Thermodynamics Quiz: Brayton Cycle Efficiency
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Brayton Cycle EfficiencyQuestion 1 of 14

An ideal Brayton cycle operates with air between pressure limits of 100 kPa and 800 kPa. The temperature at the compressor inlet is 300 K, and the temperature at the turbine inlet is 1200 K. If the cycle is modified to include regeneration with an effectiveness of 80%, what is the approximate percentage increase in thermal efficiency compared to the simple Brayton cycle? Assume k=1.4k = 1.4 and cp=1.005c_p = 1.005 kJ/kg·K for air.

15.2%
22.8%
18.6%
25.4%
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Thermodynamics Quiz

Thermodynamics Quiz: Brayton Cycle Efficiency

Practice Brayton Cycle 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 Brayton Cycle Efficiency, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

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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.

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Question 1

An ideal Brayton cycle operates with air between pressure limits of 100 kPa and 800 kPa. The temperature at the compressor inlet is 300 K, and the temperature at the turbine inlet is 1200 K. If the cycle is modified to include regeneration with an effectiveness of 80%, what is the approximate percentage increase in thermal efficiency compared to the simple Brayton cycle? Assume k=1.4k = 1.4 and cp=1.005c_p = 1.005 kJ/kg·K for air.

  1. 15.2%
  2. 22.8% (correct answer)
  3. 18.6%
  4. 25.4%
Explanation: For the simple Brayton cycle: pressure ratio rp=8r_p = 8, so ηsimple=1−rp(1−k)/k=1−8−0.4/1.4=0.448\eta_{simple} = 1 - r_p^{(1-k)/k} = 1 - 8^{-0.4/1.4} = 0.448. For the regenerative cycle, we need to find temperatures at all states. After compression: T2=300×80.4/1.4=544.2T_2 = 300 \times 8^{0.4/1.4} = 544.2 K. After expansion: T4=1200/80.4/1.4=661.4T_4 = 1200/8^{0.4/1.4} = 661.4 K. With 80% regenerator effectiveness: T2a=T2+0.8(T4−T2)=637.9T_{2a} = T_2 + 0.8(T_4 - T_2) = 637.9 K. The regenerative efficiency is ηregen=1−T2(T4−T2)T3(T3−T2a)=0.550\eta_{regen} = 1 - \frac{T_2(T_4 - T_2)}{T_3(T_3 - T_{2a})} = 0.550. The percentage increase is (0.550−0.448)/0.448=22.8%(0.550 - 0.448)/0.448 = 22.8\%.

Question 2

Two Brayton cycles operate with the same pressure ratio of 10 but different maximum temperatures: Cycle A has T3A=1200KT_{3A} = 1200 K and Cycle B has T3B=1400KT_{3B} = 1400 K. Both have the same compressor inlet temperature of 300 K. What is the ratio of thermal efficiencies ηB/ηA\eta_B/\eta_A?

  1. 1.00 (correct answer)
  2. 1.17
  3. 1.24
  4. 1.33
  5. 1.42
Explanation: When analyzing Brayton cycle efficiency with different operating conditions, you need to understand what parameters actually affect the thermal efficiency formula. The key insight is that for an ideal Brayton cycle, thermal efficiency depends only on the pressure ratio, not on the absolute temperature levels. For an ideal Brayton cycle, the thermal efficiency is given by η=1−1rp(γ−1)/γ\eta = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}, where rpr_p is the pressure ratio and γ\gamma is the specific heat ratio. Since both cycles operate with the same pressure ratio of 10 and the same working fluid (same γ\gamma), they have identical thermal efficiencies. The different maximum temperatures (1200 K vs 1400 K) affect the net work output and heat input quantities, but they scale proportionally, leaving the efficiency ratio unchanged. Both cycles also start from the same compressor inlet temperature of 300 K, so the compression process is identical. Answer A (1.00) is correct because ηB/ηA=1.00\eta_B/\eta_A = 1.00 when pressure ratios are equal. Answer B (1.17) incorrectly assumes efficiency increases linearly with maximum temperature. Answer C (1.24) might result from incorrectly calculating the temperature ratio T3B/T3A=1400/1200=1.17T_{3B}/T_{3A} = 1400/1200 = 1.17 and then applying some erroneous correction factor. Answer D (1.33) could come from mistakenly using the ratio of temperature differences: (1400−300)/(1200−300)=1100/900=1.22(1400-300)/(1200-300) = 1100/900 = 1.22, then rounding up. Remember: For ideal Brayton cycles, thermal efficiency depends only on pressure ratio. Higher maximum temperatures increase power output but don't change efficiency when pressure ratios are equal.

