The method that would be used to allocate general overhead associated with the inventory of materials is "Usage of overhead." The usage of overhead method of allocating general overhead costs involves determining a standard overhead rate based on the anticipated usage of the overhead costs.
This method involves dividing the overhead expenses by a practical activity measure such as direct labor hours or machine hours, which are expected to incur overhead costs in a company's operations.
The overhead costs are allocated to products or services based on the predetermined rate multiplied by the actual amount of activity.
Incremental general overhead costs and material costs are not methods used to allocate general overhead associated with the inventory of materials. Percentage of a company's revenues is also not used as a method to allocate general overhead costs to inventory of materials since it does not consider the actual usage of overhead costs in the production process.
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3.3 What is the primary use of synthetic graphite and how is
it produced?
Synthetic graphite serves as a crucial anode material in lithium-ion batteries and finds applications in various electronic products, fuel cells, solar panels, refractory materials, and lubricants. While it is more expensive than natural graphite, it offers the advantage of producing finer and more consistent powders.
The production of synthetic graphite involves high-temperature methods such as graphitization of carbonaceous materials, pyrolysis, and chemical vapor deposition. Carbonaceous materials are subjected to temperatures ranging from 2200°C to 3000°C in vacuum or inert gas environments to facilitate the formation of graphite.
Unlike natural graphite, which occurs naturally as a mineral, synthetic graphite is artificially manufactured. It serves as a viable substitute and is created by subjecting carbonaceous raw materials to high temperatures for carbonization. The heating process involves materials such as petroleum coke, coal tar pitch, and carbon black. The conversion into graphite requires temperatures of 2200–3000 °C.
Synthetic graphite is primarily available in two primary forms: carbon fibers and carbon black. Carbon fibers are characterized by their long and thin structure, while carbon black is fine-grained. These two forms differ in their respective applications and methods of production.
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Which of the following statement is false about long columns? Select the correct response a. None of the statements b. The critical stress rapidly decreases as the slenderness ratio increases. c. Columns may undergo inadmissibly large deformations under loads below the critical value d. Critical buckling loads is critical to the axis with highest slenderness ratio. Drovis
The statement "The critical stress rapidly decreases as the slenderness ratio increases" is false about long columns.
What are long columns?A long column is a structural element that experiences compressive forces. It possesses a slenderness ratio exceeding 12, which represents the relationship between its length and its narrowest lateral dimension.
The pivotal stress denotes the maximum load that a column can sustain before succumbing to buckling. The slenderness ratio signifies the proportion between the column's length and its most diminutive cross-sectional dimension.
As the slenderness ratio amplifies, the critical stress diminishes. This phenomenon arises due to the heightened susceptibility of longer columns to buckle when subjected to external loads.
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Given an irrigation regulator with the following properties:
The difference between U/S and D/S water levels (AH) = 0.15m
The approach velocity (Va)=0.7 m/sec The tail water depth (y) = 2.1m The water way width (S) = 6m Use the acceleration of gravity (g)=9.81m/sec²
Then the discharge (Q) of this regulator is approximately equal to:
A regulator is a mechanism that controls the flow of fluids in an irrigation system. The rate of flow is a critical aspect of irrigation systems because it influences the amount of water delivered to crops. The flow rate of an irrigation regulator can be calculated by using various parameters.
Given the parameters, we can use them to determine the discharge of this regulator:
[tex]AH = 0.15 mVa = 0.7 m/secy = 2.1 mS = 6 mg = 9.81 m/s²[/tex]
The discharge (Q) of a regulator is estimated using the following formula:
[tex]Q = C √2gAH (2/3) S (3/2)/√(y+AH/2)[/tex]
The equation for the discharge is as follows:
[tex]Q = C √(2gAH (2/3) S (3/2))/(y+AH/2)[/tex]
Where:C is the regulator's coefficient of discharge, which is equal to 0.62 in this case.
We have all of the variables in the formula, and we can substitute them to get the answer.
[tex]Q = 0.62 √(2 × 9.81 × 0.15 (2/3) × 6 (3/2))/(2.1 + 0.15/2)Q = 0.62 × 2.876Q = 1.78 m³/s[/tex]
Therefore, the regulator's discharge is roughly [tex]1.78 m³/s.[/tex]
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What are the advantages and disadvantages of using a Raw
Wastewater Treatment System or a Settled Wastewater Treatment
System, when looking at sludge management?
Activated sludge process is a commonly used biological wastewater treatment method that uses microorganisms to break down organic pollutants in wastewater.
Advantages:
High treatment efficiency: Activated sludge process is effective in removing a high percentage of organic pollutants, including nitrogen and phosphorus, from wastewater.
Flexibility: Activated sludge process can be easily adapted to different wastewater flows and strengths, making it a versatile treatment option.
Low maintenance: Activated sludge process requires relatively low maintenance compared to other wastewater treatment methods.
Disadvantages:
Sludge production: Activated sludge process generates a large volume of sludge, which requires proper disposal to prevent environmental contamination.
Energy consumption: Activated sludge process requires a significant amount of energy for aeration and mixing, making it an energy-intensive process.
Process instability: Activated sludge process can be sensitive to changes in wastewater flow and strength, leading to process instability and reduced treatment efficiency.
High capital cost: The construction and implementation of an activated sludge process can be expensive, requiring a significant investment in infrastructure and equipment.
The choice of activated sludge process type depends on the specific requirements of each wastewater treatment facility, and the trade-off between the advantages and disadvantages of each process type.
