Variable area flow meters and rotameters

Why it matters

The rotameter is the flow meter you see on almost every purge line, analyser sample system, dosing skid, gas cylinder panel and laboratory column. It is cheap, needs no power and shows flow at a glance. It is also a neat example of a constant pressure drop, variable area meter, the opposite of the orifice plate (constant area, variable pressure drop), and GATE likes to test that contrast and the fluid-density correction.

Key ideas

Construction. A vertical tube, tapered so that it is narrowest at the bottom, carries a float (bob) that is denser than the fluid. Flow enters at the bottom. The flow passes through the annulus between the float's largest diameter (the metering edge) and the tube wall. Glass or plastic tubes allow direct reading; armoured metal tubes with a magnetic follower are used for high pressure, high temperature or opaque fluids and can carry a transmitter.

Force balance. At a steady float position:

• downward: float weight Vf·ρf·g;
• upward: buoyancy Vf·ρ·g plus the pressure-drop (drag) force ΔP·Af. So ΔP = Vf·(ρf − ρ)·g / Af, which does not depend on flow. When flow increases, the float rises to a wider part of the tube, increasing the annular area until the pressure drop is back to this same value.

Flow equation. The annulus acts like an orifice of area Aa: Q = Cd·Aa·√(2·ΔP/ρ). Since ΔP is constant, Q ∝ Aa. For a gentle linear taper, Aa grows almost linearly with float height, so the scale is nearly linear — a big advantage over the square-root scale of DP meters. Typical turndown is about 10:1.

Viscosity. Cd depends on Reynolds number. Floats with sharp metering edges are designed to make Cd nearly constant over a wide viscosity range; even so, a rotameter calibrated on water should not be used on a viscous oil without a fresh calibration.

Fluid density and temperature. The calibration depends on ρ through both ΔP (via ρf − ρ) and √(1/ρ). For a fluid other than the calibration fluid, readings must be corrected. Gas rotameters are calibrated at stated pressure and temperature; a change in either changes gas density and the reading.

Density-compensated design. Mass flow ṁ = ρQ ∝ √((ρf − ρ)·ρ). This is least sensitive to small changes of ρ when ρf = 2ρ, so a float of about twice the fluid density makes the meter nearly independent of small density changes in mass-flow terms.

Installation and limits. Must be vertical with flow upward; pulsating flow makes the float bounce; glass tubes are limited in pressure and temperature; accuracy is typically 1–5 % of full scale; the reading is taken at the float's metering edge as specified by the manufacturer.

Formulas

ΔP = Vf·(ρf − ρ)·g / Af (constant pressure drop)

Aa = (π/4)·(Dt² − Df²) (annular area at the float position)

Q = Cd·Aa·√(2·ΔP/ρ) = Cd·Aa·√(2·g·Vf·(ρf − ρ)/(Af·ρ))

Q₂/Q₁ = √[ (ρf − ρ₂)·ρ₁ / ((ρf − ρ₁)·ρ₂) ] (same float position, two fluids, volumetric)

ṁ₂/ṁ₁ = √[ (ρf − ρ₂)·ρ₂ / ((ρf − ρ₁)·ρ₁) ] (same float position, mass flow)

Symbols: ΔP = pressure drop across the float (Pa); Vf = float volume (m³); ρf = float density, ρ = fluid density (kg/m³); g = 9.81 m/s²; Af = float cross-section at the metering edge (m²); Dt = tube bore at the float position, Df = float diameter (m); Aa = annular flow area (m²); Q = volumetric flow (m³/s); Cd = discharge coefficient (from calibration). The correction formulas assume Cd is unchanged, i.e. similar viscosity.

Worked examples

Example 1 (standard): flow at a given float position. Given: stainless float Vf = 6.0 cm³, ρf = 8000 kg/m³, Df = 20 mm; water ρ = 1000 kg/m³; tube bore at the float = 24 mm; Cd = 0.70.

1. Af = π × 0.020²/4 = 3.142 × 10⁻⁴ m².
2. ΔP = 6.0 × 10⁻⁶ × (8000 − 1000) × 9.81 / 3.142 × 10⁻⁴ = 1311.5 Pa (the same at every float height).
3. Aa = (π/4) × (0.024² − 0.020²) = (π/4) × 1.76 × 10⁻⁴ = 1.382 × 10⁻⁴ m².
4. √(2ΔP/ρ) = √(2 × 1311.5/1000) = 1.620 m/s.
5. Q = 0.70 × 1.382 × 10⁻⁴ × 1.620 = 1.567 × 10⁻⁴ m³/s.
6. Answer: Q ≈ 1.57 × 10⁻⁴ m³/s ≈ 9.4 L/min.

Example 2 (GATE level): using a water-calibrated rotameter on another liquid. Given: rotameter calibrated on water (ρ₁ = 1000 kg/m³) with a stainless float ρf = 8000 kg/m³. It is used on a liquid of ρ₂ = 800 kg/m³ of similar viscosity and reads 10.0 L/min on the water scale. Find the actual volumetric and mass flow.

1. Same float position ⇒ same Aa, so Q₂/Q₁ = √[(ρf − ρ₂)·ρ₁ / ((ρf − ρ₁)·ρ₂)].
2. = √[(7200 × 1000)/(7000 × 800)] = √(7.2 × 10⁶ / 5.6 × 10⁶) = √1.2857 = 1.1339.
3. Q₂ = 10.0 × 1.1339 = 11.34 L/min.
4. ṁ₂ = 800 kg/m³ × 11.34 × 10⁻³ m³/min = 9.07 kg/min (water at the same reading would be 10.0 kg/min).
5. Answer: actual flow ≈ 11.3 L/min (≈ 9.07 kg/min). The lighter liquid gives less buoyancy and needs more velocity to hold the float, so the water scale reads low by about 12 %.

Example 3 (pressure drop). A float of 5 cm³, density 8000 kg/m³ and metering area 2 cm² in water: ΔP = 5 × 10⁻⁶ × 7000 × 9.81 / 2 × 10⁻⁴ = 1717 Pa, whatever the flow.

Common mistakes

• Using the tube bore area instead of the annular area (Dt² − Df²) in the flow equation.
• Thinking the pressure drop rises with flow, as in an orifice; in a rotameter it is constant.
• Forgetting buoyancy: the net weight uses (ρf − ρ), not ρf.
• Applying the density correction the wrong way round (inverting the ratio).
• Using a liquid-calibrated meter on gas, or a gas meter at a different pressure or temperature, without correction.
• Saying a rotameter scale is exactly linear; it is nearly linear only for a small taper and constant Cd.

For GATE IN

Expect conceptual questions on constant ΔP versus constant area meters, the forces on the float and why the scale is nearly linear. Numericals ask for ΔP across the float, flow through the annulus, or the corrected flow when the fluid density differs from the calibration fluid. Practise deriving the force balance yourself so that the (ρf − ρ) term and the ρ under the square root are never lost.

Quick check

1. Which quantity stays constant in a rotameter as flow changes?
2. Why is the tube tapered, and which end is narrow?
3. A rotameter's flow doubles. What happens to the annular area (constant Cd)?
4. What float density makes mass-flow reading least sensitive to small fluid-density changes?
5. Why must a rotameter be mounted vertically?

Answers: 1. The pressure drop across the float. 2. So the annular area increases as the float rises; the narrow end is at the bottom. 3. It doubles. 4. About twice the fluid density. 5. The float is balanced against gravity, so the force balance and calibration assume vertical upward flow.