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A natural gas containing 82.0 mole \(\% \mathrm{CH}_{4}\) and the balance \(\mathrm{C}_{2} \mathrm{H}_{6}\) is burned with \(20 \%\) excess air in a boiler furnace. The fuel gas enters the furnace at \(298 \mathrm{K}\), and the air is preheated to 423 \(\mathrm{K}\). The heat capacities of the stack-gas components may be assumed to have the following constant values: $$\begin{aligned}\mathrm{CO}_{2}: & C_{p}=50.0 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): & C_{p}=38.5 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) \\\\\mathrm{O}_{2}: & C_{p}=33.1 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \mathrm{N}_{2}: & C_{p}=31.3 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\end{aligned}$$ (a) Assuming complete combustion of the fuel, calculate the adiabatic flame temperature. (b) How would the flame temperature change if the percent excess air were increased? How would it change if the percentage of methane in the fuel increased? Briefly explain both of your answers.

Short Answer

Expert verified
The adiabatic flame temperature would be found by utilizing a heat balance equation, taking into account the molecular composition of the gas and the number of moles of each component in the reacting gases along with their respective heat capacities. Changes in the percentage of excess air and the composition of the fuel would affect the flame temperature - increasing excess air decreases the flame temperature and increasing methane in the fuel increases the flame temperature.

Step by step solution

01

Calculate the moles of fuel

First, find the number of moles of CH4 and C2H6 in one mole of the gas. Since the fuel contains 82.0 mole% CH4 and the remainder is C2H6, there are 0.82 moles of CH4 and 0.18 moles of C2H6.
02

Perform Stoichiometric Calculations

Use the chemical equations to find the number of moles of oxygen required for complete combustion. For CH4, the equation is \(CH4 + 2O2 \rightarrow CO2 + 2H20\). Thus, each mole of CH4 requires 2 moles of O2 for combustion. Similarly, the equation for C2H6 is \(2C2H6 + 7O2 \rightarrow 4CO2 + 6H20\). So, each mole of C2H6 requires 7/2 moles of O2 for combustion. Given that 20\% excess air is used, calculate the number of moles of O2 supplied. 1 mole of air is 21 mol% O2 and 79 mol% N2.
03

Calculate the stack-gas composition

Calculate the moles of each gas in the combustion products or stack gas. For complete combustion, all carbon in the fuel goes to CO2 and all hydrogen goes to H2O. Oxygen in excess of the stoichiometric requirement exits as O2. The balance of the air, N2, also exits unchanged.
04

Apply the heat balance equation

Apply the heat balance equation (adiabatic flame, no heat loss or gain, \(\sum \Delta H_{reactants} = \sum \Delta H_{products}\)). The contribution of each component to \(\Delta H_{products}\) is (number of moles) \(\times (Cp)\) \(\times (T - T_{reference})\). Equate heat produced by combustion of reactants (there is no energy change for incoming air as its components are also in the stack gas). Set this equal to the heat capacity of the stack gas multiplied by the temperature rise, and solve for the flame temperature, \(T\).
05

Discuss the effect of increasing the percentage of excess air

Increasing the percentage of excess air would decrease the flame temperature. This is because the excess air, primarily nitrogen, absorbs some of the heat generated during combustion, but does not participate in the reaction. Hence, more excess air causes more heat to be absorbed, lowering the flame temperature.
06

Discuss the effect of increasing the percentage of methane in the fuel

Increasing the percentage of methane (CH4) in the fuel would increase the flame temperature. This is because methane has a higher heat of combustion than C2H6, such that for each mole of methane burned, more heat is produced than for a mole of C2H6. Thus, greater methane percentage leads to a higher flame temperature.

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Key Concepts

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

Adiabatic Flame Temperature
In the context of combustion analysis, the adiabatic flame temperature is essential. It refers to the temperature a flame reaches under the assumption that no heat is lost to the surroundings. The temperature depends on several factors, such as the type of fuel used and the amount of excess air introduced.