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Assume the MOC is Stainless Steel and the permissible stress is 1350 kg/cm 2
. Welded joint efficiency is 75%. A cylindrical thermic fluid storage made from an SS tank has a 2.5 m inside diameter and it is subjected to a pressure of 5 kg/cm 2
. The weight of the vessel with all its contents is 5000 kg. The torque exerted due to offset piping is 50 kg−m. The thermic fluid storage tank has torispherical heads. The crown radius of the head is equal to the diameter of the vessel. The knuckle radius of the head is 8% of the diameter of the vessel. Calculate the minimum thickness required for the cylindrical shell and torispherical head of the thermic fluid storage tank. Also calculate the equivalent resultant stress induced in the Vessel (2+3+5=10M)
The minimum thickness required for the torispherical head of the thermic fluid storage tank is 74.91 mm.
The allowable stress is given as 1350 kg/cm2. The welded joint efficiency is 75%.A cylindrical thermic fluid storage made from an SS tank has a 2.5 m inside diameter and it is subjected to a pressure of 5 kg/cm2. The weight of the vessel with all its contents is 5000 kg. The torque exerted due to offset piping is 50 kg−m. The thermic fluid storage tank has torispherical heads. The crown radius of the head is equal to the diameter of the vessel. The knuckle radius of the head is 8% of the diameter of the vessel. We have to determine the minimum thickness required for the cylindrical shell and torispherical head of the thermic fluid storage tank. We also have to calculate the equivalent resultant stress induced in the Vessel.Minimum thickness required for the cylindrical shell of the thermic fluid storage tank:Calculate the hoop stress in the cylindrical shell of the thermic fluid storage tank.σ = (pr)/tWhere, p = Internal Pressure in N/m2, r = Internal Radius in m, and t = thickness of the shell in m.σ = (5 × 10^4 × 1.25)/tσ = 62,500/tThe allowable stress is given as 1350 kg/cm2. So, the maximum permissible stress is σp = 1350 × 9.81 = 13,243.5 N/m2We know that, the welded joint efficiency is 75%.Therefore, the actual permissible stress (σpa) will be given by:σpa = (0.75) × (σp) = (0.75) × (13,243.5)σpa = 9932.625 N/m2Let the minimum thickness of the shell be 't'.Then,9932.625 = 62,500/tt = 62,500/9932.625t = 6.3 mmTherefore, the minimum thickness required for the cylindrical shell of the thermic fluid storage tank is 6.3 mm. Let us now calculate the minimum thickness required for the torispherical head of the thermic fluid storage tank.Minimum thickness required for the torispherical head of the thermic fluid storage tank:Calculate the stresses due to the applied loads in the torispherical head of the thermic fluid storage tank.σh = (pd)/(4tc) + (M)/(A)Where, p = Internal Pressure in N/m2, d = diameter of the head in m, t = thickness of the head in m, c = crown radius of the head in m, M = bending moment due to torque in N-m, and A = section modulusσh = (5 × 10^4 × 1.25)/(4 × t × 0.167) + (5000 × 9.81 × 1.25)/(π × 0.167^2 × t) + (50 × 9.81)/(π × 0.167^2 × t)σh = (939,522.75)/tThe crown radius of the head is equal to the diameter of the vessel. Therefore, c = 1.25 m.The knuckle radius of the head is 8% of the diameter of the vessel. Therefore, r = 0.08 × 1.25 = 0.1 m.A = (π/64) × (d^3 − (d − 2t)^3) + (π/6) × t^3 + (π/32) × c^3 + (π/2) × (0.1^3)A = (π/64) × (2.5^3 − (2.5 − 2t)^3) + (π/6) × t^3 + (π/32) × 1.25^3 + (π/2) × (0.1^3)Let the minimum thickness of the torispherical head be 't1'.Then,σh = (pd)/(4t1c) + (M)/(A)9932.625 = (5 × 10^4 × 1.25)/(4 × t1 × 1.25) + (5000 × 9.81 × 1.25)/(π × 1.25^2 × t1) + (50 × 9.81)/(π × 1.25^2 × t1) + (M)/(A)The maximum equivalent stress should be less than the allowable stress.The allowable stress is given as 1350 kg/cm2.So, the maximum permissible stress is σp = 1350 × 9.81 = 13,243.5 N/m2Let the equivalent resultant stress induced in the Vessel be 'σr'.σr = √((σh^2) + 3(τ^2))Where, τ = shear stressτ = (0.289 × pd)/(tc) + (0.375 × M)/(tc × c)τ = (0.289 × 5 × 10^4 × 1.25)/(t1 × 1.25) + (0.375 × 50)/(t1 × 1.25 × 1.25)τ = (1816.40625)/t1σr = √((σh^2) + 3(τ^2))σr = √((939,522.75/t1)^2 + 3((1816.40625/t1)^2))σr = (991350.47)/t1σr should be less than σp. Therefore,σr ≤ σp(991350.47)/t1 ≤ 13,243.5t1 ≥ 74.91 mm.
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Which of the following is an example of a categorical variable? color of car time to \( 60 \mathrm{mph} \) from a complete stop speed in which the air bag deploys force in which the air bag deploys
The color of a car is an example of a categorical variable. A categorical variable is a variable that takes on discrete values and can be grouped into categories based on some shared characteristic.
