Let's consider the next turboprop arrangement: Energy-Balance-Shaft-Turboprop

What I am looking for

I am looking for an equation that involves $\dot{W}_T$, $\dot{W}_C$ and $\dot{W}_P$ (i.e. the power done by each component). This needs to be done through an energy balance in the shaft.

What we know

We know that the shaft has an efficiency of $\eta_{\text{shaft}}$, and the gearbox has an efficiency of $\eta_{\text{box}}$.

My try (#1)

Since energy is derived from the turbine, we adopt the convention $\dot{W}_T>0$, therefore $\dot{W}_P, \dot{W}_C < 0$. In other words, energy is required to move the compressive devices.

Following the definition of mechanical efficiency: $$\boxed{\eta_{\text{shaft}}=\dfrac{-\dot{W_C}-\dot{W_P}}{\dot{W_T}}}$$

However, I don't know how to incorporate the efficiency $\eta_{\text{box}}$ into this equation. This led me to try and find another expression.

My try (#2)

Following an hydraulic analogy for the power, I thought that whatever power the turbine generates needs to be invested in:

  • Increasing the temperature of the bearings of the main shaft (because the shaft is not ideal and has an efficiency of $\eta_{\text{shaft}}$).
  • Moving the compressor.
  • Increasing the temperature of the gearbox (because the box is not ideal and has an efficiency of $\eta_{\text{box}}$).
  • Moving the propeller.

Following this thinking, I can intuitively find this next equation (which I don't know if is correct or not): $$\boxed{\eta_{\text{shaft}}\times \dot{W}_T = -\dot{W}_C-\dfrac{\dot{W}_P}{\eta_{\text{box}}}}$$

The thing is I don't know if I assigned the efficiency $\eta_{\text{box}}$ correctly or not. Is there an easier way (i.e. less intuitive/more mechanical way) to apply the energy equation? What if we had 2 or more shafts?

  • $\begingroup$ The shaft is generally pretty stiff and doesn't directly have moving parts. So I assume by shaft efficiency you are thinking about bearing losses? $\endgroup$
    – OSUZorba
    May 24, 2016 at 0:48
  • $\begingroup$ Exactly OSUZorba. That's the energy lost due to friction in the bearings (unconveniently used to increase its temperature ) $\endgroup$ May 24, 2016 at 1:21

1 Answer 1


Since the compressor loses a fraction of the turbine's work through the shaft, but is unaffected by the gearbox, but the prop experiences losses from the shaft and the gearbox, this equation should be what you need:


Which is the same as your attempt two (with some algebra).

As you add more efficiencies to a component, you would just continue to multiple the all $\eta$'s together to get the total efficiency of that component, since efficiencies are multiplicative and the associative property of multiplication.

  • $\begingroup$ Happy to see that our solutions agree! $\endgroup$ May 24, 2016 at 11:10

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