Introduction & Context

This sheet compares the filtration time required to obtain a target filtrate volume under vacuum and pressure driving forces. The calculation is based on the classical constant-pressure cake-filtration model (also known as the Ruth equation). It is a first-pass screening tool used by process engineers to decide whether a vacuum filter (≈ 80 kPa) or a pressure filter (≈ 5 bar) is the faster, more economical, or mechanically feasible option for a given slurry.

Typical applications include clarification of fermentation broths, removal of precipitated solids from crystallisers, dewatering of mineral slurries, and polishing of edible oils. The result is the total cycle time required to reach the specified filtrate volume; this time can be directly compared with the plant’s batch schedule or continuous-filter throughput target.

Methodology & Formulas

Convert practical inputs to SI units
\[ \mu = \mu_{\text{cP}} \times 0.001 \quad [\text{Pa·s}] \] \[ \Delta P_{\text{vac}} = \Delta P_{\text{vac,kPa}} \times 1000 \quad [\text{Pa}] \] \[ \Delta P_{\text{press}} = \Delta P_{\text{press,kPa}} \times 1000 \quad [\text{Pa}] \]
Vacuum filtration time
The Ruth equation integrated at constant pressure gives: \[ t_{\text{vac,cake}} = \frac{\mu\, r_{\text{cake}}\, c\, V^{2}}{2\, A^{2}\, \Delta P_{\text{vac}}} \] \[ t_{\text{vac,medium}} = \frac{\mu\, R_{\text{m}}\, V}{A\, \Delta P_{\text{vac}}} \] \[ t_{\text{vac,total}} = t_{\text{vac,cake}} + t_{\text{vac,medium}} \]
Pressure filtration time
Because the same cake and medium resistances apply, the time scales inversely with the applied pressure: \[ t_{\text{press,total}} = t_{\text{vac,total}} \times \frac{\Delta P_{\text{vac}}}{\Delta P_{\text{press}}} \]

Empirical validity limits for the Ruth correlation
Parameter	Lower bound	Upper bound	Units
Slurry concentration, c	5	500	kg m⁻³
Viscosity, μ	0.5	100	cP
Pressure drop, ΔP	—	1	MPa

Vacuum filtration is preferred when:

The slurry is relatively dilute and the solids load is low to moderate.
The cake is compressible or fragile and low differential pressure minimizes damage.
Lower capital and operating costs are a priority.
Batch operation is acceptable and the solvent has a high vapor pressure at the operating temperature.

Start with a laboratory pressure-to-filtrate test:

Incrementally increase pressure in 0.2 bar steps while recording filtrate volume vs. time.
Stop when cake moisture stops improving or solids begin extruding through the medium.
Use the plateau pressure as the design ceiling; scale up with a 0.3 bar safety margin.

Only after a mechanical review:

Verify the vessel is rated for positive pressure per ASME VIII or local code.
Replace the vacuum pump with a clean-air compressor and add a pressure-relief valve set ≤ the vessel MAWP.
Upgrade the shaft seal from a simple lip seal to a double mechanical seal rated for the new pressure.
Re-rate the filter medium support plate for the higher compressive loads.

Pressure filtration usually wins:

Higher driving force pushes wash liquor through the cake faster, reducing wash time by 30–70%.
Uniform pressure distribution minimizes channeling, giving more efficient displacement.
Counter-current washing schemes are easier to implement in enclosed pressure filters.

However, if the cake is highly compressible, vacuum may still be faster because the cake remains more permeable.

Worked Example – Choosing Between Vacuum and Pressure Filtration

A specialty-chemical plant must clarify 3 m³ of an aqueous suspension containing 20 kg m⁻³ of inert solids. The suspension viscosity is 1 cP, the specific cake resistance is 1×10¹¹ m kg⁻¹, and the medium resistance is 1×10¹⁰ m⁻¹. A 5 m² filter is available for one 1-hour (3600 s) batch. Management wants to know whether an 80 kPa vacuum pump or a 5 bar (500 kPa) pressure system will finish the job faster.

Knowns

Batch volume, V = 3 m³
Filter area, A = 5 m²
Solids concentration, c = 20 kg m⁻³
Liquid viscosity, μ = 1 cP = 0.001 Pa·s
Specific cake resistance, r = 1×10¹¹ m kg⁻¹
Medium resistance, R_m = 1×10¹⁰ m⁻¹
Target batch time, t_target = 3600 s
Vacuum option: ΔP_vac = 80 kPa = 80,000 Pa
Pressure option: ΔP_press = 5 bar = 500,000 Pa

Step-by-Step Calculation

Convert the constant-rate/constant-pressure equations into a single quadratic form for time. For incompressible cake, the governing relation is: \[ \frac{t}{V} = \frac{\mu r c}{2 A^2 \Delta P} \frac{V}{A} + \frac{\mu R_m}{A \Delta P} \]
Solve for the total filtration time t for each driving force. Rearranging gives: \[ t = \frac{\mu r c}{2 A^2 \Delta P} V^2 + \frac{\mu R_m}{A \Delta P} V \]
Vacuum case (ΔP = 80,000 Pa):
- Cake term: 0.001 × 1×10¹¹ × 20 / (2 × 5² × 80,000) × 3² = 4500 s
- Medium term: 0.001 × 1×10¹⁰ / (5 × 80,000) × 3 = 75 s
- t_vac,total = 4500 + 75 = 4575 s
Pressure case (ΔP = 500,000 Pa):
- Cake term: 0.001 × 1×10¹¹ × 20 / (2 × 5² × 500,000) × 3² = 720 s
- Medium term: 0.001 × 1×10¹⁰ / (5 × 500,000) × 3 = 12 s
- t_press,total = 720 + 12 = 732 s
Compare with the 3600 s window. Vacuum needs 4575 s (27% overtime), whereas pressure needs only 732 s (80% under-time).

Final Answer

Pressure filtration at 5 bar completes the 3 m³ batch in 732 s, well within the 1-hour target, while vacuum filtration at 80 kPa requires 4575 s and misses the schedule. Select the 5 bar pressure system for this duty.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle