MEGR 3171 — Flow & Displacement Measurement

Learning Objectives

Apply the Bernoulli equation to derive flow from differential pressure across a restriction.
Compare operating principles of turbine, vortex, electromagnetic, Coriolis, and ultrasonic flow meters.
Decode quadrature encoder signals to determine position and velocity.
Explain LVDT operating principle and advantages for linear displacement measurement.
Select an appropriate position sensor for a given accuracy, range, and environment requirement.

1. Differential Pressure Flow Meters

Applying a restriction (orifice, venturi, nozzle) in a pipe creates a pressure drop proportional to the square of the flow velocity. Combining Bernoulli's equation with continuity gives the volumetric flow rate.

Orifice / Venturi Flow Equation

Q = C_d · A_2 · √(2·ΔP / (ρ · (1 - (A_2/A_1)²)))

where:
  Q   = volumetric flow rate (m³/s)
  C_d = discharge coefficient (0.6 to 0.99)
  A_2 = throat/orifice area (m²)
  A_1 = pipe area (m²)
  ΔP = differential pressure (Pa)
  ρ   = fluid density (kg/m³)

Nonlinear Relationship Flow rate is proportional to the square root of ΔP. This makes differential pressure meters inherently nonlinear. A 2:1 flow turndown corresponds to a 4:1 pressure range, requiring high-accuracy low-pressure sensors at low flows.

Other Flow Meter Types

Turbine Flow Meter

Rotating blades; frequency proportional to flow velocity. High accuracy, good rangeability. Moving parts limit life in dirty flows.

Vortex Shedding

Vortices shed from a bluff body at frequency ∝ velocity (Strouhal number). No moving parts. Good for gas and steam.

Electromagnetic

Faraday induction: V ∝ v for conductive fluids. No moving parts, no pressure drop, excellent for slurries. Requires conductive fluid.

Coriolis

Measures mass flow directly via Coriolis-induced tube twist. Most accurate technology (<0.1%). High cost. Also measures density.

Ultrasonic (Transit-Time)

Speed of sound difference upstream vs. downstream ∝ flow velocity. Non-invasive clamp-on version possible. Requires clean, bubble-free fluid.

2. Rotary Encoders

Incremental encoders output pulses per revolution (PPR) on two quadrature channels (A and B, 90° apart). Quadrature decoding provides 4× resolution and direction sensing.

Encoder Calculations

Angular resolution (quadrature): θ_res = 360° / (4 × PPR)

Velocity (from pulse period T_p):
  ω = 2π / (4 × PPR × T_p)   [rad/s]

Absolute encoder: Gray code output; no initialization required.
  Multi-turn: extends range with additional revolution counter.

Direction: A leads B → clockwise; B leads A → counterclockwise. The index pulse (Z channel) provides one pulse per revolution for absolute position reference.

3. Linear Variable Differential Transformer (LVDT)

An LVDT uses electromagnetic induction to convert linear displacement to an AC voltage. A movable ferromagnetic core inside a coil assembly (one primary, two secondary coils) produces a differential output proportional to core position.

Truly frictionless and contactless — excellent for high-cycle fatigue testing
Operates under extreme temperature, pressure, and radiation environments
Requires AC excitation (typically 1-10 kHz) and a phase-sensitive demodulator to recover the DC position signal
Null position (zero output) at mechanical center; polarity indicates direction

Practice Problems

Problem 1 — Encoder Resolution A 1000 PPR incremental encoder is used with quadrature decoding on a motor shaft. What is the angular resolution in degrees? If the motor runs at 1200 RPM, what is the pulse frequency on each channel?

Angular resolution = 360° / (4 × 1000) = 360 / 4000 = 0.09°

Pulse frequency per channel = 1000 PPR × 1200 RPM / 60 s/min = 1000 × 20 = 20,000 Hz = 20 kHz

Resolution = 0.09°; Pulse frequency = 20 kHz per channel

Problem 2 — Orifice Meter Water (ρ = 1000 kg/m³) flows through an orifice plate. The pipe diameter is 50 mm (A₁ = 1.963×10&sup-₃ m²), the orifice diameter is 25 mm (A₂ = 4.909×10&sup-⁴ m²), C_d = 0.61, and ΔP = 5000 Pa. Calculate Q.

(A_2/A_1)² = (4.909×10&sup-⁴ / 1.963×10&sup-₃)² = (0.25)² = 0.0625

Q = 0.61 × 4.909×10&sup-⁴ × √(2 × 5000 / (1000 × (1 - 0.0625)))

= 0.61 × 4.909×10&sup-⁴ × √(10000 / 937.5) = 0.61 × 4.909×10&sup-⁴ × √10.667

= 0.61 × 4.909×10&sup-⁴ × 3.266 = 9.77×10&sup-⁴ m³/s

Q ≈ 9.77 × 10&sup-⁴ m³/s ≈ 0.977 L/s