Benefit of a deswirl vane for a turboshaft helicopter engine

Question

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?

Answer 1

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.

Answer 2

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