Thermodynamics Quiz: Thermal Efficiency

20 questions · exam conditions

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Thermal EfficiencyQuestion 1 of 20

An Otto cycle heat engine has a compression ratio of 8. If the temperatures at the beginning and end of the adiabatic compression process are 300 K and 600 K respectively, and the engine produces 2 kJ of work while receiving 8 kJ of heat input, what is its thermal efficiency?

56.5%

25.0%

33.3%

50.0%

75.0%

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Thermodynamics Quiz

Thermodynamics Quiz: Thermal Efficiency

Practice Thermal Efficiency in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Thermal Efficiency, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

56.5%
25.0% (correct answer)
33.3%
50.0%
75.0%

Explanation: When you encounter Otto cycle problems, remember that thermal efficiency has a straightforward definition: it's the ratio of useful work output to heat energy input. Don't get distracted by the extra information about temperatures and compression ratio. The thermal efficiency formula is:

\eta = \frac{W_{out}}{Q_{in}}

Given that the engine produces 2 kJ of work and receives 8 kJ of heat input:

\eta = \frac{2 \text{ kJ}}{8 \text{ kJ}} = 0.25 = 25.0\%

This confirms answer (B) 25.0% is correct. Now let's see why the other options are wrong: (A) 56.5% likely comes from incorrectly using the theoretical Otto cycle efficiency formula

\eta = 1 - \frac{1}{r^{\gamma-1}}

with compression ratio r=8 and γ=1.4. This gives the ideal efficiency, but the problem asks for the actual efficiency based on given work and heat values. (C) 33.3% might result from incorrectly calculating

\frac{8-2}{8} \times \frac{1}{2} = 0.375 \times \frac{2}{3}

, representing a computational error or misapplication of efficiency concepts. (D) 50.0% could come from mistakenly using the temperature ratio:

\frac{600-300}{600} = 0.5

, but temperature differences don't directly give thermal efficiency. Study tip: For thermodynamic cycles, always check whether the problem gives you actual performance data (work and heat values) or asks for theoretical analysis. When work output and heat input are explicitly given, use the basic efficiency definition rather than cycle-specific formulas.

Question 2

A Carnot heat engine operates between reservoirs at 500°C and 25°C. A real heat engine operating between the same reservoirs has an actual thermal efficiency of 45%. What is the ratio of the actual efficiency to the Carnot efficiency?

0.73 (correct answer)
0.58
1.37
0.45
0.61

Explanation: When you encounter Carnot engine problems, you're dealing with the theoretical maximum efficiency any heat engine can achieve between two thermal reservoirs. This sets the benchmark for comparing real engines. First, calculate the Carnot efficiency using absolute temperatures. Convert to Kelvin: 500°C = 773 K and 25°C = 298 K. The Carnot efficiency is

\eta_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}} = 1 - \frac{298}{773} = 0.614

or 61.4%. The ratio of actual to Carnot efficiency is simply

\frac{\eta_{actual}}{\eta_{Carnot}} = \frac{0.45}{0.614} = 0.73

, confirming answer A. Answer B (0.58) represents a common error where students accidentally calculate the ratio backwards or make arithmetic mistakes in the temperature conversion. Answer C (1.37) would imply the real engine is more efficient than the Carnot engine, which violates the second law of thermodynamics—impossible for any real engine. Answer D (0.45) is just the actual efficiency itself, showing the student missed that the question asks for a ratio, not the efficiency value. Remember that Carnot efficiency always uses absolute temperatures (Kelvin), and real engines must always be less efficient than the Carnot limit. When you see efficiency comparison problems, immediately convert temperatures to Kelvin and calculate the theoretical maximum first—this prevents impossible answers and guides your reasoning.

Question 3

A heat pump operating in heating mode has a coefficient of performance (COP) of 3.5. If this same system were operated as a heat engine between the same temperature reservoirs, what would be its thermal efficiency?

