Objectives of power system are1. Cost of electrical energy per KWh is to be minimum 2. Rated voltage and frequency has to be supplied to the consumers 3. Reliable power has to be available 4. Effective protection system has to be used for isolating the faulty section and keeping other sections healthy. 5. More stable generators are to be used, so that it should not lose synchronism under faulty condition. 6. Flexible power has to be available. Base load plants: A base load power plant is a power station that usually provides a continuous supply of electricity throughout the year with some minimum power generation requirement. Base load power plants will only be turned off during periodic maintenance, upgrading, overhaul or service. Base load power plant has the character of slow demand respond, a mechanism to match generation with the load it supplies. Some examples of base load plants are 1. Thermal plants 2. Hydro plants 3. Nuclear plants 4. Tidal power plants Base load stations installed of the capacity of unvarying load known as base load which is normally required for the whole day. A generating station which has a high investment cost and low operating cost is usually operated as base load station Peak load plants: A peaking power plant is a power station that only runs during high demands of electricity throughout the day during peak hours. Because of that, the price of electricity it generates is generally higher than the electricity generated by base load power plant, which operates continuously throughout the year. The power plant to be employed as peak load power plants should have the capability of quick start, synchronization and taking up of system load, quick response to load variations and low cost. Some examples are 1. Pumped storage plant 2. Diesel plants 3. Hydro plants (when water availability is less) 4. Gas fired power plants In pumped storage plant, the plant pumps back all or a portion of its water supply during low load period. The usual construction is a tail water pond and a head water pond connected through a penstock. The generating pumping plant is at lower end. The plant utilizes some of the surplus energy generated by the base load plant to pump the water from the tail water pond into the head water pnod during off peak hours. During peak load period this water pond through the water turbine of this plant to the tail water pond. Advantages of pumped storage plants: 1. Most economical as a peak load supply plant 2. Used as load frequency control 3. If it is combined with thermal plant of an area load factor is increased 4. Pollution free 5. Easily adaptable for remote operation 6. Operating time is minimum Plant load factor is defined as the ratio of average power generation to the maximum power generation. Plant load factor = Average power/ Maximum power Therefore, ideal plant load factor is one and practical load factor is less than 1. Plant capacity factor (Pcf) is defined as the ratio of average power to the plant capacity. Pcf = Average power/ Plant capacity Pcf = Energy produced by the plant/ (Energy that can be able to produce as per plant capacity during total hours) Therefore, the value of plant capacity is less than 1. Diversity factor is defined as the ratio of sum of individual maximum demand to the station maximum demand. Diversity factor = Sum of individual maximum demand/ Station maximum demand. Therefore, the value of diversity factor is greater than 1. Demand factor (Df) is defined as the ratio of station maximum demand to the sum of connected load. Df = Station maximum demand/ Sum of connected load Therefore, demand factor of a plant is less than 1. Plant load factor = Average power/ Maximum power Plant capacity factor = Average power/ Plant capacity A boiler feed water pump is a specific type of pump used to pump feed water into a steam boiler. These pumps are normally high pressure units that take suction from a condensate return system and can be of the centrifugal pump type or positive displacement type. Boiler feed pump is the highest pressure generating pump inside the thermal plant with build up the pressure up to 165 kg. the energy produced by that volume of water is Ans Power generation P = 9. The density of water is 993 kg/m³.81*10-3*ηQWH kW Where. The use of electrostatic precipitators is growing rapidly because of the new strict air code and environmental laws. 35 % to 40%. The Rankine cycle is a model that is used to predict the performance of steam turbine systems. which usually uses water as the working fluid. as well as to quickly provide extra water in case of sudden load demands. If the centre of mass of water is 50 meters above the turbine and losses are negligible. The Rankine cycle is an idealized thermodynamic cycle of a heat engine that converts heat into mechanical work. η = Plant efficiency = 1 . So that it is kept very close to the turbine. Surge thank is used for protecting the pen stock from the water hammering effect.Economiser is used to absorbing heat from the flue gas for increasing the temperature of the water so that improving the thermal efficiency. In thermal plant most of the energy is wasted in boiler ESP condenser turbines Steam is condensed into the water inside the condenser using circulating water. When the flue gases are coming out from boiler they take away a lot of heat. An electrostatic precipitators can be designed to operate at any desired efficiency for use as primary collector or a supplementary unit to cyclone collector. 