# Difference between revisions of "Generator Efficiency"

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+ | Here several useful generator equations are derived. | ||

+ | =Electrical Efficiency= | ||

− | + | Often, because maximum power transfer occurrences when impedance is match, there are designs where the generator load equals the internal source resistance. Sure maximum possible power is transferred, but 1/2 of the mechanical input becomes heat inside the generator. Instead, oversize the generator, and use a fraction of the possible power to keep efficiency. | |

− | + | Electrical efficiency as a fraction input mechanical power that is available as electricity a the output terminals. Of course the voltage of the available power might not be optimal - but thats what switching power supplies are for. | |

− | + | <pre> | |

− | + | Rl = load resistance | |

− | + | Rm = motor series (internal) resistance | |

− | |||

− | |||

− | + | E = Pout / (Pout + Ploss) (efficiency) | |

− | |||

− | + | Pout = I^2 * Rl and Ploss = I^2 * Rm thus | |

− | |||

− | k = 4 -> 80% | + | E = Rl / (Rl+Rm) |

− | + | ||

+ | let Rl = k * Rm then (define the load resistance as a fraction of the motor series resistance) | ||

+ | |||

+ | E = k/(1+k) | ||

+ | |||

+ | Thus to keep the efficiency high, then keep the load resistance | ||

+ | higher than the motor source resistance, or: | ||

+ | |||

+ | Rl >= Rm * E/(1-E) | ||

+ | |||

+ | some results: | ||

+ | k = 4 -> 80% | ||

+ | k = 9 -> 90% | ||

+ | </pre> | ||

+ | |||

+ | =Efficiency resulting from under loaded generator= | ||

+ | |||

+ | For a load resistance that is a multiple of the generator resistance, a fraction of the maximum <strong>possible</strong> generator output power is obtained. If the load equals the internal generator resistance, then the maximum output power (Pmax) is obtained (but 1/2 of the input power becomes heat.) Here we calculate k (from above) based on how undersized the load is. | ||

+ | |||

+ | Let C be the generator load rating reduction factor. | ||

+ | C = Pload/Pmax | ||

+ | |||

+ | below it is shown the resulting 'k' is : | ||

+ | k = [2-C +- 4sqrt(1-C)] / C | ||

+ | |||

+ | Found a mistake in the original solution so I rework below. Lots of math, probably I'm missing the easy method.... | ||

+ | |||

+ | |||

+ | <pre> | ||

+ | |||

+ | load power : Pload = Pin - Ploss (1) | ||

+ | |||

+ | Pin = Vemf * Im (2) | ||

+ | Ploss = Im^2 * Rm (3) | ||

+ | |||

+ | substituting into (1) | ||

+ | Pload = Vemf * Im - Rm*Im^2 (4) | ||

+ | |||

+ | Im = Vemf / (Rm + Rload) (5) | ||

+ | |||

+ | let Rload = k*Rn (6) | ||

+ | |||

+ | substituting 6 into 5: | ||

+ | |||

+ | Im = Vemf / Rm (1+k) (7) | ||

+ | |||

+ | substituting 7 into 4 | ||

+ | |||

+ | Pload = Vemf * Vemf / Rm(1+k) - Rm * Vemf^2 / [Rm(1+k)]^2 | ||

+ | |||

+ | Pload = Vemf^2/Rm(1+k) - Vemf^2/Rm(1+k)^2 (8) | ||

+ | |||

+ | collecting common factors : | ||

+ | Pload = Vemf^2/Rm * [ (1+k)^-1 = (1+k)^-2] (9) | ||

+ | |||

+ | knowing Vemf = Ks * omega (10) (Ks units are volts/RPM) | ||

+ | |||

+ | replace Vemf in 9 with 10: | ||

+ | Pload = Ks^2 omega^2 / Rm * [ (1+k)^-1 = (1+k)^-2] (11) | ||

+ | |||

+ | From elsewhere, the equation for maximum possible generator power: | ||

+ | Pmax = omega^2 Ks^2 / 4*Rm (12) [note in derivation of Pmax below I have Ks units of RPM/volt] | ||

