24

Discounting the V-22 Osprey and the other tiltrotors, the drawbacks are: Wasted cabin space L-duct inlet(s) needed to guide the air in With that potential problems with getting the air in as the air prefers fewer turns A reduction gearbox is still needed Gearbox still needed for the tail rotor Harder maintenance access as it will be buried into the fuselage ...


14

No a turboshaft cannot directly drive the rotor without a reduction gear, the rotor torque is too high for the ungeared turbine torque. The rotor blades are much longer than those of a prop or fan (relatively) and the rotor turns slower, a definite case for torque gearing. Mounting the engine vertically saves a 90 deg gearing assembly, but places the engine ...


9

The first figure, 1 litre per mile, is very wrong. The second, 39 to start, is most probably the fuel required for a warm up from a cold start. The Abrams tank weighs 62 tonnes, so you won't get 1 mile on a litre. Its fuel consumption is quoted as 0.6 mpg, or 1.66 gallons per mile, which is 6.3 litres per mile. Quite a difference. The engine is the ...


7

Yes, the exhaust does provide some thrust. The amount of thrust is in the range of 4%-15% of prop thrust for turboprops. For a helicopter, it's only 2%-4% of rotor lift, but it's directed backwards. The helicopter in your picture would get 100-150 lbf of thrust from the engine exhaust, compared to 3,000-5,000 lbf of lift provided by the rotor. This thrust ...


5

The main way to convert a gas turbine to provide thrust rather than shaft horsepower is to attach a large fan or propeller in place of the previous load. For example, the GE CF6, which powers aircraft such as the 747, 767, and A330, has a gas turbine variant called the LM6000 that provides around 50 MW of power. If you want to provide thrust from the ...


3

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


3

The reason a Diesel engine has a torque figure and a horsepower power figure is because max torque and max horsepower occur at different engine RPMs. Have a look at any performance curves of a piston engine and this will be evident. Hence, to define in a few numbers the maximum performance of the engine, you need to give both figures. However, turboprop ...


2

I suppose the F135-PW-400 powerplant on the F-35B does just that. A turbine engine is used to create high enthalpy gas which can either pass through a turbine to produce mechanical work or pass through a diffuser to accelerate it and create a reaction impulse. General Electric has designed power generation gas turbines which use the gas core from the GE ...


2

Most turbine engines designed to drive some sort of mechanical system such as a hydraulic pump and electrical generators are not designed to provide thrust because their fuel controller can only run it at one speed. In other words, there is no throttle. Also, the internal components may not be designed for that type of application. So while it is in theory ...


2

You can only use for shaft horse power what is available after the energy to drive the compressor is subtracted. This subtraction is done by the high pressure turbine. Any energy left in the exit flow after that can be extracted to drive a propeller, a fan or a generator. What happens in your arrangement if the power turbine extracts too much energy to keep ...


1

I think your friend is playing a joke on you. I have worked with helicopters since the late 1970's and there is no way to distinguish military from civilian simply by engine noise. You may be able to distinguish between models of helicopters, and in so doing identify a model used by the military versus a civilian one, but that's about it.


1

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 ...


1

Found it here https://www.manualslib.com/manual/812510/Pratt-And-Whitney-Canada-Pt6a-Turboprop.html?page=1 First hit on google for "Pratt & Whitney Canada PT6 operator's manual"


1

A slower compressor will lower the compression ratio, and therefore will hurt the fuel economy of the engine. A faster free turbine will require a much stronger and bigger reduction gearbox (remember the prop rotates much slower than the turbine). There is no easy way to have the concentric (or otherwise) shafts run the LPC (or prop) via the HPT (pictured ...


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