Report

Generator Application and Sizing By: Arnie De Guzman Generator Duty Cycle/Application • Power rating - The power of the generating set is the power output available for consumer loads at the generating set terminals excluding the electrical power absorbed by the essential independent auxiliaries. Limited-Time running Power (LTP) • Limited-time running power is defined as the maximum power available, under the agreed operating conditions, for which the generating set is capable of delivering for up to 500 h of operation per year. Emergency Standby Power (ESP) • Typical usage of 50 hours per year with a maximum of 200 hours per year with varying loads. Average variable load factor is 70% of ESP rating. Emergency standby power is the maximum power available during a variable load sequence in the event of a utility power outage or under test conditions for up to 200 hours a year while maintaining per the manufacturer’s specified intervals. The permissible average power output over a 24 hour period shall not exceed 70% of the ESP rating. • No overload is available. • Not for maintained utility paralleling applications. Standby Power (CAT) • Typical usage 200 hours per year, with a maximum of 500 hours per year with varying loads. Average variable load factor is 70% of Standby rating. • No overload is available. • Not for maintained utility paralleling applications. Prime Power (PRP) • Unlimited hours of usage. Average variable load factor is 70% of the Prime power rating. 10% overload No overload is available limited to 1 in 12 hours but not to exceed 25 hours per year. The 10% overload is available in accordance with ISO 3046-1 (2002). Life to overhaul of the engine is dependent on operation as outlined in ISO8528 (2005) and time spent operating above the rating guidelines will reduce the hours to engine overhaul. Continuous Power (COP) • Unlimited hours of usage. Non-varying load factor is 70% to 100% of the published Continuous Power. Typical peak demand is 100% of continuous rating for 100% of operating hours. Load Factor • Load factor of a generator set is used as one criterion for rating a genset. It is calculated by finding the product of various loads: % = % = =% time x % load Example • For example, assume a facility has a genset rated at 1056 kW and runs it two ours a week. During those two hours, it runs at 950 kW for 1.5 hours. Find the load factor. % = 950 1056 = 0.90 % = 90 120 = 0.75 =0.90 x 0.75 = 67.5% Mode of Operation • Island Mode • Grid Mode • Back Synchronization • t = time • P = power • a = Emergency Standby Power or Prime Power • b = Permissible Average Power output in 24hrperiod(Ppp) • c = Actual Average Power in (Ppa)24hr period • d = stop Sample Scenario • We are operating 3 x 9 MW, 6 kV, 50 Hz gas turbines as part of Utilities power house of a large fertilizer complex which has a total load of 22 MW (mostly large induction motors) The power house is not connected with the public utility, i.e., we are operating our power house in an island mode. At the power house, we also have a load shedding system. In case any gas turbine trips, this load shedding system sheds the fertilizer plant load to bring the total load within the capacity of remaining gas turbine(s). The load shedding system actuates instantly on detection of either turbine trip signal or generator circuit breaker opening. All the gas turbines are being operated in droop mode (droop= 4 %). Sample Scenario • We are operating 3 x 9 MW, 6 kV, 50 Hz gas turbines as part of Utilities power house of a large fertilizer complex which has a total load of 22 MW (mostly large induction motors) The power house is not connected with the public utility, i.e., we are operating our power house in an island mode. At the power house, we also have a load shedding system. In case any gas turbine trips, this load shedding system sheds the fertilizer plant load to bring the total load within the capacity of remaining gas turbine(s). The load shedding system actuates instantly on detection of either turbine trip signal or generator circuit breaker opening. All the gas turbines are being operated in droop mode (droop= 4 %). Sample Scenario • On March 14, 2011, only two gas turbines were running normal (in droop mode) with a plant load of 15 MW. One of the gas turbines tripped on failure of turbine enclosure pressure (which occurred during ventilation fan changeover activity). Tripping of one gas turbine was followed by instant load shedding. However, the 2nd gas turbine also tripped (on generator over-frequency) resulting in total power failure. The plant management afterwards decided to always run all three gas turbines whenever the fertilizer complex total load exceeded 15 MW. The generator under frequency/over-frequency trip settings were: UnderFreq=49,0 Hz / 2.0 s, Over-Freq=51.0 Hz / 0.5 s. power failure. Sample Scenario • On May 10, 2011, we were operating all three gas turbines (in droop mode) with a plant load of 21 MW. Tripping of one gas turbine (on actuation of a gas detector installed in the turbine enclosure) was followed by load shedding and the remaining two gas turbines remained stable, i.e., saved power failure. Sample Scenario • In July 2011, the under-frequency/over-frequency trip settings of the generator were revised to have a wide band in frequency. The revised settings are: UnderFreq=47.5 Hz / 2.0 s, Over-Freq = 52.5 Hz / 3.0 s. Now referring back to the incident of March 14, 2011, the investigation team has recommended to run at least one gas turbine in Isochronous mode to prevent total power failures. You are requested to comment on whether switching one gas turbine to Isochronous mode (while the other machines shall remain in droop mode) shall help in preventing blackouts in case any gas turbine trips. Sample Scenario • In continuation of the above: Suppose a situation when plant total load is 20 MW with following distribution: GT1 (Isochronous mode)=4 MW, GT2 (Droop mode)= 8 