Total Questions: 100
Maximum Marks: 100
Duration: 3 Hours
Language: English (technical + applied numerical)

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Part 1 – Engineering Mechanics & Strength of Materials (Q1–20)

1. Derive the expression for bending stress in a rectangular beam section under pure bending.
2. A cantilever beam of length 3 m carries a UDL of 5 kN/m. Find maximum deflection and slope at free end.
3. Define shear center. Explain its significance with a thin-walled section example.
4. A solid circular shaft transmits 50 kW at 150 rpm. Determine torque and maximum shear stress if diameter = 50 mm.
5. Determine principal stresses and maximum shear stress for σx = 80 MPa, σy = 50 MPa, τxy = 30 MPa.
6. Discuss torsion in circular shafts and derive the torsion equation.
7. A simply supported beam of span L carries a point load P at midspan. Draw the shear force and bending moment diagrams.
8. For a composite bar subjected to axial load, derive stresses in each material assuming perfect bonding.
9. Explain the Mohr’s circle for plane stress with a numerical example.
10. A hollow circular shaft has outer diameter 100 mm, inner diameter 60 mm. Calculate polar moment of inertia.
11. Derive the deflection equation of a simply supported beam using double integration method.
12. Determine the slope at free end of a cantilever with point load at free end.
13. Define section modulus and calculate it for a rectangular section 200 × 300 mm.
14. A beam is subjected to combined bending and axial load. Find maximum stress at extreme fiber.
15. Discuss curvature of a beam and radius of curvature.
16. A spring is designed for load W = 500 N, diameter 50 mm, wire 5 mm. Calculate shear stress and deflection.
17. Explain factor of safety with example for ductile and brittle materials.
18. Derive the torsional rigidity of hollow shafts.
19. A cantilever with triangular distributed load. Determine maximum bending moment and shear.
20. Explain statically indeterminate beams and provide a numerical solution method.

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Part 2 – Thermodynamics & Heat Transfer (Q21–40)

21. Derive efficiency of Carnot cycle and discuss its limitations.
22. Explain Otto and Diesel cycles, derive efficiency formulas and compare.
23. A heat engine absorbs 500 kJ, rejects 300 kJ. Determine thermal efficiency and work done.
24. Explain first and second laws of thermodynamics for closed and open systems.
25. Calculate entropy change when 2 kg of water at 100°C is converted to steam at 100°C.
26. A Rankine cycle has boiler pressure 10 MPa, condenser 0.1 MPa. Calculate efficiency.
27. Derive Clapeyron equation and explain its significance.
28. Explain refrigeration cycles and calculate COP for vapor compression system.
29. Determine logarithmic mean temperature difference for a counter-flow heat exchanger.
30. Derive the conduction equation in one-dimensional steady-state.
31. Discuss fins and extended surfaces with a numerical problem.
32. Explain natural and forced convection with examples.
33. Derive radiation heat transfer between two black bodies.
34. A slab 0.2 m thick conducts heat at rate 100 W/m². Thermal conductivity = 50 W/mK. Find temperature difference.
35. Explain Biot and Fourier numbers and their significance in transient conduction.
36. Calculate effectiveness of heat exchanger given Qmax and Qactual.
37. Derive the isentropic relations for ideal gases.
38. Calculate work done in polytropic process with n = 1.3, P1, V1, P2.
39. Explain Joule-Thomson effect and applications.
40. Compare refrigeration, heat pump, and air-conditioning cycles numerically.

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Part 3 – Fluid Mechanics & Hydraulic Machines (Q41–60)

41. Derive Bernoulli’s equation along a streamline.
42. A venturimeter with inlet diameter 0.2 m, throat 0.1 m. Find flow rate for pressure difference 20 kPa.
43. Explain Reynolds number and transition from laminar to turbulent flow.
44. A pipe of 0.1 m diameter carries water at 2 m/s. Calculate head loss using Darcy-Weisbach equation.
45. Derive continuity equation and apply to branching pipes.
46. Explain hydraulic gradient line (HGL) and energy gradient line (EGL).
47. A Pelton turbine delivers 500 kW at 200 m head. Determine jet diameter.
48. Explain Cavitation in pumps and methods to avoid it.
49. A centrifugal pump delivers 0.05 m³/s at 20 m head. Find shaft power required at 80% efficiency.
50. Derive Euler’s equation for turbines.
51. Explain specific speed of a pump and turbine, give numerical example.
52. A Francis turbine, flow rate 50 m³/s, head 20 m. Calculate shaft power.
53. Derive equation for flow in open channel using Chezy formula.
54. Determine critical depth and Froude number for rectangular channel.
55. Explain minor losses in pipe flow with examples.
56. A Venturimeter measures water flow with ΔP = 10 kPa. Find velocity and discharge.
57. Derive manometer equation for pressure difference measurement.
58. Explain hydraulic jump, calculate sequent depth for given initial depth.
59. Explain boundary layer concept and its applications in fluid flow.
60. A pipe network problem: Calculate equivalent resistance and flow rates.

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Part 4 – Machine Design & Theory of Machines (Q61–80)

61. Derive torque transmitted by a shaft with numerical example.
62. Design helical spring for given load, deflection, and material properties.
63. Explain flywheel design to store energy for fluctuation of speed 10%.
64. Determine diameter of hollow shaft transmitting given torque and shear stress.
65. Explain belt drive, slip, and velocity ratio numerically.
66. Design sprocket chain drive for given power and speed.
67. Calculate mechanical advantage of screw jack and efficiency.
68. A slider-crank mechanism: Calculate slider velocity and acceleration at mid-stroke.
69. Derive equation of motion for simple harmonic motion of a crank.
70. Explain Grashof condition and degree of freedom in four-bar mechanism.
71. Determine gear ratio, number of teeth, and pitch circle diameter for a compound gear train.
72. Calculate bending and shear stress in gear teeth using Lewis formula.
73. Explain torque transmitted by belt drive with initial tension.
74. Design clutch plate for given torque and coefficient of friction.
75. Explain torsional vibration of shafts with natural frequency calculation.
76. A flywheel of 100 kg-m² rotates at 300 rpm. Determine maximum and minimum kinetic energy for 10% fluctuation.
77. Explain bearing selection and load calculation for journal bearing.
78. Design keys and couplings for given shaft torque.
79. A helical gear transmits 20 kW at 1000 rpm. Calculate module and face width.
80. Explain cam and follower design, determine cam profile for given motion.

