GB2534840A

GB2534840A - Static supercharger

Info

Publication number: GB2534840A
Application number: GB1500045.8A
Authority: GB
Inventors: Leslie Winter Christopher
Original assignee: Individual
Current assignee: Individual
Priority date: 2015-01-04
Filing date: 2015-01-04
Publication date: 2016-08-10

Abstract

A piston engine supercharging device wherein high pressure gas is directed from the cylinder a or the exhaust port n into the flow of low pressure gas, increasing the energy of the lower pressure gas. Preferably the exhaust gas is injected by way of an eductor e into the lower pressure gas upstream of a manifold or inlet pipe c. The pipe containing only the high pressure gas may have a non-return valve k. To allow the gasses to mix faster and create a more compact design a wire mesh may be placed downstream of the eductor. Preferably this wire mesh is of non-uniform thickness and may be shaped to introduce a rotational movement to the flow of gas.

Description

Static supercharger This invention relates to piston engines. It relates to increasing the mass flow into or out of a cylinder on the inlet and/or outlet stroke without the energy for the increased flow being derived from a rotating blade compressor (supercharger or turbo).

In accordance with the present invention there is provided the process of use of an amount of compressed gas originating from within a cylinder and/or the exhaust gas pipe or manifold and using said gas as the motive power for an educator in the inlet and/or exhaust pipe(s) (sometimes called the inlet or exhaust ports -upstream/downstream of their respective valves) or manifolds of an engine that increases the flow in the inlet and/or exhaust pipe(s) or manifolds of said engine. (Eductors are sometimes called ejectors.) And: In accordance with the present invention there is provided a device comprising a source of high pressure gas, a pipe to take some of the high pressure gas leading to a jet through which the high pressure gas flows into another gas, the first gas giving up some of its energy to increase the energy of the second gas: the high pressure gas being sourced from the working fluid of a piston engine. Preferably said pipe having a device to reduce backflow; and preferably the source of high pressure * * * * * * gas is a cylinder of the piston engine. -* * * * * Advantages of this invention are that it can have a fast response time (no 'turbo la(); it eliminates the back-pressure required by a turbo; need have no moving parts; its frequency is automatically aligned * to engine RPM; is cheap to make; is very light weight -it can increase the power to weight ratio of an * * * * * * engine, and that it may be applied as a modification to existing engines. Applied to the discharge port * -it can also remove more mass from the cylinder on the exhaust stroke -increasing the pressure drop available for the cylinder charging stroke.

* * * **** ** * Diagrammatically, the idea of the invention is given in Figure 1, where: a) is a cylinder, b) is a piston, c) Is an Inlet port, d) is an open inlet valve, e) is a working eductor, f) is a pipe carrying compressed gas from an originating source such as shown as h), l), and j); k) is an optional partially acting non-return device, I) is a compressed gas reservoir, m) is a pipe from the reservoir to the educator. n) is an exhaust port, o) is an eductor in the exhaust port, p) is a pipe from the reservoir to the exhaust eductor, q) is a closed exhaust valve.

* . Part of the theory behind this invention is that the flow of gas through a restriction (e.g. an inlet valve) will not increase once the pressure downstream of the restriction is less than about 50% of the upstream absolute pressure. This is because the flow reaches sonic velocity of the gas in such circumstances. In these circumstances, the only way of increasing the flow through said restriction is by increasing the pressure upstream of the restriction: increasing the upstream pressure being an advantage of this invention.

* ** The eductor could also be located in the exhaust port/manifold. (Aside from tuned outlets -it would only work during the times when the pressure differential across the exhaust valve is less than 50% of the absolute pressure upstream of the exhaust valve.) This invention could also be used to increase the flow of exhaust gas by placing an eductor in the cylinder head just upstream of the exhaust port valve, the eductor blowing into the 'orifice' created when the exhaust valve opens.