Categorical variables are a type of variable that takes on discrete values and can be grouped into categories based on some shared characteristic. This type of variable is often used in statistics to group data into meaningful categories and to help analyze patterns and trends in the data. The color of a car is an example of a categorical variable because it can be classified into different categories like red, blue, black, etc. Other examples of categorical variables include gender, race, and education level. These variables are used to group people or things into categories based on some shared characteristic. For example, gender can be used to group people into male and female categories, while education level can be used to group people into categories like high school, college, and graduate school. In conclusion, the color of a car is an example of a categorical variable. This type of variable is used in statistics to group data into meaningful categories and to help analyze patterns and trends in the data. Other examples of categorical variables include gender, race, and education level.
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CE-3610
Design Basic
1. The building is to be designed according to the provisions for structural requirement for LRFD
for steel as set forth in ASCE 7/IBC.
2. Material Specifications:
2.1 Light weight Concrete floor over deck
2.2 ASTM A992 (Gr.50) for all W shape Beams, Girders and Columns
2.3 Braces HSS (ASTM A500) or W shapes (ASTM A992, Gr.50)
3. Dead Loads:
3.1 Roof:
3.1.1 Roofing Materials (Water Proofing etc.) = 4 psf
3.1.2 18" Gage deck= 3 psf
3.1.3 Light weight concrete 5 in thick
3.1.4 Framing & Fire proofing= 8 psf
3.1.5 Suspended ceiling= 4 psf
3.1.6 Mechanical & Electrical= 4 psf
3.1.7 Solar panels & assembly= 9 psf
3.2 Floor:
3.2.1 Tile including assembly= 9.5 psf
3.2.2 18" Gage deck
3.2.3 Light weight concrete 6 1/4 "= 3 psf
3.2.4 Framing & Fire proofing = 15 psf
3.1.5 Suspended ceiling= 5 psf
3.1.6 Mechanical & Electrical= 5 psf
3.3 Wall:
3.3.1 Parapets on roof (outer boundary only) = 25 psf (3.5 ft high)
3.3.2 Glazed walls (outer boundary only) = 18 psf (ground to roof level)
Floors require 2 hour fire rating.
Live Load considers partition loads as appropriate.
4. The deflection requirements are as follows:
4.1 L/360 due to live load deflection in all interior Beams and Girders
4.2 L/180 due to total load for all spandrel Beams and Girders
5. For Seismic design, the values of spectral response acceleration parameters for the given location should be found using the USGS website
The primary objective of this project is to design a building in accordance with the structural requirements for LRFD for steel as defined in ASCE 7/IBC. The building's materials are defined in the following terms:
Light-weight concrete flooring over deck with a thickness of 5 inches.
ASTM A992 (Gr.50) for all W shape Beams, Girders, and Columns.
HSS Braces (ASTM A500) or W shapes (ASTM A992, Gr.50).
Dead Loads: The building's dead load will be made up of a variety of elements, including:
Roof: Roofing Materials (Water Proofing, etc.) = 4 psf.
18" Gage deck = 3 psf.
Light-weight concrete 5 in thick.
Framing & Fire proofing = 8 psf.
Suspended ceiling = 4 psf.
Mechanical & Electrical = 4 psf.
Solar panels & assembly = 9 psf.
Floor: Tile including assembly = 9.5 psf.
18" Gage deck.
Light-weight concrete 6 1/4 "= 3 psf.
Framing & Fire proofing = 15 psf.
Suspended ceiling = 5 psf.
Mechanical & Electrical = 5 psf.
Wall: Parapets on roof (outer boundary only) = 25 psf (3.5 ft high).
Glazed walls (outer boundary only) = 18 psf (ground to roof level).
The live load of partitions is taken into consideration as appropriate. Flooring requires a two-hour fire rating. The following deflection requirements are in effect:
L/360 due to live load deflection in all interior Beams and Girders.
L/180 due to total load for all spandrel Beams and Girders.
the building's seismic design should consider the values of spectral response acceleration parameters for the given location, which can be found using the USGS website.
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A toll bridge across XYZ river in USA is being considered to replace the current bridge. Because this bridge, if approved, will become a part of the U.S. Interstate Highway system, the B-C ratio method must be applied in the evaluation. Initial costs of the bridge are estimated to be US$ 17,500,000, and annual operating and maintenance costs of US$ 325,000 are anticipated. In addition, the bridge must be resurfaced every 5th year of its 30-year projected life at a cost of US$ 1,250,000 per occurrence (no resurfacing cost in year 30). Revenues generated from the toll are anticipated to be US$ 2,500,000 in its first year of operation, with a projected annual rate of increase of 2.25% per year. MARR (Minimum acceptable rate of return) is 10% per year. (a) Assume that there will be no salvage value for the bridge at the end of 30 years. (i) Find the Present Worth (PW) of the benefits. Round off your final answer to the NEAREST thousand. (ii) Find the Present Worth (PW) of the costs. Round off your final answer to the NEAREST thousand. (iii) Calculate the conventional B-C ratio using PW. Round off your final answer to two decimal places. (iv) Should the toll bridge be constructed? Why? (b) Assume that the salvage value for the bridge at the end of 30 years will be US$ 100,000. (i) Find the Present Worth (PW) of the benefits. Round off your final answer to the NEAREST thousand. (ii) Will it alter your decision in Part (a) (iv)? Why?
(a)(i) Calculation of the Present Worth of Benefits: In order to calculate the present worth (PW) of benefits, the revenues generated from the tolls must be calculated.
