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Chapter 6 • Theory & Derivations

Hydraulic Turbines and Renewable Power

Comprehensive university textbook unit covering fundamental principles, governing relationships, and engineering applications of Hydraulic Turbines and Renewable Power.

§6.1 Gross versus net hydraulic head

**Outcome.** Use conservation laws and appropriate reference stations to analyze **gross versus net hydraulic head**; solve a traceable numerical example; expose model limitations; and justify a provisional engineering choice. This is Unit 06, **Hydraulic Turbines and Renewable Power**.

§6.2 Hydroelectric turbine power and efficiency

**Outcome.** Use conservation laws and appropriate reference stations to analyze **hydroelectric turbine power and efficiency**; solve a traceable numerical example; expose model limitations; and justify a provisional engineering choice. This is Unit 06, **Hydraulic Turbines and Renewable Power**.

§6.3 Impulse turbine jet and runner speed

**Outcome.** Use conservation laws and appropriate reference stations to analyze **impulse turbine jet and runner speed**; solve a traceable numerical example; expose model limitations; and justify a provisional engineering choice. This is Unit 06, **Hydraulic Turbines and Renewable Power**.

§6.4 Francis and Kaplan reaction runners

**Outcome.** Use conservation laws and appropriate reference stations to analyze **francis and kaplan reaction runners**; solve a traceable numerical example; expose model limitations; and justify a provisional engineering choice. This is Unit 06, **Hydraulic Turbines and Renewable Power**.

§6.5 Runner specific speed and selection

**Outcome.** Use conservation laws and appropriate reference stations to analyze **runner specific speed and selection**; solve a traceable numerical example; expose model limitations; and justify a provisional engineering choice. This is Unit 06, **Hydraulic Turbines and Renewable Power**.

THREE WORKED QUESTIONS

Step-by-Step Turbomachinery Solutions

Three original questions connect calculations, definitions and interpretation. Open each solution to follow the reasoning.

Solved Problem Example 6.1: Calculation & Verification: Gross versus net hydraulic head

Problem Statement: An engineering practitioner analyzes a mechanical system governed by **Gross versus net hydraulic head** under steady operational loading.

Worked Engineering Derivation:

**Teaching case:** A water scheme with 43 m gross head and 5 m penstock losses has 38 m net head at that flow. **Explicit substitution:** H_net=43−5; therefore **38 m**. Penstock friction depends on discharge and must be reassessed off-design. Check arithmetic independently. Then perturb one input by 10%: linear, quadratic and cubic factors become 1.1, 1.21 and 1.331, respectively. Recompute nonlinear property relationships. An algebraic change is not evidence of a new real-machine operating point.

Solved Problem Example 6.2: Calculation & Verification: Hydroelectric turbine power and efficiency

Problem Statement: An engineering practitioner analyzes a mechanical system governed by **Hydroelectric turbine power and efficiency** under steady operational loading.

Worked Engineering Derivation:

**Teaching case:** A plant using Q=2.4 m3/s, H=28 m, rho=1000 kg/m3 and eta_total=0.87 supplies about 573.7 kW. **Explicit substitution:** P=1000×9.81×2.4×28×0.87/1000; therefore **573.55 kW**. Here total efficiency includes generator loss and is only an assumed value. Check arithmetic independently. Then perturb one input by 10%: linear, quadratic and cubic factors become 1.1, 1.21 and 1.331, respectively. Recompute nonlinear property relationships. An algebraic change is not evidence of a new real-machine operating point.

Solved Problem Example 6.3: Calculation & Verification: Impulse turbine jet and runner speed

Problem Statement: An engineering practitioner analyzes a mechanical system governed by **Impulse turbine jet and runner speed** under steady operational loading.

Worked Engineering Derivation:

**Teaching case:** With H=80 m, ideal water jet speed is about 39.6 m/s before nozzle losses. **Explicit substitution:** V=sqrt(2×9.81×80); therefore **39.62 m/s ideal**. Actual jet velocity is reduced by nozzle losses. Check arithmetic independently. Then perturb one input by 10%: linear, quadratic and cubic factors become 1.1, 1.21 and 1.331, respectively. Recompute nonlinear property relationships. An algebraic change is not evidence of a new real-machine operating point.

Solved Problem Example 6.4: Calculation & Verification: Francis and Kaplan reaction runners

Problem Statement: An engineering practitioner analyzes a mechanical system governed by **Francis and Kaplan reaction runners** under steady operational loading.

Worked Engineering Derivation:

**Teaching case:** With outlet velocity 6 m/s, recoverable ideal velocity head is 1.84 m before losses. **Explicit substitution:** V²/(2g)=6²/(2×9.81); therefore **1.835 m of velocity head**. Recoverable portion is lower because draft tubes dissipate energy. Check arithmetic independently. Then perturb one input by 10%: linear, quadratic and cubic factors become 1.1, 1.21 and 1.331, respectively. Recompute nonlinear property relationships. An algebraic change is not evidence of a new real-machine operating point.

Solved Problem Example 6.5: Calculation & Verification: Runner specific speed and selection

Problem Statement: An engineering practitioner analyzes a mechanical system governed by **Runner specific speed and selection** under steady operational loading.

Worked Engineering Derivation:

**Teaching case:** Using n=300 rpm, P=500 kW and H=30 m gives a numerical metric about 95.6 in rpm-kW-m convention. **Explicit substitution:** N_s=300×sqrt(500)/30^1.25; therefore **95.54 (rpm–kW–m convention)**. Specific-speed values from different numerical conventions must not be compared directly. Check arithmetic independently. Then perturb one input by 10%: linear, quadratic and cubic factors become 1.1, 1.21 and 1.331, respectively. Recompute nonlinear property relationships. An algebraic change is not evidence of a new real-machine operating point.