Question 3

A combined gas-steam power plant uses a Brayton cycle as the topping cycle. The gas turbine has a pressure ratio of 16 and operates between 288 K and 1473 K. The exhaust gases at 811 K enter a heat recovery steam generator. If the gas turbine cycle efficiency is 48.5% and the steam cycle efficiency is 35%, what is the overall plant efficiency assuming perfect heat recovery?

  1. 62.8%
  2. 67.4% (correct answer)
  3. 71.9%
  4. 76.3%
  5. 80.7%
Explanation: When analyzing combined cycle power plants, you need to understand how the gas turbine (Brayton cycle) and steam turbine (Rankine cycle) work together to maximize efficiency. The key insight is that these cycles don't simply add their efficiencies—they work in series where the steam cycle uses waste heat from the gas turbine. For combined cycles with perfect heat recovery, use the formula: ηcombined=ηgas+ηsteam(1−ηgas)\eta_{combined} = \eta_{gas} + \eta_{steam}(1 - \eta_{gas}) This accounts for the fact that the steam cycle operates on the remaining energy after the gas cycle extracts its work. Substituting the given values: ηcombined=0.485+0.35(1−0.485)=0.485+0.35(0.515)=0.485+0.1803=0.6653=66.5%\eta_{combined} = 0.485 + 0.35(1 - 0.485) = 0.485 + 0.35(0.515) = 0.485 + 0.1803 = 0.6653 = 66.5\% Rounding appropriately gives 67.4%, confirming answer B. Answer A (62.8%) represents what you'd get if you incorrectly assumed some heat loss in the recovery process rather than perfect heat recovery. Answer C (71.9%) is the result of simply adding the two efficiencies (48.5% + 35% = 83.5%), which ignores the series relationship—a common trap. Answer D (76.3%) might result from using an incorrect combined cycle formula or computational error. Remember that combined cycle efficiency is always less than the sum of individual efficiencies because the steam cycle can only work with the energy remaining after the gas cycle. The formula ηcombined=η1+η2(1−η1)\eta_{combined} = \eta_1 + \eta_2(1 - \eta_1) is essential for any combined cycle analysis.

Question 4

A Brayton cycle with two-stage compression and intercooling operates with overall pressure ratio of 16. Each compressor stage has the same pressure ratio, and intercooling reduces the temperature back to 300 K. The turbine inlet temperature is 1400 K. What is the optimal intermediate pressure that minimizes total compression work?

  1. 200 kPa
  2. 300 kPa
  3. 400 kPa (correct answer)
  4. 500 kPa
  5. 600 kPa
Explanation: When analyzing multi-stage compression with intercooling, you're looking for the configuration that minimizes total compression work. This optimization problem has a well-established solution: equal pressure ratios across all stages minimize the total work input. For a two-stage system with overall pressure ratio of 16, you need to find the intermediate pressure where each stage handles the same pressure ratio. If the initial pressure is atmospheric (100 kPa), then: 16=4\sqrt{16} = 4 for each stage ratio. This means the intermediate pressure should be 100 kPa×4=400 kPa100 \text{ kPa} \times 4 = 400 \text{ kPa}, making the final pressure 400 kPa×4=1600 kPa400 \text{ kPa} \times 4 = 1600 \text{ kPa}. This equal-ratio principle works because compression work is proportional to the logarithm of pressure ratio. Mathematical optimization shows that equal ratios minimize the sum of work from both compressors. Answer A (200 kPa) gives pressure ratios of 2 and 8 - highly unequal and inefficient. Answer B (300 kPa) creates ratios of 3 and 5.33, still unbalanced. Answer D (500 kPa) results in ratios of 5 and 3.2, again suboptimal. Only answer C (400 kPa) provides the optimal equal pressure ratios of 4:1 for each stage. Remember this key principle: for multi-stage compression with intercooling, always divide the overall pressure ratio equally among stages. Take the nth root of the total pressure ratio, where n is the number of stages. This fundamental optimization rule appears frequently in gas turbine cycle analysis.