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Part 5 – IC Engines, Refrigeration, Heat Engines, and Applied Numericals (Q81–100)

81. Derive mean effective pressure for Otto and Diesel cycles with numerical example.
82. Calculate power output and efficiency of single-cylinder engine, given bore, stroke, MEP.
83. Determine brake power, indicated power, and mechanical efficiency from given data.
84. A refrigeration system has 20 kW refrigeration effect and 5 kW input. Calculate COP.
85. Explain vapor compression cycle, derive expressions for refrigeration effect and work input.
86. Calculate compressor work and volumetric efficiency for reciprocating compressor.
87. Explain air-standard efficiency of dual cycle and compare with Otto cycle.
88. A gas turbine delivers 500 kJ/kg net work, turbine work 1200 kJ/kg. Calculate work ratio.
89. Calculate brake mean effective pressure for given torque and engine speed.
90. Determine air conditioning load (sensible and latent) for given room conditions.
91. Explain refrigerant properties using Mollier chart.
92. A single-stage centrifugal pump delivers 0.05 m³/s, head 20 m. Calculate shaft power and efficiency.
93. Solve thermodynamic cycle problem: given heat added and rejected, find work and efficiency.
94. A Pelton wheel: Head = 200 m, velocity coefficient = 0.98, diameter = 1.5 m. Calculate power developed.
95. Explain turbine efficiency, hydraulic efficiency, and mechanical efficiency numerically.
96. Calculate pressure and velocity distribution in pipe network.
97. A refrigeration system: refrigerant absorbs 50 kJ/kg, rejects 60 kJ/kg. Calculate work input.
98. A gas expands polytropically, n = 1.3. Find work done given P1, V1, P2.
99. Calculate flow through nozzle and exit velocity using energy equation.
100. A single-cylinder engine: bore = 0.1 m, stroke = 0.12 m, MEP = 1.2 MPa. Find work per stroke.

BPSC AEO Mechanical Main Exam – Solutions

Part 1 – Engineering Mechanics & Strength of Materials (Q1–20)

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Q1. Derive the expression for bending stress in a rectangular beam section under pure bending.

Solution:

• Consider a beam subjected to pure bending moment $M$M.
• Assumptions:
  1. Material is homogeneous and isotropic.
  2. Beam obeys Hooke’s law.
  3. Plane sections remain plane.
  4. Beam experiences pure bending only (no shear).

Derivation:

• Bending stress, $\sigma = \frac{M y}{I}$σ=IMy​
  Where:
  • $M =$M= bending moment
  • $y =$y= distance from neutral axis
  • $I =$I= moment of inertia of cross-section about neutral axis
• For rectangular section:
  $I = \frac{b h^3}{12}$I=12bh3​
  Maximum stress occurs at $y = h/2$y=h/2:
  $\sigma_\text{max} = \frac{M (h/2)}{I} = \frac{6M}{b h^2}$σmax​=IM(h/2)​=bh26M​

✅ Answer: $\boxed{\sigma = \frac{M y}{I}, \; \sigma_\text{max} = \frac{6M}{b h^2}}$σ=IMy​,σmax​=bh26M​​

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Q2. A cantilever beam of length 3 m carries a UDL of 5 kN/m. Find maximum deflection and slope at free end.

Given:

• $L = 3\,\text{m}, w = 5\,\text{kN/m} = 5000\,\text{N/m}$L=3m,w=5kN/m=5000N/m
• Cantilever, uniform load
• Modulus of elasticity $E$E, moment of inertia $I$I (symbolic)

Formulas for cantilever with UDL:

• Maximum deflection at free end:
  $\delta_\text{max} = \frac{w L^4}{8 E I}$δmax​=8EIwL4​
• Slope at free end:
  $\theta_\text{max} = \frac{w L^3}{6 E I}$θmax​=6EIwL3​

Calculation:
$\delta_\text{max} = \frac{5000 \times 3^4}{8 E I} = \frac{5000 \times 81}{8 E I} = \frac{405000}{8 E I} = \frac{50625}{E I}$δmax​=8EI5000×34​=8EI5000×81​=8EI405000​=EI50625​ m

$\theta_\text{max} = \frac{5000 \times 27}{6 E I} = \frac{135000}{6 E I} = \frac{22500}{E I}$θmax​=6EI5000×27​=6EI135000​=EI22500​ rad

✅ Answer: $\delta_\text{max} = 50625 / (E I), \; \theta_\text{max} = 22500 / (E I)$δmax​=50625/(EI),θmax​=22500/(EI)

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Q3. Define shear center. Explain its significance with a thin-walled section example.

Solution:

• Shear center: Point in cross-section through which applied transverse load produces bending only, without twisting.
• Significance: Avoids torsional rotation in thin-walled beams like channels, angles, or I-sections.
• Example:
  • For a C-channel, shear center lies outside the web, along symmetry axis.
  • If load passes through centroid but not shear center → beam twists.

✅ Answer: Shear center = point where load causes bending without twisting. Critical for thin-walled open sections.

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Q4. A solid circular shaft transmits 50 kW at 150 rpm. Determine torque and maximum shear stress if diameter = 50 mm.

Given:

• $P = 50\,\text{kW}, N = 150\,\text{rpm}, d = 0.05\,\text{m}$P=50kW,N=150rpm,d=0.05m

Step 1: Torque$$\omega = \frac{2 \pi N}{60} = \frac{2 \pi \times 150}{60} = 15.71\,\text{rad/s}$$ω=602πN​=602π×150​=15.71rad/s $$T = \frac{P}{\omega} = \frac{50000}{15.71} \approx 3180\,\text{N·m}$$T=ωP​=15.7150000​≈3180N\cdotpm

Step 2: Maximum shear stress
$\tau_\text{max} = \frac{16 T}{\pi d^3}$τmax​=πd316T​$$\tau_\text{max} = \frac{16 \times 3180}{\pi (0.05)^3} \approx \frac{50880}{3.927 \times 10^{-4}} \approx 129.5\,\text{MPa}$$τmax​=π(0.05)316×3180​≈3.927×10−450880​≈129.5MPa

✅ Answer: $T \approx 3180\,\text{N·m}, \tau_\text{max} \approx 129.5\,\text{MPa}$T≈3180N\cdotpm,τmax​≈129.5MPa

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Q5. Determine principal stresses and maximum shear stress for σx=80 MPa,σy=50 MPa,τxy=30 MPa\sigma_x = 80\text{ MPa}, \sigma_y = 50\text{ MPa}, \tau_{xy} = 30\text{ MPa}σx​=80 MPa,σy​=50 MPa,τxy​=30 MPa.

Formulas:

• Principal stresses:
  $\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x – \sigma_y}{2}\right)^2 + \tau_{xy}^2}$σ1,2​=2σx​+σy​​±(2σx​−σy​​)2+τxy2​​

$$\frac{\sigma_x + \sigma_y}{2} = \frac{80 + 50}{2} = 65$$2σx​+σy​​=280+50​=65 $$\frac{\sigma_x – \sigma_y}{2} = \frac{80 – 50}{2} = 15$$2σx​−σy​​=280−50​=15 $$\sqrt{15^2 + 30^2} = \sqrt{225 + 900} = \sqrt{1125} \approx 33.54$$152+302​=225+900​=1125​≈33.54 $$\sigma_1 = 65 + 33.54 = 98.54\,\text{MPa}, \quad \sigma_2 = 65 – 33.54 = 31.46\,\text{MPa}$$σ1​=65+33.54=98.54MPa,σ2​=65−33.54=31.46MPa

• Maximum shear stress:
  $\tau_\text{max} = \sqrt{\left(\frac{\sigma_x – \sigma_y}{2}\right)^2 + \tau_{xy}^2} = 33.54\,\text{MPa}$τmax​=(2σx​−σy​​)2+τxy2​​=33.54MPa

✅ Answer: $\sigma_1 \approx 98.54\,\text{MPa}, \sigma_2 \approx 31.46\,\text{MPa}, \tau_\text{max} \approx 33.54\,\text{MPa}$σ1​≈98.54MPa,σ2​≈31.46MPa,τmax​≈33.54MPa

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Q6. Discuss torsion in circular shafts and derive the torsion equation.

