Hydraulics — Closed Conduit · PE Civil

A. Energy and/or continuity equation

The Energy Equation, EGL and HGL

Continuity and the steady-flow energy equation between two points, the velocity/pressure/elevation heads, pump and turbine head, and how to sketch the energy and hydraulic grade lines.

Every closed-conduit problem on the PE Civil WRE exam is, at bottom, an accounting of energy: water carries pressure head, velocity head, and elevation head, a pump adds head, a turbine removes it, and friction quietly drains it. Master that one balance and the whole topic — pipe friction, pumps, networks — becomes bookkeeping on a single equation. This is also where points are quietly lost: mixing gauge and absolute pressure, forgetting the velocity head when a pipe changes diameter, or sketching the grade lines backwards. The methods here are the NCEES PE Civil Reference Handbook §6.2.1 Principles of One-Dimensional Fluid Flow (Bernoulli §6.2.1.3, energy line §6.2.1.5) and §6.3 Closed Conduit Flow and Pumps.

Continuity: conservation of mass

For steady, incompressible flow, whatever volume of water enters a control volume must leave it: the discharge

Q

is the same at every section of a single pipe. Discharge is velocity times area,

Q = A\,v

, so where the pipe narrows the water must speed up. This is the lever behind every velocity-head change — halve the diameter and the area drops by four, so the velocity quadruples and the velocity head jumps by sixteen. Always compute

v = Q/A

at each section before touching the energy equation.

Q = A_1 v_1 = A_2 v_2 = \text{constant}

The steady-flow energy equation

Between an upstream section 1 and a downstream section 2, the energy equation states that the total head at 1, plus any pump head added, equals the total head at 2, plus the head removed by a turbine, plus the friction and minor losses between them. Each term is an energy per unit weight, carried in feet (or meters) of water. When there is no machine and no loss and the pipe is uniform, this collapses to Bernoulli's equation — a useful idealization, but on the exam you almost always keep

h_L

.

\frac{p_1}{\gamma} + \frac{v_1^2}{2g} + z_1 + h_p = \frac{p_2}{\gamma} + \frac{v_2^2}{2g} + z_2 + h_t + h_L

The three heads

Pressure head

p/\gamma

is the height a column of the fluid would rise in a piezometer tapped into the pipe; with

\gamma = 62.4\,\text{lbf/ft}^3

for water, remember that

1\,\text{psi} \approx 2.31\,\text{ft}

of water. Velocity head

v^2/2g

is the kinetic energy per unit weight — the height the stream would climb if its kinetic energy converted fully to elevation. Elevation head

z

is just the centerline height above an arbitrary datum. Use gauge pressure throughout unless cavitation or absolute conditions force otherwise, and keep one consistent datum for the whole problem.

H = \frac{p}{\gamma} + \frac{v^2}{2g} + z

Pump and turbine head

A pump injects energy, raising the grade line by

h_p

; a turbine extracts it, dropping the line by

h_t

. The hydraulic power transferred to or from the water is

\gamma Q h

, and for a pump the shaft (brake) power is larger by the inverse of efficiency. Keep the sign convention straight: pump head goes on the upstream side of the energy equation (it helps the flow), turbine and loss terms go on the downstream side (they cost the flow). Confusing the two is a classic sign error.

\dot{W}_{\text{fluid}} = \gamma\,Q\,h, \qquad \dot{W}_{\text{shaft,pump}} = \frac{\gamma\,Q\,h_p}{\eta}

Energy and hydraulic grade lines

The energy grade line (EGL) plots the total head

H

along the pipe; the hydraulic grade line (HGL) plots only the piezometric head

p/\gamma + z

. They are separated everywhere by exactly the velocity head,

\text{EGL} - \text{HGL} = v^2/2g

, so in a constant-diameter pipe they run parallel. The EGL always slopes downward in the direction of flow (friction), stepping down abruptly at fittings and jumping up across a pump. Where the HGL falls below the pipe centerline the gauge pressure is negative — a warning sign for cavitation and air release.

\text{EGL} = \frac{p}{\gamma} + \frac{v^2}{2g} + z, \qquad \text{HGL} = \frac{p}{\gamma} + z = \text{EGL} - \frac{v^2}{2g}

Sketching the grade lines

A reliable procedure: start the EGL at the upstream reservoir surface (there the velocity head is negligible, so EGL and HGL meet at the water surface). Step the EGL down by each entrance, fitting, and friction loss as you move downstream; jump it up by

h_p

across a pump. Draw the HGL one velocity head below the EGL — closer in large-diameter, slow reaches and farther below in fast, narrow ones. At a free discharge to atmosphere the HGL meets the pipe centerline because the gauge pressure there is zero.