It is projected that the first year will generate revenues of US$2,500,000, with a projected increase of 2.25 percent per year. We will calculate the present worth of the benefits by using the formula; PV = A/(1+i) n = 2,500,000/(1+0.1)1 + 2,500,000(1+0.1)-2.25/(0.1-0.0225) (1+0.1)30 + [2,500,000(1+0.1)-2.25/0.0225] [(1+0.1)30 -1] PV = 2,500,000 + 45,018,130 + 31,975,580PV = 79,493,710 ≈ 79,494,000Ans:
(i) Present Worth of the benefits is US$ 79,494,000(ii) Calculation of the Present Worth of Costs: Annual operating and maintenance costs of US$325,000 are anticipated, and the cost of resurfacing the bridge every fifth year is US$1,250,000. To find the present worth (PW) of costs, we will use the following formula:
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Rayshawn leaves his car at Taylor's Auto Service center for an oil change. When the service is complete, he is obligated to pay the Auto Service for the work done. This type of contract is best characterized as:
an express contract.
a mechanic’s lien contract.
an executive contract.
an implied in fact contract.
Rayshawn leaves his car at Taylor's Auto Service center for an oil change. When the service is complete, he is obligated to pay the Auto Service for the work done. This type of contract is best characterized as an express contract.
An express contract is a type of agreement that is explicitly stated in words, whether written or oral. This agreement is created when two or more parties agree to the terms of the contract and express their willingness to enter into a contract. In an express contract, there is a clear agreement between the parties involved. The following are the characteristics of an express contract:It is a written or oral contract. It is a contract in which both parties have mutually agreed on specific terms and conditions. It is a contract in which the agreement is clear and concise, with no ambiguity in its language or meaning. In conclusion, an express contract is a type of agreement that is explicitly stated in words, whether written or oral. When two or more parties agree to the terms of the contract and express their willingness to enter into a contract, an express contract is formed. In the given case, Rayshawn leaves his car at Taylor's Auto Service center for an oil change. When the service is complete, he is obligated to pay the Auto Service for the work done. This type of contract is best characterized as an express contract.
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1. Sustainable buildings provide many types of benefits to stakeholders and the environment. Despite so, some building developers still have some concerns over implementing sustainable designs to their new building developments.
(a) Suggest reasons why some building developers still have some concerns over implementing sustainable designs to their new building developments.
(b) Some city governments favor granting 'density bonus' while some others favor giving out tax rebates or credits' for encouraging sustainable building developments. Compare and contrast the underlying rationales of these two approaches.
Some building developers still have concerns over implementing sustainable designs to their new building developments due to the following reasons:Cost: Implementing sustainable designs.
often requires more money than traditional designs. This is due to the fact that the materials required for sustainable building designs are often more expensive than traditional materials.Lack of awareness: Some building developers may not be aware of the long-term benefits of sustainable designs.
They may view it as an extra cost to their project without fully understanding how sustainable designs can save them money and enhance their project's value.Lack of regulations: Sustainable designs are still not a requirement in some areas. Without regulations that make it mandatory for may not see the need to invest more money in their projects.
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1) Does hydropower contribute to climate change? Justify.
2) Explain the design consideration of mini-hydro for Renewable Energy. 3) What is the challenge of climate change to hydropower and its impact on the safety of the community downstream of the dam?
4) The role of hydropower in climate change mitigation and adaptation.
Yes, hydropower contributes to climate change. While it is true that hydropower is a renewable energy source, it is not free from negative environmental impacts.
Hydroelectric power plants usually require the construction of dams, which can lead to the destruction of natural habitats and the displacement of communities. These emissions can also occur during the transportation of the necessary equipment and materials.
Mini-hydro systems are small hydroelectric power plants that generate power from the flow of small streams and rivers. Head and flow rate of the water: Mini-hydro systems must be designed with consideration for the head and flow rate of the water. The head is the vertical distance between the water source and the turbine. The type of turbine to be used is determined by the head and flow rate.
The challenge of climate change to hydropower is that it can lead to changes in the water cycle, which affects the availability of water for power generation. Climate change can lead to changes in rainfall patterns, increasing the frequency of floods and droughts.
The role of hydropower in climate change mitigation and adaptation. Hydropower can play a significant role in mitigating climate change by providing a renewable energy source. It is an alternative to fossil fuels and can reduce greenhouse gas emissions.
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Explain the mechanism of SAP concretes. For shotcrete applications, which type of fibers would be recommended (steel or polymer). Explain why, in detail. Fibers are mainly active in which stage? Pre-crack or post-crack? What is the reason, explain in detail.
SAP (Super Absorbent Polymer) concrete is a new type of concrete that is reinforced by the addition of super absorbent polymer. The concrete is created by blending an absorbent polymer with Portland cement, sand, and water, which is then cured in a conventional manner.
SAP concrete's mechanisms are based on the swelling characteristics of the super absorbent polymer. The fibers recommended for shotcrete applications are polymer fibers, not steel fibers. Polymer fibers are recommended because they are less prone to corrosion than steel fibers. polymer fibers are less expensive than steel fibers and they can be easily added to the shotcrete mix.
The polymer fibers are also more effective in resisting plastic shrinkage than steel fibers. Since steel fibers are prone to corrosion, they are not recommended for shotcrete applications. This can lead to concrete damage and reduce its lifespan.
Fibers are mainly active in the post-crack stage rather than the pre-crack stage. This is due to the fact that fibers in concrete do not prevent cracks from forming; instead, they work to hold the concrete together after it has already started to crack. This ensures that the concrete remains stable and durable, even after experiencing some damage due to cracking.
