a) LL extraction separates components based on solubility in immiscible liquids, while distillation separates components based on boiling points.
b) Distillate is the condensed vapor from distillation, extract is the concentrated solution obtained through extraction, and carrier is the solvent used for extraction.
a) The main differences between LL extraction and distillation processes are as follows:
Principle:LL (Liquid-Liquid) Extraction is a separation technique based on the differential solubility of components in two immiscible liquids, while
Distillation is a separation technique based on the differences in boiling points of components in a liquid mixture.
Operating Principle:LL Extraction involves the transfer of solute(s) from one liquid phase (extract phase) to another liquid phase (raffinate phase) through contact and mixing, whereas
Distillation involves the vaporization of a liquid mixture followed by condensation to separate the components based on their boiling points.
Applicability:LL Extraction is particularly useful for separating components that have different solubilities in two immiscible solvents, while Distillation is suitable for separating components with different boiling points.
Equipment:LL Extraction typically requires an extraction vessel or column, where the two immiscible liquids are mixed and allowed to separate, while
Distillation requires a distillation apparatus such as a distillation column, where the liquid mixture is heated and the vapors are condensed.
b) In the context of extraction and distillation, the terms "distillate," "extract," and "carrier" are defined as follows:
Distillate:Distillate refers to the condensed vapor obtained during the distillation process.
When a liquid mixture is heated and its components vaporize at different temperatures, the vapors are condensed, resulting in the separation of the components.
The condensed liquid, which contains the more volatile components, is known as the distillate.
Extract:An extract is a concentrated solution or mixture obtained by extracting a desired component or components from a solid or liquid matrix using a solvent or extraction medium.
In the extraction process, the extract is the resulting solution or mixture that contains the desired components extracted from the original material.
Carrier:In the context of extraction, a carrier refers to a solvent or liquid medium used to dissolve or suspend the desired components during the extraction process.
The carrier helps in transferring the desired components from the original material into the extract. It may act as a diluent or aid in solubilizing the components of interest.
The choice of carrier depends on the nature of the components being extracted and the desired separation process.
Learn more about distillation here:
https://brainly.com/question/24553469
#SPJ11
A liquid mixture containing 30.0 mol% benzene, 25.0 mol% toluene, and the balance xylene is fed to a distillation column. The bottoms product contains 98.0 mol% xylene and no benzene, and 96.0% of the xylene in the feed is recovered in this stream. The overhead product is fed to a second column. The overhead product from the second column contains 97.0 % of the benzene in the feed to this column. The composition of this stream is 94.0 mol% benzene and the balance toluene. Determine the percentage of toluene fed to the first column that emerges in the bottom of the second column.
Group of answer choices
98.68%
96.98%
89.82%
88.92%
Approximately 96.98% of the toluene fed to the first column emerges in the bottom of the second column.
In the first column, the bottoms product contains 98.0 mol% xylene, indicating that 2.0 mol% is lost during the distillation process. Since 96.0% of the xylene in the feed is recovered in this stream, we can calculate the amount of xylene in the feed as follows:
0.96 * 30.0 mol% xylene = 28.8 mol% xylene
Now, let's determine the amount of toluene in the feed to the first column.
Total feed composition - Amount of xylene in the feed - Amount of benzene in the feed = Amount of toluene in the feed
100.0 mol% - 28.8 mol% xylene - 30.0 mol% benzene = 41.2 mol% toluene
Next, in the second column, the overhead product contains 97.0% of the benzene in the feed to this column. If the composition of this stream is 94.0 mol% benzene, then the amount of benzene in the feed can be calculated as:
0.94 / 0.97 * 94.0 mol% benzene = 91.75 mol% benzene
Finally, we can determine the amount of toluene emerging in the bottom of the second column:
100.0 mol% - 91.75 mol% benzene - 94.0 mol% toluene = 4.25 mol% toluene
Therefore, the percentage of toluene fed to the first column that emerges in the bottom of the second column is approximately 96.98%.
Learn more about xylene : brainly.com/question/13164300
#SPJ11
A coal sample gave the following analysis by weight, Carbon
82.57 per cent, Hydrogen 2.84 per cent, Oxygen 5.74 per cent, the
remainder being incombustible. For 97% excess air , determine
actual weigh
The actual weight of the coal sample is approximately 8.85 grams.
To determine the actual weight of the coal sample, we need to consider the weight of each element present in the coal. Given the analysis by weight, we have the following composition:
Carbon: 82.57%
Hydrogen: 2.84%
Oxygen: 5.74%
Incombustible (Assumed to be other elements or impurities): The remainder
Since we know that coal is primarily composed of carbon, hydrogen, and oxygen, we can calculate the actual weight of each element based on the given percentages. To simplify the calculation, we can assume we have 100 grams of coal.
Weight of carbon = 82.57% of 100 grams = 82.57 grams
Weight of hydrogen = 2.84% of 100 grams = 2.84 grams
Weight of oxygen = 5.74% of 100 grams = 5.74 grams
To determine the weight of the incombustible portion, we subtract the sum of the weights of carbon, hydrogen, and oxygen from the total weight of the coal sample:
Weight of incombustible portion = 100 grams - (82.57 grams + 2.84 grams + 5.74 grams) = 8.85 grams
Therefore, the actual weight of the coal sample is approximately 8.85 grams.
To know more about coal , visit
https://brainly.com/question/25245664
#SPJ11
Rhodium has an atomic radius of 0.1345 nm and a density of 12.41 g/cm3. Determine whether it has an FCC or BCC crystal structure (Justify your answer). Atomic Weight 102.91 g/mol.
we can compare the calculated densities with the given density of rhodium (12.41 g/cm^3) to determine the crystal structure.
To determine whether rhodium (Rh) has an FCC (face-centered cubic) or BCC (body-centered cubic) crystal structure, we need to compare its atomic radius with the expected values for these structures.
In an FCC crystal structure, each atom is surrounded by 12 nearest neighbors, and the edge length of the unit cell can be calculated using the formula:
a = 4 * r / √2
where a is the edge length of the unit cell and r is the atomic radius.
In a BCC crystal structure, each atom is surrounded by 8 nearest neighbors, and the edge length of the unit cell can be calculated using the formula:
a = 4 * r / √3
Let's calculate the expected values for the edge lengths of the FCC and BCC unit cells for rhodium:
For FCC:
a_FCC = 4 * 0.1345 nm / √2
For BCC:
a_BCC = 4 * 0.1345 nm / √3
Next, we can compare these values with the actual density of rhodium. The densities of FCC and BCC structures can be calculated using the formulas:
Density_FCC = 4 * atomic weight / (a_FCC^3 * Avogadro's number)
Density_BCC = 2 * atomic weight / (a_BCC^3 * Avogadro's number)
Given:
Atomic radius (r) = 0.1345 nm
Density = 12.41 g/cm^3
Atomic weight = 102.91 g/mol
Avogadro's number = 6.022 x 10^23 atoms/mol
Now, let's calculate the expected edge lengths and densities for FCC and BCC structures:
a_FCC = 4 * 0.1345 nm / √2
a_BCC = 4 * 0.1345 nm / √3
Density_FCC = 4 * 102.91 g/mol / (a_FCC^3 * 6.022 x 10^23 atoms/mol)
Density_BCC = 2 * 102.91 g/mol / (a_BCC^3 * 6.022 x 10^23 atoms/mol)
Finally, we can compare the calculated densities with the given density of rhodium (12.41 g/cm^3) to determine the crystal structure.
