CELE Hydraulics & Fluid Mechanics — Flow in PipesStudy Notes
Full study notes for Flow in Pipes — built specifically for the CELE 2026. These notes cover every concept, definition, formula, and worked example you need for the Hydraulics & Fluid Mechanics subtest of the CELE, structured in the order Professional Regulation Commission (PRC) — Board of Civil Engineering typically tests them.
Exam context
On the CELE 2026, the Hydraulics & Fluid Mechanics subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Flow in Pipes lands at position 6th out of 10 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Hydraulics & Fluid Mechanics on a typical CELE paper.
Flow in Pipes - Study Notes
Flow in pipes is a fundamental topic in hydraulics and fluid mechanics that engineers encounter in water supply systems, sewerage networks, industrial piping, and irrigation projects across the Philippines. When water or other fluids flow through pipes, energy is lost due to friction along the pipe walls (major losses) and from fittings, valves, and changes in flow direction (minor losses). Understanding how to calculate these losses is essential for designing efficient piping systems, selecting appropriate pump sizes, and determining the diameter of pipes needed to carry a required discharge with acceptable head loss. This chapter builds the skills needed to analyze real-world pipe flow problems using industry-standard methods including the Darcy-Weisbach equation, Manning formula, and Hazen-Williams equation—all commonly tested in the PRC Civil Engineer Licensure Examination.
Summary
Flow in pipes is governed by the balance between driving forces (pressure, elevation) and resistance forces (friction, turbulence). The Reynolds number determines whether flow is laminar (smooth, f = 64/Re) or turbulent (chaotic, f from Moody diagram). Major losses (friction along the pipe) are calculated using the Darcy-Weisbach equation: h_f = f(L/D)(v²/2g), or empirically via Manning or Hazen-Williams equations. Minor losses at fittings, bends, and transitions are expressed as h_m = K(v²/2g), where K depends on the component type. In series pipes, the same discharge flows through all sections and head losses add; in parallel pipes, head loss is the same across all branches and discharges add. Total head loss is the sum of major and minor losses, and determines the pump head required and energy cost of the system. Practical design must respect velocity limits (0.6–1.5 m/s for water supply, 0.75–1.5 m/s for sewerage), acceptable head loss (5–10 m per 1000 m), and material-specific roughness values. Common pitfalls include confusing flow regimes, mixing equations inappropriately, ignoring minor losses, unit inconsistencies, and confusion between series and parallel configurations. The methods and equations covered in this chapter are fundamental tools used by civil engineers in the Philippines for designing water supply networks, sewerage systems, irrigation infrastructure, and industrial piping—all essential infrastructure in the country.
Sections
Pipe flow is classified into three regimes based on the Reynolds number (Re), which compares inertial forces to viscous forces in the fluid. The Reynolds number is calculated as: **Re = (ρvD) / μ = vD / ν** where: - ρ = fluid density (kg/m³) - v = mean flow velocity (m/s) - D = pipe diameter (m) - μ = dynamic viscosity (Pa·s or N·s/m²) - ν = kinematic viscosity (m²/s) = μ/ρ For water at 20°C, ν ≈ 1 × 10⁻⁶ m²/s (this value is frequently used in PRC exam problems). **Flow Regime Classification:** - **Laminar flow:** Re < 2000. Flow is smooth and orderly; fluid particles move in parallel layers. The friction factor is exactly f = 64/Re (independent of pipe roughness). - **Transitional flow:** 2000 < Re < 4000. Flow is unstable; neither fully laminar nor fully turbulent. Avoid designing systems in this range. - **Turbulent flow:** Re > 4000. Flow is chaotic with eddies and mixing. Most engineering pipe flows are turbulent. The friction factor f depends on both Re and the relative roughness ε/D, obtained from the Moody diagram. In the Philippines, most municipal water supply and wastewater systems operate in the turbulent regime due to the high flow rates and reasonable pipe diameters used in practice. **Physical Interpretation:** A low Re indicates that viscous forces dominate (typical in small-diameter pipes or with high-viscosity oils), while a high Re indicates inertial forces dominate (typical in large-diameter pipes with water at normal velocities).
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1. Introduction to Pipe Flow & Flow Regimes
Examples
Problem
Water at 20°C flows at 2 m/s in a 100 mm diameter pipe. Determine the Reynolds number and identify the flow regime.
Solution
Given: v = 2 m/s, D = 100 mm = 0.1 m, ν = 1 × 10⁻⁶ m²/s Re = vD/ν = (2)(0.1)/(1 × 10⁻⁶) = 0.2 / (1 × 10⁻⁶) = 200,000 Since Re = 200,000 > 4000, the flow is **TURBULENT**. Note: This is typical for water supply mains in residential areas.
Problem
Oil with kinematic viscosity ν = 4 × 10⁻⁵ m²/s flows at 1.5 m/s in a 50 mm pipe. Determine the flow regime.
Solution
Given: v = 1.5 m/s, D = 50 mm = 0.05 m, ν = 4 × 10⁻⁵ m²/s Re = vD/ν = (1.5)(0.05)/(4 × 10⁻⁵) = 0.075 / (4 × 10⁻⁵) = 1875 Since Re = 1875 < 2000, the flow is **LAMINAR**. This can occur in small-diameter industrial piping with viscous fluids, or in very small-diameter laboratory equipment.
Problem
Water flows in a 200 mm pipe. At what velocity does the flow transition from laminar to turbulent (Re = 2000)?
Solution
At the laminar-turbulent boundary, Re = 2000. 2000 = vD/ν 2000 = v(0.2)/(1 × 10⁻⁶) v = 2000 × 1 × 10⁻⁶ / 0.2 = 0.01 m/s = 1 cm/s So the critical velocity is v = 0.01 m/s. In practice, water supply systems operate at velocities far higher than this, ensuring fully turbulent flow.
