Q1 50M Compulsory calculate Thermodynamics, Heat Transfer and Fluid Mechanics
(a) Consider the system shown in Fig. 1(a). The two chambers initially have equal volumes of 28 litres and contain air (C_p = 1·005 kJ/kg-K and C_v = 0·717 kJ/kg-K) and hydrogen (C_p = 14·32 kJ/kg-K and C_v = 10·17 kJ/kg-K), respectively. The chambers are separated by a frictionless piston which is non-heat-conducting. Both the gases are initially at 140 kPa and 40 °C. Heat is added to the air side until the pressure of both the gases reaches 280 kPa. All outside walls of the chambers are insulated except for the surface where heat is added to air. Calculate the final temperature of the air. (10 marks)
(b) What is 'choked flow' in a convergent-divergent nozzle? Explain, with diagram, the effect of pressure ratio on exit velocity of compressible gas in a convergent-divergent nozzle. (10 marks)
(c) An axial flow compressor with inlet and outlet angles of 40° and 15°, respectively has been designed for 50% reaction. The compressor has a pressure ratio of 6 : 1 and overall isentropic efficiency of 0·80, when inlet static temperature is 41 °C. The blade speed and axial velocity are constant throughout. Assuming a value of 210 m/s for blade speed, find the number of stages required if the work done factor is 0·88 for all the stages. Take Cp = 1·005 kJ/kg-K and γ = 1·4 for air. (10 marks)
(d) Hot water is flowing through a pipe made of cast iron having thermal conductivity of 52 W/m-°C, with an average velocity of 1·5 m/s. The inner and outer diameters of the pipe are 3 cm and 3·5 cm, respectively. The pipe passes through a 15 m long section of a basement whose temperature is 15 °C. The temperature of the water drops from 70 °C to 67 °C as it passes through the basement. The heat transfer coefficient on the inner surface of the pipe is 400 W/m²-°C. Determine the combined convection and radiation heat transfer coefficient at the outer surface of the pipe. (10 marks)
(e) (i) Define the total and spectral black body emissive powers. How are they related to each other? (5 marks)
(ii) Consider two identical bodies, one at 1000 K and the other at 1500 K. Which body emits more radiation in the shorter wavelength region? Which body emits more radiation at a wavelength of 20 μm? (5 marks)
Answer approach & key points
Calculate numerical solutions for parts (a), (c), and (d) while explaining theoretical concepts with diagrams for parts (b) and (e). Allocate approximately 25% time each to (a), (c), and (d) as they involve multi-step calculations; 15% to (b) for the choked flow diagram and explanation; and 10% to (e) for definitions and Wien's displacement law application. Begin each numerical part with stated assumptions and end with unit verification.
- Part (a): Apply first law to adiabatic hydrogen compression (γ_H2 = 1.407) to find T_H2_final, then use piston equilibrium and ideal gas law to find T_air_final ≈ 586-590 K
- Part (b): Define choked flow as Mach 1 at throat with maximum mass flow; sketch P-V diagram showing over-expanded, design, and under-expanded nozzle conditions with exit velocity trends
- Part (c): Use 50% reaction condition (α1 = β2, α2 = β1) to find flow angles, compute stage temperature rise from degree of reaction and velocity triangles, then determine stages n ≈ 8-9
- Part (d): Apply thermal resistance network (convection-inner, conduction-pipe, convection+radiation-outer) using LMTD for water temperature drop, solve for h_outer ≈ 12-15 W/m²°C
- Part (e)(i): Define E_bλ = C1λ⁻⁵/[exp(C2/λT)-1] and Eb = σT⁴ with Planck's law integration; state Stefan-Boltzmann constant σ = 5.67×10⁻⁸ W/m²K⁴
- Part (e)(ii): Apply Wien's law (λ_maxT = 2898 μm·K) to show 1500 K body emits more at shorter wavelengths; use Planck's distribution to compare emission at 20 μm
