Q1 50M Compulsory calculate Thermodynamics, Heat Transfer and Fluid Mechanics
(a) A lump of ice with a mass of 1·5 kg at an initial temperature of 260 K melts at the pressure of 1 bar as a result of heat transfer from the environment. After some time, the resulting water attains the temperature of environment, 293 K. Calculate the entropy production associated with this process. The latent heat of fusion of ice is 334 kJ/kg, the specific heat of ice and water are 2·07 kJ/kg-K and 4·2 kJ/kg-K, respectively. Assume that ice melts at 273·15 K. (10 marks)
(b) Downstream of a normal shock wave, the characteristic Mach number M*_{y} = 0·5 and the stagnation pressure is 2 bar. Determine the following:
(i) The Mach number M_{y} downstream of the shock
(ii) The stagnation pressure upstream of the shock
The fluid is air and γ = 1·4, and R = 0·287 kJ/kg-K.
Normal shock table and Isentropic flow table provided at the end, may be used. (10 marks)
(c) Show, in the form of a table, how the following parameters change (increase/decrease/remain constant) across a normal shock wave:
Static pressure, Static temperature, Stagnation pressure, Stagnation temperature, Stagnation density, Mach number, Entropy and Stagnation enthalpy (10 marks)
(d) At a certain instant of time, the temperature across a large wall of thickness 50 cm is given as T(x) = 90 - 80x + 16x² + 32x³ - 25x⁴ where x is in metres and measured from the left face of the wall as shown in the figure below and T is in °C. If the area of the wall is 10 m² and there is no generation of heat in the wall, compute the following:
(i) The rate of heat entering and leaving the wall
(ii) The rate of heat energy stored in the wall
(iii) The rate of temperature change with time at x = 0 and x = 0·5 m (10 marks)
(e) A 120 W electric bulb has a filament temperature of 3005 °C. Assuming the filament to be black, calculate (i) the diameter of the filament wire if the length is 250 mm and (ii) the efficiency of the bulb based on visible radiation if the visible radiation lies in the wavelength range from 0·4 μm to 0·75 μm. Assume Stefan-Boltzmann constant (σ) as 5.67×10⁻⁸ W/m²-K⁴. The black body radiation functions are given in the table:
| λT(μm-K) | f₀→λ |
|----------|------|
| 1300 | 0.004317 |
| 1400 | 0.007791 |
| 2400 | 0.140266 |
| 2500 | 0.161366 | (10 marks)
Answer approach & key points
Calculate systematically across all five parts: (a) entropy production for ice melting requires three-stage entropy analysis; (b) normal shock uses characteristic Mach number relations and tables; (c) tabular comparison of shock parameters; (d) transient heat conduction with polynomial temperature profile; (e) blackbody radiation with fractional emission. Allocate approximately 20% time to each part, ensuring all numerical working is shown with proper units and sign conventions.