28.6% (correct answer)
35.0%
71.4%
22.2%
77.8%

Explanation: When you encounter problems involving heat pumps and heat engines operating between the same temperature reservoirs, remember that these devices are thermodynamically related through their coefficients of performance and efficiency. For a heat pump in heating mode, the coefficient of performance is defined as

COP_{HP} = \frac{Q_H}{W}

, where

Q_H

is heat delivered to the hot reservoir and

W

is work input. For the same system operating as a heat engine, thermal efficiency is

\eta = \frac{W}{Q_H}

. Notice that these relationships are mathematical reciprocals when the energy flows are considered. Since energy is conserved,

Q_H = Q_C + W

for both devices. This leads to the relationship:

\eta = \frac{1}{COP_{HP}}

. With a COP of 3.5, the thermal efficiency is

\eta = \frac{1}{3.5} = 0.286 = 28.6\%

. This confirms answer A is correct. Looking at the wrong answers: B (35.0%) incorrectly assumes the COP directly equals the efficiency percentage. C (71.4%) mistakenly calculates

1 - \frac{1}{COP}

, which has no physical meaning in this context. D (22.2%) appears to use an incorrect formula, possibly confusing relationships between different thermodynamic cycles. Study tip: Remember the reciprocal relationship between heat pump COP and heat engine efficiency for the same temperature reservoirs:

\eta_{engine} = \frac{1}{COP_{heat pump}}

. This is a direct consequence of energy conservation and appears frequently on thermodynamics exams.

Question 4

A gas turbine engine receives air at 300 K and compresses it to 900 K. After combustion, the gas temperature reaches 1400 K before expanding back to 600 K. If the engine produces 150 kJ/kg of specific work and has a specific heat input of 500 kJ/kg, what is its thermal efficiency?

30% (correct answer)
42.9%
70%
57.1%
25%

Explanation: When analyzing gas turbine engine performance, thermal efficiency is the fundamental measure of how effectively the engine converts heat input into useful work output. The thermal efficiency is defined as the ratio of net work output to heat input. The calculation is straightforward using the given values. Thermal efficiency

\eta = \frac{\text{Net Work Output}}{\text{Heat Input}} = \frac{150 \text{ kJ/kg}}{500 \text{ kJ/kg}} = 0.30 = 30\%

. This confirms that answer A) 30% is correct. The temperature values provided (300 K, 900 K, 1400 K, 600 K) represent the thermodynamic cycle states but are not needed for this efficiency calculation since you're given the actual work and heat values directly. B) 42.9% likely results from incorrectly using temperature ratios in efficiency formulas meant for ideal cycles. C) 70% represents an unrealistically high efficiency that would violate practical thermodynamic limitations for gas turbines. D) 57.1% might come from confusion between different efficiency definitions or improper manipulation of the given data. Remember that thermal efficiency problems can be solved two ways: either using the fundamental definition (work output divided by heat input) when these values are given directly, or through temperature-based relationships for ideal cycles. Always check which approach the given data supports. Gas turbine efficiencies in practice typically range from 25-45%, making 30% a realistic value that should immediately seem reasonable.

Question 5

A refrigeration cycle operating as a heat engine would have a thermal efficiency of 35%. When operating as intended (as a refrigerator), what is its coefficient of performance for cooling?

1.86 (correct answer)
2.86
0.65
1.54
0.35

Explanation: When you encounter a problem connecting heat engine efficiency to refrigerator performance, you're dealing with the fundamental relationship between these two thermodynamic cycles operating between the same temperature reservoirs. A refrigeration system can theoretically operate in reverse as a heat engine. The key insight is that both the thermal efficiency (

\eta

) of a heat engine and the coefficient of performance for cooling (

COP_c

) of a refrigerator are related through the same temperature limits. For any reversible cycle, these quantities satisfy:

COP_c = \frac{1-\eta}{\eta}

Given that the thermal efficiency is 35% (0.35), we can calculate:

COP_c = \frac{1-0.35}{0.35} = \frac{0.65}{0.35} = 1.86

Looking at the wrong answers: B) 2.86 results from incorrectly using

COP_c = \frac{1}{\eta}

, which would be the relationship for a heat pump's coefficient of performance, not cooling. C) 0.65 simply takes (1-η) without dividing by η, missing the fundamental relationship entirely. D) 1.54 appears to come from using an incorrect formula or calculation error. The correct answer is A) 1.86. Remember this key relationship: for reversible cycles operating between the same reservoirs,

COP_c = \frac{1-\eta}{\eta}

. This connects heat engine efficiency to refrigerator performance and frequently appears on thermodynamics exams. Always verify which coefficient of performance is being asked for—cooling or heating—as they have different formulas.

