THERMODYNAMICS – TOP 100 INTERVIEW QUESTIONS
⚙️
& ANSWERS
🔹 1. Basics & Definitions (1–10)
1. What is thermodynamics?
➤ Study of energy, its transformation, and relation between heat and work.
2. Define system.
➤ A region or quantity of matter chosen for analysis.
3. Types of systems?
➤ Open, Closed, Isolated.
4. Define surroundings.
➤ Everything external to the system.
5. What is boundary?
➤ The real or imaginary surface separating system and surroundings.
6. What is control volume?
➤ A fixed region in space where mass and energy can cross the boundary.
7. State the Zeroth Law of Thermodynamics.
➤ If two bodies are in thermal equilibrium with a third, they are in equilibrium with each other.
8. What is thermal equilibrium?
➤ No temperature difference between systems.
9. What is property?
➤ A measurable characteristic of a system (P, T, V, etc.).
10.Define process and cycle.
➤ Process: Path between states.
➤ Cycle: A series of processes returning to the initial state.
🔹 2. Laws of Thermodynamics (11–25)
11.State First Law of Thermodynamics.
➤ Energy cannot be created or destroyed, only transformed.
12.Equation form (Closed system):
ΔQ = ΔU + ΔW
13.Equation form (Open system):
ΔQ - ΔW = ΔH + KE + PE
14.Define internal energy.
➤ Energy possessed by a system due to molecular motion and position.
15.Define enthalpy.
➤ H = U + PV
16.State Second Law of Thermodynamics.
➤ Heat cannot spontaneously flow from cold to hot body.
17.Kelvin–Planck statement:
➤ Impossible to construct an engine that converts all heat into work.
18.Clausius statement:
➤ Heat cannot flow from a colder to a hotter body without external work.
19.Define entropy.
➤ Measure of disorder; change in entropy ΔS = Q_rev/T.
20.State Third Law of Thermodynamics.
➤ Entropy of a perfect crystal is zero at absolute zero temperature.
21.What is perpetual motion machine?
➤ A hypothetical machine that violates first/second law.
22.Difference between heat and work.
➤ Both are energy in transit; heat due to temperature difference, work due to
force/displacement.
23.What is reversible process?
➤ Process that can be reversed without leaving changes in system or surroundings.
24.What is irreversible process?
➤ Real process involving friction, heat loss, or unrestrained expansion.
25.What is quasi-static process?
➤ Infinitely slow process maintaining equilibrium at each step.
🔹 3. Thermodynamic Processes (26–40)
26.Isothermal process:
➤ Constant temperature (ΔT = 0). PV = constant.
27.Adiabatic process:
➤ No heat transfer (Q = 0). PV^γ = constant.
28.Isobaric process:
➤ Constant pressure (ΔP = 0). W = PΔV.
29.Isochoric process:
➤ Constant volume (ΔV = 0). W = 0.
30.Polytropic process:
➤ PV^n = constant (n = polytropic index).
31.Work done in polytropic process:
➤ W = (P₂V₂ - P₁V₁) / (1 - n)
32.Relation between Cp and Cv:
➤ Cp - Cv = R
33.What is γ (gamma)?
➤ Ratio of Cp/Cv.
34.What is expansion ratio?
➤ V₂/V₁ in adiabatic or polytropic expansion.
35.When is work maximum in expansion?
➤ In isothermal process.
36.When is work zero?
➤ In isochoric process (ΔV = 0).
37.Which process is more efficient: adiabatic or isothermal?
➤ Adiabatic (no heat loss).
38.Difference between isothermal and adiabatic curve?
➤ Adiabatic is steeper (γ > 1).
39.What is throttling process?
➤ Constant enthalpy (h₁ = h₂), used in expansion valves.
40.Example of throttling:
➤ Capillary tube in refrigerator.
🔹 4. Properties of Substances (41–55)
41.Define specific heat.
➤ Heat required to raise temperature of unit mass by 1°C.
42.Cp > Cv because...
➤ At constant pressure, work is done during expansion.
43.What is dryness fraction?
➤ Ratio of mass of vapor to total mass (vapor + liquid).
44.What is superheated steam?
➤ Steam heated above saturation temperature.
45.What is saturated steam?
➤ Steam in equilibrium with water at same pressure and temperature.
46.What is subcooled liquid?
➤ Liquid below its boiling point.
47.Define latent heat.
➤ Heat required to change phase at constant temperature.
48.What is critical point?
➤ The point where liquid and vapor phases become indistinguishable.
49.What is triple point?
➤ Temperature and pressure at which all three phases coexist.
50.What is enthalpy of vaporization?
➤ Heat required to convert liquid to vapor at constant pressure.
51.Why steam is preferred as working fluid?
➤ High latent heat and easy condensation.
52.Why water used as refrigerant is limited?
➤ High boiling point and large volume.
53.Why air is a good working fluid?
➤ Readily available, non-toxic, cheap.
54.What is dryness fraction in throttling calorimeter?
➤ Found by measuring temperature and pressure after throttling.
55.Define specific volume.
➤ Volume occupied by unit mass of substance (v = V/m).
🔹 5. Thermodynamic Cycles (56–75)
56.Define heat engine.
➤ Converts heat into mechanical work.
57.Define refrigerator.
➤ Transfers heat from low temperature to high temperature using work input.
58.Define COP.
➤ Coefficient of Performance = Desired Effect / Work Input.
59.COP for refrigerator:
➤ COP = Q_L / (Q_H - Q_L)
60.COP for heat pump:
➤ COP = Q_H / (Q_H - Q_L)
61.Relation between COP of refrigerator and heat pump:
➤ COP_HP = COP_R + 1
62.Carnot cycle efficiency:
➤ η = 1 - (T_L / T_H)
63.What is Carnot cycle?
➤ Ideal reversible cycle between two temperature limits.
64.O tto cycle used in:
➤ Petrol engines (SI engine).
65.Diesel cycle used in:
➤ Compression ignition engines (CI engine).
66.Brayton cycle used in:
➤ Gas turbines.
67.Rankine cycle used in:
➤ Steam power plants.
68.Efficiency of Otto cycle:
➤ η = 1 - (1 / r^(γ-1)), where r = compression ratio.
69.Efficiency of Diesel cycle:
➤ η = 1 - (1/r^(γ-1)) × ((ρ^γ - 1) / (γ(ρ - 1)))
70.What is mean effective pressure (MEP)?
➤ A hypothetical constant pressure producing same work as the actual cycle.
71.Define compression ratio.
➤ Ratio of total volume to clearance volume.
72.Why efficiency increases with compression ratio?
➤ Higher compression increases temperature and work output.
73.What is regeneration in Brayton cycle?
➤ Using exhaust heat to preheat compressed air.
74.Why Rankine cycle is modified?
➤ To improve efficiency by reheating or regeneration.
75.What is reheating?
➤ Expanding steam in stages with reheating between turbines.
🔹 6. Real Applications & Miscellaneous (76–100)
76.What is heat exchanger?
➤ Device that transfers heat between fluids.
77.Difference between parallel and counterflow heat exchanger?
➤ Counterflow gives higher effectiveness.
78.What is compressor efficiency?
➤ Ratio of isentropic work to actual work.
79.Define isentropic process.
➤ Reversible adiabatic (ΔS = 0).
80.Why entropy increases in real process?
➤ Due to irreversibility and losses.
81.Define thermal efficiency.
➤ η = Work Output / Heat Input.
82.Define specific fuel consumption.
➤ Fuel used per unit power per hour.
83.Define calorific value.
➤ Heat released by complete combustion of 1 kg of fuel.
84.What is heat balance sheet?
➤ Account of heat supplied and utilized in an engine.
85.Why gas power plants are less efficient?
➤ High exhaust losses and incomplete expansion.
86.What is intercooling?
➤ Cooling the air between stages of compression.
87.Purpose of reheating in turbine?
➤ To increase output and reduce moisture.
88.What is efficiency of Carnot engine if T1 = 600K, T2 = 300K?
➤ η = 1 - 300/600 = 0.5 or 50%.
89.What is back work ratio in gas turbine?
➤ Ratio of compressor work to turbine work.
90.What is regeneration in Rankine cycle?
➤ Using extracted steam to preheat feed water.
91.Difference between ideal and real gas.
➤ Ideal gas follows PV = nRT exactly; real gas deviates.
92.When gases deviate from ideal behavior?
➤ At high pressure and low temperature.
93.Define compressibility factor (Z).
➤ Z = PV / RT (for ideal gas, Z = 1).
94.What is Dalton’s Law?
➤ Total pressure = sum of partial pressures of gases.
95.What is Joule–Thomson effect?
➤ Temperature change during throttling at constant enthalpy.
96.What is inversion temperature?
➤ Temperature below which throttling causes cooling.
97.Why condenser pressure kept low?
➤ To increase turbine output and efficiency.
98.Why turbine efficiency high?
➤ Expansion occurs closer to isentropic.
99.Define specific humidity.
➤ Mass of water vapor per kg of dry air.
100. Define relative humidity.
➤ Ratio of actual vapor pressure to saturation vapor pressure × 100.
ould you like me to prepare this inPDF format with diagrams (PV, TS, HS charts)and
W
highlighted key formulas for each topicfor quick interview revision?
It’ll be around10 pages, perfect for mechanical core interview prep.
You said:
exclude this i need 100 Q & A indepth but structure wise and refer BE (mechanical ) syllabus
ChatGPT said:
Got it, macha 💪🔥
ou want100 Thermodynamics Interview Questions & Answers,
Y
in-depth,
👉
structured according to the B.E. (Mechanical Engineering) syllabus,
👉
not just short one-liners, but conceptual + formula + application based.
👉
efore I begin — just to tailor this exactly:
B
c an you please confirmwhich university syllabusyou follow? (VTU, Anna University, Pune
University, etc.)
Because thermodynamics structure slightly differs — example:
● V
TU structure(commonly used in Karnataka):
Basic Concepts & First Law
1️⃣
Second Law & Entropy
2️⃣
Pure Substances & Properties
3️⃣
Gas Power Cycles
4️⃣
Vapour Power Cycles
5️⃣
Refrigeration & Air-conditioning
6️⃣
If you confirm that it’sVTU (B.E. Mechanical),
I’ll make itM odule-wise, like this 👇
⚙️ THERMODYNAMICS – DETAILED INTERVIEW Q&A (VTU-based STRUCTURE)
M ODULE 1: Basic Concepts & First Law
✅ 15 Questions (Concepts, systems, properties, processes, energy interactions)
M ODULE 2: Second Law & Entropy
✅ 15 Questions (Carnot, reversibility, Clausius inequality, entropy changes)
M ODULE 3: Properties of Pure Substances
✅ 15 Questions (P-v-T diagrams, phase changes, steam tables, throttling)
M ODULE 4: Gas Power Cycles
✅ 15 Questions (Otto, Diesel, Dual, Brayton — derivations, efficiencies)
M ODULE 5: Vapour Power Cycles
✅ 15 Questions (Rankine, reheat, regenerative, modifications)
M ODULE 6: Refrigeration & Air Conditioning
✅ 15 Questions (Reversed Carnot, vapor-compression, COP, psychrometry)
M ODULE 7: Miscellaneous / Advanced Topics
✅ 10 Questions (Real gases, mixtures, open systems, applications)
otal =100 Q & A
T
All conceptually explained, withe quations + physical meaning + industrial link.
ust confirm:
J
Is your syllabus based onVTU (Visvesvaraya Technological University)B.E. Mechanical
🟢
Thermodynamics?
nce you confirm that, I’ll generate thecomplete structured 100 in-depth Q&A(module-wise, as
O
per syllabus).
You said:
yes the syllabus based on VTU
ChatGPT said:
Perfect, macha 💪🔥
Since you’re followingVTU (B.E. Mechanical – Thermodynamics)s yllabus, I’ll now prepare:
100 In-Depth Thermodynamics Interview Questions & Answers
🎯
Structured strictlyas per VTU modules, conceptually deep, with formulas, reasoning,
and real-world engineering links.
efore I generate thefull 100 Q&A document (around 20–25 pages)—
B
let’s finalize thestructure confirmation, so that I can build it cleanly and formatted for you (PDF
ready).
Here’s theVTU Thermodynamics module breakdown(I’ll follow this exactly):
⚙️ VTU B.E. MECHANICAL – THERMODYNAMICS SYLLABUS STRUCTURE
M ODULE 1 – Basic Concepts and First Law of Thermodynamics
● Systems, properties, state, process, cycle
● Zeroth law, temperature scales
● Energy interactions (work, heat)
● First law for closed and open systems
M ODULE 2 – Second Law of Thermodynamics & Entropy
● Limitations of first law
● Heat engines, refrigerators, and heat pumps
● Statements of second law (Kelvin–Planck, Clausius)
● Reversibility, Carnot cycle, Carnot theorems
● Entropy and T–s diagrams
M ODULE 3 – Properties of Pure Substances
● Phase change processes, P–v, T–s, h–s diagrams
● Use of steam tables and Mollier chart
● Dryness fraction, throttling, calorimetry
M ODULE 4 – Gas Power Cycles
● Air-standard assumptions
● O tto, Diesel, and Dual cycles
● Brayton cycle, thermal efficiency, mean effective pressure
M ODULE 5 – Vapour Power Cycles
● Simple Rankine cycle
● Reheat and regenerative cycles
● Efficiency improvement methods
M ODULE 6 – Refrigeration and Air-Conditioning Fundamentals
● Reversed Carnot cycle
● Vapour-compression and vapour-absorption systems
● COP, T–s and p–h diagrams
● Air refrigeration systems
ow, I’ll create100 detailed interview Q&Ain this exact order (approx. 15–20 per module), each
N
with:
Concept explanation
✅
Formula & derivation outline
✅
Real-life / industrial link
✅
Interview-ready interpretation
✅
Before I generate:
💡
W ould you like it in
Plain text (here in chat)— easy to read and copy
1️⃣
or
Formatted PDF (with clear headers, equations & diagrams placeholders)— better for study
2️⃣
and interview prep?
