17/04/2020 ELK474E POWER SYSTEM CONTROL & COMMUNICATIONS Chapter 4 LOAD SHEDDING Chapter 4 Copyright @ Under no circumstances this document can copies be made for sale without the permition of following name. ELK474E, O Usta, ITU Electrical Engineering 1 Load Shedding and restoration Load shedding schemes have began quite important in present day systems, where there is a lack of adequate spinning reserve margins , and a shortage of tie-line capacity to make up for the loss of generations. It is the function of under frequency load shedding relays, to detect the onset of the decay in system frequency, and shed approptiate amounts of system load, so that the generation and load are again in balanced, ans power system can return to operating frequency. ELK474E, O Usta, ITU Electrical Engineering 2 Tie line AREA 1 AREA 2 Under normal conditions, in this two area system there is a balance between total lad and generation, and the power system frequency is within the operation limits. When there is a load increase in one of the area (say area 1), The system frequency will decay (before the contro system in action). • Firstly if there is enough spinning reserve in area 1, the generation in area 1 will be increase by the primery control system, and a balance be provided between the total load and generation in the system, and frequency will be come around 50Hz. The supplimentary control brings ELK474E,to O Usta, ITU Electrical Engineering 3 the frequency exactly 50Hz. • O. Usta, ITU Electrical Engineering 1 17/04/2020 • When there is no enough spinning reserve in area 1, the required power should be taken from area 2 to reach a balanced, if there is enough tie-line capacity. • When there is no change to import power from area 2, a load shedding process will be in action in area 1 to match the load and generation to bring the frequency to around 50Hz. ELK474E, O Usta, ITU Electrical Engineering 4 Swing (Motion )Equation of Synchonous Machines According to the Newton’s law, the torque of acceleration is given by the moment of inertia J times the angular acceleration of the rotor. Remember F=ma for direct motion and T=J(dω/dt) for a rotting mass). Where; J: moment of inertia in kg. ω(t): the angular velocity in rad/s Tm: mechanical torque (generator input) N.m Te: Electrical torque (generator output) N.m Td: damping torque, Ta: accelerating torque ELK474E, O Usta, ITU Electrical Engineering 5 Swing (Motion )Equation of Synchonous Machines Remember: The angular moment M=Jω (Joule.s/rad) For directional motion P=F V (W) For rotational motion P=T ω (W) ELK474E, O Usta, ITU Electrical Engineering O. Usta, ITU Electrical Engineering 6 2 17/04/2020 Eq(2) The Definition of Inertia Constant H H=Ek/S=Kinetix Energy/ Rated MVA H=(1/2) MωR/S⇒M=2HS/ωR Then the swing equation: ELK474E, O Usta, ITU Electrical Engineering 7 The change in the rotor speed is directionally proportional to the acceleration power (Pa=Pm-Pe) and inversely proportional to inertia constant of the machine. If the power system is modeled by a large generator, the change in the system frequency, from the motion equation, will be: ELK474E, O Usta, ITU Electrical Engineering 8 System Frequency Dynamics • From the Swing Equation Where, dīF/dt = Rate of change in frequency Fo = Base frequency, 50Hz H = Inertia constant of system Pgen = Active power generation Pload = Active power demand S = Total MVA rating of rotating plant Pa=īP =Pgen-Pload, overload. ELK474E, O Usta, ITU Electrical Engineering O. Usta, ITU Electrical Engineering 9 3 17/04/2020 Overload Influence on frequency • Overload is the difference between the active power demand and the active power generated, as a percentage of power generation, i.e. (Pgen – Pload) x100 Pgen • The greater the overload, the greater the rate of frequency decay • Small islanded systems: overloads tend to be larger. • ΔP=Pa= ELK474E, O Usta, ITU Electrical Engineering Δω =Δđš = ΔPa 10 Change in the system load in pu With control Frequency dependent load (no control) Frequency independent load (no control) ELK474E, O Usta, ITU Electrical Engineering 11 Inertia (H) Influence on frequency 50 D=2, īP=50% 49 Frequency (Hz) 48 47 46 45 44 H = 7s H = 4s 43 H = 2s 42 41 40 0 2 4 6 8 10 12 ELK474E, O Usta, ITU Electrical Engineering Time (seconds) O. Usta, ITU Electrical Engineering 14 16 18 20 12 4 17/04/2020 Load Self-regulation Influence on frequency • As system frequency decays, the system demand also decreases due to the frequency dependent load. • This effect due primarily to the dynamic behaviour of motor loads • Load Reduction Factor : ELK474E, O Usta, ITU Electrical Engineering 13 Load Self-regulation Influence on frequency 50 49 Frequency (Hz) 48 47 D=d=2 46 45 d = 3.0 44 43 d = 2.0 42 d = 1.2 41 40 0 2 4 6 8 10 12 14 16 18 ELK474E, O Usta, ITU Electrical Engineering Time (seconds) 20 14 System Response, 2 MW Generation Loss With Primary Control vs LSD load self response (no control) 50.2 Frequency 50 49.8 49.6 49.4 49.2 Seconds ELK474E, O Usta, ITU Electrical Engineering O. Usta, ITU