Key Points
- Reynolds number determines the flow regime: laminar (Re < 2000), transitional (2000 < Re < 4000), turbulent (Re > 4000)
- For laminar flow, friction factor f = 64/Re exactly; for turbulent flow, f depends on both Re and relative roughness ε/D
- Most practical pipe flows in civil engineering (water supply, sewerage, irrigation) are turbulent
- Kinematic viscosity ν for water at 20°C is approximately 1 × 10⁻⁶ m²/s—memorize this for exam calculations
- The transition between regimes occurs around Re = 2000–4000; this is a critical threshold in pipe design
Major losses (or friction losses) represent the energy lost due to friction between the flowing fluid and the pipe wall over the length of the pipe. The most widely used and scientifically rigorous method is the Darcy-Weisbach equation: **h_f = f × (L/D) × (v²/2g)** where: - h_f = friction head loss (m) - f = Darcy friction factor (dimensionless) - L = pipe length (m) - D = pipe diameter (m) - v = mean flow velocity (m/s) - g = gravitational acceleration (9.81 m/s²) **Determining the Friction Factor f:** **For Laminar Flow (Re < 2000):** f = 64/Re (exact—no empiricism required) Example: If Re = 1000, then f = 64/1000 = 0.064 **For Turbulent Flow (Re > 4000):** The friction factor depends on both the Reynolds number and the relative roughness (ε/D), where ε is the absolute roughness of the pipe material. The Moody diagram (also called the Moody chart) is the graphical tool used to find f for turbulent flow. Key steps: 1. Calculate Re using v, D, and ν. 2. Estimate the relative roughness ε/D. Typical values for common pipe materials: - Commercial steel: ε ≈ 0.045 mm (ε/D = 0.045/D in mm) - Cast iron (new): ε ≈ 0.25 mm - Cast iron (old, corroded): ε ≈ 2.5 mm - Plastic (PVC): ε ≈ 0.0015 mm (very smooth) - Concrete: ε ≈ 0.3–3 mm 3. Locate the point (Re, ε/D) on the Moody diagram to read off f. Alternatively, the **Colebrook-White equation** can be used iteratively for higher precision (though rarely required in exams): 1/√f = −2 log₁₀(ε/(3.7D) + 2.51/(Re√f)) **Practical Note for Exams:** If a friction factor value is not provided in the problem, you must either (1) be given the Moody diagram, (2) be given enough information to estimate f, or (3) be told to use a specific empirical formula like Manning or Hazen-Williams. **Key Observation:** The friction loss h_f scales with v² (and also with L/D). Doubling the velocity quadruples the loss; this is why high-velocity pipelines are penalized heavily in head loss calculations. **Units Check:** Ensure all quantities are in SI units: D in meters, v in m/s, L in meters, and g = 9.81 m/s². The result h_f will be in meters of head (or equivalently, N/kg or J/kg when multiplied by g).
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2. Major Losses — Darcy-Weisbach Equation
Examples
Problem
Water flows at 3 m/s in a 200 mm diameter steel pipe that is 100 m long. Calculate the friction head loss using the Darcy-Weisbach equation. Assume f = 0.02 (turbulent).
Solution
Given: v = 3 m/s, D = 0.2 m, L = 100 m, f = 0.02, g = 9.81 m/s² h_f = f(L/D)(v²/2g) h_f = 0.02 × (100/0.2) × (3²/(2 × 9.81)) h_f = 0.02 × 500 × (9/19.62) h_f = 0.02 × 500 × 0.4588 h_f = 4.59 m The friction head loss is **4.59 m**. This means the pressure drop is 4.59 × 9.81 = 45.0 kPa, or water level drops 4.59 m over the 100 m length of pipe.
Problem
Oil (ν = 4 × 10⁻⁵ m²/s) flows at 1.5 m/s in a 50 mm pipe that is 200 m long. The pipe is commercial steel (ε ≈ 0.045 mm). Calculate the friction head loss.
Solution
Step 1: Calculate Reynolds number. Re = vD/ν = (1.5 × 0.05) / (4 × 10⁻⁵) = 0.075 / (4 × 10⁻⁵) = 1875 Since Re = 1875 < 2000, the flow is **LAMINAR**. Step 2: For laminar flow, f = 64/Re. f = 64/1875 = 0.0341 Step 3: Apply Darcy-Weisbach. h_f = f(L/D)(v²/2g) h_f = 0.0341 × (200/0.05) × (1.5²/(2 × 9.81)) h_f = 0.0341 × 4000 × (2.25/19.62) h_f = 0.0341 × 4000 × 0.1148 h_f = 15.65 m The friction head loss is **15.65 m**. Note: The head loss is much higher than in the turbulent water case because the oil is more viscous, creating more friction.
Problem
A PVC pipe (ε ≈ 0.0015 mm) carries water at 2.5 m/s. The pipe diameter is 150 mm and length is 250 m. Use f = 0.018. Find the friction head loss.
Solution
Given: v = 2.5 m/s, D = 0.15 m, L = 250 m, f = 0.018 h_f = f(L/D)(v²/2g) h_f = 0.018 × (250/0.15) × (2.5²/(2 × 9.81)) h_f = 0.018 × 1666.67 × (6.25/19.62) h_f = 0.018 × 1666.67 × 0.3186 h_f = 9.58 m The friction head loss is **9.58 m**. This is a typical scenario for medium-sized water distribution pipes.
Key Points
- The Darcy-Weisbach equation h_f = f(L/D)(v²/2g) is the most fundamental and reliable method for calculating friction head loss
- For laminar flow (Re < 2000), the friction factor is exactly f = 64/Re—no need for a diagram
- For turbulent flow, the friction factor f depends on both Reynolds number and relative roughness; use the Moody diagram or iterative equations
- Friction loss is proportional to v² and to L; doubling velocity quadruples the loss
- Common pipe material roughness values: steel ~0.045 mm, cast iron (new) ~0.25 mm, PVC ~0.0015 mm, concrete 0.3–3 mm
- Always maintain SI units: D in m, v in m/s, L in m, g = 9.81 m/s²
While the Darcy-Weisbach equation is the most theoretically sound method, two empirical equations are commonly used in practice, especially in older engineering work and in certain applications like sewerage and irrigation: **Manning Equation (used primarily for open channels, but also for full pipes):** v = (1/n) × R^(2/3) × S^(1/2) where: - v = mean velocity (m/s) - n = Manning's roughness coefficient (dimensionless) - R = hydraulic radius (m) = Area / Wetted Perimeter. For a full circular pipe, R = D/4. - S = friction slope = h_f / L (dimensionless) Rearranging for head loss: **h_f = (6.35 × n² × L × v²) / D^(4/3)** (This is the SI form of the Manning equation for full pipes.) Typical Manning's n values: - PVC or plastic pipes: n ≈ 0.009–0.010 - Concrete pipes: n ≈ 0.012–0.015 - Brick sewers: n ≈ 0.014–0.016 - Corroded cast iron: n ≈ 0.025–0.035 **Hazen-Williams Equation (widely used for water distribution systems):** v = 0.849 × C × R^(0.63) × S^(0.54) where: - C = Hazen-Williams roughness coefficient (dimensionless) - R = hydraulic radius (D/4 for full pipes) - S = friction slope Rearranging for head loss: **h_f = (10.67 × L × Q^1.85) / (C^1.85 × D^4.87)** where Q is discharge in m³/s (this form is often more convenient for design). Alternatively, in terms of velocity: **h_f = (6.05 × L × v^1.85) / (C^1.85 × D^1.17)** Typical Hazen-Williams C values: - PVC or plastic (new): C ≈ 150 - Steel (new): C ≈ 140 - Cast iron (new): C ≈ 130 - Cast iron (old, corroded): C ≈ 90–100 - Concrete: C ≈ 120–130 **When to Use Each Method:** - **Darcy-Weisbach:** Most accurate; required by modern standards; preferred for research and complex networks. - **Manning:** Traditional choice for sewerage design and open-channel flows; NSCP and Philippine standards reference this. - **Hazen-Williams:** Common in older water supply designs; easier hand calculations; less accurate for very small or very large diameters. **Important Note:** Do not mix coefficients between methods. If you use Manning's n, do not try to convert it to a Darcy f value directly—the equations have different forms and assumptions. **Comparison Example:** For the same pipe and flow, all three methods should yield approximately the same head loss (within 5–10%), but they will differ slightly due to the different empirical bases and exponents.
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3. Major Losses — Manning & Hazen-Williams Equations
Examples
Problem
A concrete sewerage pipe (Manning's n = 0.013) has a diameter of 300 mm and carries a flow of 0.15 m³/s. The pipe is 500 m long. Calculate the friction head loss using the Manning equation.
Solution
Step 1: Calculate velocity. Q = (π/4) × D² × v 0.15 = (π/4) × (0.3)² × v 0.15 = 0.0707 × v v = 2.122 m/s Step 2: Apply Manning's equation for head loss. h_f = (6.35 × n² × L × v²) / D^(4/3) h_f = (6.35 × 0.013² × 500 × 2.122²) / (0.3)^(4/3) h_f = (6.35 × 0.000169 × 500 × 4.503) / 0.1646 h_f = 9.636 / 0.1646 h_f = 58.6 m The friction head loss is **58.6 m** over 500 m of pipe.
Problem
A steel water main (Hazen-Williams C = 140, D = 200 mm) carries 0.1 m³/s over a distance of 1000 m. Calculate the head loss using the Hazen-Williams formula.
Solution
Use the Hazen-Williams equation in terms of discharge: h_f = (10.67 × L × Q^1.85) / (C^1.85 × D^4.87) where Q must be in m³/s and D in m. h_f = (10.67 × 1000 × 0.1^1.85) / (140^1.85 × 0.2^4.87) Calculate the numerator: Q^1.85 = 0.1^1.85 = 0.01413 Numerator = 10.67 × 1000 × 0.01413 = 150.8 Calculate the denominator: C^1.85 = 140^1.85 = 12,480 D^4.87 = 0.2^4.87 = 0.005245 Denominator = 12,480 × 0.005245 = 65.4 h_f = 150.8 / 65.4 = 2.31 m The friction head loss is **2.31 m** over 1000 m—much more modest than the sewerage pipe example because the diameter is larger and the flow velocity lower.
Problem
Compare the head loss for the same conditions using all three methods: D = 150 mm, v = 2 m/s, L = 300 m, n = 0.012, C = 130, f = 0.022.
Solution
**Darcy-Weisbach:** h_f = f(L/D)(v²/2g) = 0.022 × (300/0.15) × (2²/19.62) = 0.022 × 2000 × 0.2041 = 8.98 m **Manning:** h_f = (6.35 × 0.012² × 300 × 2²) / 0.15^(4/3) = (6.35 × 0.000144 × 300 × 4) / 0.1146 = 1.097 / 0.1146 = 9.58 m **Hazen-Williams:** First, Q = (π/4)(0.15)²(2) = 0.0353 m³/s h_f = (10.67 × 300 × 0.0353^1.85) / (130^1.85 × 0.15^4.87) = 10.67 × 300 × 0.00179 / (10,700 × 0.00376) = 5.72 / 40.2 = 8.92 m **Summary:** - Darcy-Weisbach: 8.98 m - Manning: 9.58 m - Hazen-Williams: 8.92 m All three methods yield similar results (within ~7%), confirming that when applied correctly with appropriate coefficients, they are consistent.