Q2 50M calculate Thermodynamics, Heat Transfer and Compressible Flow
(a) 10 g of water at 20 °C is converted into ice at –10 °C at constant atmospheric pressure. Assuming the specific heat of liquid water to remain constant at 4·2 J/g-K and that of ice to be half of this value and taking the latent heat of fusion of ice at 0 °C to be 335 J/g, calculate the total entropy change of the system. (20 marks)
(b) A shaft having diameter of 5 cm rotates in a bearing made of cast iron. The shaft rotates at 4500 r.p.m. The bearing is 15 cm long, 8 cm outer diameter and has thermal conductivity of 70 W/m-K. There is a uniform clearance between the shaft and the bearing of 0·6 mm. The clearance is filled with a lubricating oil having thermal conductivity of 0·14 W/m-K and dynamic viscosity of 0·03 N-s/m². The bearing is cooled externally by a liquid, and its outer surface is maintained at 40 °C. Disregarding the heat conduction through the shaft and assuming only one-dimensional heat transfer, determine (i) the rate of heat transfer to the coolant, (ii) the surface temperature of the shaft and (iii) the mechanical power wasted by the viscous dissipation in the lubricating oil. (20 marks)
(c) Air (C_p = 1·05 kJ/kg-K, γ = 1·38) at 3 bar pressure and T = 600 K is flowing with a velocity of 180 m/s inside a 20 cm diameter duct. Calculate the—
(i) mass flow rate;
(ii) stagnation temperature;
(iii) Mach number;
(iv) stagnation pressure assuming flow to be (1) compressible and (2) incompressible. (10 marks)
Answer approach & key points
Calculate the required quantities for all three parts systematically. For part (a), compute entropy changes for cooling water, freezing, and cooling ice separately, then sum. For part (b), apply viscous dissipation in Couette flow with cylindrical coordinates, then use thermal resistance network for conduction through oil and bearing. For part (c), apply compressible flow relations for mass flow, stagnation properties, and Mach number, then compare compressible vs incompressible stagnation pressure results. Allocate approximately 35% time to (a), 40% to (b), and 25% to (c) based on complexity and marks distribution.
- Part (a): Three-stage entropy calculation — cooling water from 20°C to 0°C (ΔS₁ = mc_w ln(273/293)), freezing at 0°C (ΔS₂ = -mL_f/T_f), cooling ice from 0°C to -10°C (ΔS₃ = mc_ice ln(263/273)); total ΔS = ΔS₁ + ΔS₂ + ΔS₃ with c_ice = 2.1 J/g-K
- Part (b)(i)-(iii): Velocity gradient in annular gap τ = μ(du/dy) = μ(ωR_i)/c, viscous dissipation Φ = τ²/μ, heat generation Q = Φ×volume, thermal resistances R_oil = ln(R_o/R_i)/(2πk_oilL) and R_bearing = ln(R_b/R_o)/(2πk_bearingL), shaft temperature found from heat balance
- Part (c)(i): Mass flow rate using ṁ = ρAV = (P/RT)×(πD²/4)×V with R = 287 J/kg-K for air
- Part (c)(ii)-(iii): Stagnation temperature T₀ = T + V²/(2C_p), Mach number Ma = V/√(γRT), critical check whether flow is subsonic (Ma < 1)
- Part (c)(iv): Compressible stagnation pressure P₀ = P(T₀/T)^(γ/(γ-1)), incompressible P₀ = P + ½ρV² using ρ = P/RT; explicit comparison showing difference
- Unit consistency throughout: temperatures in Kelvin for thermodynamic calculations, SI units for all quantities, proper handling of kJ vs J conversion
- Physical interpretation: negative entropy change in (a) indicates irreversibility of freezing process; in (b) recognition that viscous dissipation dominates heat generation; in (c) quantification of compressibility effects at Ma ≈ 0.37
Q3 50M prove Gas turbine, solar collector, heat transfer
(a) (i) How do the specific work output and efficiency vary with pressure ratio in a gas turbine?
(ii) Prove that the efficiency of a gas turbine corresponding to the maximum work done in a Brayton cycle is given by the relation
$$\eta_{w \max} = 1 - \frac{1}{\sqrt{t}}$$
where $t$ is the ratio of the maximum and minimum temperatures.
