Sample Problem 1: Magnesium Sulfate Crystallization

This page details a sample problem involving the crystallization of magnesium sulfate heptahydrate (MgSO₄ · 7H₂O) from a solution.

Given information:

• Feed (F) = 1000 kg
• Feed temperature (TF) = 353 K
• Solubility at 353 K = 64.2 kg MgSO₄ / 100 kg H₂O
• Solubility at 303 K = 40.8 kg MgSO₄ / 100 kg H₂O
• 10% of water from feed is evaporated

The problem requires calculating the crystal yield and other process parameters.

Vocabulary: Crystal yield refers to the amount of solid crystals produced from a solution during the crystallization process.

The solution involves material balance equations and solubility calculations to determine the amount of crystals formed and the yield percentage.

Highlight: This problem demonstrates how to use solubility data at different temperatures to calculate crystallization yields, which is a fundamental skill in crystallization process in chemistry.

Sample Problem 2: Sodium Carbonate Crystallization

This section presents a problem focusing on the crystallization of sodium carbonate decahydrate (Na₂CO₃ · 10H₂O) from a solution.

Given information:

• Feed (F) = 5000 kg of Na₂CO₃ and H₂O mixture
• 5% of water evaporates
• Solubility data provided for different temperatures

The problem requires calculating the yield of Na₂CO₃ crystals.

Example: The solubility of sodium carbonate at the crystallization temperature is 21.5 kg Na₂CO₃ / 100 kg H₂O.

The solution involves material balance equations, similar to the previous problem, but with different solubility data and molecular weights.

Highlight: This problem illustrates how the solubility of sodium carbonate at different temperatures affects the crystallization yield, which is crucial knowledge for industrial crystallization processes.

Sample Problem 3: Magnesium Sulfate Crystallization with Heat Loss

This problem expands on the previous magnesium sulfate crystallization example by incorporating heat loss calculations.

Given information:

• Feed (F) = 2000 kg MgSO₄
• Initial temperature (TF) = 330 K
• Final temperature (T) = 290 K
• Solubility and molecular weight data provided

The problem requires calculating the yield of crystals and the heat loss during the process.

Vocabulary: Heat of crystallization (Δhc) is the energy released when a substance transitions from a liquid to a solid crystalline state.

The solution involves both material balance and energy balance equations to determine the crystal yield and the heat loss during the crystallization process.

Highlight: This problem demonstrates the importance of considering both mass and energy balances in crystallization processes, which is essential for designing efficient industrial crystallizers.

Sample Problem 4: Sodium Sulfate Crystallization

This problem focuses on the crystallization of sodium sulfate decahydrate (Na₂SO₄ · 10H₂O), also known as Glauber's salt.

Given information:

• Feed composition: 500 kg Na₂SO₄ + 2500 kg H₂O
• Initial temperature (TF) = 333 K
• Vessel mass and specific heat capacity
• Latent heat of vaporization and heat of solution data

The problem requires calculating the heat loss during the crystallization process.

Vocabulary: Glauber's salt is the common name for sodium sulfate decahydrate, an important industrial chemical.

The solution involves material balance equations to determine the amount of crystals formed, followed by an energy balance to calculate the heat loss.

Highlight: This problem showcases how the heat of solution in crystallization processes affects the overall energy balance, which is crucial for designing energy-efficient crystallization systems.

Sample Problem 5: Iron(II) Sulfate Crystallization

This problem deals with the crystallization of iron(II) sulfate heptahydrate (FeSO₄ · 7H₂O), also known as copperas, in an adiabatic vacuum crystallizer.

Given information:

• Feed composition: 38.9 parts FeSO₄ per 100 parts H₂O
• Initial temperature (TF) = 343 K
• Enthalpy data provided
• Production rate: 10 tons/hr of copperas crystals

The problem requires calculating the feed rate for steady-state operation.

Vocabulary: An adiabatic process is one that occurs without heat transfer to or from the surroundings.

The solution involves material and energy balance equations to determine the feed rate necessary to produce the specified amount of crystals.

Highlight: This problem illustrates the application of crystallization process steps in an industrial setting, demonstrating how theoretical concepts are applied to real-world production scenarios.

Sample Problem 6: Sodium Phosphate Crystallization

The final problem focuses on the crystallization of sodium phosphate dodecahydrate (Na₃PO₄ · 12H₂O) in a continuous crystallizer.

Given information:

• Feed temperature (TF) = 313 K
• Specific heat capacity and latent heat data
• Overall heat transfer coefficient
• Cooling water temperatures
• Crystal production rate: 0.060 kg/s

The problem requires calculating the length of the crystallizer.

Vocabulary: The overall heat transfer coefficient (U) is a measure of the overall ability of a series of conductive and convective barriers to transfer heat.

The solution involves energy balance equations to determine the heat transfer area required, which is then used to calculate the crystallizer length.

Highlight: This problem demonstrates how heat transfer principles are applied in the design of continuous crystallization equipment, which is essential knowledge for chemical engineers working on industrial crystallization processes.

Sample Problems on Crystallization Computations

This section introduces the concept of crystallization computations with sample problems focusing on solubility and heat of solution for various compounds.

Vocabulary: Soda ash is another name for sodium carbonate (Na₂CO₃ · 10H₂O).

Example: The solubility of soda ash at 30°C in boiling water is 38.8g/100g.

Definition: Solubility is the amount of a substance that can dissolve in a given amount of solvent at a specific temperature.

The document provides solubility data for several compounds:

• Sal ammoniac (ammonium chloride, NH₄Cl) at 70°C: 60.2 g/100g water
• Epsom salt at 10°C: 20.9g/100g water
• Saltpeter (potassium nitrate, KNO₃) heat of solution: -8.633 kcal/mol
• Hydroxybenzene (phenol) heat of solution: -2,405 cal/mol

Highlight: Understanding solubility at different temperatures is crucial for designing crystallization processes in chemical engineering.