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Induction Dynamo Analysis: In this first Design Analysis section, we will consider a disk induction dynamo with a pure copper rotor 18" in diameter and 0.187" thick, which is mounted to a 1"-dia. dual-bearing drive shaft made of a beryllium/copper alloy. The disk will be secured to the shaft using press-fit CDA18135 (99%Cu;Cd/Cr) split flanges that are silver-soldered to the disk and then set-screwed both to each other and to the shaft. To allow for stator thickness, brush slip rings, bearings, and drive coupling, the rotor shaft will be 8.5" in length.
Two field piece arrays, each composed of 173 NdFeB disk magnets that are 1" in diameter and ½" thick, will be epoxy-resin-bonded into solid stator assemblies and mounted plane-parallel to the rotor with a realistic mechanical clearance (in the flux gap) of 0.0085" on each side. Silver-graphite brushes (93%Ag) will then be mounted and connected as described above, with the matching rotor-edge and shaft contact surfaces silver-plated.
 
Since multiple brushes will be absolutely necessary, the outer brushes will be parallel-connected in sets of four (4) per inner shaft brush, each separated by 90° of rotation – and more than one such set may in turn be uniformly distributed around the rotor by equal 'sectoring'. An even multi-pole number of such sets should be used, so that the corresponding inner brushes may be installed uniformly on the rotor shaft. The inner (shaft) brushes are assumed to be positive [see the 2nd following graphic].
Stator dimensions & flux density:
[i] flux gap outside radius OR = 8.2857" + ½" = 8.7857"
[ii] flux gap inside radius IR = 2.2142" – ½" = 1.7142"
[iii] flux gap radial width = OR – IR = 7.0715" = Ra = 0.180 m
[iv] mean induction radius = (OR + IR)/ 2 = 5.25" = r = 0.133 m
[v] net flux gap area = π(OR)2 – π(IR)2 = 233.26 sq.in.
[vi] total magnet area = 173 (0.7854) = 135.87 sq.in.
[vii] gap area B-factor = (135.87 / 233.26) = 0.5825 = 58.25%
[viii] disk magnet residual induction = Br = 12,900 gauss
[ix] computed gap flux density* = 7378 gauss (see graph at right)
[x] net flux density = (0.5825)(7378 gauss) = B = 0.430 T
* flux density graph courtesy of Australian Magnetic Solutions
Rotor mass & moment of inertia:
[i] disk density (pure Cu) = 0.323 lb./cu.in.
[ii] disk area A = π(Ro)2 – π(½")2 = 253.68 sq.in.
[iii] disk volume = (253.68)(0.187") = 47.44 cu.in.
[iv] wt. = (47.44)(0.323) = 15.323 lb., and mass md = 6.965 kg
[v] equivalent inertial radius = (A /π)1/2 = 8.986 in. = 0.228 m
[vi] moment of inertia = ½(6.965)(0.228)2 = 0.181 kg-m2
Shaft mass & moment of inertia:
[i] density (Be/Cu alloy) = 0.302 lb./cu.in.
[ii] volume = 8.5 [π(½")2] = 6.68 cu.in.
[iii] wt. = (6.68)(0.302) = 2.017 lb., and mass ms = 0.917 kg
[iv] moment of inertia = ½(0.917)(0.0127 m)2 = 7.40 x 10–5 kg-m2
Now that we have developed the necessary physical data for an 18" dynamo model, we need only specify a few more operating parameters before beginning a concise series of definitive performance calculations. The magnets' residual induction (or Br) has already been selected, and in this case is 12,900 gauss (1.29T) for standard grade-42 NdFeB disk magnets that have reasonable availability and justifiable price. The flux gap distance is readily figured from previous data, and is 0.204" or 5.2 mm. These criteria and the specified magnet dimensions were used to generate the flux density calculator graph provided above, from which the resultant gap flux density (quoted above) was obtained.
Next, a rotation speed f must be selected which is not only within the brushes' maximum rating but is also hopefully at or very near an industry-standard electric motor speed. Additional data (like brush spring pressure) will be furnished as needed from OEM / vendor recommendations and specifications.
