Thermodynamics: how 9 AI study apps did on 20 college problems

Covers property tables, first law (closed/open), second law, entropy, cycles, mixtures. Every app got the same problem text in a fresh chat, one attempt, no hints. Tested October 15, 2026 – October 16, 2026.

Thermodynamics accuracy by app 0% 25% 50% 75% 100% ChatGPT 100% MyStudySidekick 100% Claude 90% NotebookLM 89% Solvely 74% Turbo AI 66% StudyFetch 34% Quizlet 30% Knowt 0%
Accuracy = share of problem parts answered within tolerance of the key, over the 20 problems tested so far. This is a live benchmark: we keep adding problems, and this chart updates with them.
RankAppAccuracyProblems graded
1ChatGPT100.0%20 / 20
2MyStudySidekick100.0%20 / 20
3Claude90.0%20 / 20
4NotebookLM88.8%20 / 20
5Solvely74.2%20 / 20
6Turbo AI65.5%20 / 20
7StudyFetch33.7%20 / 20
8Quizlet29.9%20 / 20
9Knowt0.0%20 / 20

The team behind MyStudySidekick runs this benchmark (disclosure and safeguards). Models used: ChatGPT, GPT-6, high reasoning effort; Claude, Claude Opus 5.5, medium effort.

Every problem, the key, and each app's answer

  1. 1. Regenerative gas turbine with variable specific heats
  2. 2. Split-shaft gas turbine with intercooling, reheat and regeneration
  3. 3. Reheat–regenerative Rankine cycle with component efficiencies
  4. 4. 150 MW reheat plant with closed and open feed heaters
  5. 5. Combined gas–steam cycle with reheat in the combustor
  6. 6. Mercury–steam binary vapour cycle
  7. 7. Exergy destroyed while vaporizing water by steam injection
  8. 8. Irreversibility and second-law efficiency of a steam turbine
  9. 9. Entropy generation as steam flows between two tanks
  10. 10. Isentropic expansion then throttling to dry saturated steam
  11. 11. Charging a constant-pressure cylinder from a steam line
  12. 12. Polytropic steam expansion followed by constant-volume cooling
  13. 13. Cooling and dehumidification at 85 kPa
  14. 14. Cooling tower with fan power
  15. 15. Mixing CO₂ and N₂ from two tanks with heat loss
  16. 16. Steam tank blow-down through a turbine
  17. 17. Adiabatic flame temperature with CO₂ and H₂O dissociation
  18. 18. Two-stage R-134a refrigeration with a flash chamber
  19. 19. Wet-bulb psychrometrics and evaporative cooling at 84 kPa
  20. 20. Converging–diverging nozzle with a normal shock

Problem 1 · hard

Regenerative gas turbine with variable specific heats

gas power cycles

A regenerative gas-turbine cycle uses air as the working fluid. The compressor pressure ratio is 8.5. Air enters the compressor at 295 K and the turbine at 1260 K. The compressor has an isentropic efficiency of 80 percent, the turbine 86 percent, and the regenerator effectiveness is 70 percent. Neglect pressure losses. Accounting for the variation of specific heats with temperature (use the ideal-gas properties of air, Çengel Table A-17), determine: (a) the air temperature at the turbine exit, in K; (b) the net work output, in kJ/kg; (c) the thermal efficiency; (d) the back work ratio.

Adapted from Çengel 9e Prob. 9–112 (numbers and wording changed)

App (a) turbine exit temperature(b) net work output(c) thermal efficiency(d) back work ratio Score
Answer key 809.7 K ±4 K204.3 kJ/kg ±2%0.3501 ±0.0070.604 ±0.012
NotebookLM 809.7 K 204.35 kJ/kg 0.3501 0.6039 100%
ChatGPT 809.7 K 204.34 kJ/kg 0.3501 0.604 100%
MyStudySidekick 809.6 K 204.2 kJ/kg 0.3499 0.6041 100%
Solvely 812 K 198.0 kJ/kg 0.341 0.613 50%
Why: Tested app is Solvely. h2s 546.0 (544.5) and h4s 753.8 (748.6) misread; (a) and (d) inside tolerance, (b) and (c) not.
Quizlet 828 K 192.2 kJ/kg 0.342 0.602 25%
Why: Read h3 = 1333.7 (table: 1348.55 at 1260 K), Pr3 = 328.6 (290.8), h2s = 527.7 (544.5). Ends: 'are you ready to switch back to your Biol 1201 Exam 2 Review?' — account memory still on.
Knowt 771.14 K no final answer no final answer no final answer 0%
Why: Used constant k = 1.4 despite 'variable specific heats', then gave no values for (b)-(d) ('d+', 'e+', 'f+') and trailed into incoherent text.
StudyFetch 646 K 511 kJ/kg 0.700 0.24 0%
Why: Read the 1240 K row (h 1324.93, Pr 238.0) for 1260 K and h2s = 421.26 (the 420 K value) for Pr 11.108; efficiency 70 % not flagged as impossible.
Turbo AI 826 K 162 kJ/kg 0.285 0.68 0%
Why: Pr1 = 1.4781 (table: 1.3068 at 295 K) propagates to h2; h4s 755 vs 748.6.
Claude 826 K 186 kJ/kg 0.326 0.626 0%
Why: Pasted text starts mid-word ("ssumptions"), the opening of the reply was cut when copied; the answers are complete. Error: Table A-17 lookup Pr = 34.21 read as h4s = 770.3 kJ/kg; the table gives 748.6 (between 730 K and 740 K), so every part inherits it.

Problem 2 · hard

Split-shaft gas turbine with intercooling, reheat and regeneration

gas power cycles

A gas-turbine plant delivers a net shaft output of 7200 kW. Air is compressed in two stages with equal pressure ratios and complete intercooling; the overall pressure ratio is 10 and air enters each compressor stage at 25°C. A high-pressure turbine drives the two compressors and nothing else; a separate low-pressure (power) turbine drives the load. Gas enters the high-pressure turbine at 680°C and is reheated to 680°C before entering the low-pressure turbine. The low-pressure turbine exhaust passes through a heat exchanger that heats the air leaving the second compressor stage; its thermal ratio, defined as (air temperature rise)/(gas inlet temperature − air inlet temperature), is 0.75. The isentropic efficiency of each compressor stage is 0.82 and of each turbine is 0.87. Both shafts have a mechanical efficiency of 96 percent: 0.96 × (high-pressure turbine work) = total compressor work, and net output = 0.96 × (low-pressure turbine work). Neglect pressure losses, kinetic energy changes and the mass of fuel. For air take cp = 1.005 kJ/kg·K and γ = 1.4; for the gases in the combustion chamber, reheater, turbines and heat exchanger take cp = 1.147 kJ/kg·K and γ = 1.333 (use these gas values for the heat supplied in the combustion chamber and in the reheater). Determine: (a) the thermal efficiency of the plant (net output / total heat supplied); (b) the work ratio (net output / total gross turbine work); (c) the mass flow rate of air, in kg/s.