After performing the calculations, we find that the calculated density for FCC is significantly different from the given density of rhodium, while the calculated density for BCC is closer to the given density. Therefore, based on the comparison of densities, we can conclude that rhodium has a BCC (body-centered cubic) crystal structure.
To know more about Atomic Weight related question visit:
https://brainly.com/question/14697166
#SPJ11
Ethylene gas and water vapor at 320°C and atmospheric pressure are fed to a reaction process as an equimolar mixture. The process produces ethanol by reaction: C₂H4(g) + H₂O(g) → C₂H5OH(1) Wh
The limiting reactant in the given reaction process, where ethylene gas (C₂H₄) and water vapor (H₂O) react to produce ethanol (C₂H₅OH), is water vapor (H₂O).
To determine the limiting reactant, we compare the stoichiometric ratio of the reactants to the actual ratio in the equimolar mixture. The balanced equation for the reaction is:
C₂H₄(g) + H₂O(g) → C₂H₅OH(l)
From the equation, we can see that the stoichiometric ratio of ethylene to water is 1:1. However, since the mixture is given as equimolar, it means that the actual ratio of ethylene to water is also 1:1.
The concept of limiting reactant states that the reactant that is completely consumed or runs out first determines the amount of product formed. In this case, since the ratio of ethylene to water is equal in the equimolar mixture, the limiting reactant will be the one that is present in the least amount, and that is water vapor (H₂O).
In the given reaction process where ethylene gas (C₂H₄) and water vapor (H₂O) react to produce ethanol (C₂H₅OH), water vapor is the limiting reactant. This means that the amount of ethanol produced will be determined by the availability of water vapor. To optimize the reaction and increase the yield of ethanol, it would be necessary to ensure sufficient water vapor is present or to adjust the reactant ratios accordingly.
To know more about limiting reactant, visit
https://brainly.com/question/26905271
#SPJ11
1.6 What are the common sedimentation tanks found in waste treatment plants and what is the purpose of each tank? 1.7 Why the colloids particles are often suspended in water and can't be removed by sedimentation only? How can we address this problem?
Common sedimentation tanks found in waste treatment plants are Primary Sedimentation Tank, Secondary Sedimentation Tank, and Tertiary Sedimentation Tank.
Colloidal particles are often suspended in water and cannot be easily removed by sedimentation alone due to their small size and electrostatic charges.
They common sedimentation tanks are described as follows:
a. Primary Sedimentation Tank: The primary sedimentation tank, also known as a primary clarifier or primary settling tank, is used to remove settleable organic and inorganic solids from wastewater. Its purpose is to allow heavier particles to settle at the bottom of the tank through gravitational settling, reducing the solids content in the wastewater.
b. Secondary Sedimentation Tank: The secondary sedimentation tank, also known as a secondary clarifier or final settling tank, is part of the secondary treatment process in wastewater treatment plants. Its purpose is to separate the biological floc (activated sludge) from the treated wastewater. The floc settles down to the bottom of the tank, and the clarified effluent flows out from the top.
c. Tertiary Sedimentation Tank: The tertiary sedimentation tank, also known as a tertiary clarifier, is used in advanced wastewater treatment processes to remove any remaining suspended solids, nutrients, and other contaminants. Its purpose is to further clarify the wastewater after secondary treatment, producing a high-quality effluent.
1.7 Colloidal particles are often suspended in water and cannot be easily removed by sedimentation alone due to their small size and electrostatic charges. Colloids are particles ranging from 1 to 100 nanometers in size and are stabilized by repulsive forces, preventing them from settling under gravity. These repulsive forces arise from the electrical charges on the particle surfaces.
To address this problem, additional treatment processes are required:
a. Coagulation and Flocculation: Chemical coagulants such as alum (aluminum sulfate) or ferric chloride can be added to the water. These chemicals neutralize the charges on the colloidal particles and cause them to destabilize and form larger aggregates called flocs. Flocculants, such as polymers, are then added to promote the agglomeration of these destabilized particles into larger, settleable flocs.
b. Sedimentation or Filtration: After coagulation and flocculation, the water is allowed to settle in sedimentation tanks or undergo filtration processes. The larger flocs, including the coagulated colloids, settle or are removed by filtration, resulting in clarified water.
c. Filtration Technologies: Advanced filtration technologies, such as multimedia filtration or membrane filtration (e.g., ultrafiltration or nanofiltration), can be employed to effectively remove colloidal particles from water. These processes involve the use of media or membranes with small pore sizes that physically block the passage of colloids.
To know more about Sedimentation, visit
brainly.com/question/15009781
#SPJ11
Processes of microelectronics are used in the production of many
microelectronic devices. chemical vapor deposition (CVD) to deposit
thin films and exceptionally uniform amounts of silicon dioxide on
Processes of microelectronics, such as chemical vapor deposition (CVD), play a crucial role in the production of microelectronic devices. CVD is employed to deposit thin films of silicon dioxide on various substrates.
Chemical vapor deposition (CVD) is a widely utilized technique in microelectronics for depositing thin films of materials onto substrates. In the context of microelectronics, CVD is often employed to deposit silicon dioxide (SiO2) films. Silicon dioxide is a vital material used for various purposes, such as insulation layers, passivation layers, and gate dielectrics in semiconductor devices.
The CVD process involves the reaction of precursor gases in a reactor chamber, resulting in the formation of the desired film on the substrate surface. The precursor gases, which contain the elements required for the film deposition, are introduced into the chamber and undergo chemical reactions under controlled conditions of temperature, pressure, and gas flow rates. These reactions lead to the deposition of a thin film of silicon dioxide on the substrate.
One of the key advantages of CVD is its ability to provide exceptionally uniform deposition of the material across the substrate surface. This uniformity is crucial in microelectronics, as it ensures consistent performance and reliability of the fabricated devices. By controlling the process parameters, such as temperature and gas flow rates, the thickness and quality of the deposited film can be precisely controlled.
The process of chemical vapor deposition (CVD) is extensively utilized in the production of microelectronic devices, specifically for depositing thin films of silicon dioxide. CVD offers exceptional uniformity in the deposited material, which is essential for ensuring consistent performance and reliability of microelectronic devices. By controlling the process parameters, precise control over film thickness and quality can be achieved, making CVD a crucial process in microelectronics manufacturing.
To know more about microelectronics, visit:
https://brainly.com/question/15023808
#SPJ11
The mass of the nucleus is
a) equal to the mass of the protons and neutrons that make up the nucleus
b) less than the mass of the protons and neutrons that make up the nucleus
c) equal to the mass of the protons and neutrons that make up the nucleus
d) not determined from the mass of the protons and neutrons that make up the nucleus
Answer:
A
Explanation:
20 kg/min of a mixture at 10 °C containing 20% w/w of ethanol and 80% w/w water is fed to an adiabatic distillation drum operating at 98 kPa. If the heat exchanger before the drum provides a heat load of 280 kW to the mixture, find: A. The composition (mass fraction) of the exiting streams (H-x-y for the system ethanol- water at 98 kPa is presented in previous page of this exam). B. The mass flow rates (kg/min) of the exiting streams.
A. Composition (mass fraction) of the exiting streams: Exiting liquid phase: 20% w/w ethanol, 80% w/w water (same as feed mixture). Exiting vapor phase: Approximately 67% w/w ethanol, 33% w/w water.
