Turboprop torque vs. air density.
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Turboprop torque vs. air density.
Given: A PW121-powered turboprop will use the assumed temperature method for take-off.
Unlike a jet engine that - in this case, will operate at a lower N1, the TP pilots must use a reduced torque setting, e.g. 80% instead of 97,5% TQ.
Since the actual OAT is lower than the assumed OAT, the air density is higher than assumed, and the jet engine will thus produce more thrust (for a given N1) than the take-off calculations are based on. A bonus.
But what about the turboprop? RPM is constant at 1200, and 80% torque should be 80% torque, no matter what the OAT and air density. Sure, the required fuel flow might differ, but will the engines give more thrust at an OAT of 15 degrees C, that at 49C?
The only interesting (to me) variable is blade angle, and perhaps the propeller blades will be at a less efficient angle at lower air densities, in order to absorb the power from the turbine without increasing propeller RPM?
So - for a jet engine we can safely conclude that for a given N1 it will produce more thrust at 15 degrees C than at 49C. But what about the turboprop? Will there be any additional thrust due to the difference between assumed and actual OAT, when using assumed (flex) temperature?
Was that question clear as mud?
Unlike a jet engine that - in this case, will operate at a lower N1, the TP pilots must use a reduced torque setting, e.g. 80% instead of 97,5% TQ.
Since the actual OAT is lower than the assumed OAT, the air density is higher than assumed, and the jet engine will thus produce more thrust (for a given N1) than the take-off calculations are based on. A bonus.
But what about the turboprop? RPM is constant at 1200, and 80% torque should be 80% torque, no matter what the OAT and air density. Sure, the required fuel flow might differ, but will the engines give more thrust at an OAT of 15 degrees C, that at 49C?
The only interesting (to me) variable is blade angle, and perhaps the propeller blades will be at a less efficient angle at lower air densities, in order to absorb the power from the turbine without increasing propeller RPM?
So - for a jet engine we can safely conclude that for a given N1 it will produce more thrust at 15 degrees C than at 49C. But what about the turboprop? Will there be any additional thrust due to the difference between assumed and actual OAT, when using assumed (flex) temperature?
Was that question clear as mud?
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Hey Crossunder, I fly with the same type and if i understand your question correctly, the gas generator will be at lower NH for Takeoff 90% TRQ on denser and colder altitudes vice versa. Once you hit the thermodynamic limit then you TRQ reduces for the same NH %. This is my understanding.
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Given: A PW121-powered turboprop will use the assumed temperature method for take-off.
Unlike a jet engine that - in this case, will operate at a lower N1, the TP pilots must use a reduced torque setting, e.g. 80% instead of 97,5% TQ.
Since the actual OAT is lower than the assumed OAT, the air density is higher than assumed, and the jet engine will thus produce more thrust (for a given N1) than the take-off calculations are based on. A bonus.
But what about the turboprop? RPM is constant at 1200, and 80% torque should be 80% torque, no matter what the OAT and air density. Sure, the required fuel flow might differ, but will the engines give more thrust at an OAT of 15 degrees C, that at 49C?
The only interesting (to me) variable is blade angle, and perhaps the propeller blades will be at a less efficient angle at lower air densities, in order to absorb the power from the turbine without increasing propeller RPM?
So - for a jet engine we can safely conclude that for a given N1 it will produce more thrust at 15 degrees C than at 49C. But what about the turboprop? Will there be any additional thrust due to the difference between assumed and actual OAT, when using assumed (flex) temperature?
Was that question clear as mud?
Unlike a jet engine that - in this case, will operate at a lower N1, the TP pilots must use a reduced torque setting, e.g. 80% instead of 97,5% TQ.
Since the actual OAT is lower than the assumed OAT, the air density is higher than assumed, and the jet engine will thus produce more thrust (for a given N1) than the take-off calculations are based on. A bonus.
But what about the turboprop? RPM is constant at 1200, and 80% torque should be 80% torque, no matter what the OAT and air density. Sure, the required fuel flow might differ, but will the engines give more thrust at an OAT of 15 degrees C, that at 49C?
The only interesting (to me) variable is blade angle, and perhaps the propeller blades will be at a less efficient angle at lower air densities, in order to absorb the power from the turbine without increasing propeller RPM?
So - for a jet engine we can safely conclude that for a given N1 it will produce more thrust at 15 degrees C than at 49C. But what about the turboprop? Will there be any additional thrust due to the difference between assumed and actual OAT, when using assumed (flex) temperature?
Was that question clear as mud?
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Good question, and I agree the key variable is the resulting blade angle. To put the bottom line on top I believe you will get more thrust.
As stated the Tq and RPM are fixed hence blade drag and rotational airlfow are fixed. And the question is how does thrust vary as a function of rho?
For a given blade drag (ie Tq) and RPM (rotational airflow) the Cd of the blade must reduce as rho increases. Therefore the blade angle must also reduce but does that result in more or less thrust? The answer depends but if we assume that 80% Tq puts the blade at a higher angle of attack than its most efficient the answer is the thrust will increase: because under that assumption the slope of the Cd curve is steeper than the Cl curve, and/or the total reaction of the blade is more aligned with the thrust axis as the blade angle reduces. I think.
As stated the Tq and RPM are fixed hence blade drag and rotational airlfow are fixed. And the question is how does thrust vary as a function of rho?
For a given blade drag (ie Tq) and RPM (rotational airflow) the Cd of the blade must reduce as rho increases. Therefore the blade angle must also reduce but does that result in more or less thrust? The answer depends but if we assume that 80% Tq puts the blade at a higher angle of attack than its most efficient the answer is the thrust will increase: because under that assumption the slope of the Cd curve is steeper than the Cl curve, and/or the total reaction of the blade is more aligned with the thrust axis as the blade angle reduces. I think.
