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Low specific speed reaction turbines

Short answer

A low Ns Francis runner is large, heavy, less efficient and needs a costly many-pole generator; advantages are low cavitation risk, high-head capability and higher setting. Pelton efficiency with speed is a parabola in u/V1: zero at rest and runaway, maximum at u/V1 = 0.5, about 0.46 in practice.

What are the disadvantages of a very low specific speed reaction turbine? What are its advantages? How does the efficiency of the pelton wheel vary with its speed?

A very low specific speed reaction turbine is a slow Francis runner (about NsN_s = 60–100, kW units) used near the top of the Francis head range. Its runner is large in diameter with long, narrow passages and nearly radial flow.

Disadvantages

  1. Large runner diameter for the power produced, so the turbine itself is heavy and costly.
  2. High friction and disc-friction losses in long narrow passages, so peak efficiency is lower than a medium-NsN_s Francis.
  3. Low running speed means a large, costly generator with many poles (N=120f/pN = 120f/p).
  4. Leakage losses through seals rise at high head.
  5. At heads where low-NsN_s Francis is used, a Pelton may be cheaper and more efficient at part load.

Advantages

  1. Low cavitation risk: Thoma's critical σ\sigma is small at low NsN_s, so the runner can be set higher above tailwater with less excavation.
  2. Can work under high heads (up to about 350–600 m), with small discharge.
  3. Robust runner, stable operation and smaller draft tube.
  4. Less sensitive to tailwater level changes.

Efficiency of Pelton wheel with speed

For a Pelton wheel with jet velocity V1V_1, bucket speed uu, blade outlet angle ϕ\phi and friction factor kk, the hydraulic efficiency is

ηh=2u (V1−u)(1+kcos⁡ϕ)V12\eta_h = \frac{2u\,(V_1 - u)(1 + k\cos\phi)}{V_1^2}

Setting dηhdu=0\dfrac{d\eta_h}{du} = 0:

V1−2u=0u=V12ηh,max=1+kcos⁡ϕ2\begin{aligned} V_1 - 2u &= 0 \\ u &= \frac{V_1}{2} \\ \eta_{h,max} &= \frac{1 + k\cos\phi}{2} \end{aligned}
  • Efficiency is a parabola in speed ratio u/V1u/V_1: zero when the wheel is stationary (u=0u=0), maximum at u/V1=0.5u/V_1 = 0.5 (in practice about 0.46 due to losses), and zero again at runaway (u=V1u = V_1).
  • With k=1k = 1 and ϕ=15∘\phi = 15^\circ, ηh,max=(1+cos⁡15∘)/2≈0.983\eta_{h,max} = (1+\cos 15^\circ)/2 \approx 0.983 (theoretical).
 eff
  |      ___
  |    /     \
  |   /       \
  |  /         \
  | /           \
  +--------------+---> u/V1
  0     0.46    1.0

Since a Pelton runs at constant synchronous speed while the jet velocity is fixed by head, the speed ratio stays near optimum, and load changes are handled by the needle; this is why its efficiency is high over a wide load range.

✓ Answer checked by Claude Opus 5.5 (high effort)

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