Piet Lammertse on propeller theory

AXIAL MOMENTUM

5 Thrust and the mass flow

5.1 The mass flow in the actuator disk

The mass flow dropped out of our equations for the momentum loss. The efficiencies (4.14) and (4.18) depend only on the induction a, c.q. b.

By Newton’s law (4.7) however, repeated here as (5.1), the thrust does depend on the mass flow :

T = m^{•} . w = m^{•} . b . V

(5.1)

By this indirect route, the mass flow will affect the efficiency for a given thrust after all. If the mass flow m^● is for some reason reduced ( and we will find later that this is the case in the tip loss ), then to maintain a given thrust T, the induction w will have to be increased. By (4.18), this increase in w will cause a loss of efficiency.

We will loosely come to call this loss of efficiency "the tip loss", but it is really a loss of thrust, for a given efficiency.

In the ideal actuator disk of figure 4.1, the volume of air passing through the prop disk ( per second ) equals the disk frontal area π R² times the velocity ( V + v ). The mass flow m^● equals this volume flow, multiplied by the density ρ ( in kg/m³ ) of the passing air. With (4.1) and (4.19) we have :

$m^{•} = ρ . π R^{2} . (V + v)$
$= ρ . V . π R^{2} . (1 + a)$
$= ρ . V . π R^{2} . (1 + ½ b)$	(5.2)

This mass flow is slightly higher than the mass flow before induction. By (5.1), this will give slightly more thrust. Since this is a secondary effect and it complicates many of the later derivations, we will often use the so-called "light loading" approximation. This assumes that v ≪ V, or a ≪ 1, or in other words : ( V+ v ) ≅ V.

The light loading assumption simplifies the mass flow for the lightly loaded ideal actuator disk from (5.2), to :

m^{•} = ρ . V . π R^{2}

(5.3)

5.2 The non-dimensional mass flow m '

We define the non-dimensional mass flow. We will find later that the tip loss acts as a reduction factor on it.

Using the simple light loading mass flow of (5.3) as a reference, we define :

m ′ \equiv \frac{m^{•}}{ρ . V . π R^{2}}

(5.4)

The non-dimensional version of the mass flow (5.2) is :

m ′ = 1 + ½ b

(5.5)

With the light loading assumption of (5.3) this simplifies to :

m ′ = 1

(5.6)

The fact that a physical propeller can even approach the ideal of accelerating just the air inside the flow tube ( like in figure 4.1 ), and that it therefore can approach m ′ = 1 is by non means obvious. We will find a justification in the vortex view later. For the purposes of this chapter and the next ones, we will just accept it as a given.

5.3 Effective disk area

The more heavily loaded propeller of (5.2) handles slightly more mass than the lightly loaded one of (5.3). We can interpret this as a slight increase in the effective frontal disk area which captures the incoming air.

We derived the increase in mass flow in figure 4.1 from the speed increase at the prop disk. From the same figure we might say that the propeller handles the incoming air from far ahead, at the unaccelerated velocity V, through a larger frontal area of the flow tube, its frontal area before contraction. The frontal area of this incoming air is larger than the prop disk by a factor of ( 1+½ b ).

We can therefore also interpret m ′ as an effective ( non-dimensional ) disk area, slightly larger than one.

5.4 The non-dimensional thrust T '

We make the thrust (5.1) non-dimensional by applying (3.2) :

T ′ = \frac{m^{•}}{ρ V^{2} . π R^{2}} . b . V

(5.7)

The non-dimensional mass flow of (5.4) changes this into :

T ′ = m ′ . b

(5.8)

With (5.6) this yields :

T ′ = b . (1 + ½ b)

(5.9)

For light loading we can use the m ′ = 1 of (5.5), which yields :

T ′ = b

(5.10)

5.5 The ideal, light loading efficiency expressed in T '

Equation (4.18) gave us the efficiency of an ideal, lightly loaded actuator disk in terms of the far wake axial induction b. With the light loading T ′ of (5.11) we can change this into a relation with the thrust coefficient :

η_{m} = \frac{1}{1 + ½ T ′}

(5.11)

or ith the aircraft version of the thrust coefficient :

η_{m} = \frac{1}{1 + ¼ C_{T}}

(5.12)

The momentum loss ratio (4.17) becomes :

\frac{P_{m} ′}{T ′} = ½ T ′ = ¼ C_{T}

(5.13)