Question 6

A combined heat and power (CHP) system produces 40 MW of electricity and 60 MW of useful thermal energy. The system consumes natural gas providing 125 MW of chemical energy input. What is the electrical efficiency of the CHP system?

32% (correct answer)
48%
80%
67%
25%

Explanation: When analyzing combined heat and power (CHP) systems, you need to distinguish between different types of efficiency. These systems produce both electricity and useful heat, but the question asks specifically for electrical efficiency, not overall efficiency. Electrical efficiency focuses solely on how much electrical power is generated relative to the total energy input. Here, the system produces 40 MW of electricity from 125 MW of chemical energy input. The calculation is straightforward:

\text{Electrical efficiency} = \frac{\text{Electrical output}}{\text{Total energy input}} = \frac{40 \text{ MW}}{125 \text{ MW}} = 0.32 = 32\%

This confirms answer (A) 32% is correct. The wrong answers represent common calculation errors. (B) 48% likely comes from dividing electrical output by only part of the input (40/83.3 ≈ 48%). (C) 80% results from calculating overall efficiency by including both outputs: (40 + 60)/125 = 80%. While this represents the system's total useful energy efficiency, it's not what the question asks for. (D) 67% might come from incorrectly using thermal output in the calculation: 60/90 ≈ 67%. The key trap here is confusing electrical efficiency with overall system efficiency. CHP systems are attractive precisely because their overall efficiency is high (80% in this case), but their electrical efficiency alone is typically much lower. Always read carefully to determine which specific efficiency metric is being requested, as CHP problems often test your ability to distinguish between these different performance measures.

Question 7

An air-standard Brayton cycle operates with a pressure ratio of 8. The air enters the compressor at 300 K and the turbine at 1200 K. Assuming constant specific heats with γ = 1.4, what is the thermal efficiency of this cycle?

44.8% (correct answer)
75.0%
55.2%
25.0%
37.5%

Explanation: When you encounter a Brayton cycle problem, you're dealing with the idealized gas turbine cycle used in jet engines and power plants. The key insight is that thermal efficiency depends only on the pressure ratio and specific heat ratio for an air-standard cycle with constant specific heats. For the Brayton cycle, thermal efficiency is given by:

\eta = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}

, where

r_p

is the pressure ratio and

\gamma

is the specific heat ratio. With your given values:

r_p = 8

and

\gamma = 1.4

, you can calculate:

\eta = 1 - \frac{1}{8^{(1.4-1)/1.4}} = 1 - \frac{1}{8^{0.4/1.4}} = 1 - \frac{1}{8^{2/7}}

Since

8^{2/7} = (2^3)^{2/7} = 2^{6/7} \approx 1.812

\eta = 1 - \frac{1}{1.812} = 1 - 0.552 = 0.448 = 44.8\%

This confirms answer A is correct. Answer B (75.0%) likely comes from using an incorrect formula or confusing this with a Carnot efficiency calculation. Answer C (55.2%) represents the value of

1/r_p^{(\gamma-1)/\gamma}

itself, which is the term you subtract from 1, not the final efficiency. Answer D (25.0%) might result from using the wrong exponent or misapplying the pressure ratio. Remember: Brayton cycle efficiency increases with pressure ratio but is independent of the actual temperatures. Always use the pressure ratio formula—the given temperatures are often distractors unless you need to find work or heat transfer values.

Question 8

A steam turbine receives steam at high pressure and temperature and exhausts it at low pressure. The enthalpy at the inlet is 3200 kJ/kg and at the outlet is 2400 kJ/kg. If the steam flow rate is 50 kg/s and the turbine generates 35 MW of power, what is the turbine efficiency?