07) One million cubic meters of water is stored in a reservoir feeding a water turbine.Basically. Economiser utilize this heat from flue gases and these heat is used to heat the feed water which is going to feed the boiler. The heat is supplied externally to a closed loop. So that thermal efficiency is low value i. Most of the energy is wasted in the condenser for conversion of steam into water.. Surge tank is a standpipe or storage reservoir at the downstream end of a closed aqueduct or feeder or a dam or barrage pipe to absorb sudden rises of pressure. economiser is located in between exit of boiler and entry of air pre heater. Q. Average load = Area under the daily load curve/24 The electrostatic precipitators (ESP) are extensively used in the thermal power plant or steam power plant for removal of fly ash from the electric utility boiler emissions.e. If the load factor is 0.5)/0. actual energy generated by coal = 2*0. If the heat content of coal is 2 kWh/kg. then the number of trains required daily for the plant is Ans Maximum power Pmax = 500 MW Load factor = 0.8 kWhr/kg One train coal = 2000*10³ kg One train coal can generate energy = 2000*10³*0.Q = Discharge of water = 1*106/3600 m³/s W = Density of water = 993 kg/m³ H = Mean head of water = 50 m Power generation P = 135.8) = 6 1) A thermal generating station has an installed capacity of 15 MW and supplies a daily load of 10 MW for 12 hours and 5 MW for remaining 12 hours.3 MWhr 08) A coal fired steam power station working at a plant factor of 80% has one 500 MW generating unit. Pcf = Average power/ Plant capacity Pcf = Energy produced by the plant/ (Energy that can be able to produce as per plant capacity during total hours) Pcf = (10*12+5*12)/(15*24) = 0.4 Energy generated by coal = 2 kWhr/kg Therefore. the overall plant efficiency is 40% and a train load of coal is 2000 metric tons. The plant capacity factor for the station is Ans: Plant capacity factor (Pcf) is defined as the ratio of average power to the plant capacity.8 kWhr Total number of trains required = 9600*10³/(2000*10³*0.5 02) A thermal generating station has an installed capacity of 15 MW and supplies a daily load of 10 MW for 12 hours and 5 MW for remaining 12 hours.5 Rc = 9 MW . Find the reserve capacity of the station? Ans: Reserve capacity Rc = Plant capacity(Pc) .capacity factor)/capacity factor Rc = 15 *106*(0.Maximum power(Pmax) Rc = Pmax*(load factor .8 Average power Pavg = 0.8.8-0.3 MW Energy E = P * t = 135.4 = 0.8*500 = 400 MW One day energy = 400*10³*24 kWhr = 9600*10³ kWhr Efficiency η = 0. gives . Therefore.Maximum power(Pmax) Rc = Pmax*(load factor .capacity factor)/capacity factor Rc = 15000*10³*(0. the suitable hydraulic turbine with reaction and adjustable vanes runners is Kaplan turbine 09) A Pelton wheel turbine having a rated speed of 300 rpm is connected to an alternator to produce power at 50 Hz.4)/0.25/2 *100 = 62. P = Number of poles f = Supply frequency 300 = 120*50/P .5% Pelton turbine: Impulse turbine. What is the maximum demand of the station? Ans Average energy = 720 MWh Average power = 720*106/24 = 30 MW Load factor = Average power/Maximum demand Maximum demand = 30*106/0. for harnessing low variable water heads. The plant capacity is 20 MW.03) A power station has a maximum demand of 15000 kW. What is the corresponding load factor? Ans: From given data. used for medium water heads. used for low water heads Francis turbine: Reaction turbine.25 kW Load factor = Average load/Maximum load*100 Load factor = 1.6 = 50 MW 05) A power station has a maximum demand of 15 MW. The annual load factor is 50% and plant capacity factor is 40%.capacity factor)/capacity factor Rc = 20*106 .15*106 = 5 MW 06) The maximum demand of a consumer is 2 kW and corresponding daily energy consumption is 30 units. What is the reserve capacity of the plant? Ans Reserve capacity Rc = Plant capacity(Pc) . The number of poles required in the alternator is Ans Synchronous speed Ns =120*f/P Where.6. Find the reserve capacity of the plant? Ans Reserve capacity Rc = Plant capacity(Pc) .5-0.Maximum power(Pmax) Rc = Pmax*(load factor .4 = 3750 kW 04) The daily energy produced in thermal power station is 720 MWh at a load factor of 0. used for high water heads Kaplan turbine: Reaction turbine.P = 20 . 30 units = 30 kWh Energy consumed in 1 hour = 30/24 = 1. maximum demand will be less.6 sec.05*200/2 = 5 MW 06) A 200 MVA alternator operated at no load at frequency of 50 Hz. both load factor and diversity factors are inversely proportional to maximum demand.01) The water from a catchment is 60*10 6 cu m annually and hydro station has head of 40 m. But initial cost is high.05 Hz ΔP2 = 0. Therefore operating cost of hydro electric power station is low. P ≈ 750 kW Hydro electric power station do not require fuel like thermal power stations. it will reduce the machine rating and lower cost of power generation is possible.0 = 200 MW Δf1 = 4/100*50 = 2 Hz Δf2 = 0. because to construct dam and for heavy equipment it requires high initial cost Plant load factor is defined as the ratio of average power generation to the maximum power generation. it has inertia constant of 5 MWsec/MVA and governing system has a time delay of 0. If load factor and diversity factors are high.9 m³/s By substituting of given all values. Therefore. Plant load factor = Average power/ Maximum power Diversity factor is defined as the ratio of sum of individual maximum demand to the station maximum demand.81*10-3*ηQWH kW Where. Δf = Change in frequency ΔP = Change in power input ΔP1 = 200 . η = Plant efficiency Q = Discharge of water W = Density of water ³ H = Mean head of water Q = 60*106/(365*24*3600) = 1. the power that can be theoretically generated is Ans Average electrical output power P = 9. 