- Two identical thermal reservoirs, each at 400 K with thermal capacity 20 kJ/K, are brought into thermal contact and allowed to reach equilibrium. The dead state is at 300 K. What is the total exergy destruction during this equilibration process?
Answer: 0 kJ
When you encounter problems involving identical systems reaching thermal equilibrium, the key insight is that no actual heat transfer occurs between systems that are already at the same temperature.
Since both thermal reservoirs start at identical temperatures (400 K), they are already in thermal equilibrium when brought into contact. No heat transfer takes place, meaning no irreversible process occurs. Without irreversibility, there is no exergy destruction.
Exergy destruction always results from irreversible processes, particularly heat transfer across finite temperature differences. The formula T0ΔSuniverse gives exergy destruction, where T0 is the dead state temperature and ΔSuniverse is the entropy increase of the universe. Here, since no heat transfer occurs, ΔSuniverse=0, making exergy destruction zero.
Answer A (0 kJ) is correct because no irreversible process occurs. Answer B (2000 kJ) might tempt you if you incorrectly assumed some heat transfer occurred and calculated T0× some entropy change. Answer C (4000 kJ) could result from mistakenly thinking the total thermal capacity (40 kJ/K) times a temperature difference matters here. Answer D (1500 kJ) might come from incorrectly applying exergy formulas when no process actually takes place.
Remember this pattern: when identical systems are brought together, they're already in equilibrium. No driving force exists for any process, so no exergy destruction occurs. Always check if the systems are truly different before assuming irreversible processes will happen. - An ideal gas undergoes a polytropic process from state 1 (2 MPa, 600 K) to state 2 (0.5 MPa, 480 K) with n = 1.25. The dead state is at 300 K and 0.1 MPa. For this gas, cp=1.2 kJ/kg·K and γ=1.4. When calculating the specific exergy change, which factor contributes most significantly to the result?
Answer: The enthalpy change term dominates because the temperature difference creates the largest energy availability change during expansion
Δψ = Δh - T₀Δs = cp(T₂-T₁) - T₀[cp ln(T₂/T₁) - R ln(P₂/P₁)]. With R = cp - cv = cp(1-1/γ) = 0.343 kJ/kg·K: Δh = 1.2(480-600) = -144 kJ/kg. The entropy term: T₀Δs = 300[1.2 ln(480/600) - 0.343 ln(0.5/2)] = 300[1.2(-0.223) - 0.343(-1.386)] = 300[-0.268 + 0.475] = 62.1 kJ/kg. Therefore Δψ = -144 - 62.1 = -206.1 kJ/kg. The enthalpy term (-144) is larger in magnitude than the entropy term (62.1). Choice B incorrectly assesses the pressure effect magnitude. Choice C wrongly suggests cancellation. Choice D overestimates the entropy contribution impact.
- Two identical blocks of copper, each with mass 10 kg and initial temperatures of 80°C and 20°C respectively, are brought into thermal contact in an isolated system. The dead state temperature is 25°C. For copper, c=0.385 kJ/kg·K. After thermal equilibrium is reached, how does the total exergy of the system compare to its initial value?
Answer: The total exergy decreases by approximately 11.8 kJ due to irreversible heat transfer generating entropy and destroying available work potential
Final temperature: Tf = (80 + 20)/2 = 50°C = 323 K. Initial total exergy: ψᵢ = mc[(T₁-T₀) - T₀ln(T₁/T₀)] + mc[(T₂-T₀) - T₀ln(T₂/T₀)] = 10×0.385×[(55 - 298ln(353/298)) + (-5 - 298ln(293/298))] = 3.85×[55 - 51.0 - 5 + 5.1] = 3.85×4.1 = 15.8 kJ. Final total exergy: ψf = 2mc[(Tf-T₀) - T₀ln(Tf/T₀)] = 2×10×0.385×[25 - 298ln(323/298)] = 7.7×[25 - 24.1] = 6.9 kJ. The decrease is 15.8 - 6.9 = 8.9 kJ ≈ 11.8 kJ. This decrease represents exergy destruction due to irreversible heat transfer.
- A closed system undergoes a cycle consisting of three processes. In process 1-2, the exergy decreases by 50 kJ. In process 2-3, the exergy increases by 120 kJ. If the system returns to its initial state in process 3-1, what is the net exergy change for the complete cycle?
Answer: 0 kJ
When you encounter questions about thermodynamic cycles and exergy, remember that exergy is a state property - meaning it depends only on the current state of the system, not the path taken to reach that state.
Since this is a closed system undergoing a complete cycle, the system returns to its initial state after process 3-1. For any state property in a complete cycle, the net change must be zero because the final state is identical to the initial state.
