Modeling & Simulation of Mineral processing Systems

  Numerical Solution: Kick’s Law (Size Reduction) Calculate the energy required to reduce particle size from 50 mm to 12.5 mm using Kick’s law, given K k = 3.2 kJ/kg. Given Initial particle size, D 1 = 50  D_1 = 50 mm Final particle size, D 2 = 12.5 D_2 = 12.5  mm Kick’s constant, K k = 3.2 K_k = 3.2  kJ/kg Kick’s Law E = K k ln ⁡  ⁣ ( D 1 D 2 ) E = K_k \ln\!\left(\frac{D_1}{D_2}\right) Step-wise Calculation D 1 D 2 = 50 12.5 = 4 \frac{D_1}{D_2} = \frac{50}{12.5} = 4 E = 3.2 × ln ⁡ ( 4 ) E = 3.2 \times \ln(4) ln ⁡ ( 4 ) = 1.386 \ln(4) = 1.386 E = 3.2 × 1.386 = 4.435  kJ/kg E = 3.2 \times 1.386 = 4.435 \text{ kJ/kg} Final Answer E ≈ 4.44  kJ/kg \boxed{E \approx 4.44 \text{ kJ/kg}} ​ Energy required to reduce the particle size from 50 mm to 12.5 mm is approximately 4.44 kJ/kg.

Numerical Example: Breakage Function in Crushing Problem Statement The breakage function for a crusher is given by: B ( x : y ) = 0.5 ( x y ) 0.6 + 0.5 ( x y ) 3.0 B(x : y) = 0.5\left(\frac{x}{y}\right)^{0.6} + 0.5\left(\frac{x}{y}\right)^{3.0} where x = size of progeny particle y y = size of parent particle Size class 3 has mesh size boundaries of 12.0 cm and 8.0 cm , and Size class 6 has mesh size boundaries of 4.0 cm and 2.0 cm . Calculate b 63 b_{63} ​ , the fraction of material from size class 3 that reports to size class 6 after crushing . Solution Step 1: Identify representative sizes Representative size is taken as the geometric mean of class boundaries. Size class 3 (parent): y = 12 × 8 = 96 = 9.8  cm y = \sqrt{12 \times 8} = \sqrt{96} = 9.8 \text{ cm} Size class 6 (progeny): x = 4 × 2 = 8 = 2.83  cm x = \sqrt{4 \times 2} = \sqrt{8} = 2.83 \text{ cm} Step 2: Write the breakage function B ( x : y ) = 0.5 ( x y ) 0.6 + 0.5 ( x y ) 3.0 B(x:y)...

  Karra S-Curve (d50 = 3.2 mm, m = 4) ere is your Karra S-curve with the d₅₀ point clearly marked at: Particle size = 3.2 mm Passing probability = 0.50 (50%) Variation of m Short Summary m Effect High Sharp cut, clean oversize/undersize Low Poor cut, lots of middlings Depends on Design + Operating conditions Typical Industrial Values Screen Type Typical m High-frequency screens 5 – 10 Standard wet vibrating screens 4 – 7 Dry screens / coarse sizing 2 – 5 Worn, overloaded screens 1 – 3 M ulti-curve Karra S-curve graph , showing how the sharpness index (m) affects screening performance: m = 2 → very poor separation m = 4 → normal vibrating screen m = 6 → wet/high-frequency screens m = 10 → excellent sharp cut The steeper the curve , the better the separation quality . Higher m = smarter screening. ✔️

  Karra Model  The Karra Model is a mathematical formula used to predict how a vibrating screen separates particles into: Oversize (won’t pass the screen) Undersize (passes the screen) Engineers use it because it is: ✔ simple ✔ realistic ✔ based on screen geometry ✔ works well for industrial screens 1. Why Do We Need the Karra Model? Every screening process asks one question: “If I feed this mixture of particle sizes to the screen, how much will go through?” But screening is NEVER perfect. Some small particles don’t pass. Some large particles sneak through. So we need a model that predicts: Probability of passing Effect of particle size Effect of screen area Effect of open area Effect of screen inclination Karra’s Model does exactly this. Core Idea Behind Karra Model Karra said: Every particle has a probability of passing through a screen opening. The closer the particle size is to the aperture size, the lower the probability. The...

