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Liquid methanol is fed to a space heater at a rate of \(12.0 \mathrm{L} / \mathrm{h}\) and burned with excess air. The product gas is analyzed and the following dry-basis mole percentages are determined: \(\mathrm{CH}_{3} \mathrm{OH}=0.45 \%\) \(\mathrm{CO}_{2}=9.03 \%,\) and \(\mathrm{CO}=1.81 \%\) (a) Draw and label a flowchart and verify that the system has zero degrees of freedom. (b) Calculate the fractional conversion of methanol, the percentage excess air fed, and the mole fraction of water in the product gas. (c) Suppose the combustion products are released directly into a room. What potential problems do you see and what remedies can you suggest?

Short Answer

Expert verified
The fractional conversion of methanol, percentage excess air, and the mole fraction of water in the product gas can be calculated using the provided data and applying chemical reaction stoichiometry and balance equations. However, releasing combustion products directly into a room can pose serious health and safety issues and hence, it is recommended to ensure proper safety measures and controls.

Step by step solution

01

Drawing and labelling the flowchart

First, identify the reactants and the products of the combustion reaction. Draw a flowchart where liquid methanol and air are fed into a space heater, and the products (methanol, carbon dioxide, and carbon monoxide) exit from the other end. Since all the reactants and their flow rate along with the products and their corresponding analysis is provided, this system has zero degrees of freedom.
02

Calculate fractional conversion of Methanol

The rate of methanol fed into the system is 12.0 L/h. Let's assume that fully combusted methanol gives only CO2 as the product. Under these conditions, look at the moles of CO2 produced. Since 0.45% of the product is unburned methanol (CH3OH), and 9.03% is CO2, then the remaining 90.52% must be made up of CO and unquantified products, most likely including water. The fractional conversion of methanol is then calculated as the moles of methanol converted (which can be inferred from CO2 produced) divided by the total moles of methanol fed into the system.
03

Calculate the percentage excess air

Calculate the stoichiometric air required for full combustion of methanol and compare it with the actual air supplied. The excess air percentage can be determined by the formula: \(excess\ air\%= \frac{(actual\ air\ supplied - stoichiometric\ air) x 100}{stoichiometric\ air}\)
04

Calculate mole fraction of water

Since water is also a combustion product, determine it based on the balance of the combustion equation. The mole fraction of water in the product gas can be calculated by dividing moles of water by the total moles of products.
05

Potential problems and recommendations

If the combustion products are released directly into a room, they may lead to serious health and safety issues. The presence of carbon monoxide (a poisonous gas), unburned methanol (a flammable substance), and excess heat can pose serious threats. It is highly recommended to ensure proper ventilation, heat insulation, and combustion control measures to mitigate these risks.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fractional Conversion
Fractional conversion is a useful measure in chemical engineering and combustion processes. It tells us what proportion of a reactant is transformed into a product during a chemical reaction, such as the combustion of methanol. In the given problem, we know that liquid methanol (CH₃OH) is being fed into a space heater and burned. The analysis of the product gas shows 0.45% unburned methanol and 9.03% carbon dioxide (CO₂). Let's dive deeper:
  • The unburned methanol indicates the amount that hasn't reacted.
  • The COâ‚‚ proportion helps us understand the methanol that successfully underwent combustion.
To calculate the fractional conversion, you divide the moles of methanol converted into the intended products (mostly inferred from COâ‚‚ produced) by the total moles of methanol fed into the system. This provides a clear picture of the efficiency of the reaction.
Understanding fractional conversion is crucial as it helps in optimizing reaction conditions for better energy efficiency and reduced waste in industrial processes.
Excess Air Calculation
During combustion, air is needed to provide the oxygen required for the chemical reaction to proceed. Sometimes, more air than necessary – known as "excess air" – is supplied. This is often a strategy to ensure that all the fuel combusts.How do we calculate this? First, determine the stoichiometric air, which is the amount of air needed for complete combustion of methanol with no unburned fuel. In the problem, you'll compare this to the actual air supplied to find the excess air percentage. The formula goes like:\[excess\ air\% = \frac{(\text{actual air supplied} - \text{stoichiometric air}) \times 100}{\text{stoichiometric air}}\]This concept is important because:
  • Excess air ensures complete combustion, reducing pollutants like CO.
  • Too much excess air, however, can lead to thermal inefficiencies.
Balancing excess air in combustion systems boosts their efficiency and effectiveness – a key goal in both environmental and industrial contexts.
Health and Safety in Combustion
Combustion, while necessary for many industrial and residential heating applications, can also entail significant health and safety risks if not properly managed. Let's consider what happens when combustion products are released into a confined space like a room. The main health concern here is the presence of carbon monoxide (CO), a colorless, tasteless, and odorless gas resulting from incomplete combustion. It's poisonous and can be fatal if inhaled in large quantities. Furthermore:
  • Unburned methanol is a flammable substance and poses a fire risk.
  • Excess heat from combustion can also become a safety issue, potentially causing burns or triggering fires.
To prevent these hazards, it is crucial to ensure proper ventilation in areas where combustion occurs. This will dilute any harmful gases and lower the risk of fire. Implementing controls to maintain an appropriate combustion temperature and efficiency can also significantly reduce safety risks. Awareness and practical precautions go a long way in safeguarding health around combustion systems.

