# Tag Info

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The answer depends on factors like the bypass ratio of the engine and the design of the thrust reverser, e.g. bucket doors at the back of the engine or vanes to deflect the bypass airflow from the fan. By the OP's proposed measurement (reverse thrust / forward thrust), the efficiency is also strongly dependent on the engine speed. At maximum thrust the ...

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I think the thrust is generally applied as show on the top part of the following image. If you are trying to apply the force on the mount, then the same force is applied, however you need to add a bending moment. The bending moment should be equal to the product of the thrust T, times the distance between the axis of the axle and the mount points.

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The thrust force acts on the axle bearings that support the compressor stage shaft of the turbine. That reaction force is conveyed out to the engine mounts which are in turn bolted to the airframe.

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It's not clear, but I think you are asking whether you can control your solenoid valve with pulses instead of steady DC. Yes, above some frequency, the mechanical system actuated by the solenoid won't "see" individual pulses, just the average. Usually at a higher frequency, the solenoid itself will smooth out the pulses and maintain are more average steady ...

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Product design often involves juggling many parameters at the same time, prioritizing them, then testing a design. In sprinklers themselves, the forces associated with the thrust are much lower that the forces the sprinkler experiences during assembly or from internal pressure; so typically they are not a design priority. For irrigation sprinklers the ...

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This gives a complete analysis of how to calculate : http://web.mit.edu/13.012/www/handouts/propellers_reading.pdf

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Just think of the propeller as a mean to thrust air in one direction. The higher the angle of incidence, the more the propeller is pushing the air (up to about 45°, then the force perpendicular to the rotational direction starts to take over). Now consider the fact that air is inertial mass. So accelerating it means building up pressure behind the propeller....

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It would be slower. Compressing the air takes energy - that will slow the propeller, and the plane as a whole. Imagine a water pump, which is pumping water out of a pipe. Now, if you put your finger over the end of the pipe, the water speeds up, but the pump impeller will slow down, and, as a result, the total mass-flow rate goes down. The speed of the ...

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As a reference, consider the Meredith effect, which applies (applied) in the construction of the P-51 Mustang radiator. From the linked Wiki site: The Meredith effect occurs when air flowing through a duct is heated by a heat-exchanger or radiator containing a hot working fluid such as ethylene glycol. Typically the fluid is a coolant carrying waste ...

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Usually what happens when you are trying to determine parameters about drilling, lathing or milling a material you need to consider the following parameters: material that is being removed material of the bit that does the cutting the relative velocity between the material and the "bit". determine if a liquid medium is used to remove heat. There ...

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Unless an aircraft takes off vertically the thrust it needs is much less than its weight, roughly proportional to the sin of take-off angle. Or sometimes a plane can fly with zero trusts. such as on the top end of a sharp climb or at the pull-up of an aerobatic loop. In both cases, its speed is enough to keep the wind on the wings and maintain the lift. Just ...

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The easiest way to view the problem is to draw a control volume around the engine that is big enough so that the air pressure is constant on the boundary surface of the control volume and the upstream air velocity is just the free-stream velocity everywhere on the upstream boundary surface of the control volume. The downstream velocity is faster, and the ...

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In the 50's the DARPA designed such a craft but due to aerodynamic instability it was a complete failure and abandoned. The problem you have is maintaining 4 separate intakes which have a greater volumetric capacity than each propeller can handle at maximum power in coldest conditions. You cannot combine the intakes and expect stability, it's just not ...

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At the start-up of the motion of the airplane the thrust mostly goes to overcome the static and then dynamic friction of tires so we need to consider the moment with respect to the distance of the center of the jet engines to the runway (for your design). Then the thrust goes mostly to overcome the lift-induced drag and parasite drag. And a fraction of the ...

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What you are interested boils down to the "specific cutting force" ($N/mm^2$) parameter - which varies by material. Not just steel or aluminum - but what alloy (e.g. addition of lead acts as a lubricant and makes it easier to machine), and hardness treatment it has. In approximate terms: Aluminum 800 MPa, Iron 1500 MPa, Steel 2500-3000 MPa. So ...

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In your force/momentum balance equations you always can include the pressure of the gas. The only time you can ignore it, is if the difference in $PA$ is small compared to other forces. For example in a diffuser at the entrance you have a lower pressure and lower area than at the exit. This results in a significantly lower $PA$ at the entrance, more than ...

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the amount of energy per second you would have to dump into the air passing through the "heating chamber" in order to make the turboprop engine work would have to be at least as great as that produced by the usual combustion of jet fuel in the heating chamber. Your homework is to do a quick energy balance on the system (using the two heat sources- jet fuel ...

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This is done in a jet engine, and partly, by the cowl around the front fan on a jet airliner engine, but on a large propeller, you could only do it slightly without slowing the airflow through the propeller. The extra weight would outweigh the benefits.

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Well the best learning happens when there is a need, and the internet can answer almost any question. Easiest way to understand physics is to use SI derived units and always carry your units through your equation. velocity = 5 m/s mass flowrate = ?kg/s volumetric flowrate = 5m^3/s Thrust: ? Newtons = kg*m/s^2 I asked google what the mass of a cubic meter ...

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This is one analysis that gives you information . Source : http://naca.central.cranfield.ac.uk/reports/arc/rm/2218.pdf

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If you know the Froude efficiency $\eta_{\text{Fr}}$ and assume a constant velocity $u$ for the ship. Then you can calculate the thrust $F_{\text{T}}$ for given power at the shaft $P_{\text{S}}$ as $$F_{\text{T}}=\eta_{\text{Fr}}\frac{P_{\text{S}}}{u}.$$ There is also a dimensionless thrust coefficient $C_{\text{T}}$ which is defined as: $$C_{\text{T}}=\... 1 So I figured it out, I first took this equation. (C_F) = (CT_F) - (E)(P_a)/(P_C)_N Then I isolated (CT_f) and substituted (C_F) equation resulting in C = \sqrt{\frac{2y^2}{y-1)}*\frac{2}{y+1}^{y+1/y-1}*(1-\frac{P_e}{P_C}^{\frac{y-1}{y}})+e(\frac{P_e - P_a}{P_C})} + e\frac{P_e}{P_C} Then I solved for P_e using this equation E = \frac{(\frac{2}{y+... 1 This would be a very rough estimation.$$P = F v Now you assume that the pumps efficiency $\eta$ is 1. And for the power we assume 2.7 A x 12 V = 32.4 W. For the flow rate you assume $4.3 \mathrm{\frac{L}{min}} = 7,16 \cdot 10^{-5} \mathrm{\frac{m^3}{s}}$. This is due to the fact that flow rate and pressure are linked, as you already pointed out. ...

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I agree that reverse thrust is not nearly as effective ("efficient" might not be the correct term) as forward thrust, and I'm surprised that it could be as high as 50%. The previous responder (alephzero) did say that it is probably only that efficient at full power, and in my experience full reverse thrust is rarely used - very noisy. However, on big ...

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