Well I consulted a good old booklet I had bought a century ago:
S.L.Dixon “Fluid Mechanics, Thermodynamics of Turbomachinery”
Pergamon Press 1984, ISBN 0-08-022722-8
It says that generally the specific work done on the fluid (in an axial compressor rotational stage or impeller of a centrifugal compressor) is equal to the stagnation enthalpy rise, which expressed by Euler’s pump equation looks like:
delta W = *ctang2-U1ctang 1 = h02- h01
delta W is that specific work done per unit mass of fluid
U1, U2 are blade speeds at inlet and outlet points of a flow particle (i.e. streamline points, station 1 and 2)
ctang1 & 2 are the tangential speeds of the particle (which I do not quite understand how they can be different from blade speeds which are also tangential speeds).
h01 h02 stagnation enthalpies
Rearranging the formula, all station 1 parameters on one side, station 2 on the other and substituting stagnation enthalpy with sum of specific (static) and dynamic enthalpy we get the following equation
h1+(1/2)c1^2- U1ctang 1 = h2+(1/2)c2^2- U2ctang 2 = I
This newly invented parameter I is named rothalpy and by its virtue it is constant from station 1 to station 2. Using more mathematical hocus-pocuses, velocity triangles and as you mentioned rothalpy conservation rule the gentleman (Dixon) leads us to the following:
new thing is “w” which is a flow particle relative velocity in respect to the blade
If the streamline is at same radii before and after the impeller (axial compressor case) the first term is 0. That is the reason why the static enthalpy rise in a centrifugal compressor is so large compared with a single stage axial compressor.
And finally to quote the gentleman:
“The rothalpy is fundamental property which is constant for an adiabatic irreversible flow process, relative to a rotating component.”
I've got the same book just the fourth edition.
But unfortunately in the fourth edition, the section that covers Rothalpy is not very helpful. The formula's you've provided will definitely help.
The fact that rothalpy is a constant for a adiabatic irreversible flow process narrows the scenarios down to a few solutions.
The formula I have state that rothalpy is:
I = h+½(cr² + c² + cx² - 2Uc)
Where
cr = radial velocity
c = tangential velocity
cx = axial velocity
From velocity triangles:
w1 = U – c
together with w² =cr² + w1² + cx²
then Rothalpy
I = h + ½(w² - U²)
Since
h = h2 - h1
and
h2 – h1 =½(U2² – U1²) + ½(w1² – w2²)
Riaan de Jager
Mechanical Projects Engineer
British American Tobacco Manufacturers
South Africa
Unfortunately I was able to open it only as HTML file, meaning no pictures and occasional strange-confusing changes of formatting (the normal but inaccessible address is
But the most importantly it originates from University of Twente, Netherlands. Therefore you as a Boer may have some privilege addressing Mr Kruyt directly. Being the member of the nation that suffered Concentration Camps the first and speaking the similar language as Mr Kruyt you should get a favourable treatment from him.
Radomir, In your first equation (delta W = *ctang2-U1ctang 1 = h02- h01), c is the absolute fluid velocity at the blade tip (c2 is the exit end of the blade and c1 is the inlet end, and w2 is the fluid velocity at the blade tip). Then, ctang2 is the component of c2 in the direction of U2. Adding the vectors U2 and w2 gives you c2 as shown in fig 7.1 of Dixon 3rd ed. Also, adding vectors wtang2 and U2 will give you ctang2. Depending on the blade angle, U2 and ctang2 maybe the same, but also, they may not. I = h+½(cr² + c² + cx² - 2Uc)is the three dimension form of h1+(1/2)c1^2- U1ctang 1 = h2+(1/2)c2^2- U2ctang 2 = I. As far as your quoted statement, I can't find that exact wording in my edition, but rothalpy has the same value at blade inlet and exit and must be constant along flow lines between these two points. If you know the inlet conditions and dimensions, you can then determine the outlet conditions and dimensions for your impeller design.