Some questions about the specific speed (Ns) of hydro turbines

Question

So we all know the basics behind specific speed Ns of hydro-turbines.

$$N_s = RPM \cdot \frac{ \sqrt{(P)} }{ H^{1.25}}$$

where P is the power output and H is the head. The theory is that once you know the Ns for that model of turbine, you are able to find out either the RPM, power output or head if you are given absolute values for 2 of the 3 mentioned variables.

Let's say I have a turbine with a known Ns. If my head increases but my power output remains constant due to say, reduced flow, then the RPM will increase since Ns is fixed. So far, so good.

However, if my head remains constant but my power output increases due to increased flow, then my RPM must decrease since Ns is fixed. But if we think about it that doesn't make sense...the RPM must be increased for a larger power output, no? Is there something wrong with my understanding of Ns?

Answer 1

Judging by the question, apparently not!

Okay, this isn't a trivial thing to visualize, but when you say flow increases, head stays the same, power increases, and Ns stays the same, what has to happen to the geometry of the rotor?

Ns being constant implies here that the two rotors being talked about are geometrically similar. The one with the larger flow is bigger. To operate at the same head, it will run at a lower rpm.

Answer 2

Specific speed of a turbomachine is calculated at the point in the operating range that has the highest efficiency. For a single machine, the speed, power and head are not related to a single number $N_s$. The group of variables is created to omit the actual size so that typical design givens can be used to determine what general shape the turbine (or pump) should have.

D G Shepard in $Principles$ $of$ $ Turbomachinery$ (1956) says about $N_s$ that looking at the head and flow of one machine, it can have any value, so "As a practical parameter then, the restriction is introduced that it is evaluated at the point of best efficiency. When this is done, it becomes a parameter of great significance in defining the general type of design of turbomachine, because it is found that each different class of machine has its maximum efficiency within a relatively restricted range of specific speed, this range being different for each class."(p. 35)

Shepard has a figure that shows efficiency vs $N_s$ on page 39. He says the expression in the post is used for water turbines with units of RPM, hp and H in feet. The full expression would have density to the 1/2 in the denominator.

Answer 3

'Is there something wrong with my understanding of $N_{\textrm{s}}$?'

Two things. Firstly, the formula you've quoted does not give the true dimensionless specific speed: to correct it, you'd need the factor of the square root of fluid density in the denominator proposed by @WHG, and a factor of the $5/4$ power of the acceleration of free fall in the denominator. You'd also need to express the rotation rate in units coherent with the units you use for the other quantities, and those units are very unlikely to be RPM.

Secondly, it looks like you believe that there's a physical principle that the specific speed of a given turbine is fixed to a constant value. There is no such principle. The physical principle for which the specific speed is useful is that, for a family of geometrically similar turbines, the efficiency depends only on the values of the specific speed, the specific diameter, and one other dimensionless number built from viscosity, density, the product of head and acceleration of free fall, and power.