Key Points
- Manning equation: h_f = (6.35 n² L v²) / D^(4/3) for full pipes in SI units; commonly used in sewerage and Philippine practice
- Hazen-Williams equation: h_f = (10.67 L Q^1.85) / (C^1.85 D^4.87); widely used in water supply design
- Manning's n and Hazen-Williams C are empirical roughness coefficients that account for pipe material and age
- Typical values: Manning n ≈ 0.009–0.015 for newer pipes; Hazen-Williams C ≈ 90–150
- Do not mix equations or coefficients—each method has its own parameters and form
- All three methods (Darcy-Weisbach, Manning, Hazen-Williams) should yield similar results when applied correctly
Minor losses (also called local losses or fitting losses) occur at specific locations in the pipe system where the flow is disturbed: bends, valves, tees, expansions, contractions, entrances, and exits. These losses are typically expressed in terms of a loss coefficient K: **h_m = K × (v²/2g)** where: - h_m = head loss at the fitting (m) - K = loss coefficient (dimensionless, specific to the fitting type and geometry) - v = mean flow velocity in the pipe section (m/s) - g = 9.81 m/s² **Typical Loss Coefficients for Common Fittings:** | Fitting Type | K Value | Notes | |---|---|---| | Sharp-edged pipe entrance | 0.5 | Water enters directly from a sharp edge | | Rounded entrance | 0.05–0.1 | Smooth, rounded inlet | | Pipe exit (into reservoir) | 1.0 | All kinetic energy is dissipated | | 90° elbow (threaded) | 0.9–1.0 | Standard plumbing elbow | | 45° elbow | 0.4–0.5 | Less severe than 90° | | Tee junction (flow through main) | 0.6 | Flow continues straight | | Tee junction (flow into branch) | 1.0–1.5 | Flow makes a 90° turn | | Globe valve (fully open) | 10 | Highly restrictive | | Gate valve (fully open) | 0.2 | Minimal resistance | | Check valve | 2–3 | Allows flow one direction only | | Sudden expansion from D₁ to D₂ | (1 − D₁²/D₂²)² | See derived formula below | | Sudden contraction from D₁ to D₂ | 0.5(1 − D₂²/D₁²) | Approximate for sharp contraction | **Special Case: Sudden Expansion (Borda-Carnot):** When flow suddenly expands from diameter D₁ (velocity v₁) to diameter D₂ (velocity v₂), the head loss is: **h_m = ((v₁ − v₂)² / (2g)) = ((v₁)² / (2g)) × (1 − D₁²/D₂²)²** This arises because some of the kinetic energy in the smaller section is dissipated as turbulence in the larger section. **Special Case: Sudden Contraction:** For a sudden contraction from D₁ to D₂, the loss coefficient is approximately: **K ≈ 0.5 × (1 − D₂²/D₁²)** Note that contraction losses are generally smaller than expansion losses because the flow is able to form a vena contracta (a narrower jet) before re-expanding. **Equivalent Pipe Length Method:** Some engineers use an "equivalent length" L_eq approach: instead of calculating h_m directly, they find the length of pipe that would produce the same friction loss: **h_m = K(v²/2g) = f(L_eq/D)(v²/2g)** Solving for L_eq: **L_eq = K × D / f** This is useful when comparing the relative importance of major vs. minor losses. For example, a 90° elbow (K = 0.9) in a 100 mm pipe with f = 0.02 is equivalent to approximately 4.5 m of straight pipe. **Combining Minor Losses:** When multiple fittings are in series, the total minor head loss is: **h_minor_total = (K₁ + K₂ + K₃ + ...) × (v²/2g)** Note that all K coefficients are summed before multiplying by (v²/2g). This is valid only when the velocity is the same throughout (i.e., the pipe diameter does not change between fittings). **Important:** If the pipe diameter changes between fittings, you must use the velocity corresponding to each section and apply losses separately.
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4. Minor Losses
Examples
Problem
Water flows at 3 m/s in a 100 mm pipe and enters through a sharp-edged inlet (K = 0.5). Calculate the entrance loss.
Solution
Given: v = 3 m/s, K = 0.5 h_m = K(v²/2g) = 0.5 × (3²/(2 × 9.81)) = 0.5 × (9/19.62) = 0.5 × 0.4588 = 0.229 m The entrance head loss is **0.229 m** or approximately **23 cm**.
Problem
A 150 mm diameter pipe contains a 90° elbow (K = 0.9), a gate valve (K = 0.2), and exits into a reservoir (K = 1.0). Water flows at 2.5 m/s. Calculate the total minor losses.
Solution
Given: v = 2.5 m/s, K_total = 0.9 + 0.2 + 1.0 = 2.1 h_m_total = K_total(v²/2g) = 2.1 × (2.5²/(2 × 9.81)) h_m_total = 2.1 × (6.25/19.62) = 2.1 × 0.3186 = 0.669 m The total minor head loss is **0.669 m** or approximately **67 cm**.
Problem
Water flows from a 100 mm pipe (v₁ = 4 m/s) into a sudden expansion to a 200 mm pipe (v₂ = 1 m/s). Calculate the expansion loss using the Borda-Carnot formula.
Solution
Given: D₁ = 0.1 m, D₂ = 0.2 m, v₁ = 4 m/s, v₂ = 1 m/s Note: v₂ is determined by continuity: A₁v₁ = A₂v₂ v₂ = (π/4)(0.1)² × 4 / [(π/4)(0.2)²] = 1 m/s ✓ Using the direct formula: h_m = (v₁ − v₂)²/(2g) = (4 − 1)²/(2 × 9.81) = 9/19.62 = 0.459 m Alternatively, using the diameter-ratio formula: h_m = (v₁²/2g) × (1 − D₁²/D₂²)² = (4²/19.62) × (1 − 0.1²/0.2²)² h_m = 0.8165 × (1 − 0.25)² = 0.8165 × 0.5625 = 0.459 m ✓ The expansion head loss is **0.459 m**. This energy is dissipated as heat due to the turbulent eddies formed in the expanding section.
Problem
A 100 mm pipe (f = 0.02) contains a 90° elbow (K = 0.9). Compare the head losses: 1 m of straight pipe vs. the elbow at v = 2 m/s.
Solution
**Straight pipe (1 m):** h_f = f(L/D)(v²/2g) = 0.02 × (1/0.1) × (2²/19.62) = 0.02 × 10 × 0.2041 = 0.0408 m = 4.08 cm **90° Elbow:** h_m = K(v²/2g) = 0.9 × (2²/19.62) = 0.9 × 0.2041 = 0.184 m = 18.4 cm **Equivalent length of straight pipe for the elbow:** L_eq = K × D / f = 0.9 × 0.1 / 0.02 = 4.5 m Interpretation: A 90° elbow in this system causes as much head loss as **4.5 m of straight pipe**. So the elbow is significant and should not be ignored in system design.