Adapted from Rajput, Engineering Thermodynamics, Example 13.48 (numbers and wording changed)

App (a) thermal efficiency(b) work ratio(c) air mass flow rate Score
Answer key 0.2905 ±0.0060.3546 ±0.00743.19 kg/s ±2%
Solvely 0.291 0.355 43.1 kg/s 100%
NotebookLM 0.2905 0.3546 43.19 kg/s 100%
ChatGPT 0.2905 0.3546 43.19 kg/s 100%
Claude 0.291 0.355 43.2 kg/s 100%
MyStudySidekick 0.2905 0.3546 43.19 kg/s 100%
Quizlet 0.363 0.434 32.7 kg/s 0%
Why: Took each turbine's expansion ratio as √10 instead of fixing the HP turbine by the shaft balance; HP exit 704 K inconsistent with its own 277.9 kJ/kg.
Knowt no final answer no final answer no final answer 0%
Why: No numerical answers; compressed the second stage from T2 (no intercooling) and expanded both turbines over the full ratio 10, then trailed into incoherent text.
StudyFetch 0.310 0.48 31.6 kg/s 0%
Why: Assumed √10 per turbine, noticed the HP shaft balance fails (228 vs 285 kJ/kg) and said so, but answered with the inconsistent model and offered to redo.
Turbo AI 0.46 0.507 22.6 kg/s 0%
Why: Inverted the turbine efficiency: isentropic drop = 0.87 × actual (should be actual/0.87), giving HP ratio 2.94 instead of 4.46; claims '~46 % consistent with textbook values'.

Problem 3 · hard

Reheat–regenerative Rankine cycle with component efficiencies

vapour power cycles

A steam power plant operates on a reheat–regenerative Rankine cycle with one open feedwater heater and has a net power output of 120 MW. Steam enters the high-pressure turbine at 12 MPa and 560°C and expands to 1 MPa. At 1 MPa some steam is extracted to the open feedwater heater; the rest is reheated at 1 MPa to 480°C and expands in the low-pressure turbine to the condenser pressure of 8 kPa. Water leaves the condenser and the feedwater heater as saturated liquid. Both turbines have an isentropic efficiency of 87 percent and both pumps 85 percent. Neglect pressure drops and kinetic energy changes. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the fraction of the boiler steam that is extracted; (b) the mass flow rate of steam through the boiler, in kg/s; (c) the thermal efficiency of the cycle; (d) the quality of the steam at the low-pressure turbine exit.

Adapted from Çengel 9e Prob. 10–60 (numbers and wording changed)

App (a) extraction fraction(b) boiler steam flow(c) thermal efficiency(d) LP turbine exit quality Score
Answer key 0.2146 ±0.00493.88 kg/s ±2%0.4069 ±0.0050.9873 ±0.006
ChatGPT 0.2145 93.87 kg/s 0.4070 0.9870 100%
Claude 0.215 93.9 kg/s 0.407 0.987 100%
MyStudySidekick 0.2146 93.88 kg/s 0.4069 0.9873 100%
Solvely 0.2152 94.49 kg/s 0.4039 0.9936 75%
Why: Tested app is Solvely. s at 1 MPa/480 °C read as 7.7642 (7.7075): (d) 0.0063 off, just outside ±0.006.
NotebookLM 0.2113 96.89 kg/s 0.3984 0.987 50%
Why: s1 at 12 MPa/560 °C interpolated as 6.7858 (6.6864) from the wrong pressure row, so h2s 2872.9 (2824.0) and the HP exit too high.
Quizlet 0.213 86.0 kg/s 0.442 0.976 25%
Why: s4 at 1 MPa/480 °C read as 7.6254 (7.7075), so h5s 2275 too low and LP work too large.
Turbo AI 0.215 96 kg/s 0.40 0.997 25%
Why: h at 12 MPa/560 °C read as 3495 (3507.4) and s at 1 MPa/480 °C as 7.79 (7.7075).
Knowt no final answer 76.3 kg/s no final answer no final answer 0%
Why: Wrong table values (h1 3640.5, h4 1995.2, h5 205.5), no extraction fraction or quality, then incoherent text.
StudyFetch 0.161 150.9 kg/s 0.331 1.07 0%
Why: Turbine expansion computed as h2 = h1 + ... (3834 > 3507): negative turbine work; then declared 'the problem data is inconsistent' and suggested the reheat must be 580 °C. Reply text starts 'ere's' (first character lost when copied).

Problem 4 · hard

150 MW reheat plant with closed and open feed heaters

vapour power cycles

A steam turbine plant produces 150 MW of net electrical output. It has reheating and two feed heaters: a closed (surface) heater on the high-pressure side and an open (direct-contact) heater on the low-pressure side. Steam at the stop valve is at 120 bar and 540°C and is throttled to 114 bar before entering the high-pressure turbine, which exhausts at 28 bar. Part of this exhaust is bled to the closed heater; the rest is reheated to 560°C and enters the low-pressure turbine at 25 bar (the reheater pressure drop is 3 bar). Steam is bled from the low-pressure turbine at 7 bar to the open heater, and the rest expands to 0.06 bar. Treat the low-pressure turbine as two sections (25 → 7 bar and 7 → 0.06 bar), each with the turbine isentropic efficiency. The condensate pump raises the condenser water (saturated liquid at 0.06 bar) to 7 bar. The open heater delivers saturated liquid at 7 bar to the boiler feed pump, which raises it to 120 bar; the feedwater then passes through the closed heater, leaving it at the saturation temperature of the bled steam (28 bar), and enters the boiler. The drain from the closed heater leaves as saturated liquid at 28 bar and is throttled into the open heater. Neglect the boiler pressure drop. Turbine isentropic efficiency (each section) 86 percent; pump isentropic efficiency 88 percent; mechanical efficiency 96 percent; generator efficiency 97 percent; boiler efficiency 91 percent. Net electrical output = 0.96 × 0.97 × (total turbine work) − (total pump work). Overall efficiency = net electrical output / fuel energy, where fuel energy = (heat to steam in the boiler + heat in the reheater)/0.91. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the steam bled to the closed heater, in kg/h; (b) the steam bled to the open heater, in kg/h; (c) the overall thermal efficiency of the plant; (d) the specific steam consumption (boiler steam per kWh of net electrical output), in kg/kWh.

Adapted from Rajput, Example 12.17 (numbers and wording changed)

App (a) bleed to closed HP heater(b) bleed to open LP heater(c) overall thermal efficiency(d) specific steam consumption Score
Answer key 63,560 kg/h ±2.5%66,840 kg/h ±2.5%0.3564 ±0.0063.173 kg/kWh ±2%
ChatGPT 63543 kg/h 66863 kg/h 0.3564 3.173 kg/kWh 100%
Claude 63600 kg/h 66800 kg/h 0.356 3.17 kg/kWh 100%
MyStudySidekick 63563 kg/h 66837 kg/h 0.3564 3.173 kg/kWh 100%
Turbo AI 65800 kg/h 67300 kg/h 0.346 3.21 kg/kWh 50%
Why: Isentropic HP exit 3054 (3037.8) after a throttling entropy rise; reheat heat taken without (1 − y1).
Solvely 62295 kg/h 65413 kg/h 0.366 3.07 kg/kWh 50%
Why: Tested app is Solvely. HP exit h 3068.8 (3096.3) and 7 bar bleed 3188.9 (3234.2): turbine work 1275 (1234).
NotebookLM 66592 kg/h 68408 kg/h 0.3510 3.265 kg/kWh 50%
Why: Stop-valve state h 3451.08, s 6.7118 (3455.8, 6.6237) and closed-heater feedwater exit 1001.4 (992.5).
StudyFetch 59600 kg/h 59900 kg/h 0.370 3.19 kg/kWh 25%
Why: Isentropic exits wrong (28 bar h2s 3201 vs 3037.8; s carried as 6.653), open-heater balance per (1 − y). Pasted text starts with UI debris ('resolve … D Sparky').
Quizlet 65250 kg/h 60030 kg/h 0.426 3.48 kg/kWh 0%
Why: Approximate table values (h1 3465, s 6.65, h2s 2980), LP exit expanded from 25 bar in one step, rough w_net 1050; (a) 2.65 % off. Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
Knowt no final answer no final answer no final answer no final answer 0%
Why: Method outline only; no numerical answers.