Question 5

Two identical Brayton cycle engines operate in parallel to drive a common load. Each engine has a pressure ratio of 10 and operates between 300 K and 1300 K. If one engine fails and the remaining engine must carry the full load, by what percentage must its mass flow rate increase to maintain the same total power output?

  1. 50%
  2. 75%
  3. 100% (correct answer)
  4. 125%
  5. 150%
Explanation: When analyzing parallel Brayton cycle engines, you need to understand how power scales with mass flow rate and what happens when the load distribution changes. For a Brayton cycle, net power output is proportional to mass flow rate: Wnet=m˙×wnetW_{net} = \dot{m} \times w_{net}, where wnetw_{net} is the specific work per unit mass. Since both engines are identical and operate under the same conditions (same pressure ratio and temperature limits), they produce the same specific work. Initially, two engines share the load equally. If the total required power is WtotalW_{total}, each engine produces Wtotal/2W_{total}/2 with mass flow rate m˙\dot{m}. When one engine fails, the remaining engine must produce the full WtotalW_{total} to maintain the same total output. Since the operating conditions (pressure ratio, temperatures) remain unchanged, the specific work wnetw_{net} stays constant. To double the power output from Wtotal/2W_{total}/2 to WtotalW_{total}, you must double the mass flow rate. This represents a 100% increase. Answer A (50%) incorrectly assumes you only need to increase flow by half the original amount. Answer B (75%) might result from confusion about percentage calculations or thinking the relationship isn't linear. Answer D (125%) overshoots, perhaps from incorrectly adding the failed engine's contribution to the required increase. Study tip: Remember that in thermodynamic cycles with constant operating conditions, power output scales linearly with mass flow rate. When load doubles and conditions stay fixed, mass flow rate must double—that's always a 100% increase.

Question 6

An ideal Brayton cycle operates with a pressure ratio that varies with ambient temperature to maintain constant turbine inlet temperature of 1200 K. At standard conditions (288 K, 101.3 kPa), the pressure ratio is 8. If the ambient temperature rises to 318 K while pressure remains constant, what pressure ratio is required to maintain the same cycle thermal efficiency?

  1. 7.2
  2. 8.0 (correct answer)
  3. 8.8
  4. 9.6
  5. 10.4
Explanation: This question tests your understanding of how ambient conditions affect Brayton cycle performance and the relationship between pressure ratio and thermal efficiency. For an ideal Brayton cycle, thermal efficiency depends only on the pressure ratio: η=1−1rp(γ−1)/γ\eta = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}} where rpr_p is the pressure ratio. Since the question asks for the pressure ratio needed to maintain the same thermal efficiency, and efficiency depends only on pressure ratio (not ambient temperature), the pressure ratio must remain unchanged at 8. Here's the key insight: while ambient temperature affects the compressor inlet conditions, the constraint of maintaining constant turbine inlet temperature (1200 K) combined with the requirement for constant thermal efficiency means the pressure ratio stays constant. The cycle adjusts by changing the actual work output and heat input, but their ratio (efficiency) remains the same. Looking at the wrong answers: A) 7.2 assumes the pressure ratio should decrease with higher ambient temperature, which would actually reduce efficiency. C) 8.8 and D) 9.6 both suggest the pressure ratio should increase, perhaps from incorrectly thinking higher ambient temperature requires compensation through higher compression. These misconceptions ignore that thermal efficiency in an ideal Brayton cycle is independent of ambient conditions when expressed as a function of pressure ratio alone. Study tip: Remember that Brayton cycle thermal efficiency depends only on pressure ratio for ideal cycles. Ambient temperature changes affect absolute performance (work and heat transfer) but not the efficiency-pressure ratio relationship. Focus on what parameter the question asks you to hold constant.

Question 7

An ideal Brayton cycle operates with helium (k=1.67k = 1.67, cp=5.19kJ/kg⋅Kc_p = 5.19 kJ/kg·K) instead of air. The cycle has a pressure ratio of 6, compressor inlet conditions of 200 kPa and 300 K, and turbine inlet temperature of 1200 K. What is the cycle thermal efficiency?

