This set of Thermal Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Introduction to Cogeneration and Combined Cycles”.

1. Which of the following processes uses the concept of utilizing waste heat energy to process heat?

a) Regeneration

b) Cogeneration

c) Reheat

d) Superheating

View Answer

Explanation: In the cogeneration process, the waste heat energy of the power plant, such as heat rejected from the condenser and cooling tower, is utilized for process heating. Many industries such as chemical, paper, oil, textile, steel, etc. use this process heat for their industrial requirements.

2. Topping and bottoming cogeneration cycles are the types of cogeneration cycle.

a) True

b) False

View Answer

Explanation: Cogeneration or combined heat and power (CHP) cycles are classified into two types, the Topping cogeneration cycle and Bottoming cogeneration cycle. Topping cogeneration cycles are further classified as combined cycle cogeneration plant, steam cycle cogeneration plant, the internal combustion engine, and gas turbine.

3. Utilization factor (ε_{u}) in case of co-generative cycle is given by which of the following?

a) ε_{u} = \(\frac{Net \,work\, output + Process\, heat}{Total\, heat\, input}\)

b) ε_{u} = \(\frac{Net \,work\, output + Total \,heat \,input}{Process\, heat}\)

c) ε_{u} = \(\frac{Net \,work\, output + Total \,heat \,input}{Net \,work\, output}\)

d) ε_{u} = \(\frac{Net\, work \,output + Total \,heat\, input}{Total \,heat\, input}\)

View Answer

Explanation: Utilization factor (ε

_{u}) is defined as the fraction of energy that is utilized for process heating as well as power generation from the total heat input. An ideal or best cogeneration power plant will have an utilization factor equal to 1.

4. In which of the following co-generative cycle electric power is produced first and then the remaining heat is used for the manufacturing sector?

a) Topping cycle

b) Bottoming cycle

c) Reheat cycle

d) Regeneration cycle

View Answer

Explanation: In the topping co-generative cycle, the fuel is burnt and high-temperature steam is produced. The high-temperature steam is primarily used for power production with the help of a turbine. The low-temperature steam from the exit of the turbine is used for process heating in the manufacturing sector.

5. In which of the following co-generative cycles, heat energy is produced first, and then the remaining heat is used for power production?

a) Regeneration cycle

b) Reheat cycle

c) Bottoming cycle

d) Topping cycle

View Answer

Explanation: In the bottoming co-generative cycle, the heat produced by the combustion of fuel is primarily used for process heating in the manufacturing sector. Then the waste heat is utilized for electrical power production.

6. In a co-generative steam power plant, the values of enthalpy at different points in the diagram are h_{4} = 3200kJ/kg, h_{5} = 2675kJ/kg, and h_{6} = 2315kJ/kg. The mass flow rate of steam from the boiler is 15kg/sec. If the amount of steam extracted for process heating is 5kg/sec, then which of the following is the power output in KW? (Neglecting pump work)

a) 10000KW

b) 11000KW

c) 12000KW

d) 11475KW

View Answer

Explanation:

Given: \(\dot{m}\) = 15kg/sec; h

_{4}= 3200kJ/kg; h

_{5}= 2675kJ/kg; h

_{6}= 2315kJ/kg;

Power output from turbine is given by,

W

_{T}= \(\dot{m}\) (h

_{4}– h

_{5}) + (\(\dot{m}\) – 5)(h

_{5}– h

_{6})

W

_{T}= 15(3200 – 2675) + (15 – 5)(2675 – 2315)

W

_{T}= 11475KW

7. In a co-generative steam power plant the values of enthalpy at different points in the diagram are h_{5} = 2830kJ/kg and h_{7} = 720kJ/kg. The mass flow rate of steam from the boiler is 20kg/sec. If the amount of steam extracted for process heating is 8kg/sec, then which of the following is the amount of process heat energy utilized in KW?

a) 16880KW

b) 15000KW

c) 12300KW

d) 12175KW

View Answer

Explanation:

Given: \(\dot{m}\) = 20kg/sec; h

_{5}= 2830kJ/kg; h

_{7}= 720kJ/kg

Process heat energy utilized is given by;

Q

_{p}= 8(h

_{5}– h

_{7}); where 8 is the the mass flow rate of steam entering the process heater

Q

_{p}= 8(2830 – 720)

Q

_{p}= 16880KW

8. In a co-generative steam power plant the values of enthalpy at different points in the diagram are h_{2} = 188kJ/kg and h_{8} = 876kJ/kg. The mass flow rate of steam from the boiler is 12kg/sec. If the amount of steam extracted for process heating is 4kg/sec, then which of the following is the enthalpy of water coming out from the mixing chamber in kJ/kg?

a) 417.35kJ/kg

b) 188kJ/kg

c) 876kJ/kg

d) 400kJ/kg

View Answer

Explanation:

Given: \(\dot{m}\) = 12kg/sec; h

_{2}= 188kJ/kg; h

_{8}= 876kJ/kg;

h

_{3}can be calculated by using heat balance in the mixing chamber

\(\dot{m}\)h

_{3}= (\(\dot{m}\)– 4) h

_{2}+ 4h

_{8};

where 4 is the the mass flow rate of steam entering the process heater

12h

_{3}= [(12 – 4) × 188] + (4 × 876)

h

_{3}= 417.35kJ/kg

9. In a co-generative steam power plant the values of enthalpy at point 4 and 3 are h_{4} = 2990kJ/kg and h_{3} = 185kJ/kg respectively. The turbine produces 8270KW of power output with 8kg/sec of the mass flow rate of steam from the boiler. Which of the following is the thermal efficiency of the cycle in percentage? (Neglecting pump work)

a) 36.85%

b) 100%

c) 49.86%

d) 26.42%

View Answer

Explanation:

Given: \(\dot{m}\) = 8kg/sec; h

_{4}= 2990kJ/kg;

W

_{T}= 8270KW; h

_{3}= 185kJ/kg

Thermal efficiency is given by; η

_{th}= \(\frac{W_T}{Q_{in}}\); → (1)

where Q

_{in}is the heat supplied by the boiler to water, which is given by,

Q

_{in}= \(\dot{m}\)(h

_{4}– h

_{3});

Q

_{in}= 8(2990 – 185)

Q

_{in}= 22440KW

and from equation (1)

η

_{th}= 8270/22440

η

_{th}= 36.85%

10. In a co-generative steam power plant the boiler consumes 18952KW of heat to produce steam which is then used by a turbine to produce 7846KW of power output. The process heater absorbs 6453KW of heat. Which of the following is the effectiveness of the cycle in percentage?

a) 75.44%

b) 100%

c) 49.86%

d) 26.42%

View Answer

Explanation:

Given: Turbine work, W

_{T}= 7846KW;

Process heat, Q

_{P}= 6453KW;

Total heat input, Q

_{in}= 18952;

Effectiveness is given by;

ε = \(\frac{W_T + Q_P}{Q_{in}}\);

ε = \(\frac{7846 + 6453}{18952}\)

ε = 75.44%

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