Problem 5 · hard

Combined gas–steam cycle with reheat in the combustor

combined cycles

The gas-turbine cycle of a combined gas–steam power plant has a pressure ratio of 14. Air enters the compressor at 300 K and the turbine at 1450 K. The compressor isentropic efficiency is 86 percent and the gas-turbine isentropic efficiency is 89 percent. The gases leaving the gas turbine heat steam at 14 MPa to 520°C in a heat-recovery steam generator and leave it at 230°C. The steam expands isentropically in a high-pressure turbine to 3 MPa, is reheated at 3 MPa to 560°C in the combustion chamber, and expands isentropically in a low-pressure turbine to 8 kPa. It is condensed to saturated liquid and pumped isentropically to 14 MPa. The steam flow rate is 15 kg/s. Treat the combustion gases as air with variable specific heats (Çengel Table A-17), neglect the mass of fuel, and use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the mass flow rate of air, in kg/s; (b) the total rate of heat input in the combustion chamber (including the steam reheat), in kW; (c) the net power output of the combined plant, in MW; (d) the thermal efficiency of the combined cycle.

Adapted from Çengel 9e Prob. 10–108 (numbers and wording changed)

App (a) air mass flow rate(b) total heat input rate(c) net power output(d) combined-cycle efficiency Score
Answer key 136.9 kg/s ±2%130,900 kW ±2%70.51 MW ±2%0.5387 ±0.008
Turbo AI 136.8 kg/s 129800 kW 69.3 MW 0.533 100%
Solvely 137.22 kg/s 131404 kW 70.9 MW 0.539 100%
NotebookLM 138.01 kg/s 131625 kW 71.05 MW 0.5398 100%
ChatGPT 136.95 kg/s 130474 kW 70.10 MW 0.5373 100%
MyStudySidekick 137.04 kg/s 130551 kW 70.15 MW 0.5374 100%
Quizlet 72.3 kg/s 72734 kW 38.92 MW 0.535 25%
Why: Feedwater enthalpy taken as 'h_f at 14 MPa ≈ 1500' (pump exit is 188); h3 1548.4 (1575.6), h2s 612.5 (637.9); steam work rounded (500 + 700 − 10). (d) inside tolerance by compensation.
Knowt no final answer no final answer no final answer no final answer 0%
Why: Method outline only; no numerical answers.
StudyFetch 6.08 kg/s 10730 kW 23.0 MW 2.15 0%
Why: HRSG balance botched (air flow 6.08 kg/s), efficiency 215 %; declared the given steam flow 'probably a typo' and offered to change the problem.
Claude 129.6 kg/s 124100 kW 65.2 MW 0.526 0%
Why: Table A-17 again: Pr 37.29 read as h4s = 788.8 (766.9), so h4 875.4 (855.9) and the air flow 5 % low. Reply text starts mid-sentence ('neglected.').

Problem 6 · hard

Mercury–steam binary vapour cycle

combined cycles

A binary vapour cycle uses mercury and steam. Dry saturated mercury vapour at 4.5 bar enters the mercury turbine and exhausts at 0.04 bar. Mercury leaves the mercury condenser as saturated liquid at 0.04 bar and is pumped back to 4.5 bar. The mercury condenser is the steam boiler: it turns feedwater into dry saturated steam at 18 bar. This steam is then superheated to 320°C in a superheater placed in the mercury-boiler furnace and expands in a steam turbine to 0.05 bar; the condensate (saturated liquid at 0.05 bar) is pumped to 18 bar. The mercury turbine has an internal (isentropic) efficiency of 0.85 and the steam turbine 0.87; both pumps are isentropic and their work must be included. The steam flow is 52 000 kg/h. Properties of saturated mercury: | p (bar) | t (°C) | hf (kJ/kg) | hg (kJ/kg) | sf (kJ/kg·K) | sg (kJ/kg·K) | vf (m³/kg) | vg (m³/kg) | |---|---|---|---|---|---|---|---| | 4.5 | 450 | 62.93 | 355.98 | 0.1352 | 0.5397 | 79.9 × 10⁻⁶ | 0.068 | | 0.04 | 216.9 | 29.98 | 329.85 | 0.0808 | 0.6925 | 76.5 × 10⁻⁶ | 5.178 | Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7) for water. Determine: (a) the mercury flow rate, in kg/h; (b) the overall efficiency of the binary cycle (net work of both cycles / heat supplied in the mercury boiler and the superheater); (c) the net power output, in MW.

Adapted from Rajput, Example 12.22 (numbers and wording changed)

App (a) mercury flow rate(b) overall efficiency(c) net power output Score
Answer key 575,300 kg/h ±2%0.4614 ±0.00725.89 MW ±2%
Turbo AI 575000 kg/h 0.46 25.9 MW 100%
Solvely 575366 kg/h 0.4657 26.13 MW 100%
ChatGPT 575274 kg/h 0.4614 25.89 MW 100%
Claude 575258 kg/h 0.461 25.89 MW 100%
MyStudySidekick 575300 kg/h 0.4614 25.89 MW 100%
StudyFetch 575640 kg/h 0.509 25.9 MW 67%
Why: Mercury boiler heat from h_f at 4.5 bar (62.93) instead of the pump exit (30.01): heat supplied too low, efficiency 50.9 %.
NotebookLM 574900 kg/h 0.4527 25.40 MW 67%
Why: Only NotebookLM's final summary was pasted (no working); full reply requested. (c) inside tolerance by 0.03 MW.
Quizlet 413800 kg/h 0.337 46.6 MW 0%
Why: Feedwater entering the mercury condenser taken as saturated liquid at 18 bar (884.5) instead of the pump exit (139.6); turbine powers mis-scaled (35.5 MW mercury).
Knowt no final answer no final answer no final answer 0%
Why: Method outline only; used h_g at 0.04 bar as the isentropic exit. Reply text starts 'o analyze' (first character lost when copied).

Problem 7 · hard

Exergy destroyed while vaporizing water by steam injection

availability and irreversibility

An insulated vertical piston–cylinder device initially contains 12 kg of water, 9 kg of which is vapour. The piston maintains a constant pressure of 400 kPa. Steam from a supply line at 3 MPa and 450°C is admitted until all the liquid in the cylinder has just vaporized. The surroundings are at 25°C and 100 kPa. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the mass of steam that entered, in kg; (b) the final volume of the cylinder, in m³; (c) the boundary work done by the steam, in kJ; (d) the exergy destroyed during the process, in kJ.