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Calculate the total live load on a dairy barn floor for calves. The Calves apply a load of 50 PSF. The floor is 15' by 49' and has a density of 150 PSF. The walls and roof apply a load of 200 PSF. What is the total load?
Live load is a term used to describe the weight that is temporarily added to a structure, such as a building, and then removed again. It is a measure of the weight that is placed on a structure due to its intended use. In this case, we are trying to calculate the total live load on a dairy barn floor for calves. Here are the steps to find the total load:
Step 1: Calculate the load applied by the calves. Calves apply a load of 50 PSF. The floor is 15' by 49'.Area of the floor = length × breadth = 15 × 49 = 735 sq. ftLoad applied by calves = load per sq. ft × area of the floor= 50 × 735= 36750 pounds
Step 2: Calculate the density of the floor. The density of the floor is given as 150 PSF. This means that every square foot of the floor weighs 150 pounds. Density of the floor = 150 pounds per sq. ft.
Step 3: Calculate the weight of the floor. Weight of the floor = area of the floor × density of the floor= 735 × 150= 110250 pounds.
Step 4: Calculate the load applied by the walls and roof. The walls and roof apply a load of 200 PSF.Load applied by the walls and roof = 200 pounds per sq. ft.
Step 5: Calculate the total load. Total load = weight of the floor + load applied by the walls and roof+ load applied by the calves= 110250 + (200 × 735) + 36750= 246750 pounds Therefore, the total live load on a dairy barn floor for calves is 246750 pounds.
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A two-cell intake tower located in a cold climate reservoir is being designed for a winter design flow rate of 6,000 rr/d. The tower will have three ports at three different elevations in each cell. Each port must be able to deliver the design flow rate operating alone. Determine the area of each port opening, A two-cell intake tower located in a cold climate reservoir is being designed for a winter design flow rate of 6,000 rr/d. The tower will have three ports at three different elevations in each cell. Each port must be able to deliver the design flow rate operating alone. Determine the area of each port opening,
The required area of each port opening is 1666.67 sq.m. A two-cell intake tower located in a cold climate reservoir is being designed for a winter design flow rate of 6,000 rr/d.
The tower will have three ports at three different elevations in each cell.
Each port must be able to deliver the design flow rate operating alone.
Determine the area of each port opening.
Q = Flow rate = [tex]6000 cu.m/d[/tex]
Number of ports = 3
Design flow rate of each port =[tex]6000 / 3 = 2000 cu.m/d[/tex]
Let, the area of each port opening be 'A'
Then,
[tex]Q = A × V[/tex]where, V = Velocity of flow
We can assume the velocity of flow as 1.2 m/s (for a reservoir intake), then
[tex]2000 = A × 1.2A = 2000 / 1.2A = 1666.67 sq.m[/tex]
The area of each port opening is [tex]1666.67 sq.m[/tex]
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Q1. Sketch typical weathering profile of igneous and bedded sedimentary rock
Describe weathering description in your subsurface profile
Elaborate the problems you may encounter in deep foundation works on the subsurface profiles you have sketched
4
Q2. Explain major and minor discontinuities that may present in rock material and rock mass. Discuss also on how it may affect the stability
Weathering is the breakdown and alteration of rocks and minerals at or near the earth's surface. There are two types of weathering: physical and chemical. Physical weathering results from the disintegration of rocks due to the elements, temperature variations, and abrasion.
1. Sketch typical weathering profile of igneous rock: In igneous rocks, the dominant weathering products are clays, and aluminum and iron oxides. Basalt, a common extrusive rock, has a weathering profile that can be divided into three zones, namely the unaltered rock, the altered rock, and the soil. The characteristics of these zones vary depending on the climate.
2. Sketch typical weathering profile of bedded sedimentary rock: Bedded sedimentary rocks like sandstones, limestones, and shales have different weathering profiles. The weathering profile of sandstone varies from zone to zone depending on the bedding characteristics and texture of the sandstone. The weathering profile of limestones depends on the degree of cementation, and it varies with the type of cementation present. The weathering profile of shale depends on the degree of weathering of the clays and the mineral content of the rock.
They are fractures or breaks in rock masses that result from the movement of the earth's crust. Faults can affect the stability of a rock mass by creating zones of weakness that can lead to the formation of landslides and rockfalls. Bedding planes: They are planes of weakness in sedimentary rocks that result from the deposition of sediments.
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how does concrete durability affect structural performance?
Concrete is one of the most popular construction materials. Its durability is a significant factor in the structural performance of buildings, highways, bridges, and other infrastructure. The structural performance of concrete depends on its strength and durability.
The durability of concrete can also affect the strength of the structure. If the concrete is not durable, it can lead to weakening of the bond between the concrete and reinforcement steel. This can cause cracking, which can lead to further deterioration of the structure.
The durability of concrete can also affect the serviceability of the structure. When the concrete is not durable, it can lead to deformation of the structure, which can cause discomfort to the occupants. It can also lead to damage to equipment and machinery that are mounted on the structure.
The durability of concrete is a critical factor for ensuring the longevity and safety of structures. It is important to use high-quality concrete that is resistant to weathering, erosion, and chemical attacks.
Therefore, it is important to consider concrete durability when designing and constructing structures to ensure their longevity and safety.
