10.1 Continuity, Energy, and HGL/EGL

Energy Bookkeeping Before Formula Hunting

The NCEES PE Civil: Water Resources and Environmental (WRE) exam is an 80-question, 9-hour computer-based test (CBT) delivered year-round at Pearson VUE test centers. It is scored by NCEES against a scaled passing standard, so there is no published fixed cut score or fixed pass rate to chase. Closed-conduit hydraulics and open-channel hydraulics are the two largest content blocks on the specification, each carrying roughly 7 to 11 of the 80 questions, so this chapter routinely decides pass or fail.

The exam is open-resource: you work from the searchable NCEES PE Civil Reference Handbook (PDF on-screen), not your own notes, so knowing where the energy equation, loss equations, and Manning's equation live matters as much as knowing them.

Treat continuity and energy as the foundation for every pipe, pump, culvert, spillway, and channel item that follows. The exam rarely rewards memorizing an equation without first identifying what each head term represents.

Core Head Terms

Continuity is the first check: Q = VA. If a pipe diameter doubles, area increases by four (A = piD^2/4), so velocity drops to one-fourth for the same flow. Because area is proportional to D^2, any diameter error is squared in the velocity and again in velocity head, so a 10 percent diameter error becomes a ~21 percent area error. In an open channel, area changes with depth, so normal-depth and critical-depth calculations are usually iterative or geometry-dependent.

General Energy Setup

Pick one consistent datum and write every term as a length of water:

1. Start with upstream total head: z1 + p1/gamma + V1^2/(2g).
2. Add pump head (h_p) if a pump adds energy between sections.
3. Subtract turbine head (h_t) if energy is extracted.
4. Account for downstream z2, p2/gamma, and V2^2/(2g).
5. Add all friction losses (h_f) and minor losses (h_m) in the flow direction.

A reliable balance is: z1 + p1/gamma + V1^2/(2g) + h_p = z2 + p2/gamma + V2^2/(2g) + h_t + h_f + h_m. Reservoir and tank surfaces usually have negligible velocity head and zero gauge pressure. Pipe sections need pressure head and velocity head. Channel sections use depth and velocity head in the specific-energy term.

HGL and EGL Sanity Checks

The HGL in a pressurized pipe is the level water would rise to in a piezometer tapped into the pipe. If the HGL is 30 ft above the pipe centerline, the gauge pressure head is 30 ft (about 13 psi, since 1 psi is about 2.31 ft of water). If the HGL falls below the centerline, gauge pressure is negative. Negative gauge pressure is not impossible, but it is a warning flag near summits, inverted siphons, and pump suctions where vapor pressure may be reached.

The EGL equals the HGL plus velocity head. Because velocity head is nonnegative, the EGL can never be below the HGL. Through a constant-diameter pipe with only friction, HGL and EGL are parallel and both fall. Through a sudden expansion velocity head drops and pressure head may rise, but the EGL still drops by the loss. Across a pump the EGL jumps upward by the pump head.

Open-channel problems hide the same logic. Specific energy is depth plus velocity head from the channel bottom; critical flow is the minimum specific energy for a given flow and shape. A hydraulic jump converts shallow, fast supercritical flow into deeper, slower subcritical flow and dissipates energy, so the downstream EGL is lower even though depth rises.

Common Traps

• Sketch sections and datum before substituting numbers.
• Convert pressure to head: 1 psi is about 2.31 ft of water; 1 atm is about 33.9 ft.
• Convert diameter to area carefully; diameter errors are squared.
• Put pump head in as a gain and losses as losses, never reversed.
• Reject any passive (no-pump) answer that shows pressure rising in the flow direction, or any EGL below the HGL.

Worked Mini-Example

Water flows from Reservoir A (surface elevation 500 ft) through a pipe to Reservoir B (surface elevation 460 ft). Both surfaces are open to the atmosphere, so p/gamma is zero at each and the surface velocities are negligible. Writing the energy balance from A to B: 500 + 0 + 0 = 460 + 0 + 0 + h_L. Therefore the total headloss between the reservoirs equals the surface elevation difference, h_L = 40 ft. No pump is present, so every foot of that 40 ft is consumed by friction and minor losses. If the problem then asks for the flow, you set 40 ft equal to f(L/D)V^2/(2g) plus the summed minor losses and solve for V, then Q = VA.

This pattern, where the static head difference is simply the available driving head, recurs in gravity transmission mains, siphons, and pond-to-pond conveyance.

A classic trap variant places a high point (summit) along the pipe well above both reservoir surfaces. Even though the endpoints have positive pressure, the HGL can dip below the pipe crown at the summit, producing negative gauge pressure. If that pressure approaches the vapor pressure of water (about -33 ft gauge at sea level, where absolute pressure reaches zero), the flow will cavitate or the siphon will break. The exam rewards recognizing that the controlling check at a summit is the local HGL elevation versus the pipe crown, not the endpoint pressures.