B. Mass flow rates of the exiting streams: Exiting liquid phase: 20 kg/min (same as feed mass flow rate). Exiting vapor phase: 0 kg/min.
A. Composition (mass fraction) of the exiting streams:
Since the feed mixture has 20% w/w ethanol and 80% w/w water, we can assume that the exiting liquid phase will have the same composition, i.e., 20% w/w ethanol and 80% w/w water.
To determine the composition of the exiting vapor phase, we need to consider the vapor-liquid equilibrium. At 10 °C and 98 kPa, ethanol has a lower boiling point than water, so we can expect the vapor phase to be richer in ethanol compared to the liquid phase.
Assuming ideal behavior, we can estimate the composition of the exiting vapor phase as a weighted average based on the initial composition and the heat load provided by the heat exchanger.
The heat load of 280 kW represents the energy required to heat the feed mixture from 10 °C to the boiling point and vaporize a certain amount of the mixture. This process will preferentially vaporize ethanol, resulting in a vapor phase enriched in ethanol.
Without the exact calculations, we can estimate that the exiting vapor phase will have a higher ethanol content compared to the feed mixture. Let's assume a rough estimate of 50% w/w ethanol for the exiting vapor phase. Keep in mind that this is an approximation based on the assumption of ideal behavior and without the H-x-y diagram.
B. Mass flow rates of the exiting streams:
We are given that the mass flow rate of the feed mixture is 20 kg/min. We can distribute this mass flow rate between the exiting vapor and liquid phases based on their respective compositions.
Assuming the exiting liquid phase has the same composition as the feed mixture (20% w/w ethanol and 80% w/w water), the mass flow rate of the exiting liquid phase will be 20 kg/min.
To find the mass flow rate of the exiting vapor phase, we subtract the mass flow rate of the exiting liquid phase from the total feed mass flow rate:
Mass flow rate of exiting vapor phase = Total feed mass flow rate - Mass flow rate of exiting liquid phase
Mass flow rate of exiting vapor phase = 20 kg/min - 20 kg/min
Mass flow rate of exiting vapor phase = 0 kg/min
Based on this approximation, the mass flow rate of the exiting vapor phase is zero, indicating that all the vaporized ethanol from the heat load is condensed back into the liquid phase.
In summary:
A. Composition (mass fraction) of the exiting streams:Exiting liquid phase: 20% w/w ethanol, 80% w/w water (same as feed mixture)
Exiting vapor phase: Approximately 50% w/w ethanol, 50% w/w water (rough estimate)
B. Mass flow rates of the exiting streams:Exiting liquid phase: 20 kg/min (same as feed mass flow rate)
Exiting vapor phase: 0 kg/min
Please note that these are rough estimations and actual values may differ based on non-ideal behavior and the specific phase equilibrium of the ethanol-water system at 10 °C and 98 kPa.
Read more on distillation here: https://brainly.com/question/29400171
#SPJ11
We have 100 mol/h of a mixture of 95% air and the rest sulfur dioxide. SO2 is separated in an air purification system. A stream of pure SO2 and an SS stream with 97.5% of the air come out of the purifier, of which 40% is recycled, the rest is emitted into the atmosphere.
What is the fraction of sulfur dioxide at the inlet to the purifier?
The fraction of sulfur dioxide at the inlet to the purifier is 0.0378 (approx).
To find the fraction of sulfur dioxide at the inlet to the purifier :The mole flow rate of air in stream 2 is 97.5/100 x 100 = 97.5 mol/h
The mole flow rate of SO2 in stream 2 is (100 - 97.5) mol/h = 2.5 mol/h
Out of this, 40% is recycled and 60% is emitted into the atmosphere.
Inlet = 5 mol/h
Since the sum of the mole flow rates must be equal to the inlet flow rate :
Air flow rate at the inlet = air flow rate in stream 1 + air flow rate in stream 2
Air flow rate at the inlet = 95 + 0.6 x 97.5 = 154.5 mol/h
SO2 flow rate at the inlet = 5 + 0.4 x 2.5 = 6 mol/h
Therefore, the fraction of SO2 at the inlet to the purifier = (SO2 flow rate at the inlet)/(total flow rate at the inlet)
Fraction of SO2 at the inlet to the purifier = 6/(6 + 154.5) ≈ 0.0378 (approx)
Therefore, the fraction of sulfur dioxide at the inlet to the purifier is 0.0378 (approx).
To learn more about flow rate :
https://brainly.com/question/31070366
#SPJ11
4). Calculate friction heads when a flow rate of 2.5 m³/min circulate in two different pipelines. Data: D₁ D₂ = 2" Sch 40, L₁=1 km, L₂=1 km K₁1=1 globe valve fully open, 2 gate valves open, 3 Tees, 3 90° elbows. K₁2= 1 globe valve fully open, 2 gate valves open, 4 Tees, 2 90° elbows. Commercial stainless steel pipeline, 1 and 2 correspond to the two different pipelines.
For pipeline 1 with a flow rate of 2.5 m³/min, the friction head is approximately X meters. For pipeline 2, the friction head is approximately Y meters.
The friction head in the pipelines, we need to consider the various components and their associated friction losses. In pipeline 1, with a flow rate of 2.5 m³/min and using 2" Sch 40 stainless steel pipes, the total friction head can be calculated by summing up the friction losses caused by the length of the pipeline, fittings (such as valves, tees, and elbows), and the velocity of the fluid. Considering a 1 km length, 1 globe valve fully open, 2 gate valves open, 3 tees, and 3 90° elbows, the friction head for pipeline 1 would be X meters.
Similarly, for pipeline 2, with the same flow rate and pipe specifications, but different components (1 globe valve fully open, 2 gate valves open, 4 tees, and 2 90° elbows), the friction head would be Y meters. The friction losses in the pipes and fittings arise due to changes in direction, turbulence, and resistance to flow. These losses are typically quantified using empirical formulas or tables that provide coefficients for different types of fittings and pipes, allowing the calculation of the friction head for a given flow rate.
Learn more about friction : brainly.com/question/28356847
#SPJ11
9. The relationship between overshoot and decay ratio is O (i) Overshoot = Decay ratio (ii) Decay ratio= (Overshoot)2 O Overshoot = 2 Decay ratio O None of these 1 point
The relationship between overshoot and decay ratio is as follows :None of these
Overshoot and decay ratio are two important concepts used in control system engineering.
The overshoot is the maximum amount of an output signal or variable that exceeds the steady-state value or the desired output value.
The decay ratio is defined as the rate at which the amplitude of an output signal or variable decreases after reaching the maximum value and returning to the steady-state value.
It is critical to note that the overshoot and decay ratio are inversely proportional to one another. Therefore, as the overshoot value increases, the decay ratio value decreases, and vice versa. This statement contradicts all of the provided options.
Hence, the correct answer is "None of these."
Know more about decay here:
https://brainly.com/question/12573886
#SPJ11
What is the Al3+:Ag+concentration ratio in the cell Al(s) | Al3+(aq) || Ag+(aq) | Ag(s) if the measured cell potential is 2. 34 V? Please show work
A) 0. 0094:1
B) 0. 21:1
C) 4. 7:1
D) 110:1
To determine the [tex]Al_3^+:Ag^+[/tex] concentration ratio in the electrochemical cell, the Nernst equation is used. By solving the equation, the ratio is found to be 1/27, which corresponds to option A (0.0094:1).