Solution:

• Torsion: Twisting of shaft under applied torque $T$T.
• Assumptions: Circular, homogeneous, isotropic, linear elastic.
• Torsion equation:

$$\tau = \frac{T r}{J}$$τ=JTr​

Where:

• $T =$T= applied torque
• $r =$r= radius at point
• $J =$J= polar moment of inertia ($J = \frac{\pi d^4}{32}$J=32πd4​ for solid shaft, $J = \frac{\pi}{32}(d_o^4 – d_i^4)$J=32π​(do4​−di4​) for hollow shaft)

✅ Answer: $\tau = \frac{T r}{J}$τ=JTr​, linear variation from center to surface.

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Q7. Simply supported beam with point load at midspan – Draw SFD and BMD.

Solution:

• Reactions: $R_A = R_B = P/2$RA​=RB​=P/2
• Shear force:

$$V = +P/2 \text{ from left support to midspan}, \; V = -P/2 \text{ from midspan to right support}$$V=+P/2 from left support to midspan,V=−P/2 from midspan to right support

• Bending moment:

$$M = R_A x = \frac{P}{2} x, \quad 0 \le x \le L/2$$M=RA​x=2P​x,0≤x≤L/2 $$M_\text{max} = P L/4 \text{ at midspan}$$Mmax​=PL/4 at midspan

Diagrams:

• SFD: Step from +P/2 to -P/2 at midspan
• BMD: Triangular, peak at midspan P L/4

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Q8. Composite bar subjected to axial load – derive stresses.

Solution:

• Bar made of two materials in series, perfectly bonded. Axial load $P$P.
• Strain compatibility: $\varepsilon_1 = \varepsilon_2$ε1​=ε2​
• Stress in each material:

$$\sigma_1 = \frac{P}{A_1} \frac{E_1}{E_1 + E_2}, \quad \sigma_2 = \frac{P}{A_2} \frac{E_2}{E_1 + E_2}$$σ1​=A1​P​E1​+E2​E1​​,σ2​=A2​P​E1​+E2​E2​​

✅ Stress is proportional to modulus and cross-section.

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Q9. Mohr’s circle for plane stress – numerical example.

Given: $\sigma_x = 80\text{ MPa}, \sigma_y = 50\text{ MPa}, \tau_{xy} = 30\text{ MPa}$σx​=80 MPa,σy​=50 MPa,τxy​=30 MPa

• Center: $C = (\sigma_x + \sigma_y)/2 = 65$C=(σx​+σy​)/2=65
• Radius: $R = \sqrt{( (\sigma_x – \sigma_y)/2)^2 + \tau_{xy}^2} = 33.54$R=((σx​−σy​)/2)2+τxy2​​=33.54

✅ Principal stresses: $\sigma_1 = 98.54, \sigma_2 = 31.46$σ1​=98.54,σ2​=31.46
✅ Maximum shear: $\tau_\text{max} = 33.54$τmax​=33.54

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Q10. Hollow circular shaft: d_o = 100 mm, d_i = 60 mm – polar moment of inertia.

$$J = \frac{\pi}{32}(d_o^4 – d_i^4) = \frac{\pi}{32}(0.1^4 – 0.06^4)$$J=32π​(do4​−di4​)=32π​(0.14−0.064) $$J = \frac{\pi}{32}(1e-4 – 1.296e-5) = \frac{\pi}{32}(8.704e-5) \approx 8.54 \times 10^{-6}\, \text{m}^4$$J=32π​(1e−4−1.296e−5)=32π​(8.704e−5)≈8.54×10−6m4

✅ Answer: $J \approx 8.54 \times 10^{-6}\, \text{m}^4$J≈8.54×10−6m4

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Q11. Deflection of simply supported beam using double integration method.

• Equation: $EI \frac{d^2 y}{dx^2} = M(x)$EIdx2d2y​=M(x)
• Integrate twice: $\frac{d^2 y}{dx^2} = \frac{M(x)}{EI} \Rightarrow \frac{dy}{dx} = \int \frac{M(x)}{EI} dx + C_1 \Rightarrow y = \int \frac{dy}{dx} dx + C_2$dx2d2y​=EIM(x)​⇒dxdy​=∫EIM(x)​dx+C1​⇒y=∫dxdy​dx+C2​
• Use boundary conditions: $y = 0$y=0 at supports to solve $C_1, C_2$C1​,C2​

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Q12. Slope at free end of cantilever with point load P at free end.

$$\theta_\text{free} = \frac{P L^2}{2 E I}, \quad \delta_\text{free} = \frac{P L^3}{3 E I}$$θfree​=2EIPL2​,δfree​=3EIPL3​

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Q13. Section modulus for rectangular section 200 × 300 mm.

$$Z = \frac{I}{y_\text{max}} = \frac{b h^3 / 12}{h/2} = \frac{b h^2}{6}$$Z=ymax​I​=h/2bh3/12​=6bh2​ $$Z = \frac{0.2 \times 0.3^2}{6} = \frac{0.2 \times 0.09}{6} = 0.003\,\text{m}^3$$Z=60.2×0.32​=60.2×0.09​=0.003m3

✅ Answer: $Z = 3 \times 10^{-3} \, \text{m}^3$Z=3×10−3m3

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Q14. Beam subjected to combined bending and axial load.

• Maximum stress:

$$\sigma_\text{max} = \frac{P}{A} \pm \frac{M y_\text{max}}{I}$$σmax​=AP​±IMymax​​

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Q15. Curvature of beam and radius of curvature.

$$\kappa = \frac{1}{\rho} = \frac{d^2 y}{dx^2} / \left(1 + (dy/dx)^2\right)^{3/2} \approx \frac{d^2 y}{dx^2} = \frac{M}{EI}$$κ=ρ1​=dx2d2y​/(1+(dy/dx)2)3/2≈dx2d2y​=EIM​

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Q16. Spring design – load 500 N, diameter 50 mm, wire 5 mm.

• Maximum shear stress: $\tau = \frac{8 W D}{\pi d^3} K$τ=πd38WD​K (K = Wahl factor for curvature)
• Deflection: $\delta = \frac{8 W D^3 N}{G d^4}$δ=Gd48WD3N​

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Q17. Factor of safety – ductile/brittle examples.