Electrical Engineering 46 42 39 36 33 29 26 23 20 16 13 7 3 0 10 49 48.8 15 5 17/04/2020 Under-frequency Load Shedding (UFLS) Load shedding is required, when there is lack of adequate spinning reserve margins, and a shortage of tie-line power flow capacity to make up for the loss of generation. It is the function of under frequency load shedding relays, to detect the onset of the decay in frequency, and shed approptiate amount of system load, so that the generation and loads are ones again in balance and power system can return to normal operating frequency. ELK474E, O Usta, ITU Electrical Engineering 16 •UFLS is a last action, emergency action to balance the system load and generation. •In small isolated systems: -Overloads tend to be larger -System Inertia is smaller, so more drastic, faster action is required ELK474E, O Usta, ITU Electrical Engineering 17 Under-frequency Load Shedding • Design Considerations: • Spinning Reserves • Level • Distribution • Load Shedding Parameters • Maximum anticipated overload • Amount of load to be shed • Frequency thresholds • Step size and number of steps • Time delays • Priorities and distribution ELK474E, O Usta, ITU Electrical Engineering O. Usta, ITU Electrical Engineering 18 6 17/04/2020 Under-frequency Load Shedding • Generating plant is highly sensitive to frequency drop • Low pressure turbine stages • Motor driven auxiliaries • Mechanical fatigue is cumulative • Transient minimum frequency thresholds • Recovery frequency band • Governor action - Spinning reserves • System dispatcher action • Manual increase of load shedding • Start-up additional generation ELK474E, O Usta, ITU Electrical Engineering 19 Under-frequency Load Shedding 53 Maximum Recovery Frequency 52 Frequency (Hz) 51 Example of over-shedding 50 Example of under-shedding 49 48 Minimum Recovery Frequency 47 46 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Time (seconds) ELK474E, O Usta, ITU Electrical Engineering 20 Under-frequency Load Shedding • Number and size of shedding stages: • Many & Small vs. Few & Large • Many Small Stages • Avoids over-shedding • Too slow for large overloads • Tends to inhibit system recovery • Few Large Stages • Over-shedding may occur • Faster corrective action for large overloads ELK474E, O Usta, ITU Electrical Engineering O. Usta, ITU Electrical Engineering 21 7 17/04/2020 Under-frequency Load Shedding H=4s, D=2, īP=20% 50 Frequency (Hz) 49 48 47 46 0 1 2 3 4 5 6 7 8 9 ELK474E, O Usta, ITU Electrical Engineering Time (seconds) 10 22 Under-frequency Load Shedding H=4s, D=2, īP=50% 50 Frequency (Hz) 49 48 47 46 0 1 2 3 4 5 6 7 8 9 Time (seconds) ELK474E, O Usta, ITU Electrical Engineering 10 23 Example: For a system with 10 000MW connected load, a generation plant delivering 1500MW is lost due to a contingency. The system inertia is 5 second. Determine the settings of under frequency relays which will accomplish a load shedding plan to drop 1500MW of load in two steps of 750MW each. Assume that time delay for tripping is 15 cycle and safety margin is 0.2 Hz. 8500MW X X Power plant G 1500MW Initially PL=PG= 10000MW=10 GW and F=50Hz Load, 10 000MW Allowable change in frequency ΔF=±1% ELK474E, O Usta, ITU Electrical Engineering O. Usta, ITU Electrical Engineering 24 8 17/04/2020 Solution: Remember: Rate of change of frequency, â = = xFo, Hz/s for frequency independent load. â ( ) [ x đšđ, đģđ§/đ for frequency dependent load ] Immediately after switching off 1500MW of load: âđ = x100=17.65% over loading after load switching ELK474E, O Usta, ITU Electrical Engineering 25 First load shedding will be happen just after 49.5 Hz. = â ( ) [ x đšđ = . ( . . [ ] ) ] x 50 = −0.879 This means frequency decreases by 0.879 Hz in a second. With a tripping delay of td= 15x20 = 300ms=0.3s Within 1s Within 0.3 s 0.879Hz drop 0.264Hz drop, Then actual load shedding happens at (49.5-0.264)= 49.24Hz 0.879Hz in 1s then 0.76hz in 0.865 s, 0.865s ahter the load switching operation of 1500MW, the first step of load shedding happens. ELK474E, O Usta, ITU Electrical Engineering 26 Since the safety margin is 0.2Hz, the next shedding of 750MW will be at the frequency of (49.5-0.264-0.2)= 49.04 Hz. The rate of change of frequency at this point will be: âđ = Then x100=8.8% = . ( [ . . . . ] ) x 50= -0.434 Hz/s This means that frequency decrease by 0.434Hz in a second. The second step of load shedding will be within: (49.24-49.04)/0.434 = 0.46 second. Due to the 0.3s time delay, the second shedding occurs at the frequency of 49.91 Hz. ELK474E, O Usta, ITU Electrical Engineering O. Usta, ITU Electrical Engineering 27 9 17/04/2020 The two steps of load setting will be completed [(50-49.24)/0.879]+[(49.24-48.92)/0.434= 0.865+0.761 =1.626 second at the point of PL=PG, and the network frequency will begin to return to its nominal value of 50hz. ELK474E, O Usta, ITU Electrical Engineering REFERENCE: Roger Blackman Barbados Light & Power REFERENCE: O Usta Lecture Notes ELK474E, O Usta, ITU Electrical Engineering O. Usta, ITU Electrical Engineering 28 29 10