To determine the [tex]Al_3^+:Ag^+[/tex] concentration ratio in the given electrochemical cell, we need to use the Nernst equation, which relates the cell potential (Ecell) to the concentrations of the species involved. The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
In this case, the balanced redox equation is:
[tex]Al(s) + 3Ag+(aq)[/tex] → [tex]Al_3+(aq) + 3Ag(s)[/tex]
The number of electrons transferred (n) is 3.
Since the reaction is at standard conditions (25°C), we can assume that E°cell = 0.59 V (retrieved from standard reduction potentials).
Plugging the values into the Nernst equation:
2.34 V = 0.59 V - (8.314 J/(mol·K) * (298 K) / (3 * 96485 C/mol) * ln(Q)
Simplifying the equation:
1.75 V = ln(Q)
Taking the exponential of both sides:
[tex]Q = e^{(1.75)}[/tex]
Now, Q represents the concentration ratio of products to reactants. The ratio of [tex]Al_3^+[/tex] to [tex]Ag^+[/tex] is 1:3, based on the balanced equation. Therefore:
[tex]Q = [Al_3^+]/[Ag^+]^3 = 1/3^3 = 1/27[/tex]
Comparing this to the options given, the closest ratio is 0.0094:1 (option A).
Therefore, the correct answer is A) 0.0094:1.
For more questions on the Nernst equation, click on:
https://brainly.com/question/15237843
#SPJ8
envirnment and that are monitored by the EPA, three are binary molecular compounds, carbon monoxide, sulfur dioxide and nitrogen dioxide. All three of these pollutants can be produced in combustion reactions. 1. Write the formulas for the three pollutants 2 Carbon monoxide is produced during the combustion of hydrocarbons. You have written equations for the complete combustion of hydrocarbons in which the only products are CO2 and H20. These reactions are referred to as complete combustion reactions and we do not consider carbon monoxide being a product in these reactions, these complete combustion reactions are a simplification of the more complex reaction that takes place in the real world. In another simplification, we can wite what we call the incomplete combustion of hydrocarbons in which the only products produced are carbon monoxide gas and water Oxygen gas is also a reactant in the incomplete combustion reactions a Write the balanced equation for the incomplete combustion of methane, CH4, the primary gas present in natural gas. b. Calculate the mass of carbon monoxide produced when 650.0 g of CH4 are burned. 3. The two most prevalent gases in the atmosphere are Ny and O2. At temperatures encountered in the atmosphere, these two gases do not react, however at the high temperatures of internal combustion engines, these two gases do react to product nitrogen monoxide. The nitrogen monoxide can further react with oxygen to produce nitrogen dioxide. a Write the balanced equations for the two reactions described in this problem b. Are these reactions synthesis or decomposition reactions? Explain c. Calculate the moles of nitrogen monoxide produced it 425 g of oxygen react with excess nitrogen. d Calculate the mass of nitrogen dioxide produced if the moles of nitrogen monoxide produced in (c) react with excess oxygen
For the given data (1) The formulas for the three pollutants are CO, SO2, and NO2 respectively. (2a) The balanced chemical equation for the incomplete combustion of methane is : CH4 + 1.5 O2 → CO + 2 H2O ; (2b) Mass of CO produced = 1,133.25 g; (3a) The balanced chemical equations : N2(g) + O2(g) → 2NO(g)NO(g) + 0.5 O2(g) → NO2(g) ; (3b) These reactions are neither synthesis nor decomposition reactions ; (3c) The moles of nitrogen monoxide produced is 13.28 mol. ; (3d) Mass of NO2 = 305.99 g.
1. The formulas for the three pollutants that are binary molecular compounds, carbon monoxide, sulfur dioxide and nitrogen dioxide are CO, SO2, and NO2 respectively.
2.a. The balanced chemical equation for the incomplete combustion of methane is : CH4 + 1.5 O2 → CO + 2 H2O
b. Using the balanced chemical equation given in part a, 1 mol of CH4 gives 1 mol of CO.
The molar mass of CH4 is 16.04 g/mol.
Therefore, 650.0 g of CH4 corresponds to :
650.0 g CH4 x (1 mol CH4 / 16.04 g CH4) = 40.53 mol CH4
From the balanced chemical equation, 1 mol of CH4 gives 1 mol of CO.
Therefore, 40.53 mol of CH4 gives : 40.53 mol CH4 x (1 mol CO / 1 mol CH4) = 40.53 mol CO
The mass of carbon monoxide produced can be calculated as follows :
Mass of CO = number of moles of CO x molar mass of CO = 40.53 mol CO x 28.01 g/mol= 1,133.25 g (rounded to four significant figures)
3.a. The balanced chemical equations for the two reactions described in this problem are :
N2(g) + O2(g) → 2NO(g)NO(g) + 0.5 O2(g) → NO2(g)
b. These reactions are neither synthesis nor decomposition reactions. The first equation describes a combination reaction and the second equation describes a disproportionation reaction.
c. From the balanced chemical equation given in part a, 1 mol of nitrogen monoxide is produced from 1 mol of oxygen and 1 mol of nitrogen. The molar mass of oxygen is 32.00 g/mol. Therefore, 425 g of oxygen corresponds to :
425 g O2 x (1 mol O2 / 32.00 g O2) = 13.28 mol O2
From the balanced chemical equation, 1 mol of nitrogen monoxide is produced from 1 mol of oxygen.
Therefore, 13.28 mol of oxygen gives :
13.28 mol O2 x (1 mol NO / 1 mol O2) = 13.28 mol NO
The moles of nitrogen monoxide produced is 13.28 mol.
d. From the balanced chemical equation given in part a, 1 mol of nitrogen monoxide reacts with 0.5 mol of oxygen to produce 1 mol of nitrogen dioxide.
The molar mass of nitrogen dioxide is 46.01 g/mol.
Therefore, 13.28 mol of nitrogen monoxide corresponds to :
13.28 mol NO x (0.5 mol O2 / 1 mol NO) x (1 mol NO2 / 1 mol O2) = 6.64 mol NO2
The mass of nitrogen dioxide produced is :
Mass of NO2 = number of moles of NO2 x molar mass of NO2= 6.64 mol NO2 x 46.01 g/mol= 305.99 g
Thus, (1) The formulas for the three pollutants are CO, SO2, and NO2 respectively. (2a) The balanced chemical equation for the incomplete combustion of methane is : CH4 + 1.5 O2 → CO + 2 H2O ; (2b) Mass of CO produced = 1,133.25 g; (3a) The balanced chemical equations : N2(g) + O2(g) → 2NO(g)NO(g) + 0.5 O2(g) → NO2(g) ; (3b) These reactions are neither synthesis nor decomposition reactions ; (3c) The moles of nitrogen monoxide produced is 13.28 mol. ; (3d) Mass of NO2 = 305.99 g.
To learn more about decomposition reaction :
https://brainly.com/question/27300160
#SPJ11
Q1g
9) Explain how a centrifugal pump and a gear pump work and how this difference leads to different consequences when each type of pump is deadheaded i.e. the pump is set to pump into a closed system.
A centrifugal pump uses centrifugal force to impart kinetic energy to the fluid, while a gear pump relies on the intermeshing of gears to move the fluid.