• Ductile: $FS = \frac{\sigma_\text{yield}}{\sigma_\text{working}}$FS=σworking​σyield​​
• Brittle: $FS = \frac{\sigma_\text{ultimate}}{\sigma_\text{working}}$FS=σworking​σultimate​​

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Q18. Torsional rigidity of hollow shafts.

$$GJ/L, \quad J = \frac{\pi}{32}(d_o^4 – d_i^4)$$GJ/L,J=32π​(do4​−di4​)

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Q19. Cantilever with triangular UDL – max bending moment and shear.

• Max bending: $M_\text{max} = \frac{w L^2}{6}$Mmax​=6wL2​
• Max shear at fixed end: $V_\text{max} = \frac{w L}{2}$Vmax​=2wL​

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Q20. Statically indeterminate beams.

• Beams with more reactions than equations of equilibrium.
• Solution: Use force method or displacement method (compatibility equations + bending moment equations).

Part 2 – Thermodynamics & Heat Transfer (Q21–40)

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Q21. Derive efficiency of Carnot cycle and discuss its limitations.

Solution:

• Carnot cycle: Reversible cycle with two isothermal and two adiabatic processes.
• Heat absorbed at high temperature: $Q_H$QH​
• Heat rejected at low temperature: $Q_L$QL​
• Efficiency:

$$\eta = \frac{W_\text{net}}{Q_H} = \frac{Q_H – Q_L}{Q_H} = 1 – \frac{Q_L}{Q_H}$$η=QH​Wnet​​=QH​QH​−QL​​=1−QH​QL​​

• For reversible isothermal processes, $Q/T = \text{constant}$Q/T=constant →

$$\frac{Q_L}{T_L} = \frac{Q_H}{T_H} \Rightarrow \eta = 1 – \frac{T_L}{T_H}$$TL​QL​​=TH​QH​​⇒η=1−TH​TL​​

• Limitations:
  1. Idealization – requires reversible processes.
  2. No practical cycle is completely reversible.
  3. Efficiency depends only on temperature limits.

✅ Answer: $\eta_\text{Carnot} = 1 – T_L/T_H$ηCarnot​=1−TL​/TH​

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Q22. Explain Otto and Diesel cycles, derive efficiency formulas and compare.

Otto cycle:

• Constant-volume heat addition
• Efficiency:

$$\eta_\text{Otto} = 1 – \frac{1}{r^{\gamma-1}}, \quad r = \text{compression ratio}$$ηOtto​=1−rγ−11​,r=compression ratio

Diesel cycle:

• Constant-pressure heat addition
• Efficiency:

$$\eta_\text{Diesel} = 1 – \frac{1}{r^{\gamma-1}} \cdot \frac{\rho^\gamma – 1}{\gamma (\rho – 1)}, \quad \rho = V_2/V_1 = cut-off ratio$$ηDiesel​=1−rγ−11​⋅γ(ρ−1)ργ−1​,ρ=V2​/V1​=cut−offratio

• Comparison:
  • Diesel cycle has lower efficiency for same $r$r because heat addition at constant pressure.
  • Otto more efficient for same compression ratio.

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Q23. Heat engine absorbs 500 kJ, rejects 300 kJ. Determine thermal efficiency and work done.

$$\eta = \frac{W_\text{net}}{Q_\text{in}} = \frac{Q_\text{in} – Q_\text{out}}{Q_\text{in}} = \frac{500 – 300}{500} = 0.4 = 40\%$$η=Qin​Wnet​​=Qin​Qin​−Qout​​=500500−300​=0.4=40% $$W_\text{net} = Q_\text{in} – Q_\text{out} = 500 – 300 = 200 \, \text{kJ}$$Wnet​=Qin​−Qout​=500−300=200kJ

✅ Answer: $\eta = 40\%, W_\text{net} = 200\text{ kJ}$η=40%,Wnet​=200 kJ

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Q24. Explain first and second laws of thermodynamics for closed and open systems.

• First law (energy conservation):
  • Closed system: $\Delta U = Q – W$ΔU=Q−W
  • Open system (control volume): $\Delta E_\text{cv} = \Delta E_\text{in} – \Delta E_\text{out}$ΔEcv​=ΔEin​−ΔEout​
• Second law:
  • Heat cannot spontaneously flow from cold to hot.
  • Entropy generation $S_\text{gen} \ge 0$Sgen​≥0
  • Clausius inequality: $\oint \frac{\delta Q}{T} \le 0$∮TδQ​≤0

✅ Answer: First law → energy balance, Second law → irreversibility and direction of processes.

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Q25. Calculate entropy change when 2 kg of water at 100°C is converted to steam at 100°C.

Given:

• Mass $m = 2\,\text{kg}$m=2kg, latent heat $L_v = 2257\,\text{kJ/kg}$Lv​=2257kJ/kg

$$\Delta S = \frac{Q_\text{rev}}{T} = \frac{m L_v}{T} = \frac{2 \times 2257}{373} \approx 12.1\,\text{kJ/K}$$ΔS=TQrev​​=TmLv​​=3732×2257​≈12.1kJ/K

✅ Answer: $\Delta S \approx 12.1\,\text{kJ/K}$ΔS≈12.1kJ/K

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Q26. Rankine cycle with boiler P = 10 MPa, condenser P = 0.1 MPa. Calculate efficiency.

Solution:

• Use steam tables:
  • Boiler (superheated) enthalpy $h_1$h1​
  • Condenser (saturated) enthalpy $h_4$h4​
• Net work: $W_\text{net} = h_1 – h_2 – (h_4 – h_3)$Wnet​=h1​−h2​−(h4​−h3​)
• Heat added: $Q_\text{in} = h_1 – h_4$Qin​=h1​−h4​
• Efficiency: $\eta = W_\text{net}/Q_\text{in}$η=Wnet​/Qin​
• Typical result for 10 MPa / 0.1 MPa: $\eta \approx 42\%$η≈42%

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Q27. Derive Clapeyron equation and explain significance.

$$\frac{dP}{dT} = \frac{L}{T \Delta v}$$dTdP​=TΔvL​

• $L =$L= latent heat per unit mass
• $\Delta v =$Δv= specific volume change during phase change

Significance:

• Describes slope of phase boundary in P–T diagram
• Used to calculate boiling point, vapor pressure.

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Q28. Explain refrigeration cycles and calculate COP for vapor compression system.

• Vapor-compression cycle:
  1. Evaporation → absorbs heat
  2. Compression → raises pressure
  3. Condensation → rejects heat
  4. Expansion → throttling
• COP:

$$\text{COP}_\text{R} = \frac{Q_L}{W_\text{input}} = \frac{h_1 – h_4}{h_2 – h_1}$$COPR​=Winput​QL​​=h2​−h1​h1​−h4​​

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Q29. Determine LMTD for a counter-flow heat exchanger.

$$\Delta T_m = \frac{\Delta T_1 – \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$$ΔTm​=ln(ΔT1​/ΔT2​)ΔT1​−ΔT2​​

Where:
$\Delta T_1 = T_{h,\text{in}} – T_{c,\text{out}}, \; \Delta T_2 = T_{h,\text{out}} – T_{c,\text{in}}$ΔT1​=Th,in​−Tc,out​,ΔT2​=Th,out​−Tc,in​

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Q30. Derive conduction equation in 1D steady-state.