MW, GT3 (Droop mode)=8 MW and without any gas turbine tripping, a big block of load like 8 MW is shed (for example, due to a feeder tripping). What would be the response of Isochronous machine in this case? Will it go into reverse power in an attempt to maintain bus frequency? Please keep in mind that there is no communication link between the three turbines for load sharing and the power house is in island mode (i.e., not connected with public utility). De-rate factor and Ambient Condition • Site conditions and power requirements will also play a key role in generator selection. Careful consideration must be given to the environment that the package will be operating in. Voltage • Voltage plays a key role in generator rating and must be considered. In some cases, generator voltage will not match the preferred operating voltage. A voltage regulator can provide voltage adjustment capability, however, when “dialing down” generator voltage, the current will increase for a given rating. This will increase generator heat and may require generator derating. An alternative to generator derating is to use a larger generator to maintain the standard rating. The standard set by NEMA allows a generator to be adjusted up or down by five percent (― 5%) as installed. Environmental Condition • Altitude and temperature most heavily influence engine ratings. • The higher the altitude, the lower the air density. Clean dense air is needed for efficient combustion. • Likewise, an increase in temperature lowers air density. Sample altitude/temperature derate curve Source: http://www.cummins.co.kr/new/images/product/ndata/C1250D6/3.%20 fr6250.pdf Specifications Required • • • • • • • Voltage Frequency Fuel Type Installation type Application Duty Cycle Phase • • • • • • Island Grid Sycnhronised Peak Shaving Peak Lopping Motor Starting The Load Considerations • Type of load in terms of application. • PF to be maintained. • Voltage Dip Consideration Type of Loads • Single-phase loads and load imbalance: Single phase loads should be distributed as evenly as possible between the three phases of a three-phase generator set in order to fully utilize generator set capacity and limit voltage imbalance. • Peak loads: Peak loads are caused by loads that cycle on and off—such as welding equipment, medical imaging equipment, or motors. Taking cyclic loads into account can significantly increase the size of the recommended generator set despite painstaking efforts to place loads in a step starting sequence. Type of Loads • Motor loads: Calculating specific motor loads is best handled by sizing software which will convert types of motors into load starting and running requirements. • None-Linear Loads: Due to harmonics caused by electronic rectifiers, larger alternators are required to prevent overheating and to limit system voltage distortion by lowering alternator reactance. Type of Load • Lighting loads: In addition to lamp wattages, ballast wattages and starting and running power factors should be considered. • Regenerative loads: For loads such as elevators, cranes and hoists, the power source is often relied upon for absorbing power during braking. That is usually not a problem when the utility is supplying power because it can be considered as an infinite power source with many loads. A generator set, in comparison, is able to absorb far less power, especially with no other loads connected. Generally, the regeneration problem can be solved by making sure there are other connected loads which can absorb the regenerative power. Excessive regenerative load can cause a generator set to over-speed and shut down. Voltage Dip Calculation • Here are some reasons why voltage dip calculation is important; 1. Flickering Light - The human eye is sensitive to slight lighting fluctuations. Even a decrease of 1/2 volt on a 110 volt incandescent bulb is noticeable. A one volt dip, if repeated, becomes objectionable. 2. Medical Imaging Equipment – to maintain high quality imaging, the voltage dip is limited to 10%. 3. Equipment Undervoltage protection may trip 4. Motors to be started may not start as torque is proportional to voltage, induced motors typically designed to start at terminal voltage >80% Why do the calculation? • This calculation is more or less done to verify that the largest motor does not cause system wide problems upon starting. Therefore it should be done after preliminary system design is complete. Step 1. G1 C1 Sg1 = 4000kVA Vg1 = 11,000V X’d = 0.33pu cos ᶿ = 0.85 Size = 500mm2 Length = 50m R = 0.0522 Ω/km X = 0.0826 Ω/km 1 1 kV B u s Size = 35mm2 Length = 150m R = 0.668 Ω/km X = 0.115 Ω/km C2 M1 S1 SS1 = 950kVA Vg1 = 11,000V cos ᶿ = 0.84 PM1 = 250kW Vg1 = 11,000V ILRC = 106.7 A ILRC/ IFLC = 6.5pu cos ᶿm = 0.85 cos ᶿs = 0.30 C3 Size = 120mm2 Length = 60m R = 0.196 Ω/km X = 0.096Ω/km T x1 0 .4 kV B u s STx1 = 1600kV Vt1 = 11,000V Vt2 = 400V Uk = 0.06pu Pkt = 12700W tap= 0.0% SS1 = 600kVA Vg1 = 400V cos ᶿ = 0.80 S2 G1 C1 Sg1 = 4000kVA Vg1 = 11,000V X’d = 0.33pu cos ᶿ = 0.85 Step 2. Calculate all Impedances Synchronous Generator Impedance G1 G1 C1 Sg1 = 4000kVA Vg1 = 11,000V X’d = 0.33pu cos ᶿ = 0.85 Size = 500mm2 Length = 50m R = 0.0522 Ω/km X = 0.0826 Ω/km Cable Impedances C1,C2,C3 1 1 kV B u s C3 T x1 STx1 = 1600kV Vt1 = 11,000V Vt2 = 400V Uk = 0.06pu Pkt = 12700W tap= 0.0% 0 .4 kV B u s Transformer Impedances TX1 Base Load Impedances S1,S2 1 1 kV B u s SS1 = 950kVA Vg1 = 11,000V cos ᶿ = 0.84 S1 0 .4 kV B u s SS1 = 600kVA Vg1 = 400V cos ᶿ = 0.80 S2 Motor Impedances M1 C2 M1 Computed Impedances Step 3. Referring the LV Impedances to HV side 415V Standing Load Referred to 11kV Impedance R = 161.33333 Ω and X = 121.00 Ω. Step 4. Z C1 Equivalent Circuit Z C3 1 1 kV Bus Z TX1 Z C2 Z g1 0 .4 kV Bus Z S1 Z S2 Z M1 G1 PCC Z g1 G1 Z Eq Step 5. Calculate Initial Source Voltage Z C1 Z C3 1 1 kV Bus Z g1 Z S1 G1 M o to r not c o n n e cte d Z TX1 0 .4 kV Bus Z S2 Step 6. Calculate Voltage Dip during stating