The source of pressurised gas is preferably taken from a cylinder, but may be taken from any part of the engine having a pressure above that required to allow the eductor to increase the flow of gas in the prevailing circumstances: e.g. in the (eductor) locations indicated in Figure 1.

Preferably the peak pressure in the source of pressurised gas is greater than 1.1 bar absolute, more preferably greater than 1.5 bars absolute, more preferably greater than 2 bars absolute, more preferably greater than 5 bars absolute, more preferably greater than 10 bars absolute, more preferably greater than 15 bars absolute, more preferably greater than 20 bars absolute, more preferably greater than 2S bars absolute, more preferably greater than 30 bars absolute, more preferably greater than 50 bars absolute The eductor motive power gas (compressed gas) may flow, from its source via a pipe, into a reservoir. The pipe and/or reservoir may contain a device to restrain backflow e.g. a device having a lower internal Cd (Drag Coefficient) for flow into the reservoir than for reverse flow -thereby aiding the net flow towards the eductor and reducing the flow back towards the source of the eductor motive power gas: e.g. as k) in Figure 1.

Whilst a continuous flow of eductor motive power gas into the eductor may be used, the introduction "°' . of a valve to time the flow, e.g. to make the motive power flow flow when it can be best used to **** * supercharge the engine into the eductor may be used. For instance, based on engine RPM, a rotating * * port valve may be used, as may a valve timed either by mechanical or electronic means also based * upon the stoke type and crankshaft angle. A simple flap valve could also be used. * *

Given that some sources of eductor motive power gas could be very hot, the material of this invention * * * * must be such as to withstand such temperatures and pressures. Therefore it is desirable to make the " hot parts of this invention from high nickel and/or chromium alloys, or from a high molybdenum * * * * * ** alloys. Furthermore in view of a desire for vibration cracking resistance, micro-alloy materials are also * * * * * * desirable. Preferably the mass of the materials of this invention are kept to a minimum consistent with * * the mechanical conditions under which they are operating, so as to reduce the stresses at points where pipes enter fixed locations (e.g. at the points where pipes enter into manifolds). Springs may also be used to dampen vibration.

The eductor compressed gas (motive power gas) may be cooled prior to it being used as eductor motive power. However, cooling the gas takes energy from the system; and calculations show that a hotter gas can have a better effect. This is perhaps not surprising in that at the lower density, the velocity out of the eductor is higher, thus its momentum (mvA2) is higher.

Cooling, say from 1200 C to 600 C may be achieved by any normal means such as air cooling. Preferably the cooling should not take the compressed gas below its dew point: supported 'Microporous' insulation might be useful as these temperatures.

Whilst cooling the compressed gas will decrease its volume, its pressure may be maintained simply by the use of the said partially known return device (e.g. "k" in figure 1), or by a non-return valve which may be as simple as a flap valve or sprung ball arrangement -forward flow compressing the spring and allowing the ball to move from its seat and thus allowing gas to flow past the ball in the desired direction, whilst when the pressure of the incoming gas falls below that at which the spring can overcome such pressure the spring causes the ball to reseat thereby stopping reverse flow.

Preferably, to avoid vibration damage to any narrow tubes carrying the eductor motive power gas, bellows may be used to support the narrow tubes, and/or the tubes may be supported (for e.g. exhaust port applications) by a suitable elastomeric material -the selection of which has to take into account the maximum and minimum operating temperatures of the gas and the engine in the areas where the pipes carrying the eductor motive power gas are located.

The source of eductor motive power gas need not come from the cylinder with which the eductor is associated in its inlet port/manifold, but may come from any other cylinder, or even from the outlet manifold at times when its pressure is suitably high.

One embodiment of this invention may contain a tuned volume, or, it may be tuneable in the sense that it may have a piston so as to make the pressure of the compressed gas at the eductor to be maximised at the time when the flow from the eductor is required.