- (a) Three-stage entropy calculation: ice warming 260→273.15 K, melting at 273.15 K, water warming 273.15→293 K; entropy production = ΔS_system + ΔS_surroundings with Q/T_boundary for surroundings
- (b)(i) Characteristic Mach number M* = √[(γ+1)/2 * M²/(1+(γ-1)M²/2)] inverted to find M_y = 0.534 downstream
- (b)(ii) Stagnation pressure ratio across shock from normal shock tables: p_0x/p_0y = 1.113, giving p_0x = 2.226 bar upstream
- (c) Tabular comparison: static pressure ↑, static temperature ↑, stagnation pressure ↓, stagnation temperature →, stagnation density ↓, Mach number ↓, entropy ↑, stagnation enthalpy →
- (d) Heat rates using Fourier's law at x=0 and x=0.5 m: q = -kA(dT/dx); energy storage from ∂T/∂t = α(∂²T/∂x²) with no generation
- (e)(i) Stefan-Boltzmann law Q = σAT⁴ with T = 3278 K to find filament diameter d = 0.457 mm
- (e)(ii) Blackbody radiation function interpolation: f(0.4μm) and f(0.75μm) at λT values, visible efficiency η = (f_0.75 - f_0.4) × 100 = 6.53%
- Consistent units throughout: kJ/kg-K for specific heats, bar for pressure, μm-K for λT products, W/m²-K⁴ for σ
Q2 50M calculate Thermodynamics, Turbomachinery and Convective Heat Transfer
(a) (i) A reversible power cycle engine is used to drive a reversible heat pump. The power cycle takes in Q₁ heat units at temperature T₁ and rejects heat Q₂ at temperature T₂. The heat pump abstracts heat Q₄ from the sink at temperature T₄ and discharges heat Q₃ at temperature T₃. Develop an expression for the ratio Q₄/Q₁ in terms of the four temperatures T₁, T₂, T₃ and T₄. (10 marks)
(ii) Argon gas expands adiabatically in a turbine from 2 MPa, 1000 °C to 350 kPa. The mass flow rate of argon is 0·5 kg/s and the turbine develops power at the rate of 120 kW. Determine the following:
(1) The temperature of argon at the turbine exit
(2) The irreversibility rate
(3) The second law efficiency
Neglect kinetic and potential energy effects and take T₀ = 20 °C and P₀ = 1 atm. Take molecular weight of argon as 40 kg/kmol and γ = 1·67. (T₀ and P₀ are the environment temperature and pressure, respectively) (10 marks)
(b) Compressed air is transported in an industrial pipeline of 50 mm internal diameter. The stagnation conditions at the inlet are P₀ = 10 bar and T₀ = 400 K. The average Fanning friction factor f̄ = 0·002. If the Mach number changes from 3 at the entry to 1 at the exit, determine the following:
(i) The length of the pipe
(ii) The velocity at the exit
(iii) The change in the stagnation temperature
(iv) The change in the stagnation pressure
(v) The change in the entropy
(vi) The mass flow rate
Assume the flow to be adiabatic. For air, take γ = 1·4 and R = 287 J/kg-K.
Isentropic flow table and Fanno flow table attached at the end, may be used. (20 marks)
(c) A thin flat plate, that is 0·2 m×0·2 m on a side, is oriented parallel to an atmospheric air stream having a velocity of 40 m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 120 °C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0·075 N. What is the rate of heat transfer from both sides of the plate to the air?
(Assume ρair = 0·995 kg/m³, νair = 20·92×10⁻⁶ m²/s, Prair = 0·7, kair = 30×10⁻³ W/m-K) (10 marks)
Answer approach & key points
Calculate and derive expressions for all nine sub-parts systematically, allocating approximately 20% time to (a)(i) reversible cycle derivation, 20% to (a)(ii) argon turbine analysis, 45% to (b) Fanno flow calculations using attached tables, and 15% to (c) Reynolds analogy application. Begin each part with governing equations, substitute given data with proper unit conversions, and conclude with numerical answers and physical interpretation.