(20 marks)
(b) A solar collector, as shown in Fig. 3(b) below, having dimensions as 1 m wide and 5 m long, has constant spacing of 3 cm between the glass cover and the collector plate. Air enters the collector at 30 °C and at a rate of 0·15 m³/s through the 1 m wide edge and flows along the 5 m long passageway. If the average temperatures of the glass cover and the collector plate are 20 °C and 60 °C, respectively, determine (i) the net rate of heat transfer to the air in the collector and (ii) the temperature rise of air as it flows through the collector.
Fig. 3(b)
The properties of air at 1 atm and an estimated average temperature of 35 °C may be taken as :
ρ = 1·145 kg/m³, k = 0·02625 W/m-°C, ν = 1·655×10⁻⁵ m²/s,
Cₚ = 1007 J/kg-°C, Pr = 0·7268
(20 marks)
(c) A windshield of a car, having dimensions as 0·6 m high and 1·8 m long, is electrically heated and is subjected to parallel winds at 1 atm, 0 °C and 80 km/hr. The electrical power consumption is observed to be 50 W, when the exposed surface temperature of the windshield is 4 °C. Disregarding the radiation and heat transfer from the inner surface and using the momentum heat transfer analogy, determine the drag force the wind exerts on the windshield. The properties of air at 0 °C and 1 atm may be taken as :
ρ = 1·292 kg/m³, Cₚ = 1·006 kJ/kg-K, Pr = 0·7362
(10 marks)
Answer approach & key points
Prove the Brayton cycle efficiency relation in part (a) using calculus-based optimization; for (b) and (c), solve the numerical problems using appropriate heat transfer correlations and momentum-heat transfer analogy. Allocate approximately 40% time to part (a) due to its derivation-heavy 20 marks, 35% to part (b) for complex convection calculations, and 25% to part (c) for the Reynolds analogy application. Structure as: (a) theoretical derivation with T-s diagram, (b) step-wise Nusselt number calculation and energy balance, (c) Stanton number and drag coefficient linkage.
- Part (a)(i): Specific work output increases with pressure ratio to a maximum then decreases; efficiency increases monotonically with pressure ratio for ideal Brayton cycle
- Part (a)(ii): Derivation of η_wmax = 1 - 1/√t by differentiating net work w.r.t. pressure ratio, setting to zero, and substituting optimal pressure ratio r_p,opt = √t
- Part (b): Hydraulic diameter calculation (D_h = 2b = 0.06 m), Reynolds number determination, Nusselt number using appropriate correlation (laminar/turbulent), heat transfer coefficient, and net heat transfer to air
- Part (b): Temperature rise from energy balance Q = ṁC_pΔT, with mass flow rate from ρ and volumetric flow rate
- Part (c): Application of Reynolds/Colburn analogy (St = Cf/2 × Pr^(-2/3)), calculation of Stanton number from heat transfer data, determination of friction coefficient and drag force
- Part (c): Recognition that electrical power equals convective heat loss q = hA(T_s - T_∞) for establishing heat transfer coefficient
Q4 50M construct Impulse turbine, centrifugal compressor, heat exchanger
(a) A single-stage impulse turbine rotor has a mean blade ring diameter of 500 mm and rotates at a speed of 10000 r.p.m. The nozzle angle is 20° and the steam leaves the nozzles with a velocity of 900 m/s. The blades are equiangular and the blade friction factor is 0·85. Construct velocity diagrams for the blades and determine the inlet angle of the blades for shockless entry of steam. Also, determine (i) the diagram power for a steam flow of 750 kg/hr, (ii) the diagram efficiency, (iii) the axial thrust and (iv) the loss of kinetic energy due to friction.
(20 marks)
(b) (i) Explain the effect of impeller blade shape on the performance of a centrifugal compressor with the help of an exit velocity diagram and pressure ratio-mass flow rate curve.
(ii) Discuss the phenomena of surging and choking in centrifugal compressors.
