Linear ( streamline) Flow and Flow Power Amplification
Abstract: Part 1 is a brief review of air flow fundamentals which lead to a way of enormously increasing the power of an air flow by efficiently accelerating it. This leads to two new inventions for extracting the increased flow power to do useful work. The new methods use ordinary air and involve no fossil fuel heat input or noxious exhaust gases. If desired, the new air motor can potentially provide a self-sustaining mode of operation ( Perpetual Motion).
1.1 Fluid Flow Principles: Linear (streamline) flow
1.2 Flow Losses and Inefficiencies
1.3 Flow Velocity Changes/ Acceleration and Pressure Changes
1.4 Flow Power Amplification: Accelerating the Linear Flow of a Gas to Increase its Flow Power
1.5 Accelerating the Basic Linear Mass Flow
1.1 Fluid Flow Principles: Linear ( streamline) flow:
Before we describe the new acceleration and energy production process, we need a few basic fluid flow principles. The general subject of air flow is a part of fluid mechanics [Ref.1] which deals with both incompressible and compressible flows. Air is compressible, so that its density is not constant but varies with the flow speed V, the pressure p and the temperature T. The changes in air density ρ with flow speed are quite small up to speeds of about Mach 0.3 (about 90 m/s), so that for low wind speeds, air is often treated as being incompressible with the density ρ being taken as constant (at about 1.2 kg per cu. m. at m.s.l.).With the density constant, only the pressure and temperature changes need be considered. In this case, for the low speed incompressible case, the two principal flow equations are the Bernoulli equation and the equation of mass continuity which are as follows: ( units are SI, that is to say, kilogram/meter/second units)
First, the Bernoulli equation, which is related to the energy, is:
∫dp/ρ + ˝ V2 = constant along a streamline (1)
For incompressible flow, this becomes
p/ ρ + ˝ V2 = constant along a streamline; in energy units (joules) (1a)
or p + ˝ ρV2 = constant along a stream line; in pressure units (Pascals) (1b)
The second flow equation is the Equation of Continuity of Mass:
ρ1 V1 A1 = ρ2 V2 A2 = dm/dt = m-dot = constant ( kilograms/sec) (2)
where the subscript numerals refer to values of the density ρ, velocity V and cross-sectional area A at different cross sections along the flow path.
It is seen that Equation (2) represents the mass m of air flowing per second (dm/dt), which in SI units would be expressed as kilograms of air passing through any given area per second. Thus, a reduced cross-sectional area A, means an increased velocity V and vice versa.. The velocity increase in a flow passing through a converging conical nozzle can thus be formulated from (2) as
V2 = V1 [A1/ A2] [ ρ1/ ρ2] (3)
From this we can see that, if the air density ρ were to be constant, the velocity increase in a converging nozzle is inversely proportional to the cross-sectional area decrease. This simplifying constant density assumption could apply for example to air speeds of less than about Mach 0.3 ( 90 m/s) . Above that speed compressibility becomes important and density changes must explicitly be taken into account through the isentropic relationships.
In an isentropic flow [Ref. 1, 2, 3] the changes in the thermodynamic variables of the gas, i.e. in the pressure p, density ρ, temperature T, speed of sound c, and flow velocity V, all take place according to the isentropic relationships:
(p/po)(k-1)/k = ( ρ/ ρo)k-1 = T/To = (c/co )2 = 1 – (V/co)2 /n (4)
where n = 2/(k-1) and k is the ratio of specific heats. Expressed in terms of the number of ways n that the energy is divided ( n = 2/(k -1)) we also have
(p/po)2/(n+2) = ρ/ ρo)2/n = T/To = (c/co )2 = [1 – (V/co)2 /n ] (4a)
Thus, for example, a drop in any one of pressure, density or temperature brings about an increase in the flow velocity. The subscripted values are for stagnation or zero velocity initial conditions. For air, the value of k, the ratio of specific heats ( k = cp/cv) , has the value 1.4 . Here c is the speed of sound and co is its stagnation ( no flow) speed.
As an example of the magnitude of the isentropic effects, we may note that that at the flow speed of approximately 313 m/s ( i.e. at the sonic speed), the pressure will have dropped by 47.2% to almost half an atmosphere, the density will have dropped by 37 % and the temperature by 17%,. The isentropic values are readily available from tables ( Ref. 1,2,3] or from various commercially available computer programs for isentropic flow calculations.
