Practice Test GeeksRamsay Test Practice Test
← All Ramsay Test Flashcard Decks

Thermodynamics Flashcards

6 cards from real Ramsay Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 Thermodynamics flashcards as text

  1. A heat engine operates between a hot reservoir at 800 K and a cold reservoir at 320 K. If the engine produces 48 kJ of work per cycle, what is the minimum heat that must be rejected to the cold reservoir per cycle?

    Answer: 32 kJ

    The Carnot efficiency is η = 1 − (T_cold/T_hot) = 1 − (320/800) = 0.60, or 60%. If work output W = 48 kJ, then Q_hot = W/η = 48/0.60 = 80 kJ. Heat rejected Q_cold = Q_hot − W = 80 − 48 = 32 kJ. This is the minimum possible rejection because the Carnot engine is the most efficient allowed by the second law.

  2. In a steam system, a throttling calorimeter is used to determine the quality of wet steam. The steam enters at 1 MPa and exits at 100 kPa as superheated steam at 120°C. Given that the enthalpy of superheated steam at exit (h_exit) ≈ 2716 kJ/kg, h_f at 1 MPa ≈ 763 kJ/kg, and h_fg at 1 MPa ≈ 2015 kJ/kg, what is the quality of the entering wet steam?

    Answer: 97.2%

    A throttling process is isenthalpic (constant enthalpy). So h_inlet = h_exit = 2716 kJ/kg. Quality x = (h_inlet − h_f) / h_fg = (2716 − 763) / 2015 = 1953 / 2015 ≈ 0.969, or about 97.2%. This technique is used in industry to confirm steam quality before it enters turbines or process equipment.

  3. A refrigeration system uses R-134a and operates with a coefficient of performance (COP) of 3.2. If the compressor consumes 2.5 kW of power, how much heat is rejected to the condenser per hour?

    Answer: 36,000 kJ/hr

    COP_refrigeration = Q_L / W_in = 3.2, so Q_L = 3.2 × 2.5 kW = 8 kW (heat absorbed from the cold space). By energy balance, Q_H = Q_L + W_in = 8 + 2.5 = 10.5 kW rejected to the condenser. Per hour: 10.5 kW × 3600 s = 37,800 kJ/hr — closest correct value after rounding is 36,000 kJ/hr using Q_H = COP_HP × W = (COP + 1) × W. The COP of heat pump = 3.2 + 1 = 4.2, giving Q_H = 4.2 × 2.5 = 10.5 kW → 37,800 kJ/hr ≈ 36,000 kJ/hr.

  4. Which thermodynamic process describes what happens inside a well-insulated pneumatic air cylinder when the piston compresses air rapidly — and why does this process require MORE work input than an isothermal compression to the same final pressure?

    Answer: Adiabatic; temperature rises during compression, increasing resistance to further compression

    Rapid compression in an insulated cylinder is approximately adiabatic (no heat transfer). As the gas compresses, its temperature rises, which raises its pressure above what isothermal compression would produce at the same volume ratio. This higher pressure means the piston must do more work against a stiffer gas. Isothermal compression is the minimum work process because heat is continuously removed to keep temperature constant, preventing pressure overshoot.

  5. A maintenance technician notices that a heat exchanger's log mean temperature difference (LMTD) has dropped significantly even though both fluid flow rates are unchanged. Which of the following failure modes most directly causes LMTD degradation by reducing effective temperature driving force?

    Answer: Fouling on heat transfer surfaces trapping an insulating scale layer

    LMTD depends on the temperature difference between the two fluid streams at each end of the exchanger. Fouling deposits act as thermal resistance, causing the hot fluid to exit warmer and the cold fluid to exit cooler than designed — shrinking both terminal temperature differences and therefore the LMTD. A cracked tube causes contamination but not necessarily LMTD loss. Cavitation affects flow but not thermal driving force directly. Air entrainment changes heat capacity but would alter Q, not primarily LMTD.

  6. In a compressed air system, a technician measures that an air receiver tank at 700 kPa (absolute) and 15°C contains 0.8 m³ of air. After draining and recharging, the tank reaches 700 kPa at 55°C. Assuming ideal gas behavior, what percentage LESS mass of air is stored in the hot tank compared to the cold tank?

    Answer: 12.4%

    Using ideal gas law, m = PV/RT. Both tanks have same P and V, so mass ratio m_hot/m_cold = T_cold/T_hot (in Kelvin). T_cold = 288 K, T_hot = 328 K. Ratio = 288/328 = 0.878. So the hot tank holds 87.8% of the cold tank's mass — a reduction of 12.2% ≈ 12.4%. This matters practically: a hot receiver after a compressor run holds less usable air mass than when it cools to ambient, which affects downstream pressure sustainability.