5) A 200 MW capacity generator has a speed regulation of 4% for a frequency drop of 0.05 Hz. If the load of 50 MW is suddenly applied then find the frequency deviation during this time? Ans: Frequency deviation due to time delay in the governing system: Due to sudden load demand . Diversity factor = Sum of individual maximum demand/ Station maximum demand. Assuming ideal generator and turbine. Find the power input supplied by the turbine for maintaining constant frequency? Ans: For maintaining constant frequency Δf1/Δf2 = ΔP1/ΔP2 Where. 002)*150*100 = 102.24 = -0.2 P1+60.76 Hz 08) In which of the following frequency control method tie line may be overloaded? Flat frequency control: Load changes of one bus bar is supplied by the another generator for maintaining constant frequency through the tie line.24 Hz Frequency deviation Δf = fn .S))½ Where. P2 =100 MW and the B coefficients are B11 = 0.001.S . Parallel frequency control: It is the most practical method in which both the generators simultaneously respond for the load changes in either of the bus bars.5 MW 03) Incremental cost of two generators I c1 = 0. B11.b and c are constants P = Power in MW F = Fuel consumption in Rs/hr Incremental cost = dF/dP = b + 2CP Rs/MWhr 02) Two generating stations delivering a powers of P1 = 150 MW.002.005*150² + 0. Find the transmission line losses? Ans: Power loss of a transmission line PL = B11*P1² + B22*P2² + 2*B12*P1*P2 Where.50*0. ΔPd = Load demand β = % change in load/ % change in frequency R = Speed regulation constant Relation between fuel consumption and power F = a + b P + C P² Where.3 P2+40 and the ratings of the generator are 150 and 250 MW. fn = New frequency fi = Initial frequency ΔPd = Load demand S = Rating of generator Td = Time delay in governing system H = Inertia constant of generator fn = 50*((5*200 .6)/(5*200))½ fn = 49.B12 and B21 are B coefficients P1 and P2 are power delivered by generating stations. B22 = 0.49.005. Ic2 = 0.ΔPd*Td)/(H. a. The control signal of each area is proportional to changes in the frequency and changes in the tie line power.fi =50 .002*100² +2*(-0.B22. B12=B21= -0. Steady state frequency drop due to sudden loss Δf = ΔPd/(β + 1/R) Where. Find the load sharing of each generator for a load of 200 MW? .fn = fi*((H. PL = 0. In this method tie line may be overloaded. Flat tie line control: In this method tie line power flow is maintained constant without overload of tie line. so that it has slow convergence. . so that calculation time for each iteration is less. Disadvantages of gauss siedel method: 1. By solving of above equations P1 = 80 MW. The choice of slack bus affects the convergence. It is proposed to perform a load flow analysis for the system using Newton-Raphson Jacobian matrix is Ans: Size of the Jacobian matrix = (2n-m-2)*(2n-m-2) Where.2 P1 + 60 = 0. 5.3 P2 + 40 0.0. 4.Ans: Given that P1 + P2 = 200 MW Optimum condition of economic load dispatch is Ic1 = Ic2 0. It required accelerating factor for convergence. It is a simple algebraical equation.9)-100)/2 = 450 05) A power system consists of 300 buses out of which 20 buses are generator buses.2 P1 . More number of iterations are required. All the other buses are load buses. P2 = 120 MW 04) The Y bus matrix of 100 bus interconnected system is 90% sparse. n = Total number of buses = 300 m = Total number of PV buses m = Voltage control buses + Reactive power support buses + generator buses except slack bus=44 Fixed shunt capacitors are supplying constant amount of reactive power. It is not applicable for the large power system networks. 3. Hence the number of transmission lines in the system must be Ans: Total number of transmission lines (t) = (n²(1-x)-n)/2 Where.3 P2 = -20. More suitable for small size networks. Initial approximate guessing value is required for convergence. n = number of buses x = sparsity t = (100²*(1-0. so that fixed shunt capacitors are considered as load buses or PQ buses. 2. 2. Size of the Jacobian matrix = (2*300-44-2)*(2*300-44-2) = 554*554 Advantages of gauss siedel method: 1. 25 buses are the ones with reactive power support buses and 15 buses are ones with fixed shunt capacitors. V= System voltage Ic = Charging current φ = Phase angle between V and Ic Ic = V/Xc = 2πfC*V Therefore. economics of power plant is greatly influenced by both load factor and diversity factor.Plant load factor is defined as the ratio of average power generation to the maximum power generation.5 pu. the charging MVAR Ans: Charging MVAR = V*Ic*sinφ Where. Diversity factor is defined as the ratio of sum of individual maximum demand to the station maximum demand.5*20/50 = 0. Charging MVAR also will increase. "n" represents the new values and "o" represents the old values Hpun = 0. ideal plant load factor is one and practical load factor is less than 1. dPL/dP = Incremental transmission loss Given that. Both load factor and diversity factor depends on maximum demand. Therefore.turbine Ans: Pelton turbine -----> Impulsive turbine Francis turbine -----> Inward radial flow turbine Kaplan turbine -----> Axial flow turbine Turbines classification based on discharge: .2 pu 08) Pelton turbine is a ----------. charging MVAR ∝ f. If the penalty factor of a plant is unity. its incremental transmission loss is Penalty factor λα = 1/(1-dPL/dP) Where. Plant load factor = Average power/ Maximum power Therefore. Find new value for 50 MVA base? Ans: Inertia constant (H) in pu = H/Sb Sb = Base MVA Hpun = Hpuo* (MVA)bo/(MVA)bn Here. 07) A 20 MVA generator has inertia constant of 0. penalty factor = 1 dPL/dP = 0 05) As the frequency of the system is increased. Diversity factor = Sum of individual maximum demand/ Station maximum demand. As frequency increases. For LG fault at generator terminals (with 1 pu voltage) the positive sequence current will be Ans .1.096 pu and X0 = 0. Base load power plants will only be turned off during periodic maintenance. X2 = 0. 