Free PE CSE Practice Questions

Step 1: Recall the orifice flow equation

From Bernoulli's equation applied across the restriction:

Q = C_d × A × √(2ΔP / ρ)

Step 2: Identify the relationship

Since C_d, A, and ρ are constants for a given installation, Q ∝ √ΔP.

This is why DP flow meters require square-root extraction to linearize the output signal.

Why not the others? Q ∝ ΔP (linear) and Q ∝ ΔP² are incorrect relationships. Q is never independent of ΔP.

Step 1: Recall PID controller actions

P (Proportional) — Output proportional to current error. Cannot eliminate offset alone.

I (Integral) — Accumulates error over time. Drives steady-state error to zero.

D (Derivative) — Responds to rate of change of error. Anticipates future error.

Step 2: Answer

Integral action eliminates steady-state error (offset) by continuing to increase its output as long as any error persists.

Why not the others? Reducing oscillations is more associated with D action. Anticipating future error is D action. Increasing bandwidth is not a primary PID function.

Step 1: Identify the Seebeck effect

The Seebeck effect produces a voltage (EMF) when two dissimilar metals are joined and their junctions are at different temperatures.

Step 2: Match to sensor type

A thermocouple is the only sensor that generates its own voltage — no external excitation required.

Why not the others?

• RTD — Resistance changes with temperature (requires excitation current)

• Thermistor — Resistance changes with temperature (requires excitation)

• Infrared pyrometer — Measures emitted thermal radiation (non-contact), not based on Seebeck effect

Step 1: Determine the percent of span

Span = 0–100 psig. Reading = 50 psig → 50 / 100 = 50% of span.

Step 2: Apply the 4–20 mA linear scaling formula

I = 4 + (% span) × (20 − 4)

I = 4 + 0.50 × 16 = 4 + 8 = 12 mA

Quick check: 4 mA = 0%, 12 mA = 50%, 20 mA = 100%. ✓

Step 1: Understand the fail action

A fail-closed (FC) valve is also called air-to-open (ATO). Air pressure pushes the valve open against a spring.

Step 2: Determine failure position

On loss of instrument air, the spring returns the valve to its fully closed position.

Why not the others?

• Fully open = fail-open (FO) / air-to-close (ATC) valve

• Last position held = requires a lock-in-place mechanism (not standard)

• 50% open = not a standard failure mode

Step 1: Recall the hydrostatic pressure formula

P = ρ × g × h → h = P / (ρ × g)

Step 2: Calculate density from specific gravity

ρ = SG × ρ_water = 0.85 × 1,000 kg/m³ = 850 kg/m³

Step 3: Solve for height

h = 25,000 Pa / (850 kg/m³ × 9.81 m/s²)

h = 25,000 / 8,338.5 = 3.0 m

Step 1: Recall the standard first-order transfer function form

G(s) = K / (τs + 1)

Step 2: Compare with the given transfer function

G(s) = 5 / (10s + 1)

Matching terms: K (steady-state gain) = 5, τ (time constant) = 10 s.

Physical meaning: The system reaches 63.2% of its final value in 10 seconds and settles to a gain of 5 at steady state.

Step 1: Recall the SIL / PFD_avg table (IEC 61511)

• SIL 1: PFD_avg ≥ 10⁻² to < 10⁻¹

• SIL 2: PFD_avg ≥ 10⁻³ to < 10⁻²

• SIL 3: PFD_avg ≥ 10⁻⁴ to < 10⁻³

• SIL 4: PFD_avg ≥ 10⁻⁵ to < 10⁻⁴

Step 2: Answer

SIL 2 requires a PFD_avg of ≥ 10⁻³ to < 10⁻² (i.e., the SIF must have a risk reduction factor of 100–1,000).

Reference: IEC 61511, Table 3 — Safety Integrity Levels.

Step 1: Recall the liquid valve sizing equation

Q = C_v × √(ΔP / SG)

Step 2: Substitute values

Q = 50 × √(25 / 1.0) = 50 × √25 = 50 × 5 = 250 GPM

Reference: ISA/IEC 60534 — Control Valve Sizing, Liquid Flow.

Step 1: Construct the Routh array

Characteristic equation: s³ + 4s² + 5s + 2 = 0

Row s³: | 1 | 5 |

Row s²: | 4 | 2 |

Row s¹: | (4×5 − 1×2) / 4 = 18/4 = 4.5 | 0 |

Row s⁰: | 2 |

Step 2: Check first column for sign changes

First column: 1, 4, 4.5, 2 — all positive, no sign changes.

Conclusion: The system is stable. The number of sign changes equals the number of RHP roots (zero in this case).

Step 1: Recall the Wheatstone bridge output formula

V_out = V_exc × (R4/(R3 + R4) − R2/(R1 + R2))

Step 2: Substitute values

V_out = 10 × (120.5/(120 + 120.5) − 120/(120 + 120))

V_out = 10 × (120.5/240.5 − 120/240)

V_out = 10 × (0.50104 − 0.50000)

V_out = 10 × 0.00104 = 10.4 mV

Note: This small voltage is then amplified by an instrumentation amplifier for measurement.

Step 1: Recall the three required SIS elements (IEC 61511)

A Safety Instrumented System must include all three functional subsystems:

• A — Logic solver (safety PLC) ✅ — Processes inputs and determines trip action

• B — Sensor/input element ✅ — Detects the hazardous condition (e.g., pressure transmitter)

• C — Final control element ✅ — Takes the safe action (e.g., shutdown valve, trip relay)

Why not D and E?

• D — Historian database: Useful for diagnostics but NOT a required SIS element

• E — Operator HMI display: Important for operations but NOT part of the SIS safety function

Step 1: Understand cascade control architecture

Cascade control uses two controllers: a primary (outer/master) loop and a secondary (inner/slave) loop.

Step 2: Signal flow

Primary controller output → Setpoint of the secondary controller → Secondary controller output → Final control element

The secondary loop rejects disturbances faster because it has a shorter time constant.

Why not the others?

• B — The PV of the secondary comes from its own sensor, not the primary output

• C — The feedback to the primary is its own PV measurement

• D — The primary never directly drives the final element in cascade control

Step 1: Find the phase crossover frequency (ω_pc)

Set ∠G(jω) = −180°. For G(s) = 10/[(s+1)(s+2)(s+5)]:

∠G(jω) = −[arctan(ω/1) + arctan(ω/2) + arctan(ω/5)] = −180°

Solving numerically: ω_pc ≈ 3.32 rad/s.

Step 2: Calculate gain at ω_pc

|G(jω_pc)| = 10 / [√(1+3.32²) × √(4+3.32²) × √(25+3.32²)]

= 10 / [3.47 × 3.87 × 5.50] = 10 / 73.8 ≈ 0.1355

Step 3: Calculate gain margin

GM = −20 log₁₀(0.1355) = −20 × (−0.868) ≈ 17.4 dB ≈ 17 dB

Positive gain margin → system is stable with ~17 dB of margin.

Proportional (P) → Output proportional to current error magnitude. Provides immediate corrective action but leaves a steady-state offset.

Integral (I) → Accumulates error over time to eliminate steady-state offset. The longer the error persists, the larger the correction.

Derivative (D) → Responds to the rate of change of error, effectively anticipating future error. Provides damping but is sensitive to noise.

Feedforward (FF) → Measures disturbances directly and compensates before they affect the process variable. Unlike feedback, it acts proactively.