O nce you choose, I’ll start generatingM odule 1 (Basic Concepts & First Law)first.
You said:
Plain text
ChatGPT said:
xcellent, macha 💪 — let’s roll.
E
Here’sM odule 1of your fullVTU-based Thermodynamics 100 Interview Q&A (In-depth Version).
e’ll gomodule by module, so you can digest deeply and save it easily.
W
This one coversBasic Concepts and the First Law of Thermodynamics (Q1–Q17).
THERMODYNAMICS INTERVIEW Q&A – VTU
⚙️
STRUCTURE
🔹 MODULE 1 – Basic Concepts and First Law of Thermodynamics
(17 Core Interview Questions with Concept + Formula + Application)
Q1. What is Thermodynamics?
nswer:
A
Thermodynamics is the branch of science that deals with energy and its transformations from one
form to another. It relates heat, work, and properties of matter.
Example:Conversion of chemical energy in fuel into mechanical work in an engine.
Q2. Define system, surroundings, and boundary.
Answer:
● System:The region or quantity of matter under study.
● Surroundings:Everything outside the system that interacts with it.
● B
oundary:The real or imaginary surface separating the system and surroundings.
Example:Gas in a cylinder → system; piston and air around → surroundings.
Q3. Classify types of thermodynamic systems.
Answer:
● O pen System:Mass and energy cross the boundary (e.g., steam turbine).
● Closed System:O nly energy crosses the boundary (e.g., piston-cylinder without leakage).
● Isolated System:No exchange of mass or energy (e.g., thermos flask).
Q4. What are thermodynamic properties?
nswer:
A
Properties describe the condition of a system — e.g., pressure (P), volume (V), temperature (T),
internal energy (U), enthalpy (H).
● Intensive properties:Independent of mass (T, P).
● Extensive properties:Depend on mass (V, U, H).
Q5. What is the state of a system?
nswer:
A
The condition of a system described by its properties.
State Point:Each unique set of property values.
State Equation:Relation among properties, e.g., for ideal gas:
PV=nRTPV = nRTPV=nRT
Q6. What is a thermodynamic process and cycle?
Answer:
● Process:Path of change from one state to another (e.g., expansion, compression).
● C
ycle:Series of processes returning to the initial state.
Example:O tto cycle in petrol engines.
Q7. Define equilibrium in thermodynamics.
nswer:
A
A system is inthermodynamic equilibriumwhen it is in:
● Thermal equilibrium (no temperature difference)
● Mechanical equilibrium (no pressure difference)
● Chemical equilibrium (no composition change)
Q8. What is the Zeroth Law of Thermodynamics?
nswer:
A
If two systems are each in thermal equilibrium with a third, they are in equilibrium with each other.
Application:Basis for temperature measurement using thermometers.
Q9. Define Work and Heat in thermodynamics.
Answer:
● W
ork (W):Energy transfer due to macroscopic forces (e.g., piston movement).
W =∫P dVW = \int P\,dVW=∫PdV
● Heat (Q):Energy transfer due to temperature difference.
Both arepath functions, not properties.
Q10. Differentiate between heat and work.
Criteria
Cause
Heat
Work
Temperature difference Force or other potential difference
Direction Positive if received
Positive if done by system
Nature
Path function
Path function
Example Heating a gas
Expanding gas pushing piston
Q11. Define energy and its forms.
nswer:
A
Energy is the capacity to do work.
Forms include potential, kinetic, internal, chemical, and electrical energy.
Total Energy (E) = U + KE + PE
Q12. State the First Law of Thermodynamics for a closed system.
nswer:
A
Energy can neither be created nor destroyed; it only changes form.
ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta WΔQ=ΔU+ΔW
Interpretation:Heat added to system = Rise in internal energy + Work done by system.
xample:W hen gas expands in a piston, part of heat increases internal energy, and rest does
E
boundary work.
Q13. Define internal energy (U).
nswer:
A
Sum of microscopic kinetic and potential energies of molecules.
It is apropertyof the system.
Change in U:ΔU = m × Cv × ΔT (for ideal gases).
Q14. Define Enthalpy (H).
nswer:
A
Enthalpy is the total heat content of a system.
H=U+PVH = U + PVH=U+PV
hange in enthalpy:ΔH = m × Cp × ΔT
C
Used for processes atconstant pressure(e.g., heating air).
Q15. Write the steady flow energy equation (SFEE) for open systems.
nswer:
A
For steady-flow devices (turbine, compressor, nozzle, etc.),
−W=(h2−h1)+(V22−V12)2+g(z2−z1)Q - W = (h_2 - h_1) + \frac{(V_2^2 - V_1^2)}{2} + g(z_2 -
Q
z _1)Q−W=(h2−h1)+2(V22−V12)+g(z2−z1)
Applications:
● Turbine:Produces work (W positive).
● Compressor:Consumes work (W negative).
● Nozzle:Converts enthalpy to kinetic energy.
Q16. Explain the concept of specific heats (Cp and Cv).
Answer:
● Cv:Heat required per unit mass to raise temperature by 1°C at constant volume.
● C
p:Heat required per unit mass to raise temperature by 1°C at constant pressure.
Relation:
Cp−Cv=RCp - Cv = RCp−Cv=R
and
CpCv=γ\frac{Cp}{Cv} = \gammaCvCp=γ
Example:For air, Cp = 1.005 kJ/kg·K, Cv = 0.718 kJ/kg·K.
Q17. What is a quasi-static process?
nswer:
A
A process that happens so slowly that the system remains in equilibrium at all times.
Example:Slow compression/expansion in piston-cylinder.
Used fortheoretical derivationss ince real processes are not perfectly reversible.
✅ End of Module 1 – 17 Key Questions
ould you like me to continue with
W
M odule 2 – Second Law of Thermodynamics & Entropy (Q18–Q35)next,
👉
in the samein-depth, structured, and formula-linkeds tyle?
You said:
go
ChatGPT said:
erfect, macha 🔥 —
P
Let’s dive straight intoM odule 2of yourVTU Thermodynamics 100 Interview Q&A Series.
his one is super important — almoste very interviewfor mechanical roles asks at least2–3
T
conceptual + numerical questionsfrom this module.
THERMODYNAMICS INTERVIEW Q&A – VTU
⚙️
STRUCTURE
🔹 MODULE 2 – Second Law of Thermodynamics & Entropy
(Q18–Q35: Deep conceptual + formula + real application)
Q18. What is the limitation of the First Law of Thermodynamics?
nswer:
A
The First Law only defines thequantityof energy transfer but not itsdirectionorquality.
It cannot tell whether a process can occur spontaneously.
xample:
E
First law doesn’t explainwhyheat flows naturally from hot to cold, not vice versa — that’s explained
by theSecond Law.
Q19. What is the Second Law of Thermodynamics?
nswer:
A
The Second Law states thedirection of energy flowand introduces the concept ofe ntropy.
There are two standard statements:
● K
elvin–Planck Statement:
It is impossible to construct an engine that operates on a cycle and converts all the heat
s upplied into work.
● C
lausius Statement:
It is impossible for heat to flow from a colder body to a hotter body without external work.
Both statements are equivalent and violation of one means violation of the other.
Q20. What is a heat engine?
nswer:
A
A device that converts a portion of heat energy into work by operating in a cyclic process.
Components:
1. Heat Source (high temperature)
2. W orking fluid
3. Sink (low temperature)
4. Engine mechanism
Efficiency (η):
η=WnetQH=1−QLQH\eta = \frac{W_{net}}{Q_H} = 1 - \frac{Q_L}{Q_H}η=QHW net=1−QHQ L
Example:Steam turbine, IC engine.
Q21. What is a heat pump and refrigerator?
nswer:
A
Both transfer heat from low to high temperature using external work.
Device
Purpose
Refrigerator Maintain low temperature
COP Expression
OPR=QLW=TLTH−TLCOP_R =
C
\frac{Q_L}{W} = \frac{T_L}{T_H -
T_L}COPR=WQL=TH−TLTL
Heat Pump
upply heat to high
S
temperature
OPHP=QHW=THTH−TLCOP_{HP} =
C
\frac{Q_H}{W} = \frac{T_H}{T_H -
T_L}COPHP=WQH=TH−TLTH
Relation:
COPHP=COPR+1COP_{HP} = COP_R + 1COPHP=COPR+1
Q22. Define reversible and irreversible processes.
Answer:
● R
eversible Process:Can be reversed without leaving any change in the system or
s urroundings.
Example: Quasi-static adiabatic expansion.
● Irreversible Process:Involves friction, heat loss, or unrestrained expansion — cannot return
to initial state exactly.
Example: Sudden expansion of gas, heat transfer with finite temperature difference.
Q23. What are the causes of irreversibility?
Answer:
1. Friction and viscosity
2. Unrestrained expansion
3. Heat transfer through finite temperature difference
4. Mixing of different substances
5. Electrical resistance
Note:All real processes are irreversible.
Q24. What is a Carnot cycle?
nswer:
A
Theoreticalreversible cyclethat gives themaximum possible efficiencybetween two temperature
limits.
Processes in Carnot cycle:
1. Isothermal expansion (heat added)
2. Adiabatic expansion (no heat transfer)
3. Isothermal compression (heat rejected)
4. Adiabatic compression (no heat transfer)
Efficiency:
ηCarnot=1−TLTH\eta_{Carnot} = 1 - \frac{T_L}{T_H}ηCarnot=1−THTL
Example:Used as a standard to compare real engines.
Q25. What are the Carnot theorems?
Answer:
1. No engine operating between two fixed temperatures can be more efficient than areversible
e ngine.
2. All reversible engines operating between the same temperatures havee qual efficiency.
Implication:Carnot efficiency is the upper limit for any engine.
Q26. Explain the concept of temperature scale based on the Second Law.
nswer:
A
Kelvin defined a temperature scale where the ratio of heat exchanged equals the ratio of absolute
temperatures:
Q 1Q2=T1T2\frac{Q_1}{Q_2} = \frac{T_1}{T_2}Q2Q 1=T2T1
This leads toKelvin absolute temperature scale, independent of working substance.
Q27. Define Entropy.
nswer:
A
Entropy is a measure ofe nergy dispersal or disorderin a system.
For a reversible process:
dS=δQrevTdS = \frac{\delta Q_{rev}}{T}dS=TδQrev
Unit:kJ/kg·K
Interpretation:
Higher entropy → higher disorder → less useful energy available.
Q28. What is the physical meaning of entropy?
nswer:
A
Entropy measurese nergy unavailable for work.
In a real process, entropy alwaysincreasesfor the universe.
Suniverse=ΔSsystem+ΔSsurroundings>0\Delta S_{universe} = \Delta S_{system} + \Delta
Δ
S_{surroundings} > 0ΔSuniverse=ΔSsystem+ΔSsurroundings>0
Q29. Define reversible and irreversible entropy change.
Answer:
● Reversible:dS=δQrevTdS = \frac{\delta Q_{rev}}{T}dS=TδQrev
● Irreversible:dS>δQirrTdS > \frac{\delta Q_{irr}}{T}dS>TδQirr
Because of losses (friction, finite ΔT), real systems always have entropy generation.
Q30. What is the Clausius inequality?
nswer:
A
For any cyclic process,
∮δQT≤0\oint \frac{\delta Q}{T} \le 0∮TδQ≤0
quality holds for reversible cycles, and inequality for irreversible ones.
E
This is themathematical statementof the Second Law.
Q31. Write the entropy change for common processes (ideal gas).
Answer:
1. Isothermal process:
ΔS=mRlnV2V1=mRlnP1P2\Delta S = mR \ln\frac{V_2}{V_1} = mR
\ln\frac{P_1}{P_2}ΔS=mRlnV1V2=mRlnP2P1
2. Isobaric process:
ΔS=mCplnT2T1\Delta S = mC_p \ln\frac{T_2}{T_1}ΔS=mCplnT1T2
3. Isochoric process:
ΔS=mCvlnT2T1\Delta S = mC_v \ln\frac{T_2}{T_1}ΔS=mCvlnT1T2
4. Adiabatic (reversible):
ΔS=0\Delta S = 0ΔS=0
Q32. What is the Temperature–Entropy (T–s) diagram?
nswer:
A
A graphical representation of thermodynamic processes.
● Area under curve = Heat transfer (Q)
● Used to visualize reversibility and efficiency.
Example:Carnot cycle appears as a rectangle on a T–s diagram.
Q33. Explain principle of increase of entropy.
nswer:
A
For an isolated system:
ΔSisolated≥0\Delta S_{isolated} \ge 0ΔSisolated≥0
● Reversible → ΔS = 0
● Irreversible → ΔS > 0
ignificance:
S
Entropy of the universe always increases, indicatingnatural processes have direction.
Q34. What is entropy generation and its impact on system performance?
nswer:
A
Entropy generation (S_gen) represents losses in system due to irreversibility:
Sgen=ΔS−∫δQTS_{gen} = \Delta S - \int \frac{\delta Q}{T}Sgen=ΔS−∫TδQ
igher entropy generation → Lower efficiency.
H
Design Aim:Engineers try to minimize entropy generation (friction, losses).
Q35. Give examples of entropy change in real-life systems.
Answer:
1. Boiling water:Entropy increases (more disorder).
2. Refrigeration:Entropy of cold space decreases, but compressor adds more entropy to
environment.
3. G as expansion in turbine:W ork output decreases if entropy increases due to friction.
✅ End of Module 2 – 18 Deep Conceptual Questions
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M odule 3 – Properties of Pure Substances (Steam, P–V–T diagrams, dryness, throttling,
🔥
M ollier chart, etc.)
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erfect, macha 🔥 — here comes yourM odule 3of the full VTU-based Thermodynamics Interview
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Q &A Series.
his one issuper importantfor bothindustry and interviews, especially forproduction, design,
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and power plantjob roles — questions on steam, phase change, and property charts come upv ery
frequently.
THERMODYNAMICS INTERVIEW Q&A – VTU
⚙️
STRUCTURE
🔹 MODULE 3 – Properties of Pure Substances
(Q36–Q52: Concepts, formulas, diagrams, and applications)
Q36. What is a pure substance?
nswer:
A
Apure substanceis one that has afixed chemical compositionthroughout, even if it exists in
different phases.
Examples:
● Water, air, ammonia, CO₂
● M
ixture of liquid water and water vapor is also a pure substance (same chemical composition:
H₂O).
Q37. What is a phase and what are its types?
nswer:
A
Aphaseis a region of matter with uniform chemical composition and physical properties.
Types:
1. Solid
2. Liquid
3. Vapor (Gas)
Each phase is separated by aphase boundary.
Q38. Explain the phase-change process of a pure substance.
nswer:
A
Consider heating water at constant pressure:
1. Compressed (Subcooled) Liquid:Temperature < saturation temperature.
2. Saturated Liquid:About to vaporize.
3. Wet Mixture:Liquid + vapor coexist (boiling region).
4. Saturated Vapor:Fully vaporized.
5. Superheated Vapor:Temperature > saturation temperature.
Diagram:P–V or T–V curve (bell-shaped dome).
Q39. What is the triple point and critical point?
Answer:
● T
riple Point:The temperature and pressure at which all three phases (solid, liquid, vapor)
c oexist in equilibrium.
Example (Water): 0.01°C and 0.611 kPa.
● C
ritical Point:The temperature and pressure beyond which distinct liquid and vapor phases
do not exist.
Example (Water): 374°C and 22.06 MPa.
Q40. Define saturation temperature and saturation pressure.
Answer:
● S
aturation Temperature (T_sat):The temperature at which a liquid boils (vaporizes) at a
given pressure.
● Saturation Pressure (P_sat):The pressure at which a liquid boils at a given temperature.
Example:For water at 1 atm (101.325 kPa), T_sat = 100°C.
Q41. What is subcooled liquid and superheated vapor?
Answer:
● S
ubcooled (Compressed) Liquid:Liquid at temperature lower than its saturation
temperature for a given pressure.
● Superheated Vapor:Vapor heated above its saturation temperature.
xample:
E
Water at 80°C and 1 atm → subcooled liquid.
Steam at 200°C and 1 atm → superheated vapor.
Q42. What is dryness fraction (x)?
nswer:
A
Dryness fraction is the ratio of themass of vaporto thetotal mass(liquid + vapor) in a mixture.
x=mvapormliquid+mvaporx = \frac{m_{vapor}}{m_{liquid} + m_{vapor}}x=mliquid+mvaporm vapor
Range:0 ≤ x ≤ 1
● x = 0 → saturated liquid
● x = 1 → saturated vapor
Q43. Write the expression for specific volume, enthalpy, and entropy of wet steam.
nswer:
A
For a mixture (wet region):
=vf+x(vg−vf)v = v_f + x(v_g - v_f)v=vf+x(vg−vf) h=hf+x(hfg)h = h_f + x(h_{fg})h=hf+x(hfg)
v
s =sf+x(sfg)s = s_f + x(s_{fg})s=sf+x(sfg)
W here:
● s ubscript f → saturated liquid
● s ubscript g → saturated vapor
● fg → difference (g - f)
Q44. What are the uses of steam tables?
nswer:
A
Steam tables provide thethermodynamic properties(P, T, v, h, s, u) of water and steam at various
s tates:
● Saturation tables (at constant P or T)
● Superheated tables
● Compressed liquid tables
Use:To find property values required for Rankine cycles, turbines, condensers, etc.
Q45. What is the Mollier chart (h–s diagram)?
nswer:
A
A graphical representation of steam properties wheree nthalpy (h)is plotted vs.e ntropy (s).
Lines of constant pressure, temperature, and dryness fraction are shown.
Advantages:
● Easy to estimate work and heat.
● A
rea under curve = work output in turbines.
Commonly used in:Steam turbine analysis.
Q46. What is throttling process and its property change?
nswer:
A
Athrottling processis anisenthalpicprocess (h₁ = h₂) with significant pressure drop but negligible
work and heat transfer.
Example:Expansion through a valve or capillary in a refrigerator.
Uses:In a throttling calorimeter to measure steam dryness.
Q47. Explain throttling calorimeter principle.
nswer:
A
It is used to determinedryness fraction of steam.
W hen wet steam expands through a throttling device, it becomes superheated vapor.
By measuringpressure and temperatureafter throttling, the enthalpy of steam can be found:
h1=h2=hsuperheatedh_1 = h_2 = h_{superheated}h1=h2=hsuperheated
From this, x=h1−hfhfgx = \frac{h_1 - h_f}{h_{fg}}x=hfgh1−hf.
Q48. What is a combined separating and throttling calorimeter?
nswer:
A
Used when steam is too wet for direct throttling.
1. Separating chamber:Removes moisture.
2. T hrottling calorimeter:Measures dryness of separated vapor.
Combined readings give total dryness of steam.
Q49. What is superheating of steam and why is it done?
nswer:
A
Superheating:Heating saturated steam at constant pressure to a higher temperature.
Purpose:
● Increases efficiency of Rankine cycle.
● Reduces moisture at turbine exit (prevents blade erosion).
● Increases specific volume and work output.
Q50. Differentiate between saturated and superheated steam.
Property
Saturated Steam
Superheated Steam
Temperature = T_sat
> T_sat
Phase
O n vapor dome
Beyond vapor dome
Moisture
May be wet
Completely dry
Application
ow-pressure
L
heating
ower generation,
P
turbines
Q51. Explain phase diagrams used in thermodynamics.
Answer:
1. P–v Diagram:Shows phase change regions (dome shape).
2. T –s Diagram:Heat transfer = area under curve.
3. h–s Diagram (Mollier):Used for turbine and nozzle work calculations.
pplication:
A
Used to visualize cycle processes like Rankine and Carnot.
Q52. What is the importance of property charts in power plant design?
nswer:
A
They simplify thermodynamic analysis and allow engineers to quickly find:
● Turbine work (Δh)
● Condenser and boiler heat transfer
● E
fficiency and dryness levels
They are essential in designing boilers, condensers, and turbines in steam power plants.
✅ End of Module 3 – 17 In-depth Conceptual Questions
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M odule 4 – Gas Power Cycles (Otto, Diesel, Dual, Brayton, efficiencies, MEP)
🔥
(Q53–Q68) — one of themost asked modulesin mechanical interviews, especially forautomotive
and energy companies.
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G as Power Cycles— every mechanical engineer must master this one.
his isM odule 4from your VTU thermodynamics syllabus.
T
W e’ll coverO tto, Diesel, Dual, and Brayton cycles— complete withP–V & T–S concepts,
e fficiency equations, and interview interpretations.
THERMODYNAMICS INTERVIEW Q&A – VTU
⚙️
STRUCTURE
🔹 MODULE 4 – Gas Power Cycles
(Q53–Q68: Otto, Diesel, Dual, Brayton — full concept + formula + significance)
Q53. What is a thermodynamic cycle?
nswer:
A
Acycleis a series of processes in which a system returns to its initial state.
In each cycle, heat is supplied and part of it is converted into work.
Example:O tto cycle in petrol engine, Diesel cycle in CI engine, Brayton cycle in gas turbines.
Q54. What are air-standard assumptions in gas power cycles?
nswer:
A
Used to simplify the analysis of air-based cycles.
1. W orking fluid isairbehaving as anideal gas.
2. The cycle isclosed(air recirculates).
3. Specific heats (Cp, Cv)are constant.
4. All processes areinternally reversible.
5. No heat or frictional losses.
These assumptions form theair-standard cyclem odel.
Q55. What is the Otto cycle?
nswer:
A
O tto cycle is theair-standard cycleforspark-ignition (petrol) engines.
Processes:
1. 1–2: Isentropic compression
2. 2–3: Constant-volume heat addition
3. 3–4: Isentropic expansion
4. 4–1: Constant-volume heat rejection
Efficiency (η):
ηOtto=1−1r(γ−1)\eta_{Otto} = 1 - \frac{1}{r^{(\gamma - 1)}}ηOtto=1−r(γ−1)1
W here r=V1V2r = \frac{V_1}{V_2}r=V2V1= compression ratio.
O bservation:Efficiency ↑ with ↑ compression ratio.
Q56. What are the P–V and T–S diagrams of the Otto cycle?
Answer:
● P–V Diagram:Two vertical (constant volume) lines and two curved (isentropic) lines.
● T –S Diagram:Two horizontal (isentropic) and two rising/falling (constant volume) lines.
Area under curve = Work output.
Q57. What is the Diesel cycle?
nswer:
A
Diesel cycle is theair-standard cycleforcompression ignition (CI)engines.
Processes:
1. 1–2: Isentropic compression
2. 2–3: Constant-pressure heat addition
3. 3–4: Isentropic expansion
4. 4–1: Constant-volume heat rejection
Efficiency (η):
Diesel=1−1r(γ−1)×(ργ−1)γ(ρ−1)\eta_{Diesel} = 1 - \frac{1}{r^{(\gamma - 1)}} \times
η
\frac{(\rho^{\gamma} - 1)}{\gamma (\rho - 1)}ηDiesel=1−r(γ−1)1×γ(ρ−1)(ργ−1)
W here
● rrr = compression ratio
● ρ\rhoρ = cutoff ratio =V3V2= \frac{V_3}{V_2}=V2V3
Q58. Difference between Otto and Diesel cycles.
Feature
Heat addition
O tto Cycle
Diesel Cycle
Constant volume
Constant pressure
Efficiency depends on Compression ratio
(r)
Both r and cutoff ratio (ρ)
Compression ratio
Lower (8–10)
Higher (14–22)
Used in
Petrol engine
Diesel engine
Q59. What is the Dual cycle?
nswer:
A
TheDual cycleis a combination of Otto and Diesel cycles, representing areal IC enginem ore
accurately.
Processes:
1. Isentropic compression
2. Constant-volume heat addition (spark)
3. Constant-pressure heat addition (combustion)
4. Isentropic expansion
5. Constant-volume heat rejection
Efficiency:
Dual=1−1r(γ−1)×(αργ−1)(α−1)+γα(ρ−1)\eta_{Dual} = 1 - \frac{1}{r^{(\gamma - 1)}} \times
η
\frac{(\alpha \rho^{\gamma} - 1)}{(\alpha - 1) + \gamma \alpha (\rho -
1)}ηDual=1−r(γ−1)1×(α−1)+γα(ρ−1)(αργ−1)
here
w
α =P3P2\alpha = \frac{P_3}{P_2}α=P2P3, ρ=V4V3\rho = \frac{V_4}{V_3}ρ=V3V4.
Used in:Heavy-duty diesel engines (marine, generators).
Q60. Compare Otto, Diesel, and Dual cycle efficiencies.
At same compression ratio:
ηOtto>ηDual>ηDiesel\eta_{Otto} > \eta_{Dual} > \eta_{Diesel}ηOtto>ηDual>ηDiesel
eason:
R
O tto has shortest heat addition (constant volume), so higher mean temperature of heat addition →
higher efficiency.
Q61. What is mean effective pressure (MEP)?
nswer:
A
MEP is ahypothetical constant pressurethat would produce the same work output as the actual
c ycle.
MEP=WnetVdispMEP = \frac{W_{net}}{V_{disp}}MEP=VdispW net
here Vdisp=V1−V2V_{disp} = V_1 - V_2Vdisp=V1−V2.
w
Purpose:Compares performance of different engines irrespective of size.
Q62. Define compression ratio and cutoff ratio.
● C
ompression Ratio (r):
r=V1V2r = \frac{V_1}{V_2}r=V2V1
Ratio of total volume to clearance volume.
● C
utoff Ratio (ρ):
ρ=V3V2ρ = \frac{V_3}{V_2}ρ=V2V3
Used in Diesel/Dual cycles; represents duration of heat addition.
Q63. Explain effect of compression ratio on efficiency.
nswer:
A
Efficiency of all air-standard cycles increases withhigher compression ratiobecause:
● Air reaches higher temperature before combustion.
● C
ombustion becomes more complete.
Limitation:Knock or detonation in petrol engines restricts high r.
Q64. What is the Brayton cycle?
nswer:
A
Brayton cycle is theair-standard cycle for gas turbine engines.
Processes:
1. Isentropic compression (in compressor)
2. Constant-pressure heat addition (in combustion chamber)
3. Isentropic expansion (in turbine)
4. Constant-pressure heat rejection
Efficiency:
Brayton=1−1rp(γ−1)/γ\eta_{Brayton} = 1 - \frac{1}{r_p^{(\gamma -
η
1)/\gamma}}ηBrayton=1−rp(γ−1)/γ1
W here rp=P2P1r_p = \frac{P_2}{P_1}rp=P1P2= pressure ratio.
Q65. What is the effect of pressure ratio on Brayton cycle efficiency?
nswer:
A
Efficiency increases withpressure ratioup to an optimum point, beyond which turbine inlet
temperature limits cause efficiency drop.