Key Points
- Minor losses are expressed as h_m = K(v²/2g), where K is a fitting-specific loss coefficient
- Typical K values: sharp entrance 0.5, exit 1.0, 90° elbow 0.9, gate valve 0.2, globe valve 10
- For sudden expansion: h_m = ((v₁ − v₂)²/2g) where v₁ and v₂ are velocities before and after expansion
- Multiple fittings in series: sum all K values, then multiply once by (v²/2g)
- Equivalent pipe length method: L_eq = K × D / f—useful for comparing major and minor losses
- Exit losses account for all remaining kinetic energy being dissipated; entrance losses depend on the sharpness of the inlet
In a complete piping system, the total head loss is the sum of all major (friction) losses and all minor losses: **h_total = h_f_major + h_m_minor** or **h_total = f(L/D)(v²/2g) + Σ[K(v²/2g)]** **h_total = [(v²/2g)] × [f(L/D) + ΣK]** If multiple pipes with different diameters and velocities are involved, each section must be analyzed separately because the velocity varies. **System Head-Flow Relationship (H-Q Curve):** For a given piping system with specified geometry (lengths, diameters, fittings), the relationship between total head loss and discharge can be expressed as: **h_total = a × Q² + b × Q** or more generally, **h_total = c × Q^n** where the exponent n is typically between 1.8 and 2 depending on the flow regime and equation used. For **turbulent flow** with Darcy-Weisbach, the relation is approximately quadratic (n ≈ 2). This H-Q relationship is crucial because: 1. It characterizes the "resistance" of the piping system. 2. When plotted on a graph, it is the **system curve** against which a pump's performance curve is overlaid to find the operating point. 3. Systems in series have their H-Q curves added vertically (same Q, head losses add). 4. Systems in parallel have their H-Q curves added horizontally (same head loss, flows add). **Energy Balance in Pipes (General System Equation):** Between two points in a piping system, the energy balance (Bernoulli equation with losses) is: **(P₁/ρg) + (v₁²/2g) + z₁ = (P₂/ρg) + (v₂²/2g) + z₂ + h_L** where: - P = static pressure (Pa) - ρ = fluid density (kg/m³) - v = velocity (m/s) - z = elevation (m) - h_L = total head loss (m) If the pipe diameter and fluid are the same, v₁ = v₂, and the equation simplifies to: **ΔP = ρ × g × h_L** The pressure drop ΔP (in Pa or kPa) is directly proportional to the head loss. **Practical Example:** A 1 m head loss in a water system corresponds to a pressure drop of 1 m × 9.81 kN/m³ = 9.81 kPa (or approximately 10 kPa, often rounded to 0.1 bar). **Iterative Solution for Unknown Velocity (or Diameter):** Often in pipe design, the velocity (or diameter) is unknown but the discharge Q is specified. The procedure is: 1. Assume an initial velocity (or diameter). 2. Calculate Re and determine f from the Moody diagram or the appropriate equation. 3. Calculate the total head loss h_L using the Darcy-Weisbach equation and minor loss coefficients. 4. If h_L is not close to the allowable head loss (or if the pressure drop exceeds the available pump head), revise the assumed diameter and repeat. 5. Converge on a diameter that satisfies both the discharge requirement and the acceptable head loss. Alternatively, use iterative software or graphical methods (system curve intersection with pump curve).
Heading
5. Total Head Loss & System Equations
Examples
Problem
A 150 mm diameter steel pipe (f = 0.018, ε = 0.045 mm) carries 0.05 m³/s of water over 500 m. The line includes an entrance (K = 0.5), two 90° elbows (K = 0.9 each), and an exit (K = 1.0). Calculate the total head loss.
Solution
**Step 1: Calculate velocity.** A = (π/4) × (0.15)² = 0.01767 m² v = Q/A = 0.05/0.01767 = 2.828 m/s **Step 2: Major loss (Darcy-Weisbach).** h_f = f(L/D)(v²/2g) = 0.018 × (500/0.15) × (2.828²/19.62) h_f = 0.018 × 3333.3 × 0.408 = 24.56 m **Step 3: Minor losses.** K_total = 0.5 + 0.9 + 0.9 + 1.0 = 3.3 h_m = K_total(v²/2g) = 3.3 × (2.828²/19.62) = 3.3 × 0.408 = 1.346 m **Step 4: Total head loss.** h_total = h_f + h_m = 24.56 + 1.346 = 25.91 m ≈ **26 m** Pressure drop: ΔP = ρgh = 1000 × 9.81 × 25.91 = 254 kPa
Problem
A pump must deliver 0.1 m³/s through a 200 m system consisting of 200 mm and 150 mm pipes in series. Estimate the required pump head (assume both pipes have f = 0.02, and the 150 mm section has 40 m, the 200 mm section has 160 m; ignore minor losses for simplicity).
Solution
**For the 200 mm section (160 m, Q = 0.1 m³/s):** A₁ = (π/4)(0.2)² = 0.0314 m² v₁ = 0.1/0.0314 = 3.18 m/s h_f1 = 0.02 × (160/0.2) × (3.18²/19.62) = 0.02 × 800 × 0.515 = 8.24 m **For the 150 mm section (40 m, Q = 0.1 m³/s):** A₂ = (π/4)(0.15)² = 0.01767 m² v₂ = 0.1/0.01767 = 5.66 m/s h_f2 = 0.02 × (40/0.15) × (5.66²/19.62) = 0.02 × 266.67 × 1.631 = 8.70 m **Total head loss (series):** h_total = h_f1 + h_f2 = 8.24 + 8.70 = 16.94 m ≈ **17 m** Note: The smaller diameter (150 mm) section, although much shorter, contributes almost as much loss because velocity is higher (v ∝ 1/D² effect).
Problem
Two identical 150 mm pipes (each 200 m long, f = 0.02) are connected in parallel and carry a total of 0.1 m³/s. Find the discharge in each pipe and verify that the head loss is the same in both branches.