- (a)(i) Derivation: COP_heat_pump = Q₃/W = T₃/(T₃-T₄), η_power_cycle = W/Q₁ = 1-T₂/T₁, combine to get Q₄/Q₁ = (T₁-T₂)(T₃-T₄)/(T₁(T₃-T₄)) with algebraic manipulation for Q₄ = Q₃ - W
- (a)(ii) Argon turbine: T₂ = T₁(P₂/P₁)^((γ-1)/γ) = 1273×(0.35/2)^0.402 = 773 K; actual work from power = 120/0.5 = 240 kJ/kg; actual T₂ = 1273 - 240/0.312 = 503 K; irreversibility = T₀(s₂-s₁) - q = 293×0.312×ln(773/503) = 38.6 kJ/kg; η_II = (h₁-h₂_actual)/(h₁-h₂_isentropic)
- (b) Fanno flow: Use tables at M₁=3, M₂=1 to find 4fL*/D = 0.522; L = 0.522×0.05/(4×0.002) = 3.26 m; V_exit = a* = √(γRT*) = 347 m/s; T₀ constant (adiabatic) so ΔT₀=0; P₀₂/P₀₁ from tables = 0.408; Δs = Rln(P₀₁/P₀₂) = 287×ln(2.45) = 258 J/kg-K; ṁ = ρ*A*a*
- (c) Reynolds analogy: St = Cf/2 = h/(ρCpU); from drag F_D = 2×(1/2)ρU²×Cf×A, find Cf = 0.075/(0.995×1600×0.08) = 0.000588; h = St×ρCpU = (Cf/2)×ρCpU = 0.000294×0.995×1005×40 = 11.7 W/m²K; Q = 2×h×A×(T_s-T_∞) = 2×11.7×0.04×100 = 93.6 W
- Unit consistency: Convert all temperatures to Kelvin, pressures to consistent units (Pa or kPa), and verify dimensional homogeneity in each equation
- Table usage in (b): Explicitly state Fanno flow parameter values extracted from tables (e.g., P/P*, T/T*, P₀/P₀* at M=3 and M=1) and show interpolation if needed
- Second law analysis: Clearly distinguish between isentropic, actual, and dead states for exergy and irreversibility calculations in (a)(ii)
- Physical verification: Check that results satisfy thermodynamic constraints (e.g., entropy generation ≥ 0, Mach number trends in Fanno flow, heat transfer direction in (c))
Q3 50M calculate Heat transfer and thermodynamics
(a) A long cylindrical rod of radius 10 cm consists of a nuclear reacting material (k = 0·5 W/m-K) generating 24000 W/m³ uniformly throughout its volume. The rod is encapsulated within another cylinder whose outer radius is 20 cm and that has a thermal conductivity of 4 W/m-K. The outer surface is surrounded by a fluid at 100 °C, and the convection coefficient between the surface and the fluid is 20 W/m²-K. Find the temperatures of the interface between the two cylinders, at the outer surface and the maximum temperature under steady-state condition. (20 marks)
(b) A centrifugal compressor is to be designed for an industrial application handling air. The inlet stagnation conditions are P₀₁ = 1·1 bar and T₀₁ = 295 K. The air enters the eye of the impeller axially, without any prewhirl. The axial velocity is uniform throughout the eye and is equal to 143 m/s. The eye tip and root diameters are 0·30 m and 0·15 m, respectively. Calculate the mass flow rate of air.
The overall diameter of the impeller is 0·50 m. The power input factor is 1·04 and the slip factor is 0·9. The rotational speed of the compressor is 290 revolutions/second. The isentropic efficiency of the compressor (based on total head) is 0·78. The radial velocity at the impeller tip is 143 m/s. Assume that 'half the total losses' occurs in the impeller. Determine the pressure ratio and the power required to drive the compressor. Also, determine the axial depth of the impeller channels at the periphery of the impeller. Draw the T-s diagram showing the variations of both static and stagnation pressures and temperatures in the impeller and the diffuser.
For air, γ = 1·4, Cp = 1·005 kJ/kg-K. (20 marks)
(c) The velocity and temperature profiles for laminar flow in a tube of radius r0 = 10 mm have the form at a particular axial location
u(r) = 0·1(1 - r²/r0²)
T(r) = 344·8 + 75 r²/r0² - 18·8 r⁴/r0⁴
with units of m/s and K, respectively. Determine the corresponding value of the mean (or bulk) temperature Tm at this axial position. (10 marks)
Answer approach & key points
Calculate systematically across all three sub-parts: for (a) apply cylindrical heat conduction with internal heat generation and composite wall resistance; for (b) solve centrifugal compressor thermodynamics with velocity triangles, slip factor, and efficiency corrections; for (c) integrate velocity-weighted temperature for bulk mean temperature. Allocate approximately 35% time to part (a), 45% to part (b) due to its multi-step complexity and diagram requirement, and 20% to part (c). Present derivations with clear boundary condition statements and conclude each part with physically verified numerical answers.
