--- title: "Bernoulli's equation limitations and application" subject: "Hydro Power" chapter: 1 key: "semester_6/hydro_power:bernoulli-limitations-application" source: https://www.bctengineer.com/q/semester_6/hydro_power/bernoulli-limitations-application site: BCT Engineering --- # Bernoulli's equation limitations and application Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics - Asked in: 2081 Asoj, 5+2+1 marks - Video: [Calculating the power a turbine can generate on a hydroelectric scheme using Bernoulli’s equation](https://www.youtube.com/watch?v=1mXO1jmCI1Q) — Fluids Explained - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/bernoulli-limitations-application - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1 ## Question Derive Bernoulli's equation. Explain its limitations and provide an example of its application in hydropower engineering. ## Short answer Bernoulli: p/ρg + V²/2g + z = constant. Limitations: ideal fluid, steady, incompressible, along streamline, no energy addition/removal. Hydropower application: net head Hnet = Hgross - hL, power P = ηρgQHnet, jet velocity V = Cv√(2gH) for Pelton wheel. ## Answer **Bernoulli's equation** states that, for steady, ideal, incompressible flow along a streamline, the sum of pressure head, velocity head and datum head is constant. #### Derivation (from Euler's equation) Consider a small element of length $ds$ and area $dA$ along a streamline. Forces in the flow direction: - net pressure force $= p\,dA - (p + dp)\,dA = -dp\,dA$; - weight component $= -\rho g\,dA\,ds\,\dfrac{dz}{ds} = -\rho g\, dA\, dz$. Mass × acceleration $= \rho\,dA\,ds \cdot V\dfrac{dV}{ds} = \rho\, dA\, V\,dV$ (steady flow). ```text (p+dp)dA <-- [ element ds ] <- streamline p dA --> | v rho g dA ds ``` Newton's second law: $$ -dp\,dA - \rho g\,dA\,dz = \rho\,dA\,V\,dV $$ $$ \frac{dp}{\rho} + g\,dz + V\,dV = 0 \quad \text{(Euler's equation)} $$ Integrating with $\rho$ constant and dividing by $g$: $$ \frac{p}{\rho g} + \frac{V^2}{2g} + z = \text{constant} $$ #### Limitations 1. It is valid only for **ideal (non-viscous)** fluid. Real flows have friction losses, so a loss term $h_L$ must be added. 2. It assumes **steady flow**; it does not apply directly to unsteady flow such as water hammer or surge. 3. It assumes **incompressible flow**; not valid for high-speed gas flow. 4. It applies along a single streamline (or to irrotational flow); it assumes uniform velocity across the section, so for real pipes a kinetic energy correction factor $\alpha$ is needed. 5. It does not include energy added or removed by pumps or turbines, or heat transfer, unless those terms are added. 6. It ignores forces other than gravity and pressure, such as surface tension and centrifugal forces in bends. #### Application in hydropower **Net head and power of a plant:** applying Bernoulli's equation (with losses) between the reservoir surface (1) and the turbine inlet (2): $$ z_1 = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_L $$ The head available at the turbine is $H_{net} = H_{gross} - h_L$, and power $P = \eta\rho g Q H_{net}$. For a Pelton wheel, the jet velocity $V = C_v\sqrt{2gH}$ also comes from Bernoulli's equation. It is also used to design the draft tube (pressure at its inlet) and to size intakes and flow-measuring devices (venturimeter) in the plant. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the derivation, what is the net pressure force on the element in the flow direction? A. -dp dA B. +dp dA C. p dA D. (p + dp) dA Answer: A. p dA - (p + dp) dA = -dp dA. (Key point: Force balance) 2. Which limitation is correct? A. It applies only to ideal fluid, so real flows need a loss term h_L B. It works for unsteady surge C. It applies to compressible high-speed gas D. It includes pump energy Answer: A. Bernoulli's equation neglects viscosity; real flows add h_L. It also needs steady, incompressible flow, and excludes pumps/turbines unless added. (Key point: Limitation: friction) 3. Why does Bernoulli's equation not apply directly to water hammer? A. It assumes rotational flow B. It assumes viscous flow C. It assumes compressible flow D. It assumes steady flow Answer: D. Water hammer is unsteady, violating the steady-flow assumption. (Key point: Limitation: unsteady flow) 4. How is Bernoulli used for net head in a hydropower plant? A. H_net = H_gross + h_L B. H_net = h_L only C. H_net = V^2/2g D. H_net = H_gross - h_L between reservoir and turbine inlet Answer: D. Applying Bernoulli with losses between reservoir and turbine gives H_net = H_gross - h_L and P = eta rho g Q H_net. 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