UPSC ESE Mechanical Engineering Thermodynamics Flashcards

24 sample cards from the Thermodynamics deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

1. Write the air-standard efficiency of the Otto cycle in terms of compression ratio.

   $$\eta_{Otto} = 1 - \frac{1}{r^{\gamma - 1}}$$ where $r = \frac{V_1}{V_2}$ is the compression ratio. Efficiency increases with $r$ and with $\gamma$.

2. List the four processes of the air-standard Diesel cycle and state where heat is added.

   (1) Isentropic compression; (2) Constant-pressure heat addition; (3) Isentropic expansion; (4) Constant-volume heat rejection. Heat is added at constant pressure and rejected at constant volume. It models the compression-ignition (diesel) engine.

3. Write the air-standard efficiency of the Diesel cycle using compression ratio $r$ and cut-off ratio $\rho$.

   $$\eta_{Diesel} = 1 - \frac{1}{r^{\gamma - 1}}\left[\frac{\rho^{\gamma} - 1}{\gamma(\rho - 1)}\right]$$ where $r$ is the compression ratio and $\rho = \frac{V_3}{V_2}$ is the cut-off ratio.

4. For the same compression ratio, compare the efficiencies of the Otto and Diesel cycles.

   For the same compression ratio and heat rejection, $\eta_{Otto} > \eta_{Diesel}$, because the Diesel correction factor $\frac{\rho^{\gamma}-1}{\gamma(\rho-1)} > 1$. However, Diesel engines use much higher compression ratios in practice, so actual diesel efficiency is usually higher.

5. For the same maximum pressure and temperature (same peak conditions), rank the efficiencies of Otto, Diesel, and Dual cycles.

   For the same maximum pressure and heat rejection: $\eta_{Diesel} > \eta_{Dual} > \eta_{Otto}$. (This ordering is the reverse of the same-compression-ratio comparison, where Otto leads.)

6. Describe the Dual (limited-pressure / mixed) cycle and where heat is added.

   The Dual cycle adds heat in two stages: first at constant volume, then at constant pressure, followed by isentropic expansion and constant-volume heat rejection. It models modern high-speed CI engines and lies between the Otto and Diesel cycles. It uses both a pressure ratio $\alpha$ and a cut-off ratio $\rho$.

7. Write the air-standard efficiency of the Dual cycle.

   $$\eta_{Dual} = 1 - \frac{1}{r^{\gamma - 1}}\left[\frac{\alpha \rho^{\gamma} - 1}{(\alpha - 1) + \gamma \alpha (\rho - 1)}\right]$$ where $r$ = compression ratio, $\alpha$ = constant-volume pressure ratio, $\rho$ = cut-off ratio.

8. List the four processes of the ideal Rankine cycle in order with the component for each.

   (1) Isentropic compression in the pump (liquid); (2) Constant-pressure heat addition in the boiler; (3) Isentropic expansion in the turbine; (4) Constant-pressure heat rejection in the condenser. It is the basic vapor power cycle.

9. Write the thermal efficiency of the Rankine cycle in terms of enthalpies.

   $$\eta_{Rankine} = \frac{W_{turbine} - W_{pump}}{Q_{boiler}} = \frac{(h_1 - h_2) - (h_4 - h_3)}{h_1 - h_4}$$ where $h_1$ = turbine inlet, $h_2$ = turbine exit, $h_3$ = pump inlet (sat. liquid), $h_4$ = pump exit.

10. Write the reversible (isentropic) pump work for the Rankine cycle and why it is small.

    $$W_{pump} = v_f (p_2 - p_1)$$ where $v_f$ is the specific volume of saturated liquid at pump inlet. It is small because liquid is nearly incompressible ($v_f$ is very small compared to vapor specific volume).

11. How do superheating, reheating, and regeneration improve the Rankine cycle?

    Superheating raises mean temperature of heat addition (more work, drier turbine exit). Reheating expands steam in stages with re-heating between, raising efficiency and reducing turbine moisture. Regeneration (feedwater heating) preheats feedwater using bled steam, raising the mean temperature of heat addition and thus efficiency.

12. Why is the Rankine cycle preferred over the Carnot cycle for vapor power plants?

    The Carnot cycle is impractical for vapor power because: (1) isentropic compression of a wet liquid–vapor mixture (pumping a two-phase mixture) is difficult, and (2) isothermal heat addition limits the maximum temperature. The Rankine cycle condenses fully to saturated liquid, allowing easy pumping and superheating.

13. List the four processes of the ideal Brayton (Joule) cycle with components.

    (1) Isentropic compression in the compressor; (2) Constant-pressure heat addition in the combustor; (3) Isentropic expansion in the turbine; (4) Constant-pressure heat rejection. It is the air-standard cycle for gas turbines.

