Question

Consider an aqueous feedstock with \(10 \mathrm{wt} \%\) organics (take glucose for the calculations). This solution is gasified at \(600^{\circ} \mathrm{C}\) and \(25.0 \mathrm{MPa}\) (abs.), and the reactor effluent is heat exchanged with the feed at an efficiency of \(70 \%\). Estimate the amount of energy that has to be provided to the reactor per \(\mathrm{kg}\) of (wet) feed. You may assume that the reaction reaches equilibrium.

Short answer

Question: Estimate the amount of energy required to gasify an aqueous feedstock containing 10 wt% organics (glucose) at 600°C and 25 MPa, with a heat exchange efficiency of 70%. Answer: To estimate the amount of energy required, first determine the glucose gasification reaction, calculate the equilibrium constants and moles of each component, determine the energy required for the reaction to reach equilibrium, and finally calculate the provided energy per kg of wet feed considering the heat exchange efficiency.

Step-by-step solution

Determine the reaction for glucose gasification

The gasification reaction of glucose (C6H12O6) can be represented as: C6H12O6 + nH2O -> mCO2 + nH2 where n and m are the stoichiometric coefficients.

Calculate the equilibrium constants and moles of each component

At the given conditions (600°C and 25 MPa), we can determine the equilibrium constants for the reaction (use available thermodynamic data or software like FACTSAGE). With the equilibrium constants, we can calculate the values of m and n, giving us the moles of each component (CO2 and H2) at equilibrium.

Determine the energy required for the reaction

The energy required for the glucose gasification reaction to reach equilibrium is the change in the Gibbs free energy (ΔG) times the moles of glucose in the feed: ΔG = -RT ln(K_eq) where R is the universal gas constant, T is the temperature in Kelvin (873 K), and K_eq is the equilibrium constant. We can determine the amount of energy required per mole from the Gibbs free energy using the mass of glucose in the feedstock (10 wt% or 0.1 kg/kg feed).

Calculate the provided energy per kg of wet feed

Given that the heat exchange efficiency is 70%, we can determine the amount of energy required per kg of wet feed as: Energy_required = ΔG * moles_of_glucose * (1/0.7) This will give us the amount of energy that has to be provided to the reactor per kg of wet feed.

Key concepts

Equilibrium Constants

When dealing with chemical reactions like glucose gasification, the concept of equilibrium constants comes into play. Equilibrium constants help us understand how a reaction behaves at a certain temperature and pressure. They tell us the ratio of products to reactants when the chemical reaction is balanced and stable.

For glucose gasification, the equilibrium constant, denoted as \( K_{eq} \), allows us to determine the amount of carbon dioxide (CO2) and hydrogen gas (H2) formed. Given the reaction conditions of 600°C and 25 MPa, it is crucial to use thermodynamic data to find this constant. By doing so, you can ascertain the values of the stoichiometric coefficients for products and reactants.

The equilibrium constant is vital because it directly influences the calculation of energy requirements. It indicates the extent to which glucose is converted to gases in the reaction. This helps in designing reactors more efficiently, ensuring that maximum possible conversion occurs with the right conditions in place.

Gibbs Free Energy

Gibbs free energy is a concept that can feel intimidating, but it is essential in chemical thermodynamics. It helps us quantify the energy changes during a chemical reaction, such as glucose gasification.

In essence, the Gibbs free energy change, denoted \( \Delta G \), indicates whether a reaction can occur spontaneously. For our scenario, \( \Delta G \) is calculated using the formula: \( \Delta G = -RT \ln K_{eq} \). Here, \( R \) represents the universal gas constant, \( T \) stands for temperature in Kelvin (in this case, 873 K), and \( K_{eq} \) embodies the equilibrium constant that we previously determined.

What this tells us is whether we need to add energy to make the reaction happen. A negative \( \Delta G \) implies that the reaction may occur without external energy, while a positive \( \Delta G \) means energy input is required. In the context of glucose gasification, understanding \( \Delta G \) allows us to calculate the necessary energy input, making it a crucial step.

Heat Exchange Efficiency

Heat exchange efficiency is a fascinating topic that comes into play when we want to conserve energy in industrial processes. It describes how effectively heat is transferred from one medium to another, in this instance, from the reactor effluent to the feedstock.

In the glucose gasification process, we've assumed a heat exchange efficiency of 70%. This means that 70% of the heat from the reactor can be recycled back to preheat the incoming feed. This is valuable because it allows for energy savings, which means less external energy is required to maintain the reaction at the desired conditions.

The efficiency impacts the calculation of the energy that must be supplied to the reactor. With a 70% efficiency, we calculate the needed energy as:\[ \text{Energy equired} = \Delta G \times \text{moles aisebox{-1ex}{of aisebox{-1ex}{glucose}}} \times \left(\frac{1}{0.7}\right) \]
By recycling a portion of the energy, the process becomes more sustainable and cost-effective, highlighting the significance of optimizing heat exchange in chemical processes.