+ | |||

+ | Let C be the generator load rating reduction factor. | ||

+ | C = Pload/Pmax | ||

+ | or | ||

+ | C * Pmax = Pload (13) | ||

+ | |||

+ | substituting (12) into (13) LHS, and (11) into (13) RHS. | ||

+ | |||

+ | C*omega^2 Ks^2 / 4*Rm = Ks^2 omega^2 / Rm * [(1+k)^-1 - (1+k)^-2] (14) | ||

+ | |||

+ | we can can cancel : omega^2, Ks^2 and Rm | ||

+ | |||

+ | C/4 = [ (1+k)^-1 - (1+k)^-2] (15) | ||

+ | |||

+ | making RHS denominator in (15) the same | ||

+ | |||

+ | C/4 = [(1+k) - 1]/(1+k)^2 | ||

+ | |||

+ | simplify and multiply by 4 | ||

+ | C/4 = k /(1+k)^2 (16) | ||

+ | |||

+ | C*(1+k)^2 = 4k (17) | ||

+ | |||

+ | C*(k^2 + 2k + 1) -4k = 0; | ||

+ | C*k^2 + 2Ck + C - 4k = 0; | ||

+ | |||

+ | C*k^2 + k(2C-4) + C = 0; (18) | ||

+ | |||

+ | solve for k with quadratic equation with : | ||

+ | |||

+ | k = -b +- sqrt(b^2 - 4ac) / 2a; (19) | ||

+ | |||

+ | a = C | ||

+ | b = 2C-4 | ||

+ | c = C | ||

+ | |||

+ | |||

+ | k = [4-2C +- sqrt(4C^2 - 8C -8C + 16 - 4*C*C)] / 2C | ||

+ | |||

+ | simplify : | ||

+ | k = [2-C +- sqrt(16-16C)] | ||

+ | |||

+ | </pre> | ||

+ | |||

+ | <strong> | ||

+ | k = [2-C +- 4sqrt(1-C)] / C (20) | ||

+ | </strong> | ||

+ | |||

+ | Comments : | ||

+ | *k is the ratio Rload to Rm (6) | ||

+ | *If C > 1 (load greater than maximum possible power, imaginary solution) | ||

+ | *The negative root results in a negative resistance for Rload in (6) | ||

+ | |||

+ | =Maximum Power= | ||

+ | |||

+ | Its also possible to size a generator based on two fundamental properties of the generator. Here we derive the maximum generator power as a function of rotational velocity omega. | ||

+ | |||

+ | <pre> | ||

+ | For a generator / motor: | ||

+ | |||

+ | omega = Ks * Vemf (Vemf is back emf, and Ks is the speed constant) | ||

+ | or Vemf = omega/Ks | ||

+ | |||

+ | Vm = Vemf - I * Rm (Vm is voltage at terminals) | ||

+ | |||

+ | I = Vm / Rl | ||

+ | let Rl = Rm for max output power | ||

+ | |||

+ | then Vm = Vemf - (Vm/Rm)*Rm | ||

+ | or | ||

+ | Vm = omega/Ks - Vm | ||

+ | |||

+ | so at max output power | ||

+ | Vm = omega/2Ks | ||

+ | |||

+ | output power is V^2/R so | ||

+ | |||

+ | Pmax = Vm^2 / Rm | ||

+ | or | ||

+ | Pmax = omega^2 / 4 Rm Ks^2 | ||

+ | |||

+ | lets define the power constant as | ||

+ | |||

+ | Pk = 1/4RmKs^2 | ||

+ | |||

+ | so Pmax = Pk * omega^2 | ||

+ | </pre> | ||

+ | |||

+ | <pre> | ||

+ | For example: | ||

+ | A generator might produce 70 volts at 2500 rpm. The internal resistance is 7.7 ohms. | ||

+ | |||

+ | ks = 2500/70 = 35.7 (rpm per volt) | ||

+ | |||

+ | Pk = 25e-6 units are v^2/(ohm rpm^2) | ||

+ | at 1000 rpm maximum power is 25.5 watts | ||

+ | at 2000 rpm 101.8 watts. | ||

+ | </pre> | ||

+ | |||

+ | |||

+ | A 2nd example is [http://www.aeromicro.com/Catalog/e-flite_power_110_bl_outrunner_motor__295kv_5387650.htm this 490g outrunner motor]: | ||