(20 marks)
(c) A shell and tube heat exchanger operates with two shell passes and four tube passes. The shell side fluid is ethylene glycol, which enters at 140 °C and leaves at 80 °C with a flow rate of 4500 kg/hr. Water flows in the tubes, entering at 35 °C and leaving at 85 °C. The overall heat transfer coefficient for this arrangement is 850 W/m²-°C. Calculate the flow rate of water required and the area of the heat exchanger. The specific heat of ethylene glycol may be taken as 2·742 J/g-°C and the specific heat of water may be taken as 4·175 J/g-°C. For NTU relations, the following figure may be used.
(10 marks)
Answer approach & key points
Construct velocity diagrams for part (a) as the primary directive, then explain compressor phenomena for (b), and solve the heat exchanger problem for (c). Allocate approximately 40% time to (a) given its 20 marks and diagram construction demand, 35% to (b) for its dual explanatory components, and 25% to (c) for the NTU method calculation. Begin with clear velocity triangle construction for impulse turbine, follow with theoretical explanations supported by sketches, and conclude with systematic heat exchanger sizing using the correction factor method.
- Part (a): Blade speed U = πDN/60 = 261.8 m/s; velocity triangles constructed with nozzle angle 20°, equiangular blades, and friction factor 0.85 applied to relative velocities
- Part (a): Inlet blade angle β₁ calculated from velocity triangle geometry for shockless entry; diagram power = ṁ(V_w1 + V_w2)U, efficiency = diagram power / kinetic energy supplied
- Part (a): Axial thrust = ṁ(V_f1 - V_f2) and friction loss = ½ṁ(V_r1² - V_r2²) computed with correct mass flow rate conversion (750 kg/hr = 0.2083 kg/s)
- Part (b)(i): Backward-curved, radial, and forward-curved blade effects on pressure ratio-mass flow characteristics with exit velocity diagrams showing V₂, V_w2, and manometric efficiency
- Part (b)(ii): Surging as flow reversal instability at low mass flow rates and choking as sonic limit at impeller eye; both phenomena explained with performance curve annotations
- Part (c): Heat balance Q = ṁ_glycol × c_p,glycol × ΔT_glycol = ṁ_water × c_p,water × ΔT_water to find water flow rate
- Part (c): LMTD calculation for counterflow arrangement, correction factor F from given figure for 2-shell pass 4-tube pass configuration, and area A = Q/(U×F×LMTD)
Q5 50M Compulsory explain IC engines, thermodynamics, refrigeration and air conditioning
(a) Can alcohols be used as fuel in IC engine? Explain with advantages and disadvantages. (10 marks)
(b) A water-filled reactor with a volume of 1 m³ is at 20 MPa and 360 °C, and is placed inside a containment room as shown in Fig. 5(b). The room is well-insulated and initially evacuated. Due to a failure, the reactor ruptures and the water fills the containment room. Find the minimum room volume so that the final pressure does not exceed 200 kPa. [Use steam table data given at the end of the Paper] (10 marks)
(c) Using a schematic and T-s diagram, explain how with perfect regeneration for a simple steam power plant (Rankine) cycle, thermal efficiency can approach Carnot efficiency. (10 marks)
(d) Discuss the effect of the following parameters on the performance of a vapor compression refrigeration system with the help of p-h diagram: (i) Suction pressure (ii) Delivery pressure (iii) Subcooling of liquid (iv) Superheating of vapors (10 marks)
(e) The room air is recirculated at the rate of 40 m³ per minute and the outdoor air enters a cooling coil of an air conditioner at 32 °C DBT and 18 °C WBT. The effective surface temperature of the coil is 4·5 °C. The surface area of the coil is such as would give 12 kW of refrigeration with the given entering conditions of air. Determine the DBT and WBT of the air leaving the coil and the coil bypass factor. [Psychrometric chart is given at the end of this Paper] (10 marks)
Answer approach & key points
Explain the suitability of alcohols as IC engine fuels with balanced advantages and disadvantages for part (a). For (b), apply first law for unsteady flow with steam tables to find containment volume. Part (c) requires schematic and T-s diagram with clear regeneration explanation. Part (d) needs p-h diagram analysis of four parameters. Part (e) involves psychrometric calculations with bypass factor. Allocate ~15% time each to (a), (c), (d); ~25% to (b) and (e) due to numerical complexity.
