CELE Hydraulics & Fluid Mechanics — Hydrodynamics and Fluid MachineryStudy Notes

The foundation of hydrodynamic force analysis is Newton's second law applied to a fluid stream. When a fluid jet changes direction or speed, it exerts a force on whatever changes its momentum. **Momentum Equation (General Form):** $$\sum F = \dot{m}(v_{\text{out}} - v_{\text{in}}) = \rho Q(v_{\text{out}} - v_{\text{in}})$$ where: - $F$ = net force on the fluid (N) - $\dot{m} = \rho Q$ = mass flow rate (kg/s) - $\rho$ = fluid density, typically 1000 kg/m³ for water at 15°C - $Q$ = volumetric flow rate (m³/s) - $v$ = velocity (m/s) By Newton's third law, the force the plate exerts on the fluid equals the force the fluid exerts on the plate (in opposite direction). **Case 1: Jet on a Stationary Flat Plate (Normal Impact)** When a jet strikes a flat plate perpendicularly and the fluid leaves parallel to the plate: - Incoming momentum: $p_1 = \rho Q v$ (normal to plate) - Outgoing momentum: $p_2 \approx 0$ (splits along the plate) - Force exerted by jet on plate: $$F = \rho Q v = \rho A v^2$$ This is one of the most fundamental formulas for jet impact. The force is proportional to the square of velocity and the jet area—doubling velocity quadruples the force. **Case 2: Jet on an Inclined Flat Plate** When a jet strikes a flat plate at an angle $\alpha$ (measured between the jet direction and the plate surface): - Component of velocity normal to plate: $v_n = v \sin\alpha$ - Normal force on plate: $$F_n = \rho Q v \sin\alpha$$ Note: When $\alpha = 90°$ (jet perpendicular to plate), $F_n = \rho Q v$, confirming the flat-plate formula. **Case 3: Jet on a Smooth Curved Vane (Stationary)** When a jet strikes a curved vane and is deflected through angle $\theta$ (the deflection angle): - Incoming velocity component along initial direction: $v$ - Outgoing velocity component along initial direction: $v \cos\theta$ (because the vane changes the jet direction by $\theta$) - Change in momentum along initial direction: $\Delta(mv) = \rho Q(v \cos\theta - v) = \rho Q v(\cos\theta - 1)$ - Force in the direction of initial jet motion: $$F_x = \rho Q v(1 - \cos\theta)$$ This force acts on the vane in the direction opposite to the original jet (the vane exerts this force on the jet to redirect it). Special cases: - $\theta = 0°$: $F = 0$ (no deflection, no force) - $\theta = 90°$: $F = \rho Q v$ (perpendicular turn) - $\theta = 180°$: $F = 2\rho Q v$ (complete reversal, maximum force) **Case 4: Jet on a Moving Curved Vane** For a vane moving at velocity $u$ in the direction of the incoming jet: - The **relative velocity** of jet with respect to vane is $(v - u)$ - The **relative discharge** (flow rate hitting the vane in its reference frame) is $Q_{\text{rel}} = A(v - u)$ - Force on the vane in the direction of jet motion: $$F = \rho A(v - u)^2(1 - \cos\theta)$$ This principle is critical for Pelton wheel (impulse turbine) analysis. As the buckets move faster, the relative velocity decreases, reducing the force. Maximum power occurs at $u = v/2$ (approximately).

When flow passes through a pipe bend, both the direction and the pressure may change. The anchoring force required to hold the pipe in place must balance both the momentum change and the pressure forces. **General Momentum Equation for a Pipe Bend:** Consider a control volume encompassing the bend. Applying the momentum equation in the $x$ and $y$ directions: $$\sum F_x = \rho Q(v_{2x} - v_{1x}) + (p_1 A_1)_x - (p_2 A_2)_x$$ $$\sum F_y = \rho Q(v_{2y} - v_{1y}) + (p_1 A_1)_y - (p_2 A_2)_y$$ where: - $v_{1x}, v_{1y}$ = velocity components at section 1 - $v_{2x}, v_{2y}$ = velocity components at section 2 - $p_1 A_1$ = pressure force at section 1 (acts on the fluid in the positive direction) - $p_2 A_2$ = pressure force at section 2 (acts on the fluid in the direction of the outflow) - The term $\sum F_x$ represents the net external force on the fluid (from the pipe walls) The anchoring force (force exerted by the supports to hold the pipe) is equal and opposite to $\sum F_x$ and $\sum F_y$: $$F_{\text{anchor}} = \sqrt{F_x^2 + F_y^2}$$ **Special Case: 90° Horizontal Bend with Uniform Velocity** For a 90° bend where the pipe diameter and velocity remain constant (smooth bend, no friction): - At section 1: velocity is horizontal ($v_1 = v$, $v_{1y} = 0$) - At section 2: velocity is vertical ($v_2 = v$, $v_{2x} = 0$) - If pressures are atmospheric at both sections ($p_1 = p_2 = 0$ gauge): $$F_x = -\rho Q v \quad (\text{force in negative } x \text{ direction})$$ $$F_y = \rho Q v \quad (\text{force in positive } y \text{ direction})$$ $$F_{\text{anchor}} = \sqrt{(\rho Q v)^2 + (\rho Q v)^2} = \rho Q v\sqrt{2}$$ The resultant force acts at 45° (or 225° depending on convention). **Sign Convention:** - Pressure forces point in the direction the fluid is leaving the control volume. - Momentum terms use the velocity vectors with their proper signs. - The anchoring force is the reaction force required from the pipe supports.

A pump adds energy to fluid by rotating impellers, increasing the pressure and/or elevation (head) of the water. Understanding pump performance, power requirements, and operating characteristics is essential for design of water supply, irrigation, and drainage systems in the Philippines. **Pump Fundamentals:** A pump converts mechanical energy (from a motor or engine) into fluid energy in the form of: 1. **Pressure energy** — increases pressure (used for distribution networks) 2. **Potential energy** — raises water elevation (used in irrigation and water storage) 3. **Kinetic energy** — increases velocity (used in some applications) **Head and Power:** The **pump head** $H$ (in meters) represents the energy added per unit weight of fluid: $$H = \frac{\Delta p}{\gamma} + \Delta z + \frac{\Delta v^2}{2g}$$ where: - $\Delta p / \gamma$ = pressure head (m) - $\Delta z$ = elevation gain (m) - $\Delta v^2 / (2g)$ = kinetic energy head (m) - $\gamma = \rho g = 9.81$ kN/m³ for water at 15°C For most practical pump applications, kinetic energy change is small, so: $$H \approx \frac{\Delta p}{\gamma} + \Delta z$$ **Water (Hydraulic) Power Output:** The **water power** or **output power** delivered to the fluid is: $$P_{\text{water}} = \gamma Q H = \rho g Q H$$ with units: - $\gamma$ = 9.81 kN/m³ (or 9810 N/m³) - $Q$ = m³/s - $H$ = m - Result: **kW** (if $\gamma$ in kN/m³) or **W** (if $\gamma$ in N/m³) **Pump Efficiency:** Not all input mechanical power goes to the water. Losses occur due to: - Friction in bearings and seals - Turbulence and recirculation in the pump - Leakage past the impeller **Pump efficiency** $\eta$ is defined as: $$\eta = \frac{P_{\text{water}}}{P_{\text{input}}} = \frac{\gamma Q H}{P_{\text{input}}}$$ Typical pump efficiencies: 60–85% for centrifugal pumps, depending on design and operating conditions. **Input (Brake) Power:** The **input power** required from the motor is: $$P_{\text{input}} = \frac{\gamma Q H}{\eta} = \frac{P_{\text{water}}}{\eta}$$ This is the mechanical power that must be supplied by the pump motor. **Common Exam Problem Setup:** "A pump lifts 0.08 m³/s through a vertical height of 45 m at 70% efficiency. What motor size (in kW) is required?" $$P_{\text{input}} = \frac{9.81 \times 0.08 \times 45}{0.70} = \frac{35.388}{0.70} = 50.55 \text{ kW}$$ **Affinity Laws (Speed, Discharge, and Head Relations):** When a pump operates at different speeds, its performance scales predictably: 1. **Discharge scales linearly with speed:** $$\frac{Q_2}{Q_1} = \frac{N_2}{N_1}$$ 2. **Head scales with the square of speed:** $$\frac{H_2}{H_1} = \left(\frac{N_2}{N_1}\right)^2$$ 3. **Power scales with the cube of speed:** $$\frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^3$$ where $N$ is the pump rotational speed (rpm). **Example Application of Affinity Laws:** A pump at 1450 rpm delivers a head of 25 m. At what head will it operate if the speed increases to 1750 rpm? $$H_2 = H_1 \left(\frac{N_2}{N_1}\right)^2 = 25 \times \left(\frac{1750}{1450}\right)^2 = 25 \times (1.207)^2 = 25 \times 1.457 = 36.4 \text{ m}$$ **NPSH (Net Positive Suction Head):** The **NPSH required** is the minimum absolute pressure head that must exist at the pump inlet to prevent cavitation (vapor bubble formation): $$\text{NPSH}_{\text{req}} \approx 0.5 \text{ to } 3.0 \text{ m (depends on pump design)}$$ The **NPSH available** at the inlet: $$\text{NPSH}_{\text{avail}} = \frac{p_{\text{atm}} - p_v}{\gamma} - z_s - f_s$$ where: - $p_{\text{atm}}$ = atmospheric pressure (≈ 101.3 kPa at sea level) - $p_v$ = vapor pressure of water at operating temperature (≈ 2.3 kPa at 20°C) - $z_s$ = suction lift (if inlet is above water surface) - $f_s$ = friction losses in suction line For safe operation: **NPSH$_{\text{avail}}$ > NPSH$_{\text{req}}$** This is critical in Philippine applications where pumps are often located above water sources (suction lift) or in high-temperature industrial settings. **Pump Types and Operating Ranges:** 1. **Centrifugal Pumps** — most common; handle large flows at moderate heads; efficient at design point; prone to cavitation if NPSH insufficient. 2. **Reciprocating (Piston) Pumps** — high pressure, small displacement; used in high-head applications. 3. **Gear Pumps** — used for viscous fluids and high pressures. 4. **Jet (Ejector) Pumps** — no moving parts; used in deep wells.

A turbine extracts energy from flowing or falling water, converting fluid energy into mechanical (shaft) energy. This chapter focuses on the energy analysis and power calculations for turbines, which is essential for hydroelectric projects in the Philippines. **Turbine Fundamentals:** Unlike a pump, a turbine removes energy from the fluid. The water loses pressure, elevation, and/or velocity as it passes through the turbine. **Net Head and Gross Head:** - **Gross head** $H_g$ = total elevation difference between intake and tailwater levels (m) - **Losses** = friction in penstocks, spillways, and approaches (m) - **Net head** $H_n$ = $H_g$ - losses (m) The net head $H_n$ is what actually drives the turbine. **Turbine Power Output:** The **mechanical power output** (shaft power or useful power) from a turbine is: $$P_{\text{output}} = \eta \times \gamma Q H_n$$ where: - $\eta$ = turbine efficiency (typically 0.85–0.95 for well-designed turbines) - $\gamma$ = 9.81 kN/m³ - $Q$ = discharge (m³/s) - $H_n$ = net head (m) - Result: power in **kW** **Turbine Efficiency:** Turbine efficiency is the ratio of output power to available water power: $$\eta = \frac{P_{\text{output}}}{\gamma Q H_n}$$ Typical values: - **Pelton wheel (impulse):** 85–90% (high head) - **Turgo impulse:** 80–85% (medium-high head) - **Crossflow (Banki):** 75–85% (low-to-medium head) - **Francis turbine (reaction):** 85–92% (medium head) - **Kaplan turbine (reaction):** 85–94% (low head, high flow) **Turbine Classification:** 1. **Impulse Turbines** — use a high-velocity jet striking buckets or vanes. - **Pelton wheel:** Best for very high heads (>300 m); horizontal or vertical axis; used in mountain regions. One or more jets strike hemispherical buckets. - **Turgo:** Similar to Pelton but with a simpler design; heads 50–250 m. - **Crossflow (Banki):** Jet passes through a cylindrical runner twice; heads 10–200 m; simple and robust. 