Exam strategy

Label section 1 and section 2 deliberately — usually two reservoir surfaces or a surface and a discharge — and write the full energy equation before substituting. At a free water surface or a free jet, set

p = 0

(gauge); at a large reservoir, set

v \approx 0

. Pick a datum at the lowest point to keep elevations positive. Confirm units agree: psi to feet via

\times 2.31

, and

g = 32.2\,\text{ft/s}^2

or

9.81\,\text{m/s}^2

. If the question asks for the pressure at an interior point, the fastest route is often the HGL:

p/\gamma = \text{HGL} - z

.

Key equations

Continuity (incompressible)

Q = A_1 v_1 = A_2 v_2

Discharge $Q$ (cfs or m³/s) is constant along a single pipe; $A$ = cross-sectional area, $v$ = mean velocity. Velocity rises where area falls.

Steady-flow energy equation

\frac{p_1}{\gamma} + \frac{v_1^2}{2g} + z_1 + h_p = \frac{p_2}{\gamma} + \frac{v_2^2}{2g} + z_2 + h_t + h_L

Head balance between sections 1 and 2. $h_p$ = pump head added, $h_t$ = turbine head removed, $h_L$ = total head loss, $\gamma$ = specific weight.

Bernoulli equation (ideal)

\frac{p_1}{\gamma} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\gamma} + \frac{v_2^2}{2g} + z_2

Energy equation with no machine and no loss (steady, inviscid, incompressible). Total head is constant.

Total head at a section

H = \frac{p}{\gamma} + \frac{v^2}{2g} + z

Sum of pressure, velocity, and elevation heads (ft or m). This is the height of the EGL above the datum.

Energy and hydraulic grade lines

\text{EGL} - \text{HGL} = \frac{v^2}{2g}

EGL plots total head; HGL plots piezometric head $p/\gamma + z$ . They are separated by the velocity head everywhere.

Pressure head conversion

h = \frac{p}{\gamma}, \qquad 1\ \text{psi} = 2.31\ \text{ft H}_2\text{O}

Converts gauge pressure to feet of water with $\gamma = 62.4\,\text{lbf/ft}^3$ . In SI, $p/\gamma$ with $\gamma = 9.81\,\text{kN/m}^3$ .

Fluid power of a machine

\dot{W}_{\text{fluid}} = \gamma\,Q\,h

Power added by a pump or extracted by a turbine; $h$ = head change across the machine. Divide by $\eta$ for pump shaft power.

Velocity head

h_v = \frac{v^2}{2g}

Kinetic energy per unit weight (ft or m). Grows with the square of velocity; the gap between EGL and HGL.

Worked examples

Pump head and power between two reservoirs

Problem. A pump lifts water from a lower reservoir to an upper reservoir whose surface is $40\,\text{m}$ higher. The total head loss in the piping is $6.5\,\text{m}$ at the design flow $Q = 0.18\,\text{m}^3/\text{s}$ . The pump-motor combination is $78\%$ efficient. Find the required pump head and the electrical input power. Use $\gamma = 9{,}810\,\text{N/m}^3$ .

Solution. Take section 1 at the lower surface and section 2 at the upper surface. Both are open to atmosphere (

p_1 = p_2 = 0

gauge) and large (

v_1 \approx v_2 \approx 0

). Energy equation:

z_1 + h_p = z_2 + h_L

, so

h_p = (z_2 - z_1) + h_L = 40 + 6.5 = 46.5\,\text{m}

. Fluid power:

\gamma Q h_p = 9{,}810 (0.18)(46.5) = 82{,}100\,\text{W}

. Input power:

\dot{W} = 82{,}100 / 0.78 = 105{,}000\,\text{W} = 105\,\text{kW}

. Sanity check: the input (

105\,\text{kW}

) exceeds the fluid power (

82.1\,\text{kW}

) by

1/0.78 = 1.28

, as it must. Final:

h_p \approx 46.5\,\text{m}

,

\dot{W} \approx 105\,\text{kW}

.

Pressure to EGL and HGL elevation

Problem. At a point on a water main the centerline elevation is $100\,\text{ft}$ above datum, the gauge pressure is $50\,\text{psi}$ , and the velocity is $8\,\text{ft/s}$ . Find the pressure head, velocity head, and the elevations of the HGL and EGL at this point. Use $\gamma = 62.4\,\text{lbf/ft}^3$ , $g = 32.2\,\text{ft/s}^2$ .