Isentropic velocity increases readily take place, for example, in a converging nozzle through which the flow is directed; the velocity reaches its maximum value at the narrowest cross-section of the nozzle called the ‘throat’. If the flow is then passed on through a diverging nozzle, it will decelerate efficiently without losses and drop back to normal velocity, pressure and density value at the exit. The governing equations for compressible flow of air in a nozzle are the isentropic flow relationships as given in Eqns.4 and 4a plus the equation of continuity of mass in Eqn.2
1.2 Flow Losses and Inefficiencies
The Bernoulli equation, mass continuity equation and isentropic equations given above for linear flow all assume that no flow turbulence, viscosity or friction is present This amounts to assuming that the flow is 100 % efficient. In practice, of course, some turbulence, heat flow ,viscosity and so on will be present, but the losses can still with care be kept as low as a few percentage points. We also note that fluid acceleration can quite easily be made nearly 100% efficient; deceleration, on the other hand, is more difficult to achieve efficiently, that is to say without losses.[ Ref. 1]. We shall consider such accelerations in detail shortly.
1.3 Flow Velocity Changes/ Acceleration and Pressure Changes
From isentropic considerations, we have [Ref. 2]:
dp = −ρV dV
Also, in incompressible flow the familiar Bernoulli equation
p +1/2 ρ V2 = constant
gives the same result, namely dp = − ρV dV
Thus, an acceleration in flow velocity must result in a decrease in pressure, whereas deceleration must bring about an increase in pressure.
We can now apply the above flow principles to explain several interesting experimental problems.
Since an acceleration increases the kinetic energy of the flow, it is natural to consider how the kinetic energy and power of a flow may be increased and then extracted to do useful work.
Energy can be added to a fluid in various ways. Heating a gas adds internal energy and the temperature and pressure are thereby raised. This process is the basis for the operation of steam engines and internal combustion engines of all sorts, and its thermodynamics and dynamics are well understood.
Another way of adding energy is to increase the flow velocity, so as to add kinetic energy to the flow. This process is also very well understood, for example in rocket nozzle propulsion, gas dynamics, and so on.
Here we shall be concerned with adding kinetic energy and power to a flow of gas ( specifically to air) and then extracting a portion of the power increase to do useful work. The basic mass flow of gas can be supplied (1) by a vacuum motor or an air compressor ( and this leads to a first invention to be described in Section 2, or (2) by the wind or a moving vehicle ( which is the basis for a second invention to be described in Section 3).
1.4 Flow Power Amplification: Accelerating the Linear Flow of a Gas to Increase its Power
The starting point here is a basic mass flow of air. This is supplied to a fluid in several ways, for example by a “pull” or vacuum source, a “push” or compressor source, by the wind, or by the air flowing past a moving vehicle on which the flow apparatus is mounted.
Mass flow rate = ρ1 V1 A1 = ρ2 V2 A2 = dm/dt = m-dot = constant ( kg/s)
(which is the Equation of Mass Continuity ( Eq. 2) of Section 1.1 above)
The power of this basic mass flow at the source is usually low since the flow velocity initially is low. The formula for air flow power is
P = ˝ x (mass flow rate ) x V2
Thus a mass flow of 0.5 kg/s at a velocity of 10 m/s has a power of P = ˝ x 0.5 x 102 = 25 watts
In connection with supplying and maintaining a given mass flow, it will be well to point out that any flow of air from whatever source through an apparatus will involve the pressure difference Δp = pin – pout across the apparatus from inlet to outlet; it is this pressure difference which drives and maintains the mass flow. In the case of a vacuum pump flow source (Fig.1), the pressure difference will be that between the ambient pressure at the inlet to the apparatus and the vacuum pressure maintained at the vacuum pump. In the case of a compressor, it will be the difference between the compressor exit pressure and the ambient pressure at the apparatus flow exit port. In the case where the source of the mass flow is the wind, the pressure differences across the apparatus from entrance to exit are very much smaller, being just the dynamic or flow pressure of the air, p = 1/2 ρ V2 .
For example, a flow speed of V = 10 m/s will have a dynamic pressure of p = ˝ ρ V2 = ˝ x 1.2 x 102 = 60 Pascals. This is the pressure differential Δp across the apparatus from entrance port to exit port needed to maintain the flow of 10 m/s through the apparatus. Inefficiencies in the flow through the apparatus, such as turbulence and frictional losses will lower the pressure differential, flow speed, flow power and mass flow rate.