• Fourier’s law: $q = -k \frac{dT}{dx}$q=−kdxdT​
• Energy balance in dx → $\frac{d}{dx}(-k A \frac{dT}{dx}) + q”’ A dx = 0$dxd​(−kAdxdT​)+q′′′Adx=0
• Steady state, no generation:

$$\frac{d^2 T}{dx^2} = 0$$dx2d2T​=0

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Q31. Discuss fins and extended surfaces with numerical problem.

• Purpose: Increase heat transfer area
• Fin equation (1D, straight fin):

$$\frac{d^2 \theta}{dx^2} – m^2 \theta = 0, \quad m = \sqrt{\frac{h P}{k A_c}}$$dx2d2θ​−m2θ=0,m=kAc​hP​​

• Numerical: Calculate tip temperature for given fin dimensions.

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Q32. Explain natural and forced convection with examples.

• Natural: Driven by density differences, e.g., air around heated wall
• Forced: Driven by external means, e.g., fan blowing over pipe

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Q33. Derive radiation heat transfer between two black bodies.

$$Q = \sigma A (T_1^4 – T_2^4)$$Q=σA(T14​−T24​)

• $\sigma = 5.67 \times 10^{-8} \text{ W/m²K⁴}$σ=5.67×10−8 W/m²K⁴

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Q34. Slab conduction: t = 0.2 m, k = 50 W/m·K, q = 100 W/m² → find ΔT.

$$q = k \frac{\Delta T}{L} \Rightarrow \Delta T = \frac{q L}{k} = \frac{100 \times 0.2}{50} = 0.4\,\text{K}$$q=kLΔT​⇒ΔT=kqL​=50100×0.2​=0.4K

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Q35. Explain Biot and Fourier numbers.

• Biot number: $\text{Bi} = \frac{h L_c}{k}$Bi=khLc​​, ratio of surface to conduction resistance
• Fourier number: $\text{Fo} = \frac{\alpha t}{L^2}$Fo=L2αt​, dimensionless time in transient conduction

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Q36. Calculate effectiveness of heat exchanger given Qmax and Qactual.

$$\varepsilon = \frac{Q_\text{actual}}{Q_\text{max}}$$ε=Qmax​Qactual​​

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Q37. Derive isentropic relations for ideal gases.

• For isentropic: $PV^\gamma = \text{constant}$PVγ=constant, $T V^{\gamma-1} = \text{constant}$TVγ−1=constant, $T P^{(\gamma-1)/\gamma} = \text{constant}$TP(γ−1)/γ=constant

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Q38. Polytropic process work with n = 1.3.

$$W = \frac{P_2 V_2 – P_1 V_1}{1 – n}$$W=1−nP2​V2​−P1​V1​​

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Q39. Joule-Thomson effect and applications.

• Definition: Temperature change during throttling process at constant enthalpy
• Applications: Refrigeration, liquefaction of gases

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Q40. Compare refrigeration, heat pump, air-conditioning cycles numerically.

• COP_R = Q_L / W
• COP_HP = Q_H / W = COP_R + 1
• Air-conditioning is similar to refrigeration but maintains comfort temp + humidity

Part 3 – Fluid Mechanics & Hydraulic Machines (Q41–60)

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Q41. Derive Bernoulli’s equation along a streamline.

Solution:

• Consider incompressible, non-viscous, steady flow along a streamline.
• Work-energy principle: Work done by pressure + gravity = kinetic energy change

$$\frac{P}{\rho} + \frac{v^2}{2} + g z = \text{constant along streamline}$$ρP​+2v2​+gz=constant along streamline

Where:

• $P =$P= pressure, $\rho =$ρ= density, $v =$v= velocity, $z =$z= elevation

✅ Answer: Bernoulli’s equation: $\boxed{\frac{P}{\rho} + \frac{v^2}{2} + g z = \text{constant}}$ρP​+2v2​+gz=constant​

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Q42. Venturimeter: d1 = 0.2 m, d2 = 0.1 m, ΔP = 20 kPa → flow rate.

Given: ΔP = 20 kPa, ρ = 1000 kg/m³$$Q = C_d A_2 \sqrt{\frac{2 \Delta P}{\rho (1 – (A_2/A_1)^2)}}$$Q=Cd​A2​ρ(1−(A2​/A1​)2)2ΔP​​

• $A_1 = \pi d_1^2 /4 = 0.0314$A1​=πd12​/4=0.0314 m²
• $A_2 = \pi d_2^2 /4 = 0.00785$A2​=πd22​/4=0.00785 m²

$$1 – (A_2/A_1)^2 = 1 – (0.00785/0.0314)^2 = 0.9375$$1−(A2​/A1​)2=1−(0.00785/0.0314)2=0.9375 $$Q = 0.98 \times 0.00785 \sqrt{\frac{2 \times 20000}{1000 \times 0.9375}} = 0.0077 \times \sqrt{42.67} = 0.0077 \times 6.54 \approx 0.0504\,\text{m³/s}$$Q=0.98×0.007851000×0.93752×20000​​=0.0077×42.67​=0.0077×6.54≈0.0504m³/s

✅ Answer: $Q \approx 0.05\,\text{m³/s}$Q≈0.05m³/s

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Q43. Reynolds number and laminar/turbulent transition.

$$\text{Re} = \frac{\rho v D}{\mu}$$Re=μρvD​

• Laminar: Re < 2000
• Transitional: 2000 < Re < 4000
• Turbulent: Re > 4000

Significance: Determines flow regime and friction factor.

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Q44. Pipe: d = 0.1 m, v = 2 m/s, head loss using Darcy-Weisbach.

$$h_f = f \frac{L}{D} \frac{v^2}{2g}$$hf​=fDL​2gv2​

• Friction factor $f$f depends on Re and roughness.
• Plug values: Example L = 50 m, f = 0.02:

$$h_f = 0.02 \frac{50}{0.1} \frac{2^2}{2 \cdot 9.81} = 10 \cdot \frac{4}{19.62} \approx 2.04\,\text{m}$$hf​=0.020.150​2⋅9.8122​=10⋅19.624​≈2.04m

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Q45. Continuity equation for branching pipes.

$$Q_\text{in} = Q_1 + Q_2 + \dots$$Qin​=Q1​+Q2​+…

• For incompressible flow: $A v = \text{constant}$Av=constant
• Apply at junctions for branched network.

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Q46. Hydraulic gradient line (HGL) and energy gradient line (EGL).

• HGL: Represents pressure head + elevation head
• EGL: Represents total energy (pressure + velocity + elevation)
• Relation: EGL lies above HGL by velocity head (v²/2g)

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Q47. Pelton turbine: 500 kW, head 200 m → jet diameter.

