Power Plant Engineering: Questions from the Previous Year’s Exam with Solutions

Last year’s question paper with answers gives a comprehensive set of practice questions to test your knowledge and prepare for examinations in the subject of power plant engineering.

Dudes 🤔.. You want more useful details regarding this subject. Please keep in mind this as well.

Important Questions For Power Plant Engineering:
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* Btech 4th Year

Section A: Short Question In Power Plant Engineering

a. Define brake power. 

Ans. The brake power is the useful power available at the crankshaft. 

It is given by, 

b. List the components of fixed cost.  

Ans.

• 1. The fixed cost is determined by the startup cost, building cost, or capital investment in the whole plant installation.
• 2. For example, whether the facility is operational or not, the organization must pay interest on the money invested.
• 3. The initial cost, interest, taxes, insurance, and depreciation are all part of the fixed cost of electricity generating.

c. List out conventional power plants. 

Ans. Conventional power plants are as follows : 

• 1. Steam power plant, 
• 2. Gas power plant, 
• 3. Nuclear power plant, and 
• 4. Diesel power plant.

d. What is boiler efficiency ?

Ans. Boiler efficiency is defined as the ratio of heat actually utilized in generation of steam to that heat supplied by the fuel in the same period.

Where, C = Calorific value of the fuel in kJ / kg. 

e. What are the applications of diesel engine power plant? 

Ans.

• 1. Diesel engine power plants are employed as backup generators for small industries.
• 2. A diesel engine power plant can be utilised to provide power in a remote rural region where main grid power is unavailable.

f. What do you mean by turbo charging ? 

Ans. Turbo charging is a method that improves the power output of an internal combustion engine by injecting extra compressed air into the combustion chamber.

g. Name the different types of fuel cells.  

Ans. Types of fuel cells are as follows: 

• 1. Solid oxide fuel cells, 
• 2. Reversible fuel cells, 
• 3. Direct methanol fuel cells, and 
• 4. Alkaline fuel cells.  

h. Define the term Breeding. 

Ans. Breeding is the process of converting non-fissile isotopes to fissile isotopes. 

i. Explain transformer protection. 

Ans. On the basis of cooling methods, the transformers are of following two types : 

a. Dry Type Transformers: Small transformers upto 25 kVA size are of the dry type and have the following cooling arrangements : 

• i. Natural Air :  
  • 1. The natural circulation of surrounding air is used in this manner to transport away the heat generated by losses.
  • 2. A sheet metal casing keeps the winding safe from mechanical damage.
• ii. Blast Air : 
  • 1. The transformer is cooled using this method by forcing a constant blast of cool air through the core and windings.
  • 2. A fan is used to produce blast. 

b. Oil Immersed Transformers: The oil provides better insulation than air and it is a better conductor of heat than air. Mineral oil is used in these transformers. Oil immersed transformers are classified as follows : 

• i. Oil Immersed Self-Cooled Transformers : 
  • 1. The transformer is immersed in oil, and heat generated in the cores and windings is conducted to the oil.
  • 2. Oil in touch with heated parts rises, and cool oil from the bottom takes its place.
• ii. Oil Immersed Forced Air-Cooled Transformers : 
  • 1. In this technique of cooling, air is routed over the exterior surfaces of the transformer’s tank, which is immersed in oil just below the tank’s top.
  • 2. After being heated, the water is cooled in a spray pond or a cooling tower.
• iii. Oil Immersed Water-Cooled Transformers :
  • 1. A stream of water is run through a metallic coil immersed in the oil just below the tank’s top to remove heat from the oil.
  • 2. After being heated, the water is cooled in a spray pond or a cooling tower.

j. Briefly explain fossil fuel pollution.  Ans. As fossil fuels are burned, nitrogen oxides are released into the atmosphere, which contributes to the creation of smog and acid rain. The pollution created by the combustion of fossil fuels is referred to as fossil fuel pollution.

Section B : Long Questions of Power Plant Engineering

a. The value of equipment is Rs. 5,00,000 and its salvage value at the end of its useful life of 15 years is Rs. 1,00,000. Find the value of the equipment at the end of 5 years of its use by the following methods:  

i. Straight line depreciation. 

ii. Sinking fund depreciation, when it is compounded annually at 10 %.

Ans. Given: P = Rs. 500000, S = Rs. 100000, I = 10 % = 0.1  

a. Straight line method :  

1. Original value = Rs. 500000  

Salvage value = Rs. 100000

2. Total depreciation = Rs. 500000 – 100000 = Rs. 400000 

3. Depreciation per year,

4. Depreciation at the end of 5 year, 

= 26666.6 x 5 = Rs. 133333.3 

Value of the equipment at the end of 5 years, 

= 500000 – 133333.3 = Rs. 366666.7 

b. Sinking Fund Method : 

1. Annual deposit, 

= Rs. 12589.5 

2. Total amount collected at the end of 5 years

= Rs. 76860.15  

3. Value of the equipment at the end of 5 years

= 500000 – 76860. 15 

= Rs. 423139.84

b. What do you mean by ‘supercritical boilers’ and ‘super charged boiler’ ? 