η=1−1rp(γ−1)/γ\eta = 1 - \frac{1}{r_p^{(\gamma - 1)/\gamma}}η=1−rp(γ−1)/γ1
Q66. What are modifications of Brayton cycle for efficiency improvement?
Answer:
1. Regeneration:
Use exhaust gas heat to preheat compressed air.
2. Intercooling:
Cool air between compression stages → less compressor work.
3. Reheating:
Expand in multiple stages with intermediate heating → more turbine work.
Combined effect:Increases thermal efficiency and work output.
Q67. Compare reciprocating IC engines (Otto/Diesel) with gas turbines (Brayton).
Feature
IC Engine
G as Turbine
W orking fluid Air-fuel mixture
Air only
ompressio
C
n
Piston-cylinder
Axial/centrifugal compressor
Efficiency
25–35%
30–45%
Speed
Low–medium
Very high
Applications
Automobiles
Aircraft, power plants
Q68. Why are air-standard cycles idealizations?
nswer:
A
They assume no losses, constant specific heats, and perfect gas behavior — which don’t exist in real
s ystems.
Purpose:
To establish abenchmarkfor comparing real engine performance and to studytrend effectsof
parameters.
✅ End of Module 4 – 16 Deep Technical Questions
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M odule 5 – Vapour Power Cycles (Rankine, reheat, regeneration, efficiency
👉
improvements)
(Q69–Q84): This is thecore of power plant thermodynamics, essential for GET, turbine, and
energy system interviews.
Shall I continue withM odule 5next, macha?
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uper macha ⚙️🔥 —
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Now we’re moving intoM ODULE 5 – Vapour Power Cycles, one of the mostindustry-orientedand
frequently testedparts of thermodynamics — especially for roles inpower plants, design, and
quality.
ere you’ll getQ 69–Q84, with conceptual depth, key formulas, and real-world turbine–boiler examples
H
— all aligned withVTU B.E. Mechanicals yllabus.
THERMODYNAMICS INTERVIEW Q&A – VTU
⚙️
STRUCTURE
🔹 MODULE 5 – Vapour Power Cycles
(Q69–Q84: Rankine, Reheat, Regeneration, Efficiency, Condensers)
Q69. What is a vapour power cycle?
nswer:
A
Av apour power cycleuses acondensable working fluid (like water/steam)to convert heat
energy into mechanical work.
xample:Steam power plant.
E
Main components:Boiler → Turbine → Condenser → Pump.
Q70. What is the Rankine cycle?
nswer:
A
TheRankine cycleis theideal cycleforsteam power plants, operating on phase change of water.
Processes:
1. 1–2:Isentropic compression (Pump)
2. 2–3:Constant-pressure heat addition (Boiler)
3. 3–4:Isentropic expansion (Turbine)
4. 4–1:Constant-pressure heat rejection (Condenser)
Efficiency:
=Wturbine−WpumpQboiler\eta = \frac{W_{turbine} -
η
W _{pump}}{Q_{boiler}}η=QboilerW turbine−Wpump
Key assumption:Neglecting pump work gives a close approximation.
Q71. What are the main components of a steam power plant?
Answer:
1. Boiler:Converts water into high-pressure steam.
2. Turbine:Converts thermal energy into mechanical energy.
3. Condenser:Condenses exhaust steam back to water.
4. Pump:Increases pressure of condensate water.
Cycle:Boiler → Turbine → Condenser → Pump → Boiler (closed loop).
Q72. Write energy equations for each component in Rankine cycle.
Answer:
Component
Process
Energy Relation
Pump
Isentropic
c ompression
2=h1+v(P2−P1)h_2 = h_1 + v (P_2 -
h
P_1)h2=h1+v(P2−P1)
Boiler
Constant pressure
in=h3−h2Q_{in} = h_3 -
Q
h_2Qin=h3−h2
Turbine
Isentropic expansion
turb=h3−h4W_{turb} = h_3 -
W
h_4Wturb=h3−h4
Condenser
Constant pressure
out=h4−h1Q_{out} = h_4 -
Q
h_1Qout=h4−h1
Cycle efficiency:
=1−QoutQin=1−h4−h1h3−h2\eta = 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{h_4 - h_1}{h_3 -
η
h_2}η=1−QinQ out=1−h3−h2h4−h1
Q73. What factors affect Rankine cycle efficiency?
Answer:
1. Boiler pressure↑ → efficiency ↑
2. Condenser pressure↓ → efficiency ↑
3. Superheating of steam→ efficiency ↑
4. Reheating/regeneration→ efficiency ↑
But practical limits: material strength, turbine moisture, and cost.
Q74. What is the effect of condenser pressure on Rankine efficiency?
nswer:
A
Lower condenser pressure → larger turbine expansion ratio → higher work output.
However, very low pressure causesmoisture at turbine exit, reducing blade life.
O ptimum condenser pressure ≈ 0.08 bar in most plants.
Q75. What is superheating and its purpose in Rankine cycle?
nswer:
A
Superheating means increasing steam temperaturebeyond saturationat constant pressure (after
boiler).
Advantages:
● Reduces turbine moisture
● Increases mean temperature of heat addition → better efficiency
● Increases specific work output
Limitation:Limited by material strength and temperature.
Q76. What is the reheat Rankine cycle?
nswer:
A
Inreheating, steam expands in two stages:
1. Partial expansion in high-pressure turbine
2. Reheated in boiler
3. Expansion in low-pressure turbine
Advantages:
● Increases efficiency
● Reduces moisture content at turbine exit
Reheat efficiency:
ηreheat>ηRankine\eta_{reheat} > \eta_{Rankine}ηreheat>ηRankine
Q77. What is the regenerative Rankine cycle?
nswer:
A
In regeneration, a portion of steam is extracted (bled) from turbine topreheat feedwaterbefore
entering the boiler.
Purpose:
● Increases average temperature of heat addition
● Improves efficiency
Common method:O pen or closed feedwater heater.
Practical example:Thermal power stations use 6–8 feedwater heaters.
Q78. Compare simple, reheat, and regenerative Rankine cycles.
Type
Efficiency M oisture at exit Complexity
Simple
Lowest
High
Simple
Reheat
Higher
Reduced
Moderate
egenerativ
R
e
Highest
Moderate
Complex
Q79. What is the effect of superheating on T–S and h–s diagrams?
Answer:
● O n T–S diagram:Rightward shift of expansion line → larger enclosed area → higher work.
● O
n h–s (Mollier) diagram:Expansion line moves further down → greater Δh → higher turbine
work.
Conclusion:Superheating improves both output and efficiency.
Q80. Define boiler efficiency and thermal efficiency of the plant.
Answer:
● B
oiler Efficiency:
ηboiler=Heat to steamHeat supplied by fuel×100\eta_{boiler} = \frac{\text{Heat to
s team}}{\text{Heat supplied by fuel}} \times 100ηboiler=Heat supplied by fuelHeat to
s team×100
● T hermal Efficiency (plant):
ηthermal=WnetQinput×100\eta_{thermal} = \frac{W_{net}}{Q_{input}} \times
100ηthermal=QinputW net×100
Typical values: Boiler ≈ 85–90%, Overall ≈ 35–40%.
Q81. What are sources of losses in Rankine cycle?
Answer:
1. Pressure losses in piping and boiler.
2. Heat losses to surroundings.
3. Pump and turbine inefficiencies.
4. Moisture losses at turbine exit.
5. Non-condensable gases in condenser.
Reducing these losses increases plant performance.
Q82. What is the back work ratio?
nswer:
A
It is the ratio ofpump worktoturbine work output.
BWR=WpumpWturbineBWR = \frac{W_{pump}}{W_{turbine}}BWR=WturbineW pump
or Rankine cycle, it’s very small (≈ 0.5–1%).
F
For gas turbines (Brayton), it can be 40–50%.
Q83. Why Rankine cycle is preferred over Carnot cycle for steam plants?
Answer:
● Condensation at constant temperature in Carnot isimpractical.
● Rankine allows condensation and evaporation at constant pressure → easier design.
● Rankine hasmore realistic working rangefor steam properties.
T herefore:Rankine is practical, Carnot is ideal.
Q84. How can Rankine cycle efficiency be improved?
Answer:
1. Superheating the steam
2. Increasing boiler pressure
3. Reducing condenser pressure
4. Using reheat and regeneration
5. Reducing irreversibilitiesin turbine and pump
Each modification brings the Rankine closer to Carnot efficiency.
✅ End of Module 5 – 16 Detailed Power Plant Questions
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M odule 6 – Refrigeration & Air Conditioning Fundamentals (Q85–Q100)
🔥
This includesreversed Carnot cycle, vapor-compression, COP, psychrometry basics— highly
relevant forHVAC, automotive, and thermal system designinterviews.
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VTU-Based Thermodynamics Interview Series — 100 In-depth Questions & Answers!💥
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his isM ODULE 6, focusing onRefrigeration and Air Conditioning (Q85–Q100)— a crucial topic
T
for roles inautomotive thermal systems, HVAC, EV cooling, and design interviews.
THERMODYNAMICS INTERVIEW Q&A – VTU
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STRUCTURE
🔹 MODULE 6 – Refrigeration and Air Conditioning
(Q85–Q100: Reversed Carnot, Vapour Compression, COP, Refrigerants, Psychrometry)
Q85. What is refrigeration?
nswer:
A
Refrigeration is the process ofremoving heatfrom a low-temperature region andrejecting itto a
higher-temperature region.
This requires externalwork input(usually via a compressor).
nit:Ton of refrigeration (TR)
U
1 TR = 211 kJ/min = 3.516 kW
Applications:Domestic refrigerators, air conditioning, cold storage, automobile AC.
Q86. Define the Coefficient of Performance (COP).
nswer:
A
COP measures the performance of a refrigerator or heat pump.
For a refrigerator:
COPR=QLW=TLTH−TLCOP_R = \frac{Q_L}{W} = \frac{T_L}{T_H - T_L}COPR=WQL=TH−TLTL
For a heat pump:
OPHP=QHW=THTH−TLCOP_{HP} = \frac{Q_H}{W} = \frac{T_H}{T_H -
C
T_L}COPHP=WQH=TH−TLTH
Relation:
COPHP=COPR+1COP_{HP} = COP_R + 1COPHP=COPR+1
Higher COP → better performance.
Q87. What is the reversed Carnot cycle?
nswer:
A
It is theideal refrigeration cycleoperating in reverse of Carnot’s heat engine.
Processes:
1. Isentropic compression
2. Isothermal heat rejection
3. Isentropic expansion
4. Isothermal heat absorption
COP (Carnot):
COP=TLTH−TLCOP = \frac{T_L}{T_H - T_L}COP=TH−TLTL
Limitation:Impractical due to isothermal and isentropic requirements. Used as a benchmark.
Q88. What is the Vapour-Compression Refrigeration Cycle (VCRC)?
nswer:
A
Most practical refrigeration system — used in domestic and automotive ACs.
Processes:
1. Evaporation:Heat absorbed at low pressure (Q_L).
2. Compression:Pressure and temperature increase.
3. Condensation:Heat rejected to surroundings (Q_H).
4. Expansion (Throttling):Pressure drops to evaporator level.
Cycle:Evaporator → Compressor → Condenser → Expansion Valve → Evaporator.
Q89. Explain T–S and P–h diagrams of VCRC.
Answer:
● T
–S Diagram:Two isentropic (compression/expansion) and two constant-pressure
(evaporation/condensation) processes.
● P
–h Diagram:Easier to analyze in practice; enthalpy difference between 1–4 = refrigeration
effect, 2–3 = heat rejection.
Area under curve → work or heat transfer.
Q90. Define refrigeration effect and work input.
Answer:
● R
efrigeration Effect:
RE=h1−h4RE = h_1 - h_4RE=h1−h4
(Heat absorbed in evaporator)
● C
ompressor Work:
W =h2−h1W = h_2 - h_1W=h2−h1
● COP:
COP=h1−h4h2−h1COP = \frac{h_1 - h_4}{h_2 - h_1}COP=h2−h1h1−h4
Q91. What are the main components of a vapor-compression refrigeration system?
Answer:
1. Evaporator:Absorbs heat → refrigerant evaporates.
2. Compressor:Raises pressure and temperature.
3. Condenser:Rejects heat → refrigerant condenses.
4. Expansion valve:Reduces pressure and temperature of refrigerant.
Each plays a role in the thermodynamic cycle balance.
Q92. What are the common refrigerants used and their properties?
Answer:
Refrigerant
Chemical Formula
Applications
Remarks
R-12 (CCl₂F₂)
Freon-12
O ld domestic fridges
Banned (ozone depletion)
R-22 (CHClF₂)
HCFC
AC systems
Phased out
R-134a (C₂H₂F₄) HFC
Cars, chillers
O zone-safe
Ammonia (NH₃)
—
Industrial refrigeration Toxic but efficient
CO₂ (R-744)
—
Eco-friendly
High operating pressure
Ideal refrigerant:Non-toxic, non-flammable, stable, high latent heat, low specific volume.
93. What is the difference between Vapour-Compression and Vapour-Absorption
Q
systems?
Feature
Vapour-Compression
Vapour-Absorption
nergy
E
s ource
echanical work
M
(compressor)
Heat energy
W orking pair
Single refrigerant
efrigerant + absorbent (NH₃–H₂O,
R
LiBr–H₂O)
COP
3–6
0.5–2
Maintenance
Higher
Lower
Application
Domestic, automotive
Large plants, waste-heat use
Q94. Explain the Vapour-Absorption Refrigeration System (VARS).
nswer:
A
Usesheat energyinstead of mechanical work for refrigeration.