87.5% (correct answer)
75.0%
25.0%
112.5%
62.5%

Explanation: When analyzing steam turbine performance, you're dealing with energy conversion efficiency - how well the turbine converts available thermal energy into useful work output. The key is comparing actual work produced to the maximum theoretical work available. First, calculate the maximum theoretical work output. This equals the enthalpy change multiplied by mass flow rate:

(h_1 - h_2) \times \dot{m} = (3200 - 2400) \times 50 = 800 \times 50 = 40,000 \text{ kJ/s} = 40 \text{ MW}

This represents the ideal work if the turbine converted all available enthalpy drop into mechanical work. The actual power output is given as 35 MW. Turbine efficiency is:

\eta = \frac{\text{Actual Work Output}}{\text{Theoretical Work Available}} = \frac{35 \text{ MW}}{40 \text{ MW}} = 0.875 = 87.5\%

Therefore, A (87.5%) is correct. B (75.0%) likely results from calculation errors or using wrong values in the efficiency formula. C (25.0%) suggests major conceptual confusion, possibly confusing efficiency with loss percentage or inverting the calculation. D (112.5%) is impossible since efficiency cannot exceed 100% - this violates the first law of thermodynamics and indicates the actual output was incorrectly placed in the denominator. Remember this pattern: turbine efficiency problems always require you to find the theoretical maximum work from enthalpy drop, then compare it to actual output. Efficiencies above 100% are physically impossible and signal calculation errors.

Question 9

A heat engine operates on a cycle where it receives 800 kJ of heat from a source at 600 K and rejects heat to a sink at 300 K. If the engine produces 350 kJ of work, what is the ratio of its actual thermal efficiency to the maximum possible (Carnot) efficiency for these temperature limits?

87.5% (correct answer)
50.0%
43.8%
56.3%
75.0%

Explanation: When you encounter heat engine problems involving efficiency comparisons, you need to calculate both the actual efficiency and the theoretical maximum (Carnot) efficiency, then find their ratio. First, let's find the actual thermal efficiency. Thermal efficiency is the ratio of useful work output to heat input:

\eta_{actual} = \frac{W}{Q_H} = \frac{350 \text{ kJ}}{800 \text{ kJ}} = 0.4375 = 43.75\%

Next, calculate the maximum possible efficiency using the Carnot formula:

\eta_{Carnot} = 1 - \frac{T_C}{T_H} = 1 - \frac{300 \text{ K}}{600 \text{ K}} = 1 - 0.5 = 0.5 = 50\%

The ratio of actual to maximum efficiency is:

\frac{\eta_{actual}}{\eta_{Carnot}} = \frac{43.75\%}{50\%} = 0.875 = 87.5\%

This confirms answer A) 87.5% is correct. B) 50.0% is the Carnot efficiency itself, not the ratio. C) 43.8% is the actual thermal efficiency, which students might mistakenly think is the final answer. D) 56.3% doesn't correspond to any meaningful calculation in this problem and likely results from computational errors. Study tip: Always identify what the question is asking for in efficiency problems. Is it asking for actual efficiency, maximum efficiency, or a comparison between them? Set up your calculations accordingly, and remember that the ratio of actual to Carnot efficiency tells you how close a real engine comes to theoretical perfection.

Question 10

A regenerative Rankine cycle has a turbine that produces 100 MW and a pump that consumes 2 MW. The cycle receives 180 MW of heat input in the boiler. An additional 15 MW of heat is added in the feedwater heater using extracted steam. What is the thermal efficiency of this regenerative cycle?

