Constant-pressure and constant-rate filtration

Integrated filtration equations at constant pressure (t/V vs V plot) and constant rate (linear Δp–t), combined cycles, and extracting α and R_m from test data.

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Why it matters

Every filter runs either at (nearly) constant pressure — a vacuum filter or a pressure filter fed from a header — or at constant rate — a positive-displacement pump feeding a press — and many press cycles do both in sequence. Integrating the basic filtration equation for each mode gives the filtration time, the volume of filtrate per cycle, and a straight-line plot that turns laboratory data into the cake and medium resistances.

Key ideas

Starting point. From the previous topic, dt/dV = (μ/(A·Δp))·(α·c·V/A + R_m). Write this as dt/dV = K_p·V + 1/q₀, where K_p = μ·c·α/(A²·Δp) and 1/q₀ = μ·R_m/(A·Δp). Here q₀ is the initial filtrate rate, when there is no cake.

Constant-pressure filtration. Δp is fixed, so K_p and 1/q₀ are constants. Integrating from t = 0, V = 0 gives a quadratic: t = (K_p/2)·V² + V/q₀. Dividing by V: t/V = (K_p/2)·V + 1/q₀. A plot of t/V against V is a straight line with slope K_p/2 and intercept 1/q₀. From the slope you get α; from the intercept, R_m. The rate falls continuously as the cake grows. If R_m is negligible, t ∝ V²: collecting twice the filtrate takes four times as long.

Constant-rate filtration. The rate q = V/t is fixed, so V = q·t, and the pressure must rise as the cake thickens: Δp = (μ·α·c·q²/A²)·t + μ·R_m·q/A. For an incompressible cake, Δp rises linearly with time; the intercept is the pressure drop across the clean medium. The run ends when the pump or filter reaches its pressure limit.

Constant rate followed by constant pressure. A common practical cycle: start at constant rate (gentle, so fines do not blind the cloth and a good cake structure forms) until the pressure reaches its set value, then continue at that constant pressure. For the constant-pressure part, integrate dt/dV from V₁ (end of constant-rate period) to V₂: t − t₁ = (K_p/2)(V₂² − V₁²) + (V₂ − V₁)/q₀.

Compressible cakes. Use the α appropriate to the operating pressure (α = α₀Δp^s). Plots of t/V vs V at several pressures give α at each pressure and hence s.

Washing. After filtration, wash liquid follows the same path as the final filtrate at constant cake resistance, so the washing rate equals the final filtration rate (for the same pressure and viscosity) in filters where wash and filtrate take the same path; in plate-and-frame presses with through-washing the rate is lower. Details are in the equipment topic.

Formulas

dt/dV = K_p·V + 1/q₀ K_p = μ·c·α / (A²·Δp) (s/m⁶), 1/q₀ = μ·R_m / (A·Δp) (s/m³) Constant pressure: t = (K_p/2)·V² + V/q₀, t/V = (K_p/2)·V + 1/q₀ From the plot: α = 2·slope·A²·Δp / (μ·c), R_m = intercept·A·Δp / μ Constant rate q: Δp = (μ·α·c·q²/A²)·t + μ·R_m·q/A Constant pressure after a constant-rate period: t − t₁ = (K_p/2)·(V² − V₁²) + (V − V₁)/q₀

  • t: time (s); V: filtrate volume (m³); A: area (m²); Δp: pressure drop (Pa); μ: viscosity (Pa·s); c: dry cake per filtrate volume (kg/m³); α: specific cake resistance (m/kg); R_m: medium resistance (m⁻¹); q: constant filtrate rate (m³/s).

Worked examples

Example 1 (standard): resistances from a constant-pressure test. A leaf filter of area A = 0.05 m² at Δp = 100 kPa gives 0.5 L of filtrate in 18 s and 1.0 L in 56 s. μ = 10⁻³ Pa·s, c = 20 kg/m³. Find α, R_m and the time to collect 3 L.

