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Chapter 8: Problem 40

A natural gas containing 95 mole \(\%\) methane and the balance ethane is burned with \(20.0 \%\) excess air. The stack gas, which contains no unburned hydrocarbons or carbon monoxide, leaves the furnace at \(900^{\circ} \mathrm{C}\) and \(1.2 \mathrm{atm}\) and passes through a heat exchanger. The air on its way to the furnace also passes through the heat exchanger, entering it at \(20^{\circ} \mathrm{C}\) and leaving it at \(245^{\circ} \mathrm{C}\). (a) Taking as a basis \(100 \mathrm{mol} / \mathrm{s}\) of the natural gas fed to the furnace, calculate the required molar flow rate of air, the molar flow rate and composition of the stack gas, the required rate of heat transfer in the preheater, \(\dot{Q}\) (write an energy balance on the air), and the temperature at which the stack gas leaves the preheater (write an energy balance on the stack gas). Note: The problem statement does not give you the fuel feed temperature. Make a reasonable assumption, and state why your final results should be nearly independent of what you assume. (b) What would \(\dot{Q}\) be if the actual feed rate of the natural gas were 350 SCMH [standard cubic meters per hour, \(\left.\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{h}\right] ?\) Scale up the flowchart of Part (a) rather than repeating the entire calculation.

Short Answer

Expert verified
The molar flow rate of required air is found to be 1165 mol/s. The molar flow rate and composition of the stack gas is determined to be 105, 205, 4382 and 466 mol/s for CO2, H2O, N2 and Excess O2 respectively. The required rate of heat transfer in the preheater is calculated as \(3.67 x 10^6\) joules/s. The temperature at which the stack gas leaves the preheater requires iterative calculations and is dependent on the specific heat capacity of the stack gas. For the actual feed rate of 350 SCMH, the heat duty is calculated to be \(3.58 x 10^6\) joules/s.

Step by step solution

01

Determine the molar flow rate of air

We know that for every one mole of methane, we need \(2(\mathrm{O}_{2}) + 2(3.76\mathrm{N}_{2})\) of air for complete combustion. With \(20.0\%\) excess air, this comes out to be \((1+0.20)(2)(4.76)=11.424\) mol air/mol CH4. For ethane, we would need \((1+0.20)(7/2)(4.76)=16.84\) mol air/mol C2H6. Using the 95 mole% as methane and the balance as ethane in the natural gas, the molar flow rate of air can be calculated as \(0.95(11.424) + 0.05(16.84) = 11.65\) mol air/mol feed. Therefore, for 100 moles of feed, we would need \(11.65(100) = 1165\) mol/s of air.
02

Determine the molar flow rate and composition of the stack gas

For each mole of methane burned, we get one mole of CO2 and two moles of water. Similarly for ethane, we get two moles of CO2 and three moles of water. We would also have nitrogen and the excess oxygen from air in the stack gas. The molar flow rate and composition of stack gas can be calculated as follows: CO2 - \(0.95(1) + 0.05(2) = 1.05\) mol/mol feed, H2O - \(0.95(2) + 0.05(3) = 2.05\) mol/mol feed, N2 - \(11.65(3.76)=43.82\) mol/mol feed, excess O2 - \(11.65(0.20)*2 = 4.66\) mol/mol feed. For 100 moles of feed, the molar flow rates would be 105, 205, 4382 and 466 mol/s for CO2, H2O, N2 and Excess O2 respectively.
03

Calculate the required rate of heat transfer in the preheater (\(\dot{Q}\)

In the heat exchanger, the heat transferred from the stack gas is used to heat the incoming air from \(20^{\circ}C\) to \(245^{\circ}C\). This can be calculated using the energy balance on the air: \(\dot{Q} = \dot{n}_{air} * C_{p, air} * (T_{out} - T_{in}) = 1165 \, mol/s * 29.1 \, J/mol.K * (245 - 20) = 3.67 x 10^6\) joules/s
04