14. Write the thermal efficiency of the ideal Brayton cycle in terms of the pressure ratio.

    $$\eta_{Brayton} = 1 - \frac{1}{r_p^{\frac{\gamma - 1}{\gamma}}}$$ where $r_p = \frac{p_2}{p_1}$ is the pressure ratio. Efficiency depends only on pressure ratio and $\gamma$.

15. What is the optimum pressure ratio for maximum net work output in a Brayton cycle between temperature limits $T_{min}$ and $T_{max}$?

    $$r_{p,opt} = \left(\frac{T_{max}}{T_{min}}\right)^{\frac{\gamma}{2(\gamma - 1)}}$$ At this ratio, net specific work is maximum and the compressor and turbine outlet temperatures are equal.

16. Define the back-work ratio for a gas turbine (Brayton) cycle and why it matters.

    Back-work ratio is the fraction of turbine work consumed by the compressor: $$BWR = \frac{W_{compressor}}{W_{turbine}}$$ Gas turbines have a high BWR (often 40–80%), so small inefficiencies in compressor/turbine greatly affect net output — unlike Rankine cycles where pump work is negligible.

17. How do intercooling, reheating, and regeneration affect the Brayton cycle?

    Intercooling (cooling between compressor stages) reduces compressor work. Reheating (between turbine stages) increases turbine work. Both increase net work but alone reduce thermal efficiency unless combined with regeneration. Regeneration uses hot turbine exhaust to preheat compressed air, raising efficiency when turbine exit is hotter than compressor exit.

18. List the four processes of the ideal vapor-compression refrigeration cycle with components.

    (1) Isentropic compression in the compressor (sat. vapor to superheated); (2) Constant-pressure heat rejection in the condenser (to sat. liquid); (3) Throttling (isenthalpic) through the expansion valve; (4) Constant-pressure heat absorption in the evaporator. It is the most common refrigeration cycle.

19. Write the COP of the ideal vapor-compression refrigeration cycle in terms of enthalpies.

    $$\text{COP}_R = \frac{q_L}{w_{comp}} = \frac{h_1 - h_4}{h_2 - h_1}$$ where $h_1$ = evaporator exit / compressor inlet, $h_2$ = compressor exit, $h_4$ = evaporator inlet ($= h_3$, the condenser exit, due to throttling).

20. Why is a throttling valve used instead of a turbine (isentropic expansion) in vapor-compression refrigeration?

    The work recoverable from expanding a liquid is very small and does not justify the cost and complexity of a turbine. A throttling valve is simple, cheap, and reliable, and conveniently controls refrigerant flow. The throttling is isenthalpic: $h_3 = h_4$, increasing irreversibility but accepted for practicality.

21. Compare the Bell–Coleman (reverse Brayton) air refrigeration cycle with the vapor-compression cycle.

    The Bell–Coleman cycle uses air as refrigerant in a reverse Brayton cycle (compression, cooling, expansion in a turbine, heat absorption). It has a low COP and bulky equipment but no phase change; it is used in aircraft because air is freely available and equipment is lightweight. Vapor-compression has higher COP and is standard for most applications.

22. Describe the working principle of a vapor-absorption refrigeration system and a common refrigerant–absorbent pair.

    A vapor-absorption system replaces the mechanical compressor with a generator, absorber, pump, and heat input. Low-grade heat (steam, waste heat) drives the cycle. Common pairs: ammonia–water ($\ce{NH3}$ refrigerant, water absorbent) and lithium bromide–water (water refrigerant, LiBr absorbent). It is quiet and uses heat rather than large work input.

23. Define one tonne of refrigeration (TR).

    One tonne of refrigeration is the rate of heat removal required to freeze one short tonne (2000 lb) of water at $0^{\circ}\text{C}$ into ice at $0^{\circ}\text{C}$ in 24 hours. $$1\ \text{TR} = 3.5\ \text{kW} = 211\ \text{kJ/min} = 3024\ \text{kcal/h}$$

24. Compare a heat pump and a refrigerator: same hardware, different purpose.

    Both use the same vapor-compression cycle. A refrigerator's useful effect is heat removed from the cold space ($Q_L$), so $\text{COP}_R = \frac{Q_L}{W}$. A heat pump's useful effect is heat delivered to the warm space ($Q_H$), so $\text{COP}_{HP} = \frac{Q_H}{W}$. They are related by $\text{COP}_{HP} = \text{COP}_R + 1$, so a heat pump COP always exceeds 1.

What this deck covers

Thermodynamics flashcards FAQ