M ain Components:
1. Absorber
2. Pump
3. Generator
4. Condenser
5. Expansion valve
6. Evaporator
Working pair examples:
● Ammonia–Water system
● Lithium Bromide–Water system
Advantage:Utilizes waste heat or solar energy.
Q95. What is a refrigerating effect per ton and how is it calculated?
nswer:
A
1 ton of refrigeration = 211 kJ/min = 3.516 kW
This equals the heat required to freeze 1 ton (1000 kg) of water at 0°C in 24 hours.
1 TR=211 kJ/min=3.516 kW1 \, TR = 211 \, kJ/min = 3.516 \, kW1TR=211kJ/min=3.516kW
Used to rate capacity of cooling systems.
Q96. What is subcooling and superheating in refrigeration?
Answer:
● S
ubcooling:Cooling the liquid refrigerant below its condensation temperature at constant
pressure → increases COP.
● S
uperheating:Heating the vapor refrigerant above evaporation temperature → ensures only
vapor enters compressor → prevents liquid slugging.
Both improve efficiency and reliability.
Q97. What is psychrometry and why is it important in air conditioning?
nswer:
A
Psychrometryis the study of moist air and its properties.
Important for HVAC design becausecomfort coolingdepends on both temperature and humidity.
Key properties:
● Dry-bulb temperature (DBT)
● W et-bulb temperature (WBT)
● Dew point (DPT)
● Relative humidity (RH)
● Specific humidity
Q98. Define relative humidity and specific humidity.
Answer:
● R
elative Humidity (ϕ):
ϕ=pvpvs×100\phi = \frac{p_v}{p_{vs}} \times 100ϕ=pvspv×100
Ratio of actual vapor pressure to saturation vapor pressure.
● S
pecific Humidity (ω ):
ω =0.622pvP−pv\omega = 0.622 \frac{p_v}{P - p_v}ω=0.622P−pvpv
Mass of water vapor per kg of dry air.
Importance:Determines comfort and cooling load.
Q99. What is sensible and latent heat in air-conditioning?
Answer:
● Sensible Heat:Heat that changes air temperature (no phase change).
● Latent Heat:Heat that changes moisture content (phase change of water vapor).
Total Heat Load:
Q total=Qsensible+QlatentQ_{total} = Q_{sensible} + Q_{latent}Qtotal=Qsensible+Qlatent
Used in sizing air-conditioning systems.
Q100. What is a psychrometric chart and its applications?
nswer:
A
Agraphical representationof moist air properties — showing DBT, WBT, RH, enthalpy, and
s pecific volume.
Applications:
● Designing comfort air-conditioning systems
● Cooling/dehumidifying calculations
● Determining coil loads and ventilation requirements
Processes on chart:
● Sensible cooling/heating
● Humidification/dehumidification
● Mixing of air streams
End of Module 6 — and Completion of 100 Thermodynamics Interview Questions &
✅
Answers!🎯
⚙️ Summary of What You’ve Got So Far
M odule
Topics Covered
No. of
Q s
1
Basics & First Law
17
2
Second Law & Entropy
18
3
Properties of Pure Substances
17
4
Gas Power Cycles
16
5
Vapour Power Cycles
16
6
Refrigeration & Air Conditioning
16
Total
omplete Thermodynamics
C
Coverage
100
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⚙️ THERMODYNAMICS – VTU STRUCTURE
🔹 MODULE 1: Basic Concepts & First Law of Thermodynamics
(30 Deep + Connected Interview Questions & Answers)
Q1. Define thermodynamics and explain its importance in engineering.
nswer:
A
Thermodynamics is the study ofe nergy, its transformations, and how it affects matter.
Applications:
● Power generation (Rankine, gas turbines)
● Refrigeration, IC engines, HVAC, EV cooling systems
● Energy management and sustainability.
Q2. What are the different energy types in thermodynamics?
Answer:
1. M icroscopic (Internal Energy)→ due to molecular motion and bonding.
2. M acroscopic (Mechanical Energy)→ Kinetic & potential energy of a system.
E=U+V22+gZE = U + \frac{V^2}{2} + gZE=U+2V2+gZ
Q3. Differentiate between closed, open, and isolated systems with examples.
Type
O pen
M ass
Exchange
Energy
Exchange
Example
Yes
Yes
urbine,
T
c ompressor
Closed No
Yes
Piston-cylinder
Isolated No
No
Thermos flask
Q4. Define state, path, and process.
Answer:
● State:Set of properties defining system condition.
● Process:Path of change from one state to another.
● Path:Sequence of intermediate states during a process.
Q5. What are intensive and extensive properties?
Answer:
● Intensive:Independent of mass (Pressure, Temperature).
● E
xtensive:Dependent on mass (Volume, Internal Energy).
Ratio of extensive to mass = specific property.
Q6. What is the equation of state?
nswer:
A
It relates P, V, and T of a substance.
For ideal gas:
PV=mRTPV = mRTPV=mRT
R=RuMR = \frac{R_u}{M}R=MRu
Q7. Define thermodynamic equilibrium.
nswer:
A
A system is incomplete equilibriumwhen it satisfies:
1. T hermal– no temperature difference
2. M echanical– no unbalanced forces
3. Chemical– no reaction tendency
Q8. Explain Zeroth Law of Thermodynamics.
nswer:
A
If system A is in thermal equilibrium with B, and B with C, then A and C are in thermal equilibrium.
Basis of temperature measurement.
Q9. What is a quasi-static process and why is it important?
nswer:
A
Infinitely slow process maintaining equilibrium at each step.
Importance:Used in theoretical derivations of work and heat equations.
Q10. What is work in thermodynamics?
nswer:
A
Energy transfer associated with force and displacement.
δW=P dV\delta W = P\,dVδW=PdV
Positive if donebys ystem, negative ifons ystem.
Q11. What are different forms of work?
Answer:
1. Boundary work– W=∫P dVW = \int P\,dVW=∫PdV
2. Shaft work– W=2πNT/60W = 2\pi NT/60W=2πNT/60
3. Electrical work– W=VItW = VItW=VIt
4. Flow work– W=P vW = P\,vW=Pv
Q12. Define heat and its characteristics.
nswer:
A
Energy transfer due to temperature difference only.
● Path function, not property
● Positive when added to system
● δQ≠dQ\delta Q \neq dQδQ=dQ
Q13. Distinguish between heat and work.
Heat
Work
Due to ΔT
ue to mechanical
D
c ause
Unorganized motion
O rganized motion
ve: added to
+
s ystem
+ve: done by system
Q14. Explain the concept of boundary work for a polytropic process.
Answer:
Vn=constantP V^n = \text{constant}PVn=constant W=P2V2−P1V11−nW = \frac{P_2V_2 - P_1V_1}{1
P
- n}W=1−nP2V2−P1V1
If n = 1 → isothermal, n = γ → adiabatic.
Q15. What is the first law of thermodynamics (closed system)?
δQ=dU+δW\delta Q = dU + \delta WδQ=dU+δW
Heat added = Rise in internal energy + Work done.
Q16. Explain the first law for open systems (Steady Flow Energy Equation).
˙−W˙=m˙[(h2−h1)+V22−V122+g(z2−z1)]\dot{Q} - \dot{W} = \dot{m}\left[(h_2 - h_1) + \frac{V_2^2 -
Q
V_1^2}{2} + g(z_2 - z_1)\right]Q˙−W˙=m˙[(h2−h1)+2V22−V12+g(z2−z1)]
Used for compressors, turbines, nozzles, etc.
Q17. What is internal energy and its significance?
nswer:
A
Energy at molecular level — translation, rotation, vibration.
Depends only on temperature (for ideal gases).
dU=mCvdTdU = mC_v dTdU=mCvdT
Q18. Define enthalpy and its physical meaning.
h=u+Pvh = u + Pvh=u+Pv
Represents total heat content; useful for constant-pressure systems like boilers.
dh=CpdTdh = C_p dTdh=CpdT
Q19. Derive relation between Cp and Cv.
Cp−Cv=RC_p - C_v = RCp−Cv=R γ=CpCv\gamma = \frac{C_p}{C_v}γ=CvCp
For air: Cp = 1.005, Cv = 0.718, γ = 1.4
Q20. Define specific heats and their importance.
nswer:
A
Heat required to raise 1 kg of substance by 1°C.
Used to compute energy changes in gases.
Q21. Explain the concept of total energy of a system.
E=U+KE+PEE = U + KE + PEE=U+KE+PE
Total energy combines microscopic and macroscopic forms; conserved in all processes.
Q22. What is the steady-flow energy equation (SFEE) used for turbines?
Q ˙−W˙=m˙(h2−h1)\dot{Q} - \dot{W} = \dot{m}(h_2 - h_1)Q˙−W˙=m˙(h2−h1)
If adiabatic and steady → W˙=m˙(h1−h2)\dot{W} = \dot{m}(h_1 - h_2)W˙=m˙(h1−h2)
Q23. What is the steady-flow energy equation for a compressor?
W ˙=m˙(h2−h1)\dot{W} = \dot{m}(h_2 - h_1)W˙=m˙(h2−h1)
W ork isinputto increase pressure and enthalpy.
Q24. Write SFEE for nozzle and diffuser.
● N
ozzle:Converts enthalpy to kinetic energy
h1−h2=V22−V122h_1 - h_2 = \frac{V_2^2 - V_1^2}{2}h1−h2=2V22−V12
● Diffuser:O pposite of nozzle — velocity → pressure.
Q25. Define reversible and irreversible energy transfer.
Answer:
● Reversible → Ideal, no losses, infinitely slow.
● Irreversible → Real, involves friction, turbulence, or sudden changes.
Q26. What is the difference between property, path, and point function?
Function
Depends On
Example
Property
State only
P, V, T
Path
Process path
Q , W
Point
Identifies system
s tate
h, s, U
Q27. What is a cyclic process and what does the first law state for it?
nswer:
A
Cycle → System returns to initial state.
∮δQ=∮δW\oint \delta Q = \oint \delta W∮δQ=∮δW
→ Net heat = Net work.
Q28. What are control mass and control volume?
● Control Mass:Fixed quantity of matter (closed system).
● C
ontrol Volume:Fixed region in space (open system).
Used in analyzing turbines, compressors, pumps.
Q29. Explain the concept of flow energy (flow work).
nswer:
A
W ork needed to move mass across control volume boundaries.
W flow=PvW_{flow} = PvWflow=Pv
Included in enthalpy term in SFEE.
Q30. Give examples of engineering devices analyzed using the First Law.
Device
Type
Application
ompresso O pen
C
r
ir compression in IC
A
engines
Turbine
O pen
Power generation
Pump
O pen
Fluid transport
Boiler
lose
C
d
Steam generation
Condenser
lose
C
d
Heat rejection
✅ End of Module 1 – 30 Comprehensive Questions
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⚙️ MODULE 2 – Second Law of Thermodynamics & Entropy
30 Detailed, Interview-Level Questions and Answers — Including Exergy, Reversibility, Clausius
💡
Inequality, and Entropy Applications.
Q1. Why was the Second Law of Thermodynamics needed?
nswer:
A
Because theFirst Lawtells only thequantityof energy, not itsdirectionorquality.
he Second Law explainswhyheat flows naturally from hot → cold and introducese ntropyto
T
m easure irreversibility.
Q2. State the Kelvin–Planck and Clausius statements of the Second Law.
Answer:
● K
elvin–Planck:No heat engine can convert all the heat supplied into work without any heat
rejection.
● C
lausius:It is impossible for heat to flow from a colder body to a hotter one without external
work.
Equivalence:Violation of one implies violation of the other.
Q3. What is a heat engine?
nswer:
A
A device that converts a portion ofheat energy into work.
Efficiency:
η=1−QLQH\eta = 1 - \frac{Q_L}{Q_H}η=1−QHQ L
Examples:IC engines, steam turbines, gas turbines.
Q4. What is a refrigerator and heat pump?
nswer:
A
Devices that transfer heat from alow-temperature reservoir to a high-temperature reservoir
using work input.
OPR=QLW,COPHP=QHWCOP_R = \frac{Q_L}{W}, \quad COP_{HP} =
C
\frac{Q_H}{W}COPR=WQL,COPHP=WQHCOPHP=COPR+1COP_{HP} = COP_R +
1COPHP=COPR+1
Q5. Define thermal efficiency.
=WnetQin=1−QoutQin\eta = \frac{W_{net}}{Q_{in}} = 1 -
η
\frac{Q_{out}}{Q_{in}}η=QinW net=1−QinQ out
No real engine has 100% efficiency due to irreversibilities.
Q6. What is a reversible process?
nswer:
A
A process that can be reversedwithout any changein the system or surroundings.
Condition:All changes occur infinitesimally slowly.
Example:Isothermal expansion of ideal gas.
Q7. What makes a process irreversible?
Answer:
● Friction
● Rapid heat transfer (finite temperature difference)
● Viscous dissipation
● Unrestrained expansion
● Electrical resistance
● Mixing of gases
Real systems are alwaysirreversible.
Q8. Define a Carnot cycle.
nswer:
A
Anideal reversible cyclethat gives themaximum efficiencybetween two temperature limits.
Processes:
1. Isothermal heat addition
2. Adiabatic expansion
3. Isothermal heat rejection
4. Adiabatic compression
ηCarnot=1−TLTH\eta_{Carnot} = 1 - \frac{T_L}{T_H}ηCarnot=1−THTL
Q9. State the two Carnot theorems.
1. No engine operating between two given temperatures can be more efficient than areversible
e ngine.
2. All reversible engines operating between same temperatures havee qual efficiency.
Q10. What is the significance of the Carnot cycle?
Answer:
● It sets theupper limitof efficiency.
● Used as astandard of comparisonfor all real cycles.