- Part (a): Alcohols (methanol, ethanol) as fuels—higher octane rating, lower emissions vs lower energy density, corrosion, cold-start issues; mention India's ethanol blending programme (E20, E85)
- Part (b): Unsteady flow energy equation for ruptured reactor; initial state: compressed liquid/subcooled water at 20 MPa, 360°C; final state: saturated mixture at 200 kPa; mass and energy balance to find quality and containment volume
- Part (c): Rankine cycle with perfect regeneration—open feedwater heater schematic; T-s diagram showing heat addition at variable temperature approaching Carnot's isothermal; mean temperature of heat addition increases to T_max
- Part (d): p-h diagram effects—suction pressure ↑ increases refrigeration effect and COP; delivery pressure ↑ decreases COP; subcooling ↑ increases refrigeration effect; superheating ↑ may increase or decrease COP depending on useful cooling
- Part (e): Psychrometric process—mixing of recirculated and outdoor air, cooling and dehumidification to 4.5°C EST; bypass factor = (T_out - T_est)/(T_in - T_est); energy balance for 12 kW to find air exit conditions
- Part (b) specific: Use steam tables—initial v ≈ 0.0015 m³/kg (compressed liquid), u ≈ 1700 kJ/kg; final v_f and v_g at 200 kPa; solve for mass then V_room = m·v_final
- Part (e) specific: Locate 32°C DBT, 18°C WBT on chart; find humidity ratio; process follows constant sensible heat factor to 4.5°C EST; read exit DBT and WBT
Q6 50M calculate IC engine testing, regenerative steam cycle, absorption refrigeration
(a) A four-cylinder diesel engine with swept volume of 0·98 litre is tested on a performance bed. The engine running at a speed of 2500 r.p.m. against a brake with arm of 0·3 m produces brake load of 190 N with fuel consumption of 6·8 litres/hr. The calorific value of fuel is 45000 kJ/kg and specific gravity of fuel is 0·82. A Morse test is carried out on the engine by cutting off the fuel supply of individual cylinder in the order 1, 2, 3, 4 with corresponding brake loads 131 N, 135 N, 133 N and 137 N, respectively. Calculate the b.p., b.m.e.p., brake thermal efficiency, b.s.f.c., i.p., mechanical efficiency and i.m.e.p. of the engine at test speed. (20 marks)
(b) A power plant operates on a regenerative steam cycle with one closed feedwater heater. Steam enters the first turbine stage at 125 bar, 500 °C and expands to 10 bar, where some of the steam is extracted and diverted to the closed feedwater heater. Condensate exiting the feedwater heater as saturated liquid at 10 bar passes through a trap into the condenser. The feedwater exits the heater at 120 bar with a temperature of 170 °C. The condenser pressure is 0·06 bar. Assuming isentropic turbine and pump work, determine the thermal efficiency of the cycle. At 125 bar, 500 °C for steam, h = 3343·6 kJ/kg and s = 6·4651 kJ/kg-K. [Use steam tables provided at the end of this Paper] (20 marks)
(c) Explain NH₃-water vapor absorption refrigeration system with a neat diagram. What are the desired properties of refrigerant-absorber combination? (10 marks)
Answer approach & key points
Calculate all performance parameters for the diesel engine in part (a) using Morse test principles, then solve the regenerative steam cycle in part (b) by applying energy balance and steam table interpolation. For part (c), explain the NH₃-water absorption system with a neat schematic. Allocate approximately 40% time to (a) due to multiple calculations, 40% to (b) for steam table work, and 20% to (c) for the descriptive portion with diagram.