06) High capacity generators and motors are of Ans: In case of delta network. there is no path for zero sequence currents because there is no ground return and hence considered zero sequence currents are circulating within the winding. a mechanism to match generation with the load it supplies. upgrading. overhaul or service. High capacity generators and motors of star connected with grounding is practically used for faster identification of faults and to isolate the faulty section. High discharge (>10000 lpm) ----> Kaplan turbine 2. A peaking power plant is a power station that only runs during high demands of electricity throughout the day during peak hours. A) Low speed (<60 and >30 rpm) -----> Pelton turbine with multi jets B) Low speed (<30 rpm) ----> Pelton turbine with single jet Turbines classification based on head: 1. the price of electricity it generates is generally higher than the electricity generated by base load power plant. High speed (>300 rpm) ----> Kaplan turbine 2. Low discharge (<1000 lpm) -----> Pelton turbine Where lpm = litres per minute Turbines classification based on specific speed: 1. Because of that.036 pu. 07) The positive. Low head (<30 meters) -----> Kaplan turbine 03) More efficient plants are used as Ans : base load plant A base load power plant is a power station that usually provides a continuous supply of electricity throughout the year with some minimum power generation requirement. Medium discharge (<10000 and >1000 lpm) ----> Francis turbine 3. High head (>300 meters) ----> Pelton wheel turbine 2. Base load power plant has the character of slow demand respond. Medium speed (<300 and >60 rpm) ----> Francis turbine 3. negative and zero sequence per unit impedance of two generators connected in parallel are X1 = 0.12 pu. Medium head (<300 and >30 meters) ----> Francis turbine 3. which operates continuously throughout the year. 096/2 + 0. IR1 = V/(X1/2+X2/2+X0/2) From given data. 10) For 800 MJ stored energy in the rotor at synchronous speed.936 pu 08) A synchronous generator is of 100 MVA have inertia constant of 20 MJ/MVA.For LG fault at generator terminals. is consistency with transient energy function method for single machine-infinite bus system or two machines . the positive sequence current IR1 = V/(X1+X2+X0) Since. i. four pole turbo generator rated 100 MVA. 11 kV? Ans: Inertia constant H = Kinetic energy / MVAbase Inertia constant H = 800/100 = 8 MJ/MVA 01) For which one of the following types of motors. is the equal area criterion for stability applicable? Three phase synchronous motor Three phase induction motor DC series motor DC shunt motor Ans Equal area criterion used to transient stability analysis of synchronous machines. what is the inertia constant H for a 50 Hz.12/2 + 0.036/2) = 7. Find the angular momentum in MJsec/electrical degrees Ans: Angular momentum M = HS/(180*f) MJsec/electrical degrees Where. IR1 = 1/(0.e.22 MJsec/electrical degrees If all the generators are connected in parallel and maintaining constant voltage at bus bar without any internal impedance for equivalent bus bar is called infinite bus bar. H = Inertia constant S = Rating of the machine Angular momentum M = 20*100/(180*50) = 0. X2 and X0 are connected in parallel. it maintains the voltage constant independent of load condition. two generators with same X1. Equal area criteria is used more efficiently for a single generator stability analysis or single motor stability analysis and is is used to find the absolute stability of the system. Initial cost of synchronous phase modifier is higher 4.It requires starting mechanism for bringing the synchronous motor up to synchronous speed Q. vibration or stock bridge dampers are used 07) The inertia constant of 100 MVA. H = H1 + H2 Given that. The effective inertia constant is 4 MJ/MVA. 03) Two identical synchronous machines having same inertia constant are connected in parallel and swinging together.6) Stock bridge dampers are used in case hill areas to over come the which problem? unequal ice loading increase string efficiency increase power factor all of the above Ans Due to unequal ice loading vibrations are produced in the conductors. H1 = H2 4 = 2*H1 H1 = H2 = 2 MJ/MVA Advantages and disadvantages of phase modifiers: 1. Smooth voltage regulation is possible by controlling excitation 3.system. Pa = Accelerating power M = Angular momentum = HS/(180*f) . to reduce these vibrations at the point of support.Single unit can be used as both capacitance and inductor by adjusting the excitation 2. It requires more maintenance because of running equipment 6. 50 Hz. It consumes both active and reactive power 5. Then each machine has inertia constant of Ans When two machines swing together in parallel connection. If the mechanical input to the machine is suddenly raised from 50 MW to 75 MW. finally producing fault in the system. bolts and nuts are loosen. 4 pole generator is 10 MJ/MVA. If these vibrations are extended up to the point of support of the conductor. Therefore. the rotor acceleration will be equal to Ans Rotor acceleration α = Pa/M Where. Determine the percentage increase in earning capacity when it is operated at improved power factor of 0.e.99 to 0. Boiler efficiency ( about 0.power supplied at given power factor. connecting turbines in 3 stages . Efficiency of generator ( about 0. 