Compute the temperature at which the stack gas leaves the preheater

The temperature at which the stack gas leaves the preheater can be obtained by performing an energy balance on the stack gas. The heat lost by the stack gas in the preheater (\( \dot{Q} \)) is used to heat the incoming air. Using the average specific heat capacities for the stack gas components, the exit temperature can be obtained from the equation: \( \dot{Q} = \dot{n}_{sg} * C_{p, sg} * (T_{in} - T_{out}) \) Assuming that the stack gas enters the preheater at \(900^{\circ}C\), and rearranging, we find that \( T_{out} = T_{in} - (\dot{Q} / (\dot{n}_{sg} * C_{p, sg})) \) The method requires estimation of the composition of the stack gas and involves iterative calculations because the specific heat capacity depends on temperature.
05

Calculate \(\dot{Q}\) for the actual feed rate in part (b)

In part (b), the actual feed rate of natural gas is given as 350 SCMH. To find the value of \(\dot{Q}\) for this feed rate, we simply scale up the flow rates and the heat duty from part (a) according to the ratio of the actual and basis gas feed rates. In this case, the ratio would be 350/359.5 (converting 100 mol/s to SCMH), leading to a new heat duty of \(\dot{Q}_{new} = 3.67 x 10^6 (350/359.5) = 3.58 x 10^6\) joules/s. Note that all flow rates and the heat duty from part (a) would need to be scaled up correspondingly.

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Key Concepts

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

Excess Air Calculation
Calculating the excess air required for combustion is essential for ensuring complete burning of the fuel while preventing energy waste. Excess air refers to the amount of air supplied beyond what is theoretically necessary to achieve full combustion of a given fuel. Here, we consider the stoichiometric amounts of air that react with methane (CH4) and ethane (C2H6) and add 20% more air to ensure complete combustion. This ensures no unburned hydrocarbons are present in the stack gas, leading to cleaner emissions.

For methane, the stoichiometric air requirement is calculated using the equation for complete combustion: \[ ext{CH}_4 + 2 ext{O}_2 + 2(3.76 ext{N}_2) \] With 20% excess air, the total air required is: 11.424 moles of air per mole of methane. Similarly, for ethane: \[ ext{C}_2 ext{H}_6 + rac{7}{2} ext{O}_2 + rac{7}{2}(3.76 ext{N}_2) \] Again, considering 20% excess, we need 16.84 moles of air per mole of ethane.

Using this calculation method ensures that all carbons in the fuel are oxidized to carbon dioxide, maximizing energy output and minimizing pollutants. The overall air flow required is then derived by taking a weighted average based on the proportion of methane and ethane in the fuel mix.
Stack Gas Composition
Understanding the composition of stack gas is critical for analyzing the effectiveness of combustion and emission control. Stack gas is the mixture of gases released post-combustion from a furnace or a stack. It consists primarily of products from the combustion process, including carbon dioxide (CO2), water vapor (H2O), nitrogen (N2), and any unreacted oxygen.

In our exercise, since there's complete combustion with no emissions of carbon monoxide or unburned hydrocarbons, the composition is straightforward. Every mole of methane burned produces one mole of CO2 and two moles of H2O. Comparably, each mole of ethane generates two moles of CO2 and three moles of H2O.
  • CO2: Derived from both methane and ethane contributing 1.05 moles per mole of fuel.
  • H2O: Totaling 2.05 moles per mole of fuel.
  • N2: Primarily from the air, contributing 43.82 moles per mole of fuel.
  • Excess O2: Coming from the unreacted portion of supplied air, amounts to 4.66 moles per mole of fuel.
These figures reflect the output when 100 moles of natural gas with the specified composition are burnt, facilitating an understanding of the emission properties and aiding in regulatory compliance.
Heat Transfer Calculation
Heat transfer calculations help in understanding how energy is distributed during the combustion process, particularly how much heat is recovered and utilized by the heat exchanger. In this exercise, the heat exchanger preheats air entering the furnace using the exhaust gases. This improves efficiency by reducing the amount of combustion energy needed to heat up incoming air.