- Part (a): Apply heat conduction equation in cylindrical coordinates with uniform heat generation; use thermal resistance network for composite cylinder with convection at outer surface; identify maximum temperature at centerline (r=0)
- Part (a): Calculate interface temperature T₁, outer surface temperature T₂, and T_max using continuity of heat flux and temperature at r=10 cm interface
- Part (b): Calculate mass flow rate using continuity equation with variable annulus area at impeller eye; determine blade tip speed U₂ = πND₂
- Part (b): Apply slip factor and power input factor to find actual work input; use isentropic efficiency to find pressure ratio; determine impeller channel depth from radial velocity and through-flow area
- Part (b): Draw complete T-s diagram showing: stagnation and static states at impeller inlet, impeller outlet (with losses), diffuser outlet; label pressure rise and temperature rise correctly
- Part (c): Apply definition of bulk mean temperature T_m = (∫ρuC_p T dA)/(ṁC_p) = (∫₀^r₀ uT r dr)/(∫₀^r₀ u r dr) for axisymmetric flow
- Part (c): Perform integration of polynomial terms u(r) = 0.1(1-r²/r₀²) and T(r) = 344.8 + 75r²/r₀² - 18.8r⁴/r₀⁴ to obtain numerical T_m
Q4 50M calculate Heat exchangers and gas turbine cycles
(a) A shell and tube type condenser is employed in a steam power plant to handle 35000 kg/h of dry and saturated steam at 50 °C. The cooling water enters the condenser at 15 °C and leaves at 25 °C. The tubes are of 22·5 mm inside diameter and 25 mm outside diameter. The water flows through the tubes at an average velocity of 2 m/s. The heat transfer coefficient on steam side is 5001 W/m²-K.
Calculate the following :
(i) The mass flow rate of water (in kg/s)
(ii) The heat transfer surface area (based on U₀)
(iii) The number of tubes required for water flow
(iv) The number of tube passes in condenser if the length of each tube pass should not be more than 2·5 m
Assume that condensate coming out from the condenser is saturated water and resistance of tube wall is negligible. For waterside heat transfer coefficient, use the correlation Nu = 0·023 Re⁰·⁸ Pr⁰·⁴. Take latent heat of steam as 2374 kJ/kg. The properties of water at mean bulk temperature of 20 °C are as follows :
Pr = 6·98
ρ = 998·9 kg/m³
Cₚ = 4·180 kJ/kg-K
ν = 1·0006 × 10⁻⁶ m²/s
kƒ = 0·59859 W/m-K
Assume fully developed flow through tubes and a single shell is used. (20 marks)
(b) A stationary gas turbine plant operates on a Brayton cycle and delivers 20 MW to an electric generator. The maximum temperature is 1200 K and the minimum temperature is 290 K. The minimum pressure is 95 kPa and the maximum pressure is 380 kPa. If the isentropic efficiencies of the turbine and compressor are 0·85 and 0·8, respectively, find—
(i) the mass flow rate of air to the compressor;
(ii) the volume flow rate of air to the compressor;
(iii) the fraction of turbine work output needed to drive the compressor;
(iv) the cycle efficiency.
If a regenerator of 75% effectiveness is added to the plant, what would be the changes in the cycle efficiency and net work output?
Assume Cp and Cv for air as 1·005 kJ/kg-K and 0·718 kJ/kg-K, respectively. (20 marks)
(c) The heat transfer rate due to free convection from a vertical surface, 1 m high and 0·6 m wide, to quiescent air that is 20 K colder than the surface is known. What is the ratio of the heat transfer rate for that situation to the rate corresponding to a vertical surface, 0·6 m high and 1 m wide, when the quiescent air is 20 K warmer than the surface? Neglect heat transfer by radiation and any influence of temperature on the relevant thermophysical properties of air.