2. **Reaction Turbines** — entire runner submerged; pressure decreases through the runner. - **Francis:** Most versatile; heads 10–250 m; fixed vanes (guide vanes) direct flow radially inward. Runners are complex, precision-cast designs. - **Kaplan:** Axial-flow; low heads (2–20 m) but high discharge; adjustable blades for variable speed operation. Suitable for dams with low elevation differences. - **Deriaz:** Oblique-flow; medium heads; combines features of Francis and Kaplan. **Theoretical vs. Actual Head Delivery:** In a **Pelton wheel**, the jet velocity is nearly equal to the theoretical velocity from Bernoulli (assuming frictionless penstock): $$v = \sqrt{2gH_n}$$ The jet loses speed due to friction; typical velocity coefficient $C_v \approx 0.97–0.99$. **Specific Speed:** The **specific speed** $N_s$ is a dimensionless parameter that characterizes turbine type and operating envelope: $$N_s = \frac{N \sqrt{Q}}{H_n^{0.75}}$$ or in rpm units (common in Philippine practice): $$N_s = \frac{n \sqrt{Q}}{(H_n)^{0.75}}$$ where $n$ is speed in rpm, $Q$ in m³/s, $H_n$ in m. Typical ranges: - $N_s < 20$: Pelton wheel (high head, low flow) - $20 < N_s < 50$: Turgo or crossflow - $50 < N_s < 100$: Francis (reaction) - $N_s > 100$: Kaplan (low head, high flow) **Affinity Laws for Turbines:** Similar to pumps, turbines also follow affinity laws when operating at different speeds: 1. $Q_2 / Q_1 = N_2 / N_1$ 2. $H_2 / H_1 = (N_2 / N_1)^2$ 3. $P_2 / P_1 = (N_2 / N_1)^3$ **Cavitation in Turbines:** Like pumps, turbines can cavitate if local pressures fall below vapor pressure. This is more critical in reaction turbines (where pressure is reduced throughout) than in impulse turbines. Cavitation is controlled by the **sigma coefficient** (Thoma coefficient): $$\sigma = \frac{H_b - H_v}{H_n}$$ where $H_b$ is the barometric head and $H_v$ is the vapor head. A smaller $\sigma$ indicates greater cavitation risk.

This section synthesizes hydrodynamics and fluid machinery concepts into practical problem-solving strategies for the PRC Civil Engineer Licensure Examination. **Common Exam Question Types:** 1. **Jet Force Problems** — Calculate force on a stationary or moving plate/vane using momentum equation. 2. **Pipe Bend Problems** — Find anchoring force considering momentum and pressure terms. 3. **Pump Selection and Power** — Given discharge, head, and efficiency, determine motor size (kW). 4. **Affinity Law Problems** — Scale pump or turbine performance with speed changes. 5. **Cavitation/NPSH Problems** — Check if available NPSH exceeds required NPSH. 6. **Turbine Power and Efficiency** — Calculate output power given head, discharge, and efficiency. 7. **Specific Speed and Turbine Type Selection** — Use $N_s$ to recommend Pelton, Francis, Kaplan, etc. **Exam Problem-Solving Checklist:** **For Jet and Momentum Problems:** - Identify whether the vane is stationary or moving. - If moving, use **relative velocity** $(v - u)$ in all formulas. - Determine the deflection angle $\theta$. - Apply $F = \rho Q v(1 - \cos\theta)$ for curved vanes or $F = \rho Q v$ for perpendicular plates. - Watch units: $\rho$ is in kg/m³, force comes out in Newtons. **For Pipe Bend Problems:** - Draw a clear diagram showing the inlet and outlet sections. - Resolve all vectors (velocity and pressure) into $x$ and $y$ components. - Apply momentum equation in both directions. - Include pressure forces: $(p_1 A_1)_x$, $(p_2 A_2)_y$, etc. - Calculate resultant: $F = \sqrt{F_x^2 + F_y^2}$. - Remember: anchoring force is the reaction from the pipe supports. **For Pump Problems:** - Water power (output): $P_w = \gamma Q H$ (do NOT divide by efficiency). - Input power: $P_{in} = \gamma Q H / \eta$ (divide by efficiency). - Always use $\gamma = 9.81$ kN/m³ for easy kW calculation. - Total head includes elevation and friction losses: $H = \Delta z + h_f + \Delta p / \gamma$. - For affinity laws, remember the exponents: $Q \propto N^1$, $H \propto N^2$, $P \propto N^3$. - NPSH check: available must exceed required. **For Turbine Problems:** - Output power: $P = \eta \times \gamma Q H_n$ (multiply by efficiency). - Use net head (gross head minus losses). - Specific speed: $N_s = n\sqrt{Q} / H_n^{0.75}$ determines turbine type. - Affinity laws apply to turbines too. **Common Mistakes to Avoid:** | Mistake | Correct Approach | |---------|------------------| | Using $\rho$ (kg/m³) in power formula instead of $\gamma$ | Always use $\gamma = 9.81$ kN/m³ for power (kW); use $\rho g$ in SI for watts | | Dividing by efficiency for turbine output | Multiply by efficiency: $P = \eta \gamma Q H$ | | Forgetting to use relative velocity for moving vanes | Always check: is the vane/bucket moving? If yes, use $(v - u)$ | | Pressure forces in pipe bends | Include $pA$ terms; don't forget to resolve them into components | | Mixing up deflection angle $\theta$ and incline angle $\alpha$ | $\theta$ is turn angle for curved vanes; $\alpha$ is angle to plate surface | | Not checking NPSH > NPSH$_{\text{req}}$ | Always verify cavitation will not occur | | Forgetting total head in pump problems | Head = elevation + friction + pressure differences | | Applying affinity laws incorrectly | Check: does discharge depend on speed? (Yes, linearly.) Does head? (Yes, squared.) | **Step-by-Step Example: A Complete Pump System Design Problem** **Problem (Realistic for Board Exam):** An irrigation system must lift water from a reservoir (elevation 100 m) to an elevated tank (elevation 180 m), with a required discharge of 0.15 m³/s. The piping system has friction losses equivalent to 6 m of head. A centrifugal pump with 80% efficiency is being considered. (a) What total head must the pump provide? (b) What is the water power output? (c) What is the motor power input (in kW)? (d) If a 40 kW motor is available and the speed can be adjusted (via VFD), what maximum discharge can be achieved? **Solution:** **Part (a): Total Head** $$H = \Delta z + h_f = (180 - 100) + 6 = 80 + 6 = 86 \text{ m}$$ **Part (b): Water Power** $$P_w = \gamma Q H = 9.81 \times 0.15 \times 86 = 126.65 \text{ kW}$$ **Part (c): Input Motor Power** $$P_{in} = \frac{P_w}{\eta} = \frac{126.65}{0.80} = 158.3 \text{ kW}$$ So a **158 kW motor** is required. (A 160 kW motor would be specified with margin.) **Part (d): Maximum Discharge with 40 kW Motor** With a 40 kW motor and 80% efficiency: $$P_w = \eta \times P_{in} = 0.80 \times 40 = 32 \text{ kW}$$ $$Q = \frac{P_w}{\gamma H} = \frac{32}{9.81 \times 86} = \frac{32}{843.66} = 0.0379 \text{ m}^3/\text{s} \approx 0.038 \text{ m}^3/\text{s}$$ Or approximately **136.4 m³/h**, which is far less than the required 540 m³/h (0.15 m³/s). This confirms a 40 kW motor is insufficient for the design requirement. **Key Takeaway:** Always calculate the motor power before specifying the equipment. In exam problems, pump sizing is a straightforward application of $P_{in} = \gamma Q H / \eta$. **Exam Strategy Tips:** 1. **Read Carefully:** Distinguish between gross head (total elevation) and net head (after losses). 2. **Unit Consistency:** Always check units. If $\gamma$ is in kN/m³, result is in kW. If in N/m³, result is in W. 3. **Diagram It:** Sketch the system (pipe bend, jet path, pump/turbine). Label velocities, pressures, and directions. 4. **Momentum vs. Energy:** - Use momentum equations ($\rho Q v$) for forces on objects. - Use energy equations ($\gamma Q H$) for power. 5. **Efficiency Direction:** - Pump: divide by efficiency (input > output). - Turbine: multiply by efficiency (output < available water power). 6. **Double-Check Calculations:** In exam conditions, small arithmetic errors propagate. Verify each step. 7. **Know the Standards:** Philippine codes (NSCP) don't directly govern hydrodynamics, but design should follow international pump/turbine standards and local water regulations (RA 9275 on Clean Water Act, etc.).