• Power: $P = \eta \rho g Q H$P=ηρgQH, assume $\eta = 0.9$η=0.9

$$Q = \frac{P}{\eta \rho g H} = \frac{500000}{0.9 \times 1000 \times 9.81 \times 200} \approx 0.283\,\text{m³/s}$$Q=ηρgHP​=0.9×1000×9.81×200500000​≈0.283m³/s

• Jet diameter: $Q = \pi d^2 v/4$Q=πd2v/4, $v = \sqrt{2 g H} = 62.6$v=2gH​=62.6 m/s

$$d = \sqrt{\frac{4Q}{\pi v}} = \sqrt{\frac{4 \cdot 0.283}{3.1416 \cdot 62.6}} \approx 0.076\,\text{m} = 76\,\text{mm}$$d=πv4Q​​=3.1416⋅62.64⋅0.283​​≈0.076m=76mm

✅ Answer: $d \approx 76\,\text{mm}$d≈76mm

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Q48. Cavitation in pumps and avoidance.

• Cavitation: Formation of vapor bubbles due to low pressure → collapse → damage.
• Avoidance: Increase suction head, reduce velocity, proper pump selection.

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Q49. Centrifugal pump: Q = 0.05 m³/s, H = 20 m, η = 80%. Shaft power.

$$P_\text{shaft} = \frac{\rho g Q H}{\eta} = \frac{1000 \cdot 9.81 \cdot 0.05 \cdot 20}{0.8} = 12.26\,\text{kW}$$Pshaft​=ηρgQH​=0.81000⋅9.81⋅0.05⋅20​=12.26kW

✅ Answer: $P_\text{shaft} \approx 12.3\,\text{kW}$Pshaft​≈12.3kW

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Q50. Derive Euler’s equation for turbines.

• Euler equation for rotating impeller:

$$W_s = \dot{m} (V_u2 r_2 – V_u1 r_1)$$Ws​=m˙(Vu​2r2​−Vu​1r1​)

• $V_u =$Vu​= tangential velocity component of fluid

✅ Governs energy transfer from fluid to rotor.

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Q51. Specific speed of pump and turbine, numerical example.

$$N_s = N \sqrt{P}/H^{3/4} \quad (\text{metric})$$Ns​=NP​/H3/4(metric)

• Determines pump/turbine type selection.

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Q52. Francis turbine: Q = 50 m³/s, H = 20 m → shaft power.

$$P = \rho g Q H \eta = 1000 \cdot 9.81 \cdot 50 \cdot 20 \cdot 0.9 \approx 8.83\,\text{MW}$$P=ρgQHη=1000⋅9.81⋅50⋅20⋅0.9≈8.83MW

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Q53. Flow in open channel – Chezy formula.

$$V = C \sqrt{R S}, \quad Q = A V$$V=CRS​,Q=AV

• $C =$C= Chezy coefficient, $R =$R= hydraulic radius, $S =$S= slope

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Q54. Critical depth and Froude number.

$$V^2 / (g y) = 1 \Rightarrow F_r = \frac{V}{\sqrt{g y}} = 1$$V2/(gy)=1⇒Fr​=gy​V​=1

• Rectangular channel: $y_c = (Q^2 / g b^2)^{1/3}$yc​=(Q2/gb2)1/3

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Q55. Minor losses in pipe flow.

• Losses due to bends, valves, expansions
• $h_L = K v^2 / 2g$hL​=Kv2/2g
• Example: elbow K = 0.3

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Q56. Venturimeter: ΔP = 10 kPa → velocity and discharge.

• Same formula as Q42, plug ΔP = 10 kPa, find $v, Q$v,Q

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Q57. Derive manometer equation.

$$\Delta P = \rho g h$$ΔP=ρgh

• For U-tube: $\Delta P = (\rho_m – \rho) g h$ΔP=(ρm​−ρ)gh

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Q58. Hydraulic jump, sequent depth.

• Energy conservation and momentum:

$$y_2 = \frac{y_1}{2} (\sqrt{1 + 8 F_1^2} – 1), \quad F_1 = \frac{V_1}{\sqrt{g y_1}}$$y2​=2y1​​(1+8F12​​−1),F1​=gy1​​V1​​

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Q59. Boundary layer concept.

• Definition: Thin region near wall with velocity gradient
• Applications: Drag calculation, heat transfer enhancement, separation control

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Q60. Pipe network: Calculate equivalent resistance and flow rates.

• Series: $R_\text{eq} = R_1 + R_2 + \dots$Req​=R1​+R2​+…
• Parallel: $1/R_\text{eq} = 1/R_1 + 1/R_2 + \dots$1/Req​=1/R1​+1/R2​+…
• Flow: $Q = \Delta H / R_\text{eq}$Q=ΔH/Req​

Part 4 – Machine Design & Theory of Machines (Q61–80)

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Q61. Derive torque transmitted by a solid circular shaft (numerical example).

Torsion equation:$$\tau_\text{max} = \frac{16 T}{\pi d^3} \quad \Rightarrow T = \frac{\pi d^3 \tau_\text{max}}{16}$$τmax​=πd316T​⇒T=16πd3τmax​​

Example:

• $d = 50 \text{ mm} = 0.05\text{ m}$d=50 mm=0.05 m, $\tau_\text{max} = 80 \text{ MPa}$τmax​=80 MPa

$$T = \frac{3.1416 \cdot 0.05^3 \cdot 80 \times 10^6}{16} \approx 3140 \text{ N·m}$$T=163.1416⋅0.053⋅80×106​≈3140 N\cdotpm

✅ Answer: $T \approx 3.14 \text{ kN·m}$T≈3.14 kN\cdotpm

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Q62. Design helical spring for given load, deflection, and material properties.

• Formulas:
  • Shear stress: $\tau = \frac{8 W D_m K}{\pi d^3}$τ=πd38WDm​K​, K = Wahl factor
  • Deflection: $\delta = \frac{8 W D_m^3 N}{G d^4}$δ=Gd48WDm3​N​
• Solve for wire diameter ddd, mean diameter DmD_mDm​, and number of coils NNN using given $W, \delta, G, \tau_\text{allow}$W,δ,G,τallow​.

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Q63. Explain flywheel design to store energy for fluctuation of speed 10%.

• Fluctuation energy: $\Delta E = I \omega^2 \frac{\delta^2}{2}$ΔE=Iω22δ2​
• Moment of inertia: $I = \frac{\Delta E}{\omega^2 (\delta/2)}$I=ω2(δ/2)ΔE​
• Design rim and hub based on maximum stress.

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Q64. Determine diameter of hollow shaft transmitting torque.

$$\tau_\text{max} = \frac{16 T}{\pi (d_o^4 – d_i^4)} d_o \quad \Rightarrow d_o \text{ or } d_i$$τmax​=π(do4​−di4​)16T​do​⇒do​ or di​

• Use torsion formula, assume hollow ratio $d_i/d_o$di​/do​, solve numerically.

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Q65. Explain belt drive, slip, and velocity ratio numerically.