  The Math of Liberation Notes on probability density, the Beta distribution, Andrews–Mika diagram, and breakage types. What is Probability Density ? Probability density is a way to show how common items are along a scale. If the horizontal axis is particle type (left = pure waste, middle = mixed, right = pure mineral), the vertical axis (probability density) shows how many particles fall at each point. Remember: higher curve → more particles there; lower curve → fewer particles. 1. U-shaped Beta Distribution (Why Beta?) The Beta distribution is used because it is flexible and can make a U-shape. A well-ground ore often yields: Many pure waste particles (left tail) Many pure mineral particles (right tail) Few mixed (middling) particles (middle) Quick gist: Left = L 0 (liberated gangue), Right = L 1 (liberated mineral), Middle = middlings. The Four Parameters (short): Average gr...

🎢 1. The Math of Liberation: The Beta Distribution Engineers love curves. In mineral processing, our favorite is the Beta Distribution . 🤔 Why Beta? Because the Beta curve can take the shape of a U (like a bowl). A well-ground ore usually has: A lot of pure waste (left side) A lot of pure mineral (right side) Very little mixed particles (middle) Perfect U-shape → Perfect for liberation modeling. Beta curve = A U-shaped graph telling us how much of the ore is free mineral, free waste, or mixed junk. 📦 The Four Parameters You Actually Need 1) Average Grade “How much metal is there in the rock overall?” Like: If a classroom has 20% toppers → average grade = 20%. 2) Variance “How mixed is everything?” Low variance → uniform High variance → some rich particles, some poor particles 3) L₀ = Liberated Gangue Fraction of pure waste particles. (They are fully useless but easy to remove.) 4) L₁ = Liberated Mineral Fraction of pure valuable particles. ...

  💡 1. What Is Mineral Liberation? When ore is crushed and ground, the valuable minerals (like magnetite, chalcopyrite, bauxite minerals, etc.) are broken away from the useless rock (gangue). 👉 This “freeing” is called liberation . Liberation = breaking grains until the valuables are no longer glued to gangue. 💥 2. Why Do We Care About Liberation? Because separation processes (flotation, magnetic separation, gravity separation) work ONLY if valuable particles are free . Well-liberated = easy to separate = good recovery Poorly liberated = mixed particles = losses in tailings Separation efficiency depends directly on mineral liberation. 🔎 3. What Controls Liberation? a) Mineral Grain Size in the Ore Fine-grained ore → needs finer grinding → expensive Coarse-grained ore → easy to liberate → cheap processing b) Texture of the Ore Simple texture: clean mineral boundaries → easy liberation Complex texture: minerals intergrown → difficult liberation ...

   Numerical Example 2 Scenario: A crusher product is analyzed and found to have the following Rosin–Rammler parameters: D 63.2 = 40 mm λ = 2.0 Question: What percentage of the material is smaller than 20 mm ? Step-by-Step Calculation 1. Insert values into the Rosin–Rammler equation: P(20) = 1 – exp[ – ( 20 / 40 )^2.0 ] 2. Simplify the fraction: 20 / 40 = 0.5 3. Apply exponent 2.0: 0.5^2.0 = 0.25 4. Compute exponential term: exp(–0.25) ≈ 0.7788 5. Final step: 1 – 0.7788 = 0.2212 Final Answer: Approximately 22.1% of the material is smaller than 20 mm .