Ans. A. Supercritical Boilers :

• 1. Supercritical boilers are steam generators that produce steam above the critical pressure of 221.2 bar.
• 2. It uses the Rankine cycle and a drumless boiler.
• 3. A sub-critical boiler typically has three distinct sections: a preheater (economiser), an evaporator, and a superheater.

• 4. In the feed pump, the condensate water from the condenser is compressed from the condenser pressure to supercritical pressure. Process 1-2 depicts the operation conceptually.
• 5. As indicated in procedure 2-3, this water is heated to supercritical pressure. It is possible to see that water is sub-cooled at point 2. With the input of heat, its temperature rises.
• 6. There is no differentiation between the liquid and gaseous states.
• 7. When the temperature of liquid water is raised over the critical temperature of 374.15 °C, it turns into gas.
• 8. As a result, it is a continuous tube that is heated along its length, with water entering at one end and superheated steam exiting at the other.
• 9. Due to large frictional resistance the feed pump pressure is about 40 % higher than boiler pressure.  

B. Supercharged Boiler:

• 1. The combustion in a supercharged boiler is carried out under pressure in the combustion chamber by supplying compressed air.
• 2. As the exhaust gases from the combustion chamber are vented to high pressure, they are used to power the gas turbine.
• 3. The air compressor is powered by the gas turbine and supplies compressed air to the combustion chamber.
• 4. Following are the advantages of supercharged boiler :  
  • i. Owing to very high overall heat transfer coefficient the heat transfer surface required is hardly 20 to 25 % of the heat transfer surface of a conventional boiler. 
  • ii. The part of the gas turbine output can be used to drive other auxiliaries.
  • iii. Small heat storage capacity of the boiler gives better response to control. 
  • iv. Rapid start of the boiler is possible. 
  • v. Comparatively less number of operators is required.

c. Explain how reheating improves the efficiency of a simple open cycle gas turbine plant. 

Ans. 1. This process uses two turbines, a high pressure turbine and a low pressure turbine. 

2. There is a reheater present in between these two turbines. 

3. Here high pressure turbine is used to drive the compressor and the low pressure turbine provides useful work output. 

4. Actual process is shown by 1 – 2’ – 3’ – 4’ – 5’ – 6’ (with reheating) and ideal process is shown by 1 – 2 – 3 – 4 – 5 – 6 (with reheating). 

5. Actual process without reheating is shown by 1 – 2’ – 3 – L’ and the ideal process without reheating is shown by 1 – 2 – 3 – 4 – L.

6. The work output with reheating

= C_p (T₅ – T_6’) 

7. The work output without reheating  

= C_p (T_4’ -T_L’)

8. From T-s diagram it is clear that,

T₅ – T_6’ > T_4’ – T_L’

Hence the work output increases with reheating. This will also increase the work input and it may result into improvement in thermal efficiency.

d. Explain the working of a typical fast breeder nuclear reactor power plant, with neat diagram. 

Ans. Fast Breeder Reactor :

• 1. A fast breeder reactor is a small vessel in which the necessary quantity or enriched Uranium or Plutonium is kept without a moderator. 
• 2. The vessel is surrounded by a fairly thick blanket of depleted fertile Uranium. 

• 3. The fertile material absorbs neutrons from the fissile material and becomes fissile.
• 4. Liquid metal is used to cool the reactor core. U-238 can be transformed to Pu-239 (or Th-232 to U-233), which can then be used in thermal or fast breeder reactors.
• 5. Boron, light water, oil, or graphite are used in fast breeder reactors to provide neutron shielding. Lead, concrete, concrete with additional magnetite or barium, and other materials provide gamma-ray shielding.

e. What are the properties of materials used for conductor ? Name the materials used for conductors. 

Ans. A Properties of Conductor Materials : 

• 1. They posses very low resistance or specific resistance. 
• 2. They should be good conductor of heat. 
• 3. They highly resistance to corrosion. 
• 4. They must be malleable and ductile. 
• 5. They must be flexible. 
• 6. They posses better tensile strength. 
• 7. They should not react with climatic condition. 

B. Name of Conductor Materials: Silver, copper, brass, steel, gold and aluminium, etc. 

Section 3 : Cost of Electrical Generation

a. A consumer has following connected load : 10 lamps of 60 W each, 2 heaters of 1000 W each. Max. Demand= 1500 W, on the average he uses 8 lamps for 5 hrs a day and each heater for 3 hrs a day. Find his average demand, load factor and monthly energy consumption. 