The flow of entrained gas may itself flow into a tuned inlet manifold. * * * * * *

* * * * * * * * * * * * * * * * * * * * * * * One means of reducing the possibility of clogging the compressed gas (source) pipe is to ensure that the velocity in the pipe is high. This may be achieved by taking a significant pressure drop across the source pipe, for instance greater than 0.1 bar, more preferably greater than 0.2 bar, more preferably greater than 0.5 bar, more preferably greater than 1 bar. In view of the short length, say below 10 feet, of the compressed gas source pipe these pressure drops infer that a high velocity will occur in said pipe, said high velocity fluidising/dislodging any deposits/dew that might form in the pipe.

It may be desirable to include a particulate filter e.g. in the compressed gas source pipe or in the reservoir so as to reduce the possibility of clogging the eductor nozzles -this especially in the case of diesel engines, and being particularly useful times when an engine has just been started or is warming up. In order for any such particulate filter to have a chance of burning off any particulates it catches, Its location should preferably be where the gas flowing through it is as hot as possible -that is to say it should preferably be upstream of any compressed gas cooling device.

It may be possible to route the pipes through, or make the pipes part of, the engine block and cylinder head to avoid the need for external piping thereby eliminating the possibility of vibration fracture of any external pipe.

Another advantage of this invention is that it may be used to effect exhaust gas recycle.

In order to have a more compact design, downstream of the eductor there may be placed a wire mesh. The purpose of this wire mesh is to faster mix the high velocity gas emanating from the eductor with the gas to be entrained. Preferably this wire mesh should be of non-uniform thickness -this to reduce the probability of shock waves lining up downstream of the mesh. Additionally, the mesh might be in the form of a non-uniformly thick disc but one which is split in the manner of a solid (in the sense of covering the entire gas flow area) split spring washer: the purpose of which is to induce a rotational component into the flow of the gas. Such rotational flow may then be carried into the cylinder to induce swirl therein.

Where used in a large engine, an eductor may have a number of holes through which the compressed gas flows -so as to increase the surface area between the high speed gas emanating from the eductor and the gas to be entrained -the better to entrain the gas to be entrained.

Figure 2 and the calculation given below are used to demonstrate the theory of steady-state flow of an eductor to give an indication of the quantitative performance of this invention. In practice, due to the reciprocating nature of the engine, the flows are not steady-state and therefore additional calculations are required to take into account the acceleration and deceleration of both the compressed gas, and the gas flow it is acting upon as it leaves the eductor. Also of any tuned chamber(s) included in the design.

Such calculations would best be carried out by the use of computational fluid dynamics programs and are significantly affected by and dependent upon the mechanical design and parameters of the source of the compressed gas, it's piping to the eductor, any cooling and/or reservoir devices incorporated, the mechanical design of the eductor, and the ports into which the bulk gas from the eductor flows.

In Figure 2, r) is an inlet port, and s) is an eductor, whilst the other letters with subscripts show the positions of the various parameters used in the following calculation.

In figure 2, the compressed gas high velocity fluid enters the port at the exit of the eductor and mixes with the secondary, bulk, low velocity fluid (e.g. the fresh charge going into the cylinder); mixing is assumed to be complete at Point 2. * .* * * *

* * ** Through this mixing, a major portion of the momentum of the educted compressed gas is imparted to ^ * the fluid to be entrained, resulting in a static pressure at Point 2 greater than at Point 1. This increase in static pressure is given by equating the pressure and momentum forces along the port. This * increase in static pressure (P2 minus P1) may be found from the equation: * * *** (P2-P1)GcA = Wp(Vp-Vm) + Ws(Vs-Vm) * * ** * where P1 equals the pressure at Point 1, pounds force per square foot; P2 equals pressure at Point 2, pounds force per square foot; Gc equals dimensional constant 32.17 pound feet per pound force second squared; A equals cross-sectional area of the port, square feet Wp equals weight rate of flow, pounds per second; Vp equals velocity feet per second of the primary (eductor) fluid; Ws equals weight rate of flow, pounds per second and Vs equals velocity feet per second of the secondary fluid; Vm equals velocity of the combined fluids, feet per second.