Solution. Pressure head:

p/\gamma = (50)(144)/62.4 = 7{,}200/62.4 = 115.4\,\text{ft}

(equivalently

50 \times 2.31

). Velocity head:

v^2/2g = 8^2/(2 \cdot 32.2) = 64/64.4 = 0.994\,\text{ft}

. HGL elevation:

z + p/\gamma = 100 + 115.4 = 215.4\,\text{ft}

. EGL elevation:

\text{HGL} + v^2/2g = 215.4 + 0.994 = 216.4\,\text{ft}

. Sanity check: the velocity head is under

1\,\text{ft}

— tiny next to the

115\,\text{ft}

pressure head, so EGL and HGL nearly coincide, typical of a slow large main. Final: HGL

\approx 215\,\text{ft}

, EGL

\approx 216\,\text{ft}

.

\text{HGL} = z + \frac{p}{\gamma}, \qquad \text{EGL} = \text{HGL} + \frac{v^2}{2g}

Pressure change across a contraction

Problem. A horizontal pipe contracts from $D_1 = 0.30\,\text{m}$ to $D_2 = 0.15\,\text{m}$ while carrying $Q = 0.060\,\text{m}^3/\text{s}$ . The upstream gauge pressure is $200\,\text{kPa}$ . Neglecting losses, find the downstream pressure. Use $\rho = 1{,}000\,\text{kg/m}^3$ .

Solution. Continuity:

A_1 = \tfrac{\pi}{4}(0.30)^2 = 0.0707\,\text{m}^2

,

v_1 = 0.060/0.0707 = 0.849\,\text{m/s}

;

A_2 = \tfrac{\pi}{4}(0.15)^2 = 0.01767\,\text{m}^2

,

v_2 = 0.060/0.01767 = 3.40\,\text{m/s}

. Horizontal, lossless Bernoulli (

z_1 = z_2

):

p_2 = p_1 + \tfrac{\rho}{2}(v_1^2 - v_2^2)

.

p_2 = 200{,}000 + \tfrac{1000}{2}(0.849^2 - 3.40^2) = 200{,}000 + 500(0.721 - 11.53) = 200{,}000 - 5{,}400 = 194{,}600\,\text{Pa}

. So

p_2 \approx 195\,\text{kPa}

. Sanity check: the pipe narrows, so velocity rises and pressure must fall — the

5.4\,\text{kPa}

drop converts to the gain in velocity head. Final:

p_2 \approx 195\,\text{kPa}

.

Common pitfalls

•Mixing pressure units. $p/\gamma$ needs consistent units: in USCS, multiply psi by 144 to get $\text{lbf/ft}^2$ before dividing by $62.4$ (or just use $1\,\text{psi} = 2.31\,\text{ft}$ ). Forgetting the 144 is the most common factor error.
•Dropping the velocity head when the diameter changes. EGL and HGL diverge wherever the pipe narrows; if you ignore $v^2/2g$ at a contraction you mislocate the pressure.
•Putting the pump head on the wrong side. $h_p$ adds to the upstream total head (helps flow); losses and turbine head go downstream. A sign flip doubles the error.
•Mixing gauge and absolute pressure. Use gauge throughout the energy equation; switch to absolute only for cavitation/NPSH and vapor-pressure work.
•Putting efficiency in the numerator for pump power. Shaft power is fluid power divided by efficiency; multiplying understates the motor size.
•Sketching the HGL above the EGL, or letting the EGL rise in the flow direction without a pump. The EGL only ever goes down (friction) except across a pump.
•Forgetting $v \approx 0$ and $p = 0$ at a large open reservoir surface — carrying a nonzero reservoir velocity head into the balance.

References

NCEES PE Civil Reference Handbook — §6.2.1 Principles of One-Dimensional Fluid Flow (Bernoulli §6.2.1.3, Energy Line §6.2.1.5)
NCEES PE Civil Reference Handbook — §6.3 Closed Conduit Flow and Pumps
Munson, Young & Okiishi, Fundamentals of Fluid Mechanics — energy equation and grade lines background

B. Pressure conduit

Pipe Friction and Minor Losses (Hazen-Williams and Darcy-Weisbach)

Darcy-Weisbach with the Moody friction factor, the Hazen-Williams C-factor form, minor losses by K-factor and equivalent length, and the laminar/turbulent and relative-roughness distinctions.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

C. Pump application and analysis

Pumps, System Curve, NPSH, Wet Wells and Lift Stations

The operating point where the pump curve meets the system curve, total dynamic head, affinity laws, NPSH available vs required and cavitation, power and efficiency, and wet-well cycling.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

D. Pipe network analysis

Pipe Network Analysis (Series, Parallel and Hardy Cross)

Series and parallel pipe combination, the node-continuity and loop-energy conditions, the Hardy Cross flow-balancing method, and equivalent-pipe reduction.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders