CELE Hydraulics & Fluid Mechanics — Fundamentals of Fluid FlowDetailed Explanation
The Fundamentals of Fluid Flow chapter rewards slow, careful thinking over quick pattern matching, especially on Professional Regulation Commission (PRC) — Board of Civil Engineering's scenario-based CELE items. This detailed explanation walks through the full derivation of every core idea, then links each one to a worked example pulled from recent CELE Hydraulics & Fluid Mechanics papers.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Hydraulics & Fluid Mechanics section sits under a "Core" weighting, and Fundamentals of Fluid Flow is the 5th chapter in the 10-chapter CELE Hydraulics & Fluid Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Hydraulics & Fluid Mechanics.
Fundamentals of Fluid Flow - Detailed Explanation
Fluid flow is the backbone of hydraulic engineering. Whether you are designing a water supply network for a Philippine municipality, sizing a pump for a wastewater treatment plant, or analyzing the force on a pipe bend, three master conservation principles govern every calculation: conservation of mass (continuity), conservation of energy (Bernoulli/energy equation), and conservation of momentum. This chapter develops each principle systematically, links them through worked board-style problems, and highlights the pitfalls that consistently trap examinees in the PRC Civil Engineer Licensure Examination. Mastery here directly supports succeeding chapters on pipe flow, open channels, pumps, and turbines.
Concepts
Continuity Equation — Conservation of Mass
For any steady, incompressible flow, mass can neither be created nor destroyed within a control volume. Because water is effectively incompressible (ρ ≈ 1000 kg/m³ at 20 °C), conservation of mass reduces to conservation of volume flow rate Q (m³/s). The continuity equation states: Q = A₁v₁ = A₂v₂ where A is the cross-sectional area (m²) and v is the mean velocity (m/s). For a circular pipe of diameter D: A = πD²/4 v₂ = v₁(D₁/D₂)² This means velocity is inversely proportional to the square of the diameter. Halving the diameter quadruples the velocity — a critical intuition for pipe-reducer problems. For branching systems (one inlet, two outlets): Q₁ = Q₂ + Q₃ For unsteady flow or compressible flow, the full continuity equation involves the time derivative of density, but these cases are beyond the PRC exam scope for hydraulics.
Examples
The area ratio is (D₁/D₂)² = 4, so the velocity quadruples. Q is constant — the same volume per second passes through both sections.
Scenario
A pipe reduces from D₁ = 300 mm to D₂ = 150 mm. The upstream velocity is v₁ = 2 m/s. Find Q and v₂.
Solution
Step 1 — Compute upstream area: A₁ = π(0.3)²/4 = 0.07069 m² Step 2 — Compute Q: Q = A₁v₁ = 0.07069 × 2 = 0.1414 m³/s Step 3 — Compute downstream velocity using the diameter-squared ratio: v₂ = v₁(D₁/D₂)² = 2 × (300/150)² = 2 × 4 = 8.0 m/s Verification: A₂ = π(0.15)²/4 = 0.01767 m² A₂v₂ = 0.01767 × 8.0 = 0.1414 m³/s ✓
The junction acts as a mass-balance node. The total flow in must equal the total flow out, applied branch by branch.
Scenario
A 250 mm main pipe splits into a 150 mm branch (Q₂ = 0.03 m³/s) and a 200 mm branch. The velocity in the 250 mm pipe is 1.5 m/s. Find v₃ in the 200 mm branch.
Solution
Step 1 — Compute Q₁: A₁ = π(0.25)²/4 = 0.04909 m² Q₁ = 0.04909 × 1.5 = 0.07363 m³/s Step 2 — Apply junction continuity: Q₃ = Q₁ − Q₂ = 0.07363 − 0.03 = 0.04363 m³/s Step 3 — Find v₃: A₃ = π(0.20)²/4 = 0.03142 m² v₃ = Q₃/A₃ = 0.04363/0.03142 = 1.39 m/s
Applications
- Sizing reducer fittings and transition sections in water mains.
- Determining velocities in multi-branch distribution networks.
- Checking flow balance in irrigation canal bifurcations.
- Computing discharge through venturi meters and orifice meters (area changes create measurable velocity/pressure differences).
- Hydraulic model scaling — prototype and model Q relationships.
Misconceptions
- Using the diameter ratio (not squared) for velocity: v₂ = v₁(D₁/D₂) is WRONG — the correct form uses (D₁/D₂)².
- Assuming velocity is the same on both sides of a reducer — velocity MUST change when area changes.
- Mixing units — using mm for D and m for v in Q = Av gives nonsensical results.
- Forgetting that continuity applies to volume flow rate, not velocity alone.
Related Concepts
- Bernoulli's Equation (pressure changes follow from velocity changes via continuity)
- Venturi Meter and Orifice Meter (applications of combined continuity + Bernoulli)
- Pipe Network Analysis (Hardy-Cross method uses continuity at each node)
- Open Channel Flow (continuity applies to cross-sectional area × mean velocity)
Common Exam Questions
Example
Given D₁ = 400 mm, D₂ = 200 mm, v₁ = 1 m/s → v₂ = 1×(400/200)² = 4 m/s.
Approach
Use v₂ = v₁(D₁/D₂)². Never forget to square the ratio — this is the single most common arithmetic error.
Question Type
Direct velocity calculation in a pipe reducer
Example
D = 300 mm, v = 3 m/s → Q = (π/4)(0.3)²(3) = 0.2121 m³/s.
Approach
Q = (π/4)D²v. Convert D to metres first.
Question Type
Flow rate from velocity and diameter
Example
Q₁ = 0.10 m³/s enters; Q₂ = 0.04 m³/s exits one branch → Q₃ = 0.06 m³/s in the other.
Approach
ΣQ_in = ΣQ_out. Identify the unknown branch and solve directly.
Question Type
Junction problem — find unknown branch flow
Key Points To Remember
- Q = Av = constant along a streamtube for steady, incompressible flow.
- For circular pipes: v₂ = v₁(D₁/D₂)² — velocity scales with the SQUARE of the diameter ratio.
- At branching junctions: ΣQ_in = ΣQ_out.
- Larger area → smaller velocity; smaller area → larger velocity.
- Units: Q in m³/s, A in m², v in m/s — always verify unit consistency.
- Convert all diameters to metres before computing areas.
Energy Equation — Bernoulli's Principle with Head Loss and Machine Heads
The energy equation is the most powerful tool in hydraulics. It is derived from the first law of thermodynamics applied to steady, incompressible flow along a streamline, expressed in units of metres (head = energy per unit weight). General form between points 1 and 2: p₁/γ + v₁²/(2g) + z₁ + hA = p₂/γ + v₂²/(2g) + z₂ + hE + hL where: • p/γ = pressure head (m) — energy from fluid pressure • v²/(2g) = velocity head (m) — kinetic energy per unit weight • z = elevation head (m) — potential energy per unit weight • hA = head added by a pump (m) — energy input • hE = head extracted by a turbine (m) — energy output • hL = total head loss due to friction and minor losses (m) The SUM (p/γ + v²/(2g) + z) at any cross-section is the total head H. Ideal Bernoulli (no machines, no losses, hA = hE = hL = 0): p₁/γ + v₁²/(2g) + z₁ = p₂/γ + v₂²/(2g) + z₂ Sign convention (critical for exams): • hA is on the LEFT (same side as point 1) — a pump adds energy upstream of point 2. • hE and hL are on the RIGHT (same side as point 2) — they remove available energy. Alternate form (everything on one side): H₁ + hA − hE − hL = H₂ Head lines: • Energy Grade Line (EGL): plots total head H = p/γ + v²/(2g) + z along the pipe. It always slopes downward (due to losses) except where a pump raises it. • Hydraulic Grade Line (HGL): plots p/γ + z (piezometric head). HGL = EGL − v²/(2g). When HGL drops below the pipe centreline, the gauge pressure is negative (subatmospheric).
Examples
Pressure dropped from 200 kPa to 120.9 kPa because (a) the velocity head increased from 0.204 m to 3.262 m and (b) the elevation increased by 5 m. Both consumed available pressure head.
Scenario
At section 1 (300 mm pipe): p₁ = 200 kPa, v₁ = 2 m/s, z₁ = 0. At section 2 (150 mm pipe, from continuity v₂ = 8 m/s): z₂ = 5 m. Neglect losses. Find p₂.
Solution
Step 1 — Write ideal Bernoulli: p₁/γ + v₁²/(2g) + z₁ = p₂/γ + v₂²/(2g) + z₂ Step 2 — Compute each head term: p₁/γ = 200/9.81 = 20.387 m v₁²/(2g) = 2²/(2×9.81) = 0.204 m z₁ = 0 m v₂²/(2g) = 8²/(2×9.81) = 3.262 m z₂ = 5 m Step 3 — Solve for p₂/γ: 20.387 + 0.204 + 0 = p₂/γ + 3.262 + 5 20.591 = p₂/γ + 8.262 p₂/γ = 12.329 m Step 4 — Convert to pressure: p₂ = 12.329 × 9.81 = 120.9 kPa
Setting p and v terms to zero at both free surfaces (large tank assumption) simplifies the equation greatly. The pump head equals the static lift plus all losses.
Scenario
A pump draws water from a sump (z₁ = 0, p₁ = 0 gauge, v₁ ≈ 0) and delivers it to a tank at z₂ = 15 m (p₂ = 0 gauge, v₂ ≈ 0) with a total head loss hL = 3.5 m. Find the pump head hA.
Solution
Step 1 — Write energy equation (hE = 0, no turbine): 0 + 0 + 0 + hA = 0 + 0 + 15 + 0 + 3.5 Step 2 — Solve: hA = 18.5 m Interpretation: The pump must supply 15 m to lift the water plus 3.5 m to overcome friction losses.
As velocity increases through the nozzle (continuity), pressure drops — the classic Bernoulli trade-off. This is also the principle of the venturi meter.
Scenario
In a horizontal nozzle: p₁ = 300 kPa, v₁ = 3 m/s, z₁ = z₂ = 0 (horizontal). v₂ = 12 m/s. Neglect losses. Find p₂.
Solution
Step 1 — Bernoulli (horizontal, no machines, no loss): p₁/γ + v₁²/(2g) = p₂/γ + v₂²/(2g) Step 2 — Compute: 300/9.81 + 3²/19.62 = p₂/9.81 + 12²/19.62 30.581 + 0.459 = p₂/9.81 + 7.339 31.040 − 7.339 = p₂/9.81 p₂/9.81 = 23.701 m p₂ = 232.5 kPa
Applications
- Pump system design: determining required pump head = static lift + friction losses + velocity head change.
- Turbine analysis: head available to the runner = gross head − pipe losses.
- Venturi and orifice meters: Bernoulli + continuity give the differential pressure–flow rate relationship.
- Pitot tube: stagnating the flow locally measures velocity head → velocity.
- Siphon design: checking that HGL does not fall more than ~7.5 m below the pipe (cavitation limit).
- Sprinkler system hydraulics in buildings (reference: NFPA 13, adopted in the Philippines).
Misconceptions
- Placing hA on the wrong side of the equation — hA must be on the UPSTREAM (point 1) side.
- Forgetting to include velocity head when diameters change between the two reference points.
- Using absolute pressure for one point and gauge for another — be consistent throughout one equation.
- Assuming EGL is horizontal — it is horizontal ONLY for ideal flow (no losses). In real flow it always drops.
- Neglecting the velocity head at a pipe exit into a reservoir — at the reservoir surface v≈0, but inside the pipe the velocity head is real and must be counted as a minor loss.
- Using γ = 9.81 m/s² (g, not γ) — always check: γ_water = 9.81 kN/m³ or 9810 N/m³.
Related Concepts
- Continuity Equation (provides velocities needed in Bernoulli)
- Darcy-Weisbach Equation (computes hL for substitution into energy equation)
- Pump and Turbine Power (P = γQhA or γQhE)
- Venturi Meter (combines continuity + Bernoulli to measure Q)
- Cavitation (occurs when absolute pressure falls below vapour pressure — HGL analysis)
Common Exam Questions
Example
D₁=200mm, p₁=150kPa, v₁=2m/s, z₁=3m; D₂=100mm, z₂=3m. Find p₂ (no loss). v₂=8m/s by continuity. p₂/γ = (150/9.81+0.204+3)−(3.262+3) = 18.548 m → p₂=181.9 kPa... (work through carefully).
Approach
Apply continuity first to get v₂, then substitute all known heads into Bernoulli. Keep track of elevation changes.
Question Type
Find pressure at a constriction given upstream conditions
Example
Pump lifts water 20 m with 5 m of head loss → hA = 25 m.
Approach
Write the full energy equation. At large-tank surfaces, set p=0 gauge and v≈0. The unknown machine head appears as the only variable.
Question Type
Find pump head or turbine head
Example
In a constant-diameter pipe, EGL and HGL are parallel, both sloping downward at the friction slope Sf.
Approach
EGL drops along flow. HGL = EGL − v²/(2g). At a constriction, HGL dips sharply because v²/(2g) increases.
Question Type
Sketch or interpret EGL and HGL
Key Points To Remember
- Every term has units of metres (head). Divide pressures by γ = 9.81 kN/m³ to get metres.
- hA (pump) is added on the UPSTREAM side of the equation; hE and hL are on the DOWNSTREAM side.
- Ideal Bernoulli applies only when there are no losses and no machines.
- EGL always drops in the direction of flow (losses); pumps cause a step UP in EGL.
- HGL = EGL − velocity head. When the pipe is horizontal and velocity is constant, HGL and EGL are parallel.
- Gauge pressure is zero where HGL intersects the pipe centreline — useful for locating cavitation risk.
- For a horizontal pipe with constant diameter (no machines, no loss): p₁ = p₂ (Bernoulli trivially).
- γ_water = 9.81 kN/m³ = 9810 N/m³. Use kN/m³ with kPa to get metres directly.
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
The EGL and HGL are graphical representations of the energy equation along the length of a pipe or conduit. They are essential for visual interpretation of flow conditions and frequently appear in board examination problems asking you to 'sketch' or 'describe' grade lines. Definitions: • Total Head (EGL ordinate): H = p/γ + v²/(2g) + z • Piezometric Head (HGL ordinate): h = p/γ + z • Relationship: EGL = HGL + v²/(2g) Key behaviours: 1. The EGL always slopes DOWNWARD in the direction of flow (head loss is positive). 2. At a pump, the EGL steps UP by hA (sudden energy addition). 3. At a turbine, the EGL steps DOWN by hE (sudden energy extraction). 4. When pipe diameter decreases, v²/(2g) increases, so HGL drops faster than EGL. 5. When pipe diameter increases, v²/(2g) decreases, HGL rises toward EGL. 6. At a pipe entry from a reservoir: total head = reservoir surface elevation (if v in reservoir ≈ 0). 7. At a free discharge (pipe exit into air): p = 0 gauge, so HGL = pipe centreline elevation at exit. 8. If HGL falls BELOW the pipe centreline: gauge pressure is NEGATIVE (subatmospheric). This is not necessarily problematic unless it drops below −p_atm (absolute zero pressure → cavitation). 9. At a sudden expansion, minor losses cause a step down in EGL. 10. For a constant-diameter pipe, EGL and HGL are PARALLEL straight lines (uniform friction slope).
Examples
This illustrates why EGL and HGL must be drawn carefully. A pipe 'running full' can still have subatmospheric pressure if EGL losses are large.
Scenario
Describe the EGL and HGL for a horizontal pipe of constant 200 mm diameter, 50 m long, carrying water from a reservoir (water surface at z = 20 m) to atmosphere. Total head loss = 4 m. Velocity = 2 m/s.
Solution
At the reservoir (point 1): p₁ = 0, v₁ ≈ 0, z₁ = 20 m → EGL₁ = HGL₁ = 20 m. At the pipe exit (point 2): p₂ = 0 (atmospheric), v₂ = 2 m/s, z₂ = 20 m (horizontal). v²/(2g) = 2²/(2×9.81) = 0.204 m EGL₂ = HGL₂ + 0.204 m But EGL₂ = EGL₁ − hL = 20 − 4 = 16 m (total head at exit). HGL₂ = EGL₂ − v²/(2g) = 16 − 0.204 = 15.796 m Since the pipe is at z = 20 m and HGL₂ = 15.796 m, the HGL is 4.204 m BELOW the pipe → subatmospheric pressure throughout. This pipe would require positive upstream pressure or siphon action.
Applications
- Locating the critical (minimum pressure) point in a siphon — where HGL is lowest.
- Checking for cavitation risk in pipelines and pump suction lines.
- Designing pressure-class pipe segments based on maximum and minimum pressures.
- Understanding the behaviour of water distribution networks — where pressure is adequate.
Misconceptions
- Thinking EGL and HGL are the same line — they differ by the velocity head.
- Assuming HGL must always be above the pipe — it can be below in suction lines.
- Forgetting the velocity head 'gap' doubles when diameter halves (area quarters, v quadruples, v²/2g ×16).
Related Concepts
- Energy Equation (EGL/HGL are the graphical form of the energy equation)
- Cavitation (critical when absolute pressure < vapour pressure)
- Pipe Friction (Darcy-Weisbach sets the slope of EGL)
- Pump and Siphon Design
Common Exam Questions
Example
A pipe goes from reservoir A (z=30m) through a reducer, then a pump (hA=20m), then to reservoir B (z=40m). Sketch the EGL.
Approach
Start at the upstream reservoir (EGL = HGL = water surface). Apply head losses along the pipe. Mark machine head jumps. At exit, EGL endpoint = HGL + velocity head.
Question Type
Sketch EGL and HGL for a given pipe system
Example
HGL at section X = 18 m, pipe centreline at z = 10 m → p/γ = 8 m → p = 78.5 kPa gauge.
Approach
p/γ = HGL − z (pipe centreline elevation). Multiply by γ to get kPa.
Question Type
Determine gauge pressure at a point given EGL
Key Points To Remember
- EGL − HGL = velocity head v²/(2g). If diameter is constant, this gap is constant.
- EGL always decreases in flow direction; machines can locally raise or lower it.
- HGL above the pipe centreline → positive gauge pressure.
- HGL below the pipe centreline → negative gauge pressure (suction).
- At any large open reservoir, EGL and HGL both coincide at the water surface.
- Slope of EGL = friction head loss per unit length = Sf (hydraulic slope).
- A pump creates a vertical jump in EGL equal to hA.
Momentum Equation — Force on Pipe Bends, Nozzles, and Vanes
Newton's second law applied to a steady-flow control volume gives the linear momentum equation: ΣF = ρQ(v₂ − v₁) [scalar, single direction] For 2-D problems (x and y components separately): ΣFx = ρQ(v₂x − v₁x) ΣFy = ρQ(v₂y − v₁y) The forces in ΣF include: • Pressure forces at inlet and outlet (p₁A₁ and p₂A₂ — watch sign/direction) • Body force (weight of fluid in control volume, usually negligible for pipes) • Reaction force R (what we solve for — the force the pipe fitting exerts on the fluid; the force on the fitting is −R by Newton's 3rd law) Procedure for pipe bend problems: 1. Draw a free-body diagram of the fluid inside the control volume (the bend). 2. Define positive x and y directions. 3. Write momentum equations in x and y. 4. Include pressure forces at inlet (pushing in the flow direction) and outlet (pushing against the fluid in the pipe). 5. Solve for Rx and Ry. 6. The force on the PIPE FITTING (from the fluid) is equal and opposite: Fpipe = −R. 7. Resultant: F = √(Fx² + Fy²), direction θ = arctan(Fy/Fx). Note: For a straight pipe reducer with no change in direction, the momentum equation gives the net axial force (important for expansion joints and thrust blocks in Philippine water supply projects).
Examples
The pressure forces at inlet and outlet dominate over the momentum change (ρQΔv) for typical pipe velocities. This is why high-pressure bends need large thrust blocks — the primary load is the pressure-times-area term.
Scenario
A horizontal 90° pipe bend (elbow) carries Q = 0.05 m³/s. Inlet: D₁ = 150 mm, p₁ = 200 kPa, flow in +x direction. Outlet: D₂ = 150 mm (same diameter), p₂ = 180 kPa, flow in +y direction. Find the force on the bend. Neglect weight of fluid.
Solution
Step 1 — Velocities (same diameter, same v): A = π(0.15)²/4 = 0.01767 m² v = Q/A = 0.05/0.01767 = 2.829 m/s Inlet velocity: v₁x = 2.829 m/s, v₁y = 0 Outlet velocity: v₂x = 0, v₂y = 2.829 m/s Step 2 — Pressure forces on control volume: F_p1x = p₁A₁ = 200,000 × 0.01767 = 3534 N (in +x, pushing fluid) F_p2y = −p₂A₂ = −180,000 × 0.01767 = −3180.6 N (outlet pressure acts against +y on CV boundary) [Sign: at the outlet boundary, the pressure force on the CV fluid acts in the −y direction because the surroundings push back.] Step 3 — Momentum equation in x: F_p1x + Rx = ρQ(v₂x − v₁x) 3534 + Rx = 1000 × 0.05 × (0 − 2.829) 3534 + Rx = −141.45 Rx = −141.45 − 3534 = −3675.5 N Step 4 — Momentum equation in y: −F_p2y + Ry = ρQ(v₂y − v₁y) −(−3180.6) + Ry = 1000 × 0.05 × (2.829 − 0) 3180.6 + Ry = 141.45 Ry = 141.45 − 3180.6 = −3039.2 N Step 5 — Force on FLUID from bend = (Rx, Ry) = (−3675.5, −3039.2) N Force on BEND from fluid = (+3675.5, +3039.2) N Step 6 — Resultant: F = √(3675.5² + 3039.2²) = √(13,509,300 + 9,236,700) = √22,746,000 = 4769 N ≈ 4.77 kN θ = arctan(3039.2/3675.5) = 39.6° from x-axis
Applications
- Design of thrust blocks and concrete anchor blocks at bends in Philippine water distribution systems.
- Force on nozzle flanges — must be resisted by bolts or restraints.
- Pelton wheel bucket force analysis (momentum change of water jet).
- Reaction forces in sprinkler heads and fire-fighting nozzles.
- Analysis of forces on bridge piers due to flowing river (fluid momentum).
Misconceptions
- Forgetting pressure forces — applying only ρQΔv and ignoring p₁A₁ and p₂A₂ gives a completely wrong answer.
- Wrong sign for outlet pressure force — at the outlet face of the control volume, the surroundings push BACK (opposing the outflow direction).
- Confusing force on FLUID with force on PIPE — they are equal and opposite.
- Using gauge pressure incorrectly — for momentum equation, use gauge pressure consistently (atmospheric cancels on all free surfaces).
Related Concepts
- Continuity (provides velocities needed in momentum equation)
- Bernoulli (provides pressures if not given directly)
- Impulse-Momentum Theorem (general form of ΣF = d(mv)/dt)
- Moving Vanes and Turbine Buckets (extension of fixed-vane momentum analysis)
Common Exam Questions
Example
Standard board question: 150 mm pipe, 90° bend, p and v given → find resultant force on the bend.
Approach
Set up x-y components. Inlet flow in one direction, outlet in the perpendicular direction. Apply ΣF = ρQΔv in each direction. Include pressure forces.
Question Type
Force on a 90° pipe bend
Example
Nozzle discharges to atmosphere (p₂=0), so only p₁A₁ and momentum change drive the result.
Approach
Only one direction (axial). ΣFx = p₁A₁ − p₂A₂ + Rx = ρQ(v₂−v₁). Solve for Rx.
Question Type
Force on a nozzle (straight, axial)
Example
Pelton bucket moving at u m/s, jet at v₁ m/s → relative velocity = v₁−u. F = ρA(v₁−u)²(1−cosβ).
Approach
Replace Q with Q_relative = A(v_jet − v_vane). Apply momentum using relative velocity.
Question Type
Force on a moving vane (in a fixed reference frame)
Key Points To Remember
- ΣF = ρQ(v_out − v_in) — applied component by component.
- Include pressure forces: F_pressure = p × A, directed INTO the control volume at inlet, OUT at outlet.
- Fluid density ρ = 1000 kg/m³ for water at standard conditions.
- The force on the fluid from the pipe (reaction R) is solved from the momentum equation.
- The force on the PIPE from the fluid is EQUAL AND OPPOSITE to R (Newton's 3rd law).
- For a 90° bend with equal inlet/outlet areas: Fx and Fy are both non-zero.
- Resultant force = √(Rx² + Ry²). Always report direction as well.
- Weight of fluid in control volume is often neglected for small fittings (check the problem statement).
Power of a Flowing Stream — Pump and Turbine Power
The power P transported by a flowing fluid across a cross-section with total head H and flow rate Q is: P = γQH (watts if γ in N/m³, Q in m³/s, H in m) P = γQH/1000 (kilowatts if γ in N/m³) or equivalently, P = 9.81QH kW (using γ = 9.81 kN/m³) For a PUMP: Power delivered to fluid (water power, hydraulic power): P_w = γQhA Power input to pump shaft (brake power): P_input = γQhA / η_pump where η_pump = pump efficiency (decimal, e.g., 0.85 for 85%) For a TURBINE: Power extracted from fluid: P_w = γQhE Power output at turbine shaft: P_output = η_turbine × γQhE where η_turbine = turbine efficiency Overall plant efficiency (turbine + generator): η_overall = P_electrical / (γQH_gross) Important: The head H in the power formula is the NET head — after subtracting pipe losses (for turbines) or before subtracting losses (for pumps, where hA is the NET head added). Unit conversions: • 1 kW = 1 kN·m/s • 1 hp (metric horsepower) = 0.7457 kW • 1 hp (British) = 0.7457 kW (same numerically; Philippine exams use kW) Common board exam setup: A pump–pipeline system where you find hA from energy equation, then compute P_input = γQhA/η.
