A quarter-scale turbine model is tested under a head of \(10.8 \mathrm{~m}\). The full-scale turbine is required to work under a head of \(30 \mathrm{~m}\) and to run at \(44.86 \mathrm{rad} \cdot \mathrm{s}^{-1}(7.14 \mathrm{rev} / \mathrm{s}) .\) At what speed must the model be run? If it develops \(100 \mathrm{~kW}\) and uses water at \(1.085 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) at this speed, what power will be obtained from the full-scale turbine, its efficiency being \(3 \%\) better than that of the model? What is the power specific speed of the full-scale, turbine?

The similarity laws, or affinity laws, are essential for understanding how models of turbines relate to their full-scale counterparts. These laws help us predict the performance of a full-scale turbine based on tests conducted on a scaled-down model. These laws assume geometric and dynamic similitude between the model and the prototype.

The key relationship used here is: \ \ \(\frac{N_m}{N_f} = \frac{\text{(Head of Model)}^{1/2}}{\text{(Head of Full Scale)}^{1/2}}\) \ \ This formula allows us to determine the rotational speed of the model (\(N_m\)) if we know the rotational speed of the full-scale turbine (\(N_f\)) and the heads under which both are operating.

To calculate the speed at which a scaled-down model turbine must be run, we use the similarity laws. Given the full-scale turbine operates at a specific head and speed, we can calculate the appropriate model speed. Here's how we do it:

Switching from heads to speeds: \ \ \(N_m = N_f \times \left( \frac{\text{(Head of Model)}^{1/2}}{\text{(Head of Full Scale)}^{1/2}} \right)\) \ \ With a full-scale speed of 44.86 rad/s and heads of 10.8 m (model) and 30 m (full-scale), we substitute and solve: \ \ \(N_m = 44.86 \times \left( \frac{10.8^{1/2}}{30^{1/2}} \right) \) \ \ Simplifying leads to: \ \ \(N_m = 44.86 \times \frac{3.29}{5.48} \approx 26.94 \text{ rad} \cdot \text{s}^{-1}\) \ \ Thus, the model should run at 26.94 rad/s.

The power scale factor helps us to translate the power output from a model turbine to what a full-scale turbine would produce. This scaling takes into account the difference in both the operating head and the size scale. The power scales with the product of the heads and the fifth power of the geometric scale ratio:

Using the formula: \ \ \(\frac{P_m}{P_f} = \frac{\text{(Head of Model)} \times \text{(Linear Scale)}^2}{\text{(Head of Full Scale)} \times \text{(Linear Scale)}^5}\) \ \ Given a quarter-scale model (linear scale = 1/4) and heads: \ \ \(\frac{P_m}{P_f} = \frac{10.8 \times (1/16)}{30 \times (1/1024)}\) \ \ This simplifies further to: \ \ \(P_f = 100 \times \frac{30 \times 1024}{10.8 \times 16} \approx 17812.5 \text{kW}\)

When transferring results from the model to the full-scale turbine, we need to account for any efficiency differences. In our case, the full-scale turbine's efficiency is 3% better than the model’s. Assuming the model develops a power of 17812.5 kW without efficiency adjustments, the formula for the final power is:

\(\text{Power} = P_f \times (1 + 0.03)\) \ \ Substituting gives: \ \ \(\text{Power} = 17812.5 \times 1.03 \) \ \ Which simplifies to roughly: \ \ \(\text{Power } \approx 18347.875 \text{kW}\)

The power specific speed of a turbine (\(n_s\)) is a crucial non-dimensional number that helps in comparing different designs. It combines turbine speed, flow rate, and head. The equation is:

\(n_s = N \times \frac{Q^{1/2}}{H^{5/4}}\) \ \ For the full-scale turbine, using the given full-scale values: \ \ \(n_s = 44.86 \times \frac{(1.085)^{1/2}}{(30)^{5/4}}\) \ \ Upon simplifying, we have: \ \ \(n_s \approx 0.602 \text{ rad} \cdot \text{s}^{-1} \cdot (\text{m}^3/\text{s})^{1/2} \cdot (\text{m})^{-5/4}\)