+ | <pre> | ||

+ | Rs = 0.03, ks = 295 rpm per volt | ||

+ | Pk = 1 / 4*0.03*295^2 = 96e-6 | ||

+ | at 1000 rpm, max power is 95 watts | ||

+ | </pre> | ||

+ | |||

+ | 3rd example : [http://www.maxxprod.com/mpi/mpi-2601.html Himax HC6332-230 Outrunner]: | ||

+ | <pre> | ||

+ | Rs = 0.043, ks = 230 rpm per volt | ||

+ | Pk = 109e-6 | ||

+ | at 1000 rpm, max power is 109 watts | ||

+ | </pre> | ||

+ | |||

+ | Powermax EC30-48V : | ||

+ | <pre> | ||

+ | Rs = .386, ks = 348 | ||

+ | Pk = 5.3e-6 | ||

+ | at its peak speed of 25,000 rpm - this little motor could produce 3300 watts | ||

+ | (and overheat in a few millisconds!) thus this is a great high speed motor | ||

+ | to use at fraction of its possible power. | ||

+ | </pre> | ||

+ | |||

+ | |||

+ | *[http://www.bphobbies.com/view.asp?id=B0658308&pid=AXI112 AXI Gold 5330/24 Brushless Outrunner Motor Kv=197, Rs=0.057 = Pk 111e-6] | ||

+ | *[http://www.aeromicro.com/Catalog/e-flite_power_160_bl_outrunner_motor__245kv_5387653.htm E-flite Power 160 BL Outrunner Motor, 245Kv 0.03ohm, Pk=139e-6] | ||

+ | *[http://www.modelmotors.cz/index.php?page=61&product=5345&serie=18&line=GOLD AXI 5345/18 GOLD LINE , Kv = 171, Rs = 42mO, Pk=203e-6] | ||

+ | *[http://www.modelmotors.cz/index.php?page=61&product=5360&serie=20&line=GOLD AXI Gold 5360/20 Kv=120, Rs=.068, Pk = 255e-6] | ||

+ | *[http://www.micromo.com/servlet/com.itmr.waw.servlet.Anzeige?fremdaufruf=ja&kdid=40929&sprachid=1&htdigurl=/n112053/i125705.html GNM70130 Rs=26mO->5.7ohm, Kv=9.47->139 Pk=~500e-6] [http://www.micromo.com/uploadpk/e_GNM70130_ENG.pdf PDF] |

## Latest revision as of 19:19, 7 October 2018

Here several useful generator equations are derived.

# Electrical Efficiency

Often, because maximum power transfer occurrences when impedance is match, there are designs where the generator load equals the internal source resistance. Sure maximum possible power is transferred, but 1/2 of the mechanical input becomes heat inside the generator. Instead, oversize the generator, and use a fraction of the possible power to keep efficiency.

Electrical efficiency as a fraction input mechanical power that is available as electricity a the output terminals. Of course the voltage of the available power might not be optimal - but thats what switching power supplies are for.

Rl = load resistance Rm = motor series (internal) resistance E = Pout / (Pout + Ploss) (efficiency) Pout = I^2 * Rl and Ploss = I^2 * Rm thus E = Rl / (Rl+Rm) let Rl = k * Rm then (define the load resistance as a fraction of the motor series resistance) E = k/(1+k) Thus to keep the efficiency high, then keep the load resistance higher than the motor source resistance, or: Rl >= Rm * E/(1-E) some results: k = 4 -> 80% k = 9 -> 90%

# Efficiency resulting from under loaded generator

For a load resistance that is a multiple of the generator resistance, a fraction of the maximum **possible** generator output power is obtained. If the load equals the internal generator resistance, then the maximum output power (Pmax) is obtained (but 1/2 of the input power becomes heat.) Here we calculate k (from above) based on how undersized the load is.

