Is there a formula to calculate venturi pressure drop if you know IAS, Input diameter and constricted diameter? I am sure there are better terms I just don't know them. Lady Luscombe uses a single venturi to create vacuum for the turn and slip. That is its only vacuum instrument. It is also in the prop wash along the left side of the fuselage which must be why the gyro works on the ground. I ask because I don't have a vacuum gauge and I'm curious as to how much vacuum is being generated. Thanks

Bernoulli is the most abused equation in fluid dynamics but it is actually the correct answer here, assuming incompressible flow (Mach <0.3) and ignoring any swirl. You also use conservation of mass for closure (relating velocity at input to throat). where p1 is atmospheric pressure and p2 is the pressure at the venturi throat, rho is atmospheric density, u1 is velocity at input (airspeed if prop contribution is neglected which it probably shouldn't be), D1 is input diameter, D2 is throat diameter. Use SI units if you want to be sane (pressure in pascals, density in kg/m^3, and velocity in m/s).

Thank you and if my dad were still with us he would have given me the same answer. I'm wondering if there is a rule of thumb type of equation I can use since I want to solve for p2 and can't measure p1?

You want to solve for for the venturi pressure drop which is P1-P2. And, the equation is already arranged for that.

use the ideal gas law equation....to derive Bernoulli's EQ. http://www.calctool.org/CALC/chem/c_thermo/ideal_gas

Correct. If you only care about the pressure differential, the magnitude of p1 is irrelevant (again for incompressible flow, when you get into the compressible regime absolute magnitudes matter but the equation provided was derived assuming incompressible flow). If for whatever reason you want the absolute magnitude of P2, P1 for all intents and purposes is going to be atmospheric pressure, so if you know the altimeter setting you can use that. You will need both atmospheric pressure and temperature to compute density via ideal gas law rho = P/(R*T). You can use standard day conditions below if you want a first guess/basic rule of thumb. Standard day pressure: 101,325 Pa Standard day density: 1.229 kg/m^3 Standard day temperature: 293.15 Kelvin

I prefer to derive Bernoulli's equation from Newton's laws. (Just for those who claim that Bernoulli's law explains the low pressure on top of a wing and Newton's laws don't.)

The momentum equation of Navier-Stokes *is* Newton's second law F=ma, just written out for a Newtonian fluid. Bernoulli's equation is nothing but a conservation statement that falls out of the momentum equation under the assumption of a steady, incompressible, inviscid, irrotational flow without heat transfer, external work, or body forces beyond gravity. As Capt. Thorpe said, it is nonsensical to distinguish between Bernoulli's equation and Newton's second law when the former is just a simplification of the latter that is valid under certain conditions.

Godwin's Law applies here as well. Someone will find a way to segway this into a BBQ thread. @AKiss20 's is about as good as one can give. I haven't wielded Bernoulli's since the undergraduate ME program.

Venturis generally come in one of 3 sizes, 2", 4", or 9", the numbers referring to the level of vacuum they produce (in inches of mercury) at 100 mph. If the venturi doesn't have a smaller venturi inside it, it's probably a 2", which is all that's needed to run a single turn & bank. To calculate the pressure using Bernoulli's formula you need to know the entry and throat diameters and the speed of the air entering the venturi (something less than the aircraft's airspeed).

How much pressure from your hand must apply to the bottle of sauce to get an even spread over the **insert favorite meat here.

Last pedantic note I swear! I just finished my dissertation, so I am at peak pedantic right now. Isentropic is not a descriptor of materials but rather of thermodynamic processes. A material cannot be isentropic, but rather processes that act on them can be (at least in theory, in reality basically no typical engineering process is truly isentropic even if we idealize it to be so). Entropy is a state variable of the material/system in question and thermodynamic processes describe changes in those state variables. [/pedantry]

well then....in that case, it's not adiabatic either. lol specially if there's major "Q'n" involved....