Engineering graduate exams are no walk in the park. As students move into advanced coursework—especially in fields like Thermodynamics and Control Systems—they encounter increasingly complex problems that test not only conceptual understanding but also mathematical rigor and real-world application. For students under time pressure or juggling responsibilities, it’s no surprise that many search online wondering, “Can I hire someone to take my engineering exam?” At LiveExamHelper.com, we provide not only top-tier academic assistance but also expertly solved sample questions to reinforce your understanding of these challenging subjects.

In this post, we showcase two real sample questions—one from Thermodynamics and one from Control Systems—solved by our subject-matter experts. These examples reflect the depth and structure you can expect in a master’s level engineering exam.

Thermodynamics: Second Law Analysis in a Combined Cycle

Question:

A combined Brayton-Rankine cycle power plant is proposed with the following parameters:

• The Brayton cycle operates with air, where the compressor pressure ratio is 10:1, with inlet at 300 K and outlet pressure of 1000 kPa. The turbine inlet temperature is 1400 K.

• The Rankine cycle uses water as the working fluid. Steam is generated using the exhaust gases from the Brayton turbine, which exit at 700 K.

• Assume isentropic efficiencies of 85% for the Brayton compressor and turbine.

• The heat exchanger is 100% efficient for heat transfer between the Brayton exhaust and Rankine boiler.

Determine the total thermal efficiency of the combined cycle and quantify the entropy generation per unit mass of air due to irreversibilities.

Solution:

Step 1: Brayton Cycle Analysis

Using air-standard assumptions and assuming constant specific heats at reference conditions:

Compressor work (ideal):
Wc_ideal = Cp * (T2 - T1)
Where:
T1 = 300 K
Pressure ratio (rp) = 10
Using isentropic relations: T2s = T1 * rp^((γ-1)/γ)

For air (γ ≈ 1.4), T2s = 300 * 10^((0.4)/1.4) ≈ 300 * 2.5119 ≈ 753.6 K

Actual T2 = T1 + (T2s - T1)/η_c = 300 + (753.6 - 300)/0.85 ≈ 826.6 K

Compressor work:
Wc = Cp * (T2 - T1) = 1.005 * (826.6 - 300) ≈ 530.5 kJ/kg

Turbine inlet T3 = 1400 K
Isentropic exit temperature T4s = T3 * (1/rp)^((γ-1)/γ) ≈ 1400 * (1/10)^0.2857 ≈ 557.4 K

Actual T4 = T3 - η_t * (T3 - T4s) ≈ 1400 - 0.85*(1400 - 557.4) ≈ 627.1 K

Turbine work:
Wt = Cp * (T3 - T4) ≈ 1.005 * (1400 - 627.1) ≈ 777.5 kJ/kg

Net Brayton work = Wt - Wc ≈ 777.5 - 530.5 = 247 kJ/kg

Heat input to Brayton cycle:
Qin = Cp * (T3 - T2) = 1.005 * (1400 - 826.6) ≈ 577.4 kJ/kg

Brayton thermal efficiency = Net Work / Heat In = 247 / 577.4 ≈ 42.8%

Step 2: Rankine Cycle Input

The Brayton exhaust heat = Cp * (T4 - T_exhaust) = 1.005 * (700 - 627.1) ≈ 73.1 kJ/kg

Assuming 100% heat recovery to generate steam, this heat input becomes Rankine’s boiler heat.

Assuming Rankine cycle has ideal efficiency of 30% (approximate for subcritical cycle):

Rankine output = 0.3 * 73.1 ≈ 21.93 kJ/kg

Total net work = 247 + 21.93 = 268.93 kJ/kg

Combined heat input = 577.4 kJ/kg

Combined cycle efficiency = 268.93 / 577.4 ≈ 46.6%

Entropy generation:

ΔS_gen = (Qin / T_source) - (Qout / T_sink)
Assuming T_source = 1400 K and T_sink = 300 K, use:
S_gen ≈ Qin*(1/T_sink - 1/T_source) = 577.4*(1/300 - 1/1400) ≈ 1.66 kJ/kg·K

This quantifies inefficiencies due to finite temperature differences and component inefficiencies.

Control Systems: State-Space Feedback Control

Question:

A multivariable dynamic system is represented by the following state-space model:

dx/dt = Ax + Bu
y = Cx

Where:
A = [[0, 1], [-6, -5]],
B = [[0], [1]],
C = [1, 0]

Design a state feedback controller u = -Kx such that the closed-loop system has eigenvalues at -2 and -3. Also, determine the steady-state value of y when the system is subjected to a unit step reference using state feedback with integral control.

Solution:

Step 1: Determine desired characteristic polynomial

Desired poles: λ = -2 and -3 → Characteristic polynomial:
(s + 2)(s + 3) = s² + 5s + 6

Step 2: Use Ackermann’s formula for state feedback

A = [[0, 1], [-6, -5]]
B = [[0], [1]]

Controllability matrix:
? = [B, AB] = [[0, 1], [1, -5]]

Confirm det(?) ≠ 0 → System is controllable

Ackermann’s formula:
K = [0, 1] * ?⁻¹ * (A² + 5A + 6I)

First compute A²:
A² = A*A = [[0, 1], [-6, -5]] * [[0, 1], [-6, -5]]
= [[-6, -5], [30, 21]]

Now compute:
A² + 5A + 6I = [[-6, -5], [30, 21]] + 5*[[0, 1], [-6, -5]] + 6*I
= [[-6, -5], [30, 21]] + [[0, 5], [-30, -25]] + [[6, 0], [0, 6]]
= [[0, 0], [0, 2]]

Now K = [0, 1] * ?⁻¹ * [[0, 0], [0, 2]]

?⁻¹ = (1/det) * adjugate
? = [[0, 1], [1, -5]] → det = -1
?⁻¹ = [[5, 1], [1, 0]]

So:
K = [0, 1] * [[5, 1], [1, 0]] * [[0, 0], [0, 2]]
First compute product of matrices:
[[5, 1], [1, 0]] * [[0, 0], [0, 2]] = [[0, 2], [0, 0]]
Then multiply with [0, 1]:
K = [0, 1] * [[0, 2], [0, 0]] = [0, 0]

This suggests a misstep in handling matrix operations. Instead, directly solve by comparing:

A - BK = [[0, 1], [-6 - k1, -5 - k2]]

Characteristic polynomial = s² + (5 + k2)s + (6 + k1)

Set equal to desired: s² + 5s + 6
Then:
5 + k2 = 5 → k2 = 0
6 + k1 = 6 → k1 = 0

Contradiction arises → Suggests A already has desired poles.

So no feedback is needed (or the system is already in desired state), but this is often rare in design. Assuming miscalculation in system matrix, proceed with augmentation for integral control.

Step 3: Augment with integral control

Define ei = r - y = r - Cx
Add new state: xi_dot = ei = r - Cx
Augmented system:

X_aug = [