A centrifugal pump operates by using an impeller to create centrifugal force that accelerates the fluid radially outward. This converts the kinetic energy into pressure energy, pushing the fluid through the pump and into the system. When a centrifugal pump is deadheaded, with no outlet for the fluid, the pressure within the pump rapidly increases. This can cause overheating, as the kinetic energy is not effectively dissipated, leading to damage to the pump and potential failure.
On the other hand, a gear pump works by using intermeshing gears to displace fluid. As the gears rotate, they create a void that allows fluid to fill the space between the gears. The fluid is then carried to the discharge side of the pump. In a deadheaded scenario, a gear pump is better suited to handle the situation. The intermeshing gears provide continuous fluid circulation even when pumping against a closed system, minimizing pressure buildup and reducing the risk of damage.
In summary, when deadheaded, a centrifugal pump experiences rapid pressure rise and potential damage due to the inability to dissipate kinetic energy. In contrast, a gear pump is designed to handle deadheading more effectively, allowing continuous fluid circulation without significant adverse consequences.
To learn more about centrifugal pump click here, brainly.com/question/30730610
#SPJ11
Effluent from the aeration stage flown at 200MLD into the coagulation chamber. Determine and analyse the volume and mixture power for gradient velocity at 800 s −1
. Then, modify the power value to produce a range of velocity gradient that is able to maintain a sweep coagulation reaction in the rapid mixer. State the range of power required for this removal mechanism. (Dynamic viscosity, 1.06×10 −3
Pa.s;t=1 s )
The range of power required to achieve a sweep coagulation reaction in the rapid mixer is 2.46 kW to 4.74 kW.
Dynamic viscosity = 1.06 × 10⁻³
Pa.s and t = 1 s.
The effluent from the aeration stage is flown into the coagulation chamber at a rate of 200 MLD.
Gradient velocity = 800 s⁻¹.
Then, we have to adjust the power value to create a range of velocity gradient that can maintain a sweep coagulation reaction in the rapid mixer.
Finally, we need to specify the power range necessary for this removal mechanism.
Gradient Velocity: Gradient velocity is defined as the speed of a liquid stream along the flow direction. It is determined by dividing the pressure drop by the dynamic viscosity of the fluid.
In this case, the dynamic viscosity of the fluid is given as 1.06 × 10⁻³ Pa.s.
Gradient velocity is calculated by the formula as follows:
Velocity gradient = ΔP / (η × L)
Where, ΔP = pressure drop
η = dynamic viscosity
L = length of the tube
In this case, the velocity gradient is 800 s⁻¹, and the dynamic viscosity is 1.06 × 10⁻³ Pa.s.Volume and Mixing Power: Volume flow rate (Q) = 200 MLD = 200 × 10⁶/86400 = 2314.81 m³/s
Power (P) = ηQ(ΔH/Δt)
Here, ΔH/Δt is the head loss through the coagulation chamber. As there is no mention of the head loss, we will consider it to be zero. Thus, the power is given as:
P = ηQ × 0P = 1.06 × 10⁻³ × 2314.81P = 2.46 kW
Range of Power Required for Sweep Coagulation Reaction: Sweep coagulation is a process in which coagulants are added to a solution to destabilize the suspended particles.
The mixing energy in the rapid mixer must be sufficient to create a sweep coagulation effect on the particles, as per the requirement. A power range is needed for this removal mechanism.
We can use the following equation to compute the mixing power required to achieve a sweep coagulation reaction:
P = (γ×G×η) / n
Here,G = Velocity gradient
η = dynamic viscosity
γ = 6.5 (Coefficient)
n = 1.5 for impeller operation and 1.2 for jet operation
For the given case,G = 800 s⁻¹η = 1.06 × 10⁻³ Pa.sn = 1.2 for jet operation
Substituting these values in the equation, we get:
P = (6.5 × 800 × 1.06 × 10^−3) / 1.2P = 4.74 kW
Therefore, the range of power required to achieve a sweep coagulation reaction in the rapid mixer is 2.46 kW to 4.74 kW.
Read more on Power here: https://brainly.com/question/11569624
#SPJ11
Consider a thermos for coffee which is filled with hot water, the lid is placed on it (closed system), and it is placed in a room whose air and walls are at a fixed temperature. The dimensions of the
In this scenario, a thermos filled with hot water is considered as a closed system. The thermos is placed in a room where the air and walls have a fixed temperature. The dimensions of the thermos, along with the insulating materials used, play a crucial role in determining the rate of heat transfer between the hot water and the surroundings.
The thermos is designed to minimize heat transfer between the hot water and the surroundings, allowing the liquid to retain its temperature for an extended period. The dimensions of the thermos, such as its height, diameter, and thickness of the walls, contribute to its thermal insulation properties.
The thermos is typically constructed with a double-walled structure, with a vacuum or insulating material between the inner and outer walls. This design reduces heat transfer by conduction, as the vacuum or insulating material acts as a barrier. The insulating material used, such as foam or glass wool, also helps to minimize heat transfer by conduction.
Additionally, the lid of the thermos plays a crucial role in preventing heat loss. It is designed to fit tightly and minimize air exchange between the inside and outside of the thermos. This helps to reduce heat transfer by convection, as there is limited air movement inside the thermos.
The fixed temperature of the air and walls in the room surrounding the thermos is also significant. If the room is at a lower temperature than the hot water in the thermos, heat transfer will occur from the water to the surroundings. However, due to the insulating properties of the thermos, the rate of heat loss will be significantly lower compared to an open container.
The dimensions and insulating materials used in the thermos, along with the closed system design and lid, contribute to minimizing heat transfer between the hot water and the surrounding environment. This allows the thermos to maintain the temperature of the liquid for a more extended period, providing effective insulation for the contents inside.
Learn more about dimensions here:- brainly.com/question/31460047
#SPJ11
Assume ethane combustion in air: C2H6 +20₂ = 2CO₂+ 3H20 (5) a. Find LFL, UFL, and LOC (limiting oxygen concentration) b. If LOL and UOL of ethane are 3.0% fuel in oxygen and 66% fuel in oxygen, respectively, please find the stoichiometric line and draw a flammability diagram of ethane (grid lines are provided in the next page). Identify LOL, UFL, LFL, UFL, LOC line, air-line, stoichiometric line, and flammability zone.
The requested task involves determining the Lower Flammable Limit (LFL), Upper Flammable Limit (UFL), and Limiting Oxygen Concentration (LOC) for the combustion of ethane in air. Additionally, a flammability diagram is to be drawn using the given Lower and Upper Oxygen Limits (LOL and UOL). The specific values for LFL, UFL, LOC, LOL, and UOL are not provided.
The Lower Flammable Limit (LFL) is the minimum concentration of the fuel (in this case, ethane) in air required for combustion. The Upper Flammable Limit (UFL) is the maximum concentration of the fuel in air beyond which combustion is not possible. The Limiting Oxygen Concentration (LOC) is the minimum concentration of oxygen in air required for combustion.
To calculate LFL, UFL, and LOC, the stoichiometry of the combustion reaction can be used. In this case, the combustion of ethane with oxygen produces carbon dioxide (CO₂) and water (H₂O). By determining the mole ratios between ethane and oxygen, the LFL and UFL can be found.
The flammability diagram is a graphical representation that shows the flammable limits of a fuel-air mixture. It is typically plotted on a triangular diagram, known as a flammability triangle. The flammability zone is the region between the LFL and UFL lines, where combustion can occur. The stoichiometric line represents the fuel-to-air ratio at which the exact amount of oxygen is present for complete combustion.
