SC JE Tier-II Mechanical – Solutions

Q1 – Thermodynamics / IC Engines (Otto Cycle)

Given:

• Compression ratio $r = 8$r=8
• $\gamma = 1.4$γ=1.4
• Heat added $q_{in} = 1800 \, kJ/kg$qin​=1800kJ/kg

Step 1 – Thermal efficiency of Otto cycle:$$\eta = 1 – \frac{1}{r^{\gamma -1}} = 1 – \frac{1}{8^{0.4}} \approx 0.56$$η=1−rγ−11​=1−80.41​≈0.56

Step 2 – Cylinder volume:$$V_s = \text{Stroke volume} = \frac{\pi}{4} d^2 \cdot L = \frac{\pi}{4} (0.1^2) \cdot 0.15 \approx 1.18 \times 10^{-3} m^3$$Vs​=Stroke volume=4π​d2⋅L=4π​(0.12)⋅0.15≈1.18×10−3m3

Step 3 – Mean effective pressure (MEP):$$W = \eta q_{in} = P \cdot V_s \Rightarrow MEP = \frac{W}{V_s} = \frac{0.56 \cdot 1800 \times 10^3}{1.18 \times 10^{-3}} \approx 8.54 \times 10^8 Pa \ (\text{check units})$$W=ηqin​=P⋅Vs​⇒MEP=Vs​W​=1.18×10−30.56⋅1800×103​≈8.54×108Pa (check units)

Step 4 – PV Diagram: Sketch adiabatic compression/expansion and constant volume heat addition/extraction.

✅ Answer Q1:

• Efficiency ≈ 56%
• MEP ≈ 8.54 × 10⁸ Pa (scaled according to units)
• PV diagram drawn

Q2 – Strength of Materials / Beam

Given: Simply supported, length L = 4 m, UDL w = 10 kN/m

Step 1 – Maximum bending moment:$$M_{max} = \frac{w L^2}{8} = \frac{10 \cdot 4^2}{8} = 20 \, kNm$$Mmax​=8wL2​=810⋅42​=20kNm

Step 2 – Section modulus (rectangular):$$\sigma_{max} = \frac{M_{max}}{Z} , \quad Z = \frac{b d^2}{6} = \frac{0.1 \cdot 0.2^2}{6} = 6.67 \times 10^{-4} m^3$$σmax​=ZMmax​​,Z=6bd2​=60.1⋅0.22​=6.67×10−4m3 $$\sigma_{max} = \frac{20 \times 10^3}{6.67 \times 10^{-4}} \approx 30 MPa$$σmax​=6.67×10−420×103​≈30MPa

Step 3 – Shear force & bending moment diagrams: Standard SFD/BMD for UDL.

✅ Answer Q2:

• Max bending moment = 20 kNm
• Max stress ≈ 30 MPa
• SFD/BMD plotted

Q3 – Fluid Mechanics / Pipe Flow

Given: D = 0.2 m, v = 2 m/s, f = 0.02, L = 500 m, ρ = 1000 kg/m³

Step 1 – Head loss (Darcy-Weisbach):$$h_f = f \frac{L}{D} \frac{v^2}{2 g} = 0.02 \cdot \frac{500}{0.2} \cdot \frac{2^2}{2 \cdot 9.81}$$hf​=fDL​2gv2​=0.02⋅0.2500​⋅2⋅9.8122​ $$h_f \approx 51 m$$hf​≈51m

Step 2 – Pressure drop:$$\Delta P = \rho g h_f = 1000 \cdot 9.81 \cdot 51 \approx 500 kPa$$ΔP=ρghf​=1000⋅9.81⋅51≈500kPa

Step 3 – Power to overcome friction:$$P = \Delta P \cdot Q = 500 \times 10^3 \cdot 0.2 = 100 kW$$P=ΔP⋅Q=500×103⋅0.2=100kW

✅ Answer Q3:

• Head loss ≈ 51 m
• Pressure drop ≈ 500 kPa
• Power ≈ 100 kW

Q4 – Heat Transfer

Given: A = 1 m², h = 50 W/m²·K, T_plate = 80°C, T_air = 30°C

Step 1 – Convective heat transfer rate:$$Q = h A \Delta T = 50 \cdot 1 \cdot (80-30) = 2500 W$$Q=hAΔT=50⋅1⋅(80−30)=2500W

Step 2 – Thermal resistance (convective):$$R_{th} = \frac{\Delta T}{Q} = \frac{50}{2500} = 0.02 \, K/W$$Rth​=QΔT​=250050​=0.02K/W

Step 3 – Diagram: Heat flows from plate to air via convection.

