Puzzle: converting thrust to force

[bina]

I have a funny situation here - hope my physics teacher from school doesn't read it...

I am reading this presentation on jet aircraft performance from a seemingly reliable source. Particularly, the equation for power available (Pa) on slide 12 shows:

Pa = V * Thrust, where T = 567,000 Newtons = V * 567,000 in Watts

Now I'm confused as far as the conversion from Newtons to Watts is concerned since the former represents a value of force while the latter is a quantity of job performed in a definite period of time. In this case, however, they both have the same value...

He then uses the conversion value in calculating excess power (Pex):

Pex = Pa - Pr

But since Pr (see: slide 9) is a function of weight (W) in Newtons, any weights in the first part of equation (i.e., to find Pa) must also be converted to Newtons to ensure compatibility.

So, what is right - thrust in Newtons or N-m/s (or, lbs-ft/s)?!
And is this the proper method of converting HP to force? (eg., C172R): 160 BHP * 550 ft-lbs/s * 0.454 kg/ft * 9.81 m/s2 ~ 392 kN

Thanks!

 

[mgh]

I think you may have understood the initial equation.

Power is the product of speed and thrust and so they do have the same value

In SI units:

Speed = [m / s]
Thrust = [N] or [kg * m /s^2]

So

Power = [m /s] * [N] or [m / s] * [kg * m / s^2] which gives [kg * m^2 / s^3]

In Imperial units (if you must use them!) the units are:

P = [ft / s]
T = [lbf]

So

Power = [lbf] * [ft /s]

Both power equations are both effective the same, only depending on the units.

Note that in this case the equations are for propulsive power and not for engine power. A staionery aircraft with speed = 0 develops no propulsive power.

 

[bina]

Thank you!

Note that in this case the equations are for propulsive power and not for engine power. A staionery aircraft with speed = 0 develops no propulsive power.

If I understand you correctly, I have to play with thrust units to solve the excess power equation (and any other aerodynamic equation, in general) for non-propulsive engines, in my case it's C172R's Lycoming. The question is WHICH unit I have to use other than Newtons to express thrust?! Maybe I have to employ propeller pitch to derive thrust?

I've tried Watts to no avail yet - the power available curve extends linearly never touching the power required curve...

I'm attaching my Excel sheet to this post in case it's helpful.

Thanks!

 

[mgh]

You can use Newtons, ft lb/s or even horse power [provided ttou use consistent units - they all should give the same result.
ly estimate it otherwise.

Esentiall,y at a constant throttle setting the engine speed will inrcease until until its RPM increases and shaft output torque equals the torque needed to turn the propellor at the same RPM.