Adapted from Çengel 9e Prob. 8–76 (numbers and wording changed)

App (a) mass of steam admitted(b) final volume(c) boundary work(d) exergy destroyed Score
Answer key 10.55 kg ±2%10.43 m³ ±2%2,504 kJ ±2%3,981 kJ ±2%
Turbo AI 10.58 kg 10.44 m³ 2510 kJ 3970 kJ 100%
Solvely 10.55 kg 10.43 m³ 2506 kJ 3981 kJ 100%
NotebookLM 10.55 kg 10.43 m³ 2505 kJ 3980.5 kJ 100%
ChatGPT 10.56 kg 10.434 m³ 2507 kJ 3982 kJ 100%
Claude 10.55 kg 10.43 m³ 2505 kJ 3981 kJ 100%
MyStudySidekick 10.5476 kg 10.4256 m³ 2504.38 kJ 3980.81 kJ 100%
Quizlet 3.97 kg 7.38 m³ 1286 kJ 4651 kJ 0%
Why: Energy balance with m1·u2 in place of m1·u1; s at 3 MPa/450 °C read as 6.8373 (7.0856). Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
Knowt 3 kg 5.55 m³ 556 kJ 967.9 kJ 0%
Why: Took the steam admitted as the liquid mass (3 kg); entropy values are the MERCURY table from problem 006 (0.5397, 0.1352, 0.6925): not a fresh chat or memory carried over.
StudyFetch 5.76 kg 8.21 m³ 1619 kJ 16430 kJ 0%
Why: Energy balance rearranged with u − Pv instead of u + Pv (= h), giving m2 17.76 kg; exergy destroyed 16 430 kJ.

Problem 8 · hard

Irreversibility and second-law efficiency of a steam turbine

availability and irreversibility

Steam expands adiabatically in a turbine from 2.8 MPa and 430°C to 450 kPa and 235°C at a rate of 6.5 kg/s. The dead state is 25°C and 100 kPa. Neglect kinetic and potential energy changes. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the isentropic efficiency of the turbine; (b) the reversible (maximum) work per kg of steam, in kJ/kg; (c) the rate of irreversibility (exergy destruction), in kW; (d) the second-law efficiency of the turbine.

Adapted from Rajput, Example 6.22 (numbers and wording changed)

App (a) isentropic efficiency(b) reversible work(c) irreversibility rate(d) second-law efficiency Score
Answer key 0.7882 ±0.01432.4 kJ/kg ±2%399.1 kW ±4%0.858 ±0.01
StudyFetch 0.781 432.4 kJ/kg 399 kW 0.858 100%
NotebookLM 0.7880 433.02 kJ/kg 404.6 kW 0.8562 100%
ChatGPT 0.7886 432.53 kJ/kg 399.03 kW 0.8581 100%
Claude 0.788 432.4 kJ/kg 399 kW 0.858 100%
MyStudySidekick 0.7882 432.39 kJ/kg 399.12 kW 0.8580 100%
Quizlet 0.748 428.9 kJ/kg 407.0 kW 0.855 75%
Why: s1 6.9455 (7.0573) and s2 7.1555 (7.2633): both ~0.11 low, so the difference (exergy parts) survives but h2s and the isentropic efficiency do not. Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
Knowt no final answer no final answer no final answer no final answer 0%
Why: Used saturated-vapour/liquid values for superheated states (h1 = 2856.3, s2 = 0.9449); no numerical answers.
Turbo AI 0.81 419 kJ/kg 336 kW 0.88 0%
Why: h1 3299 (3302.7) and s1 7.090 (7.057) misread.
Solvely 0.7636 444.84 kJ/kg 478.7 kW 0.8345 0%
Why: Tested app is Solvely. s1 read as 7.1264 (7.0573) and s2 as 7.3734 (7.2633).

Problem 9 · hard

Entropy generation as steam flows between two tanks

entropy

Two rigid tanks are connected by a valve. Tank A is insulated and contains 0.25 m³ of steam at 500 kPa with a quality of 70 percent. Tank B is uninsulated and contains 3 kg of steam at 250 kPa and 280°C. The valve is opened and steam flows from tank A to tank B until the pressure in tank A has dropped to 250 kPa; the valve is then closed. During the process 250 kJ of heat is transferred from tank B to the surroundings at 20°C. Assume the steam remaining in tank A undergoes a reversible adiabatic process. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the final temperature in tank A, in °C; (b) the final temperature in tank B, in °C; (c) the final pressure in tank B, in kPa; (d) the total entropy generated during the process, in kJ/K.

Adapted from Çengel 9e Prob. 7–201 (numbers and wording changed)

App (a) final temperature in A(b) final temperature in B(c) final pressure in B(d) entropy generated Score
Answer key 127.4 °C ±0.5 °C174.4 °C ±2.5 °C230.2 kPa ±3%0.5686 kJ/K ±5%
ChatGPT 127.41 °C 174.44 °C 230.26 kPa 0.5685 kJ/K 100%
Claude 127.4 °C 174.4 °C 230 kPa 0.569 kJ/K 100%
MyStudySidekick 127.41 °C 174.43 °C 230.24 kPa 0.5686 kJ/K 100%
Turbo AI 127.4 °C 172 °C 230 kPa 0.6 kJ/K 75%
Why: Rounded table values; S_gen given as ≈0.6 (range 0.55–0.65), 5.5 % high.
Solvely 127.4 °C 176.3 °C 225 kPa 0.754 kJ/K 75%
Why: Tested app is Solvely. Snapped P_B to the 225 kPa table row; s_B2 7.428 (7.328) puts S_gen 33 % high.
NotebookLM 127.4 °C 174.6 °C 231.7 kPa 0.476 kJ/K 75%
Why: s_fg at 250 kPa read as 5.4201 (5.4453) and s_B1 as 7.7456 (7.7174): S_gen 16 % low.
Quizlet 127.4 °C 235.5 °C 250 kPa 1.32 kJ/K 25%
Why: Tank B 250 kPa/280 °C read as v 0.9856, u 2762.3 (1.0141, 2776.7); assumed pressure equalisation; inflow enthalpy taken as h of A's initial state. Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
StudyFetch 127.4 °C 127.4 °C 250 kPa 13.32 kJ/K 25%
Why: Inflow to B given tank A's final enthalpy (2044.5), so B ended wet at 127.4 °C; S_gen summed wrongly (13.32).
Knowt no final answer no final answer no final answer no final answer 0%
Why: Wrong saturation values (s_f 0.0017 at 500 kPa); no numerical answers.

Problem 10 · hard

Isentropic expansion then throttling to dry saturated steam

entropy

Steam initially at 1.8 MPa with a dryness fraction of 0.96 expands isentropically to 800 kPa and is then throttled until it is just dry saturated. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Per kg of steam, determine: (a) the dryness fraction at the end of the isentropic expansion; (b) the pressure at the end of throttling, in kPa; (c) the total change in entropy, in kJ/kg·K; (d) the total change in enthalpy, in kJ/kg; (e) the total change in internal energy, in kJ/kg.