Voltage: For the brush material grade selected, the maximum suggested contact speed is 5,000 fpm. While higher speeds are possible, brush wear will become excessive in a continuous-duty application. Therefore, our tentative design operating speed will be 850 rpm, with f = 14.167 rps, using an 8-pole AC drive motor. This yields a comfortably-high brush speed of 4,006 fpm. For comparison's sake, we'll also calculate a peak output voltage based on a rotor speed of 1150 rpm (or 19.167 rps), using a 6-pole motor for a maximum brush speed of 5,419 fpm.
Therefore, at 850 rpm, Vo = (6.283)(0.133)(0.180)(14.167)(0.430) = 0.9163 volts. And, at 1150 rpm, peak Vo would equal 1.240 volts. top of page
Resistance: From the previous discussion of resistance, it should be realized that we can quickly derive a theoretical baseline rotor circuit resistance, and a corresponding maximum possible output current, by ignoring the trivial disk, brush, and shaft resistances and considering at first just the interface resistance of a set of 4 negative brushes load-connected to one positive brush. The combined parallel resistances of a couple even-numbered multiples of such brush sets can then be easily figured, to provide a means of increasing the output current. There will then be 1/4th as many positive brushes as negative brushes, and for convenience we will consider the number of dynamo/generator poles to be equal to the number of positive brushes. preceding subsection
For optimum performance, current must be drawn from the rotor disk in as radially uniform a manner possible. Therefore, the 2-pole (2-set) brush arrangement shown in the preceding graphic is much better than the basic 1-pole (single-set) connection, and a 4-pole connection having 4 separate 1-pole systems in parallel is better yet. Based only on the ~0.003-ohm per-brush interface resistance, these connection options result in the following minimum rotor circuit resistances and corresponding maximum currents:
– in the 1-pole circuit, Rmin = 1 [(0.003 / 4) + 0.003] = 0.003750 ohm , and I max = 244.35 amps;
– in the 2-pole circuit, Rmin = ½ [(0.003 / 4) + 0.003] = 0.001875 ohm , and I max = 488.69 amps;
– in the 4-pole circuit, Rmin = ¼ [(0.003 / 4) + 0.003] = 0.000938 ohm , and I max = 977.39 amps.
Obviously, it is decidedly to our advantage to use the largest practical number of "pickup" brushes, although extensive mathematical modeling suggests that the total outer brush contact width should not be more than 50% of the disk's circumference in a disk induction dynamo intended for use as a motor. Accordingly, the brush current density rating will serve to limit the number of outer brushes used and the maximum possible current. The current density limit of the 93%Ag brushes we've specified is 300 A/in2. In practice, total brush ampacity should be strictly matched to the highest I max figure derived that does not exceed the calculated safe ampacity of the rotor disk (as discussed further below) and total pickup brush width should be absolutely maximized, approaching 100% of the disk's entire circumference, in all generator (non-motoring) variants.
All things considered, we will assume that the 4-pole/16-point pickup brush circuit just described will be used in our 'prototype' model and in the calculations to follow.
To get a better figure for our model dynamo's actual total circuit resistance Rt, we will now calculate the resistances of the brushes, rotor shaft, and disk.
Given the OEM specs for brush current density (300 A/sq.in.) and resistivity (2.0 x 10–6 ohm-cm), we may simply divide the value for I max by 16 to find the ampacity of each negative (pickup) brush and then divide the result by the current density limit to find the contact area required. Therefore, the area of each outer brush An = (977.39 / 16) / 300 = 0.2036 sq.in. or 1.314 cm2. Outer brush thickness should be > ½ and < 2/3 of the disk thickness, as the disk's outer edges should be slightly chamfered. So, the width of each pickup brush will be 0.2036 / 0.125" = 1.63". The rotor's circumference is equal to 2πRo = 56.55", the total pickup brush width is 16 (1.63) = 26.08", and so the edge-width 'coverage' ratio is 26.08 / 56.55 or 46% (which is nearly optimum for a motor variant).