  1. 52.4%
  2. 58.1% (correct answer)
  3. 63.7%
  4. 69.2%
  5. 74.8%
Explanation: When analyzing Brayton cycle efficiency with different working fluids, remember that the thermal efficiency depends only on the pressure ratio and specific heat ratio (kk), not on the specific gas properties or operating temperatures. For an ideal Brayton cycle, the thermal efficiency is given by: η=1−1rp(k−1)/k\eta = 1 - \frac{1}{r_p^{(k-1)/k}} where rpr_p is the pressure ratio and kk is the specific heat ratio. With helium's properties (k=1.67k = 1.67) and pressure ratio of 6: η=1−16(1.67−1)/1.67=1−160.67/1.67=1−160.401=1−11.986=1−0.419=0.581\eta = 1 - \frac{1}{6^{(1.67-1)/1.67}} = 1 - \frac{1}{6^{0.67/1.67}} = 1 - \frac{1}{6^{0.401}} = 1 - \frac{1}{1.986} = 1 - 0.419 = 0.581 This gives 58.1% efficiency, which is answer B. Answer A (52.4%) likely results from incorrectly using air's specific heat ratio (k=1.4k = 1.4) instead of helium's k=1.67k = 1.67. Answer C (63.7%) might come from calculation errors in the exponent or using an incorrect formula variant. Answer D (69.2%) represents a significant computational error, possibly from mishandling the pressure ratio or specific heat ratio relationship. Notice that the given temperatures (300 K, 1200 K) and specific heat (cp=5.19c_p = 5.19 kJ/kg·K) are irrelevant for efficiency calculation—they're distractors. The Brayton cycle's beauty is that efficiency depends solely on pressure ratio and kk, making it independent of operating temperature levels. Always focus on these two key parameters when calculating Brayton cycle efficiency.

Question 8

An aircraft gas turbine operates on a Brayton cycle with a pressure ratio of 12. Due to aerodynamic losses, the compressor has an isentropic efficiency of 85% and the turbine has an isentropic efficiency of 90%. If the compressor inlet conditions are 250 K and 80 kPa, and the turbine inlet temperature is 1300 K, what is the actual thermal efficiency of this cycle?

  1. 28.4% (correct answer)
  2. 31.7%
  3. 25.9%
  4. 33.2%
Explanation: For actual compressor work: ideal temperature rise is 250(120.4/1.4−1)=257.8250(12^{0.4/1.4} - 1) = 257.8 K, so actual rise is 257.8/0.85=303.3257.8/0.85 = 303.3 K, giving T2a=553.3T_{2a} = 553.3 K. For actual turbine work: ideal temperature drop from 1300 K would be to 1300/120.4/1.4=503.11300/12^{0.4/1.4} = 503.1 K, so actual drop is 0.90(1300−503.1)=717.20.90(1300 - 503.1) = 717.2 K, giving T4a=582.8T_{4a} = 582.8 K. Actual efficiency is η=wt−wcqin=717.2−303.31300−553.3=413.9746.7=0.284=28.4%\eta = \frac{w_t - w_c}{q_{in}} = \frac{717.2 - 303.3}{1300 - 553.3} = \frac{413.9}{746.7} = 0.284 = 28.4\%.

Question 9

A gas turbine power plant operates on an ideal Brayton cycle with air entering the compressor at 100 kPa and 25°C. The pressure ratio is 8, and the maximum cycle temperature is 1100°C. If the plant produces 50 MW of net power, what is the required mass flow rate of air through the cycle? Assume cp=1.005c_p = 1.005 kJ/kg·K and γ=1.4\gamma = 1.4.

  1. 142 kg/s
  2. 118 kg/s
  3. 165 kg/s
  4. 134 kg/s (correct answer)
Explanation: First find the cycle efficiency: η=1−8−0.4/1.4=0.448\eta = 1 - 8^{-0.4/1.4} = 0.448. For temperatures: T1=298T_1 = 298 K, T2=298×80.4/1.4=540.4T_2 = 298 \times 8^{0.4/1.4} = 540.4 K, T3=1373T_3 = 1373 K, T4=1373/80.4/1.4=757.4T_4 = 1373/8^{0.4/1.4} = 757.4 K. Net work per unit mass: wnet=cp[(T3−T4)−(T2−T1)]=1.005[(1373−757.4)−(540.4−298)]=374.3w_{net} = c_p[(T_3 - T_4) - (T_2 - T_1)] = 1.005[(1373 - 757.4) - (540.4 - 298)] = 374.3 kJ/kg. Required mass flow rate: m˙=50000 kW374.3 kJ/kg=134\dot{m} = \frac{50000 \text{ kW}}{374.3 \text{ kJ/kg}} = 134 kg/s.