54.4%
50.3% (correct answer)
55.6%
65.3%
44.4%

Explanation: When analyzing regenerative Rankine cycles, you need to carefully identify what constitutes the total heat input to calculate thermal efficiency. The key insight is that regeneration uses extracted steam that was already heated by the primary heat source, so you shouldn't double-count this energy. Thermal efficiency is defined as net work output divided by total heat input from the external source. The net work output is the turbine work minus the pump work:

W_{net} = 100 - 2 = 98 \text{ MW}

For the heat input, you only count the heat added in the boiler (180 MW) because this is the external energy input to the cycle. The 15 MW added in the feedwater heater comes from steam that was already heated by the boiler, so including both would be double-counting. Therefore:

\eta_{th} = \frac{W_{net}}{Q_{in}} = \frac{98}{180} = 0.503 = 50.3\%

This confirms answer (B) 50.3%. (A) 54.4% likely results from incorrectly calculating net work or making an arithmetic error in the efficiency calculation. (C) 55.6% probably comes from adding the feedwater heater input to the total heat input:

\frac{98}{180-15} = 0.594

, which is conceptually wrong. (D) 65.3% might result from adding both heat inputs in the denominator:

\frac{98}{180-15} = 0.594

or other calculation errors. Study tip: In regenerative cycles, always use only the external heat input (boiler) in efficiency calculations. Regeneration improves efficiency by reducing external heat needs, but the extracted steam energy was already counted in the boiler input.

Question 11

A heat engine operates between two thermal reservoirs at temperatures of 800 K and 300 K. The engine receives 1200 kJ of heat from the hot reservoir and rejects 500 kJ to the cold reservoir. What is the thermal efficiency of this heat engine?

41.7%
58.3% (correct answer)
62.5%
37.5%
66.7%

Explanation: When you encounter heat engine problems, focus on the fundamental definition of thermal efficiency: it's the ratio of useful work output to heat input from the hot reservoir. For any heat engine, thermal efficiency is calculated as

\eta = \frac{W_{out}}{Q_H}

, where

W_{out}

is the net work output and

Q_H

is heat input from the hot reservoir. Since energy is conserved, the work output equals the difference between heat absorbed and heat rejected:

W_{out} = Q_H - Q_C

. In this problem,

Q_H = 1200 \text{ kJ}

and

Q_C = 500 \text{ kJ}

, so

W_{out} = 1200 - 500 = 700 \text{ kJ}

. Therefore:

\eta = \frac{700}{1200} = 0.583 = 58.3\%

This confirms answer B is correct. A (41.7%) represents a common error where students mistakenly calculate

\frac{Q_C}{Q_H} = \frac{500}{1200}

, which has no physical meaning in efficiency calculations. C (62.5%) occurs when students incorrectly use

\frac{Q_C}{W_{out}} = \frac{500}{800}

, mixing up the relationship between rejected heat and work. D (37.5%) results from calculating

\frac{Q_H - Q_C}{Q_H + Q_C} = \frac{700}{1900}

, incorrectly adding both heat quantities in the denominator. Study tip: Always remember that thermal efficiency equals work output divided by heat input from the hot source. The maximum possible efficiency (Carnot efficiency) provides an upper bound:

\eta_{Carnot} = 1 - \frac{T_C}{T_H} = 62.5\%

for these temperatures, so any real engine efficiency must be less than this value.

Question 12

An Otto cycle engine has a compression ratio of 9 and operates with air as the working fluid. Assuming an air-standard analysis with γ = 1.4, what is the thermal efficiency of this engine?

58.5% (correct answer)
41.5%
88.9%
11.1%
75.0%

Explanation: When you encounter Otto cycle problems, you're dealing with the idealized thermodynamic cycle that models spark-ignition engines. The key insight is that thermal efficiency depends only on the compression ratio and the specific heat ratio γ. The thermal efficiency of an Otto cycle is given by:

\eta = 1 - \frac{1}{r^{\gamma-1}}

, where r is the compression ratio and γ is the specific heat ratio. With r = 9 and γ = 1.4, you calculate:

\eta = 1 - \frac{1}{9^{1.4-1}} = 1 - \frac{1}{9^{0.4}}

. Since

9^{0.4} = 2.41

, the efficiency becomes

\eta = 1 - \frac{1}{2.41} = 1 - 0.415 = 0.585

or 58.5%. Looking at the wrong answers: Choice B (41.5%) represents the fraction of energy that's rejected, not converted to work—this is a common mix-up between efficiency and waste heat fraction. Choice C (88.9%) would result from incorrectly using

\eta = 1 - \frac{1}{r}

instead of the proper formula with the γ-1 exponent. Choice D (11.1%) might come from calculation errors or misapplying the formula entirely. Remember that Otto cycle efficiency increases with compression ratio but is limited by the properties of the working fluid. Always use the complete formula

\eta = 1 - \frac{1}{r^{\gamma-1}}

, and double-check that your final answer makes physical sense—real engines typically have efficiencies between 25-60%.