* * * The velocity of the primary fluid may be found by using the equation for choked mass flow as given in the calculations below.

** ** The increase in the flow through the valve caused by the increase in pressure caused by the use of the eductor, may be found by assuming that the flow is turbulent -and therefore the new flow is equal to: ** * (the square root of the (new pressure difference between the port just upstream of the inlet valve and the pressure just downstream of the inlet valve, and said pressure difference if the eductor were not in operation)) times the flow that would obtain if the eductor were not present..

* * * The percentage of the maximum mass of gas in a cylinder used as compressed gas in an eductor may be: 1%, 2%, 4%, 8%,10%, 14%, 18%, 22%, 26%, 30%, 35%, 40%, 50%, 60%, greater than 60%, or any percentage between these numbers. * **

(Note the calculations that follow use a percentage of the non-eductor design flow. Sorry for the mixed units!) In Figure 3, t) Is the flow of e.g. fresh air charge, e) is the eductor, s) is the flow of combined gases flowing towards the cylinder inlet valve. The expanded flow area of the pipe is to allow the combined gases to expand due to the increase in temperature, particularly if there is no r) -an optional (flap) valve. The axial flat surfaces may be replaced by e.g. convex parabolic surfaces to guide any reverse flow back towards the s) flow direction. * * * * * *

* * * * * * * * * * * ** * * * * * * . * * * ** ** * * * * * ** *

ONE EMBODIMENT -CRITICAL FLOW CALCULATION FOR FOLLOWING EDUCTOR CALCULATION

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In this embodiment, the pressure of the compressed gas fed to the eductor nozzle may be taken as more than about twice the pressure downstream of the nozzle, thus the flow through the eductor nozzle is choked flow (and, for a given gas composition, nozzle area and temperature, the flow is only a function of the compressed gas pressure). (

The eductor calculation requires a knowledge of the mass flow and velocity of the compressed gas (motive power gas) as it leaves the eductor. These are now calculated: (Metric) Formula for choked mass flow: Mass flow Kg/sec = Cd*A*(k*density*P *(2/(k 1))^((k+1)/(k-1)))^0.5 Equation 1 Cd = discharge coefficient, dimensionless Cd for critical flow is high. Say 0.95 A = eductor discharge hole cross -sectional area, square metres Say Eductor hole diameter mm = 1 mm so, area Sq metre = 7.86E-07 psia to Pa k = weighted gamma ratio of gas specific heats 6895 Po = Absolute upstream gas pressure in Pa Say 4137000 Pa = 600 psia ]bra Absolute upstream temperature degrees Kelvin: ** * * Assume a cooled temperature of compressed gas from power stroke TDC of **** Deg F = 600 C -above the dew point = 873 Deg K * 1112 NOTE: all other parameters being kept the same, the higher the temperature of the compressed gas, the higher the pressure of the (combined) entrained flow.

-*-*_* _-_ * 1 1 Ciltfulation of weighted gamma (ratio of specific heats) and molecular weight: * * *** I I ak$ composition to be as per exhaust composition -for a slightly lean burn. F--,

Rough Composition Gamma Weighted Mol Weighted Volume % Gamma weight Mol Wt CO 2 1.281 0.10248 44 3.52 Steam 1.324 0.11916 18 1.62 Nitrogen 83 1.401 1.16283 28 23.24 1.38447 28.38 Density = upstream gas dens ty Kg/cubic metre from weighted Mol weight of 28.38 _I = 16.18 Kg per cubic meter So, at the above conditions, the density calculates as l Critical flow -applying the above conditions to Equation 1 So, the critical mass flow from Equation 1= 0.004164 Kg/sec for a mm diam hole Critical Mass flows table for other hole diameters mm is proportional to the hole area. 1 I

If the absolute pressure of the compressed gas doubles, the mass flow doubles.

