1. Lecture 21Electro Mechanical System1  A major disturbance on a system(called contingency) creates a state of emergency and immediate steps must be taken to prevent it from spreading to other regions.  The sudden loss of an important load or a permanent short-circuit on a transmission line constitutes a major contingency.  If a big load is suddenly lost, all the turbines begin to speed up and the frequency increases everywhere on the system.  If a generator is disconnected, the speed of remaining generators decreases because they suddenly have to carry the entire load.  Frequency starts to decrease sometimes at the rate of 5 Hz /s.  One or more loads must be dropped.  Load shedding is done by frequency-sensitive relays that open selected circuit breakers as the frequency falls.  A 60 Hz system may be set to shed 15% load at 59.3 Hz, another 15% at 58.9 Hz, and a final 30% at 58 Hz.  Load shedding must be done in less than one second to save the loads judged to be of prime importance. Conditions during an outage

2. Lecture 21Electro Mechanical System2  For the disconnected customers:  Such an outage creates serious problems. Elevators stop between floors, arc furnaces start to cool down, paper tears as it moves through a paper mill, traffic lights stop functioning.  It is in everyone's interest to provide uninterrupted service.  Most system short-circuits are caused by lightning, by polluted insulators, by falling trees, or by overvoltages created when circuit breakers open and close.  Such disturbances usually produce a short-circuit between two phases or between one phase and ground.  Three-phase short-circuits are very rare.  Because line short-circuits are, in general, very brief,  A major outage can usually be prevented by simply opening a short-circuited line and reclosing it very quickly.  Such fast switching of circuit breakers is done automatically because it all happens in a matter of a few cycles. Conditions during an outage

3. Lecture 21Electro Mechanical System3  Hydropower generating stations convert the energy of moving water into electrical energy by means of a hydraulic turbine coupled to a synchronous generator. Hydropower generating stations  The power that can be extracted from a waterfall depends upon its height and rate of flow.  Size and physical location of a hydropower station depends, on these two factors. Available hydro power can be calculated by: P = 9.8qh Where: P = available water power [kW] q = water rate of flow [m 3 /s] h = head of water [m] 9.8 = coefficient to take care of units  Due to friction losses in the water conduits, turbine itself, the power output of the turbine is less than calculated above.  Efficiency of large hydraulic turbines is between 90 and 94%. Available hydro power

4. Lecture 21Electro Mechanical System4 A large hydropower station has a head of 324 m and an average flow of 1370 m 3 /s. The reservoir of water behind the dams and dikes is composed of a series of lakes covering an area of 6400 km 2. Calculate a)The available hydraulic power b)The number of days this power could be sustained if the level of the impounded water were allowed to drop by 1 m (assume no precipitation or evaporation and neglect water brought in by surrounding rivers and streams) Solution a)The available hydropower is P = 9.8 qh = 9.8 x 1370 x 324 = 4 350 000 kW = 4350 MW b)A drop of 1 m in the water level corresponds to 6400 X 10 6 m 3 of water. Because the flow is 1370 m 3 /s, the time for all this water to flow through the turbines is t = 6400 x 10 6 /1370 = 4.67 X 10 6 s = 1298 h = 54 days Example