Using brushholders which are only 5/8" 'tall', a good minimum brush length L is 1.0" or 2.54 cm. So, each pickup brush's resistance will be (2.0 x 10–6)(2.54) / 1.314 = 3.866 x 10–6 ohm, or Rn = ~ 4 μohm. Applying the same method to the 4 positive shaft brushes, with an assigned thickness of 0.50" the area Ap = 4An = 0.8144 sq.in. (5.254 cm2) and width again equals 1.63" or 4.14 cm. With length once again of 1.0", each inner brush resistance Rp = ~ 1 μohm. Finally, it will be important in practice to use the heaviest and shortest brush shunts (buss bar leads) feasible.
The rotor shaft alloy that is greatly to be preferred is CDA17200 1.9% beryllium copper, with volume resistivity of 7.733 x 10–6 ohm-cm but the highest tensile strength of any copper-base alloy. To figure a liberal resistance for the rotor shaft, we will allocate 5" or 12.7 cm as its "electrical" length by mounting the positive brushes inboard of the bearings and drive coupling. The axial end area of the shaft is equal to π(½")2 = 0.7854 sq.in. = 5.067 cm2, and the shaft resistance Rs = ~ 19 μohm.
The rotor disk's greatest electrical resistance is expressed through the edge of the 1"-diameter center shaft hole, the circumference of which is 3.1416". Given a thickness of 0.187", this inner edge area then equals 0.5875 sq.in. or 3.790 cm2. The volume resistivity of pure copper is 1.724 x 10–6 ohm-cm, and the disk's radial conduction length L is equal to Ro – ½" = 8.5" = 0.216 m = 21.6 cm. Accordingly, we find the resulting maximum possible rotor disk resistance Rd = ~ 10 μohm.
Finally, we can now make the best possible projection of the rotor circuit Rt in milli-ohms as follows:
Rt = ¼ [(Rz + Rn) / 4 + (Rz + Rp + Rs + Rd)]
= ¼ [(3 + 0.004) / 4 + (3 + 0.001 + 0.019 + 0.010)]
= 3.781 / 4 = 0.945 m ohm = Rt = 945 μohm.
Thus, it's conclusively demonstrated just how little effect the electrical 'hardware' really has on the total rotor circuit resistance of a Faraday disk dynamo (if properly designed), since the figure we just derived with fair effort differs from the quick estimate we made earlier by only 7 micro-ohms! Our revised figure for the "nominal" output current is then equal to 0.9163 / 0.000945 = nom. Io = 969.6 A.
preceding subsection top of page
Output Power: The output power of our model dynamo at the given nominal rotation speed of 850 rpm (14.167 rps) is equal to I2R = Po = 888 watts. But, at the 'peak' rotation speed of 1150 rpm (19.167 rps), the output power would be equal to V2/R = (1.24)2 / 0.000945 = Ppeak = 1,627 watts. It can therefore be seen that in raising the operating speed by 1150 / 850 or 35.3%, the output current possible would rise to peak Io = 1,312 amps and the available power would nearly double! Of course, every brush's contact area would also have to be increased by 35.3%, and in the case of the outer brushes by increasing the brush width to 2.02". Total pickup brush contact width then increases to 35.29" or a reasonable 62.4% of disk circumference.
Unfortunately, it can be shown by rather involved numerical analysis of existing copper wire data that the safe ampacity of the 0.187"-thick rotor disk is 'only' 1,129 amps around its inner circumference. By increasing the shaft size to 1¼" the safe ampacity could be raised to 1,376 amps, thereby enabling the dynamo to be operated at its peak Io. Rather than effect such a substantial device redesign, though, we will continue our analysis on the assumption that the brush size and operating speed shall be adjusted such that max. Io = 1,129 amps.
Accordingly, the pickup brush width will have to be increased by 1,129 / 969.6 or 16.44% (to 1.90"). The total pickup brush width is 16(1.90) = 30.37", and so the edge-width coverage ratio is 30.37/ 56.55 or 54%. The new value for max. Vo = 1.067 volts, and the corresponding maximum rotor speed is then 990 rpm or f = 16.497 rps.