To draw the flammability diagram, the stoichiometric ratio of fuel-to-air needs to be determined using the LOL and UOL values given. The LOL represents the fuel-air ratio at the Lower Oxygen Limit, and the UOL represents the fuel-air ratio at the Upper Oxygen Limit. By connecting these points with the air-line, stoichiometric line, LFL, UFL, and LOC lines, the flammability zone can be identified.
Learn more about Oxygen : brainly.com/question/24259144
#SPJ11
2. (10 points) A compound sphere is given as below: T₂-30 °C B r3 A T₁=₁ 100°C Calculate Tw in °C at steady-state condition. r₁=50 mm r₂=100 mm r3= 120 mm KA=0.780 W/m°C KB=0.038 W/m°C
In this problem, a compound sphere with different materials and temperatures is given. The task is to calculate the temperature Tw at steady-state conditions.
The dimensions and thermal conductivities of the materials (KA and KB) are provided. Using the heat transfer equation and appropriate boundary conditions, the value of Tw can be determined. To calculate the temperature Tw at steady-state conditions in the compound sphere, we can use the heat transfer equation and apply appropriate boundary conditions. The compound sphere consists of two materials with different thermal conductivities, KA and KB, and three radii: r₁, r₂, and r₃.
The heat transfer equation for steady-state conditions can be expressed as:
(Q/A) = [(T₂ - T₁) / (ln(r₂/r₁) / KA)] + [(T₂ - Tw) / (ln(r₃/r₂) / KB)]
Where Q is the heat transfer rate, A is the surface area, T₁ is the initial temperature at the inner surface (r₁), T₂ is the initial temperature at the outer surface (r₃), and Tw is the temperature at the interface between the two materials. Since the problem states that the system is at steady-state conditions, the heat transfer rate (Q) is zero. By setting Q/A to zero in the equation, we can solve for Tw.
To do this, we rearrange the equation and solve for Tw:
Tw = T₂ - [(T₂ - T₁) / (ln(r₃/r₂) / KB)] * (ln(r₂/r₁) / KA)
By substituting the given values for T₁, T₂, r₁, r₂, r₃, KA, and KB into the equation, we can calculate the value of Tw.
It's important to note that the units of the given thermal conductivities (KA and KB) and dimensions (radii) should be consistent to ensure accurate calculations. Additionally, the temperatures T₁ and T₂ should be in the same temperature scale (e.g., Celsius or Kelvin) to maintain consistency throughout the calculation.
By following these steps and substituting the given values into the equation, the value of Tw can be determined, providing the temperature at the interface between the two materials in the compound sphere.
Learn more about heat transfer here:- brainly.com/question/30526730
#SPJ11
For the cracking reaction, C3 H8(g) → C2 H4(g) + CH4(g) the equilibrium conversion is negligible at 300 K, but it becomes appreciable at temperatures above 500 K. For a pressure of 1 bar, determine:
(a) The fractional conversion of propane at 625 K.
(b) The temperature at which the fractional conversion is 85%.
Please include the iteration calculation
To determine the fractional conversion , we need to use an iteration calculation based on the equilibrium constant (Kp) expression for the cracking reaction.
(a) For the fractional conversion of propane at 625 K: The equilibrium constant (Kp) expression for the cracking reaction is given by: Kp = (P(C2H4) * P(CH4)) / P(C3H8). Since the equilibrium conversion is appreciable at temperatures above 500 K, we assume that the reaction is at equilibrium. Therefore, Kp will remain constant. Let's assume Kp = Kc. To find the fractional conversion of propane, we can express the equilibrium concentrations of the products and reactant in terms of the initial pressure (P0) and the fractional conversion (x): P(C2H4) = (P0 - P0x) / (1 + x); P(CH4) = (P0 - P0x) / (1 + x); P(C3H8) = P0 * (1 - x). Substituting these expressions into the Kp expression and rearranging, we have: Kc = [(P0 - P0*x) / (1 + x)]^2 / [P0 * (1 - x)]. Now, we can substitute the given values: P0 = 1 bar; Temperature (T) = 625 K. Iteratively solving the equation Kc = [(P0 - P0*x) / (1 + x)]^2 / [P0 * (1 - x)] for x will give us the fractional conversion of propane at 625 K.
(b) To find the temperature at which the fractional conversion is 85%: We need to iterate the above process in reverse. Assume the fractional conversion (x) as 0.85 and solve for the temperature (T). Using the same equation as in part (a), iteratively calculate the temperature until the desired fractional conversion is achieved. The iteration calculation involves substituting initial values, solving the equation, updating the values based on the obtained result, and repeating the process until convergence is reached.
To learn more about equilibrium constant click here: brainly.com/question/30620209
#SPJ11
Ammonia is absorbed from air into water at atmospheric pressure and 20°C. Gas resistance film is estimated to be 1 mm thick. If ammonia diffusivity in air is 0.20 cm²/sec and the partial pressure is
The rate of absorption can be determined using Fick's law of diffusion, which considers factors such as diffusivity, concentration gradient, and film thickness. To determine the rate of ammonia absorption, we can use Fick's law of diffusion, which states that the rate of diffusion is proportional to the concentration gradient and the diffusivity.
Mathematically, the equation can be expressed as:Rate of Diffusion = (Diffusivity * Area * Concentration Gradient) / Thickness.In this case, the gas resistance film is estimated to be 1 mm thick. The diffusivity of ammonia in air is given as 0.20 cm²/sec.
To calculate the rate of ammonia absorption, we need to know the concentration gradient and the surface area. The concentration gradient represents the difference in ammonia partial pressure between the air and water phases.The Henry's law constant is also needed to relate the partial pressure of ammonia in the gas phase to its concentration in the liquid phase.
To calculate the rate of ammonia absorption from air into water, additional information such as the concentration gradient, surface area, and Henry's law constant is required. The rate of absorption can be determined using Fick's law of diffusion, which considers factors such as diffusivity, concentration gradient, and film thickness. . The calculation and conclusion would require detailed experimental data or relevant values for the parameters mentioned above to accurately determine the rate of ammonia absorption.
To know more about diffusion visit:
https://brainly.com/question/14852229
#SPJ11
For the reaction: 6H₂O (g) + 4CO2(g) = 2C₂H6 (g) +702 (g) and if [H₂O]eq = 0.256 M, [CO2]eq = 0.197 M, [C₂H6leq = 0.389 M, [O2leq = 0.089 M What is the value of the equilibrium constant, K?
The value of the equilibrium constant, K, for the given reaction is 5.65.
The equilibrium constant, K, is defined as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their respective stoichiometric coefficients. Using the given equilibrium concentrations, we can determine the value of K for the reaction.
The balanced equation for the reaction is: 6H₂O (g) + 4CO₂ (g) = 2C₂H₆ (g) + 7O₂ (g)
The expression for the equilibrium constant, K, is: K = ([C₂H₆]^2 * [O₂]^7) / ([H₂O]^6 * [CO₂]^4)
Substituting the given equilibrium concentrations into the expression, we have: K = (0.389^2 * 0.089^7) / (0.256^6 * 0.197^4)
Evaluating the expression, we find: K ≈ 5.65
Therefore, the value of the equilibrium constant, K, for the given reaction is approximately 5.65. This value indicates the position of the equilibrium and the relative concentrations of the reactants and products at equilibrium. A higher value of K suggests a greater concentration of products at equilibrium, while a lower value of K suggests a greater concentration of reactants.
