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What is the equation needed to determine the maximum forward airspeed of a propeller driven plane in an unaccelerated level flight? I need an example with units of measurements. The engine is rated at 55-75 hp.

enter image description here

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  • $\begingroup$ @quietflyer does this help? $\endgroup$ Commented Jan 5, 2023 at 21:30
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    $\begingroup$ Thrust reduces with speed and drag (normally!) increases with speed. The maximum speed is the one at which the thrust and drag curves intersect (T = D). Sophit's answer explains well how to mathematically determine that speed. $\endgroup$ Commented Jan 5, 2023 at 23:13

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EDIT: I added a calculation at the end of the answer



Thrust $T$ produced by a propeller is:

$T=\eta \frac{P}{V}$

where $\eta$ is the propeller's efficiency, $P$ the power supplied to the propeller and $V$ the flying speed.

This thrust wins drag, which can be as usual expressed as:

$D=½\rho V²SC_d$

where $\rho$ is air's density, $S$ wing surface and $C_d=C_{d_0}+\frac{C²_l}{\pi Ae}$.

$C_l$ can be derived from the usual equation for lift $L$ that equals weight $W$:

$L=½\rho V²SC_l=W \Rightarrow C_l=\frac{W}{½ \rho V²S} \Rightarrow C²_l=\frac{W²}{(½ \rho V²S)²}$

Substituting and equating $T=D$ we get:

$\eta \frac{P}{V}=½\rho V²S(C_{d_0}+\frac{W²}{(½\rho V²S)²(\pi Ae)})$


So far so good.

This equation can be obviously solved mathematically but a graphical representation is easier; we just have to plot that equation after having rewritten it as:

$0=-\eta \frac{P}{V}+½\rho V²S(C_{d_0}+\frac{W²}{(½\rho V²S)²(\pi Ae)})$

This equation is 0 where it intersect the x-axis and that intersection is the $V$ we are looking for.

Let's use then the values given in the question:

  • $\rho=0.96kg/m³$; I really couldn't convert the given $q$ in a standard number so I use the ISA density at 1'500m (5'000ft).
  • $S=3.5m²$.
  • $C_{d_0}=0.021$
  • $\eta= 0.8$; this value is ok if we consider a variable-pitch propeller or a propeller working at its design point; otherwise it can be as low as 0.4 or even less.
  • $P=56'000W$ (75hp).
  • $W=3'700N$.
  • $A=7.65$.
  • $e=0.821$.

Substituting we get:

$0.0355V²-44'800/V+110.81/V²$

whose graphical solution is some 108m/s 210kts (general shape in the first picture and zoom of the solution in the second picture):

enter image description here

enter image description here

Just out of curiosity, with an $\eta=0.4$ the speed reduces to some 85.8m/s 167kts.

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  • $\begingroup$ To be clear: V is measured knots? P is measured in Horse Power? p (air density) is measured in... This is why I need an example with the unit of measurements. If I strickly use SI units will it come out correct? +1 $\endgroup$ Commented Jan 6, 2023 at 2:27
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    $\begingroup$ @Justintimeforfun Here's a suggestion: Convert everything in SI units - V in m/s, P in watts, ρ in kg/m³, S in m² and so on.. After calculating the V, you may convert it back in the units you would desire (knots). Yes, if you strictly use SI units, your answer will be correct. But! this V is TAS. To convert it into IAS (which I believe is what you want), you will first need to calculate the dynamic pressure (the SI unit of which is N/m² or pascals). Then from there you can finally convert it into IAS. $\endgroup$ Commented Jan 6, 2023 at 4:06
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    $\begingroup$ @AdityaSharma is correct. Just an example about power: horse power is a unit of measurement which is not tied to anything else; on the other hand, Watt is 1kgm²/s³ and it does show a tie with the units it derives from (supposedly a mass, an acceleration and a speed) $\endgroup$
    – sophit
    Commented Jan 6, 2023 at 10:34
  • $\begingroup$ @RobertDiGiovanni: because I think it's against the policy of Stackexchange doing someone else's homework $\endgroup$
    – sophit
    Commented Jan 7, 2023 at 0:13
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    $\begingroup$ even when using metric system it's good to do a units check.. often times you end up with things like gas constant for air being R=0.287 or 287 "units" depending on where you look it up, and in results like reduced turbine mass flow being sqrt(R)*xysomething .. you're off by a factor of 31.something, not necessarily obvious you missed a kJ vs J $\endgroup$
    – Apfelsaft
    Commented Jan 7, 2023 at 9:41
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Your "givens" include a table of Cl/Cd versus airspeed.

Therefore it is simple to prepare a graph of Cd/Cl versus airspeed.

From this, you can prepare a graph of (Weight * (Cd/Cl)) versus airspeed.

Thrust Required = Weight * (Cd/Cl). So this is your graph of Thrust Required versus airspeed.

You'll also need to prepare a graph of Available Thrust versus airspeed. For this, you'll need to know more information than you've given us. For example, you might have a curve of engine horsepower versus RPM, along with some indication of prop efficiency. Or you might even have some way of measuring thrust delivered by the motor to the prop in actual flight. Without knowing more about what your "knowns" are, it's hard to be specific about how you should prepare the graph of Available Thrust versus airspeed. (It's going to make a tremendous difference whether the propeller is fixed-pitch or variable-pitch (constant speed), and if the former, for what airspeed it is optimized -- i.e. "climb prop" versus "cruise prop".)

An absolute upper bound for your Available Thrust at any given airspeed would simply be Power / airspeed, assuming that the full nominal value of (75 hp? It's unclear what "55-75 hp" means) is available regardless of airspeed. You'll never have this much thrust available in actual practice, but it's a starting point.

At any rate, the airspeed (well to the right on your graphs, i.e. well above the minimum-thrust-required airspeed) where the Thrust Required curve crosses the Available Thrust curve, is the maximum sustainable airspeed. If you've used the theoretical "upper bound" expression for Available Thrust given above, this airspeed figure will be overly optimistic. (But if this was a homework problem, perhaps this is the approach you were expected to take.)

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    $\begingroup$ Can you do this in Horse power (propeller) not thrust (turbine)? +1 $\endgroup$ Commented Jan 5, 2023 at 21:27
  • $\begingroup$ An example would help so much. $\endgroup$ Commented Jan 5, 2023 at 21:27
  • $\begingroup$ Thrust of a propeller is power divided by speed. The power being the effective power delivered by the propeller, so engine power times any transmission and propeller efficiencies (which vary with speed to some extent). $\endgroup$ Commented Jan 5, 2023 at 21:53

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