And so, the maximum allowable output power is equal to (1.067)(1,129) or max. Po = 1,205 watts, and the new maximum brush speed is 4,665 fpm.
preceding subsection top of page
Input Torque & Power: Although it may seem that we now have good final figures for operating speed and output power, we have yet to determine if the rotation speed we just derived is acceptably close to that of an off-the-shelf electric motor (as a practical source of input torque) which will provide adequate start and run torque at a given available horsepower rating. This will not be nearly as difficult as it might sound, with the aid of a simple yet indispensable electric motor formula that relates speed (rpm), torque (T), and power (Hp): T = (Hp x 5252) / (rpm) , where the constant 5252 is equal to 33,000 ft.lbs./min./Hp divided by 2π radians/rev., and T is the torque in ft.lbs. [The equivalent metric expression is: T = P / ω , where ω = 2πf , P is power in watts, and T is torque in N-m.] preceding subsection top of page
For the 'classical' case:
[i] primary full-load counterforce = Fa = B I Ra = (.43)(1,129)(.180) = 87.39 N
[ii] primary back-torque = Ta = Fa (r) = (87.39)(.133) = 11.62 N-m = 8.57 ft.lb.
[iii] neg. brush counterforce (ea.) = Fnb = pkA = (4)(0.2)[.125 x (1.1644 x 1.63)] = 0.190 lb. (run)
[iv] pos. brush counterforce (ea.) = Fpb = pkA = (4)(0.2)[(1.1644 x .500) x 1.63] = 0.759 lb. (run)
[v] total brush retarding torque = Tb = 16(Fnb)(Ro) + 4(Fpb)(½") = 28.88 in.lb. = 2.41 ft.lb.
[vi] total load torque T = Ta + Tb = 8.58 + 2.41 = 10.98 ft.lb. (treating the ~ 0.5% rolling losses as negligible).
Correctly matching a standard electric motor to the mechanical load in this unusual application can be a complex and challenging task. All things considered, there are a number of reasons for selecting a permanent magnet or shunt-wound (wound field) DC motor, since the required AC input converter (as the power source) is usually also an economical variable speed control. This feature is especially desirable in situations like the present case, where our tentative operating speed falls right between the 'standard' 1150 and 850 rpm motor speeds. Also, in certain cases it may be inadvisable (although less expensive) to use an AC capacitor-start motor in this application, since they're designed to start under fully-loaded conditions and can briefly draw over 300% of normal running amps to do so. [In an average-load starting situation like the present case, this will also put a huge and unnecessary strain on the rotor assembly, due to its large moment of inertia.]
Referring to a motor selection and ordering guide (such as any recent Grainger catalog), we find that a 2 Hp 1150 rpm PM motor (with 180vdc armature; FL amps = 9.8) is available [GE 5CD125TP002B] that develops full-load torque of 9.133 ft.lb. (109.6 in.lb.). Even though this is rather less than the 10.98 ft.lb. that would be needed to operate our model dynamo at its maximum capability, it must be remembered that (according to the torque formula above) this motor's output torque will climb as its speed is reduced – and it may be that just enough torque will be available at or very near the maximum 990 rpm operating speed we desire.
The most straightforward way of determining that point on the selected motor's output power / torque curve where its operating speed is maximized for this particular application's input torque requirement is by repeated spreadsheet calculations to assemble tabular data, from which the best 'solution' of such a complex covariable problem is obtained when the net motor torque developed (Tm) just exceeds the total load torque (Tn) required at the reduced speed.
Starting at 985 rpm and figuring output voltages in descending 5-rpm increments at first, the following optimal resolution was found at 970 rpm (f = 16.167 rps):
– net Vo = 1.0457 volts, net Io = 1,106.5 amps, and net Po = 1,157 watts ;
– net Fa = B I Ra = (.43)(1,106.5)(.180) = 85.643 N ;
– net Ta = Fa(r) = (85.643)(.133) = 11.39 N-m = 8.40 ft.lb. ; and
– total load torque = Ta + Tb = 8.40 + 2.41 = Tn = 10.81 ft.lb. [treating the ~ 0.5% rolling losses as negligible].