Question 13

A heat engine undergoes a thermodynamic cycle with the following energy transfers per cycle: receives 1000 kJ from a hot reservoir, produces 380 kJ of useful work, and transfers energy to auxiliary systems requiring 50 kJ. What is the thermal efficiency of this heat engine?

38.0% (correct answer)
43.0%
33.0%
57.0%
95.0%

Explanation: When analyzing heat engine problems, you need to understand that thermal efficiency measures how well an engine converts heat input into useful work output. The key is identifying what counts as "useful work" versus total energy transfers. For this heat engine, you have three energy transfers: 1000 kJ received from the hot reservoir (heat input), 380 kJ of useful work produced, and 50 kJ transferred to auxiliary systems. The thermal efficiency formula is:

\eta = \frac{W_{useful}}{Q_{in}} \times 100\%

Where

W_{useful}

is the work output we actually want (380 kJ) and

Q_{in}

is the heat input (1000 kJ). This gives us:

\eta = \frac{380 \text{ kJ}}{1000 \text{ kJ}} \times 100\% = 38.0\%

Choice A (38.0%) is correct because it properly uses only the useful work output in the efficiency calculation. Choice B (43.0%) incorrectly adds the auxiliary system energy to the useful work (430 kJ total) before calculating efficiency. Remember that auxiliary systems represent additional energy requirements, not useful output. Choice C (33.0%) appears to subtract the auxiliary energy from useful work (330 kJ), which misunderstands the role of auxiliary systems in the energy balance. Choice D (57.0%) likely represents the percentage of energy NOT converted to useful work, confusing efficiency with energy waste. Always remember: thermal efficiency only considers the ratio of desired work output to heat input. Auxiliary energy transfers are separate considerations in the overall energy balance, not part of the useful work calculation.

Question 14

A diesel engine has an indicated thermal efficiency of 45% and a mechanical efficiency of 85%. If the engine consumes fuel at a rate of 12 kg/hr with a heating value of 42 MJ/kg, what is the brake thermal efficiency?

38.3% (correct answer)
45.0%
52.9%
85.0%
30.0%

Explanation: When analyzing engine efficiency problems, you need to understand the relationship between different efficiency measures. Indicated thermal efficiency represents the theoretical efficiency of the combustion process, while brake thermal efficiency accounts for mechanical losses and represents the actual useful power output. The key relationship is: Brake Thermal Efficiency = Indicated Thermal Efficiency × Mechanical Efficiency. This makes intuitive sense—the brake efficiency must be lower than the indicated efficiency because mechanical losses reduce the actual power delivered to the output shaft. Given the indicated thermal efficiency of 45% and mechanical efficiency of 85%, the brake thermal efficiency equals

0.45 \times 0.85 = 0.383 = 38.3\%

. The fuel consumption rate and heating value, while important for calculating absolute power output, aren't needed for this efficiency relationship. Looking at the wrong answers: B (45.0%) represents the indicated thermal efficiency, not the brake efficiency—this ignores mechanical losses entirely. D (85.0%) is simply the mechanical efficiency, which doesn't account for combustion inefficiencies. C (52.9%) appears to result from incorrectly dividing indicated efficiency by mechanical efficiency (45%/85%), which would actually increase efficiency rather than account for losses. Remember that in engine problems, brake efficiency is always less than indicated efficiency because it includes all real-world losses. When you see multiple efficiency values given, look for multiplicative relationships—efficiencies typically compound as losses accumulate through the system.