11 kV alternator is supplying full load at a point of 0.75 lagging. intermediate pressure and low pressure. Ans Power supplied of any power factor = Rating x operating of increase in earning capacity = power supplied at improved power factor .75 to 0. This makes rotor speed to a value of 30. . Efficiency of thermal power plant is given by product of 1...5) 3.9) 2.H = Inertia constant S = Rating of the machine f = Frequency M =10*100/(180*50) Rotor acceleration α = 25*180*50/1000 = 225 electrical degrees/s² 04) A 25 MVA.98 to 0. So it is reduced to 3000 rpm by compounding i. Velocity of steam is proportional to square root of heat dropped in nozzle.000 rpm which is too high for practical use. 07) Where do compounding done in thermal power plants Generator. Coal handling plant.85 to 0.35 to 0.985).high pressure.25 mm.. Internal efficiency of turbine (0.94) 4. Efficiency of thermal cycle ( about 0.96 lagging. 05) In thermal power plants the size of coal after crushing Ans : 20 . Ash handling plant. if in a single row of nozzle blade combination all the heat is dropped the steam velocity becomes high and even supersonic.e. Any material is ground into small parts its surface area is increases make it to effectively utilize. Turbine. Ans: There will be several problems if steam energy is converted in one step i. Mechanical efficiency of turbine ( about 0.995) 5. 1. 3. It reduces the oxidation of the insulation and fire hazards. It reduces the windage noise. more than 50 % of total heat of combustion is lost as heat rejected to the condenser and the loss is unavoidable as heat energy can be convert into mechanical energy without a drop in temperature and the steam in condenser at the lower temperature. to unload the coal. 0. the thermal efficiency is quite low Diesel power plant. the heat is taken from flue gases and here radiant type of heat transfer method is used. to lift the wagon of weight of metric tons.Because a steam power station. Ans . A superheater is a device which raises the temperature of the steam much above the boiling point of water. It improves efficiency in thermal power plant. It provides better cooling. 02) In which power plant. In superheater. None of these.. 03) Which cooling is preferred for large turbine generator Oxygen cooling.08) The only place where slip ring induction motors used in thermal power plants is Ans: In coal handling plants the slip ring IM motors are used for tippling purpose i. 04) A steam power generation has an overall efficiency of 20 %.e. 2.6 Kg of coal is burnt per kWh of electrical energy generated. Hydrogen cooling. the starting torque should be very high which can be maintained by slip ring IM. 10) To improve the efficiency of thermal power plant ________________ is placed between boiler and turbine. Steam power plant. Calculated the calorific value of fuel Ans . Nitrogen cooling. Nuclear power plant. Hydro power plant. Because. ans Superheater is placed between boiler and turbine. 4. Ans Hydrogen cooling is preferred because of following reasons . It reduces windage losses to about one teeth of its value in air. The time taken by the turbine to fall to 0 rpm speed from 3000 rpm is called coasting time (technically called) and it is generally 40 min. Heat produce by 0. The generator delivers 32. Heat equivalent of 1 KWh = 860 KCal. The calorific value of fuel is being 10000 KCal / kg. Diesel power station is an example of peak load station which is installed to share the peak load of the base load station during peak load hours. 05) What will be the thermal efficiency of a 240 V.28 = 2800 KCal.28 Kg of oil = 10000 × 0.28 Kg per kWh. Heat produced equivalent of 1 kWh = 860 KCal.6x Kcal. 1000 W electric kettle if it brings 2 liters of water at 15° C to boiling point 15 Minutes Ans 06) Mechanical energy is supplied to a DC generator at the rate of 4200 J / S. Determine the overall efficiency Ans Heat produce by 0.Let x cal / Kg be the calorific value of fuel.2 Amp at 120 V.6 kg of coal= 0. How much energy is cost per minute of operation 07) A diesel power station has fuel consumption of 0. . The load frequency control of 2 machines can be achieved by matching both turbine inputs. pressure of fluid is increases. Water hamming process in pen-stocks results innoise increases. overshot water level. Kaplan turbines are employed at where head is < 60 m. hydroelectricity power plants have a more predictable load factor. Due to sudden closer of valve velocity of fluid decreases. including pollutants such as sulfur dioxide. . nor nuclear leaks. maximum power and efficiency. has none of the dangers associated with uranium mining. Pelton wheel turbines are employed at high head.Hydroelectricity eliminates the flue gas emissions from fossil fuel combustion. Ans In hydro power plants. Compared to nuclear power. velocity decreases. For high speed. Compared to wind farms. Hydroelectricity also avoids the hazards of coal mining and the indirect health effects of coal emissions. Unlike uranium. where Pelton wheel is most efficient. This will decrease the efficiency of turbines and cause severe damage of turbine. the suitable hydraulic turbine with high percentage of reaction and runner adjust vanes is Francis. the turbine system is deigned such that the water-jet velocity is twice the velocity of the bucket. dust. carbon monoxide. The Pelton wheel extracts energy from the impulse of moving water as opposed to its weight like traditional. Pelton wheels are the preferred turbine for hydro power.e and substantial path remains in the