Solution
**For parallel pipes, the head loss must be the same in both branches:** h_f1 = h_f2 f(L/D)(v₁²/2g) = f(L/D)(v₂²/2g) Since f, L, and D are identical for both pipes: v₁ = v₂ By continuity: Q₁ + Q₂ = 0.1, and since v₁ = v₂ with identical diameters: Q₁ = Q₂ = 0.05 m³/s **Velocity in each pipe:** v = 0.05 / [(π/4)(0.15)²] = 0.05 / 0.01767 = 2.828 m/s **Head loss in each pipe:** h_f = 0.02 × (200/0.15) × (2.828²/19.62) = 0.02 × 1333.3 × 0.408 = 10.88 m Verification: Both pipes have h_f = 10.88 m (same, as required for parallel configuration). ✓
Key Points
- Total head loss = major loss (friction) + minor loss (fittings): h_total = h_f + h_m
- The system H-Q relationship typically has an exponent n ≈ 2 (roughly quadratic) for turbulent flow
- Energy balance (Bernoulli with losses): (P₁/ρg) + (v₁²/2g) + z₁ = (P₂/ρg) + (v₂²/2g) + z₂ + h_L
- Pressure drop ΔP (kPa) = ρg × h_L (m); for water, 1 m head loss ≈ 10 kPa
- Pipe diameter selection requires iterative calculation or graphical methods (system curve vs. pump curve intersection)
- Series pipes: sum head losses with the same discharge; parallel pipes: same head loss, discharges add
Real-world piping systems often consist of multiple pipes connected in different configurations. Understanding how to analyze series and parallel arrangements is essential for network design and hydraulic system modeling. **PIPES IN SERIES:** When pipes are connected end-to-end such that the same discharge flows through each pipe sequentially, they are in series. Key characteristics: 1. **Same Discharge:** Q₁ = Q₂ = Q₃ = ... = Q (by continuity, mass cannot accumulate at any junction). 2. **Head Losses Add:** The total head loss is the algebraic sum of all individual losses: **h_L_total = h_L1 + h_L2 + h_L3 + ...** 3. **Different Diameters Allowed:** Each section can have a different diameter, length, and friction factor. Each must be analyzed with its own velocity: - v₁ = Q / A₁ - v₂ = Q / A₂ - etc. 4. **Pressure Drop Relationship:** **ΔP_total = ΔP₁ + ΔP₂ + ΔP₃ + ... = ρg(h_L1 + h_L2 + ...)** **Practical Example (Philippine Context):** A water supply main from a treatment plant to a residential area often passes through different diameter sections: a 300 mm trunk main, then 200 mm secondary lines, then 100 mm tertiary pipes to individual zones. All series, and head losses accumulate. **PIPES IN PARALLEL:** When multiple pipes connect the same two points (or junctions), branching out and then recombining, they are in parallel. Key characteristics: 1. **Same Head Loss:** The head loss across each branch must be identical: **h_L1 = h_L2 = h_L3 = ... = h_L** (If this were not true, water would redistribute until equilibrium is reached.) 2. **Discharges Add:** The total discharge is the sum of flows in each branch: **Q_total = Q₁ + Q₂ + Q₃ + ...** 3. **Different Velocities:** Since Q may differ across branches but h_L is the same, velocities will generally differ: v₁ = Q₁ / A₁, v₂ = Q₂ / A₂, etc. 4. **Equivalent Resistance:** The system can be represented by an equivalent single pipe with properties such that the same total Q is delivered at the same head loss h_L. **Finding Flow Distribution in Parallel Pipes:** Given that h_L is the same in all branches: h_L = f₁(L₁/D₁)(v₁²/2g) = f₂(L₂/D₂)(v₂²/2g) = ... This relationship between h_L and the pipe properties allows you to solve for the velocity (and hence discharge) in each branch. The general approach: 1. Express h_L in terms of the discharge and pipe properties for each branch. 2. Set h_L equal across all branches. 3. Solve the system of equations simultaneously with the continuity constraint Q_total = ΣQᵢ. **Practical Example:** A water distribution network with two parallel connections between points A and B (perhaps one old corroded pipe and one new smooth pipe). The flow will split based on the relative resistances—more flow goes through the new (less resistive) pipe. **COMBINED SERIES-PARALLEL NETWORKS:** Complex pipe networks often have both series and parallel sections. For example: - A main trunk (series) branches into two parallel secondary lines, each of which further branches into parallel tertiary lines (also series-parallel). Solving such networks requires: 1. **Loop Equations:** Around each closed loop, the sum of head gains (from pumps) equals the sum of head losses (Kirchhoff-like principle for hydraulics). 2. **Node Equations:** At each junction, inflow equals outflow (continuity). 3. **Iterative Methods:** The **Hardy-Cross method** is the classical approach for hand calculations, but modern software solves these systems numerically. **Hardy-Cross Method (Outline):** 1. Assume initial flow rates in each pipe (respecting continuity at nodes). 2. For each loop, calculate the algebraic sum of head losses (applying sign convention: clockwise positive). 3. If the sum is not zero, apply a correction factor to adjust flows. 4. Repeat until each loop's net head loss is approximately zero. (Full Hardy-Cross calculation is beyond typical board-exam scope, but understanding the principle is valuable.) **Pressure at Junctions:** In a series-parallel network, the pressure at a junction is the same regardless of which path the fluid took to reach it (otherwise, water would redistribute). This constraint is used to set up the equations: **P_A + (pressure drop from A to junction via path 1) = P_A + (pressure drop from A to junction via path 2)**
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6. Pipes in Series and Parallel
Examples
Problem
Three pipes in series: (1) 100 m, D = 0.2 m, f = 0.02; (2) 50 m, D = 0.15 m, f = 0.02; (3) 75 m, D = 0.25 m, f = 0.02. Water discharge is 0.08 m³/s. Calculate the total head loss.
Solution
**Pipe 1:** A₁ = (π/4)(0.2)² = 0.0314 m² v₁ = 0.08/0.0314 = 2.546 m/s h_L1 = 0.02 × (100/0.2) × (2.546²/19.62) = 0.02 × 500 × 0.330 = 3.30 m **Pipe 2:** A₂ = (π/4)(0.15)² = 0.01767 m² v₂ = 0.08/0.01767 = 4.530 m/s h_L2 = 0.02 × (50/0.15) × (4.530²/19.62) = 0.02 × 333.3 × 1.046 = 6.97 m **Pipe 3:** A₃ = (π/4)(0.25)² = 0.04909 m² v₃ = 0.08/0.04909 = 1.630 m/s h_L3 = 0.02 × (75/0.25) × (1.630²/19.62) = 0.02 × 300 × 0.136 = 0.82 m **Total head loss (series):** h_L_total = 3.30 + 6.97 + 0.82 = 11.09 m ≈ **11.1 m** Note: Pipe 2, with the smallest diameter, accounts for the largest share of loss despite its short length.
