# Benefit of a deswirl vane for a turboshaft helicopter engine

I am trying to understand the benefit of deswirl vanes on a turboshaft helicopter engine. For these type of engines, the kinetic energy of the exhaust stream is essentially wasted energy. Also, raising the static pressure at the exit plane by deswirling will reduce the shaft output power. So, why bother?

• Could you provide an example picture for better understanding? Deswirl vanes are usually a feature of a (centriugal) compressor, but it seems you are talking about the turbine/exhaust... – Zeus Sep 28 '18 at 1:52
• @Zeus. The question is asking about the benefit of de swirl vanes at the exit of a turbine, for a turboshaft engine. i.e. why not just have the last rotating stage as the final set of aerofoils, why add the weight of stationary exit guide vanes as the last stage of the LPT. – Penguin Sep 28 '18 at 9:24

This is a very good question. However, there is a slight error in your question, which is causing the confusion.

Turbine power, is determined by the ratio of total pressure across the turbine (and other factors). See the formula here on NASA’s website.

$$W_{Turbine} = \eta \cdot c_p \cdot T_{t_4} \cdot \left( 1-TPR^{\frac{\gamma-1}{\gamma}} \right)$$

Where $$\gamma = \frac{c_p}{c_v}$$

and TPR is the total pressure ratio across the turbine: $$TPR =\frac{P_{t_5}}{P_{t_4}}$$

In comparison, thrust is determined by the static pressure. See the “engine thrust equation” (the second formula) on this page here. The thrust is given by two terms; one due to the increase in velocity of the mass, the other is the “pressure thrust” term, which is due the the nozzle area x the difference in ambient static pressure and the nozzle exit static pressure.

In simple form;

$$F_{N} = \dot{m} \cdot (v_e - v_i) + A_j \cdot (p_{s_{exit}} - p_{s_{ambient}})$$

(Ignoring the mass of the fuel, which is very small compared to the air mass flow, $$\dot{m}$$).

So, increasing the exhaust nozzle static pressure at its exit increases the thrust from the exhaust gas, but will not reduce (or change) the power provided by the turbine, because that’s a function of total pressure.

That’s not to say the amount of thrust you will get is very large, and worth the weight of nozzle guide vanes at the exit. Some turboshaft engines have them, some don’t.

• Please use MathJax to format your formulae. I've converted it now, please check if I made any mistakes. – ROIMaison Sep 28 '18 at 11:43
• @ROIMaison. Thanks very much, they are all correct. – Penguin Sep 28 '18 at 21:44
• I was thinking along these lines, but it doesn't really answer why would exhaust 'deswirl' vanes be beneficial. My speculation is that they are more of a structural element than aerodynamic one. Exactly because of what you say (1st formula), and because for helicopters we don't want any jet thrust (we want the turbine to take all the available pressure), it is beneficial to expand the exhaust so that the exit total pressure equaled the ambient (or even overexpand a bit!) Hence efficient helicopter exhausts tend to be large, and need support. – Zeus Oct 1 '18 at 1:10
• Also, when we speak about jet thrust, it's worth noting that exhaust velocity (1st component of the $F_N$ formula) is the main and most efficient contributor to thrust. It normally makes sense to adjust the nozzle such that all the excess pressure turned into speed. In that case, one would think, deswirling could bring benefit of straightening the stream velocity. Yet, even when such vanes are present on jet engines, they also look more like structural elements than anything else. – Zeus Oct 1 '18 at 1:20
• @Zeus what "jet thrust" are you presuming for a turboshaft engine on a helicopter? The point of that engine design is to get as much power extracted as is feasible well before exhaust to drive the rotors. Rotor blades/rotor systems are what provides thrust for a helicopter. – KorvinStarmast Oct 2 '18 at 17:50

Thanks for the discussion - Penguin. However, for reasons below, I am not convinced of your explanation.

The formula for turbine power you quote is for a turbofan engine. For a turboshaft engine, where the exit kinetic energy is wasted, the pressure ratio and the efficiency used is inlet total to exit static. In addition, in your thrust equation above, you are assuming that the normal to the exit plane of the duct is in the flight direction which may or may not be true. Frequently, these ducts are turned upwards, sideways, etc. and in such cases, increasing the static pressure will not increase thrust.

Still at a loss on how these de-swirl vanes work for turbo shaft engines.

Sam

• Hi @sam, With your first comment, the formula isn’t specific to a turbofan, but rather is generic to any turbine. I am not sure why you think it is, maybe you could clarify? I agree completely with your second point about the direction of exhaust. It did cross my mind, so, maybe there is another reason for deswirl vanes, but what I have said isn’t technically wrong, it just might not be the complete answer to your question. Cheers. – Penguin Oct 10 '18 at 9:44