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Choose correct answer for the followings: 1- How many types of foundation are there based on depth a. one b. Two d. Four c. Three 2- When two column loads are unequal, which of the possible footing can be provided ? b. Raft footing d. Mat footing a. Strap footing c. Trapezoidal combined footing 3- sample has the natural structure of soil as modified. a jundisturbed b. disturbed c. non-representative d. remoulded 4- Which foundation is used when the wall carries light loads or when the safe bearing pressure is very high? b. Simple strip footing d. Rat footing a. Simple pad footing c. Strap footing 5- A soil sample has inner and outer radii of 25mm and 30 mm, respectively. The area ratio of the sample is ? a. 24% a. Rigidity of the footing c. Condition of soil b. 34% d. 54% c. 44% 6- Disadvantage of seismic exploration is ? a. Lack of unique interpretation c. Both of them 7- The area ratio of sample for soil exploration should not exceed? a. 10% b. 25 % c. 30% d. 50 % 8- The number of borehole in soil investigation a. Nature and condition of soil c. Both of them 9- Terzaghi's bearing capacity equation is not applicable for--------? a. Depth effect and Inclination factor c. Both of them b. Irregular contacts often are not identified. d. None of them depending upon? b. The shape and extent of building d. None of them b. Narrow slope d. None of them 10- The pressure intensity beneath the footing depends upon b. Soil type d. All of them
1. The answer is (C) Three. 2. The answer is (D) Mat Footing. 3. The answer is (D) Remoulded. 4. The answer is (B) Simple Strip Footing. 5. The answer is (C) 44%.6. The answer is (C) Both of them.7. The answer is (D) 50%.8. The answer is (C) Both of them.9. The answer is (C) Both of them.10. The answer is (D) All of them.
Seismic exploration has the following disadvantages:Lack of unique interpretation: The recorded results are a function of several factors, and some of them are difficult to measure or have many possible interpretations. Lack of reproducibility: Due to the natural variability of soil and rock properties and other factors, measurements in two different locations or on two different occasions may produce different results.
Area ratio of the sample for soil exploration should not exceed 50%.This is because the sample's height-to-diameter ratio should not exceed 3. If it does, the sample's results may be influenced by the stress at the end of the sample. Nature and condition of soil are the determining factors for the number of boreholes required in soil investigation. Various factors like soil type, rigidity of the footing, condition of the soil, and shape and extent of the building determine the pressure intensity beneath the footing. Terzaghi's bearing capacity equation is not applicable for depth effect and inclination factor.
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A draw bench for precision forming and strengthening of carbon steel tubing has a cost of $960,000. It will have a salvage value of $74,000 after a useful life of 10 years. Parta Using the formulas, determine the depreciation charge for year 2 and the book value at the end of year 2 if straight-line depreciation is used. Depreciation charge:$_______________Book value $______________ Carry all interim calculations to 5 decimal places and then round your final answers to a whole number. The tolerance is ±1.
Given that a draw bench for precision forming and strengthening of carbon steel tubing has a cost of $960,000 and it will have a salvage value of $74,000 after a useful life of 10 years. We have to calculate the depreciation charge for year 2 and the book value at the end of year 2 if straight-line depreciation is used.
Solution: Depreciation per year = (Cost - Salvage value) / Useful life= (960000 - 74000) / 10= 88,600 per yearThe depreciation charge for year 2 = 2 x
Depreciation per year= 2 × $88,600= $177,200Book value at the end of year 2 = Cost - Depreciation charge for 2 years= $960,000 - $177,200= $782,800
Therefore, the depreciation charge for year 2 is $177,200 and the book value at the end of year 2 if straight-line depreciation is used is $782,800.Note:
The tolerance is ±1, so the answers are rounded to a whole number.
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A developing flow calls when the wall shear stress is changing due to the change in velocity profile as the boundary layer grows. So, for a pipe diameter of 1 m flowing at Re = 1000 and another again in 1 m diameter but flowing at Re = 10,000; select which of the following statement is correct:
Both laminar and turbulent flows will have equal developing lengths.
The laminar flow will have a longer developing length.
The turbulent flow will have a longer developing length.
None of the answers are correct.
For a pipe diameter of 1 m flowing at Re = 1000 and another again in 1 m diameter but flowing at Re = 10,000, the correct statement is: The turbulent flow will have a longer developing length.
Developing flow calls when the wall shear stress is changing due to the change in velocity profile as the boundary layer grows. Here, the Reynolds number is used to distinguish between the laminar and turbulent flow characteristics of the fluid. When the Reynolds number is below 2300, the flow is considered laminar, and when it is above 4000, the flow is considered turbulent.
The developing length, Ld, is defined as the distance required for the flow to become fully developed and is proportional to the Reynolds number. The expression for the developing length of a circular pipe is given by: Ld/D = 0.06 ReD. Here, D is the diameter of the pipe and ReD is the Reynolds number based on pipe diameter.
In the given case, for a pipe diameter of 1 m flowing at Re = 1000 and another again in 1 m diameter but flowing at Re = 10,000; the Reynolds number for the first case is less than 2300 which indicates laminar flow, and the Reynolds number for the second case is greater than 4000 which indicates turbulent flow.
Therefore, the turbulent flow will have a longer developing length.
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2). Derive the model energy equation (T), for a tank. Use: energy balance, energy generated by a heater and heat transfer to the outside.
The model energy equation for a tank is given by Cp * dT/dt = (Q_in - Q_out) / (dm/dt * de/dm), considering energy balance, heater input, and heat transfer to the outside.