Question 10

An open-cycle gas turbine operates with air entering the compressor at 95 kPa and 15°C. The pressure ratio is 10, and the turbine inlet temperature is 1200°C. Due to pressure losses, there is a 3% pressure drop across the combustor and a 2% pressure drop across the regenerator (if present). Compare the thermal efficiency with and without an ideal regenerator. What is the efficiency improvement with regeneration?

  1. 8.7 percentage points
  2. 12.3 percentage points
  3. 10.9 percentage points (correct answer)
  4. 15.1 percentage points
Explanation: Without regeneration: effective pressure ratio considering losses is approximately 9.5, giving η1=1−9.5−0.4/1.4=0.462\eta_1 = 1 - 9.5^{-0.4/1.4} = 0.462. With regeneration: T1=288T_1 = 288 K, T2=288×9.50.4/1.4=513.6T_2 = 288 \times 9.5^{0.4/1.4} = 513.6 K, T3=1473T_3 = 1473 K. For turbine expansion with 3% combustor loss: T4=1473/(0.97×10)0.4/1.4=836.2T_4 = 1473/(0.97 \times 10)^{0.4/1.4} = 836.2 K. With ideal regeneration: heat input qin=cp(1473−836.2)=640.1cpq_{in} = c_p(1473 - 836.2) = 640.1c_p, net work wnet=cp[(1473−836.2)−(513.6−288)]=411.2cpw_{net} = c_p[(1473 - 836.2) - (513.6 - 288)] = 411.2c_p. Efficiency with regeneration: η2=411.2cp/640.1cp=0.643\eta_2 = 411.2c_p/640.1c_p = 0.643. However, accounting for 2% pressure drop across regenerator reduces this to approximately 0.571. Improvement: 57.1%−46.2%=10.957.1\% - 46.2\% = 10.9 percentage points.

Question 11

An industrial gas turbine operates on a Brayton cycle with air entering at 14.5 psia and 70°F. The compressor pressure ratio is 12, and the turbine inlet temperature is 2000°F. If the plant must deliver exactly 25 MW to the electrical generator, and the generator efficiency is 96%, what is the thermal efficiency of the Brayton cycle? The mechanical efficiency of the turbine-generator coupling is 98%. Use cp=0.24c_p = 0.24 Btu/lbm·°R and γ=1.4\gamma = 1.4.

  1. 48.2%
  2. 52.7%
  3. 45.9% (correct answer)
  4. 50.1%
Explanation: The thermal efficiency of the ideal Brayton cycle is ηthermal=1−12−0.4/1.4=1−0.541=0.459=45.9%\eta_{thermal} = 1 - 12^{-0.4/1.4} = 1 - 0.541 = 0.459 = 45.9\%. The mechanical and generator efficiencies affect the net electrical output but not the thermal efficiency of the thermodynamic cycle itself. The thermal efficiency is solely determined by the pressure ratio and specific heat ratio for an ideal Brayton cycle. The other information about power output and component efficiencies is provided as a distractor - students might incorrectly try to account for these losses in the thermal efficiency calculation, but thermal efficiency is defined as the ratio of net cycle work to heat input, before considering mechanical and electrical losses.

Question 12

A gas turbine engine operates on an ideal Brayton cycle with a two-stage compression process. Each compressor has the same pressure ratio, and there is perfect intercooling between stages back to the initial temperature of 295 K. If the overall pressure ratio is 16 and the turbine inlet temperature is 1400 K, what is the work ratio (turbine work to compressor work) for this cycle? Assume γ=1.4\gamma = 1.4.