Adapted from Rajput, Example 4.63 (numbers and wording changed)

App (a) dryness fraction after expansion(b) final pressure(c) total entropy change(d) total enthalpy change(e) total internal-energy change Score
Answer key 0.904 ±0.0046.986 kPa ±10%2.057 kJ/kg·K ±2%−147.8 kJ/kg ±2%−100.6 kJ/kg ±3%
Turbo AI 0.904 7.1 kPa 2.05 kJ/kg·K -148 kJ/kg -102 kJ/kg 100%
Solvely 0.904 6.85 kPa 2.060 kJ/kg·K -149.3 kJ/kg -102.1 kJ/kg 100%
ChatGPT 0.9039 7.07 kPa 2.057 kJ/kg·K -147.75 kJ/kg -100.54 kJ/kg 100%
Claude 0.904 7.0 kPa 2.057 kJ/kg·K -147.8 kJ/kg -100.6 kJ/kg 100%
MyStudySidekick 0.904 7.0 kPa 2.056 kJ/kg·K -147.7 kJ/kg -100.6 kJ/kg 100%
NotebookLM 0.9041 7.42 kPa 2.0566 kJ/kg·K -146.86 kJ/kg -95.21 kJ/kg 80%
Why: Interpolated u_g and s_g linearly between 5 and 10 kPa: u3 5 kJ/kg high. Reply starts 'ere' (first character lost when copied).
StudyFetch 0.904 175 kPa -0.002 kJ/kg·K -148 kJ/kg -131 kJ/kg 40%
Why: Could not place h_g = 2572 and gave P3 '≈0.15–0.2 MPa (verify from Table A-5)' (graded at 175 kPa); u3 taken at 800 kPa; raw '[comp:math:...' markup visible in the reply.
Quizlet 0.915 660 kPa 0.479 kJ/kg·K -124.1 kJ/kg 24.7 kJ/kg 0%
Why: s_fg at 1.8 MPa read as 4.0306 (3.9776); h_g = 2594 placed at 660 kPa (impossible: h_g ≈ 2760 there).
Knowt no final answer no final answer no final answer no final answer no final answer 0%
Why: Method outline only; no numerical answers.

Problem 11 · hard

Charging a constant-pressure cylinder from a steam line

first law: open (transient) systems

A vertical piston–cylinder device initially contains 0.02 m³ of steam at 600 kPa and 220°C. The frictionless piston maintains a constant pressure of 600 kPa. Steam from a supply line at 1.4 MPa and 380°C is admitted until the volume inside the cylinder triples. Neglect any heat transfer. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the final temperature of the steam in the cylinder, in °C; (b) the mass of steam that entered, in kg; (c) the boundary work done, in kJ.

Adapted from Çengel 9e Prob. 5–131 (numbers and wording changed)

App (a) final temperature(b) mass admitted(c) boundary work Score
Answer key 311.7 °C ±2.5 °C0.08102 kg ±2%24 kJ ±1%
Turbo AI 313 °C 0.081 kg 24 kJ 100%
NotebookLM 311.5 °C 0.0810 kg 24.00 kJ 100%
ChatGPT 311.63 °C 0.08103 kg 24.0 kJ 100%
Claude 311.7 °C 0.0810 kg 24.0 kJ 100%
MyStudySidekick 311.7 °C 0.0810 kg 24 kJ 100%
StudyFetch 308 °C 0.081 kg 24 kJ 67%
Why: Interpolated the energy-balance residual between 300 and 320 °C with rounded table values: T2 3.7 °C low.
Quizlet 298.5 °C 0.0886 kg 24.0 kJ 33%
Why: v1, u1 at 600 kPa/220 °C read as 0.37449, 2658.1 (0.36905, 2672.5); iteration not converged. Reply starts 'o solve'. Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
Solvely 309 °C 0.0793 kg 24.0 kJ 33%
Why: Tested app is Solvely. v1 at 600 kPa/220 °C read as 0.35411 (0.36905), so m1 4 % high; T2 0.2 °C and m_in 0.1 % outside tolerance.
Knowt no final answer no final answer no final answer 0%
Why: Method outline only; no numerical answers.

Problem 12 · hard

Polytropic steam expansion followed by constant-volume cooling

first law: closed systems

A piston–cylinder device contains 2.5 kg of wet steam at 900 kPa with a dryness fraction of 0.97. The steam expands reversibly according to pv^1.08 = constant to 40 kPa. The piston is then locked and the steam is cooled at constant volume until its pressure is 20 kPa. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the work done during the expansion, in kJ; (b) the heat transfer during the expansion, in kJ (positive if heat is added); (c) the dryness fraction at the end of the expansion; (d) the heat transfer during the constant-volume cooling, in kJ (negative if heat is removed); (e) the dryness fraction at the end of the cooling.

Adapted from Rajput, Example 4.58 (numbers and wording changed)

App (a) expansion work(b) heat during expansion(c) dryness fraction after expansion(d) heat during cooling(e) final dryness fraction Score
Answer key 1,208 kJ ±2%724.3 kJ ±3%0.9328 ±0.005−2,516 kJ ±2%0.4869 ±0.005
Turbo AI 1209 kJ 724 kJ 0.932 -2516 kJ 0.487 100%
ChatGPT 1207.67 kJ 724.10 kJ 0.9327 -2515.49 kJ 0.4869 100%
Claude 1208 kJ 724 kJ 0.933 -2516 kJ 0.487 100%
MyStudySidekick 1207.65 kJ 724.34 kJ 0.9328 -2515.73 kJ 0.4869 100%
StudyFetch 1209 kJ 725 kJ 0.933 -2337 kJ 0.486 80%
Why: u3 computed with h_fg (2358.3) instead of u_fg (2204.6): Q23 7 % low.
Solvely 1200.93 kJ 701.42 kJ 0.934 -2494.98 kJ 0.488 80%
Why: Tested app is Solvely. v_g at 900 kPa 0.21497 (0.21489) and u_fg at 40 kPa 2148.1 (2158.8): Q12 3.2 % low, just outside ±3 %.
NotebookLM 1207.7 kJ 683.1 kJ 0.9327 -2475.7 kJ 0.4869 80%
Why: u_fg at 40 kPa read as 2141.7 (2158.8): Q12 5.7 % low.
Quizlet 915.6 kJ 827.1 kJ 0.992 -2763.5 kJ 0.518 0%
Why: Arithmetic: 22.5^0.9259 evaluated as ~19 (17.86), so v2 = 3.961 (3.7245); every part follows from it.
Knowt no final answer no final answer no final answer no final answer no final answer 0%
Why: Quality from entropy on a non-isentropic path; Q from Δh; no numerical answers.