Figure 1. Linear “Pull” Flow Vacuum Motor
Also, to give a specific example, a vacuum source motor capable of delivering a pressure differential to sustain a mass flow of say, 0.074 kg/s, would typically would require a an input power of about 1700 watts to run it. However, conventional vacuum pumps are inefficient ; for example, the 1700 watts input cited would produce only 600 to 100 watts air flow power, so that the vacuum pump efficiency would only be from 35% to about 6%.
1.5 Accelerating the Basic Linear Mass Flow
This can efficiently be accomplished by passing the mass flow through a converging nozzle. The nozzle can be of various shapes, such as conical, bell- shaped, parabolic, etc., but all must be smooth walled.
Figure 2. Flow Acceleration in a converging/diverging nozzle ( De Laval Nozzle)
The appropriate mass flow equation here is, again, the Equation of Continuity of Mass
ρ1 V1 A1 = ρ2 V2 A2 = dm/dt = m-dot = constant (2)
where the subscript numerals refer to values of the density ρ, velocity V and flow duct cross-sectional area A at different cross sections along the flow path.
It is seen that Equation (5) represents the mass m of air flowing per second (dm/dt), which in SI units would be expressed as kilograms of air passing through any given area per second.
Thus, a reduced cross-sectional area A, means an increased velocity V and vice versa.. The velocity increase in a flow passing through a converging conical nozzle can thus be formulated as
V2 = V1 [A1/A2] [ ρ1/ ρ2] (3)
If the density is constant, so that ρ1 = ρ2 , then we have
V2 = V1 [A1/A2]
and we see that the velocity increase in a converging nozzle is inversely proportional to the cross-sectional area decrease. This simplifying assumption of constant density applies to air speeds of about Mach 0.3 ( 90 m/s) . Above that speed, the decrease in density with increasing velocity becomes important for compressible fluids and must be explicitly be taken into account through the isentropic relationships.
If the ratio of cross-sectional area is made large enough (that is to say, if the nozzle throat diameter is made small enough) then the throat velocity can be increased to the sonic speed (313 m/s). At this speed the flow rate ( mass flow ) cannot increase further at the given throat diameter and the flow is then said to be “choked” [ Ref. 1,2,3].
It is interesting to calculate the increase in the flow power that this acceleration to sonic speed has caused. The power of the flow is
P = ˝ (m-dot) V2 = ˝ (ρ V A) V2 (8)
Let us take as an example, a converging cone of inlet area 0.0198 m2 (diameter 15.9 cm); and throat area of .000998 m2 ( diameter 3.56 cm). Then, for an inlet flow of 10 m/s we have a mass flow of m-dot = (ρ V A) = 1.2 x 10 x .0198 = 0.2376 kg/s., which will be the same at inlet, throat and exit.
The power of the air flowing in at the inlet is then
P = ˝ x 0.2376 x 102 = 11.9 watts.
However the power at the throat, where the velocity is now sonic or 313 m/s and has the same mass flow of 0.2376 kg/s, becomes
P = ˝ x 0.2376 x 3132 = 11,639 watts
or an increase in flow power of 978 times!
Furthermore, since the process is isentropic, this all has happened without any addition of energy from the outside.
What has happened is that the internal energy of the gas itself has been lowered and has reappeared in the form of increased kinetic energy of the flow.
All that is required here is that we maintain the mass flow by maintaining the pressure differential from inlet to exit; the acceleration through the nozzle then takes place automatically. There is no input or output of outside of heat and so the process is called isentropic. Isentropic nozzle acceleration of air flow is truly an amazing phenomenon.
The natural question now is: Q. Can we extract any of this astonishing increase in flow power to do useful work? This will be answered in Sections 2 and 3.
1. Munson, Bruce, R., Donald F. Young, and Theodore H. Okiishi, Fundamentals of Fluid Mechanics, Wiley and Sons, New York, 1990.
2. Shapiro, Ascher, H., The Dynamics and Thermodynamics of Compressible Fluid Flow, 2 Vols., John Wiley & Sons, New York.1954.
3. R. Courant and K.O.
Friedrichs, Supersonic Flow and Shock Waves. Interscience,
Copyright Bernard A. Power, May 2011
Section 2: Invention No. 1: A New Isentropic Air Motor and Clean Energy Source
Section 3: To be posted in near future
Section 5: A Note on Isentropic Flow ‘Potential Motion’