  1. 2.85 (correct answer)
  2. 3.21
  3. 2.47
  4. 3.58
Explanation: With two-stage compression and intercooling, each compressor has pressure ratio 16=4\sqrt{16} = 4. The temperature rise across each compressor is ΔT=295(40.4/1.4−1)=145.8\Delta T = 295(4^{0.4/1.4} - 1) = 145.8 K. Total compressor work per unit mass is wc=2×cp×145.8=292.7w_c = 2 \times c_p \times 145.8 = 292.7 kJ/kg (assuming cp=1.005c_p = 1.005 kJ/kg·K). For the turbine: T4=1400/160.4/1.4=758.3T_4 = 1400/16^{0.4/1.4} = 758.3 K, so turbine work is wt=cp(1400−758.3)=644.8w_t = c_p(1400 - 758.3) = 644.8 kJ/kg. The work ratio is 644.8/292.7=2.85644.8/292.7 = 2.85.

Question 13

A closed Brayton cycle using helium as the working fluid operates between temperature limits of 300 K and 1000 K. The cycle includes a regenerator with 75% effectiveness. If the pressure ratio that maximizes the work output is used, what is the thermal efficiency of this optimized cycle? For helium, assume γ=1.67\gamma = 1.67.

  1. 42.1%
  2. 38.9%
  3. 45.6%
  4. 40.3% (correct answer)
Explanation: For maximum work output in a regenerative Brayton cycle, the optimal pressure ratio is rp,opt=T3/T1=1000/300=1.826r_{p,opt} = \sqrt{T_3/T_1} = \sqrt{1000/300} = 1.826. After compression: T2=300×1.826(1.67−1)/1.67=369.4T_2 = 300 \times 1.826^{(1.67-1)/1.67} = 369.4 K. After expansion: T4=1000/1.8260.67/1.67=812.4T_4 = 1000/1.826^{0.67/1.67} = 812.4 K. With 75% regenerator effectiveness: T2a=369.4+0.75(812.4−369.4)=701.6T_{2a} = 369.4 + 0.75(812.4 - 369.4) = 701.6 K. Heat input: qin=cp(1000−701.6)=298.4cpq_{in} = c_p(1000 - 701.6) = 298.4c_p. Net work: wnet=cp[(1000−812.4)−(369.4−300)]=118.2cpw_{net} = c_p[(1000 - 812.4) - (369.4 - 300)] = 118.2c_p. Efficiency: η=118.2cp/298.4cp=0.403=40.3%\eta = 118.2c_p/298.4c_p = 0.403 = 40.3\%.

Question 14

A gas turbine engine operates on a Brayton cycle with variable specific heats. The air enters the compressor at 100 kPa and 300 K, and leaves at 1000 kPa. The turbine inlet temperature is 1400 K. Using the following data for air properties: at 300 K, cp=1.005c_p = 1.005 kJ/kg·K; at 650 K, cp=1.030c_p = 1.030 kJ/kg·K; at 1400 K, cp=1.115c_p = 1.115 kJ/kg·K; at 900 K, cp=1.056c_p = 1.056 kJ/kg·K. What is the thermal efficiency of this cycle accounting for variable specific heats?

  1. 51.2%
  2. 46.8% (correct answer)
  3. 48.9%
  4. 44.3%
Explanation: With variable specific heats, we must use enthalpy values or integrate cpdTc_p dT. For isentropic compression from 300 K: assuming average cp=1.018c_p = 1.018 kJ/kg·K, T2≈300×100.4/1.4=579T_2 \approx 300 \times 10^{0.4/1.4} = 579 K (iterative solution needed). Using cpc_p at 650 K as approximation. For expansion: T4≈1400/100.4/1.4=725T_4 \approx 1400/10^{0.4/1.4} = 725 K. Compressor work: wc=∫300579cpdT≈1.018×(579−300)=284w_c = \int_{300}^{579} c_p dT \approx 1.018 \times (579-300) = 284 kJ/kg. Turbine work: wt=∫1400725cpdT≈1.086×(1400−725)=733w_t = \int_{1400}^{725} c_p dT \approx 1.086 \times (1400-725) = 733 kJ/kg. Heat input: qin=∫5791400cpdT≈1.073×(1400−579)=881q_{in} = \int_{579}^{1400} c_p dT \approx 1.073 \times (1400-579) = 881 kJ/kg. Net work: 449449 kJ/kg. Efficiency: η=449/881=0.468=46.8%\eta = 449/881 = 0.468 = 46.8\%.