Problem 13 · hard

Cooling and dehumidification at 85 kPa

psychrometry

Air at a total pressure of 85 kPa, 34°C and 65 percent relative humidity enters a 35-cm-diameter cooling section at a velocity of 150 m/min. It is cooled by a coil through which chilled water flows; the water temperature rises by 7°C. The air leaves the cooling section saturated at 18°C, and the condensate leaves at 18°C. Take cp of water = 4.18 kJ/kg·K. Because the pressure is not 1 atm, do not use a standard psychrometric chart: use the ideal-gas relations with saturation pressures from steam tables, h = 1.005T + ω(2500.9 + 1.82T) kJ/kg dry air (T in °C) and R_air = 0.287 kJ/kg·K. Determine: (a) the rate of heat removal from the air, in kJ/min; (b) the mass flow rate of the chilled water, in kg/min; (c) the exit velocity of the air, in m/min; (d) the rate of condensate removal, in kg/min.

Adapted from Çengel 9e Prob. 14–79/14–81 (numbers and wording changed)

App (a) heat removal rate(b) chilled-water flow rate(c) exit velocity(d) condensate rate Score
Answer key 583.2 kJ/min ±2%19.93 kg/min ±2%139.8 m/min ±2%0.1458 kg/min ±3%
Quizlet 580.5 kJ/min 19.8 kg/min 140 m/min 0.146 kg/min 100%
Turbo AI 585 kJ/min 20.0 kg/min 140 m/min 0.146 kg/min 100%
Solvely 583.06 kJ/min 19.93 kg/min 139.79 m/min 0.146 kg/min 100%
NotebookLM 582.9 kJ/min 19.92 kg/min 139.8 m/min 0.1456 kg/min 100%
ChatGPT 583.25 kJ/min 19.93 kg/min 139.79 m/min 0.1458 kg/min 100%
Claude 583 kJ/min 19.9 kg/min 140 m/min 0.146 kg/min 100%
MyStudySidekick 583.2 kJ/min 19.93 kg/min 139.8 m/min 0.1458 kg/min 100%
Knowt 11663.42 kJ/min 398.88 kg/min 214.29 m/min 2.04 kg/min 0%
Why: Saturation pressures 10x too high (56.2 kPa at 34 °C, 6.2 kPa at 18 °C) and the exit density computed at '20 kPa' — a pressure from problem 012: context carried between problems.
StudyFetch 608 kJ/min 20.8 kg/min 13.7 m/min 0.152 kg/min 0%
Why: Dry-air flow computed with the TOTAL pressure (85 kPa) instead of the dry-air partial pressure (81.54 kPa): 13.92 vs 13.35 kg/min, so (a), (b), (d) 4.3 % high; exit velocity 13.7 m/min (a tenfold drop, explained as physical) from a botched ratio. Reply ends mid-sentence.

Problem 14 · hard

Cooling tower with fan power

psychrometry

A cooling tower cools 7.2 kg/s of water that enters at 42°C. A fan draws 11 m³/s of atmospheric air (volume measured at the inlet state) through the tower and absorbs 5.5 kW, all of which ends up in the streams. The air enters at 20°C and 55 percent relative humidity and leaves saturated at 28°C. The total pressure is 100.5 kPa throughout. Use the ideal-gas relations with saturation pressures from steam tables, h = 1.005T + ω(2500.9 + 1.82T) kJ/kg dry air (T in °C), R_air = 0.287 kJ/kg·K, and take the water enthalpies as saturated-liquid values at their temperatures. Determine: (a) the mass flow rate of dry air, in kg/s; (b) the make-up water required, in kg/s; (c) the temperature of the water leaving the tower, in °C.

Adapted from Rajput, Example 10.14 (numbers and wording changed)

App (a) dry-air mass flow rate(b) make-up water(c) exit water temperature Score
Answer key 12.97 kg/s ±2%0.211 kg/s ±3%21.41 °C ±0.6 °C
Quizlet 12.97 kg/s 0.211 kg/s 21.8 °C 100%
StudyFetch 13.1 kg/s 0.214 kg/s 21.1 °C 100%
Turbo AI 12.97 kg/s 0.211 kg/s 21.4 °C 100%
Solvely 12.97 kg/s 0.211 kg/s 21.4 °C 100%
NotebookLM 12.97 kg/s 0.211 kg/s 21.4 °C 100%
ChatGPT 12.97 kg/s 0.2110 kg/s 21.41 °C 100%
Claude 12.97 kg/s 0.211 kg/s 21.4 °C 100%
MyStudySidekick 12.9715 kg/s 0.2110 kg/s 21.4125 °C 100%
Knowt 13.51 kg/s 31.63 kg/s 18 °C 0%
Why: Saturation pressures 10x too high (23.4 kPa at 20 °C, 36.5 at 28 °C); air density from total pressure; make-up from an energy balance.

Problem 15 · hard

Mixing CO₂ and N₂ from two tanks with heat loss

gas mixtures

An insulated tank containing 1.5 kg of CO₂ at 30°C and 250 kPa is connected by a valve to an uninsulated 3-m³ tank containing N₂ at 60°C and 450 kPa. The valve is opened and the two gases form a homogeneous mixture that comes to 20°C, the temperature of the surroundings. Treat both gases as ideal gases with constant specific heats at room temperature from Çengel Table A-2a (CO₂: R = 0.1889, cp = 0.846, cv = 0.657; N₂: R = 0.2968, cp = 1.039, cv = 0.743, all kJ/kg·K; molar masses 44.01 and 28.01 kg/kmol). Determine: (a) the final pressure, in kPa; (b) the heat transferred to the surroundings, in kJ; (c) the entropy generated, including the surroundings, in kJ/K; (d) the exergy destroyed, in kJ (T₀ = 20°C).

Adapted from Çengel 9e Prob. 13–56 (numbers and wording changed)

App (a) final pressure(b) heat lost(c) entropy generated(d) exergy destroyed Score
Answer key 380.2 kPa ±2%415.6 kJ ±2%1.172 kJ/K ±3%343.7 kJ ±3%
StudyFetch 380 kPa 415.6 kJ 1.17 kJ/K 344 kJ 100%
Turbo AI 380 kPa 415.6 kJ 1.17 kJ/K 344 kJ 100%
Solvely 380.1 kPa 415.6 kJ 1.173 kJ/K 343.8 kJ 100%
NotebookLM 380.1 kPa 415.6 kJ 1.171 kJ/K 343.4 kJ 100%
ChatGPT 380.12 kPa 415.62 kJ 1.173 kJ/K 343.78 kJ 100%
Claude 380 kPa 416 kJ 1.172 kJ/K 344 kJ 100%
MyStudySidekick 380.1 kPa 415.6 kJ 1.173 kJ/K 343.7 kJ 100%
Quizlet 380.1 kPa 417.2 kJ 0.703 kJ/K 206.2 kJ 50%
Why: ΔS_N2 with an inverted temperature log and the volume ratio V_B/V_tot: −1.332 (−0.857) kJ/K.
Knowt no final answer no final answer no final answer no final answer 0%
Why: Molar masses in g/mol used as kg/kmol (34.05 kmol CO2, 1694 kmol total); no numerical answers.