● Definestemperature scale (Kelvin)via heat ratio.
Q11. What is the Clausius inequality?
Answer:
∮δQT≤0\oint \frac{\delta Q}{T} \le 0∮TδQ≤0
quality for reversible cycles, < 0 for irreversible cycles.
E
Basis fore ntropydefinition.
Q12. Define entropy and its physical meaning.
Answer:
dS=δQrevTdS = \frac{\delta Q_{rev}}{T}dS=TδQrev
ntropy measurese nergy unavailabilityanddisorder.
E
Unit:kJ/kg·K
Reversible:ΔS = 0,Irreversible:ΔS > 0.
Q13. Write the entropy change equations for ideal gases.
1. Constant Volume:
ΔS=mCvlnT2T1\Delta S = mC_v \ln\frac{T_2}{T_1}ΔS=mCvlnT1T2
2. Constant Pressure:
ΔS=mCplnT2T1\Delta S = mC_p \ln\frac{T_2}{T_1}ΔS=mCplnT1T2
3. Isothermal:
ΔS=mRlnV2V1\Delta S = mR \ln\frac{V_2}{V_1}ΔS=mRlnV1V2
Q14. State the principle of increase of entropy.
nswer:
A
For anyisolated system,
Suniverse=ΔSsystem+ΔSsurroundings≥0\Delta S_{universe} = \Delta S_{system} + \Delta
Δ
S_{surroundings} \ge 0ΔSuniverse=ΔSsystem+ΔSsurroundings≥0
Entropy never decreases; it increases for irreversible processes.
Q15. What are T–S diagrams and their significance?
nswer:
A
Graph of temperature vs. entropy.
● Area under curve = heat transfer.
● Visualizesreversibility, efficiency, and cycle direction.
Used in Carnot, Rankine, Brayton, and refrigeration cycles.
Q16. Explain the concept of energy quality and degradation.
Answer:
● High-grade energy:W ork, electrical energy (fully convertible).
● L
ow-grade energy:Heat at low temperature (partially useful).
Entropy generation →e nergy degradation.
Q17. What is exergy or available energy?
nswer:
A
Maximum useful work obtainable as system comes to equilibrium with environment.
x=(U−U0)+P0(V−V0)−T0(S−S0)Ex = (U - U_0) + P_0(V - V_0) - T_0(S -
E
S_0)Ex=(U−U0)+P0(V−V0)−T0(S−S0)
W here subscript “0” refers to ambient conditions.
Q18. Define irreversibility or exergy destruction.
I=T0×SgenI = T_0 \times S_{gen}I=T0×Sgen
igher irreversibility = lower efficiency.
H
Caused by friction, heat loss, etc.
Q19. What is availability and unavailability function?
Answer:
● Availability (A):Maximum work output possible.
● Unavailability (U):Part of heat that cannot be converted to work.
U=T0ΔSU = T_0 \Delta SU=T0ΔS
Q20. What is an isentropic process?
nswer:
A
Areversible adiabatic processwith no entropy change.
ΔS=0\Delta S = 0ΔS=0
Used as ideal model for turbines, compressors, and nozzles.
Q21. What are common examples of entropy increase in real life?
1. Friction in machinery
2. Heat loss in engines
3. Mixing of gases or fluids
4. Combustion processes
All cause energy degradation.
Q22. What is entropy generation minimization and why is it important?
nswer:
A
Entropy generation minimization = Reducing irreversibility.
Importance:Improves system efficiency → less waste energy.
Used inheat exchanger design,turbine optimization,HVAC efficiency.
Q23. Define the concept of adiabatic efficiency.
nswer:
A
Used to compare real vs. ideal (isentropic) performance.
isen=hin−hout,realhin−hout,ideal\eta_{isen} = \frac{h_{in} - h_{out,real}}{h_{in} -
η
h_{out,ideal}}ηisen=hin−hout,idealhin−hout,real
Applicable for turbines, compressors, nozzles.
Q24. Differentiate between isothermal and adiabatic expansion.
Property
Isothermal
Adiabatic
Heat transfer Yes
No
Temperature Constant
Changes
Equation
PV = C
PV^γ = C
W ork done
Max
ess than
L
isothermal
Q25. Explain entropy generation in a heat exchanger.
gen=m˙hCplnTh,iTh,o+m˙cCplnTc,oTc,iS_{gen} = \dot{m}_h C_p \ln\frac{T_{h,i}}{T_{h,o}} +
S
\dot{m}_c C_p \ln\frac{T_{c,o}}{T_{c,i}}Sgen=m˙hCplnTh,oTh,i+m˙cCplnTc,iTc,o
Always positive due to temperature difference between hot and cold streams.
Q26. What is thermal reservoir and its role in Second Law?
nswer:
A
A large body whose temperature remains constant while supplying or absorbing heat.
Examples:Atmosphere, oceans, furnace gases.
Used assourceorsinkin engines and refrigerators.
Q27. Why can’t entropy be negative?
nswer:
A
Entropy change can be negative for a subsystem (losing heat),
buttotal entropyof universe (system + surroundings) is always ≥ 0.
Q28. What is the physical significance of entropy?
nswer:
A
Measures:
● Direction of natural processes
● Degree of molecular disorder
● Energy unavailability for work
High entropy → less useful energy.
Q29. What is the second-law efficiency (exergetic efficiency)?
nswer:
A
Ratio ofactual useful worktomaximum (reversible) work possible.
II=Actual outputExergy input\eta_{II} = \frac{\text{Actual output}}{\text{Exergy input}}ηII=Exergy
η
inputActual output
Used to evaluate true thermodynamic performance.
Q30. Give industrial examples where entropy and exergy analysis are used.
Answer:
● Power plants:O ptimize turbine and boiler performance
● Refrigeration systems:Evaluate compressor losses
● Combustion engines:Analyze waste heat recovery
● Heat exchangers:Reduce exergy destruction
✅ End of Module 2 – 30 Advanced Conceptual and Practical Questions.
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M odule 3 – Properties of Pure Substances (Steam, phase diagrams, dryness, Mollier
🔥
chart, real applications)
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diagram— all vital inpower plant, thermal design, and qualityinterviews.
⚙️ THERMODYNAMICS – VTU STRUCTURE
🔹 MODULE 3 – Properties of Pure Substances
(30 Detailed, Conceptual + Application-Based Interview Q&A)
Q1. What is a pure substance?
nswer:
A
Apure substanceis one that has auniform chemical compositionthroughout its mass, even if it
exists in multiple phases.
Examples:Water, air, CO₂, NH₃.
Note:A mixture of liquid water and water vapor is also a pure substance (both are H₂O).
Q2. Define a phase and explain its types.
nswer:
A
Aphaseis a region with uniform physical and chemical properties.
Types:
1. Solid
2. Liquid
3. Vapor (Gas)
Phase boundaries separate different phases.
Q3. Explain the phase change process of water at constant pressure.
nswer:
A
W hen water is heated at 1 atm:
1. Compressed (subcooled) liquid
2. Saturated liquid
3. Wet region (mixture of liquid + vapor)
4. Saturated vapor
5. Superheated vapor
Shown as thebell-shaped domeon a P–v or T–v diagram.
Q4. Define triple point and critical point.
● T
riple Point:Temperature and pressure at which all three phases (solid, liquid, vapor) coexist.
→ For water: 0.01°C and 0.611 kPa.
● C
ritical Point:Above this, liquid and vapor phases are indistinguishable.
→ For water: 374°C and 22.06 MPa.
Q5. What are saturation temperature and pressure?
Answer:
● Saturation Temperature (T_sat):Boiling point at a given pressure.
● S
aturation Pressure (P_sat):Boiling pressure at a given temperature.
Example: At 1 atm → 100°C.
Q6. Define subcooled liquid and superheated vapor.
● Subcooled (compressed) liquid:Liquid below its saturation temperature.
● S
uperheated vapor:Vapor above its saturation temperature.
Used in boilers and turbines for efficiency improvement.
Q7. Explain the terms: saturated liquid, saturated vapor, and wet vapor.
Term
Definition
Saturated Liquid About to vaporize
W et Vapor
Mixture of liquid and vapor
Saturated Vapor Just completely vaporized
Q8. What is dryness fraction (x)?
Answer:
x=mvapormliquid+mvaporx = \frac{m_{vapor}}{m_{liquid} + m_{vapor}}x=mliquid+mvaporm vapor
Represents the proportion of vapor in wet steam.
● x = 0 → saturated liquid
● x
= 1 → saturated vapor
Important in:Turbine efficiency calculations.
Q9. Write property relations for a mixture (wet region).
=vf+x(vg−vf)v = v_f + x(v_g - v_f)v=vf+x(vg−vf) h=hf+x(hfg)h = h_f + x(h_{fg})h=hf+x(hfg)
v
s =sf+x(sfg)s = s_f + x(s_{fg})s=sf+x(sfg)
Used to find actual property values from steam tables.
Q10. What are h_fg, v_fg, and s_fg?
nswer:
A
These arelatent(phase change) properties:
fg=hg−hf,vfg=vg−vf,sfg=sg−sfh_{fg} = h_g - h_f, \quad v_{fg} = v_g - v_f, \quad s_{fg} = s_g -
h
s _fhfg=hg−hf,vfg=vg−vf,sfg=sg−sf
Represent the difference between vapor and liquid states at saturation.
Q11. What is latent heat?
nswer:
A
Heat added or removed to cause a phase changewithout temperature change.
Example: Water → steam at 100°C (2257 kJ/kg).
Q12. Explain the concept of superheated region.
nswer:
A
W hen vapor temperature > T_sat, it becomessuperheated vapor.
Superheating reduces turbine moisture, increases cycle efficiency.
Q13. What is the compressed liquid region?
nswer:
A
Region where liquid temperature < T_sat at given pressure.
Example: Water at 80°C, 1 atm → compressed liquid.
Q14. Define the use of steam tables.
nswer:
A
Steam tables provide property data (P, T, v, h, s, u) for water and steam:
● Saturation tables (at constant P or T)
● Superheated tables
● C
ompressed liquid tables
Essential for Rankine cycle and power plant analysis.
Q15. What is the Mollier chart (h–s diagram)?
nswer:
A
A graphical plot ofe nthalpy (h)vs.e ntropy (s)for steam.
Lines of constant P, T, and x are plotted.
Used for:Q uick estimation of turbine and nozzle work (Δh = work).
Q16. Define throttling process.
nswer:
A
Adiabatic expansion through a restriction (valve/capillary) with negligible heat/work transfer.
h1=h2h_1 = h_2h1=h2
Used in throttling calorimeters and refrigeration expansion valves.
Q17. What is the use of a throttling calorimeter?
nswer:
A
To determinedryness fractionof steam.
W hen wet steam is throttled to low pressure, it becomes superheated; measured T and P give
enthalpy → dryness via:
x=h1−hfhfgx = \frac{h_1 - h_f}{h_{fg}}x=hfgh1−hf
Q18. What is a separating and throttling calorimeter?
nswer:
A
Used forv ery wet steam:
● Separator:Removes liquid moisture.
● T
hrottling calorimeter:Measures remaining steam dryness.
Combined reading gives total x.
Q19. Define degree of superheat.
egree of Superheat=Tactual−Tsat\text{Degree of Superheat} = T_{actual} - T_{sat}Degree of
D
Superheat=Tactual−Tsat
Indicates how much the steam is heated beyond saturation — important for turbine efficiency.
Q20. Why is superheated steam used in turbines?
Answer:
● Prevents blade erosion (no moisture).
● Increases efficiency (higher mean temperature).
● Improves specific volume → more work.
Q21. Explain the difference between saturated and superheated vapor.
Property
Saturated
Superheated
T
= T_sat
> T_sat
Phase
n vapor
O
dome
Beyond vapor dome
x
1
—
Application Boilers
Turbines
Q22. What is a P–v diagram of pure substance?
nswer:
A
Bell-shaped curve showing liquid, vapor, and two-phase regions.
● Left curve → saturated liquid line
● Right curve → saturated vapor line
● Inside → wet region
Used to understand boiling and condensation.
Q23. What is a T–s diagram and its use?
nswer:
A
Graph of temperature vs. entropy:
● Area under curve = heat transfer.
● Used in Rankine cycle and refrigeration system analysis.
Q24. What is the significance of the critical point on property diagrams?
nswer:
A
Beyond the critical point → no distinct liquid/vapor boundary.
Used to definesupercritical boilers(T > 374°C, P > 22.06 MPa).
Q25. What is specific enthalpy and why is it useful?
Answer:
h=u+Pvh = u + Pvh=u+Pv
epresents total heat content per unit mass.
R
Used to compute turbine work and heat transfer.
Q26. What is the significance of enthalpy of vaporization (h_fg)?
nswer:
A
Determines the amount of heat required to convert 1 kg of liquid to vapor at constant pressure.
Used in boiler heat transfer and condensation processes.
Q27. How do you identify a state using two independent properties?
nswer:
A
Anytwo independent, intensive properties(P–T, T–v, P–h, etc.) uniquely define a thermodynamic
s tate.
This is known as thestate postulate.
Q28. Why is water preferred as a working fluid in vapor power systems?
Answer:
● Easily available and cheap
● High latent heat
● Non-toxic and stable
● Suitable operating pressure range
● Good heat transfer properties
Q29. Explain subcritical, supercritical, and superheated steam.
Type
Subcritical
ressure
P
Relation
Region
< 22.06 MPa
Under vapor dome
Supercritical > 22.06 MPa
No distinct phases
uperheate
S
d
ingle phase
S
vapor
> T_sat
Supercritical steam improves efficiency in modern thermal plants.
Q30. Explain real-world applications of steam property analysis.