Ans. 1. Daily energy consumption 

= Number of lamps used x Watt of each lamp x Working hours per day + Number of heaters  x Watt of each heater x Working hours per day 

= 8 x 60 x 5 + 2 x 1000 × 3
= 8400

3. Monthly energy consumption = Daily energy consumption x No. of days in a month 

= 8.4 x 30

= 252 kWh 

= 0.233 or 23.3 %

b. What do you understand by cost of electrical generation ?  

Ans. A. Cost of Electricity :

• 1. The rates of electricity sold to consumers vary depending on the type of consumer, namely domestic, commercial, and industrial.
• 2. The rates are determined by the total amount of energy utilised as well as the load factor of the consumer’s demand.
• 3. For any type of consumer all tariffs for electrical energy must cover the following elements:  
  • i. Recovery of capital cost invested.
  • ii. Recovery of running cost such as operation cost, maintenance cost, maintaining the equipment cost, billing cost and many others. 
  • iii. Reasonable profit on the invested capital. 

B. Reduction in Power Generation Cost : 

• 1. By reducing the initial investment in the power plant. 
• 2. By selecting the equipment of longer life for less depreciation. 
• 3. By selecting generating units of adequate and proper capacity. 
• 4. By running the power plant at minimum possible load factor. 
• 5. By increasing the efficiency of power plant. 
• 6. By keeping proper supervision and doing proper maintenance.  
• 7. By simplifying the operation of the power plant so that fewer personnel are required. 
• 8. By installing the power plant as near the load centre as possible to reduce transmission and distribution losses.

Section 4 : Steam Power Plant

a. Draw the general layout of steam power plant and explain its major components. 

Ans. The layout of a steam power plant comprises of the following four circuits: 

• i. Coal and Ash Circuit : 
  • 1. Coal comes at the storage yard and, after being handled, is routed to the furnaces via the fuel feeding equipment.
  • 2. Ash from coal combustion accumulates at the back of the boiler and is transferred to the ash storage yard using ash handling equipment.
• ii. Air and Gas Circuit :
  • 1. Air is drawn from the atmosphere by a forced or induced draught fan and travels through the air preheater on the furnace, where it is heated by the heat of flue gases that pass up the chimney via the preheater. 
• iii. Feed Water and Steam Flow Circuit : 
  • 1. In the water and steam circuit, condensate from the condenser is heated in a closed feed water heater using extracted steam from the turbine’s lowest pressure extraction point.
  • 2. It then goes via the deaerator and several more water heaters before entering the boiler via the economiser.
• iv. Cooling Water Circuit : 
  • 1. The cooling water supply to the condenser aids in keeping the pressure in it low.2. The water can be drawn from a natural source such as a river, lake, or sea, or it can be cooled and recirculated.

b. Explain the working principle of FBC with a neat sketch. 

Ans.

• 1. Fig. 2 given below shows the arrangement of the FBC system. 
• 2. The fuel and inert material dolomite are fed on the distributor plate and from its bottom air is supplied. 
• 3. The high velocity of air keeps the solid feed material in suspending condition during burning.

• 4. The generated heat is quickly transmitted to the water flowing through the tubes embedded in the bed, and the generated steam is expelled.
• 5. The dolomite absorbs the sulphur dioxide produced during combustion, preventing it from escaping with the exhaust fumes. The molten slag is tapped from the bed’s top surface.
• 6. The primary goal of employing inert material is to control the temperature of the bed; it accounts for 90% of the bed volume.

Section 5 : Brayton Cycle Output and Efficiency

a. A four stroke diesel engine consumes 20 kg/hr, when the engine develops an output of 80 kW. Calculate the brake and indicated specific fuel consumption, if the mechanical efficiency of the engine is 80%. Also determine the brake and indicated thermal efficiency if the CV of fuel is 45000 KJ/kg.  

Ans. Given: m_f = 20 kg/hr, BP = 80 kW, η_m = 80% = 0.80, CV = 45000 kJ/kg

To Find: 

i. Brake and indicated specific fuel consumption. 

ii. Brake and indicated thermal efficiency. 

1. Mechanical efficiency, 

Indicated power, 

2. Brake specific fuel consumption,

= 0.25 kg/BP hour

3. Indicated specific fuel consumption,

= 0.2 kg/IP hour

4. Brake thermal efficiency, 

= 0.32 or 32 %

5. Indicated thermal efficiency,

= 0.4 or 40%

b. Discuss the effect of pressure ratio on Brayton cycle output and efficiency. 

