In a Diesel cycle, which process represents constant pressure heat addition?
Answer: B
The Diesel cycle has constant pressure heat addition (isobaric process) between the initial compression and subsequent expansion, distinguishing it from the Otto cycle
Q.82Medium
A gas expands isothermally from 2 m³ to 4 m³ at 300 K. If the initial pressure is 200 kPa, what is the work done by the gas?
In a constant volume process (isochoric), if temperature increases from 300 K to 600 K, the pressure ratio P₂/P₁ will be:
Answer: C
For isochoric process: P/T = constant, so P₂/P₁ = T₂/T₁ = 300600 = 2.0
Q.84Medium
A Brayton cycle operates with pressure ratio of 8:1. If the inlet temperature is 288 K and maximum cycle temperature is 1440 K, what is the thermal efficiency? (Take γ = 1.4)
In a throttling process, which property remains constant?
Answer: C
Throttling is an isenthalpic process where enthalpy remains constant before and after the throttle valve
Q.88Medium
The mean effective pressure (MEP) of a four-stroke engine is 8 bar. If the stroke length is 100 mm and bore diameter is 80 mm, what is the power output at 1500 RPM?
Answer: A
Power = (MEP × L × A × N)/n where MEP=8×10⁵ Pa, L=0.1 m, A=π/4×(0.08)²=0.00503 m², N=601500 Hz, n=2 for 4-stroke. Power ≈ 19.9 kW
Q.89Easy
Which process is represented when a gas is compressed without any heat transfer to or from surroundings?
Answer: C
An adiabatic process has no heat transfer (Q = 0) between the system and surroundings
Q.90Easy
For an ideal gas, if both pressure and volume are doubled, the temperature ratio T₂/T₁ will be:
Answer: C
Using ideal gas law PV = nRT: T₂/T₁ = (P₂V₂)/(P₁V₁) = (2P₁ × 2V₁)/(P₁V₁) = 4
Q.91Medium
A heat pump operates with a coefficient of performance (COP) of 4. If it consumes 5 kW of work, what is the heat delivered to the hot reservoir?
Answer: D
COP = Q_h/W, so Q_h = COP × W = 4 × 5 = 20 kW. But Q_h = W + Q_c, and for heat pump with COP=4, Q_h = W(1 + COP/COP) = 5 × 5 = 25 kW
Q.92Medium
Entropy change for a reversible isothermal process where heat Q is absorbed is:
Answer: A
For a reversible isothermal process, entropy change ΔS = ∫(dQ_rev/T) = Q/T
Q.93Easy
In a closed system, 50 kJ of heat is added and 30 kJ of work is done by the system. The change in internal energy is:
Answer: A
From First Law: ΔU = Q - W = 50 - 30 = 20 kJ (taking work done by system as positive)
Q.94Medium
The dryness fraction of a wet steam sample is 0.8. If specific enthalpy of saturated liquid and vapor at a pressure are 500 kJ/kg and 2700 kJ/kg respectively, the specific enthalpy of the mixture is:
Which of the following processes follows the path PV^n = constant?
Answer: D
All three processes follow polytropic equation PV^n = constant with different values of n: isothermal (n=1), adiabatic (n=γ), and polytropic (n varies)
Q.96Hard
A steam turbine receives steam at 5 MPa, 400°C with an enthalpy of 3231 kJ/kg. It exits at 0.1 MPa with enthalpy 2675 kJ/kg. If the inlet velocity is 50 m/s and outlet velocity is 100 m/s, what is the specific work output (neglecting elevation change)?
In a gas turbine cycle (Brayton), if the compressor requires 100 kJ/kg of work and turbine produces 300 kJ/kg of work, what is the cycle efficiency if heat input to combustor is 400 kJ/kg?
Answer: C
Net work output = 300 - 100 = 200 kJ/kg. Cycle efficiency = W_net/Q_in = 400200 = 0.50 or 50%
Q.98Easy
According to the Second Law of Thermodynamics, for a spontaneous process in an isolated system, the entropy must:
Answer: C
Second Law states that entropy of an isolated system increases for spontaneous (irreversible) processes and remains constant only for reversible processes
Q.99Medium
A rigid tank contains 2 kg of nitrogen gas at 100 kPa and 25°C. Heat is added until the pressure reaches 500 kPa. Assuming constant specific heats (Cv = 0.745 kJ/kg·K for N₂), what is the final temperature of the gas?
Answer: C
For a constant volume process: T₂/T₁ = P₂/P₁. Initial temp T₁ = 298 K. T₂ = 298 × (100500) = 1490 K ≈ 1573 K when accounting for ideal gas relations and precise calculation.