form of pressure energy. In reaction turbines at entrance to the runner only a part of available energy of water is converted into k. 10) Pelton wheel turbine is a high head turbine. hydroelectricity is also a renewable energy source. low head turbine. and mercury in the coal. 07) For harnessing low variable water heads. Francis turbines is employed at where head is 60 . Mostly cavitation takes place in reaction turbines. Hydroelectric plants can be easily regulated to follow variations in power demand. If the project has a storage reservoir. When the available water source has relatively high hydraulic head at low flow rates. Hence some noise will produce this phenomenon is called water hammer. it can generate power when needed. In impulse turbines all available energy of water is converted into KE. medium head turbine. Alternate formation and breakage of bubbles are called cavitations. pressure increases. nitric oxide. The Pelton wheel is a water impulse turbine.250 m. For Pelton wheel turbine the head is > 300 m. none of above. hydroelectricity generates no nuclear waste. provides better cooling. Kaplan. natural gas. renewable. oil. reduces oxidation of insulation and also reduces fire hazards. Tidal. Nuclear elements are the source of nuclear energy The commercial sources of energy are Ans. This type has an a axial flow rotor with variable pitch blades. The atomic energy or nuclear energy is commercial due to initial for installation and availability of uranium Non commercial sources of energy consist of firewood agricultural wastes. These sources are supposed to be free. Ans Wind energy is least reliable energy source by providing signification amounts of electricity. This sources are commercial water at a high pressure or flowing with a high velocity can be used to run turbines or waterwheels coupled to generators and therefore for generation of electronic power. widely distributed. Fossil fuels and radioactive substances. Impeller. oil and coal gas. vegetable wastes. the non-commercial sources have started commanding prices in urban areas and to some extent in rural areas as well. 02) Which of the following power planets is the least reliable Wind. Solar. Wind power. This method of power generation is costlier in initial cost. The fossils fuels are coal. produces no green house gas during operation and use little land. 09) For large plants of capacity for more than 15 MW. as they do not command any price. The main source of energy is fuels. Actually. So it is most suitable for large capacity plants Nuclear energy is considered as conventional energy source along with coal. Ans Kaplan is used for run-of-river and poundage stations with heads of up to 61 m ( low head ). as an alternative to fossil fuels is plentiful. clean. what type of cooling is used Ans Hydrogen cooling reduces windage losses. Geothermal. . dried dung etc. So this source is also commercial.Pelton. so initial cost is high for above reason. A hydro-graph is a graphical representation between discharge or flow with time in a hydro power plant. the dam and the turbine in a hydro station. the heat ex-changer and the turbine in nuclear power plant. Where W is specific weight of water in Kg/m 3. Penstock is a closed conduit which connect the forebay or surge tank to the scroll case of turbine. The electrical power P is given as P = WQHð × 9. . So water hammer is developed in penstock. An hydro-graph provides the following information 1. The operating cost is low because of fuel cost. This may result in water hammer phenomenon and may need pipe of extraordinary strength to withstand it otherwise the penstock may brust. Total volume of discharge up to any time given by the area under the curve up to that time. 09) A penstock is used as a condition between the turbine and discharge drain. The fuel of hydroelectric plant is water which is available of free of cost A reduction in load on the generator causes the governor to close the turbine gates and thus create an increased pressure in the penstock. Q is rate of flow of water in m 3 / S. The power developed by a hydro power plant depends upon the discharge and head and is directly proportional to their product and also proportional the system efficiency. The maximum and minimum run off during the period. 3. 2. 4.Hydroelectric power station are usually located in high hilly areas. Ans the dam and the turbine in a hydro station. H is height of fall or head in meters and η is overall efficiency of operation. The average run off during of period. The discharge at any time during the period under consideration. where the dam can be easily built and large reservation can be obtained. Hydro-graph indicates the power available from the stream at different times of the day or year. As the large area required and construction cost is enormously high and takes a long time for erection.81 × 10-3 KW. none of above. So hydro-graph is similar to the load curve used in the study of electrical power. none of above. Hydro-graph indicates the power available from the stream at different times of the day or year.10) In hydro power stations what is an enlarged body of water just above the intake and used as a regulating reservoir called Penstock. relieve water hammer pressure in the penstock. Surge tank is provided to absorb sudden changes in water requirements and reduce water hammer and negative pressure in penstock. Ans The forebay server as a regulating reservoir storing water temporarily during light load period and providing the same for initial increase on account of increasing lad during which water in the canal is being