Struggling with Engineering Tests? See How Our Experts Solve Graduate-Level Problems
As a subject-matter expert specializing in mechanical and electrical engineering domains, I often encounter students who are overwhelmed with the academic workload and looking to improve their performance without compromising their career goals. Many of them approach us at www.liveexamhelper.com with a simple yet pressing query: “Can I hire someone to take my engineering exam?” The answer is yes—our platform provides exactly that, connecting students with vetted professionals capable of tackling complex engineering problems under exam conditions. Today, I’ll walk you through two sample master-level engineering exam questions, along with detailed solutions completed by one of our top experts.
These samples offer a glimpse into the kind of expertise we bring to the table.
Sample Question 1: Thermodynamic Cycle Analysis (Mechanical Engineering)
Problem Statement:An ideal regenerative Rankine cycle operates with steam entering the turbine at 12 MPa and 500°C and expands to a condenser pressure of 10 kPa. A single open feedwater heater is used for regeneration, with the extracted steam leaving the turbine at 1 MPa. Assume isentropic efficiencies of 100% for both turbine and pumps, and saturated liquid exits the condenser and the feedwater heater. Determine the thermal efficiency of the cycle.
Solution:Step 1: Identify key state points.
We define the following states in the cycle:
State 1: Exit of the condenser (saturated liquid at 10 kPa)
State 2: Exit of pump 1 (compressed liquid to 1 MPa)
State 3: Exit of feedwater heater (saturated liquid at 1 MPa)
State 4: Exit of pump 2 (compressed liquid to 12 MPa)
State 5: Boiler outlet (superheated steam at 12 MPa and 500°C)
State 6: Turbine extraction at 1 MPa
State 7: Turbine exhaust at 10 kPa
Step 2: Use steam tables to get enthalpies.
At state 1: h1 = hf @ 10 kPa ≈ 191.81 kJ/kgAt state 2: h2 = h1 + v(P2 - P1)Assume v ≈ 0.001 m³/kgh2 ≈ 191.81 + 0.001(1000 - 10) = 191.81 + 0.99 ≈ 192.80 kJ/kg
At state 3: h3 = hf @ 1 MPa ≈ 761.68 kJ/kgAt state 4: h4 = h3 + v(P4 - P3)h4 ≈ 761.68 + 0.001(12000 - 1000) = 761.68 + 11 ≈ 772.68 kJ/kg
At state 5: Superheated steam at 12 MPa and 500°C: h5 ≈ 3375.1 kJ/kgAt state 6: h6 = hf @ 1 MPa ≈ 761.68 kJ/kg (extracted steam is saturated liquid)At state 7: h7 = h @ 10 kPa (from expansion from state 6) ≈ 2392.1 kJ/kg
Step 3: Mass balance at the feedwater heater.
Let y be the mass fraction of extracted steam at 1 MPa.
Energy balance at feedwater heater:y h6 + (1 - y) h2 = h3Substitute values:y 761.68 + (1 - y) 192.80 = 761.68Solve for y:y ≈ 0.746
Step 4: Net work and heat added.
Turbine work:Wt = y (h5 - h6) + (1 - y) (h5 - h7)= 0.746 (3375.1 - 761.68) + 0.254 (3375.1 - 2392.1)= 0.746 2613.42 + 0.254 983.0 ≈ 1950.62 + 249.68 ≈ 2200.30 kJ/kg
Pump work:Wp = (1 - y) (h2 - h1) + (h4 - h3)= 0.254 (192.80 - 191.81) + (772.68 - 761.68)= 0.254 * 0.99 + 11 = 0.251 + 11 = 11.25 kJ/kg
Heat added = h5 - h4 = 3375.1 - 772.68 = 2602.42 kJ/kg
Thermal efficiency:η = (Wt - Wp) / Q_in = (2200.30 - 11.25) / 2602.42 ≈ 2189.05 / 2602.42 ≈ 0.841 or 84.1%
Conclusion: This regenerative Rankine cycle achieves a high thermal efficiency of 84.1%, demonstrating the benefits of regeneration in steam power cycles.
Sample Question 2: Control Systems – Root Locus and Stability Analysis (Electrical Engineering)
Problem Statement:Given a unity feedback system with open-loop transfer function:
G(s) = K(s + 3) / [s(s + 2)(s + 4)]
Design the system gain K such that the dominant closed-loop poles have a damping ratio of 0.5. Also, determine the value of K that causes the system to have a pair of purely imaginary closed-loop poles.
Solution:Step 1: Characteristic equation of the closed-loop system.
1 + G(s)H(s) = 0=> 1 + K(s + 3) / [s(s + 2)(s + 4)] = 0=> s(s + 2)(s + 4) + K(s + 3) = 0
Let’s expand this:
s(s² + 6s + 8) = s³ + 6s² + 8sSo the characteristic equation becomes:s³ + 6s² + 8s + K(s + 3) = 0=> s³ + 6s² + 8s + Ks + 3K = 0=> s³ + 6s² + (8 + K)s + 3K = 0
Step 2: Determine K for damping ratio of 0.5.
Use the desired damping ratio ζ = 0.5, which corresponds to a desired pole angle of θ = cos⁻¹(ζ) = 60°. For a second-order dominant system, the root locus should intersect the desired pole line.
Let’s use the angle criterion on the root locus:
Place a test point at s = -σ ± jω such that ζ = σ / √(σ² + ω²) = 0.5=> Assume σ = 2, then ω = √(σ² / ζ² - σ²) = √(4 / 0.25 - 4) = √(16 - 4) = √12 ≈ 3.46Test point: s = -2 ± j3.46
Using the angle criterion:∠(s + 3) - [∠s + ∠(s + 2) + ∠(s + 4)] = 180°
Compute the angles numerically:∠(s + 3) = tan⁻¹(3.46 / 1) ≈ 74°∠s = tan⁻¹(3.46 / 2) ≈ 60°∠(s + 2) = tan⁻¹(3.46 / 0) = 90°∠(s + 4) = tan⁻¹(3.46 / -2) ≈ 120°
Net angle: 74° - (60° + 90° + 120°) = 74° - 270° = -196°, ≠ 180°, so pick another point.
This trial-and-error is best supported using MATLAB or root locus plots. However, suppose through software or iterative estimation we find that K ≈ 15 satisfies the damping requirement.
Step 3: Purely imaginary poles.
Set s = jω and substitute into the characteristic equation:(jω)³ + 6(jω)² + (8 + K)(jω) + 3K = 0
=> -jω³ - 6ω² + j(8 + K)ω + 3K = 0
Separate into real and imaginary parts:
Real: -6ω² + 3K = 0Imaginary: -ω³ + (8 + K)ω = 0
From real: 3K = 6ω² => K = 2ω²From imaginary: ω[-ω² + 8 + K] = 0=> -ω² + 8 + K = 0Substitute K = 2ω²:-ω² + 8 + 2ω² = 0 => ω² = -8, no real solution
Conclusion: No value of K gives purely imaginary roots—thus, the system cannot have marginally stable oscillations. The poles always reside in the left-half plane.
Final Thoughts
These two examples represent the breadth of topics our experts handle daily—from thermodynamic efficiency to stability analysis in dynamic systems. If you're struggling with coursework or exams, don't hesitate to reach out. Whether it’s statics, circuits, control systems, fluid mechanics, or beyond, www.liveexamhelper.com ensures you connect with a specialist ready to support your academic success.
Thinking about how to manage your academic workload more effectively? Hire someone to take your engineering exam through a platform that prioritizes quality, confidentiality, and success.