Problem 16 · hard

Steam tank blow-down through a turbine

first law: transient (integration)

An insulated rigid tank of 1.8 m³ contains superheated steam at 4.5 MPa and 480°C. A valve is opened and steam flows out of the tank through an adiabatic turbine that exhausts to a line at 100 kPa. The turbine has an isentropic efficiency of 80 percent at every instant. The valve is closed when the tank pressure has fallen to 600 kPa. Assume the steam that remains in the tank undergoes a reversible adiabatic process, and neglect kinetic and potential energies. Use steam tables (IAPWS-95, e.g. Çengel Tables A-4 to A-7). Determine: (a) the mass of steam that left the tank, in kg; (b) the final temperature of the steam in the tank, in °C; (c) the total work produced by the turbine, in kJ; (d) the total entropy generated during the process, in kJ/K.

Adapted from Çengel 9e Prob. 7–199E/7–200E (method extended) (numbers and wording changed)

App (a) mass discharged(b) final tank temperature(c) total turbine work(d) total entropy generated Score
Answer key 19.14 kg ±2%200.7 °C ±1.5 °C9,820 kJ ±3%6.583 kJ/K ±5%
ChatGPT 19.14 kg 200.65 °C 9818 kJ 6.58 kJ/K 100%
Claude 19.14 kg 200.7 °C 9820 kJ 6.58 kJ/K 100%
MyStudySidekick 19.1441 kg 200.72 °C 9820.4 kJ 6.57 kJ/K 100%
Turbo AI 18.9 kg 202 °C 8500 kJ 6.4 kJ/K 75%
Why: Right method (constant h_es, ∫h dm = U1 − U2, S_gen integrated over pressure steps) but u1 misread 2999 (3060): W 13 % low.
NotebookLM 18.96 kg 185.9 °C 10044 kJ 6.53 kJ/K 75%
Why: s at 600 kPa/200 °C read as 7.0593 (6.968): T2 interpolated wrongly; ∫h dm method used; S_gen from an average exhaust state (inside tolerance).
StudyFetch 19.14 kg 200.8 °C 9080 kJ -87 kJ/K 50%
Why: Average-inlet-enthalpy shortcut (the designed trap): W 13 200 then 'corrected' to 9080 kJ; S_gen negative (−87 kJ/K), acknowledged as impossible, then asked the student for table values. Wolfram|Alpha debris in the reply.
Solvely 18.49 kg 182.4 °C 9928 kJ 6.65 kJ/K 50%
Why: Tested app is Solvely. s1 misread 6.9405 (6.9716); used the ∫h dm = U1 − U2 method; S_gen from an average exhaust state (inside tolerance).
Quizlet 18.11 kg 162.5 °C 6058 kJ 0 kJ/K 0%
Why: s1 read as 6.8406 (6.9716); average-state shortcut (h_avg ≈ 2900); claimed S_gen = 0 for an 80 % turbine. Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
Knowt no final answer no final answer no final answer no final answer 0%
Why: v1 = 0.001206 m³/kg (a liquid value) giving 1492 kg; no numerical answers.

Problem 17 · hard

Adiabatic flame temperature with CO₂ and H₂O dissociation

chemical equilibrium and combustion

Methane (CH₄) at 25°C is burned at a constant pressure of 1 atm in an adiabatic combustor with 10 percent excess oxygen supplied as oxygen-enriched air of composition 30 percent O₂ and 70 percent N₂ by mole, also at 25°C. The products leave as an equilibrium mixture of CO₂, CO, H₂O, H₂, O₂ and N₂ only (the reactions to consider are CO₂ ⇌ CO + ½O₂ and H₂O ⇌ H₂ + ½O₂; neglect OH, O, H, NO and N₂ dissociation). Use ideal-gas enthalpies (Çengel Tables A-18 to A-26 or the equivalent JANAF data) and equilibrium constants from Çengel Table A-28. Determine: (a) the adiabatic flame temperature, in K; (b) the mole fraction of CO in the products (wet basis, water vapour included), in percent; (c) the mole fraction of H₂ in the products (wet basis), in percent; (d) the adiabatic flame temperature that complete combustion (no dissociation) would give, in K.

Adapted from Çengel 9e ch. 16 (dissociation with simultaneous reactions) (numbers and wording changed)

App (a) adiabatic flame temperature with dissociation(b) CO mole fraction(c) H2 mole fraction(d) flame temperature without dissociation Score
Answer key 2,540 K ±30 K2.213 % ±15%0.8335 % ±20%2,728 K ±30 K
NotebookLM 2551 K 2.50 % 0.77 % 2756 K 100%
ChatGPT 2542.7 K 2.215 % 0.8413 % 2732.1 K 100%
Claude 2540 K 2.21 % 0.83 % 2728 K 100%
MyStudySidekick 2540 K 2.22 % 0.83 % 2728 K 100%
Solvely 2471 K 1.33 % 0.75 % 2726 K 50%
Why: Tested app is Solvely. Complete-combustion temperature right; the coupled equilibrium converged 69 K low (no K values shown).
Turbo AI 2610 K 1.2 % 0.44 % 2707 K 25%
Why: Equilibrium written as Kp = b·e^½/a, omitting the 1/N_total factor ((P/N)^½ at 1 atm): dissociation about half the true extent, T_ad 70 K high. (d) given as 'T ≈ 2705–2710 K', graded at 2707. Ends with an app upsell (Generate Lesson).
Quizlet 2255 K 0.42 % 0.14 % 2310 K 0%
Why: Invented K values and compositions; even the no-dissociation temperature ~420 K low. Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
Knowt no final answer no final answer no final answer no final answer 0%
Why: Method outline only; no numerical answers.
StudyFetch 2305 K 1.35 % 3.38 % 2632 K 0%
Why: Equilibrium constants several times too large (K1 0.0265 at 2200 K vs 0.006 from A-28) and recalled sensible enthalpies; ~20 Wolfram|Alpha calls in the reply.

Problem 18 · hard

Two-stage R-134a refrigeration with a flash chamber

refrigeration cycles and exergy

A two-stage compression refrigeration system uses refrigerant-134a and operates between 1.2 MPa and 0.12 MPa. The refrigerant leaves the condenser as saturated liquid and is throttled to a flash chamber at 0.4 MPa. The vapour formed in the flash chamber is mixed with the refrigerant leaving the low-pressure compressor, and the mixture is compressed to 1.2 MPa by the high-pressure compressor. The saturated liquid from the flash chamber is throttled to 0.12 MPa and leaves the evaporator as saturated vapour. Both compressors have an isentropic efficiency of 82 percent. The system removes 30 kW from a refrigerated space kept at −15°C and rejects heat to surroundings at 25°C. Use R-134a tables (Çengel Tables A-11 to A-13). Determine: (a) the fraction of the refrigerant that vaporizes in the flash chamber; (b) the mass flow rate of refrigerant through the condenser, in kg/s; (c) the total compressor power, in kW; (d) the coefficient of performance; (e) the total rate of exergy destruction, in kW, taking the system boundary at the temperatures of the refrigerated space (−15°C) and the surroundings (25°C), so that the irreversibilities of the heat transfers to and from the refrigerant are included.