– net motor torque developed = Tm = (2 x 5252) / 970 = 10.83 ft.lb. > Tn (@ 10.81 ft.lb.).
– nominal motor efficiency = (Hp x 746) / (V x I) = 1,492 / 1,764 = ~ 84.6%
– nominal dynamo efficiency = Po / (Hp x 746) = 1,157 / 1,492 = ~ 77.5%
– combined system efficiency = (0.846)(0.775) = ~ 65.6%
– letting avg. brush k = ½ (0.3 + 0.2) = 0.25, avg.Tb = (0.25 / 0.2)(2.41) = 3.01 ft.lb.
– avg. no-load motor torque Tnl = Tm – avg.Tb = 10.83 – 3.01 = 7.82 ft.lb.
– no-load motor starting time min. ts = ωI / Tnl = 2πf (.181) / 7.82 = 2.35 sec.
– Pr = mπ2R2f 2 / ts
= (6.965)(9.8696)(0.052)(261.37) / 2.35 = 934.00 / 2.35 = 397.5 watts
– avg. full-load motor torque Tf = Tm – (½ Ta + avg.Tb) = 10.83 – (4.20 + 3.01) = 3.62 ft.lb.
– full-load motor starting time max. ts = ωI / Tf = 2πf (.181) / 3.62 = 5.08 sec.
– rotor mechanical power expended = Pr = mπ2R2f 2 / ts
= 934.00 / 5.08 = 183.9 watts
The start times just calculated are entirely acceptable for a PM-DC drive motor, providing for 'gentle' starting considering the disk's large moment of inertia. And it may indeed be permissible in cases with extended no-load start times of between about 2.25 and 3 seconds to use an AC capacitor-start motor as a drive if necessary and if it so happens that the factory speed of the motor selected is acceptably close to the preferred operating speed of the dynamo/generator. [Longer start times may overload the additional start windings in a capacitor-start motor, damaging or destroying the coils and/or tripping a circuit breaker, even though the effective start times will be lower by ~300%].
It's interesting to note that our model dynamo's actual induction efficiency can be found by simply refiguring its torque requirement without considering brush drag. Thus, at 953 rpm, a 1.5 Hp PM-DC motor would be adequate to power the dynamo at a voltage of 1.0273, current of 1,087.1 amps, and output power of 1,116.8 watts. With just the reasonable added proviso that the stator's outside poles must be wholly keepered at saturation, and at less than the full disk diameter, real induction efficiency is then 1,116.8 / 1.5 (746) = 99.8%! [The foregoing criteria regarding the essential use of saturated steel end-plates to keeper the outer stator poles are derived from the innovative "closed (flux) path" homopolar generator design of Trombly & Kahn (1982). A saturated material will support no further passage of flux nor any further external magnetic induction – in the form of eddy current losses, in this case.]
Purists may also notice that the motor's own armature inertia has not been considered above in the interests of clarity and brevity, having been treated as negligible since it corresponds to just 0.3% of its output torque (by OEM specs) in this particular case. The same consideration also applies to the rotor shaft's miniscule inertia in relation to that of the disk.
› And in the best-case theoretical model: The following computations are based on what we believe is a justifiable application of the 'Tesla' reduced back-torque ratio to our model dynamo in actual operation. As we developed earlier above, in this case that ratio is equal to 1 – 0.56 = 0.46.
[i] primary full-load counterforce = Fa = 46% [B I Ra] = 0.46[(.43)(1,129)(.180)] = 40.20 N
[ii] primary back-torque = Ta = Fa (r) = (40.20)(.133) = 5.35 N-m = 3.94 ft.lb.
[iii] total brush retarding torque = Tb = 16(Fnb)(Ro) + 4(Fpb)(½") = 2.41 ft.lb. (same as before)
[iv] total load torque T = Ta + Tb = 3.94 + 2.41 = 6.35 ft.lb. (treating the ~ 0.5% rolling losses as negligible).