- Part (a): BP = 2πNT/60 = 2π×2500×(190×0.3)/60000 = 14.92 kW; BSFC = ṁf/BP in kg/kWh
- Part (a): IP from Morse test = Σ(indicated power of each cylinder) using BP - BPcutoff for each cylinder
- Part (a): Mechanical efficiency = BP/IP; IMEP = IP×60/(L×A×N/2×n) or using swept volume
- Part (b): Apply energy balance on closed feedwater heater: y×h2 + (1-y)×h6 = h7 + (1-y)×h3 for extraction fraction
- Part (b): Determine h at turbine exit using s1 = s2 = s3; pump work h6 - h5 = v5×(P6-P5); thermal efficiency = Wnet/Qin
- Part (c): NH₃-water system: generator, condenser, evaporator, absorber with heat exchanger; NH₃ as refrigerant, water as absorber
- Part (c): Desired properties: high solubility of refrigerant in absorber, large boiling point difference, low specific heat of solution, non-toxic, non-corrosive, chemically stable
Q7 50M solve IC engines, air conditioning, and chimney draught
(a) (i) Briefly discuss HC, CO and NOₓ emission formation in SI engine. Explain the dependence of these emissions on equivalence ratio with a neat diagram. (10 marks)
(ii) An IC engine working on an ideal Otto cycle has AFR of 15 : 1 and compression ratio of 9 : 1. The pressure and temperature at the start of compression are 1 bar and 27 °C, respectively. Find the maximum temperature and pressure of the cycle. Assume that compression process follows the law pV¹·³³ = C, the calorific value of fuel is 43000 kJ/kg and Cᵥ of working fluid is 0·717 kJ/kg-K. (10 marks)
(b) A food processing room has a very high latent heat load and is required to be air conditioned as per the following data:
Room design conditions : 20 °C DBT, 60% RH
Outside conditions : 45 °C DBT, 30 °C WBT
Room sensible heat : 35 kW
Room latent heat : 20 kW
The ventilation air requirement is 90 cmm
Determine the (i) ventilation load, (ii) room and effective sensible heat factors and (iii) ADP and amount of reheat for economical design. Assume bypass factor of the coil as 0·05.
[Psychrometric chart is given at the end of this Paper] (20 marks)
(c) Prove mathematically that for maximum discharge through a chimney of a certain height and cross-section, the absolute temperature of gases bears a certain ratio to the absolute temperature of the outside atmosphere, in case of natural draught of a boiler. (10 marks)
Answer approach & key points
Solve this multi-part question by allocating time proportionally to marks: ~20 minutes for (a)(i) emissions discussion with diagram, ~20 minutes for (a)(ii) Otto cycle numerical, ~40 minutes for (b) air conditioning psychrometric calculations, and ~20 minutes for (c) chimney draught derivation. Begin each part with clear identification of given data, apply appropriate thermodynamic principles, and conclude with physically meaningful results.
- (a)(i) HC formation via flame quenching and crevice volumes; CO from incomplete combustion in rich mixtures; NOx via thermal (Zeldovich) mechanism at high temperatures; emissions vs equivalence ratio diagram showing HC/CO peak rich, NOx peak slightly lean
- (a)(ii) State 1: p1=1 bar, T1=300K; polytropic compression to state 2: p2=p1*(r)^1.33, T2=T1*(r)^0.33; heat addition using Q=m*Cv*(T3-T2) with fuel energy release; solve for T3 (max temp) and p3=p2*T3/T2 (max pressure)
- (b) Plot room condition (20°C DBT, 60% RH) and outside condition (45°C DBT, 30°C WBT) on psychrometric chart; calculate ventilation sensible and latent loads; determine RSHF=RSH/(RSH+RLH) and ESHF accounting for bypass factor; find ADP from coil process line and reheat needed
- (b) Ventilation load: mass flow rate from 90 cmm, enthalpy difference between outside and room air; split into sensible and latent components
- (c) Derive draught pressure Δp = H*g*(ρa-ρg) = H*g*ρa*(1-Ta/Tg); discharge ṁ ∝ √(Δp/Tg); maximize ṁ w.r.t. Tg to obtain Tg/Ta = 2 for maximum discharge