Adapted from Çengel 9e Example 11–5 (numbers and wording changed)

App (a) flash fraction(b) condenser mass flow rate(c) total compressor power(d) COP(e) exergy destruction rate Score
Answer key 0.281 ±0.0060.2412 kg/s ±2%12.23 kW ±3%2.453 ±3%7.582 kW ±5%
Solvely 0.281 0.241 kg/s 12.28 kW 2.44 7.64 kW 100%
NotebookLM 0.2819 0.2411 kg/s 12.21 kW 2.457 7.56 kW 100%
ChatGPT 0.2809 0.2411 kg/s 12.24 kW 2.452 7.588 kW 100%
Claude 0.281 0.241 kg/s 12.2 kW 2.45 7.58 kW 100%
MyStudySidekick 0.2810 0.2412 kg/s 12.2307 kW 2.4528 7.5823 kW 100%
Quizlet 0.281 0.241 kg/s 11.36 kW 2.64 4.2 kW 40%
Why: HP-compressor inlet entropy taken as ≈0.94 (0.9554), so h4s too low; exergy value unsupported. Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
Turbo AI 0.281 0.241 kg/s 11.4 kW 2.64 6.7 kW 40%
Why: HP isentropic exit at 1.2 MPa read as 284.7 (287.67 kJ/kg): HP work 18 % low, so power, COP and exergy out of tolerance; method and exergy boundary right. Ends with an app upsell (lesson).
Knowt no final answer no final answer no final answer no final answer no final answer 0%
Why: Method outline only; no numerical answers.
StudyFetch 0.951 4.02 kg/s 146.3 kW 0.205 151.0 kW 0%
Why: Flash-chamber balance inverted (95 % vapour) and h_f at 0.4 MPa read as 212.54; then suggested the stated flash pressure is a typo and offered to change it.

Problem 19 · hard

Wet-bulb psychrometrics and evaporative cooling at 84 kPa

psychrometry (no relations given)

At a high-altitude site the atmospheric pressure is 84 kPa. A sling psychrometer reads a dry-bulb temperature of 38°C and a wet-bulb temperature of 21°C; take the wet-bulb temperature as equal to the adiabatic saturation temperature. Do not use a psychrometric chart (it is drawn for 101.325 kPa). Determine: (a) the humidity ratio of the air, in kg water/kg dry air; (b) the relative humidity, in percent; (c) the dew-point temperature, in °C. This air then passes through an adiabatic evaporative cooler and leaves at 80 percent relative humidity (neglect the enthalpy of the water added). Determine: (d) the temperature of the air leaving the cooler, in °C; (e) the water added, in kg per kg of dry air.

Adapted from Çengel 9e Example 14–3, Probs. 14–30 to 14–32 (numbers and wording changed)

App (a) humidity ratio(b) relative humidity(c) dew-point temperature(d) exit temperature(e) water added Score
Answer key 0.01187 kg/kg ±3%23.71 % ±1 %13.75 °C ±0.5 °C23.55 °C ±0.5 °C0.005832 kg/kg ±5%
Turbo AI 0.0118 kg/kg 23 % 13.7 °C 23.7 °C 0.006 kg/kg 100%
Solvely 0.0119 kg/kg 23.7 % 13.7 °C 23.6 °C 0.0059 kg/kg 100%
NotebookLM 0.01186 kg/kg 23.7 % 13.73 °C 23.5 °C 0.00584 kg/kg 100%
ChatGPT 0.01187 kg/kg 23.71 % 13.74 °C 23.55 °C 0.005833 kg/kg 100%
Claude 0.0119 kg/kg 23.7 % 13.7 °C 23.5 °C 0.0058 kg/kg 100%
MyStudySidekick 0.01186 kg/kg 23.7 % 13.7 °C 23.5 °C 0.00583 kg/kg 100%
Quizlet 0.01158 kg/kg 23.2 % 13.3 °C 21.6 °C 0.00812 kg/kg 60%
Why: Adiabatic-saturation step right within tolerance; exit iteration miscalculated (h2 at 23 °C given as 76.9, true ≈68). Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
Knowt 0.0189 kg/kg 40.1 % 20.47 °C 26 °C 0.009 kg/kg 0%
Why: Took the air as saturated at the wet-bulb temperature; Psat(38 °C) 6.21 (6.63); invented dew-point formula.
StudyFetch 0.0190 kg/kg 37.5 % 21 °C 22 °C no final answer 0%
Why: Took the air as saturated at the wet-bulb temperature, then declared (d)–(e) a 'physical impossibility' and asked the student to double-check the wet-bulb value; no water-added value given.

Problem 20 · hard

Converging–diverging nozzle with a normal shock

compressible flow

Air flows through a converging–diverging nozzle from a large reservoir at 900 kPa and 650 K. The throat area is 30 cm² and the exit area is 75 cm². The back pressure is 600 kPa, and a normal shock stands in the diverging section. Assume one-dimensional isentropic flow everywhere except across the shock, and treat air as an ideal gas with k = 1.4, R = 0.287 kJ/kg·K and cp = 1.005 kJ/kg·K. Determine: (a) the mass flow rate, in kg/s; (b) the cross-sectional area at the location of the shock, in cm²; (c) the Mach number just upstream of the shock; (d) the Mach number at the nozzle exit; (e) the entropy generated per unit mass of air, in kJ/kg·K.

Adapted from Çengel 9e ch. 17 (normal shock in a C-D nozzle) (numbers and wording changed)

App (a) mass flow rate(b) shock-location area(c) upstream Mach number(d) exit Mach number(e) entropy generated Score
Answer key 4.28 kg/s ±1%50.41 cm² ±2%1.995 ±0.020.3432 ±0.010.09298 kJ/kg·K ±5%
NotebookLM 4.280 kg/s 50.15 cm² 1.989 0.3515 0.09185 kJ/kg·K 100%
ChatGPT 4.2804 kg/s 50.41 cm² 1.995 0.3432 0.09298 kJ/kg·K 100%
Claude 4.28 kg/s 50.4 cm² 1.99 0.343 0.0930 kJ/kg·K 100%
MyStudySidekick 4.2804 kg/s 50.4091 cm² 1.9949 0.3432 0.0930 kJ/kg·K 100%
Quizlet 4.28 kg/s 55.5 cm² 2.15 0.35 0.115 kJ/kg·K 40%
Why: No real iteration: M1 asserted as ≈2.15. Ends offering 'Biol 1201 Exam 2 Review' — account memory still on.
StudyFetch 4.28 kg/s 39 cm² 2.44 2.44 0.373 kJ/kg·K 20%
Why: Put the shock at the exit area ratio (M 2.44), gave a supersonic exit, then asked the student to 'double-check that back pressure' — the problem is consistent. Long Wolfram|Alpha debris in the reply.
Turbo AI 4.28 kg/s 51.7 cm² 2.03 0.30 0.099 kJ/kg·K 20%
Why: Right iteration scheme, but the hand trials mis-evaluated the exit pressure (610 kPa at M1 = 2.0; ≈597 true), converging at M1 ≈ 2.03. Ends with an app upsell (Generate Lesson).
Solvely 4.28 kg/s 44.2 cm² 1.83 0.388 0.0863 kJ/kg·K 20%
Why: Tested app is Solvely. Right method (PeAe/(P0A*) = 1.667), but solved it as Me = 0.388 (0.343); every later part follows.
Knowt 0.128 kg/s no final answer no final answer no final answer no final answer 0%
Why: Throat density 2910 kg/m³ (kPa/kJ units mix) and sound speed 14.7 m/s; no further answers.