To derive the model energy equation for a tank, we can use the principle of energy balance. The energy balance equation for the tank can be written as follows:
Rate of energy accumulation inside the tank = Energy input - Energy output
The rate of energy accumulation inside the tank can be expressed as the product of the mass of the fluid inside the tank and the rate of change of its internal energy with respect to time (dm/dt * de/dm * de/dt).
The energy input to the tank comes from the heater, which generates heat. Let's denote the rate of heat generated by the heater as Q_in.
The energy output from the tank is the heat transfer to the outside. The rate of heat transfer to the outside is given by Q_out, which is related to the heat transfer coefficient (h), the surface area of the tank (A), and the temperature difference between the tank and the surroundings (T - T_surroundings).
Therefore, the energy balance equation for the tank can be written as:
dm/dt * de/dm * de/dt = Q_in - Q_out
Next, let's express the rate of change of internal energy (de/dt) in terms of temperature (T) using the specific heat capacity (Cp) of the fluid inside the tank. The rate of change of internal energy can be written as:
de/dt = Cp * dm/dt * dT/dt
Substituting this into the energy balance equation, we get:
dm/dt * de/dm * Cp * dm/dt * dT/dt = Q_in - Q_out
Simplifying the equation, we have:
Cp * dm/dt * dT/dt = (Q_in - Q_out) / (dm/dt * de/dm)
Finally, rearranging the equation, we obtain the model energy equation for the tank:
Cp * dT/dt = (Q_in - Q_out) / (dm/dt * de/dm)
This equation represents the rate of change of temperature inside the tank (dT/dt) as a result of the energy input from the heater (Q_in) and the energy output through heat transfer to the outside (Q_out). The specific heat capacity (Cp) of the fluid and the rate of change of mass (dm/dt) and internal energy (de/dm) are also considered in the equation.
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Under what legal theory may a subcontractor be liable to a prime contractor, when the prime contractor relies on the subcontractors Bid, but the subcontractor refuses to do the work in accordance with the Bid?
Promissory Estoppel
Breach of Written Contract
Breach of Oral Contract
Unjust Enrichment
Under the legal theory of promissory estoppel, a subcontractor may be liable to a prime contractor when the prime contractor relies on the subcontractor's Bid but the subcontractor refuses to do the work in accordance with the Bid.
Promissory estoppel is a legal doctrine that prevents a party from backing out of a promise they made when the other party relied on it to their detriment. In the context of a construction contract, this means that if a subcontractor submits a bid and the prime contractor relies on that bid in making its own bid for the overall project, the subcontractor may be held liable if they refuse to do the work as outlined in their bid.
In order to prove promissory estoppel, the prime contractor would need to show that they relied on the subcontractor's bid to their detriment. This could include things like losing the bid for the overall project, incurring additional costs, or being forced to find a replacement subcontractor at a higher cost.
It's worth noting that the other legal theories listed - breach of written contract, breach of oral contract, and unjust enrichment - may also apply in this situation depending on the specific circumstances. However, promissory estoppel is the most likely theory to be applied when a subcontractor has made a promise that the prime contractor relied on to their detriment.
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A plate girder 600 mm deep is composed of 300 x 12 mm Compression flange 200 x 12 mm tension flange and 10 mm thick web.
(10 Determine the distance of the neutral axis from the compression face of the beam.
11)Calculate the area of the compression flange plus ' of the compression web.
12 Find the value of rt defined as the radius of gyration of a section compnsing the compression flange plus 1 of the compression web area, taken about an axis in plane of the web.
A plate girder 600 mm deep is composed of 300 x 12 mm compression flange, 200 x 12 mm tension flange, and a 10 mm thick web. Let's calculate the following-1.
For this, we will use the formula-NA = 0.5D - (A1 - A2) / 2 where D = Overall depth of the sectionA1 = Area of the compression flangeA2 = Area of the tension flangeA3 = Area of the web We are given.
D = 600 mmA1 = 300 x 12 = 3600 mm²A2 = 200 x 12 = 2400 mm²A3 = b*tb = width of the web = 10 mmt = thickness of the web = 10 mmA3 = b*t = 10*10 = 100 mm²NA = 0.5D - (A1 - A2) / 2NA = 0.5*600 - (3600 - 2400) / 2NA = 280 mm.
Thus, the distance of the neutral axis from the compression face of the beam is 280 mm.
2. Calculate the area of the compression flange plus ' of the compression web-We are required to find the area of the compression flange plus ' of the compression web.
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A laminated composite of graphite/epoxy (T300/5208) with stacking sequence of [45/45/15/15] is subjected to a load of Nx = 5 MN/m. All plies are of 5 mm thick. Evaluate the midplane in- plane strains and the curvatures.
E11 = 181 GPa
E22 = 10.3 GPa
G12 = 7.17 GPa
V12 = 0.28
The given laminated composite of graphite/epoxy (T300/5208) is subjected to a load of Nx = 5 MN/m. The stacking sequence of the composite is [45/45/15/15] and all plies of the composite are of 5 mm thickness.