When calculating it, one needs to apply the principle of conservation of energy. This means that the total enthalpy of reactants must equal the total enthalpy of products. During combustion, the heat from burning is used to raise the temperature of gases produced. Thus, properly determining the temperature involves knowing both the specific heat capacities of the products and the amount of thermal energy released.
  • An accurate measure of the adiabatic flame temperature is critical as it influences the efficiency of processes like power generation and heating.
  • Higher flame temperatures typically indicate more efficient combustion and energy release.
  • Variables like air preheat and ambient temperature significantly impact the final temperature.
Stoichiometric Combustion
Stoichiometric combustion involves a perfect balance between fuel and oxygen, ensuring all fuel is burned with no leftover oxygen. In our scenario, this concept helps calculate the precise amount of oxygen required for burning methane (\(\mathrm{CH}_{4}\)) and ethane (\(\mathrm{C}_{2}\mathrm{H}_{6}\)).

For methane, the reaction can be written as: \(\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\). This equation shows that each mole of methane requires 2 moles of oxygen. Similarly, ethane needs \(\dfrac{7}{2}\) moles of \(\mathrm{O}_{2}\) for every mole of \(\mathrm{C}_{2}\mathrm{H}_{6}\), as expressed by \(2\mathrm{C}_{2}\mathrm{H}_{6} + 7\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 6\mathrm{H}_{2}\mathrm{O}\).
  • In stoichiometric combustion, efficiency peaks because all fuel is used for energy production without wasting resources.
  • It forms the basis for determining the right amount of excess air needed in industrial settings.
  • Skillfully managing combustion around these principles ensures minimal emissions and energy losses.
Excess Air in Combustion
Using excess air in combustion impacts the system's temperature and efficiency. It refers to the amount of air supplied beyond the stoichiometric requirement. In our example, there's a 20% excess air used, implying more oxygen than is theoretically required.

While adding some excess air ensures thorough combustion by ensuring complete reaction with the fuel, too much excess air can cool down the reaction. That's because the extra nitrogen in the air, which does not react, absorbs energy without contributing to the combustion process.
  • Adding excess air is a common practice in industrial combustion due to its safety cushion, preventing the formation of harmful pollutants like carbon monoxide.
  • Optimal excess minimizes energy losses while ensuring full combustion.
  • However, an increase in excess air leads to a lower adiabatic flame temperature, highlighting the trade-off that must be managed.
Methane Combustion
Combusting methane is a crucial concept in this analysis. Methane (\(\mathrm{CH}_{4}\)) is a simple hydrocarbon with a high energy content, making it an efficient fuel choice.

Its combustion reaction is straightforward, producing carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). This reaction releases a significant amount of heat energy, contributing to a higher potential flame temperature compared to heavier hydrocarbons like ethane (\(\mathrm{C}_{2}\mathrm{H}_{6}\)).
  • Methane's high heat of combustion is beneficial in generating more heat per mole, leading to higher energy efficiency when used in engines or heating systems.
  • Its combustion tends to produce fewer carbon-based pollutants, making it an environmentally friendly fuel compared to other fossil fuels.
  • When methane composition in fuel increases, the adiabatic flame temperature also increases, enhancing the combustion efficiency of the system.

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Most popular questions from this chapter

A dilute aqueous solution of sulfuric acid at \(25^{\circ} \mathrm{C}\) is used to absorb ammonia in a continuous reactor, thereby producing ammonium sulfate, a fertilizer: $$2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})$$ (a) If the ammonia enters the absorber at \(75^{\circ} \mathrm{C}\), the sulfuric acid enters at \(25^{\circ} \mathrm{C}\), and the product solution emerges at \(25^{\circ} \mathrm{C}\), how much heat must be withdrawn from the unit per mol of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) produced? (All needed physical property data may be found in Appendix B.) (b) Estimate the final temperature if the reactor of Part (a) is adiabatic and the product of the solution contains 1.00 mole \(\%\) ammonium sulfate. Take the heat capacity of the solution to be that of pure liquid water [4.184 kJ/(kg.'C)]. (c) In a real (imperfectly insulated) reactor, would the final solution temperature be less than, equal to, or greater than the value calculated in Part (b), or is there no way to tell without more information? Briefly explain your answer.

A 2.00 mole \(\%\) sulfuric acid solution is neutralized with a 5.00 mole\% sodium hydroxide solution in a continuous reactor. All reactants enter at \(25^{\circ} \mathrm{C}\). The standard heat of solution of sodium sulfate is \(-1.17 \mathrm{kJ} / \mathrm{mol} \mathrm{Na}_{2} \mathrm{SO}_{4},\) and the heat capacities of all solutions may be taken to be that of pure liquid water [4.184 kJ/(kg.'C)]. (a) How much heat (kJ/kg acid solution fed) must be transferred to or from the reactor contents (state which it is) if the product solution emerges at \(40^{\circ} \mathrm{C} ?\) (b) Estimate the product solution temperature if the reactor is adiabatic, neglecting heat transferred between the reactor contents and the reactor wall.

