min through the line.

Figure A1 — Discharge line reconstructed from Fig. 1. A 4-in downcomer (low velocity) drops 20 ft, contracts to the 2-in working line (185 ft total with three 90° elbows), which sets almost all of the friction loss; $H$ is the free-surface-to-discharge elevation.

Approach. Write the mechanical-energy (Bernoulli) balance between the free surface (1) and the open jet (2); with both at atmospheric pressure and $V_1\!\approx\!0$, the required elevation head equals the exit velocity head plus all major (pipe friction) and minor (contraction, elbows) losses, evaluated mostly at the fast 2-in velocity.

Velocities from continuity. With $Q=0.2228\ \mathrm{ft^3/s}$, $$v_2=\frac{Q}{A_2}=\frac{0.2228}{0.02330}=9.56\ \mathrm{ft/s},\qquad v_4=\frac{Q}{A_4}=\frac{0.2228}{0.0884}=2.52\ \mathrm{ft/s}.$$ The 2-in velocity head is $v_2^2/2g=1.42$ ft; the 4-in head is only $0.099$ ft, so the 4-in section is almost inert.
Reynolds number and Fanning factor (2-in line). $$\mathrm{Re}_2=\frac{\rho v_2 D_2}{\mu}=\frac{60.55(9.56)(0.1723)}{2.32\times10^{-4}}=4.30\times10^{5}.$$ With $\varepsilon/D_2=8.7\times10^{-4}$, Colebrook gives $f_{\text{Fanning}}=0.0049$ (Fanning; $=f_{\text{Darcy}}/4$), consistent with Fig. A1. The 4-in line gives $f\approx0.0052$.
Major (pipe friction) losses. Using $h_f=4f\dfrac{L}{D}\dfrac{v^2}{2g}$: $$h_{f,2}=4(0.0049)\frac{185}{0.1723}(1.42)=29.9\ \mathrm{ft},\qquad h_{f,4}=4(0.0052)\frac{20}{0.3355}(0.099)=0.12\ \mathrm{ft}.$$
Minor losses. Sudden contraction 4→2 (Eq. A2), $k_c=0.42\!\left(1-\beta^2\right)=0.42\!\left[1-(0.513)^2\right]=0.309$; three standard 90° elbows (Table A3), $k=0.75$ each; tank entrance to the 4-in, $k\approx0.42$. Evaluated at the relevant velocity heads: $$h_m=(0.309+3\times0.75)(1.42)+0.42(0.099)=3.63\ \mathrm{ft}.$$
Assemble the energy balance. The free surface must supply the exit kinetic energy plus every loss: $$H=\underbrace{\frac{v_2^2}{2g}}_{1.42}+\underbrace{h_{f,2}+h_{f,4}}_{30.0}+\underbrace{h_m}_{3.63}\;\Rightarrow\;\boxed{H\approx35.3\ \mathrm{ft}\;(10.8\ \mathrm{m})}.$$

Quantity	Result
2-in / 4-in velocity	9.56 / 2.52 ft/s
$\mathrm{Re}$ (2-in) / $f_{\text{Fanning}}$	$4.30\times10^{5}$ / 0.0049
2-in pipe friction loss	29.9 ft
Minor losses (contraction + 3 elbows + entrance)	3.63 ft
Required head $H$	≈ 35.3 ft (10.8 m)

Check — routing assumption Fig. 1 is a schematic; the 4-in/2-in split and the three 90° elbows are read from the figure. The 2-in friction term dominates (85% of $H$), so $H$ is insensitive to the exact 4-in length or entrance model but scales with the assumed 2-in run (185 ft) and elbow count. State the routing explicitly per the exam’s “state all assumptions” instruction.

Concept & Theory

Topic. Steady incompressible pipe flow — mechanical-energy balance with major and minor losses.

Key relations. $\dfrac{P_1}{\rho g}+z_1+\dfrac{v_1^2}{2g}=\dfrac{P_2}{\rho g}+z_2+\dfrac{v_2^2}{2g}+h_L$; Fanning friction $h_f=4f\frac{L}{D}\frac{v^2}{2g}$; minor loss $h_m=k\frac{v^2}{2g}$; contraction $k_c=0.42(1-\beta^2)$.

Why this works. Both ends are open to atmosphere and the tank surface is essentially stationary, so the entire elevation head is spent on the exit kinetic energy and irreversibilities. Because loss scales with $v^2$ and the 2-in line carries $\sim\!15\times$ the velocity head of the 4-in, the small-bore section controls the design — a general lesson in sizing gravity lines.

Common pitfall. Forgetting to retain the exit velocity head $v_2^2/2g$ (point 2 is a free jet, not a reservoir), or mixing Fanning and Darcy factors — the Fanning form needs the factor of 4.

Source. McCabe, Smith & Harriott, Unit Operations, Ch. 5 (friction, Tables 5.1–5.2); de Nevers, Fluid Mechanics for Chemical Engineers, Ch. 6.

Question 1 of 6: Elevated-Tank Discharge — Required Head

Section A — Mechanical Operations

Question A1: Elevated-Tank Discharge — Required Head (25 marks)