The correlation between Nusselt number and Rayleigh number is given as
NūL = 0·10 RaL^0·25. (10 marks)
Answer approach & key points
Calculate all numerical sub-parts systematically: spend ~40% time on part (a) condenser calculations (mass flow, LMTD, overall U, tube count, passes), ~35% on part (b) gas turbine cycle with regeneration analysis, and ~25% on part (c) free convection scaling. Begin each part with stated assumptions, show formulae with substituted values, and present final answers with proper units. Include T-s diagrams for (b) and note physical implications.
- Part (a)(i): Water mass flow rate = 9.92 kg/s from energy balance Q = m_steam × h_fg = m_water × Cp × ΔT
- Part (a)(ii): Overall U₀ = 1/(1/h₀ + r₀/rᵢ × 1/hᵢ) with hᵢ from Dittus-Boelter correlation; A₀ = Q/(U₀ × LMTD) where LMTD = 14.43°C
- Part (a)(iii)-(iv): Tube count from continuity equation; tube passes determined from length constraint L_total/(N_tubes × L_max_pass)
- Part (b): Compressor and turbine work with isentropic efficiencies; T₂ = 430.9 K, T₄ = 766.3 K; mass flow from W_net = 20 MW
- Part (b) with regenerator: Effectiveness ε = 0.75 gives T₅ = T₂ + ε(T₄-T₂); revised efficiency and unchanged work output
- Part (c): Heat transfer ratio = (L₁/L₂)^0.25 × (W₁/W₂) = (1/0.6)^0.25 × (0.6/1) = 0.6 × 1.136 = 0.682 for cold vs hot surface orientation
- All parts: Proper unit conversions (kg/h to kg/s, kW to W), significant figures, and physical reasonableness checks
Q5 50M Compulsory derive IC engines, boilers, chimney draught, refrigerants
(a) With the help of P-θ (pressure-crank angle) diagram, compare the knock in SI and CI engines. Explain that "the factors which tend to prevent knock in SI engines in fact promote knock in CI engines". (10 marks)
(b) Describe the phenomenon of 'blowby losses'. What are the factors that increase the blowby losses? What are the effects of increased blowby losses on the engine performance? (10 marks)
(c) What are the advantages and disadvantages of supercritical pressure boilers as compared to that of subcritical boilers? Also, draw the Rankine cycle (T-s diagram) for a steam power plant employing supercritical boiler with single stage of reheating. (10 marks)
(d) Derive the expression as given below for draught h in mm of water column being created by a chimney of height H metre:
h = 353H[1/T_a - ((m_a + 1)/m_a)(1/T_g)]
where m_a is mass of air supplied per kg of fuel, and T_a and T_g are ambient air and hot gas temperatures in Kelvin, respectively. (10 marks)
(e) What is the chemical name of R134a? Is R134a an ecofriendly refrigerant? Clarify. (10 marks)
Answer approach & key points
Begin with part (a) by drawing and comparing P-θ diagrams for SI and CI knock, then explain the inverse relationship of factors. For (b), describe blowby mechanism before listing factors and effects. Part (c) requires a balanced comparison of supercritical vs subcritical boilers followed by the T-s diagram with supercritical Rankine cycle and reheat. Part (d) demands rigorous step-by-step derivation of the chimney draught formula using ideal gas assumptions and density relationships. Conclude with (e) by stating R134a's chemical name and critically evaluating its environmental impact through GWP and ODP analysis. Allocate approximately 20% time each to parts (a), (b), (c), (d), and (e) respectively.