Examples
The pump must receive 52.82 kW to deliver 41.20 kW to the water — 11.62 kW is lost to friction in the impeller, bearings, and seals.
Scenario
A pump delivers Q = 0.12 m³/s against a total head of 35 m. The pump efficiency is 78%. Find (a) water power and (b) power input to the pump motor (assume motor efficiency = 100%).
Solution
(a) Water power (hydraulic power): P_w = γQH = 9.81 × 0.12 × 35 = 41.20 kW (b) Power input (shaft/motor power): P_input = P_w / η = 41.20 / 0.78 = 52.82 kW
The fluid delivers 1226.25 kW to the turbine runner; 88% (1079.1 kW) appears as shaft power; 12% (147.15 kW) is lost to hydraulic and mechanical inefficiencies.
Scenario
A turbine receives Q = 5 m³/s under a net head of 25 m at 88% efficiency. Find the shaft power output.
Solution
P_output = η × γQH = 0.88 × 9.81 × 5 × 25 = 0.88 × 1226.25 = 1079.1 kW ≈ 1.079 MW
This is a direct application of the power formula. Note γ = 9.81 kN/m³ gives kW automatically.
Scenario
Find the power of a stream with Q = 0.1414 m³/s flowing across a head of H = 20 m.
Solution
P = γQH = 9.81 × 0.1414 × 20 = 27.74 kW
Applications
- Sizing pump motors in water supply, sewage lift stations, and irrigation systems.
- Estimating hydropower potential of rivers and reservoirs (small hydro projects in Mindanao and the Cordilleras).
- Computing generator capacity for run-of-river micro-hydro plants.
- Energy audit of pumping systems — comparing P_input to P_w gives pump efficiency.
- Life-cycle cost analysis of pump selection (energy cost = P_input × operating hours × electricity tariff).
Misconceptions
- Multiplying by efficiency for pumps — WRONG. Pump input = hydraulic power DIVIDED by η.
- Dividing by efficiency for turbines — WRONG. Turbine output = hydraulic power MULTIPLIED by η.
- Using γ = 9810 N/m³ and Q in m³/s and H in m — result is in WATTS, not kilowatts. Divide by 1000.
- Confusing gross head with net head for turbines — head losses in the penstock must be subtracted first.
- Ignoring motor efficiency when motor efficiency is given separately — overall efficiency = η_pump × η_motor.
Related Concepts
- Energy Equation (hA and hE appear directly in energy equation)
- Pump Characteristic Curves (H vs Q — operating point analysis)
- Turbine Types (Pelton, Francis, Kaplan — selection based on head and Q)
- Specific Speed (dimensionless number classifying pump and turbine types)
Common Exam Questions
Example
Q=0.08 m³/s, hA=30m, η=82% → P_input = 9.81×0.08×30/0.82 = 28.81 kW.
Approach
P_input = γQhA / η. Always divide by efficiency for pump input.
Question Type
Find pump input power given Q, head, and efficiency
Example
Q=3 m³/s, hE=20m, η=90% → P_output = 0.90×9.81×3×20 = 529.7 kW.
Approach
P_output = η × γQhE. Always multiply by efficiency for turbine output.
Question Type
Find turbine output power given Q, head, and efficiency
Example
P_input=50kW, Q=0.1m³/s, hA=40m → P_w=9.81×0.1×40=39.24kW → η=39.24/50=78.5%.
Approach
η = P_w / P_input = γQhA / P_input.
Question Type
Find efficiency given P_input and hydraulic conditions
Key Points To Remember
- P = γQH is the fundamental power-of-flow formula.
- γ = 9.81 kN/m³ → P in kW when Q in m³/s and H in m.
- Pump: P_input = γQhA / η (efficiency divides — more power IN for same hydraulic output).
- Turbine: P_output = η × γQhE (efficiency multiplies — less power OUT than fluid delivers).
- hA is the head the pump adds to the fluid; hE is the head the turbine extracts from the fluid.
- Always check if the problem gives gross head or net head for a turbine.
- 1 kW = 1000 W; 1 MW = 1000 kW. Hydropower problems often ask for MW.
Practice Problems
Using the diameter-squared ratio is faster than computing both areas and solving for v₂ directly. Always verify by checking Q₁ = Q₂.
Problem
Problem 1 (Continuity): A 250 mm main pipe carries flow at v = 1.63 m/s. It reduces to a 150 mm pipe. Find (a) Q in the main, (b) v in the 150 mm pipe, and (c) Q in litres per second.
Solution
Given: D₁ = 0.250 m, v₁ = 1.63 m/s, D₂ = 0.150 m (a) A₁ = π(0.25)²/4 = 0.04909 m² Q = A₁v₁ = 0.04909 × 1.63 = 0.08002 m³/s (b) v₂ = v₁(D₁/D₂)² = 1.63 × (250/150)² = 1.63 × 2.778 = 4.53 m/s Verify: A₂ = π(0.15)²/4 = 0.01767 m² Q = A₂v₂ = 0.01767 × 4.53 = 0.0800 m³/s ✓ (c) Q = 0.0800 m³/s × 1000 L/m³ = 80.0 L/s
Note that v₁ = v₂ ≈ 0 at the large reservoir surfaces. The pump head equals static lift (28 m) plus head loss (6 m). Free surface-to-free surface problems are the cleanest application of the energy equation.