Pv = the Pressure for choked flow velocity calculation: Pv/Po = (2/(k+1)^(k/k-1) * * * At the above conditions: Calculation for Pv gives 319 psia Gas density at critial flow conditions of 319 psia and 873 Deg K is: 8.59 Kg/cubic metre Calculation of Vp -the velocity of gas through narrowest point of the eductor at critial conditions: Vol Vp Hole Dia Compressed Mass flow flow Vel at mm A sq m gas press Kg/sec at throat throat psia M3/sec M/sec 1 7.86E-07' 600 0.004164 0.000485 617 (Used below.) We-volume of one cylinder of say a 4 cylinder 122 cubic inch (2 Litre) engine is 30.5 cu inches = 0.01765 cu feet Density (assumed of) air at say 0 C, 32 F = 0.0808 Lb per cu ft 1 I Id The mass of air required to charge one cylinder at 13.7 psia is: 0.001329 Lbs = (Allowing for inlet pressure drop, taken as 100% charging volumetric eficiency) 0.000603 Kg If xbe engine Is taken as rotating at ---- 2000 RPM, then the time for the piston to wroplete its downward, charging, stroke is 0.01500 secs Assume that the percentage of (unsupercharged) mass in the cylinder taken for powering the supercharging eductor is 60 % So the mass of gas available to drive the eductor NOT considering the supercharged mass ncrease = 0.000362 Kg If this mass eductor driving flow flows for the time for charging on the 2000 rpm case, 1 1 the rate of flow of eductor driving gas ava table under these conditions is = 0.024117 Kg/sec. 1,

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From the above calculation for a 1mm hole diameter, this would require about a 2 mm eductor hole.

In this embodiment, the mass and partial pressure of the oxygen in the cylinder remains roughly constand for the combustion of the fuel. By changing the mechanical design -e.g. by having a larger intake port, a greater mass of fresh air charge could be drawn in at the expense of a higher inlet port upstream of valve pressure. * *

ONE EMBODIMENT -EDUCIOR BASIC CALCULATION Reference FIGURE 2 I1 In Figure 2, r) is an inlet port, and s) is an eductor, whilst the other letters with subscripts show the positions of the various parameters used in Equation 2 calculation.

-1- 1 1 exit of the mixing is eductor and assumed to mixes be _I enters the port at the going into the cylinder); In Figure 2, the compressed gas high velocity primary fluid with the secondary, low velocity fluid (the fresh air charge complete at tj, Point 2.

Through this mixing, a major portion of the momentum of the primary fluid is imparted to the secondary fluid, resulting in a static pressure at Point 2 greater than at Point 1. This increase in static pressure is given by equating the pressure and momentum forces along the port. This increase in static pressure (P2 minus P1) may be found from Equation 2.

Equation 2: _ (P2-P1)GcA = Wp(Vp-Vm) + Ws(Vs-Vm) 1 1 I where: P1 equals the pressure at Point 1, pounds force per square foot; P2 equals pressure at Point 2, pounds force per square foot; Gc equals dimensional constant 32.17 pound feet per pound force second squared; A equals cross-sectional area of the port, square feet; Wp equals weight rate of flow, pounds per second and Vp equals velocity feet per second of the primary fluid; Ws equals weight rate of flow, pounds per second, and Vs equals velocity feet per second of the secondary fluid; Vm equals velocity of the combined fluids, feet per second. * * i 1

rArcTtithe above Critical Flow calculations, velocity of primary fluid from eductor is 617 M/sec = 2025 ft/sec leantider a design where: Pt the pressure in the inlet port of a cylinder is taken as 13.7 psia _*** is 32.17 Lbs.ft/lb force sec 2 te7climentional constant, ** * * The diameter of the port is 0.9 inch -so cross section area 'A' of the port 0.004418 square ft (Equivalent open area -excluding eductor area) Wp, the weigh rate flow of the primary (compressed gas) fluid is taken as a percentage of the secondary (main fresh inlet charge) fluid, prevously taken as 60 % Calculation of Ws, the weigh rate flow of the secondary (main fresh inlet charge to the cylinder:

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Volume of 1 cylinder of say a 4 cylinder 122 cubic inch (2 Litre) engine is 1 30.5 cu inches = 0.01765 cu feet Density (assumed of) air at say 32 Deg F 0.075207 Lb per cu ft at 13.7 psia From above calulations: the mass of air required to charge the cylinder is 0.001329 Lbs ---F- i If the engine is taken as rotating at 2000 RPM, then the time for the piston to complete its downward, charging, stroke is 0.01500 secs So Ws = 0.088609 Lbs/ sec So at 60 % of Ws, Wp = 0.053166 Lb/sec Re-arranging Equation 2 to obtain the increase in pressure effected by the eductor: (P2-P1) = ( Wp(Vp-Vm) + Ws(Vs-Vm)) / (Gc*A) * * * *

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From above calculation, Vp = 617 M/sec = 2025 ft/sec Calculation of Vs: In a equiv. 0.9 inch diameter port, the velocity of the fresh charge = 266 ft per sec.

Assume that the use of the eductor as in this invention, causes the pressure to increase from 13.7 psia to 20 psia So Vm = 292 ft per sec.

Using the above case igures: P2-P1 = 632.4 Lb per sq feet. = 4.39 psi difference Now, take account of the temperature change: 1 n The above calculation assumes that the temperaure of the primary f ow (compressed gas) is at the same temperature of the secondary fluid, but this is not the case in this embodiment.

The secondary fluid temperature has been taken as being 32 Deg F i The primary (compressed gas) fluid temperature has been taken as being 600 C = Deg F 1112 ** I [ :Bysimple mixing heat balance (heat given up by hot fluid = heat aquired by cold fluid -to reach same rteb;perature), the temperature of the combined flow can be calculated by:

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Zoos of secondary fluid -taken as 100 % so that the mass of the hot compressed gas in this em6odiement is 60 % -giving a combined mass of 160 % of the secondary fluid ** * If the temperature of the combined mass is "x" Deg F, then x is given by (assuming the same molar specific heats 481 both streams) I * * v. x= combined temperaure = 437 Deg F ** Now, assuming that the volume is largely fixed (using a 'swollen' pipe e.g. as per Fig 3 -where any residual gases are at the combined temperature) the backflow outlet area is small relative to the surface area, (or e.g. a flap non-return valve can be used), this increase in temperature will increase the pressure to: 39 psia.

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Say that the average pressure in the cylinder during the charging (intake) stroke = 11.25 psia Without this invention, the pressure difference between the port at the above pressure of

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13.7 psia, and the average pressure in the cylinder is 2.45 psi difference With this invention, the pressure difference is supercharged i.e. it is increase by 25.30 psi to 27.75 psi Thus, with turbulent flow, the flow will increase by a ratio of 3.365 i.e. by 236.5 96

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The calculation shows that the calculated flow increase is significantly (about 4 times) larger than the assumed increase, thus the use of the compressed gas in an eductor as per this invention will supercharge the engine.

NOTES: The pressure I * * garentering allowed for increase effected by the eductor is significantly dependant upon the temperature of the compressed the eductor. Small changes in assumed conditions (such as the port to cylinder pressure drop) can be by varying that temperature.

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Thistemperature effect is perhaps not surprising in that the hotter the compressed gas, the less energy is taken Mt bf the system, and at the lower density, the velocity out of the eductor is higher, thus the mometum( mv^2) gliigher.

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** calculation has not allowed for heat toss from the compressed gas pipe, etc. Use of e.g. Microtherm will help reduce this loss. But the margin of calculated increased flow over the assumed flow is large.

the above insulation ^-*,.-L_I

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It is accepted that the increase in inlet temperature of the combined flow will increase the compression energy requirement, and as such the temperature of the compressed gas used may be optimised.

I 1 I Higher temperature after compression will give faster ignition of fuel -good for fuel injection engines. ** * * .*