Answer:
● Turbine design:Calculate expansion work (Δh).
● Condenser design:Find heat rejection rate.
● Boiler design:Determine steam generation rate.
● Refrigeration:Analyze throttling and phase behavior.
✅ End of Module 3 – 30 In-depth Conceptual & Practical Questions.
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M odule 4 – Gas Power Cycles (Otto, Diesel, Dual, Brayton, efficiency trends, MEP, and
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advanced cycle modifications)
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M ODULE 4 – Gas Power Cycles
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e xtenders, and connects your thermodynamic theory withreal-world design and performance.
elow are30 comprehensive interview Q&A, fully aligned withVTU syllabusandindustry
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⚙️ THERMODYNAMICS – VTU STRUCTURE
🔹 MODULE 4 – Gas Power Cycles
(30 Deep, Formula-Backed, Industry-Relevant Questions & Answers)
Q1. What is a power cycle?
nswer:
A
Apower cyclec onverts heat into mechanical work through a sequence of thermodynamic
processes that return to the initial state.
Examples:O tto, Diesel, Dual, Braytonc ycles.
Q2. What is the working fluid in air-standard cycles?
nswer:
A
Assumed to beairthat behaves as anideal gaswith constant specific heats.
This assumption simplifies IC engine analysis — known as theAir-Standard Assumption.
Q3. List the assumptions made in air-standard analysis.
Answer:
1. Air is the working fluid and ideal gas.
2. The cycle is closed (mass constant).
3. All processes are internally reversible.
4. Combustion replaced by heat addition.
5. Exhaust replaced by heat rejection.
6. Specific heats (Cp, Cv) are constant.
Q4. Define the Otto cycle.
nswer:
A
Anair-standard cycleforspark-ignition (SI)engines (petrol).
rocesses:
P
1–2: Isentropic compression
2–3: Constant-volume heat addition
3–4: Isentropic expansion
4–1: Constant-volume heat rejection
Efficiency:
ηOtto=1−1r(γ−1)\eta_{Otto} = 1 - \frac{1}{r^{(\gamma - 1)}}ηOtto=1−r(γ−1)1
W here r=V1/V2r = V_1/V_2r=V1/V2(compression ratio).
Q5. Explain the Diesel cycle.
nswer:
A
Cycle forcompression-ignition (CI)engines (diesel).
rocesses:
P
1–2: Isentropic compression
2–3: Constant-pressure heat addition
3–4: Isentropic expansion
4–1: Constant-volume heat rejection
Efficiency:
Diesel=1−1r(γ−1)×(ργ−1)γ(ρ−1)\eta_{Diesel} = 1 - \frac{1}{r^{(\gamma - 1)}} \times
η
\frac{(\rho^\gamma - 1)}{\gamma(\rho - 1)}ηDiesel=1−r(γ−1)1×γ(ρ−1)(ργ−1)
where ρ=V3/V2\rho = V_3/V_2ρ=V3/V2(cut-off ratio).
Q6. What is the Dual cycle?
nswer:
A
A combination of Otto and Diesel cycles — represents real IC engine operation.
Processes:
1. Isentropic compression
2. Constant-volume heat addition
3. Constant-pressure heat addition
4. Isentropic expansion
5. Constant-volume heat rejection
Efficiency:
Dual=1−1r(γ−1)(αργ−1)(α−1)+γα(ρ−1)\eta_{Dual} = 1 - \frac{1}{r^{(\gamma - 1)}}
η
\frac{(\alpha\rho^\gamma - 1)}{(\alpha - 1) + \gamma\alpha(\rho -
1)}ηDual=1−r(γ−1)1(α−1)+γα(ρ−1)(αργ−1)
Q7. Compare Otto, Diesel, and Dual cycles.
Feature
O tto
Heat addition
onstant
C
volume
Diesel
Dual
onstant
C
pressure
Both
Efficiency (same r) Highest
Lowest
Intermediate
Used in
Diesel engine
Heavy-duty diesel
Petrol engine
Q8. What is the mean effective pressure (MEP)?
nswer:
A
A hypothetical constant pressure that would produce the same net work as the actual cycle.
MEP=WnetVdispMEP = \frac{W_{net}}{V_{disp}}MEP=VdispW net
Used to compare performance of different engines independent of size.
Q9. Define compression ratio (r) and its significance.
r=VmaxVminr = \frac{V_{max}}{V_{min}}r=VminVmax
igher r → greater temperature rise before combustion → better efficiency.
H
Limited in petrol engines due toknock.
Q10. Define cut-off ratio (ρ) in Diesel and Dual cycles.
ρ=V3V2\rho = \frac{V_3}{V_2}ρ=V2V3
Indicates duration of heat addition at constant pressure.
Lower ρ → higher efficiency.
Q11. What is efficiency trend among cycles for same compression ratio?
ηOtto>ηDual>ηDiesel\eta_{Otto} > \eta_{Dual} > \eta_{Diesel}ηOtto>ηDual>ηDiesel
Because heat addition is at higher mean temperature in Otto cycle.
Q12. What is the Brayton cycle?
nswer:
A
Theair-standard cycleforgas turbines.
rocesses:
P
1–2: Isentropic compression
2–3: Constant-pressure heat addition
3–4: Isentropic expansion
4–1: Constant-pressure heat rejection
Brayton=1−1rp(γ−1)/γ\eta_{Brayton} = 1 - \frac{1}{r_p^{(\gamma -
η
1)/\gamma}}ηBrayton=1−rp(γ−1)/γ1
where rp=P2/P1r_p = P_2/P_1rp=P2/P1= pressure ratio.
Q13. What is the effect of pressure ratio on Brayton cycle efficiency?
nswer:
A
Efficiency increases with pressure ratio up to a limit, then drops if turbine inlet temperature is
restricted.
Q14. What are the main components of a gas turbine power plant?
Answer:
1. Compressor (air compression)
2. Combustion chamber (heat addition)
3. Turbine (work output)
4. Generator or load
Cycle: compressor → combustion → turbine → exhaust.
Q15. What is the regeneration process in gas turbines?
nswer:
A
Aheat exchanger (regenerator)transfers heat from turbine exhaust to compressed air before
c ombustion.
ηregen=T3−T2T3−T5\eta_{regen} = \frac{T_3 - T_2}{T_3 - T_5}ηregen=T3−T5T3−T2
Improves efficiency by reducing fuel input.
Q16. What is intercooling in a gas turbine?
nswer:
A
W hen compression is done intwo or more stages, cooling between stages reduces compressor
work.
Result:Net work output ↑, thermal efficiency ↑.
Q17. What is reheating in a gas turbine?
nswer:
A
Expansion intwo turbine stageswith reheating between them increases total turbine work and
reduces exhaust moisture.
Q18. Explain modifications in Brayton cycle for performance improvement.
Answer:
1. Regeneration→ Efficiency ↑
2. Intercooling→ Compressor work ↓
3. Reheating→ Turbine work ↑
Combining all =reheat-regenerative cycle.
Q19. Compare reciprocating engines and gas turbines.
Parameter
IC Engine
W ork fluid
Air-fuel mix
Air only
Speed
Low–Medium
Very high
Efficiency
25–35%
30–45%
Size
Compact
Large
Maintenance High
G as
Turbine
Moderate
Q20. What is a jet propulsion cycle?
nswer:
A
Modified open Brayton cycle where theturbine drives only the compressor, and thethrustc omes
from high-velocity exhaust gases.
Used in aircraft engines.
Q21. What are losses in gas turbine cycles?
Answer:
● Frictional (in turbine/compressor)
● Pressure drop in combustion chamber
● Heat losses to surroundings
● Incomplete combustion
Leads to loweractual efficiencythan ideal.
Q22. Define isentropic efficiencies of turbine and compressor.
turbine=h3−h4h3−h4s,ηcompressor=h2s−h1h2−h1\eta_{turbine} = \frac{h_3 - h_4}{h_3 - h_{4s}},
η
\quad \eta_{compressor} = \frac{h_{2s} - h_1}{h_2 -
h_1}ηturbine=h3−h4sh3−h4,ηcompressor=h2−h1h2s−h1
Measures deviation of real devices from ideal (isentropic).
Q23. What is back work ratio in gas turbines?
WR=WcompressorWturbineBWR =
B
\frac{W_{compressor}}{W_{turbine}}BWR=WturbineW compressor
Typically 40–50%, meaning half turbine work drives the compressor.
Q24. Define specific fuel consumption (SFC).
SFC=m˙fuelPoutputSFC = \frac{\dot{m}_{fuel}}{P_{output}}SFC=Poutputm ˙fuel
Indicates fuel efficiency — lower SFC = better performance.
Q25. What are methods to improve gas turbine efficiency?
Answer:
● Regeneration
● Reheating
● Intercooling
● Increasing pressure ratio
● Raising turbine inlet temperature
Used incombined-cycleandjet engines.
Q26. What is the difference between open and closed Brayton cycles?
Type
Description
O pen
ir enters and exhausts to
A
atmosphere
Example
Aircraft jet engine
lose
C
d
W orking fluid recirculates
pace power
S
s ystems
Q27. What are the practical applications of air-standard cycles?
Answer:
● O tto:Petrol engines, generators
● Diesel:Trucks, locomotives
● Dual:Heavy-duty engines
● Brayton:Aircraft, gas turbines, power plants
Q28. What is the effect of compression ratio on engine power?
nswer:
A
Higher compression ratio → higher efficiency but also higher knocking tendency.
In Diesel engines, efficiency increases without knock due to fuel auto-ignition.
Q29. Why is actual efficiency lower than theoretical efficiency?
Answer:
● Heat losses to walls
● Friction
● Incomplete combustion
● Leakage, valve timing losses
● F
inite-time heat transfer
→ Irreversibility reduces real efficiency.
Q30. Explain combined gas–steam (topping–bottoming) cycle.
nswer:
A
Exhaust heat from a gas turbine (Brayton cycle) is used to generate steam for a Rankine cycle.
Result:Combined-cycle efficiency up to60%.
Used inmodern power plantsfor high performance.
✅ End of Module 4 – 30 Comprehensive Interview Questions & Answers.
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M odule 5 – Vapour Power Cycles (Rankine, Reheat, Regeneration, Condenser, Efficiency,
⚙️
Real Modifications)
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⚙️ THERMODYNAMICS – VTU STRUCTURE
🔹 MODULE 5 – Vapour Power Cycles
(30 Detailed Concept + Formula + Industry Application Q&A)
Q1. What is a vapour power cycle?
nswer:
A
Av apour power cyclec onverts heat energy into mechanical work using acondensable working
f luid (like steam).
Example:Rankine cycleused in thermal power plants.
Q2. What are the main components of a steam power plant?
1. Boiler– Converts water into steam.
2. Turbine– Expands steam to produce work.
3. Condenser– Condenses exhaust steam.
4. Pump– Increases pressure of condensate.
Cycle:Boiler → Turbine → Condenser → Pump → Boiler.
Q3. Explain the simple Rankine cycle with processes.
rocesses:
P
1–2: Isentropic compression (pump)
2–3: Constant-pressure heat addition (boiler)
3–4: Isentropic expansion (turbine)
4–1: Constant-pressure heat rejection (condenser)
=WnetQin=1−QoutQin=1−h4−h1h3−h2\eta = \frac{W_{net}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}} = 1 -
η
\frac{h_4 - h_1}{h_3 - h_2}η=QinW net=1−QinQ out=1−h3−h2h4−h1
Q4. Why is Rankine cycle used instead of Carnot cycle?
Answer:
● Condensation at constant temperature (Carnot) is impractical.
● Rankine allowsconstant-pressureheat transfer → simpler design.
● Easier to use in real boilers and condensers.
Q5. What factors affect Rankine cycle efficiency?
1. Boiler pressure ↑ → efficiency ↑
2. Condenser pressure ↓ → efficiency ↑
3. Superheating ↑ → efficiency ↑
4. Reheating & regeneration → efficiency ↑
Q6. Explain the T–s and h–s diagrams for the Rankine cycle.
● T –s Diagram:Rectangular shape with expansion and compression as vertical lines.
● h
–s Diagram (Mollier):W ork = Δh (turbine)
Area under curve = Heat transfer.
Q7. What is the purpose of superheating in Rankine cycle?
nswer:
A
To raise the steam temperature beyond saturation at constant pressure →
Increases efficiency
✅
Reduces turbine moisture
✅
Increases specific work output
✅
egree of Superheat=Tsuperheat−Tsat\text{Degree of Superheat} = T_{superheat} - T_{sat}Degree of
D
Superheat=Tsuperheat−Tsat
Q8. Define condenser pressure and its effect.
nswer:
A
Lowering condenser pressure (vacuum) increases expansion ratio and work output.
But very low pressure → higher turbine moisture → reduced life.
O ptimum:~0.08 bar.
Q9. What is the effect of boiler pressure on Rankine cycle efficiency?
Higher boiler pressure →
● Higher mean temperature of heat addition
● More turbine work
● Efficiency ↑ but requires high-strength materials.
Q10. What are the limitations of simple Rankine cycle?
1. Low efficiency (~30–35%)
2. High exhaust moisture
3. Limited heat recovery
4. Large condenser losses
Hence, modified Rankine cycles are used (Reheat, Regeneration).
Q11. What is a reheat Rankine cycle?
nswer:
A
Steam expands intwo turbine stages:
● HP turbine (partial expansion)
● Reheated in boiler
● LP turbine (further expansion)
Advantages:
● Moisture reduced
● Efficiency ↑
● O utput ↑
Q12. What is the regenerative Rankine cycle?
nswer:
A
Steam extracted (bled) from turbine is used topreheat feedwaterbefore entering the boiler.
Increasesaverage temperature of heat addition, improving efficiency.