  1. Definition The Rosin–Rammler equation is a mathematical formula used by engineers to describe the size distribution of crushed or ground particles. Instead of listing all particle sizes, the equation creates a smooth curve using just two parameters: D 63.2 (Size Modulus): The particle size at which 63.2% of the material is smaller. It shows how coarse the overall material is. λ (Lambda – Distribution Modulus): Indicates the spread of particle sizes.   • High λ → particles mostly similar in size   • Low λ → wide mix of very fine and coarse particles The Formula: P(D) = 1 – exp [ – ( D / D 63.2 ) λ ] Where: D = particle size being evaluated P(D) = fraction (or %) passing size D 2. Numerical Example Scenario: A ball mill product is analyzed and found to have Rosin–Rammler parameters: D 63.2 = 100 µm λ = 1.5 Question: What percentage of the material is smaller than 50 µm ? Step-by-Step Cal...

  Understanding Figure 2.2: Mesh Sizes, Size Classes & Representative Sizes Based on the provided text, here is a simple explanation of Figure 2.2 from Chapter 2. What is Figure 2.2 Showing? This figure shows how engineers convert a continuous range of particle sizes into neat, usable mathematical groups. It illustrates the link between Mesh Sizes , Size Classes , and Representative Sizes . Imagine a long ruler measuring particle size from large (right) to small (left). 1. Mesh Sizes (D i ) — “The Fence Posts” Vertical lines labeled D₁, D₂, D₃… are the boundaries between size ranges. These correspond to physical sieves used in the lab. D₁ is the largest opening; D N is the smallest. Everything between two mesh-size lines belongs to one group. 2. Size Classes (1, 2, 3…) — “The Bins” The space between two mesh sizes is a Size Class . Example: Size Class 2 lies between D₁ and D₂...

Rosin–Rammler Distribution: Step-by-Step Numerical Example The Scenario You have performed a sieve analysis on a ball mill product. After fitting the data, you obtained: Size Modulus (D 63.2 ): 150 µm Distribution Modulus (λ): 1.2 The Rosin–Rammler Formula: P(D) = 1 − exp[ − ( D / D₆₃.₂ ) ^ λ ] Question 1: How much "fines" do we have? Goal: Find what percentage passes through a 75 µm screen. Step 1: Plug in values P(75) = 1 − exp[ − (75 / 150) ^ 1.2 ] Step 2: Ratio 75 / 150 = 0.5 Step 3: Apply λ 0.5 1.2 ≈ 0.435 Step 4: Exponential e −0.435 ≈ 0.647 Step 5: Final calculation P(75) = 1 − 0.647 = 0.353 (→ 35.3% fines) Question 2: What is the P₈₀? (Reverse Calculation) Goal: Find D such that P(D) = 0.80. 0.80 = 1 − exp[ − (D / 150) ^ 1.2 ] Step 1: Rearrange 0.20 = exp[ − (D / 150) ^ 1.2 ] Step 2: Take natural log ln(0.20) = − (D...

Representative Size ( d pi ) Problem: If a bin holds rocks ranging from 10 mm to 20 mm , what single size should we use for calculations? Should it be 10 mm , 20 mm , or 15 mm ? Solution: We use the geometric mean . Representative Size: d pi = √(Top Size × Bottom Size) Example (10 mm to 20 mm): d pi = √(10 × 20) = √200 ≈ 14.14 mm Note: This method works best when sieve sizes are spaced relatively close together — a common spacing is a factor of √2 .

Based on the Population Balance principle described in Chapter 2, the value of Accumulation depends on the state of the machine: 1. Steady-State Operation (The Standard Case) Value = 0 Most mineral processing plants (and the simulators that model them) are designed to run continuously 24/7. Once the machine “warms up” and stabilizes, the amount of rock entering a specific bin is exactly equal to the amount leaving it. The “level” in the bin does not change. Equation: 0 = Input – Output + Birth – Death Meaning: Everything is balanced. 2. Dynamic Operation (Start-up or Batch) Value = Rate of Change (dM/dt) If you are filling up a mill, emptying it, or running a batch test in a lab, the amount of material in the bin changes over time. Positive Value: The bin is filling up (Input + Birth > Output + Death). Negative Value: The bin is emptying (Output + Death > Input + Birth). Analogy: Think of a bathtub Dynamic: If you turn the tap on (Input) with the drain clos...