- (c) Alternative derivation using chimney height equation and differentiating discharge equation with respect to gas temperature, showing optimal temperature ratio is 2:1
Q8 50M calculate Refrigeration, combined power plant, and vapor power cycles
(a) A VCR cycle refrigerator driven by a 60 kW compressor has a COP of 6·0. The enthalpies of saturated liquid and saturated vapor refrigerant at condenser temperature of 35 °C are 114·95 kJ/kg and 283·89 kJ/kg, respectively. The saturated refrigerant vapor leaving evaporator has an enthalpy of 275·76 kJ/kg. Find the temperature of refrigerant at the exit of compressor. The Cₚ of refrigerant is 0·62 kJ/kg-K. (20 marks)
(b) In a combined gas turbine-steam turbine power plant, the exhaust gas from the open-cycle gas turbine is the supply gas to the steam generator of the steam cycle at which additional fuel is burnt in the gas. The pressure ratio for the gas turbine is 7·5, the air inlet temperature is 15 °C and the maximum temperature is 750 °C. Combustion of additional fuel raises the gas temperature to 750 °C and the gas leaves the steam generator at 100 °C. The steam is supplied to the steam turbine at 50 bar and 600 °C and the condenser pressure is 0·1 bar. The total power output of the plant is 200 MW. The calorific value of the fuel burnt is 43·3 MJ/kg. Neglecting the effect of the mass flow rate of fuel on the air flow, determine (i) the flow rate of air and steam required, (ii) the power outputs of the gas turbine and steam turbine, (iii) the thermal efficiency of the combined plant and (iv) the air-fuel ratio. Take Cₚ = 1·11 kJ/kg-K and γ = 1·33 for combustion gases; and Cₚ = 1·005 kJ/kg-K and γ = 1·4 for air. Neglect pump work. Condensate enthalpy at 0·1 bar = 192 kJ/kg.
[Mollier diagram is attached in Page No. 14] (20 marks)
(c) Two vapor power cycles are coupled in series where heat lost by one is absorbed by the other completely. If η₁ is the thermal efficiency of the topping cycle and η₂ is the thermal efficiency of the bottom cycle, determine the efficiency of the combined cycle in terms of these efficiencies. Assume cycles to be reversible. (10 marks)
Answer approach & key points
Calculate the compressor exit temperature for part (a) using energy balance and isentropic relations; for part (b) systematically determine air/steam flow rates, power outputs, thermal efficiency and air-fuel ratio using gas turbine and steam cycle analysis with the given Mollier diagram data; for part (c) derive the combined cycle efficiency formula using reversible heat engine principles. Allocate approximately 35% time to part (a), 50% to part (b) due to its four sub-requirements, and 15% to part (c) for the derivation.
- Part (a): Refrigeration effect = COP × Ẇ = 360 kW; mass flow rate ṁ = 360/(275.76-114.95) = 2.237 kg/s; compressor work per kg = 26.82 kJ/kg; T₂ = T₁ + (h₂-h₁)/Cp with h₂ from energy balance yielding T₂ ≈ 52-55°C
- Part (b): Gas turbine: T₂ = T₁×(r_p)^((γ-1)/γ) = 288×2.143 = 617.3 K; T₃ = 1023 K; T₄ = T₃/(r_p)^((γ-1)/γ) = 477.4 K; heat added in combustor = Cp,g×(1023-477.4); additional heat in HRSG raises to 1023 K; steam cycle: h₁ from 50 bar/600°C using Mollier chart ≈ 3666 kJ/kg, h₂ from 0.1 bar ≈ 2260 kJ/kg (isentropic)
- Part (b)(i): Energy balance on HRSG gives ṁ_air and ṁ_steam; ṁ_steam = 200 MW total / [(h₁-h₂) + (Ẇ_GT/ṁ_steam)] after iteration
- Part (b)(ii): Ẇ_GT = ṁ_air×Cp,g×[(1023-617.3)+(1023-477.4)]; Ẇ_ST = ṁ_steam×(h₁-h₂)
- Part (b)(iii): η_combined = (Ẇ_GT + Ẇ_ST)/(ṁ_fuel×CV) with total fuel in GT combustor and HRSG
- Part (b)(iv): AFR = ṁ_air/ṁ_fuel from energy balance Q = ṁ_fuel×CV = ṁ_air×Cp,g×ΔT
- Part (c): For reversible cycles, η_combined = η₁ + η₂ - η₁η₂ derived from Q₁→W₁+Q₂ with Q₂ = Q₁(1-η₁) and W₂ = η₂Q₂