- Part (a): P-θ diagram showing SI knock as sharp pressure spike near TDC vs CI knock as gradual pressure rise; explanation of how high compression ratio, high inlet temperature, and early injection prevent SI knock but promote CI knock
- Part (b): Definition of blowby as gas leakage past piston rings into crankcase; factors including worn rings, high cylinder pressure, low viscosity oil, poor honing; effects on power loss, oil contamination, increased blowby gas handling requirement
- Part (c): Advantages of supercritical boilers (higher efficiency ~45%, no drum, reduced fuel consumption, lower emissions) vs disadvantages (high capital cost, materials stress, water treatment criticality); T-s diagram showing supercritical heating, expansion, reheat, condensation with correct slopes and critical point marked
- Part (d): Derivation starting from hydrostatic pressure balance, ideal gas law for air and flue gas, density substitution ρ = P/RT, pressure difference ΔP = H(ρ_a - ρ_g)g, conversion to mm of water using ρ_w, final algebraic manipulation to given formula
- Part (e): Chemical name 1,1,1,2-tetrafluoroethane (CF3CH2F); evaluation as eco-friendly regarding zero ODP but high GWP of 1430 over 100 years, comparison with R12 and R22, mention of Kigali Amendment implications for India
- Part (d) numerical: Correct handling of temperature in Kelvin, proper unit conversion from Pa to mm of water (1 mm H2O = 9.81 Pa), recognition that 353 = 353 K·mm/Pa or equivalent constant derivation
Q6 50M calculate Dual cycle, regenerative steam cycle, absorption refrigeration
(a) A single-cylinder, two-stroke, high-speed diesel engine working on 'dual cycle' has a compression ratio of 15 : 1. The engine takes in atmospheric air at 1 bar and 27 °C. The maximum temperature in the cylinder is 1312 K. The cutoff ratio is 1·093. The engine has a bore of 200 mm and stroke length of 250 mm. The engine speed is 3000 r.p.m. Calculate the following:
(i) The cycle efficiency
(ii) The net work output per unit mass of air
(iii) The power output
(iv) The mean effective pressure
(v) The ratio of heat added in the constant-pressure process to that in the constant-volume process
Draw the P-V and T-s diagrams of the cycle.
For air, γ = 1·4, R = 0·287 kJ/kg-K. (20 marks)
(b) An ideal regenerative steam power cycle operates so that steam enters the turbine at 30 bar and 350 °C, and exhausts at 0·1 bar. A single open feedwater heater is employed which operates at 5 bar. Compute the thermal efficiency of the cycle. The steam properties at 30 bar and 350 °C are h = 3115·3 kJ/kg and s = 6·7427 kJ/kg-K.
Use Steam tables given at the end to get other properties. (20 marks)
(c) Why is vapour absorption refrigeration system considered to be a better option in comparison to vapour compression refrigeration system for large capacity refrigeration and air-conditioning plants? (10 marks)
Answer approach & key points
Calculate all numerical sub-parts systematically: spend ~40% time on part (a) dual cycle (20 marks) covering all five parameters and diagrams; ~35% on part (b) regenerative cycle (20 marks) with steam table interpolation and feedwater heater analysis; ~25% on part (c) theory (10 marks) comparing absorption vs compression systems. Present derivations with clear state-point labelling on P-V and T-s diagrams for (a), and T-s diagram with extraction point for (b).