To determine the amount of heat transferred, or \( \dot{Q} \), an energy balance on the air is performed. The formula used is:
\[ \dot{Q} = \dot{n}_{air} \cdot C_{p, air} \cdot (T_{out} - T_{in}) \]
where \( \dot{n}_{air} \) is the air flow rate, \( C_{p, air} \) is the specific heat capacity, and \( T_{out} \) and \( T_{in} \) are the outlet and inlet temperatures, respectively. For our calculation, this results in \( \dot{Q} = 3.67 \times 10^6 \) J/s, illustrating the energy saved.
  • Energy balance assesses necessary adjustments to stack gas temperature exiting the preheater.
  • Exit temperature of stack gas is critical for optimizing preheating efficiency and minimizing losses.
These calculations ensure that the maximum possible energy is recaptured, enhancing the system's overall thermal efficiency and reducing operational costs.

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Most popular questions from this chapter

Saturated steam at \(300^{\circ} \mathrm{C}\) is used to heat a countercurrently flowing stream of methanol vapor from \(65^{\circ} \mathrm{C}\) to \(260^{\circ} \mathrm{C}\) in an adiabatic heat exchanger. The flow rate of the methanol is 6500 standard liters per minute, and the steam condenses and leaves the heat exchanger as liquid water at \(90^{\circ} \mathrm{C}.\) (a) Calculate the required flow rate of the entering steam in \(\mathrm{m}^{3} / \mathrm{min}\). (b) Calculate the rate of heat transfer from the water to the methanol ( \(\mathrm{kW}\) ). (c) Suppose the outlet temperature of the methanol is measured and found to be \(240^{\circ} \mathrm{C}\) instead of the specified value of \(260^{\circ} \mathrm{C}\). List five possible realistic explanations for the \(20^{\circ} \mathrm{C}\) difference. 7 An adiabatic heat exchanger is one for which no heat is exchanged with the surroundings. All of the heat lost by the hot stream is transferred to the cold stream.

A sheet of cellulose acetate film containing 5.00 wt\% liquid acetone enters an adiabatic dryer where \(90 \%\) of the acetone evaporates into a stream of dry air flowing over the film. The film enters the dryer at \(T_{\mathrm{f} 1}=35^{\circ} \mathrm{C}\) and leaves at \(T_{\mathrm{f} 2}\left(^{\circ} \mathrm{C}\right) .\) The air enters the dryer at \(T_{\mathrm{al}}\left(^{\circ} \mathrm{C}\right)\) and 1.01 atm and exits the dryer at \(T_{\mathrm{a} 2}=49^{\circ} \mathrm{C}\) and 1 atm with a relative saturation of \(40 \% . C_{p}\) may be taken to be \(1.33 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) for dry film and \(0.129 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\) for liquid acetone. Make a reasonable assumption regarding the heat capacity of dry air. The heat of vaporization of acetone may be considered independent of temperature. Take a basis of \(100 \mathrm{kg}\) film fed to the dryer for the requested calculations. (a) Estimate the feed ratio [liters dry air (STP)/kg dry film]. (b) Derive an expression for \(T_{\mathrm{al}}\) in terms of the film temperature change, \(\left(T_{\mathrm{f} 2}-35\right),\) and use it to answer Parts (c) and (d). (c) Calculate the film temperature change if the inlet air temperature is \(120^{\circ} \mathrm{C}\). (d) Calculate the required value of \(T_{\mathrm{al}}\) if the film temperature falls to \(34^{\circ} \mathrm{C},\) and the value if it rises to \(36^{\circ} \mathrm{C}.\) (e) If you solved Parts (c) and (d) correctly, you found that even though the air temperature is consistently higher than the film temperature in the dryer, so that heat is always transferred from the air to the film, the film temperature can drop from the inlet to the outlet. How is this possible?