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Most popular questions from this chapter

A gas contains 75.0 wt\% methane, \(10.0 \%\) ethane, \(5.0 \%\) ethylene, and the balance water. (a) Calculate the molar composition of this gas on both a wet and a dry basis and the ratio (mol \(\mathrm{H}_{2} \mathrm{O} /\) mol dry gas). (b) If \(100 \mathrm{kg} / \mathrm{h}\) of this fuel is to be burned with \(30 \%\) excess air, what is the required air feed rate (kmol/ h)? How would the answer change if the combustion were only \(75 \%\) complete?

The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product stream is analyzed and found to contain 51.7 mole \(\% \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\) and 17.3\% HBr. The feed to the reactor contains only ethylene and hydrogen bromide. Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is \(165 \mathrm{mol} / \mathrm{s}\), what is the extent of reaction?

Ethane is chlorinated in a continuous reactor: $$\mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{HCl}$$ Some of the product monochloroethane is further chlorinated in an undesired side reaction: $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}+\mathrm{HCl}$$ (a) Suppose your principal objective is to maximize the selectivity of monochloroethane production relative to dichloroethane production. Would you design the reactor for a high or low conversion of ethane? Explain your answer. (Hint: If the reactor contents remained in the reactor long enough for most of the ethane in the feed to be consumed, what would the main product constituent probably be?) What additional processing steps would almost certainly be carried out to make the process economically sound? (b) Take a basis of \(100 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) produced. Assume that the feed contains only ethane and chlorine and that all of the chlorine is consumed, and carry out a degree-of-freedom analysis based on atomic species balances. (c) The reactor is designed to yield a \(15 \%\) conversion of ethane and a selectivity of \(14 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} / \mathrm{mol}\) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2},\) with a negligible amount of chlorine in the product gas. Calculate the feed ratio \(\left(\mathrm{mol} \mathrm{Cl}_{2} /\right.\) mol \(\mathrm{C}_{2} \mathrm{H}_{6}\) ) and the fractional yield of monochloroethane. (d) Suppose the reactor is built and started up and the conversion is only \(14 \% .\) Chromatographic analysis shows that there is no \(\mathrm{Cl}_{2}\) in the product but another species with a molecular weight higher than that of dichloroethane is present. Offer a likely explanation for these results.

Water enters a \(2.00-\mathrm{m}^{3}\) tank at a rate of \(6.00 \mathrm{kg} / \mathrm{s}\) and is withdrawn at a rate of \(3.00 \mathrm{kg} / \mathrm{s}\). The tank is initially half full. (a) Is this process continuous, batch, or semibatch? Is it transient or steady state? (b) Write a mass balance for the process (see Example 4.2-1). Identify the terms of the general balance equation (Equation 4.2-1) present in your equation and state the reason for omitting any terms. (c) How long will the tank take to overflow?

Inside a distillation column (see Problem 4.8), a downward-flowing liquid and an upward-flowing vapor maintain contact with each other. For reasons we will discuss in greater detail in Chapter \(6,\) the vapor stream becomes increasingly rich in the more volatile components of the mixture as it moves up the column, and the liquid stream is enriched in the less volatile components as it moves down. The vapor leaving the top of the column goes to a condenser. A portion of the condensate is taken off as a product (the overhead product), and the remainder (the reflux) is returned to the top of the column to begin its downward journey as the liquid stream. The condensation process can be represented as shown below: A distillation column is being used to separate a liquid mixture of ethanol (more volatile) and water (less volatile). A vapor mixture containing 89.0 mole \(\%\) ethanol and the balance water enters the overhead condenser at a rate of \(100 \mathrm{lb}\) -mole/h. The liquid condensate has a density of \(49.01 \mathrm{b}_{\mathrm{m}} / \mathrm{ft}^{3},\) and the reflux ratio is \(3 \mathrm{lb}_{\mathrm{m}}\) reflux/lb \(_{\mathrm{m}}\) overhead product. When the system is operating at steady state, the tank collecting the condensate is half full of liquid and the mean residence time in the tank (volume of liquid/volumetric flow rate of liquid) is 10.0 minutes. Determine the overhead product volumetric flow rate (ft \(^{3}\) /min) and the condenser tank volume (gal).

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