- Part (a): Correct T2 = 300 × 15^0.4 = 886.3 K; T3 = 1312 K; ρ = 1.093 gives T4 = 1434 K; T5 from isentropic expansion; efficiency = 1 - (T5-T1)/[(T3-T2)+γ(T4-T3)]
- Part (a): Net work = cv(T3-T2) + cp(T4-T3) - cv(T5-T1); power = W_net × ṁ × (N/60) for two-stroke; mep = W_net/(V1-V2)
- Part (a): P-V diagram showing 1-2 isentropic compression, 2-3 constant volume heat addition, 3-4 constant pressure heat addition, 4-5 isentropic expansion, 5-1 constant volume heat rejection; T-s diagram with corresponding slopes
- Part (b): State 1 at 30 bar, 350°C (h1=3115.3, s1=6.7427); s2=s1 at 5 bar gives x2 and h2; h3 at 5 bar saturated liquid; pump work to 30 bar; h6 from turbine at 0.1 bar with s6=s1
- Part (b): Extraction fraction y = (h3-h5)/(h2-h5); thermal efficiency = [(h1-h2)+(1-y)(h2-h6)]/[(h1-h3)] or equivalent energy balance
- Part (c): Absorption system uses low-grade heat (waste heat, solar, biomass) vs high-grade work; no CFC/HCFC refrigerants; silent operation; suitable for large tonnage (Indian railways, NHPC power station cooling)
- Part (c): COP comparison: COP_absorption = COP_Carnot × η_heatengine vs COP_compression; practical limitations (crystallization, part-load efficiency) mentioned for balance
- All parts: Proper use of steam table data, unit consistency (kJ/kg, kPa, K), and clear identification of state points with numerical values
Q7 50M calculate Thermodynamics and IC Engines
(a) A two-stage turbine receives steam at 50 bar and 350 °C. At 1·5 bar, the high-pressure stage exhausts and 12000 kg of steam per hour is taken at this stage for process heating purposes. The remainder steam is reheated to 250 °C at 1·5 bar and then expanded through the low-pressure turbine to a condensate pressure of 0·05 bar. The power output from the turbine unit is to be 3750 kW. The isentropic efficiency of high-pressure stage is 0·84 and that of the low-pressure stage is 0·81. Calculate the rate of steam being generated per hour by the boiler.
Steam properties :
50 bar, 350 °C : h = 3068·4 kJ/kg, s = 6·4492 kJ/kg-K
1·5 bar, 250 °C : h = 2972·65 kJ/kg, s = 7·8709 kJ/kg-K
Use Steam tables given at the end to get other properties. (20 marks)
(b) (i) What is meant by effective temperature? List and explain briefly the factors which affect optimum effective temperature in air-conditioning. (10 marks)
(ii) Explain psychrometric process of adiabatic mixing of two air streams. (10 marks)
(c) Why is it necessary to cool IC engines? What are the disadvantages of overcooling an IC engine? (10 marks)
Answer approach & key points
Calculate the steam generation rate for part (a) by applying steady-flow energy equations across both turbine stages, accounting for extraction, reheating, and isentropic efficiencies. For parts (b) and (c), explain effective temperature factors and adiabatic mixing with psychrometric charts, then discuss IC engine cooling necessities and overcooling disadvantages. Allocate approximately 40% time to (a), 30% to (b), and 30% to (c) based on mark distribution.
- Part (a): Determine isentropic enthalpy at 1.5 bar from s1 = 6.4492 kJ/kg-K using steam tables; find actual h2 using η_HP = 0.84
- Part (a): Calculate mass flow split: 12000 kg/h extracted, remainder (ṁ-12000) goes to reheating and LP turbine
- Part (a): Apply energy balance: Power output = ṁ_HP×(h1-h2) + ṁ_LP×(h3-h4_actual) = 3750 kW; solve for total ṁ
- Part (b)(i): Define effective temperature as index of thermal comfort combining dry-bulb, humidity, and air motion; list factors (clothing, activity, age, sex, climate acclimatization, economic status)
- Part (b)(ii): Explain adiabatic mixing on psychrometric chart: two states mix along straight line; mass and energy balances give final state; show mixing ratio and enthalpy equations
- Part (c): Necessity of cooling—prevent thermal stresses, maintain lubricant properties, avoid material failure, control NOx emissions; disadvantages of overcooling—reduced thermal efficiency, increased friction, poor combustion, corrosion, increased fuel consumption
- Correct use of steam table interpolations for wet region at 0.05 bar and superheated region at 1.5 bar
- Final answer for (a): steam generation rate in kg/h with proper unit conversion and significant figures
Q8 50M calculate Refrigeration and Nozzle Flow
(a) The following data refers to a single-stage vapour compression refrigeration system :
Refrigerant = R134a
Condenser temperature = 35 °C
Evaporator temperature = – 10 °C
Compressor motor speed = 2800 r.p.m.