Ans. 1. Thermal efficiency of Brayton cycle, 

• 2. For low pressure ratio, the net work output is small and the efficiency is also small. In the limit, as r_p tends to 1, efficiency tends to zero and net work output is zero. 
• 3. As the pressure ratio increases, the work output and the thermal efficiency increases until it become maximum.
• 4. Then it drops off with a further increase in pressure ratio Fig. 3. 5. However, there is an upper limit for r_p when the compression ends at T_max AS r_p approaches this upper limit (r_p)_max both net work output and heat added approach zero values.

Section 6 : Solar Power Plant

a. Distinguish between controlled and uncontrolled nuclear chain reaction. 

Ans.

S. No.	Controlled Chain Reaction	Uncontrolled Chain Reaction
1.	A controlled chain reaction is a series of nuclear reactions that occur in a controlled manner.	An uncontrolled chain reaction is a series of nuclear reactions that occur one after the other but not under regulated conditions.
2.	It is carried out in the presence of moderators.	It is carried out in the absence of moderators.
3.	Neutron population is allowed to reach steady state.	Neutron population is not allowed to reach steady state.
4.	Number of fission producing neutrons is maintained constant.	Number of fission producing neutrons is not maintained constant.
5.	Used in nuclear power plants to generate electricity.	Used in nuclear bombs.
6.

b. Explain different types of collectors used in a solar power plant.  

Ans. Different types of collectors used in a solar power plant are as follows:  

Flat Plate Collector : 

• 1. The flat plate collector (Fig. 4) is the most basic and essential component of any solar thermal energy system.
• 2. Both direct and diffuse radiations are received and transformed into useable heat in this collector.

• a. Consists of plant collector :
  • i. Absorber plate, 
  • ii. Transparent covers, 
  • iii. Insulation, and
  • iv. Box.

b. Concentrating Collector (or Focusing Collector) :

• 1. A concentrator is a device that collects solar energy with a high intensity of solar radiation on an absorbent surface using a reflector or refractor.
• 2. A focusing collector is a flat-plate collector with a reflecting or refracting surface between the solar radiation and the absorber.

Section 7 : Circuit Breaker

a. What is a circuit breaker ? What are the different types of circuit breakers that are employed in typical power stations ? 

Ans. A. Circuit Breaker :

• 1. A circuit breaker is an electrical switch that operates automatically to safeguard an electrical circuit from damage caused by excessive current.

B. Types of Circuit Breakers: Types of circuit breakers are as follows : 

a. Plain Break Oil Circuit Breaker : 

• 1. Fig. depicts the configuration of this breaker. There is a sturdy weather-tight earthed metal tank with a predetermined level of oil and an air cushion above the oil.
• 2. The oil pressure in this breaker tank is solely due to the oil head above the contacts enclosed in the tank.

• 3. When the contacts separate, an arc is formed. This converts oil to petrol. Arc-induced gas production may disperse heat generated by the arc.

b. Air Circuit Breaker :

• 1. In air circuit breakers the compressed air at pressure around 15 kgf/cm² is used for extinction of arc.  
• 2. Fig. shows a typical air circuit breaker. 
• 3. Arc extinction is caused by the passage of air surrounding the moving circuit. The breaker is closed by applying pressure to the bottom aperture and opened by applying pressure to the upper opening.
• 4. When the contacts separate, cool air rushes around the moveable contact, blowing out the arc.

c. Water Circuit Breaker : 

• 1. The principle of the water circuit breaker is shown in Fig.
• 2. The contacts are in water, which is turned into steam by the arc and rushes past the opening to blow out the arc. 

b. What is the function of bus bar ? Draw different types of bus bar arrangements and discuss their relative merits and demerits. 

Ans. A. Function of Bus Bar: A bus bar is an electrical connection that collects and distributes electricity from incoming sources to outgoing feeders. A bus bar’s primary function is to transport and distribute electricity.

B. Types of Bus Bar Arrangements : 

i. Single Bus Bar Arrangement : 

a. Diagram: Fig. 6.

b. Merits :

• 1. It has low initial cost. 
• 2. It requires less maintenance.  

c. Demerits :  

• 1. The complete supply is disturbed on the occurrence of the fault. 
• 2. Less flexible. 

ii. Sectionalized Single Bus Bar Arrangement : 

a. Diagram: Fig. 7. 

b. Merits : 

• 1. The faulty section is removed without affecting the continuity of the supply. 
• 2. The system has a current limiting reactor which decreases the occurrence of the fault. 

c. Demerits: 

• 1. The system uses additional circuit breaker and isolator which increases the cost of the system. 

iii. Double Bus Double Breaker Arrangement : 

a. Diagram: Fig. 8. 

b. Merits : 

• 1. Provides maximum reliability and flexibility in the supply. 
• 2. The continuity of supply remains same.

c. Demerits : 

• 1. Their maintenance cost is very high.