Problem
Problem 2 (Energy Equation — Pump): Water is pumped from a lower reservoir (water surface at elevation 10 m) to an upper reservoir (water surface at elevation 38 m). The pipe is 200 mm diameter throughout, flow rate Q = 0.05 m³/s, and total head loss in the system is 6 m. Find (a) pump head hA, (b) water power, (c) pump input power if η = 80%.
Solution
Setup: Point 1 = lower reservoir surface (p₁=0, v₁≈0, z₁=10m); Point 2 = upper reservoir surface (p₂=0, v₂≈0, z₂=38m). Energy equation: 0 + 0 + 10 + hA = 0 + 0 + 38 + 0 + 6 hA = 38 + 6 − 10 = 34 m (a) hA = 34 m (b) P_w = γQhA = 9.81 × 0.05 × 34 = 16.68 kW (c) P_input = P_w / η = 16.68 / 0.80 = 20.85 kW
Even though the velocity at the throat is 4× the upstream velocity, the pressure drop is only about 12 kPa in this case because v₁ is modest. In a real venturi meter, this differential would drive a manometer to read Q.
Problem
Problem 3 (Bernoulli — Pressure at Constriction): A horizontal venturi has inlet diameter D₁ = 200 mm and throat D₂ = 100 mm. Upstream pressure p₁ = 180 kPa, Q = 0.04 m³/s. Neglect losses. Find throat pressure p₂.
Solution
Step 1 — Velocities: A₁ = π(0.2)²/4 = 0.03142 m² v₁ = Q/A₁ = 0.04/0.03142 = 1.273 m/s A₂ = π(0.1)²/4 = 0.007854 m² v₂ = Q/A₂ = 0.04/0.007854 = 5.093 m/s (Check: v₂ = v₁(D₁/D₂)² = 1.273×4 = 5.093 ✓) Step 2 — Bernoulli (horizontal, z₁=z₂): p₁/γ + v₁²/(2g) = p₂/γ + v₂²/(2g) 180/9.81 + 1.273²/19.62 = p₂/9.81 + 5.093²/19.62 18.349 + 0.0826 = p₂/9.81 + 1.323 18.431 − 1.323 = p₂/9.81 p₂/9.81 = 17.108 m p₂ = 167.8 kPa
Always use NET head (after pipe losses) for turbine calculations. Each stage of conversion (hydraulic→mechanical→electrical) multiplies by the respective efficiency. Overall efficiency = 0.85×0.95 = 80.75%.
Problem
Problem 4 (Turbine Power): A Francis turbine operates under a gross head of 30 m with Q = 4 m³/s. Pipe losses from reservoir to turbine amount to 3 m of head. Turbine efficiency is 85%, generator efficiency is 95%. Find (a) net head, (b) shaft power output, (c) electrical power output.
Solution
(a) Net head: H_net = H_gross − h_pipe_loss = 30 − 3 = 27 m (b) Shaft (mechanical) power: P_shaft = η_turbine × γQH_net P_shaft = 0.85 × 9.81 × 4 × 27 = 0.85 × 1059.5 = 900.6 kW (c) Electrical power: P_elec = η_generator × P_shaft = 0.95 × 900.6 = 855.5 kW ≈ 0.856 MW
Pressure forces (p×A terms, ~6 kN each) dominate the momentum change (ρQΔv ≈ 0.14 kN). This confirms that thrust blocks at bends are primarily sized for pressure loads, not flow momentum. The resultant of ~8.93 kN acts diagonally into the bend corner.
Problem
Problem 5 (Momentum — 90° Bend): A 180 mm pipe carries Q = 0.06 m³/s around a 90° horizontal bend. Inlet pressure p₁ = 250 kPa (flow in +x), outlet pressure p₂ = 235 kPa (flow in +y). Find the magnitude and direction of the resultant force on the bend. Neglect fluid weight.