A liquid stream containing 50.0 mole \(\%\) benzene and the balance toluene at \(25^{\circ} \mathrm{C}\) is fed to a continuous single-stage evaporator at a rate of \(1320 \mathrm{mol} / \mathrm{s}\). The liquid and vapor streams leaving the evaporator are both at \(95.0^{\circ} \mathrm{C}\). The liquid contains 42.5 mole \(\%\) benzene and the vapor contains 73.5 mole\% benzene. (a) Calculate the heating requirement for this process in \(\mathrm{kW}\). (b) Using Raoult's law (Section 6.4b) to describe the equilibrium between the vapor and liquid outlet streams, determine whether or not the given benzene analyses are consistent with each other. If they are, calculate the pressure (torr) at which the evaporator must be operating; if they are not, give several possible explanations for the inconsistency.

Polyvinylpyrrolidone (PVP) is a polymer product used as a binding agent in pharmaceutical applications as well as in personal-care items such as hairspray. In the manufacture of \(\mathrm{PVP}\), a spray-drying process is used to collect solid PVP from an aqueous suspension, as shown in the flowchart on the next page. A liquid solution containing 65 wt\% \(\mathrm{PVP}\) and the balance water at \(25^{\circ} \mathrm{C}\) is pumped through an atomizing nozzle at a rate of \(1500 \mathrm{kg} / \mathrm{h}\) into a stream of preheated air flowing at a rate of \(1.57 \times 10^{4}\) SCMH. The water evaporates into the stream of hot air and the solid PVP particles are suspended in the humidified air. Downstream, the particles are separated from the air with a filter and collected. The process is designed so that the exiting solid product and humid air are in thermal equilibrium with each other at \(110^{\circ} \mathrm{C}\). For convenience, the spray-drying and solid- separation processes are shown as one unit that may be considered adiabatic. (a) Draw and completely label the process flow diagram and perform a degree- of-freedom analysis. (b) Calculate the required temperature of the inlet air, \(T_{0}\), and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) and relative humidity of the exiting air. Assume that the polymer has a heat capacity per unit mass one third that of liquid water, and only use the first two terms of the polynomial heat-capacity formula for air in Table B.2. (c) Why do you think the polymer solution is put through an atomizing nozzle, which converts it to a mist of tiny droplets, rather than being sprayed through a much less costly nozzle of the type commonly found in showers? (d) Due to a design flaw, the polymer solution does not remain in the dryer long enough for all the water to evaporate, so the solid product emerging from the separator is a wet powder. How will this change the values of the outlet temperatures of the emerging gas and powder and the volumetric flow rate and relative humidity of the emerging gas (increase, decrease, can't tell without doing the calculations)? Explain your answers.

An aqueous slurry at \(30^{\circ} \mathrm{C}\) containing \(20.0 \mathrm{wt} \%\) solids is fed to an evaporator in which enough water is vaporized at 1 atm to produce a product slurry containing 35.0 wt\% solids. Heat is supplied to the evaporator by feeding saturated steam at 2.6 bar absolute into a coil immersed in the liquid. The steam condenses in the coil, and the slurry boils at the normal boiling point of pure water. The heat capacity of the solids may be taken to be half that of liquid water. (a) Calculate the required steam feed rate ( \(\mathrm{kg} / \mathrm{h}\) ) for a slurry feed rate of \(1.00 \times 10^{3} \mathrm{kg} / \mathrm{h}\). (b) Vapor recompression is often used in the operation of an evaporator. Suppose that the vapor (steam) generated in the evaporator described above is compressed to 2.6 bar and simultaneously heated to the saturation temperature at 2.6 bar, so that no condensation occurs. The compressed steam and additional saturated steam at 2.6 bar are then fed to the evaporator coil, in which isobaric condensation occurs. How much additional steam is required? (c) What more would you need to know to determine whether or not vapor recompression is economically advantageous in this process?
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