Problem
Two identical parallel pipes (both 150 mm, 200 m, f = 0.02) branch from point A and rejoin at point B. The total discharge is 0.1 m³/s. Assuming equal flow distribution, find the discharge in each pipe and the head loss.
Solution
**For identical pipes in parallel, flow splits equally:** Q₁ = Q₂ = 0.1/2 = 0.05 m³/s **Velocity in each pipe:** A = (π/4)(0.15)² = 0.01767 m² v = 0.05/0.01767 = 2.828 m/s **Head loss in each pipe:** h_L = 0.02 × (200/0.15) × (2.828²/19.62) = 0.02 × 1333.3 × 0.408 = 10.88 m Verification: Both branches have h_L = 10.88 m ✓ (equal, as required for parallel). The system head loss is **10.88 m** (same as each individual branch).
Problem
Two parallel pipes (A and B) connect points 1 and 2. Pipe A: L_A = 100 m, D_A = 0.15 m, f_A = 0.02. Pipe B: L_B = 200 m, D_B = 0.2 m, f_B = 0.018. Total discharge is 0.15 m³/s. Assuming the head loss is the same in both branches, set up the equations to find Q_A and Q_B (do not solve numerically; just set up).
Solution
Let h_L = head loss in both branches (same for parallel). **For Pipe A:** h_L = f_A(L_A/D_A)(v_A²/2g) = 0.02 × (100/0.15) × (v_A²/19.62) h_L = 0.6803 × v_A² Expressing in terms of discharge Q_A = v_A × A_A: v_A = Q_A / [(π/4)(0.15)²] = Q_A / 0.01767 h_L = 0.6803 × (Q_A/0.01767)² = 2177 × Q_A² **For Pipe B:** h_L = 0.018 × (200/0.2) × (v_B²/19.62) h_L = 0.0918 × v_B² v_B = Q_B / [(π/4)(0.2)²] = Q_B / 0.0314 h_L = 0.0918 × (Q_B/0.0314)² = 9.32 × Q_B² **System of equations:** 1) Head loss equality: 2177 × Q_A² = 9.32 × Q_B² 2) Continuity: Q_A + Q_B = 0.15 m³/s From equation (1): Q_A² / Q_B² = 9.32 / 2177 = 0.00428 Q_A / Q_B = 0.0654 Q_A = 0.0654 × Q_B Substitute into equation (2): 0.0654 × Q_B + Q_B = 0.15 1.0654 × Q_B = 0.15 Q_B = 0.1408 m³/s, Q_A = 0.0092 m³/s Interpretation: Almost all flow goes through Pipe B because it has a larger diameter and is smoother (lower f), making it much less resistive. Pipe A is a bottleneck.
Key Points
- Series pipes: same discharge Q in all sections; total head loss h_L = Σh_i; different diameters and velocities allowed
- Parallel pipes: same head loss h_L in all branches; total discharge Q_total = ΣQᵢ; different velocities in each branch
- In parallel pipes, flow distributes inversely with resistance—more flow goes through lower-resistance (larger, smoother) pipes
- Complex networks combine series and parallel sections and require node and loop equations (Kirchhoff analogy)
- Hardy-Cross method iteratively corrects assumed flows until loop head balances and node continuity are satisfied
- At any junction, pressure is the same regardless of the path followed (equilibrium condition)
When designing piping systems for Philippine water supply, wastewater, irrigation, or industrial applications, several practical considerations and common mistakes must be avoided. **DESIGN CONSTRAINTS & STANDARDS:** 1. **Velocity Limits (NSCP & Good Practice):** - **Water supply:** Typical design velocity is 0.6–1.5 m/s. Too slow (< 0.5 m/s) can cause sedimentation and taste/odor problems; too fast (> 2 m/s) causes excessive friction losses and noise. - **Sewerage:** Typical velocity is 0.75–1.5 m/s for gravity lines. Too slow causes solids deposition; too fast causes scour and erosion of concrete. - **Pressure pipes (force mains):** Can tolerate up to 2–3 m/s but higher velocities increase energy costs and pipe stress. 2. **Pressure Drop Limits:** - In long distribution networks, allowable head loss is typically 5–10 m per 1000 m (or 0.5–1% head loss per kilometer). - In irrigation systems, losses should not exceed 20% of the available head. - In industrial systems, acceptable loss depends on available pump head and energy cost. 3. **Minimum Pipe Diameter:** - Water supply networks: typically 50 mm (domestic) to 300+ mm (trunk mains). - Sewerage: typically 100 mm minimum for house connections, 150–300 mm for lateral and secondary lines, up to 600 mm+ for main collectors. - Always maintain minimum diameter to avoid blockage and ensure self-cleaning velocities. **COMMON BOARD-EXAM PITFALLS:** 1. **Mixing Equations and Coefficients:** - Do NOT use a Manning's n value in the Hazen-Williams equation, or vice versa. Each method has its own parameters. - If given n but asked for Darcy-Weisbach solution, you cannot directly convert n to f. Use Manning's h_f equation instead, or acknowledge the error. 2. **Neglecting Minor Losses:** - Minor losses are often 10–20% of major losses in long pipe runs; ignoring them can underestimate required pump head. - Always check: is the equivalent pipe length of all fittings significant compared to the actual pipe length? 3. **Unit Inconsistency:** - A common error: plugging diameter in cm instead of meters into the Darcy-Weisbach formula. - Keep all SI units: D in m, v in m/s, L in m, Q in m³/s, g = 9.81 m/s². 4. **Confusion Between Head Loss and Pressure Drop:** - Head loss h_L is expressed in meters (or equivalently, J/kg). Pressure drop ΔP = ρ × g × h_L is in pascals (Pa) or bars. - For water: 1 m head loss ≈ 10 kPa or 0.1 bar. Do not confuse "meter" with "psi" or other pressure units. 5. **Reynolds Number Mistakes:** - Forgetting that ν (kinematic viscosity) is in m²/s, not cm²/s. For water at 20°C, ν = 1 × 10⁻⁶ m²/s (not 0.01 cm²/s unless you convert consistently). - Calculating Re with the wrong viscosity value (e.g., using dynamic μ instead of kinematic ν, or vice versa). 6. **Series vs. Parallel Confusion:** - In series: same Q, head losses add (h_total = Σh_i). - In parallel: same h_L, discharges add (Q_total = ΣQᵢ). - This is often tested in PE licensure exams and is frequently misunderstood. 7. **Improper Assumption of Friction Factor:** - If the problem provides a friction factor, use it. If not, you must either: a. Calculate Re and read f from the Moody diagram (must be provided or known). b. Use an alternative method (Manning or Hazen-Williams). - Never assume f ≈ 0.02 without justification. 8. **Ignoring Changes in Pipe Diameter:** - When diameter changes, velocity changes (by continuity: v ∝ 1/D²). - Reynolds number and friction factor change too. - Each section must be analyzed with its own velocity, Re, and f. 9. **Not Checking the Flow Regime:** - Always calculate Re to confirm whether the flow is laminar or turbulent. - Using turbulent friction factor formulas (like Moody) for laminar flow, or vice versa, gives wildly incorrect results. **PRACTICAL NOTES FOR PHILIPPINE CONTEXT:** - **Water Supply:** Most municipal systems in the Philippines operate gravity-fed from elevated sources or require booster pumps to overcome topography. Head losses directly translate to required pump power and energy cost. - **Sewerage:** Gravity-driven; must maintain adequate velocity to prevent solid deposition (common problem in flat areas). Friction losses reduce driving head; slope must be sufficient. - **Corrosion:** Older cast-iron pipes in the Philippines often have roughness values far exceeding "new" estimates (ε can increase to 2–3 mm). Use historical data if available. - **Maintenance:** Sediment and calcium deposits increase effective roughness over time; systems designed for "new" conditions often underperform after 10–20 years.