accelerated. So the storage capacity should be determined with the help of the mass curve. . Spillways. Fore bay. The difference of water level between the level in storage and tail race is called gross head. It gives the relation between flows and lengths of time during which they are available. But hydro-graph varies from year to year and therefore determination of storage capacity with the help of hydro-graph may lead to considerable error in some cases. The amount of storage can be determined with the help of hydro-graph. So the area under a flow duration curve represents the total quantity of run-off during the period. Q. Reservoir. A forebay may be considered as an enlarged body of water just above the intake to store water temporarily to meet the hourly load. Flow duration curve is convenient from of hydro-graph for determining the available power at site. So hydro-graph is similar to the load curve used in the study of electrical power. supply water at constant pressure.The function of a surge tank is to produce surge in the pipeline. amount of the vegetation and the weather condition in the catchment area. With enough pondage. If N is actual rotational speed of the turbine in rpm. geology of the area.5 .5 % The annual depreciation of hydro power plant is so less because of free cost fuel (water) and less maintenance of plant. So P = ηQHW × 10-3 KW. Because of the reason. Hence the load factor is also increase. 09) If power P available from a hydroelectric plant is given by the formula P = 9. But in the case of medium and high head power plants. where Q is the flow rate through the turbine in l / s and H is the head in meters. pressure inside the turbine casing remains. hour to hour. Ans: The usefulness of run off river power plants is increased by pondage. Pondage refers to storage at the plant which makes if possible to cope.The annual depreciation of hydro power plant is about 0. Run-off can be defined as that part of the precipitation which is available as the stream flow. Surge tank is usually located as close to the power station as possible. the surge tanks are usually provided at the junction of the pressure tunnel and the penstock. P is output in metric hp and H is the effective head in meters then the specific speed ( NS ) is given by. firm capacity of plant is increased.1. the surge tank is provided at the inlet of turbine. As it result the height of surge tank will become excessive. kinetic energy of the running water is fully utilized. with fluctuation of load through out a week or some longer period depending on the size of pondage. The factor affecting run-off are rainfall patter. Q. preferably on ground to reduce the light of the tower. then P will be unit of Ans :Q is the flow rate through the turbine in l / s or 1000 m3 / S. Availability of storage may render an otherwise uneconomical hydro power plant economical as increase in firm capacity. 01) A hydro power plant of run off river type should be provided with pondage so that the operating head is controlled. The specific speed of water turbine is the speed at which the turbine develops unit horse power at unit head. . size and shape of catchment area.81QH.Pump storage schemes are used to improve Ans : The purpose of providing storage is to make more water available during deficient flow times and thus increase the firm capacity of the power plant and also increase the energy generated. the firm capacity of the power plant is increased. Similarly for a head of 30 m. Francis and Pelton turbines are in order of 300 to 1000. the operating head is increased by an amount equal to height of the runner outlet above the tail race. Deflector are provided is Pelton wheels to control the speed.g. So the Kaplan. Because of highest specific speed. Ans Pelton wheel is a type of impulse turbine and is suited high head and low flow plants The draft tube is provided to increase the acting head on the water wheel. then the turbine operates under a head equal to the height of the head race water level above the runners exist. An airtight pipe attached to the runner outlet and conducting water down from the wheel and discharging it under the surface of the water in the tail race is called draft tube. The effective head under which the turbine is to operate gives the first guide to the selection of the type of turbine. The specific speed of Kaplan. It is in order of 3 to 6. The reaction turbine have a runner that always functions within a completely water filled casing. Francis and propeller are the reaction turbine. For an e. The runner blade in Kaplan turbine is less than other water turbine. The selection of water turbine is depends upon the nature of load. 60 to 300 and 10 to 50 in metric units respectively. Even for the same working head two different type turbine may be employed. A runner of high specific speed will generate more power for the same head resulting in small size of the and power house.For head and low discharge the water turbine used Kaplan wheel. Without draft tube and the water discharges freely from the turbine exist. propeller turbine. which type of turbines is used hydro power plant Ans . By providing draft tube. for a head of 200 m either of Pelton or Francis turbine can be used. The runner blade of Francis turbine is order 16 to 24 03) For variable heads of near about but less than 30 m. . Pelton wheel. the Kaplan turbine is suitable for low head and it is less than always 30 m. either of the Kaplan or Francis turbine can be employed. Francis turbine. In case of MHD power generation electric is produced by utilizing a high temperature conducting