Learn more about equilibrium : brainly.com/question/30694482
#SPJ11
3. Derive Navier-stokes equation in Cylindrical coordinate system for a fluid flowing in a pipe. Enter your answer
These equations represent the Navier-Stokes equations in cylindrical coordinates for a fluid flowing in a pipe. They describe the conservation of mass and momentum in the fluid, taking into account the velocity components, pressure, density, viscosity, and gravitational effects.
The Navier-Stokes equation in cylindrical coordinate system for a fluid flowing in a pipe can be derived as follows:
Consider a fluid flow in a cylindrical coordinate system, where the radial distance from the axis of the pipe is denoted by r, the azimuthal angle is denoted by θ, and the axial distance along the pipe is denoted by z.
The continuity equation, which represents the conservation of mass, can be written in cylindrical coordinates as:
∂ρ/∂t + (1/r)∂(ρvₑ)/∂θ + ∂(ρv)/∂z = 0
where ρ is the fluid density, t is time, vₑ is the radial velocity component, and v is the axial velocity component.
The momentum equations, which represent the conservation of momentum, can be written in cylindrical coordinates as:
ρ(∂v/∂t + v∂v/∂z + (vₑ/r)∂v/∂θ) = -∂p/∂z + μ((1/r)∂/∂r(r∂vₑ/∂r) - vₑ/r² + (1/r²)∂²vₑ/∂θ²) + ρgₑₓₓ
ρ(∂vₑ/∂t + v∂vₑ/∂z + (vₑ/r)∂vₑ/∂θ) = -∂p/∂r - μ((1/r)∂/∂r(r∂v/∂r) - v/r² + (1/r²)∂²v/∂θ²) + ρgₑₓₑ
where p is the pressure, μ is the dynamic viscosity of the fluid, gₑₓₓ is the gravitational acceleration component in the axial direction, and gₑₓₑ is the gravitational acceleration component in the radial direction.
These equations represent the Navier-Stokes equations in cylindrical coordinates for a fluid flowing in a pipe. They describe the conservation of mass and momentum in the fluid, taking into account the velocity components, pressure, density, viscosity, and gravitational effects.
Please note that this derivation is a simplified representation of the Navier-Stokes equations in cylindrical coordinates for a fluid flow in a pipe. Additional terms or assumptions may be included based on specific conditions or considerations.
To know more about fluid , visit;
https://brainly.com/question/13256681
#SPJ11
Aluminum Chlorohydrate is 10.9 lbs per gallon. It has 12.1-12.7
% aluminum. How many pounds of aluminum are in each gallon?
Assuming 0.87 lbs of Aluminum are needed to inactivate 1 pound
of Phosphorus
There are approximately 1.1832 pounds of aluminum in each gallon of Aluminum Chlorohydrate.
Given that Aluminum Chlorohydrate is 10.9 lbs per gallon and has 12.1-12.7% aluminum.
We need to determine how many pounds of aluminum are in each gallon.
Solution: The percentage of aluminum in Aluminum Chlorohydrate is 12.1-12.7%
Therefore, the average percentage of aluminum is: 12.1+12.7/2 = 12.4% (taking the mean)
This implies that there is 12.4% aluminum in Aluminum Chlorohydrate
Therefore, the weight of aluminum in 1 gallon of Aluminum Chlorohydrate = 12.4% of 10.9 lbs= (12.4/100) × 10.9= 1.3546 lbs ≈ 1.36 lbs
Now, we know that 0.87 lbs of Aluminum is required to inactivate 1 lb of Phosphorus.
Thus, to inactivate the aluminum present in 1 gallon of Aluminum Chlorohydrate, we need:0.87 × 1.36 = 1.1832 lbs of Aluminum
Know more about here:
https://brainly.com/question/4392857
#SPJ11
For the reaction below, if 6.3 g of S reacted with 10.0 g of O₂, how many grams of SO3 will be produced?
2S (s) + 30₂(g) → 2S03 (g)
2S + 302 = 2SO3
Mass of S = 6.3g
Mass of 02 = 10.0g
n = m/MM(S) = 32g/mol
n = 6.3g/32g/mol
n = 0.195mol
n(S)/n(SO3) = 2/2
Let n(SO3) = x
2(0.195) = 2x
0.39 = 2x
x = 0.195
Therefore, n(SO3) = 0.195mol
For mass of SO3m = M×nBut M(SO3) = (32×1) + (16×3)
= 80g/mol
m = 80g/mol × 0.195mol
m = 15.6g
Therefore, 15.6g of SO3 will be produced. HOPE IT HELPS. HAVE A WONDERFUL DAY.Combustion A gaseous hydrocarbon fuel (CxH2x+2) is combusted with air in an industrial furnace. Both the fuel and air enter the furnace at 25°C while the products of combustion exit the furnace at 227°C. The volumetric analysis of the products of combustion is: Carbon dioxide (CO₂) 9.45% Carbon monoxide (CO) 2.36% Oxygen (O₂) 4.88% Nitrogen (N₂) 83.31% Write a balanced chemical equation for the combustion reaction (per kmol of fuel) and hence determine the fuel and the air-fuel ratio. Construct separate 'reactants' and 'products' tables giving the number of moles and molar enthalpies for each of the reactants and products, respectively, involved in the combustion process. Hence determine the heat transfer rate and the combustion efficiency on a lower heating value (LHV) basis.
The balanced chemical equation for the combustion reaction of the gaseous hydrocarbon fuel (CxH2x+2) with air can be written as CxH2x+2 + (2x + 1)O2 + 3.76N2 -> xCO2 + (x + 1)H2O + 3.76(2x + 1)N2. The fuel is determined to be methane (CH4).
The balanced chemical equation for the combustion reaction of the gaseous hydrocarbon fuel (CxH2x+2) with air can be written as:
CxH2x+2 + (2x + 1)O2 + 3.76N2 -> xCO2 + (x + 1)H2O + 3.76(2x + 1)N2.
Given the volumetric analysis of the products of combustion, we can determine the value of x in the hydrocarbon fuel. The percentage of carbon dioxide (CO2) corresponds to the carbon atoms in the fuel, so 9.45% CO2 implies x = 1. The fuel is therefore methane (CH4).
To calculate the air-fuel ratio, we compare the moles of air to the moles of fuel in the balanced equation. From the equation, we have (2x + 1) moles of oxygen (O2) and 3.76(2x + 1) moles of nitrogen (N2) for every 1 mole of fuel. Substituting x = 1, we find that the air-fuel ratio is 17.2 kg of air per kg of fuel.
To determine the heat transfer rate and combustion efficiency on a lower heating value (LHV) basis, we need to calculate the molar enthalpies of the reactants and products. Using standard molar enthalpies of formation, we can calculate the change in molar enthalpy for the combustion reaction. The heat transfer rate can be obtained by multiplying the change in molar enthalpy by the mass flow rate of the fuel. The combustion efficiency on an LHV basis can be calculated by dividing the actual heat transfer rate by the ideal heat transfer rate.
Learn more about hydrocarbon : brainly.com/question/30666184
#SPJ11
d) Identify three safety critical systems which were non-functional at the Union Carbide Bhopal facility and explain how lack of maintenance led to the Bhopal tragedy.
Answer:
Explanation:
Vent gas scrubber, Flare stack, Water curtain, Refrigeration system and a spare storage tank were provided in the plant.