Let C be the generator load rating reduction factor. C = Pload/Pmax

below it is shown the resulting 'k' is : k = [2-C +- 4sqrt(1-C)] / C

Found a mistake in the original solution so I rework below. Lots of math, probably I'm missing the easy method....

load power : Pload = Pin - Ploss (1) Pin = Vemf * Im (2) Ploss = Im^2 * Rm (3) substituting into (1) Pload = Vemf * Im - Rm*Im^2 (4) Im = Vemf / (Rm + Rload) (5) let Rload = k*Rn (6) substituting 6 into 5: Im = Vemf / Rm (1+k) (7) substituting 7 into 4 Pload = Vemf * Vemf / Rm(1+k) - Rm * Vemf^2 / [Rm(1+k)]^2 Pload = Vemf^2/Rm(1+k) - Vemf^2/Rm(1+k)^2 (8) collecting common factors : Pload = Vemf^2/Rm * [ (1+k)^-1 = (1+k)^-2] (9) knowing Vemf = Ks * omega (10) (Ks units are volts/RPM) replace Vemf in 9 with 10: Pload = Ks^2 omega^2 / Rm * [ (1+k)^-1 = (1+k)^-2] (11) From elsewhere, the equation for maximum possible generator power: Pmax = omega^2 Ks^2 / 4*Rm (12) [note in derivation of Pmax below I have Ks units of RPM/volt] Let C be the generator load rating reduction factor. C = Pload/Pmax or C * Pmax = Pload (13) substituting (12) into (13) LHS, and (11) into (13) RHS. C*omega^2 Ks^2 / 4*Rm = Ks^2 omega^2 / Rm * [(1+k)^-1 - (1+k)^-2] (14) we can can cancel : omega^2, Ks^2 and Rm C/4 = [ (1+k)^-1 - (1+k)^-2] (15) making RHS denominator in (15) the same C/4 = [(1+k) - 1]/(1+k)^2 simplify and multiply by 4 C/4 = k /(1+k)^2 (16) C*(1+k)^2 = 4k (17) C*(k^2 + 2k + 1) -4k = 0; C*k^2 + 2Ck + C - 4k = 0; C*k^2 + k(2C-4) + C = 0; (18) solve for k with quadratic equation with : k = -b +- sqrt(b^2 - 4ac) / 2a; (19) a = C b = 2C-4 c = C k = [4-2C +- sqrt(4C^2 - 8C -8C + 16 - 4*C*C)] / 2C simplify : k = [2-C +- sqrt(16-16C)]

**
k = [2-C +- 4sqrt(1-C)] / C (20)
**

Comments :

- k is the ratio Rload to Rm (6)
- If C > 1 (load greater than maximum possible power, imaginary solution)
- The negative root results in a negative resistance for Rload in (6)

# Maximum Power

Its also possible to size a generator based on two fundamental properties of the generator. Here we derive the maximum generator power as a function of rotational velocity omega.

For a generator / motor: omega = Ks * Vemf (Vemf is back emf, and Ks is the speed constant) or Vemf = omega/Ks Vm = Vemf - I * Rm (Vm is voltage at terminals) I = Vm / Rl let Rl = Rm for max output power then Vm = Vemf - (Vm/Rm)*Rm or Vm = omega/Ks - Vm so at max output power Vm = omega/2Ks output power is V^2/R so Pmax = Vm^2 / Rm or Pmax = omega^2 / 4 Rm Ks^2 lets define the power constant as Pk = 1/4RmKs^2 so Pmax = Pk * omega^2

For example: A generator might produce 70 volts at 2500 rpm. The internal resistance is 7.7 ohms. ks = 2500/70 = 35.7 (rpm per volt) Pk = 25e-6 units are v^2/(ohm rpm^2) at 1000 rpm maximum power is 25.5 watts at 2000 rpm 101.8 watts.

A 2nd example is this 490g outrunner motor:

Rs = 0.03, ks = 295 rpm per volt Pk = 1 / 4*0.03*295^2 = 96e-6 at 1000 rpm, max power is 95 watts

3rd example : Himax HC6332-230 Outrunner:

Rs = 0.043, ks = 230 rpm per volt Pk = 109e-6 at 1000 rpm, max power is 109 watts

Powermax EC30-48V :

Rs = .386, ks = 348 Pk = 5.3e-6 at its peak speed of 25,000 rpm - this little motor could produce 3300 watts (and overheat in a few millisconds!) thus this is a great high speed motor to use at fraction of its possible power.