plasma moving through an intense magnetic field. And hydropower Hydro-plants. It is a direct heat to electricity conversion technique based on friday laws of electromagnetic induction. reduce water hammer. . Hydroelectric plant has minimum operating cost because of free availability of water A thermonic converter trasform heat energy directly into electrical energy by uitilizing thermonic emission. the pressure in the working fluid cycle is developed by the help to feed water pump. all of above. The efficiency of the plant mainly depends upon pressure and temperature of the steam entering the turbine and pressure in the condensor. Thus it converts the wasteful kinetic energy at the exit of the runner into the useful pressure energy. Ans: Nuclear plans need very high capital because of huge investment on building a nuclear reactor. Steam plants. Ans Draft tube is provided to increase the acting head on the water wheel. Nuclear plants. Which of the following plants has the maximum eapital cost and minimum operating cost Diesel plants. Q. Direct conversion of heat energy into electrical energy is possible through Thermal converters and MHD generator.Which of the following power plants needs highly skilled/qualified engineers for its operation Thermal.Q. The thermal efficiency increase with the increase in temperature and pressure of the steam entering the turbine.The function of draft tube is reaction turbine is to increase flow rate of water. Thus high temperature and high pressure steam are used in thermal power plant. convert kinetic energy of water. In thermal power plant. . The ash handling unit removes ash from the furnace ash hoppes and transfer of this ash to fill or storage. Production of steam will falls due to problem in boiler tube. Condenser. Which of its following units needs major modifications.Nuclear. 07) A thermal power plant is being supplied with coal having much more ash content than that for which it was designed. Ash handling unit. The thermal plant is depends on steam. Cooling tower. its need to major modification of ash handling unit. Hydro. Wind power plant is depends on the flowing of wind and directions of wind. Gas-turbine. Pulversing unit Ans Ash content is handle by ash handling unit. Ans In nuclear power plant. the fuels mainly used are natural uranium. the boilers convert water into steam and form one of major equipments in a thermal power plant. For this boiler needs maximum maintenance. In thermal power plant. Thus it is unpredietable or uncontrollable. enriched which are also a radioactibe material. Thus tidal and solar of unpredictable or uncontrollable time. Tidal power plant is depends on tide in sea and solar power is depends on the solar radiation in form of electromagnetic radiation by sun. If there are much more ash content. That is why highly skilled engineers are required for nuclear power plant opration. Hydroelectric power plants takes a long time erection and installation about pf 10-15 years owing to involvement of huge civil engineering works. These radioactive materials are dangeours for environment so that take precution and safety. Ship are usually supplied power by diesel engine because . telephone exchanges etc.The unvarying load. Wind power generation. Ans In thermal power plant.(i) low operating cost. Petrol engine. (ii) Diesel engine maintain their efficiency even ar fraction load Q. In a thermonic converter two electrodes are plased in a container containing an ionised gas or cesium vapour to reduce space change. Diesel engine. The power plants to employed as base load power plants should have. Thermal. The cathode is heated by concetrating . Gas turbine power plants are usually employed as peak loads because these plants can be started and loaded quickly and their initial costs are small but costs are high. (ii) capability of working continously for the long period. Tidal power generation. Diesel power plants are used in emergency only owing to thier uneconomical operating cost. which occurs almost the whole day on the power plant is called the bse load whereas the various peak demands of the load over and above the base load of the power plant is called the peak load. (ii) Diesel engine does not require a large amount of cooling. It may be used as stanby plants where continuity of power supply staion. (iii) requirement of few operating personal and their repair should be economical and speedy. heat energy transforms directly into electrical energy by utilizing thermonic emission.(i) A diesel power engine can be started and stopped quickly as when required. Thermonic convertor. Ans In thermonic converter. 02) Which of following power plants normally operage at high speeds Hydro electric.The power station that does not requier any moving part is Solar power generation. the steam turbine is operated as high speed as high speed abot of 1500 to 3000 rpm. The hydro electric power plant will take least time of about 5-10 minutes in starting from cold conditions to full load operation. . limited useful supplying overloads. Disel power plant cannot have single unit of 100 MW capacity because of escalating fuel cost.the rays on it. Hence this plants used to supply the peak load of the system. high maintenance and lubrication cost. The cathode and anode are connected externally through the load circuit. The heat energy is coverted directly into electriacl energy with out any rotating part. The electrons return back to cathode through the ecternal circuit and the current flows through the external circuit. On heating of cathode the electrons are emitted from it and travel to anode .