The standard heat of the combustion reaction of liquid \(n\) -hexane to form \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}),\) with all reactants and products at \(77^{\circ} \mathrm{F}\) and 1 atm, is \(\Delta H_{\mathrm{r}}^{\prime}=-1.791 \times 10^{6} \mathrm{Btu} .\) The heat of vaporization of hexane at \(77^{\circ} \mathrm{F}\) is \(13,550 \mathrm{Btu} / \mathrm{b}\) -mole and that of water is \(18.934 \mathrm{Btu} / \mathrm{h}\) -mole. (a) Is the reaction exothermic or endothermic at \(77^{\circ} \mathrm{F}\) ? Would you have to heat or cool the reactor to keep the temperature constant? What would the temperature do if the reactor ran adiabatically? What can you infer about the energy required to break the molecular bonds of the reactants and that released when the product bonds form? (b) Use the given data to calculate \(\Delta H_{\mathrm{r}}^{\mathrm{r}}\) (Btu) for the combustion of \(n\) -hexane vapor to form \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\overline{\mathrm{H}}_{2} \mathrm{O}(\mathrm{g})\) (c) If \(\dot{Q}=\Delta \dot{H},\) at what rate in \(\mathrm{B}_{\text {tu } / \mathrm{s}}\) is heat absorbed or released (state which) if \(120 \mathrm{lb}_{\mathrm{n}} / \mathrm{s}\) of \(\mathrm{O}_{2}\) is consumed in the combustion of hexane vapor, water vapor is the product, and the reactants and products are all at \(77^{\circ} \mathrm{F} ?\) (d) If the reaction were carried out in a real reactor, the actual value of \(\dot{Q}\) would be greater (less negative) than the value calculated in Part (c). Explain why.

Methanol vapor is burned with excess air in a catalytic combustion chamber. Liquid methanol initially at \(25^{\circ} \mathrm{C}\) is vaporized at 1.1 atm and heated to \(100^{\circ} \mathrm{C}\); the vapor is mixed with air that has been preheated to \(100^{\circ} \mathrm{C},\) and the combined stream is fed to the reactor at \(100^{\circ} \mathrm{C}\) and 1 atm. The reactor effluent emerges at \(300^{\circ} \mathrm{C}\) and 1 atm. Analysis of the product gas yields a dry-basis composition of \(4.8 \% \mathrm{CO}_{2}\) \(14.3 \% \mathrm{O}_{2},\) and \(80.9 \% \mathrm{N}_{2}\) (a) Calculate the percentage excess air supplied and the dew point of the product gas. (b) Taking a basis of 1 g-mole of methanol burned, calculate the heat ( \(k\) J) needed to vaporize and heat the methanol feed, and the heat (kJ) that must be transferred from the reactor. (c) Suggest how the energy economy of this process could be improved. Then suggest why the company might choose not to implement your redesign.

Formaldehyde is produced commercially by the catalytic oxidation of methanol. In a side reaction, methanol is oxidized to \(\mathrm{CO}_{2}\) $$\begin{array}{l}\mathrm{CH}_{3} \mathrm{OH}+\mathrm{O}_{2} \rightarrow \mathrm{CH}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\\\\mathrm{CH}_{3} \mathrm{OH}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\end{array}$$ A mixture containing 55.6 mole \(\%\) methanol and the balance oxygen enters a reactor at \(350^{\circ} \mathrm{C}\) and \(1 \mathrm{atm}\) at a rate of \(4.60 \times 10^{4} \mathrm{L} / \mathrm{s}\). The reaction products emerge at the same temperature and pressure at a rate of \(6.26 \times 10^{4} \mathrm{L} / \mathrm{s} .\) An analysis of the products yields a molar composition of \(36.7 \% \mathrm{CH}_{2} \mathrm{O}, 4.1 \% \mathrm{CO}_{2}\) \(14.3 \% \mathrm{O}_{2},\) and \(44.9 \% \mathrm{H}_{2} \mathrm{O} .\) The required reactor cooling rate is calculated to be \(1.05 \times 10^{5} \mathrm{kW}\) (a) Is the calculated cooling rate correct for the given stream data? (b) The stream data cannot be correct. Prove it.

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