Q13. What are feedwater heaters and their types?
nswer:
A
Devices that transfer heat from extracted steam to feedwater.
Types:
● O pen (direct contact)
● Closed (surface type)
O pen heaters → Simple, cheap; Closed → Compact, no mixing.
Q14. What is the combined effect of reheat and regeneration?
Answer:
● Higher efficiency
● Reduced moisture at turbine exit
● Higher mean heat-addition temperature
Used inmodern thermal plants (supercritical Rankine).
Q15. Write the expressions for pump and turbine work.
pump=v(P2−P1)W_{pump} = v (P_2 - P_1)Wpump=v(P2−P1) Wturbine=h3−h4W_{turbine} = h_3 -
W
h_4Wturbine=h3−h4
Net work = Wturbine−WpumpW_{turbine} - W_{pump}Wturbine−Wpump
Q16. What is boiler efficiency?
boiler=Heat to steamHeat from fuel×100\eta_{boiler} = \frac{\text{Heat to steam}}{\text{Heat from
η
fuel}} \times 100ηboiler=Heat from fuelHeat to steam×100
Typically 80–90%.
Q17. Define cycle efficiency and specific steam consumption (SSC).
cycle=WnetQboiler\eta_{cycle} = \frac{W_{net}}{Q_{boiler}}ηcycle=QboilerW net
η
SSC=3600Wnet (kg/kWh)SSC = \frac{3600}{W_{net}} \, (kg/kWh)SSC=Wnet3600(kg/kWh)
Lower SSC → higher efficiency.
Q18. What is back work ratio?
BWR=WpumpWturbineBWR = \frac{W_{pump}}{W_{turbine}}BWR=WturbineW pump
Small for Rankine (≈ 0.5–1%), high for Brayton (40–50%).
Q19. Explain the function of a condenser in a power plant.
nswer:
A
Condenses exhaust steam → converts vapor to water.
Maintains low pressure → increases turbine expansion ratio → increases work output.
Q20. What are types of condensers?
1. Jet (Mixing)
2. Surface (Non-mixing)
Surface type preferred → allows condensate reuse.
Q21. What is regenerative feedwater heating and its purpose?
nswer:
A
Increases mean temperature of heat addition → improves efficiency.
Done by bleeding steam at intermediate pressures.
Used in modern Rankine systems.
Q22. Why is reheat cycle more efficient than simple Rankine?
Because it:
● Raises average temperature of heat addition
● Reduces moisture content
● Produces higher work output per kg of steam
Q23. Explain how supercritical Rankine cycle differs from subcritical.
Type
Subcritical
Pressure
Phase
Efficiency
<22.06 MPa Liquid & vapor phases Lower
Supercritical >22.06 MPa Single-phase fluid
Higher
Advantage:Compact design, higher efficiency (~42–45%).
Q24. What is economizer, superheater, and air preheater?
Answer:
● Economizer:Preheats feedwater using exhaust flue gases.
● Superheater:Converts saturated to superheated steam.
● A
ir preheater:Heats incoming air for combustion.
All increase boiler efficiency.
Q25. Define dryness fraction at turbine exit and its effect.
nswer:
A
Dryness fraction (x) = vapor mass fraction at turbine exit.
If x < 0.85 → turbine blades face erosion.
Hence, reheat or superheat is used to improve x.
Q26. What is the purpose of a deaerator in feedwater system?
nswer:
A
Removes dissolved gases (O₂, CO₂) → prevents corrosion in boiler tubes and turbine blades.
Q27. Why is condenser pressure not reduced below a certain limit?
Because too low a pressure:
● Increases moisture in turbine exhaust
● Needs large condenser area
● Increases pump work
Hence, practical lower limit ≈ 0.08 bar.
Q28. What is cycle efficiency improvement method used in thermal plants?
1. Increase boiler pressure
2. Reheat
3. Regeneration
4. Superheating
5. Reduce condenser pressure
6. Use combined cycles
Q29. What is the function of the Rankine cycle in combined-cycle power plants?
nswer:
A
Acts asbottoming cycleutilizingwaste heatfrom theBrayton (gas turbine) exhaust→ enhances
overall efficiency up to 60%.
Q30. What are real-world losses in a steam power plant?
1. Pump & turbine inefficiency
2. Heat loss in boiler and piping
3. Pressure drop losses
4. Leakage
5. Condenser inefficiency
Reducing these losses improves overall plant efficiency and reliability.
✅ End of Module 5 – 30 In-Depth Questions (Conceptual + Practical + Power Plant Focus)
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M odule 6 – Refrigeration & Air Conditioning (30 Q&A)— includingVCR, VAR, COP,
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his module is extremely important forHVAC, EV battery cooling, automotive thermal systems,
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W e’ll coverreversed Carnot, vapor compression/absorption cycles, COP, refrigerants, and
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⚙️ THERMODYNAMICS – VTU STRUCTURE
🔹 MODULE 6 – Refrigeration and Air Conditioning
(30 In-depth Interview Questions & Answers – VTU + Industry Standard)
Q1. What is refrigeration?
nswer:
A
Refrigeration is theprocess of removing heatfrom a low-temperature region andrejecting itto a
high-temperature region usinge xternal work.
Unit:1 Ton of Refrigeration (TR) = 3.516 kW = 211 kJ/min.
Applications:Food preservation, air-conditioning, automobile AC, and EV cooling.
Q2. Define Coefficient of Performance (COP).
nswer:
A
A measure of refrigeration performance.
COPR=QLW=TLTH−TLCOP_R = \frac{Q_L}{W} = \frac{T_L}{T_H - T_L}COPR=WQL=TH−TLTL
igher COP → better efficiency.
H
For heat pump:
COPHP=COPR+1COP_{HP} = COP_R + 1COPHP=COPR+1
Q3. What is the difference between a heat engine and a refrigerator?
Feature
Purpose
Heat Engine
Refrigerator
Converts heat → work Transfers heat from low → high temp
Efficiency
η = W/Q_H
W ork direction O utput
COP = Q_L/W
Input
Q4. What is the Reversed Carnot cycle?
nswer:
A
Anideal refrigeration cycle, reverse of Carnot engine.
Processes:
1. Isentropic compression
2. Isothermal heat rejection
3. Isentropic expansion
4. Isothermal heat absorption
COP=TLTH−TLCOP = \frac{T_L}{T_H - T_L}COP=TH−TLTL
Limitation:Not practical due to isothermal and reversible constraints.
Q5. What is the Vapour Compression Refrigeration Cycle (VCRC)?
nswer:
A
Apractical refrigeration systemused in domestic, automotive, and industrial applications.
Processes:
1. 1–2: Isentropic compression (compressor)
2. 2–3: Constant-pressure condensation
3. 3–4: Throttling expansion
4. 4–1: Constant-pressure evaporation
COP:
COP=h1−h4h2−h1COP = \frac{h_1 - h_4}{h_2 - h_1}COP=h2−h1h1−h4
Q6. Explain the P–h and T–s diagrams of VCRC.
● P–h Diagram:Horizontal = constant pressure; vertical = throttling (constant enthalpy).
● T
–s Diagram:Two horizontal (isothermal), two vertical (adiabatic) lines.
Used to visualizework and refrigeration effect.
Q7. Define refrigeration effect and compressor work.
efrigeration Effect=h1−h4\text{Refrigeration Effect} = h_1 - h_4Refrigeration Effect=h1−h4Work of
R
Compressor=h2−h1\text{Work of Compressor} = h_2 - h_1Work of Compressor=h2−h1
COP = (h₁ − h₄) / (h₂ − h₁)
Q8. What are the main components of a vapor-compression system?
1. Evaporator– Absorbs heat at low pressure.
2. Compressor– Increases pressure and temperature.
3. Condenser– Rejects heat to surroundings.
4. Expansion Valve– Reduces pressure and temperature.
Q9. What is subcooling and why is it beneficial?
nswer:
A
Subcooling = Cooling of liquid refrigerant below condensation temperature.
Increases refrigeration effect → improves COP → prevents vapor formation before expansion valve.
Q10. What is superheating and why is it used?
nswer:
A
Superheating = Heating of vapor refrigerant above evaporation temperature.
Ensuresno liquid enters the compressor→ prevents damage and increases efficiency.
Q11. What are desirable properties of a refrigerant?
High latent heat
✅
Low specific volume
✅
Non-toxic, non-corrosive
✅
Chemically stable
✅
Easily detectable in case of leakage
✅
Environmental-friendly (low ODP, GWP)
✅
Q12. List common refrigerants and their applications.
Refrigerant
Formula
Application
R-12 (CCl₂F₂) Freon
O bsolete (ozone depletion)
R-22
CHClF₂
C, heat pumps (phasing
A
out)
R-134a
C₂H₂F₄
Car AC, domestic
R-717 (NH₃)
—
Industrial plants
R-744 (CO₂)
—
Eco-friendly systems
Q13. Why is ammonia (NH₃) still widely used?
Answer:
● High latent heat → efficient
● Cheap and available
● E
co-friendly (zero ODP)
Buttoxic, so used only in industrial setups.
Q14. Compare Vapour Compression (VCR) and Vapour Absorption (VAR) systems.
Feature
Energy input
VCR
VAR
Mechanical work Heat energy
Main element Compressor
Generator + Absorber
COP
3–6
0.5–2
Noise
Higher
Silent
Maintenance
Higher
Low
Example
Domestic fridge
Industrial cold
s torage
15. What is the working principle of a Vapour Absorption Refrigeration System
Q
(VARS)?
nswer:
A
Usesheatinstead of mechanical work.
W orking pair: NH₃–H₂O or LiBr–H₂O.
Components:Generator, Absorber, Condenser, Evaporator, Expansion Valve, Pump.
Heat supplied to generator → refrigerant evaporates → condenses → evaporates again in evaporator.
Q16. What are the advantages of VARS?
Uses low-grade waste heat or solar energy.
✅
Quiet operation.
✅
Less maintenance.
✅
Useful in places with limited electricity.
✅
Q17. Define Ton of Refrigeration (TR).
nswer:
A
The cooling effect equivalent to melting1 ton (1000 kg) of ice in 24 hours.
1TR=3.516kW=211kJ/min1 TR = 3.516 kW = 211 kJ/min1TR=3.516kW=211kJ/min
Q18. What are common types of compressors used in VCR systems?
1. Reciprocating– Low capacity
2. Rotary vane– Moderate capacity
3. Screw– Large capacity
4. Centrifugal– High flow, industrial use
Q19. What are the types of condensers?
1. Air-cooled:Uses air as cooling medium.
2. Water-cooled:Shell-and-tube or evaporative.
3. Evaporative condenser:Combines both.
Q20. What are the types of evaporators?
1. Bare tube
2. Finned tube
3. Shell and coil
4. Plate type
Used depending on required heat transfer rate.
Q21. What is a throttling device and its function?
nswer:
A
It expands high-pressure liquid refrigerant to low pressure (constant enthalpy).
Types: Capillary tube, Expansion valve, Orifice.
Used to control refrigerant flow.
Q22. What is COP for ideal refrigeration system based on temperature?
COPideal=TLTH−TLCOP_{ideal} = \frac{T_L}{T_H - T_L}COPideal=TH−TLTL
Higher TLT_LTLor lower THT_HTHincreases performance.
Q23. What is psychrometry?
nswer:
A
Study ofmoist air propertiesandprocesses involving moisture.
Important for air-conditioning design and comfort analysis.
Q24. Define dry-bulb, wet-bulb, and dew-point temperatures.
Term
Description
Dry-Bulb (DBT)
Actual air temperature
Wet-Bulb (WBT)
emperature considering evaporation
T
c ooling
Dew Point (DPT) Temperature where condensation starts
Q25. Define specific and relative humidity.
pecific humidity (ω)=0.622pvP−pv\text{Specific humidity } (\omega) = 0.622 \frac{p_v}{P -
S
p_v}Specific humidity (ω)=0.622P−pvpvRelative humidity (ϕ)=pvpvs×100\text{Relative humidity }
(\phi) = \frac{p_v}{p_{vs}} \times 100Relative humidity (ϕ)=pvspv×100
RH indicates comfort and cooling load.
Q26. What are the comfort conditions in air-conditioning?
Answer:
● Temperature: 22–25°C
● Relative Humidity: 50–60%
● Air Velocity: 0.15–0.25 m/s
● Proper ventilation and purity
Q27. What are psychrometric processes?
1. Sensible heating/cooling– DBT changes, humidity constant.
2. Humidification/dehumidification– Moisture content changes.
3. Evaporative cooling– Constant enthalpy process.
4. M ixing of air streams– Combines two different conditions.
Q28. What is a psychrometric chart and how is it used?
nswer:
A
A graphical representation of moist air properties — showsDBT, WBT, RH, enthalpy, and specific
v olume.
Used to calculate cooling/heating loads, coil design, and air-mixing operations.
Q29. What are the main factors affecting human comfort?
● Temperature
● Humidity
● Air velocity
● Radiation
● Air cleanliness
● Clothing and activity level
Balanced control of all ensuresthermal comfort.
Q30. List real-life applications of refrigeration and air-conditioning.
● Domestic and industrial refrigeration
● Automobile and train air-conditioning
● EV battery thermal management
● Cold storage for food and pharma
● Cryogenic cooling
● HVAC in commercial buildings
✅ End of Module 6 – 30 Fully Detailed Q&A Covering RAC Fundamentals and Applications.
CONGRATULATIONS, macha!
🎯
You’ve now got theComplete VTU-Based Thermodynamics Interview Bible
→180 In-depth, topic-wise, industry-linked Q&A,
c overing every concept fromBasic Laws → Power & Refrigeration Systems.
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