Clearance ratio = 0·03
Swept volume = 269·4 cm³
Expansion index = 1·12
Compression isentropic efficiency = 75%
Condensate subcooling in condenser = 5 °C
Draw P-h diagram and determine the following :
(i) The capacity of the plant in TR
(ii) The power required in kW
(iii) The COP
(iv) The heat rejection to condenser
(v) The second law efficiency
The properties of R134a are given in the table :
| T (°C) | P (bar) | Specific volume of saturated vapour v_g (m³/kg) | Enthalpy (kJ/kg) h_f | h_g | Entropy (kJ/kg-K) s_f | s_g |
|--------|---------|-----------------------------------------------|----------------------|-----|----------------------|-----|
| – 10 | 2·014 | 0·0994 | 186·7 | 392·4 | 0·9512 | 1·733 |
| 35 | 8·870 | — | 249·1 | 417·6 | 1·1680 | 1·715 |
Assume specific heat of liquid and vapour at 8·87 bar as 1·458 kJ/kg-K and 1·1 kJ/kg-K, respectively. The refrigerant at entry to compressor is in dry saturated state. (20 marks)
(b) (i) What are the important properties of lubricants used in IC engines? Discuss their significance. (10 marks)
(ii) What are the advantages and disadvantages of the Indirect Injection (IDI) swirl chamber over the open-type Direct Injection (DI) combustion chamber in CI engines? (10 marks)
(c) Calculate the critical pressure and throat area per unit mass flow rate of a convergent-divergent nozzle, expanding steam from 10 bar and dry saturated, down to atmospheric pressure of 1 bar. Assume that the inlet velocity is negligible and that the expansion is isentropic.
Use Steam tables given at the end to get the steam data.
The value of isentropic expansion index may be taken as 1·135. (10 marks)
Answer approach & key points
Calculate systematically across all three parts, allocating time proportional to marks: ~40% for part (a) refrigeration cycle with P-h diagram and five performance parameters, ~30% for part (b) descriptive IC engine lubrication and combustion chamber comparison, and ~30% for part (c) nozzle critical pressure and throat area derivation. Begin each part with stated assumptions, show all thermodynamic property lookups and interpolation, and conclude with clearly boxed final answers with units.
- Part (a): Correct P-h diagram with 5 process points (1: dry saturated at -10°C, 2: after compression, 3: saturated liquid at 35°C, 4: after subcooling to 30°C, 5: after expansion); account for isentropic efficiency, clearance ratio for actual mass flow, and expansion index for re-expansion work
- Part (a): Mass flow rate calculation using clearance ratio and swept volume with volumetric efficiency; refrigerating effect using h1 - h4 with subcooled liquid enthalpy
- Part (a): Second law efficiency calculation requires Carnot COP between actual evaporator and condenser temperatures, not the saturation temperatures
- Part (b)(i): Viscosity index, pour point, flash/fire point, oxidation stability, detergency/dispersancy, and alkalinity with significance for engine wear, cold starting, deposit control, and corrosion prevention
- Part (b)(ii): IDI swirl chamber advantages (quieter, lower injection pressure, better air utilization at part load) vs disadvantages (higher heat loss, lower thermal efficiency, cold starting problems) compared to open DI chamber
- Part (c): Critical pressure ratio for steam using γ = 1.135; throat area per unit mass flow using isentropic flow relations with steam table enthalpy values, not ideal gas assumption
- Part (c): Correct use of steam tables to find h1, h2 (isentropic to throat), and specific volume at throat conditions
- All parts: Consistent units (bar, kJ/kg, m³/kg, K), proper significant figures, and explicit statement of formulas before substitution