Referring again to the motor selection and ordering guide, we find that a 1.5 Hp 1150 rpm PM motor (with 180vdc armature; FL amps = 7.2) is available [GE 5CD125TP001B] that develops full-load torque of 6.85 ft.lb. (82.2 in.lb.). In this case, the motor will have more than enough torque for the dynamo to be operated at the maximum allowable rotor speed (and disk ampacity limit) of 990 rpm (f = 16.5 rps):
– max. Vo = 1.067 volts, max. Io = 1,129 amps, and max. Po = 1,205 watts (from preceding subsection);
– net Ta = Fa(r) = 3.94 ft.lb. (from above); and
– total load torque Tn = 6.35 ft.lb. [treating the ~ 0.5% rolling losses as negligible].
– net motor torque developed = Tm = (1.5 x 5252) / 990 = 7.96 ft.lb. > Tn (@ 6.35 ft.lb.).
– nominal motor efficiency = (Hp x 746) / (V x I) = 1,119 / 1,296 = ~ 86.3%
– nominal dynamo efficiency = Po / (Hp x 746) = 1,205 / 1,119 = ~ 107.7% (or COP = 1.077)
– combined system efficiency = (0.863)(1.077) = ~ 92.9%
– avg. starting brush retarding torque avg.Tb = 3.01 ft.lb. (same as before)
– avg. no-load motor torque Tnl = Tm – avg.Tb = 7.96 – 3.01 = 4.95 ft.lb.
– no-load motor starting time min. ts = ωI / Tnl = 2πf (.181) / 4.95 = 3.79 sec.
– Pr = mπ2R2f 2 / ts
= (6.965)(9.8696)(0.052)(272.25) / 3.79 = 973.18 / 3.79 = 256.8 watts
– avg. full-load motor torque Tf = Tm – (½ Ta + avg.Tb) = 7.96 – (1.97 + 3.01) = 2.98 ft.lb.
– full-load motor starting time max. ts = ωI / Tf = 2πf (.181) / 2.98 = 6.30 sec.
– rotor mechanical power expended = Pr = mπ2R2f 2 / ts
= 973.18 / 6.30 = 154.5 watts
Conclusions:
We interpret the dynamo efficiency calculated above to be definitive and exciting proof [within the framework of this rigorous analytical model] that our 18" disk dynamo model could in fact be built to exhibit bona fide over-unity operation, if not yet necessarily in a self-sustaining manner. As mentioned early in the course of this study, it would require a separate bank of solid-state DC-DC step-up current converters to pre-amplify the output voltage of such a device before that output can be accepted by the vast majority of available AC inverters. Most inverters today have a reliable nominal efficiency of 93%, and there are now several extant types of suitable pre-amplifying converters having similar efficiency. So, the combined 'in-line' efficiency for our model would at best be only (1.077)(.93)(.93) = ~ 93.15%, and the integrated 'system loop' – including drive motor – would obviously not be self-sustaining.
[For an excellent and quite readable pdf technical paper (Starzyk et al.; 1999) on just one such DC current converter methodology, entitled "A DC-DC Charge Pump Design Based on Voltage Doublers", just click here.]
However, there are a number of things that can yet be done to significantly enhance the performance of our model 18"-diameter dynamo, not the least of which is to thicken the rotor disk slightly (for added rotor ampacity). We've also designed a more refined dynamo model (using many more brushes) whose inherent COP would approach 1.50, and which will in fact – if reality meets the best theoretical model – be demonstrably self-sustaining, given the typical converter, inverter, and drive motor efficiencies cited.* We therefore feel that an imperative course of research and development is plainly indicated.
* [We've also drawn up an "integrated system loop flow chart for a self-sustaining Faraday generator" having a feasible 20% net back-torque ratio and COP = 1.5, whereby it can be seen that the integrated system's net over-unity COP will be 1.12 – and thus it's more than inverter-self-running – but have been advised against its publication.] top of this subsection top of page
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