Most general aviation or private airplanes are powered by propellers. The details of how a propeller generates thrust is very complex, but we can still learn a few of the fundamentals using the simplified momentum theory presented here.

__Propeller Propulsion System__

On the slide, we show a schematic of a propeller propulsion system at the top and some of the equations that define how a propeller produces thrust at the bottom. The details of propeller propulsion are very complex because the propeller is like a rotating wing. Propellers usually have between 2 and 6 blades. The blades are usually long and thin, and a cut through the blade perpendicular to the long dimension will give an airfoil shape. Because the blades rotate, the tip moves faster than the hub. So to make the propeller efficient, the blades are usually twisted. The angle of attack of the airfoils at the tip is lower than at the hub because it is moving at a higher velocity than the hub. Of course, these variations make analyzing the airflow through the propeller a very difficult task. Leaving the details to the aerodynamicists, let us assume that the spinning propeller acts like a disk through which the surrounding air passes (the yellow ellipse in the schematic).

The engine, shown in white, turns the propeller and does work on
the airflow. So there is an abrupt change in pressure across the
propeller disk. (Mathematicians denote a change by the Greek symbol
"delta" ( ).
Across the propeller plane, the pressure changes by "delta p"
( **p**). The
propeller acts like a rotating wing. From airfoil
theory, we know that the pressure over the top of a lifting wing
is lower than the pressure below the wing. A spinning propeller sets
up a pressure lower than free stream in front of the propeller and
higher than free stream behind the propeller. Downstream of the disk
the pressure eventually returns to free stream conditions. But at
the exit, the velocity is greater than free stream because the
propeller does work on the airflow. We can apply Bernoulli's
equation to the air in front of the propeller and to the air
behind the propeller. But we cannot apply Bernoulli's equation across
the propeller disk because the work performed by the engine violates
an assumption used to derive the equation.

__Simple Momentum Theory__

Turning to the math, from the basic thrust equation, we know that the amount of thrust depends on the mass flow rate through the propeller and the velocity change through the propulsion system. Let us denote the free stream conditions by the subscript "0", the conditions at the propeller by the subscript "p", and the exit conditions by the subscript "e". The thrust (F) is equal to the mass flow rate (m dot) times the difference in velocity (V).

F = [m dot * V]e - [m dot * V]0

There is no pressure-area term because the pressure at the exit is equal to the free stream pressure. The mass flow through the propulsion system is a constant, and we can determine the value at the plane of the propeller. Since the propeller rotates, we can define an area (A) that is swept out by the propeller of blade length (L). Through this area, the mass flow rate is density (r) times velocity (Vp), times area.

m dot = r * Vp * A

Substitute this value for the mass flow rate into the thrust equation to get the thrust in terms of the exit velocity, entrance velocity, and velocity through the propeller.

F = r * Vp * A * [Ve - V0]

We can use Bernoulli's equation to relate the pressure and velocity ahead of and behind the propeller disk, but not through the disk. Ahead of the disk the total pressure (pt0) equals the static pressure (p0) plus the dynamic pressure (.5 * r * V0 ^2).

pt0 = p0 + .5 * r * V0 ^2

Downstream of the disk,

pte = p0 + .5 * r * Ve ^2

At the disk itself the pressure jumps

delta p = pte - pt0

Therefore, at the disk,

delta p = .5 * r * [Ve ^2 - V0 ^2]

The force on the propeller disk is equal to the change in pressure times the area (force/area x area = force)

F = delta p * A

If we substitute the values given by Bernoulli's equation, we obtain

F = .5 * r * A * [Ve ^2 - V0 ^2]

Combining the two expressions for the the thrust (F) and solving for Vp;

Vp = .5 [Ve + V0]

The airspeed through the propeller disk is simply the average of the free stream and exit velocities.

[Note that this thrust is an ideal number that does not account for many losses that occur in practical, high speed propellers (like tip losses). The losses must be determined by a more detailed propeller theory, which is beyond the scope of these pages. The complex theory also provides the magnitude of the pressure jump for a given geometry. The simple momentum theory, however, provides a good first cut at the answer and could be used for a preliminary design.]

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- Beginner's Guide to Aerodynamics
- Beginner's Guide to Propulsion
- Beginner's Guide to Model Rockets
- Beginner's Guide to Kites
- Beginner's Guide to Aeronautics

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*byTom
Benson
Please send suggestions/corrections to: benson@grc.nasa.gov *