• Velocity ratio: $v_1 / v_2 = N_2 / N_1 = d_2 / d_1$v1​/v2​=N2​/N1​=d2​/d1​
• Slip: $s = \frac{v_\text{driver} – v_\text{driven}}{v_\text{driver}} \times 100\%$s=vdriver​vdriver​−vdriven​​×100%
• Numerical: Calculate driven speed if slip = 2%.

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Q66. Design sprocket chain drive.

• Power transmitted: $P = \text{tension} \times \text{velocity}$P=tension×velocity
• Choose chain pitch, number of teeth, and center distance based on power and speed.

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Q67. Mechanical advantage and efficiency of screw jack.

• Torque $T = \frac{W l}{2 \pi \eta}$T=2πηWl​
• Mechanical advantage $MA = \frac{W}{F} = \frac{2 \pi r}{l}$MA=FW​=l2πr​
• Efficiency: $\eta = \frac{MA_\text{actual}}{MA_\text{theoretical}}$η=MAtheoretical​MAactual​​

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Q68. Slider-crank mechanism: Calculate slider velocity and acceleration at mid-stroke.

• Position: $x = r \cos \theta + l \cos \phi$x=rcosθ+lcosϕ
• Velocity: $v = \frac{dx}{dt} = – r \omega \sin \theta – l \frac{d\phi}{dt} \sin \phi$v=dtdx​=−rωsinθ−ldtdϕ​sinϕ
• Acceleration: $a = \frac{d^2 x}{dt^2}$a=dt2d2x​
• Mid-stroke: $\theta = 90^\circ$θ=90∘

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Q69. Equation of motion for simple harmonic motion of a crank.

• Displacement of slider: $x = r \cos \theta + l \cos \phi$x=rcosθ+lcosϕ
• SHM approximation: $x \approx r (1 – \cos \omega t)$x≈r(1−cosωt)

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Q70. Grashof condition and degree of freedom in four-bar mechanism.

• Grashof criterion:$s + l \le p + q$s+l≤p+q
  • s = shortest link, l = longest, p, q = remaining
• Degree of freedom (DoF):

$$F = 3(n-1) – 2j_1 – j_2$$F=3(n−1)−2j1​−j2​

• 4-bar: F = 1

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Q71. Gear ratio, number of teeth, pitch circle diameter for compound gear train.

• Velocity ratio: $i = \frac{N_\text{driven}}{N_\text{driver}}$i=Ndriver​Ndriven​​
• Pitch circle: $d = m \cdot N$d=m⋅N, m = module

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Q72. Bending and shear stress in gear teeth using Lewis formula.

$$\sigma = \frac{F_t}{b m Y}, \quad F_t = \text{tangential load}, b = \text{face width}, Y = \text{Lewis form factor}$$σ=bmYFt​​,Ft​=tangential load,b=face width,Y=Lewis form factor

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Q73. Torque transmitted by belt drive with initial tension.

$$T = (T_1 – T_2) r, \quad T_1/T_2 = e^{\mu \alpha}$$T=(T1​−T2​)r,T1​/T2​=eμα

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Q74. Design clutch plate for given torque and coefficient of friction.

$$T = \mu P R_\text{mean} \quad \Rightarrow \text{determine diameter and pressure}$$T=μPRmean​⇒determine diameter and pressure

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Q75. Torsional vibration of shafts – natural frequency.

$$f_n = \frac{1}{2 \pi} \sqrt{\frac{k}{I}}, \quad k = \text{torsional stiffness}, I = \text{mass moment of inertia}$$fn​=2π1​Ik​​,k=torsional stiffness,I=mass moment of inertia

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Q76. Flywheel of 100 kg-m² at 300 rpm – max/min kinetic energy for 10% fluctuation.

• $\omega = 2\pi N /60 = 31.42$ω=2πN/60=31.42 rad/s
• Kinetic energy: $E = 1/2 I \omega^2 = 0.5 \cdot 100 \cdot 31.42^2 \approx 49.3\,\text{kJ}$E=1/2Iω2=0.5⋅100⋅31.422≈49.3kJ
• Fluctuation ±10% → $E_\text{max} = 54.2\,\text{kJ}, E_\text{min} = 44.3\,\text{kJ}$Emax​=54.2kJ,Emin​=44.3kJ

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Q77. Bearing selection and load calculation for journal bearing.

• Pressure: $p = W / (L D)$p=W/(LD)
• Velocity: $V = \pi D n / 60$V=πDn/60
• PV limit: Check against allowable $PV_\text{max}$PVmax​

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Q78. Design keys and couplings for given shaft torque.

• Key design: $\tau = \frac{4 T}{d L b}$τ=dLb4T​ → choose width (b) and length (L)
• Couplings: Torque transmitted = shear stress × area

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Q79. Helical gear transmits 20 kW at 1000 rpm – module and face width.

• Tangential force: $F_t = 2 P / v$Ft​=2P/v
• Lewis formula for bending stress → calculate module (m)
• Face width $b = 10 m$b=10m typical

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Q80. Cam and follower design – determine cam profile for given motion.

• Follower displacement function: $s(\theta)$s(θ)
• Velocity: $v = ds/d\theta \cdot \omega$v=ds/dθ⋅ω
• Acceleration: $a = d^2s/d\theta^2 \cdot \omega^2$a=d2s/dθ2⋅ω2
• Use motion laws: uniform, simple harmonic, cycloidal for smooth motion

Part 5 – IC Engines, Refrigeration, Heat Engines, and Applied Numericals (Q81–100)

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Q81. Derive mean effective pressure (MEP) for Otto and Diesel cycles.

Otto cycle:$$\text{MEP} = \frac{W_\text{net}}{V_s} = \frac{1}{V_s} (Q_\text{in} – Q_\text{out})$$MEP=Vs​Wnet​​=Vs​1​(Qin​−Qout​)

• Heat addition at constant volume: $Q_\text{in} = m c_v (T_3 – T_2)$Qin​=mcv​(T3​−T2​)
• Heat rejection: $Q_\text{out} = m c_v (T_4 – T_1)$Qout​=mcv​(T4​−T1​)
• MEP formula:

$$\text{MEP}_\text{Otto} = \frac{R T_1 (r^{\gamma} – 1)}{r-1}, \quad r = \text{compression ratio}$$MEPOtto​=r−1RT1​(rγ−1)​,r=compression ratio

Diesel cycle:$$\text{MEP}_\text{Diesel} = \frac{1}{r-1} [P_3 V_3 \ln(r_c) – P_1 V_1 (r^{\gamma-1} – 1)], \quad r_c = \text{cut-off ratio}$$MEPDiesel​=r−11​[P3​V3​ln(rc​)−P1​V1​(rγ−1−1)],rc​=cut-off ratio

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Q82. Power output and efficiency of single-cylinder engine (given bore, stroke, MEP).

$$\text{BP} = \frac{p_\text{m} L A N}{2} \quad \text{(4-stroke)}, \quad \eta = \frac{BP}{Q_\text{in}}$$BP=2pm​LAN​(4-stroke),η=Qin​BP​

• $L = \text{stroke}, A = \pi d^2/4, N = \text{rev/s}$L=stroke,A=πd2/4,N=rev/s

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Q83. Determine brake power, indicated power, mechanical efficiency.

$$\eta_m = \frac{BP}{IP} \quad \Rightarrow IP = \frac{BP}{\eta_m}$$ηm​=IPBP​⇒IP=ηm​BP​

• Torque and speed used for BP: $BP = 2 \pi N T / 60$BP=2πNT/60

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Q84. Refrigeration system: 20 kW effect, 5 kW input → COP.

$$\text{COP}_R = \frac{Q_L}{W} = \frac{20}{5} = 4$$COPR​=WQL​​=520​=4

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Q85. Vapor compression cycle – refrigeration effect and work input.

$$Q_L = h_1 – h_4, \quad W = h_2 – h_1$$QL​=h1​−h4​,W=h2​−h1​

• COP = $Q_L / W$QL​/W

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Q86. Compressor work and volumetric efficiency for reciprocating compressor.

$$W = \frac{n}{\eta_s} \frac{P_2 V_2 – P_1 V_1}{n-1} \quad (\text{polytropic})$$W=ηs​n​n−1P2​V2​−P1​V1​​(polytropic) $$\eta_v = \frac{V_\text{intake}}{V_\text{cylinder}}$$ηv​=Vcylinder​Vintake​​

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Q87. Air-standard efficiency of dual cycle and comparison with Otto cycle.

• Dual cycle: combination of constant volume and constant pressure heat addition
• Efficiency:

$$\eta_\text{dual} = 1 – \frac{1}{r^{\gamma-1}} \cdot \frac{\beta \rho^\gamma – 1}{(\beta-1) + \gamma (\rho – 1)}$$ηdual​=1−rγ−11​⋅(β−1)+γ(ρ−1)βργ−1​

• Otto is more efficient for same compression ratio; Diesel higher MEP.

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Q88. Gas turbine: net work 500 kJ/kg, turbine work 1200 kJ/kg → work ratio.

$$\text{Work ratio} = \frac{W_\text{net}}{W_\text{turbine}} = \frac{500}{1200} \approx 0.417$$Work ratio=Wturbine​Wnet​​=1200500​≈0.417

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Q89. Brake mean effective pressure for given torque and engine speed.

$$\text{BMEP} = \frac{4 \pi T}{V_s} \quad (\text{4-stroke})$$BMEP=Vs​4πT​(4-stroke)

• $V_s = \text{cylinder swept volume}$Vs​=cylinder swept volume

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Q90. Air conditioning load (sensible + latent).

• Sensible load: $Q_s = m \cdot C_p \cdot \Delta T$Qs​=m⋅Cp​⋅ΔT
• Latent load: $Q_l = m \cdot h_\text{fg} \cdot \Delta w$Ql​=m⋅hfg​⋅Δw
• Total: $Q = Q_s + Q_l$Q=Qs​+Ql​

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Q91. Refrigerant properties using Mollier chart.

• Identify enthalpy, entropy, temperature, pressure from h–s chart.
• Used for work input and refrigeration effect.

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Q92. Single-stage centrifugal pump: Q = 0.05 m³/s, H = 20 m → shaft power and efficiency.

$$P_\text{shaft} = \frac{\rho g Q H}{\eta} = \frac{1000 \cdot 9.81 \cdot 0.05 \cdot 20}{0.85} \approx 11.53\,\text{kW}$$Pshaft​=ηρgQH​=0.851000⋅9.81⋅0.05⋅20​≈11.53kW

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Q93. Solve thermodynamic cycle: heat added and rejected → work, efficiency.

$$W_\text{net} = Q_\text{in} – Q_\text{out}, \quad \eta = \frac{W_\text{net}}{Q_\text{in}}$$Wnet​=Qin​−Qout​,η=Qin​Wnet​​

• Apply first law of thermodynamics.

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Q94. Pelton wheel: Head = 200 m, velocity coefficient 0.98, diameter = 1.5 m → power developed.

• Flow rate: $Q = \pi d_j^2 v /4$Q=πdj2​v/4
• Velocity: $v = \sqrt{2 g H} \cdot \text{velocity coefficient}$v=2gH​⋅velocity coefficient
• Power: $P = \rho g Q H$P=ρgQH

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Q95. Turbine efficiency: hydraulic, mechanical, overall.

• Hydraulic efficiency: $\eta_h = P_\text{water on runner}/P_\text{available}$ηh​=Pwater on runner​/Pavailable​
• Mechanical efficiency: $\eta_m = P_\text{shaft}/P_\text{runner}$ηm​=Pshaft​/Prunner​
• Overall: $\eta_o = \eta_h \cdot \eta_m$ηo​=ηh​⋅ηm​

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Q96. Pressure and velocity distribution in pipe network.

• Apply continuity and energy equations at nodes and loops
• Solve using Hardy-Cross method or equivalent resistance method.

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Q97. Refrigeration system: absorbs 50 kJ/kg, rejects 60 kJ/kg → work input.

$$W = Q_\text{out} – Q_\text{in} = 60 – 50 = 10 \text{ kJ/kg} \quad \text{COP} = Q_\text{in}/W = 5$$W=Qout​−Qin​=60−50=10 kJ/kgCOP=Qin​/W=5

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Q98. Gas expands polytropically, n = 1.3, P1, V1, P2 → work done.

$$W = \frac{P_2 V_2 – P_1 V_1}{1 – n}$$W=1−nP2​V2​−P1​V1​​

• Calculate $V_2 = V_1 (P_1/P_2)^{1/n}$V2​=V1​(P1​/P2​)1/n and plug.

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Q99. Flow through nozzle, exit velocity using energy equation.

• Energy equation: $\frac{v^2}{2} + gz + P/\rho = \text{constant}$2v2​+gz+P/ρ=constant
• Neglect height: $v_\text{exit} = \sqrt{2(P_1 – P_2)/\rho}$vexit​=2(P1​−P2​)/ρ​

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Q100. Single-cylinder engine: bore = 0.1 m, stroke = 0.12 m, MEP = 1.2 MPa → work per stroke.

$$V_s = \pi d^2/4 \cdot L = 3.1416 \cdot 0.1^2/4 \cdot 0.12 \approx 9.42 \times 10^{-4} \text{ m³}$$Vs​=πd2/4⋅L=3.1416⋅0.12/4⋅0.12≈9.42×10−4 m³ $$W = p_\text{m} V_s = 1.2 \times 10^6 \cdot 9.42 \times 10^{-4} \approx 1130\,\text{J}$$W=pm​Vs​=1.2×106⋅9.42×10−4≈1130J

✅ Answer: Work per stroke ≈ 1.13 kJ

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This mock test is for practice and educational purposes only. It is not an official BPSC exam paper, and solutions are provided to help candidates prepare and evaluate their understanding.