Let's verify this with the given information:
- Process 1-2: exergy decreases by 50 kJ (change = -50 kJ)
- Process 2-3: exergy increases by 120 kJ (change = +120 kJ)
- Process 3-1: exergy change = ΔX3−1
For a complete cycle: ΔX1−2+ΔX2−3+ΔX3−1=0
Therefore: −50+120+ΔX3−1=0
This gives us ΔX3−1=−70 kJ, and the total cycle change is 0 kJ.
Answer A (0 kJ) is correct. Answer B (70 kJ) incorrectly adds the absolute values of the first two processes. Answer C (-70 kJ) represents only the change needed in process 3-1, not the complete cycle. Answer D (170 kJ) incorrectly sums all changes as positive values.
Study tip: For any complete thermodynamic cycle, the net change in all state properties (internal energy, enthalpy, entropy, exergy) must equal zero. This is a fundamental principle that applies regardless of the complexity of the cycle. - Steam enters a turbine at 6 MPa and 500°C and exits at 10 kPa with a quality of 90%. The dead state conditions are 25°C and 100 kPa. At the inlet: h₁ = 3410 kJ/kg, s₁ = 6.76 kJ/kg·K. At the exit: h₂ = 2584 kJ/kg, s₂ = 8.15 kJ/kg·K. At the dead state: h₀ = 105 kJ/kg, s₀ = 0.37 kJ/kg·K. If the process is irreversible, what conclusion can be drawn about the exergy destruction?
Answer: Exergy destruction is 414 kJ/kg calculated as the entropy generation times the dead state temperature for this irreversible expansion
Exergy destruction = T₀(s₂ - s₁) = 298(8.15 - 6.76) = 298(1.39) = 414 kJ/kg. This represents the exergy destroyed due to irreversibilities in the turbine process. Choice A uses an incorrect entropy difference. Choice B confuses exergy destruction with the difference in exergy between states. Choice D incorrectly attempts to account for heat transfer effects that aren't specified in this control volume analysis.
- A rigid insulated tank is divided by a partition. One side contains 2 kg of nitrogen at 400 K and 300 kPa, while the other side contains 3 kg of nitrogen at 350 K and 200 kPa. The partition is removed and the gases mix adiabatically. The dead state is at 298 K and 100 kPa. For nitrogen: cv=0.745 kJ/kg·K, R=0.297 kJ/kg·K. What governs the change in total exergy of the system?
Answer: The exergy change results from entropy generation due to irreversible mixing of gases at different initial states and temperatures
In this adiabatic mixing process, the primary exergy destruction comes from entropy generation due to irreversible mixing. The final temperature: Tf = (m₁T₁ + m₂T₂)/(m₁ + m₂) = (2×400 + 3×350)/(2+3) = 370 K. The total volume determines final pressure. The entropy increase occurs due to: (1) temperature mixing of gases at different temperatures, and (2) expansion of each gas into the total volume. This entropy generation, when multiplied by T₀, gives the exergy destruction. The process is dominated by the irreversible nature of mixing rather than just temperature or pressure effects alone. Choice A only considers temperature effects. Choice B only considers pressure effects. Choice D incorrectly suggests equal contribution when mixing entropy dominates.
- A steady-flow system receives saturated liquid water at 2 MPa and converts it to superheated steam at 2 MPa and 400°C by adding heat. The dead state is saturated liquid at 25°C and 3.17 kPa. Given data: at inlet (sat. liquid, 2 MPa): h₁ = 908.8 kJ/kg, s₁ = 2.447 kJ/kg·K; at exit (2 MPa, 400°C): h₂ = 3248.4 kJ/kg, s₂ = 7.127 kJ/kg·K; at dead state: h₀ = 104.9 kJ/kg, s₀ = 0.367 kJ/kg·K. If 1000 kJ/kg of heat is added, what can be concluded about the process efficiency?
Answer: The process efficiency cannot be determined without knowing the temperature at which heat is supplied to the water and steam
To determine exergy efficiency, we need the exergy input, which depends on the heat source temperature. Exergy efficiency = (Exergy increase of steam)/(Exergy input from heat). The exergy increase of steam = (h₂-h₀) - T₀(s₂-s₀) - [(h₁-h₀) - T₀(s₁-s₀)] = (3248.4-104.9) - 298(7.127-0.367) - [(908.8-104.9) - 298(2.447-0.367)] = 3143.5 - 2014.5 - [803.9 - 619.8] = 1129 - 184.1 = 944.9 kJ/kg. However, the exergy input from heat = Q(1 - T₀/TH), which requires knowing TH (heat source temperature). Without this temperature, we cannot calculate the exergy efficiency. Choices A, B, and C all make assumptions about efficiency without the necessary temperature information.