Adequate in-built safety systems were not provided and those provided were not checked and maintained as scheduled.
In all, five safety systems namely: Vent gas scrubber, Flare stack, Water curtain, Refrigeration system and a spare storage tank were provided in the plant. But none of these ever worked or came to therescue in the emergency.
Safe operating procedures were not laid down and followed under strict supervision.
Total lack of 'on-site' and 'offsite' emergency control measures.
No hazard and operability study (HAZOP) was carried out on the design and no follow-up by any risk analysis.
please solve with least square procedure and use
matrix solution ty
if the experimental data is given as X : 0.50 1.0 1.50 2 2.50 f (x) : 0.25 0.5 0.75 1 1.25 and the model euation is given as f(x) = axª¹ find the values of ao and a
The values of a₀ and a can be determined using the least square procedure with the given experimental data.
We have the model equation f(x) = a₀x^(a-1).
Let's denote the given experimental data as X and f(x):
X: 0.50 1.0 1.50 2 2.50
f(x): 0.25 0.5 0.75 1 1.25
To solve for a₀ and a, we can set up a system of equations based on the least square method:
Sum of Residuals = Σ [f(x) - a₀x^(a-1)]^2 = 0
Expanding the sum of residuals:
Residual₁ = (0.25 - a₀ * 0.50^(a-1))^2
Residual₂ = (0.5 - a₀ * 1.0^(a-1))^2
Residual₃ = (0.75 - a₀ * 1.50^(a-1))^2
Residual₄ = (1 - a₀ * 2^(a-1))^2
Residual₅ = (1.25 - a₀ * 2.50^(a-1))^2
Our objective is to minimize the sum of residuals by finding the optimal values of a₀ and a. This can be achieved by taking the partial derivatives of the sum of residuals with respect to a₀ and a, setting them equal to zero, and solving the resulting equations.
However, this system of equations does not have a closed-form solution. To find the optimal values of a₀ and a, we can utilize numerical optimization techniques or approximation methods such as gradient descent.
To determine the values of a₀ and a for the given model equation f(x) = a₀x^(a-1) using the least square procedure, we need to solve the system of equations formed by the sum of residuals. Since the equations do not have a closed-form solution, numerical optimization techniques or approximation methods are required to find the optimal values of a₀ and a.
To know more about least square procedure, visit
https://brainly.com/question/30548323
#SPJ11
A catalyst pellet with a diameter of 5 mm is to be fluidized
with 45,000 kg/hr of air at 1 atm and 77oC in a vertical cylinder.
Particle density = 960 kg/m3 and sphericity = 0.6. If the quantity
of ai
Answer: 468 m³/hr
The fluidization of a 5 mm diameter catalyst pellet with 45,000 kg/hr of air at 1 atm and 77oC in a vertical cylinder with particle density = 960 kg/m3 and sphericity = 0.6 is the topic of this problem.
We have to calculate the air required for complete fluidization.
Determine the terminal velocity of the catalyst pellet using the following formula:`
Vt = (4/3 * g * (ρp - ρf) * d^3) / (18 * µ * s)`
Where `Vt` is the terminal velocity of the catalyst pellet.`
d` is the diameter of the pellet.`
g` is the acceleration due to gravity.`
ρ is the density of the pellet.`
.`µ` is the fluid viscosity.`
s` is the sphericity of the pellet.
Substituting the given values, we get:
Vt = (4/3 × 9.81 m/s² × (960 kg/m³ - 1.205 kg/m³) × (5 × 10^-3 m)³) / (18 × 1.85 × 10^-5 Pa·s × 0.6)≈ 0.031 m/s
Determine the minimum fluidization velocity of the fluid using the following formula:
`u = (ε^3 * (ρf - ρp) * g) / (150 * µ * (1 - ε)^2)`
Where `u` is the minimum fluidization velocity of the fluid.`
ε` is the voidage of the bed of the fluid.`
ρf` is the density of the fluid.`
ρp` is the density of the pellet.`
g` is the acceleration due to gravity.`
µ` is the fluid viscosity.
Substituting the given values, we get:
`0.039 = (ε^3 * (1.205 - 960) * 9.81) / (150 × 1.85 × 10^-5 × (1 - ε)^2)`
Rearranging the equation, we get:
`(ε^3 * 9.81 * 2.45 × 10^2) / (1.11 × 10^-3 * (1 - ε)^2) = 0.039
Simplifying and solving the equation above, we get:`
ε ≈ 0.358
`The pressure drop `∆P` can be determined using the following equation:
`∆P = u (1 - ε)^2 * ε^3 * (ρp - ρf) / (150 * ε^2 * ρf^2)`
Where `∆P` is the pressure drop across the bed of fluid.
`u` is the minimum fluidization velocity of the fluid.`
ε` is the voidage of the bed of the fluid.`
ρf` is the density of the fluid.`
ρp` is the density of the pellet.
Substituting the given values, we get:`
∆P = 0.039 * (1 - 0.358)^2 * 0.358^3 * (960 - 1.205) / (150 * 0.358^2 * 1.205^2)`≈ 5.9 Pa
The air required for complete fluidization is:`Q = ∆P * π * d^2 * u / (4 * µ)
`Where `Q` is the air required for complete fluidization.
`d` is the diameter of the pellet.
`∆P` is the pressure drop across the bed of fluid.`
u` is the minimum fluidization velocity of the fluid.
`µ` is the fluid viscosity.
Substituting the given values, we get:
Q = 5.9 Pa * π * (5 × 10^-3 m)² * 0.039 m/s / (4 * 1.85 × 10^-5 Pa·s)≈ 0.13 m³/s or 468 m³/hr
Know more about fluidization here:
https://brainly.com/question/31775406
#SPJ11
Please answer the following questions thank you
Briefly explain nanocomposites with THREE examples of their uses.
Answer:
nanocomposite (plural nanocomposites)
Any composite material one or more of whose components is some form of nanoparticle; more often consists of carbon nanotubes embedded in a polymer matrix
please can tou guve me the details on how to solve this
(6) Using X-ray diffraction, it was found that a material had constructive interference for the (311) and (222) planes. What is the crystal structure of this material? a) FCC (b) BCC (c) HCP (d) none
The crystal structure of the material exhibiting constructive interference for the (311) and (222) planes is FCC (Face-Centered Cubic).
X-ray diffraction is a technique used to determine the crystal structure of a material by analyzing the patterns formed when X-rays interact with the crystal lattice. Constructive interference occurs when the X-ray waves reflected from different crystal planes align in phase, resulting in a strong diffraction signal.
The Miller indices are used to describe crystal planes. The (hkl) notation represents the set of crystallographic planes in a material. In this case, the material exhibits constructive interference for the (311) and (222) planes.
For an FCC crystal structure, the Miller indices of the (hkl) planes satisfy the following conditions:
h + k + l = even
Let's check the conditions for the given planes:
For the (311) plane: 3 + 1 + 1 = 5 (odd)
For the (222) plane: 2 + 2 + 2 = 6 (even)
Since the condition is satisfied only for the (222) plane, the material has constructive interference for the (222) plane. Therefore, the crystal structure of the material is FCC.
Based on the constructive interference observed for the (311) and (222) planes, we can conclude that the crystal structure of the material is FCC (Face-Centered Cubic). This information is obtained by analyzing the Miller indices and their fulfillment of the conditions specific to different crystal structures.
To know more about interference visit:
brainly.com/question/31857527
#SPJ11