PROPELLER ANALYSIS
The efficiency of a piston engine-propeller combination depends on a proper match of the propeller to the engine as well as a match of the two to the airframe. Understanding propeller behavior is important, even though we are currently in the “jet age.” First, because of their cost, it is doubtful that gas turbine engines will be used in the smaller, general aviation airplanes in the foreseeable future. Second, there is currently a renewed interest in the turboprop engine because of its lower fuel consumption compared to turbojet or turbofan engines.
Momentum Theory
The classical momentum theory provides a basic understanding of several aspects of propeller performance. Referring to Figure 6.8, the propeller is approximated by an infinitely thin “actuator” disc across which the static pressure increases discontinuously. The assumptions inherent in this model are: ‘
1. The velocity is constant over the disc.
2. The pressure is uniform over the disc.
Figure 6.8 Idealized flow model for application of classical momentum theory. |
3. Rotation imparted to the flow as it passes through the propeller is neglected.
4. The flo. w passing through the propeller can be separated from the rest of the flow by a well-defined stream tube.
5. The flow is incompressible.
The transverse plane 1 is far ahead of the propeller, while plane 4 is far downstream. Planes 2 and 3 are just upstream and downstream of the propeller, respectively. In planes 1 and 4 all streamlines are parallel, so that the static pressure is constant and equal to the free-stream static pressure p0.
Let us first consider the continuity of flow in and out of the cylindrical control surface shown in the figure. This surface has a cross-sectional area of
S. The flux passing out of the surface across plane 4 minus the flux entering across plane 1 will be
Д Q = A3 V3 + (S – A3) V0 – S V0 or
AQ = A3(V3-V0) (6.2)
Assume the nontrivial case where V3 Ф V0, it follows that the flux AQ must be entering the control surface along its sides. This flux has a velocity of V0 in the direction opposite to the thrust.
Applying the momentum theorem to the cylindrical control surface and noting that the external pressures cancel out, we obtain
T = p[A3 V32 + (S – A3) V2] – pSV2 – pAQV
Substituting Equation 6.2 into the preceding equation gives
T = pA3V3(V3-V) (6.3)
A3 is the cross-sectional area in the ultimate wake of the streamtube passing through the propeller. Thus, pA3V3 is the mass flux passing through the propeller.
The thrust, T, is also equal to the pressure difference across the actuator disc multiplied by the disc area, A.
T = A(p2-Pi) (6.4)
Pv and p2 can be related by applying Bernoulli’s equation ahead of the propeller and downstream of the propeller. The equation cannot be applied through the propeller, since energy is added to the flow at the propeller.
Po + 2pV2 = pi + ipV,2 (6.5)
Po + Ip V32 = p2 + іP V22 (6.6)
Substracting Equation 6.5 from Equation 6.6 and noting that the velocity
is continuous through the propeller gives
Рг ~ Pi = 2P( V32 – V2) (6.7)
Using the fact from continuity that A} V3 = AV, and combining Equations 6.3,
6.4, and 6.7 results in the well-known relation
In words, the velocity through the propeller equals the average of the velocity far ahead of and far behind the propeller.
Let us now write
V3=V + 2w (6.9)
where w is the propeller-induced velocity. It follows that
This is easily remembered, since pA(V + w) is the mass flux through the propeller and 2w is the total increase in the velocity of the flow.
Applying the energy theorem to this system gives, for the power, P, added to the flow,
P =12PA(V+w)[(V+2w)2- V2] or
P = 2pAw(V + w)2 (6.11)
Using Equation 6.10, this becomes
P = T(V+w) (6.12)
This important result states that the power required by the propeller equals the product of its thrust and the velocity through the propeller. This can be divided into two parts. The first part is defined as the useful power.
Fuse = TV (6.13)
The second part is known as the induced power.
Pi = Tw (6.14)
Equation 6.10 can be solved for the induced velocity to give
H[-v+Vv!+©]
As an example, consider a propeller 2 m (6.6 ft) in diameter driven by a 150 kW engine (201 hp). The maximum static thrust that one might expect from this propeller can be calculated by solving Equation 6.17 for T.
T = Pf(2pA)m (6.18)
Since 1 W = l. OOm-N/s,
Pk = 151,050 m-N/s (110,550 ft-lb/sec)
For standard sea level conditions, p = 1.226 kg/m2. Thus,
T = (151,050)2/3(2 x 1.226 x 3.14)ш = 5600 N(1259 lb)
This value of T represents an upper limit that is not attainable in practice, since the momentum theory neglects profile drag of the propeller blades. Also, additional induced losses occur near the tips of the blades. Since the pressure difference across the blades must vanish at the tips, a trailing vortex system, helical in shape, is generated by the propeller in a manner similar to a finite wing.
In forward flight, an ideal efficiency, ту,, can be defined as the ratio of the useful power to the total power given by Equation 6.12.
i + Vi + tc
The thrust of a propeller divided by its disc area is referred to as the disc loading. As this loading approaches zero, the ideal efficiency is seen to approach unity.
As an example, let us again consider the Cherokee 180 having a propeller diameter of 1.88 m (6.17 ft). At a cruising speed of 60.4 m/s (135 mph) at standard sea level, its drag will equal approximately 1390 N (3121b). Thus Tc = 0.224, giving an 17, of 0.95. As we will see later, the actual propeller efficiency is more like 0.83. As with the static thrust, 17, given by the momentum theory is optimistic and represents an upper limit that is really not attainable.
Although the momentum theory is not too accurate with regard to predicting power, it is useful for estimating the induced velocity. An interesting and easily remembered relationship is the following. The dynamic pressure in the ultimate wake of a propeller is equal to the sum of the free-stream dynamic pressure and the disc loading. Proof of this statement is left to you.