Solution
Step 1 — Area and velocity: A = π(0.18)²/4 = 0.02545 m² v = Q/A = 0.06/0.02545 = 2.358 m/s (same at inlet and outlet, same pipe size) Step 2 — Velocity components: Inlet: v₁x = 2.358, v₁y = 0 Outlet: v₂x = 0, v₂y = 2.358 Step 3 — Momentum equation in x (ΣFx = ρQΔvx): p₁A − Rx = ρQ(v₂x − v₁x) [Rx is the x-force the bend exerts on fluid, opposing inlet pressure and momentum] 250,000×0.02545 − Rx = 1000×0.06×(0 − 2.358) 6362.5 − Rx = −141.5 Rx = 6362.5 + 141.5 = 6504 N (bend pushes fluid in −x; fluid pushes bend in +x) Step 4 — Momentum equation in y (ΣFy = ρQΔvy): −p₂A + Ry = ρQ(v₂y − v₁y) [outlet pressure opposes the +y outflow, acts −y on CV] −235,000×0.02545 + Ry = 1000×0.06×(2.358 − 0) −5980.8 + Ry = 141.5 Ry = 141.5 + 5980.8 = 6122.3 N (bend pushes fluid in +y; but this means fluid pushes bend in −y... let us recheck sign convention) [Recheck y-direction: at outlet, the pipe continues in +y. The outlet pressure force on the control volume acts INWARD on the CV boundary at the outlet, i.e., in the −y direction. So: −p₂A is already accounted. Ry is the y-force the bend exerts on the fluid. Result Ry = 6122.3 N means the bend pushes fluid in +y direction.] Force ON BEND from fluid (reaction): F_bend_x = −(−6504) = wait — reclarify: The bend exerts on fluid: (−Rx in x, +Ry in y) per our FBD convention. Actually from momentum: Rx = 6504 N is the bend force ON FLUID in +x direction? Let us redo clearly. Clear convention: R = (Rx, Ry) is the force the BEND exerts on the FLUID. x: p₁A + Rx = ρQ(0 − 2.358) → 6362.5 + Rx = −141.5 → Rx = −6504 N y: −p₂A + Ry = ρQ(2.358 − 0) → −5980.8 + Ry = 141.5 → Ry = 6122.3 N Force ON FLUID from bend = (−6504, +6122.3) N Force ON BEND from fluid = (+6504, −6122.3) N Step 5 — Resultant force on bend: F = √(6504² + 6122.3²) = √(42,302,016 + 37,482,529) = √79,784,545 = 8932 N ≈ 8.93 kN θ = arctan(6122.3/6504) = arctan(0.9413) = 43.3° below the +x axis
Exam Preparation Tips
- MASTER THE ENERGY EQUATION FIRST: It appears in at least 2–3 problems in every hydraulics board exam set. Practice setting it up quickly — identify points 1 and 2, list all known/unknown heads, and solve algebraically before substituting numbers.
- SQUARED DIAMETER RATIO: The single most common arithmetic error is using (D₁/D₂) instead of (D₁/D₂)² for velocity ratio. Always square the diameter ratio. Write it explicitly in every problem.
- UNIT DISCIPLINE: Use a consistent unit system. For P = γQH, using γ = 9.81 kN/m³ with Q in m³/s and H in m gives P directly in kW. Using γ = 9810 N/m³ gives P in watts — then divide by 1000 for kW.
- PUMP vs TURBINE EFFICIENCY: Pump input = water power ÷ η (more energy in). Turbine output = water power × η (less energy out). Confusing these is a guaranteed wrong answer. Use dimensional logic: a pump is less than 100% efficient, so you need MORE input, hence you DIVIDE.
- FREE-SURFACE SIMPLIFICATION: At a large open reservoir or tank surface, set p = 0 (gauge), v ≈ 0. This eliminates two head terms instantly and gives you the cleanest energy equation setup.
- EGL/HGL SKETCHING: Practice sketching grade lines for: (a) constant-diameter pipe, (b) pipe with reducer, (c) pipe with pump, (d) pipe with turbine. Know that EGL always drops; machines cause abrupt step changes.
- MOMENTUM EQUATION SETUP: Draw a proper free-body diagram EVERY time. Label inlet and outlet velocities with components. Include both pressure forces AND momentum change terms. Forgetting p×A is the #1 momentum equation error.
- PROCESS PROBLEMS IN ORDER: (1) Geometry → areas, (2) Continuity → velocities, (3) Bernoulli → pressures or head, (4) Power formula or momentum. This sequence applies to 90% of board problems.
- CAVITATION AWARENESS: If HGL drops below the pipe centreline and exceeds ~7.5 m of suction head, flag cavitation as a concern. Absolute pressure ≈ 0 when gauge pressure = −101.325 kPa = −10.34 m head.
- PRACTICE PAST BOARD PROBLEMS: The PRC regularly recycles problem structures. Solve all available past board exams in hydraulics (1990s to present) — you will find that continuity, Bernoulli, pump power, and 90° bend force account for more than half of all hydraulics questions.
- CHECK WITH ENERGY BALANCE: After solving, verify that EGL drops between points — if your computed p₂ implies EGL rose with no pump, something is wrong. Use this as a sanity check.
- SIGNIFICANT FIGURES: Report answers to 3–4 significant figures, consistent with the given data. Avoid over-rounding intermediate steps — carry at least 4 significant figures until the final answer.
In summary
The three conservation principles — continuity, energy, and momentum — together with the power formula form a complete toolkit for virtually every fluid flow problem encountered in the PRC Civil Engineer Licensure Examination. Continuity fixes the velocities; the energy equation ties pressures, elevations, and machine heads together; the momentum equation quantifies forces on fittings and structures; and P = γQH translates fluid mechanics into engineering economics. Mastery requires not just formula memorisation but disciplined setup: always draw a clear diagram, label your control volume, identify which terms are zero, and track units throughout. The worked examples and practice problems in this chapter mirror the style and difficulty of actual board exam questions. Consistent practice with these problem types — particularly reservoir-to-reservoir pump problems, pipe-reducer Bernoulli problems, and 90° bend force problems — will build both speed and confidence. Remember the critical distinctions: square the diameter ratio for velocity, place the pump head on the upstream side, divide by efficiency for pumps, multiply by efficiency for turbines, and never forget the pressure-times-area terms in the momentum equation. With these principles fully internalised, you are well-positioned to excel not only in hydraulics but also in the downstream topics of pipe networks, open-channel flow, and hydraulic machinery on the licensure examination.
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