  1. t/V: 18/0.5 = 36 s/L and 56/1.0 = 56 s/L.
  2. Slope K_p/2 = (56 − 36)/(1.0 − 0.5) = 40 s/L²; intercept 1/q₀ = 36 − 40 × 0.5 = 16 s/L.
  3. SI: K_p = 80 s/L² = 8 × 10⁷ s/m⁶; 1/q₀ = 16 s/L = 1.6 × 10⁴ s/m³.
  4. α = K_p·A²·Δp/(μ·c) = 8 × 10⁷ × 0.0025 × 10⁵/(10⁻³ × 20) = 1.0 × 10¹² m/kg.
  5. R_m = (1/q₀)·A·Δp/μ = 1.6 × 10⁴ × 0.05 × 10⁵/10⁻³ = 8.0 × 10¹⁰ m⁻¹.
  6. For V = 3 L: t = 40 × 3² + 16 × 3 = 408 s.

Example 2 (GATE level): constant rate then constant pressure. A press (A = 1 m²) filters a slurry with μ = 10⁻³ Pa·s, c = 50 kg/m³, α = 10¹¹ m/kg, R_m = 10¹⁰ m⁻¹. It is fed at a constant 2 × 10⁻⁴ m³/s until Δp reaches 200 kPa, then held at 200 kPa until a total of 0.5 m³ of filtrate is collected. Find the total time.

  1. Constant rate: Δp = (μαcq²/A²)t + μR_m·q/A = (10⁻³ × 10¹¹ × 50 × 4 × 10⁻⁸)t + 10⁻³ × 10¹⁰ × 2 × 10⁻⁴ = 200t + 2000 Pa.
  2. Δp = 2 × 10⁵ Pa at t₁ = (200 000 − 2000)/200 = 990 s; V₁ = 2 × 10⁻⁴ × 990 = 0.198 m³.
  3. Constant pressure: K_p = μcα/(A²Δp) = 10⁻³ × 50 × 10¹¹/(1 × 2 × 10⁵) = 2.5 × 10⁴ s/m⁶; 1/q₀ = μR_m/(AΔp) = 10⁻³ × 10¹⁰/(2 × 10⁵) = 50 s/m³.
  4. t − t₁ = 1.25 × 10⁴ × (0.5² − 0.198²) + 50 × (0.5 − 0.198) = 1.25 × 10⁴ × 0.2108 + 15.1 = 2635 + 15 = 2650 s.
  5. Total time = 990 + 2650 = 3640 s (about 61 min).

Common mistakes

  • Averaging initial and final rates to get the filtration time. The rate does not fall linearly; integrate the equation.
  • Forgetting the factor 2: the slope of t/V vs V is K_p/2, not K_p.
  • Starting the constant-pressure integration at V = 0 after a constant-rate period; integrate from V₁.
  • Mixing litres and m³ between slope and intercept.
  • Treating Δp at constant rate as constant; it rises linearly with time.

For GATE CH

Expect: finding α and R_m from two (t, V) points or a t/V plot; time for a new filtrate volume at constant pressure; effect of doubling the pressure or area on time; pressure–time behaviour at constant rate; combined constant-rate/constant-pressure cycles. Practise converting units of K_p and 1/q₀ cleanly.

Quick check

  1. What do you plot to get a straight line in constant-pressure filtration?
  2. With negligible medium resistance, how does filtration time change if the filtrate volume doubles at constant pressure?
  3. How does Δp vary with time at constant rate for an incompressible cake?
  4. If the filter area is doubled at constant pressure (R_m negligible), what happens to the time for a given V? Answers: 1. t/V against V. 2. It becomes four times longer. 3. Linearly. 4. It falls to one quarter, since K_p ∝ 1/A².

Interview questions

All Mechanical Operations interview questions

Try answering each one aloud before you open it.

  1. 1.What is constant-pressure filtration?Concept

    Constant-pressure filtration is a process where the pressure difference across the filter medium is kept constant throughout the filtration process. This is typically achieved by maintaining a constant pressure on the feed side or by using a vacuum on the filtrate side. The flow rate of the filtrate decreases over time as the filter cake builds up and offers more resistance.