Question 15

A steam power plant produces 50 MW of electrical power while consuming fuel at a rate that provides 125 MW of thermal input. If the plant also rejects waste heat to the environment, what is the thermal efficiency of the power plant?

40% (correct answer)
60%
25%
75%
150%

Explanation: When analyzing steam power plant performance, you're dealing with energy conversion efficiency—how well the plant converts thermal energy input into useful electrical output. Think of this as an energy balance problem where you need to account for all energy flows. To find thermal efficiency, use the fundamental definition:

\eta = \frac{\text{Useful Energy Output}}{\text{Energy Input}} \times 100\%

Here, the useful output is 50 MW of electrical power, and the energy input is 125 MW of thermal energy from fuel combustion. Therefore:

\eta = \frac{50 \text{ MW}}{125 \text{ MW}} \times 100\% = 40\%

This confirms answer A is correct. Looking at the wrong answers: B (60%) incorrectly calculates efficiency as

\frac{125-50}{125} = 60\%

, which actually gives you the percentage of energy wasted as heat, not the efficiency. C (25%) might result from incorrectly using 50 MW as the denominator:

\frac{50-25}{50}

or similar calculation errors. D (75%) could come from adding the output and input incorrectly or confusing the relationship between variables. The key insight is that thermal efficiency in power plants is always the ratio of electrical output to thermal input—never the other way around. Real steam plants typically achieve 35-45% efficiency due to thermodynamic limitations, so 40% is realistic. Remember: efficiency equals useful output divided by total input, and always expect power plant efficiencies to be well below 100% due to inevitable heat rejection requirements.

Question 16

A gas turbine power plant has a compressor that requires 200 MW of input power and a turbine that produces 550 MW of output power. The combustor adds 800 MW of thermal energy to the working fluid. What is the thermal efficiency of this power plant?

43.8% (correct answer)
68.8%
27.5%
31.3%
25.0%

Explanation: When analyzing gas turbine power plants, you need to understand that thermal efficiency measures how effectively the system converts heat input into useful work output. The key is identifying what constitutes the net work output and the total heat input. For this power plant, the net work output is the difference between what the turbine produces and what the compressor consumes: 550 MW - 200 MW = 350 MW. The heat input from the combustor is 800 MW. Therefore, the thermal efficiency is:

\eta = \frac{\text{Net Work Output}}{\text{Heat Input}} = \frac{350 \text{ MW}}{800 \text{ MW}} = 0.438 = 43.8\%

This confirms answer A is correct. Answer B (68.8%) represents a common error where students incorrectly use only the turbine output (550 MW) divided by heat input: 550/800 = 68.8%. This ignores the fact that some turbine work must drive the compressor. Answer C (27.5%) appears to use an incorrect relationship, possibly dividing compressor work by heat input (200/800 = 25%) with some variation, which has no thermodynamic meaning. Answer D (31.3%) might result from incorrectly using gross turbine output minus some other calculation error, but doesn't follow proper efficiency definitions. Remember that in gas turbine cycles, always calculate net work as turbine work minus compressor work. The compressor is essential for the cycle but consumes power, so thermal efficiency must account for this parasitic load when determining useful output.

Question 17

A heat engine operates between two thermal reservoirs at temperatures $T_H = 600\text{ K}$ and $T_C = 300\text{ K}$ . During one complete cycle, the engine absorbs $8000\text{ J}$ of heat from the hot reservoir and performs $2800\text{ J}$ of work. If the engine were modified to operate as a Carnot engine between the same reservoirs while absorbing the same amount of heat, what would be the difference in work output between the modified and original engines?

$1200\text{ J}$ more work from the Carnot engine (correct answer)
$800\text{ J}$ more work from the Carnot engine
$2000\text{ J}$ more work from the Carnot engine
$1600\text{ J}$ more work from the Carnot engine

Explanation: First, find the actual engine's efficiency:

\eta_{actual} = \frac{W}{Q_H} = \frac{2800}{8000} = 0.35

. For a Carnot engine:

\eta_{Carnot} = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.5

. If the Carnot engine absorbs the same

8000\text{ J}

, its work output would be:

W_{Carnot} = \eta_{Carnot} \times Q_H = 0.5 \times 8000 = 4000\text{ J}

. The difference is

4000 - 2800 = 1200\text{ J}

. Choice B incorrectly uses the difference in efficiencies times heat input. Choice C assumes the Carnot engine produces twice the work. Choice D miscalculates by using the rejected heat of the original engine.

Question 18

A heat engine undergoes a thermodynamic cycle where it absorbs $Q_1 = 5000\text{ J}$ from a reservoir at $T_1 = 600\text{ K}$ , absorbs $Q_2 = 3000\text{ J}$ from a reservoir at $T_2 = 400\text{ K}$ , and rejects heat to a reservoir at $T_3 = 300\text{ K}$ . If the engine produces $2500\text{ J}$ of work per cycle, what is its thermal efficiency?

$31.25\%$ based on total heat input of $8000\text{ J}$ (correct answer)
$41.67\%$ based on net heat transfer analysis
$50.00\%$ based on highest temperature reservoir only
$83.33\%$ based on work-to-heat-rejected ratio

Explanation: Thermal efficiency is defined as

\eta = \frac{W}{Q_{in}}

where

Q_{in}

is the total heat input to the system. The engine absorbs heat from two reservoirs:

Q_{in} = Q_1 + Q_2 = 5000 + 3000 = 8000\text{ J}

. Therefore:

\eta = \frac{2500}{8000} = 0.3125 = 31.25\%

. Choice B incorrectly attempts to use net heat transfer. Choice C incorrectly considers only the highest temperature reservoir. Choice D incorrectly uses the ratio of work to heat rejected rather than work to heat input.

Question 19

A Brayton cycle gas turbine operates with air entering the compressor at $300\text{ K}$ and $100\text{ kPa}$ . The pressure ratio is $8:1$ and the maximum cycle temperature is $1200\text{ K}$ . Assuming ideal gas behavior with $\gamma = 1.4$ , what is the thermal efficiency of this cycle?

$75.0\%$ using the maximum temperature relationship
$52.3\%$ using the temperature ratio relationship
$37.5\%$ using the simple temperature difference
$44.9\%$ using the pressure ratio relationship (correct answer)

Explanation: For an ideal Brayton cycle, the thermal efficiency depends only on the pressure ratio and specific heat ratio:

\eta = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}

where

r_p

is the pressure ratio. With

r_p = 8

and

\gamma = 1.4

\eta = 1 - \frac{1}{8^{(1.4-1)/1.4}} = 1 - \frac{1}{8^{0.4/1.4}} = 1 - \frac{1}{8^{2/7}} = 1 - \frac{1}{8^{0.286}} = 1 - \frac{1}{1.816} = 1 - 0.551 = 0.449 = 44.9\%

. Choice B incorrectly uses temperature ratios. Choice C uses a simplified temperature difference approach. Choice D incorrectly calculates using only the maximum temperature.

Question 20

A steam power plant operates on a Rankine cycle. The steam enters the turbine as saturated vapor at $6\text{ MPa}$ and exits at $10\text{ kPa}$ . The pump work is negligible compared to turbine work. If the turbine produces $800\text{ kJ/kg}$ of work and the heat input in the boiler is $2500\text{ kJ/kg}$ , what is the thermal efficiency of this cycle?

$68.0\%$ using the steam quality at turbine exit
$47.1\%$ after accounting for pump work corrections
$32.0\%$ based on the given work and heat input values (correct answer)
$25.6\%$ using the condenser heat rejection calculation

Explanation: Thermal efficiency is simply

\eta = \frac{W_{net}}{Q_{in}}

. Since pump work is negligible,

W_{net} \approx W_{turbine} = 800\text{ kJ/kg}

. The heat input is given as

Q_{in} = 2500\text{ kJ/kg}

. Therefore:

\eta = \frac{800}{2500} = 0.32 = 32.0\%

. Choice B incorrectly attempts to account for pump work that was stated to be negligible. Choice C incorrectly uses steam quality. Choice D incorrectly uses condenser calculations when the direct values are given.