Heading
7. Practical Design Considerations & Common Pitfalls
Examples
Problem
A municipal water main (D = 250 mm, L = 5 km, f = 0.018) must deliver 0.2 m³/s. Check if the design velocity is within acceptable limits (0.6–1.5 m/s), calculate the head loss, and estimate the required pump head assuming 10 m elevation rise and 5 m of minor losses.
Solution
**Velocity check:** A = (π/4)(0.25)² = 0.04909 m² v = 0.2/0.04909 = 4.08 m/s ✗ PROBLEM: v = 4.08 m/s > 1.5 m/s (exceeds design limit). This design is not acceptable. The system would experience: - Excessive friction losses (wasted energy) - High pressure fluctuations (water hammer risk) - Noise and vibration in pipes Recommendation: Increase diameter to reduce velocity. **If forced to proceed with D = 250 mm:** Major loss: h_f = 0.018 × (5000/0.25) × (4.08²/19.62) = 0.018 × 20,000 × 0.8492 = 305.7 m (!) Minor loss: h_m = 5 m (given) Elevation gain: z = 10 m **Total required pump head:** H_pump = h_f + h_m + z = 305.7 + 5 + 10 = 320.7 m ≈ **321 m of head** Power required: P = ρ × g × Q × H = 1000 × 9.81 × 0.2 × 321 = 630 kW This is economically infeasible. The correct approach is to select a larger diameter.
Problem
Redo the previous example with D = 400 mm.
Solution
**Velocity:** A = (π/4)(0.4)² = 0.1257 m² v = 0.2/0.1257 = 1.59 m/s ✓ Velocity is 1.59 m/s, slightly above the 1.5 m/s guideline but acceptable for a main. **Calculate Reynolds number to determine f:** Re = vD/ν = (1.59)(0.4)/(1 × 10⁻⁶) = 636,000 (highly turbulent) For D = 400 mm and commercial steel (ε ≈ 0.045 mm), ε/D = 0.045/400 = 0.000113 (very smooth relative roughness). From Moody diagram: f ≈ 0.014 **Major loss:** h_f = 0.014 × (5000/0.4) × (1.59²/19.62) = 0.014 × 12,500 × 0.1290 = 22.6 m **Minor loss:** h_m = 5 m (given) **Total required pump head:** H_pump = 22.6 + 5 + 10 = 37.6 m ≈ **38 m** Power required: P = 1000 × 9.81 × 0.2 × 38 = 74.5 kW ✓ Much more economical and practical. The larger diameter reduces losses dramatically.
Problem
A gravity sewerage line (Manning n = 0.013, D = 200 mm, slope S = 0.005, fully flowing) is to serve a neighborhood. Calculate the self-purifying velocity and the discharge capacity. Is the design adequate (target velocity ≥ 0.75 m/s)?
Solution
Using Manning's equation for a full pipe: v = (1/n) × R^(2/3) × S^(1/2) For a circular pipe: R = D/4 = 0.2/4 = 0.05 m v = (1/0.013) × (0.05)^(2/3) × (0.005)^(1/2) v = 76.923 × 0.0684 × 0.0707 v = 0.373 m/s ✗ PROBLEM: v = 0.373 m/s < 0.75 m/s (too slow). Sewage will not scour itself; solids will deposit, causing blockages and odor. **Discharge capacity:** A = (π/4)(0.2)² = 0.0314 m² Q = A × v = 0.0314 × 0.373 = 0.0117 m³/s = 11.7 L/s **Recommendation:** Either (1) increase the slope if topography permits, (2) increase the diameter, or (3) install a septic tank and low-pressure main (which is not self-purifying and requires pumping). This design is marginal and requires engineering judgment based on local conditions.
Key Points
- Design velocity for water supply: 0.6–1.5 m/s; for sewerage: 0.75–1.5 m/s gravity flow
- Allowable head loss: typically 5–10 m per 1000 m (0.5–1% per km) for long distribution systems
- Do not mix equations: Darcy, Manning, and Hazen-Williams are separate methods with their own coefficients
- Unit consistency is critical: always use SI units (m, m/s, m³/s, 9.81 m/s² for g)
- Series pipes: same Q, head losses add. Parallel pipes: same h_L, flows add. This distinction is frequently tested.
- Always calculate Reynolds number to confirm flow regime before selecting friction factor method
- Minor losses can be 10–20% of major losses; neglecting them underestimates required pump head
- Pressure drop ΔP (kPa) = 9.81 × h_L (m) for water; 1 m head loss ≈ 10 kPa
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