# degree of reacton

by ank_gl
Tags: degree, reacton
 P: 733 How is th degree of reaction defined for hydraulic turbine and a gas turbine??(i, actually, want a quantitative formula) How is it optimized for a turbine??
 Sci Advisor P: 5,095 Degree of reaction is the ratio of static enthalpy drop across the turbine nozzle to the drop across the turbine. That being said, a hydraulic turbine is a reaction turbine with no nozzle drop so its degree of reaction would be 0. A gas turbine engine will vary depending on the drops. A degree of reaction of 50% means there is an equal drop across the nozzle as the turbine. Usually losses increase greatly after a R=1, so in that aspect a R<1 is more optimal. However, that is simply looking at the R value and no other design criteria.
P: 733
 Quote by FredGarvin Degree of reaction is the ratio of static enthalpy drop across the turbine nozzle to the drop across the turbine. That being said, a hydraulic turbine is a reaction turbine with no nozzle drop so its degree of reaction would be 0. A gas turbine engine will vary depending on the drops. A degree of reaction of 50% means there is an equal drop across the nozzle as the turbine. Usually losses increase greatly after a R=1, so in that aspect a R<1 is more optimal. However, that is simply looking at the R value and no other design criteria.

Either I am completely wrong or I am completely confused.

All gas turbines have enthalpy drop in the nozzle(that means energy is been converted to kinetic energy). Right??

An Impulse turbine(eg. Curtis turbine & Rateau turbine ) has all its enthalpy drop in the nozzle(stator) and no drop in moving blades(theoretically), so degree of reaction is 1 for an impulse turbine??

A reaction turbine has enthalpy drop in both stator and rotor blades(both act as nozzle), so degree of reaction is less than 1(DoR = ΔHs/(ΔHs + ΔHr)). How can DoR be greater than 1?

I take DoR as "the ratio of enthalpy drop in rotor blades to the enthalpy drop in one stage(rotor + stator)". I studied Power plant Technology, M. M. El-Wakil, Tata McGraw Hill.

My original question was actually more inclined towards the hydraulic turbines. Head replaces the enthalpy. As previous, I took, DoR as "the ratio of head drop in rotor, to the head drop in the turbine".

In pelton(& turgo turbine) wheel, all head is converted to KE upto turbine inlet, so the DoR is 0.

In Francis and Kaplan turbine, some head converts in the rotor itself, so degree of reaction is between 0 & 1.
Total head drop, He = [(V1^2 - V2^2) + (u1^2 -u2^2) + (Vr2^2 -Vr1^2)]/(2*g)
head drop in rotor,Hr = [(u1^2 -u2^2) + (Vr2^2 -Vr1^2)]/(2*g)
therefore, DoR = Hr/He

V = absolute velocity
u = peripheral velocity of blade
1&2 = Inlet & outlet of turbine

The reason I wanted to confirm the above said definition of DoR for a hydraulic turbines is that I am getting negative values of DoR for a Francis turbine, below a particular rpm and values greater than 1 above a particular rpm. So I suppose this definition is wrong, and I am probably confused alot.

P: 5,095

## degree of reacton

An impulse turbine is a 0 DoR turbine, not 1.

I honestly can't comment on your use of head in stead of enthalpy for the DoR calculation. I personally have never heard of that, but it doesn't mean in those areas it is not used. I will have to do some looking around to verify that.

As a note, you are not limited to a DoR between 0 and 1. It is simply a ratio so to get one larger than one, you simply need the stage's higher percentage of the enthalpy drop to happen across the rotor.
P: 733
 Quote by FredGarvin I honestly can't comment on your use of head in stead of enthalpy for the DoR calculation.
That was just an analogy. Head is also a form of energy.

I am now seriously confused over its definition. Is it this??
 Quote by FredGarvin Degree of reaction is the ratio of static enthalpy drop across the turbine nozzle to the drop across the turbine
This gives DoR of an impulse turbine as 1. All the enthalpy drop is in the nozzle.
By turbine, do you mean one stage(stator + rotor) or only rotor?

Or is it this??
 the ratio of enthalpy drop in rotor blades to the enthalpy drop in one stage(rotor + stator)

 As a note, you are not limited to a DoR between 0 and 1. It is simply a ratio so to get one larger than one, you simply need the stage's higher percentage of the enthalpy drop to happen across the rotor.
Going by your definition, it is apparent(assuming turbine means rotor). But that definition also give DoR of an impulse turbine as 1.

I have searched so much for its definition, i have failed so far. Which one is the right definition?
P: 5,095
Ahh crud. I re-read that definition about 10 times and didn't catch that. The definition is:

 Degree of reaction is the ratio of static enthalpy drop across the turbine rotor to the drop across the turbine stage.
http://www.tfd.chalmers.se/~ulfh/gas...aen/sld021.htm

That makes it easy to see why an impulse turbine is DoR of 0, not 1.
 P: 733 so DoR cant be more than 1 0
 P: 733 Can I get all the design parameters(blade profile, no. of stages etc) of a gas turbine. I have written programs for single stage impulse turbine, velocity & pressure compounded impulse turbine, & reaction turbine. I need to see if the theory I understand is right.
 Sci Advisor P: 5,095 I can't say that DoR really shows up as a design criteria. I always hear our turbine guys talking about, obviously, the work extraction, losses, swirl angles, etc... For an axial, multi stage, it can vary because it is dependent on stage drops. If you are looking for design parameters, I would suggest picking up a copy of Hill and Peterson's Mechanics and Thermodynamics of Propulsion or Cohen and Roger's Gas Turbine Theory. Their chapters on turbine design are quite extensive and would take a month of Sundays to type out.
 P: 733 I am attaching two plots, showing the effect of DoR on efficiency. First plot is for a single stage turbine, while other one is for a 3 stage turbine.DoR is increasing as we go towards right. Leftmost plot is for DoR = 0, ie. impulse turbine In multistage turbine, i have assumed that all three stages have equal enthalpy drop and DoR in all stages is constant. I have neglected the effects of losses. Enthalpy drops from 3574 kJ/kg to 2238 kJ/kg. I am not very sure about this assumption, that the blades are short enough, so that blade velocity doesn't change much radially, & hence I didn't vary the blade inlet angle. I am sure, parameters will change a bit, if we get a model closer to real world, but I dont think the trend will change anyways Increasing DoR decreases maximum efficiency a little and increase the blade velocity at which maximum efficiency is attained. But Increasing DoR has a very interesting effect, it maintains high efficiency over a wide range of blade velocity, unlike an impulse turbine, which shows an upside down parabolic relation(also shown in the plot). Attached Thumbnails
 P: 1 sir, can you explain what is the degree of reaction in gas turbine ?????????????

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