  2. 2.What is constant-rate filtration?Concept

    Constant-rate filtration is a process where the rate of filtrate flow is kept constant throughout the filtration process. This is usually achieved by adjusting the pressure difference across the filter medium as the resistance increases due to cake formation. The pressure needs to be increased over time to maintain the constant flow rate.

  3. 3.Explain the main differences between constant-pressure and constant-rate filtration.Concept

    In constant-pressure filtration, the pressure difference across the filter medium is kept constant, leading to a decreasing flow rate as the cake builds up. In contrast, constant-rate filtration maintains a constant flow rate by increasing the pressure difference as the cake resistance increases. Constant-pressure filtration is simpler to operate, while constant-rate filtration requires more control to adjust the pressure.

  4. 4.Why is constant-pressure filtration often preferred in industrial applications?Application

    Constant-pressure filtration is often preferred because it is simpler to implement and control. It requires less sophisticated equipment since the pressure is maintained constant, usually by a pump or a vacuum system. This simplicity can lead to lower operational costs and easier maintenance compared to constant-rate filtration, which requires continuous adjustment of pressure.

  5. 5.What happens if the pressure is not maintained constant in a constant-pressure filtration process?Application

    If the pressure is not maintained constant in a constant-pressure filtration process, the flow rate of the filtrate will not follow the expected pattern. This can lead to inefficient filtration, with either too slow or too fast a flow rate, potentially affecting the quality of the filtrate and the efficiency of the process. It may also lead to uneven cake formation.

  6. 6.How does cake resistance affect the filtration process in constant-rate filtration?Application

    In constant-rate filtration, as the cake builds up on the filter medium, the resistance to flow increases. To maintain a constant flow rate, the pressure difference across the filter must be increased. This requires careful control and adjustment of the pressure to ensure that the desired flow rate is maintained throughout the process.

  7. 7.Where is constant-rate filtration used in practice?Application

    Constant-rate operation occurs whenever a filter is fed by a positive-displacement pump, for example a filter press fed by a diaphragm or piston pump, and in deep-bed and cartridge filtration where the flow is set by the process. It is often used for the first part of a press cycle: filtering gently at constant rate lets a good cake structure form and prevents fines from blinding the cloth. Once the pressure reaches the pump or filter limit, operation switches to constant pressure for the rest of the cycle.

  8. 8.A constant-pressure filtration test gives t/V = 2000·V + 500, with t in seconds and V in m³. How long does it take to collect 1 m³ of filtrate, and what do the two constants mean?Numerical

    Integrated constant-pressure filtration gives t/V = (K_p/2)·V + 1/q₀, so t = 2000·V² + 500·V = 2000 + 500 = 2500 s for V = 1 m³. The slope K_p/2 = 2000 s/m⁶ reflects the cake resistance (K_p = μcα/(A²Δp)), and the intercept 1/q₀ = 500 s/m³ reflects the medium resistance (μR_m/(AΔp)). You cannot simply average the initial and final rates, because the rate falls non-linearly as the cake grows.

  9. 9.In a constant-rate filtration, the pressure drop has to be doubled from 100 kPa to 200 kPa to hold the rate. What does this tell you about the resistance?Application

    At constant rate Darcy's law gives Δp = μ·q·(R_c + R_m)/A, so with μ, q and A unchanged the total resistance (cake plus medium) must have exactly doubled. Because R_m is roughly constant, the cake resistance itself has more than doubled. For an incompressible cake Δp rises linearly with time at constant rate, so this doubling also tells you how far through the cycle you are relative to the clean-medium pressure drop.

  10. 10.At constant pressure, with negligible filter-medium resistance, 2 m³ of filtrate is collected in the first hour. How long will it take to collect a total of 4 m³?Numerical

    With R_m negligible the integrated constant-pressure equation reduces to t = (K_p/2)·V², so t is proportional to V². Doubling the filtrate volume therefore needs four times the time: 4 hours in total, i.e. 3 more hours for the second 2 m³. The slowdown is because the cake keeps growing and its resistance is proportional to the filtrate already collected.

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