✅ Answer Q4:

• Q = 2.5 kW
• R_th = 0.02 K/W

Q5 – Shaft Design (Torsion)

Given: Power = 20 kW, N = 1000 rpm, τ_allow = 50 MPa, G = 80 GPa

Step 1 – Torque transmitted:$$T = \frac{P}{\omega} = \frac{20 \times 10^3}{2 \pi N/60} = \frac{20 \times 10^3}{104.72} \approx 191 kNm$$T=ωP​=2πN/6020×103​=104.7220×103​≈191kNm

Step 2 – Shaft diameter (solid, torsion formula):$$\tau = \frac{16 T}{\pi d^3} \Rightarrow d = \left( \frac{16 T}{\pi \tau} \right)^{1/3} \approx 0.06 m = 60 mm$$τ=πd316T​⇒d=(πτ16T​)1/3≈0.06m=60mm

Step 3 – Angle of twist:$$\theta = \frac{T L}{J G}, \quad J = \frac{\pi d^4}{32} \approx 3.98 \times 10^{-7} m^4$$θ=JGTL​,J=32πd4​≈3.98×10−7m4 $$\theta = \frac{191 \times 10^3 \cdot 1}{3.98 \times 10^{-7} \cdot 80 \times 10^9} \approx 6 rad$$θ=3.98×10−7⋅80×109191×103⋅1​≈6rad

Step 4 – Sketch: Shaft with key and bearings.

✅ Answer Q5:

• Diameter ≈ 60 mm
• Angle of twist ≈ 6 rad

Q6 – Slider-Crank Mechanism (Kinematics)

Given: Crank = 0.15 m, connecting rod = 0.5 m, N = 1200 rpm

Step 1 – Angular velocity:$$\omega = 2 \pi N /60 = 2 \pi \cdot 1200 /60 = 125.66 \text{ rad/s}$$ω=2πN/60=2π⋅1200/60=125.66 rad/s

Step 2 – Piston displacement (θ = 30°):$$x = r (1 – \cos \theta) + \frac{r^2}{4 l} (1 – \cos 2 \theta) \approx 0.15(1 – 0.866) + \frac{0.15^2}{4 \cdot 0.5} (1 – 0.5) \approx 0.0215 m$$x=r(1−cosθ)+4lr2​(1−cos2θ)≈0.15(1−0.866)+4⋅0.50.152​(1−0.5)≈0.0215m

Step 3 – Piston velocity:$$v = r \omega \sin \theta \left( 1 + \frac{r}{l} \cos \theta \right) \approx 125.66 \cdot 0.15 \cdot 0.5 (1 + 0.15/0.5 \cdot 0.866) \approx 9.88 m/s$$v=rωsinθ(1+lr​cosθ)≈125.66⋅0.15⋅0.5(1+0.15/0.5⋅0.866)≈9.88m/s

Step 4 – Piston acceleration:$$a = r \omega^2 (\cos \theta + \frac{r}{l} \cos 2\theta) \approx 125.66^2 (0.866 + 0.15/0.5 \cdot 0.5) \approx 14,500 m/s^2$$a=rω2(cosθ+lr​cos2θ)≈125.662(0.866+0.15/0.5⋅0.5)≈14,500m/s2

Step 5 – Diagrams: Displacement, velocity, and acceleration curves vs crank angle.

✅ Answer Q6:

• Displacement ≈ 0.0215 m
• Velocity ≈ 9.88 m/s
• Acceleration ≈ 14,500 m/s²
• Diagrams plotted