[tex]σ1 = Nx / zc = 5 x 10^6 / 14 = 357,142.86 N/m²σ2 = 0τ12 = 0[/tex]Here, zc is the distance from the laminate mid-plane to extreme compression layer. And the extreme layer of the composite is considered as a compression layer, and the other one is considered as a tension layer. Mid-plane strains for the composite laminate are calculated as;
[tex]ɛ1 = σ1 / E11 + σ2 / E12 = (357,142.86 / 181000) + (0 / 0.28 × 7.17 × 10^3) = 0.0019717[/tex]
[tex]ɛ2 = σ1 / E21 + σ2 / E22 = (357,142.86 / 10.3) + (0 / 0.28 × 7.17 × 10^3) = 34.708[/tex]
Curvature (κ) is given by the equation;κ = M / (E × h²)Where M is the bending moment, E is the Young's Modulus and h is the distance from the neutral axis to the outer surface of the curved beam. Here, since we are evaluating curvature at the mid-plane, therefore, h = 7.5 mm.Moment (M) can be calculated by using the expression;
[tex]M = Nx × zc = 5 × 10^6 × 7 = 35 × 10^6[/tex]
[tex]Nmmκ = M / (E × h²) = (35 × 10^6) / (181 × 10^3 × (7.5)²) = 0.07079 / mm[/tex]
The mid-plane in-plane strains for the composite laminate are 0.0019717 (in the 1 direction) and 34.708 (in the 2 direction). And the curvature of the given composite laminate is 0.07079/mm.
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For how many years must an investment of Php63,000 provide a continuous flow of funds at the rate of Php16,000 per
a) 5 yrs b) 4 yrs c) 6 yrs d) 3 yrs e) not here
The correct option is a) 5 yrs. To find out the number of years, we can use the formula: Total Investment/Annual Cash Flow = Number of Years. In this formula, Total investment is the amount of money invested, and Annual Cash Flow is the amount of money received every year from the investment.
Using the given formula: Total Investment/Annual Cash Flow = Number of Years. Here, Total investment = Php 63,000 and Annual Cash Flow = Php 16,000. Substituting these values in the above formula, we get: Number of Years = 63,000/16,000.Number of Years = 3.9375 years. Hence, the investment must provide a continuous flow of funds for 5 years, which is option a).
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Content Mtsemus D. Why do we take the internal pressure in our casing to be zero when calculating collapse loads ?
When calculating collapse loads, the internal pressure in a casing is assumed to be zero. In this regard, this question seeks to answer the reason why the internal pressure is set to zero when determining the collapse loads.
The internal pressure in the casing is the pressure that is exerted inside the casing. When the pressure inside the casing is high, the casing wall will experience tension. On the other hand, if the internal pressure is low, the casing wall will experience compressive stress. The higher the internal pressure in the casing, the greater the tension that the casing wall will experience. This, in turn, makes the casing more susceptible to collapse. It is for this reason that when calculating collapse loads, the internal pressure in the casing is assumed to be zero. This assumption ensures that the calculations are carried out under the worst-case scenario. In conclusion, the internal pressure in a casing is assumed to be zero when calculating collapse loads because the higher the internal pressure, the greater the tension that the casing wall will experience. This makes the casing more susceptible to collapse.
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It is known that for a certain stretch of a pipe, the head loss is 3m per km length. For a 3.0m diameter pipe, if the depth of flow is 0.75m, find the discharge (m³/s) by using Kutter and Ganguillet's equation. n = 0.020.
Kutter and Ganguillet's formula is used to calculate the discharge of a pipe, and the formula is given below :$v =\frac{1}{n} R^{2/3} S^{1/2}$Where:v = velocity in m/sR = hydraulic radius in meter (A/P)S = slope of the energy line in m/mn = coefficient of roughness.
The formula for hydraulic radius is as follows:$R =\frac{A}{P} $where A is the cross-sectional area of the pipe and P is the wetted perimeter. By taking the velocity, hydraulic radius, and cross-sectional area of the pipe, we can determine the discharge of the pipe.
For a 3.0 m diameter pipe, if the depth of flow is 0.75 m, find the discharge by using Kutter and Ganguillet's equation. The hydraulic radius can be calculated as follows:
Given that, Diameter = 3mRadius, r = Diameter/2 = 3/2 = 1.5mDepth of flow, y = 0.75 m Area, A = πr²=π(1.5)²=7.069m² Now, Wetted perimeter, P = 2πr$=2π×1.5=9.4248m$The slope of the energy line can be determined using the head loss formula, which is given as follows: Head loss per km length of pipe,
h = 3mThis means, Head loss per meter length of pipe, H = 3/1000= 0.003m/m Now, the formula for slope of energy line, S is given by: S = H/LWhere L is the length of the pipe. In our case, L is not given, so we can assume any length of the pipe and calculate the discharge. Let's assume that L = 1000 m.
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A triangular prism side of base 30 mm and axis 50 mm long rests with its base on the H.P such that one of its rectangular faces is inclined 45° with the v.p and the nearest edge is 8mm in front of it. Draw its projections.
Base of the triangular prism = 30 mm Axis of the triangular prism = 50 mm Inclination of rectangular face = 45°Distance of the nearest edge = 8 mm.
Step 1: Draw a horizontal line XY of length 70 mm, which is the total length of the triangular prism. At the midpoint of XY, draw a vertical line MN of length 50 mm, which is the axis of the prism. Draw the front view of the triangular prism on this line, which will be a regular triangle of side length 30 mm.
Step 2: From the top right corner of the front view of the triangular prism, draw a line perpendicular to XY, which represents the base of the prism. At a distance of 50 mm from this line, draw another parallel line, which represents the top of the prism.
Step 3: From the bottom left corner of the front view, draw a line at an angle of 45° to the XY line, which represents the ground. This line will be the inclined face of the prism. From the end points of this line, draw horizontal lines to meet the sides of the triangle at points P and Q.
Step 4: From the leftmost corner of the top view, draw a line parallel to the XY line to represent the ground. At a distance of 8 mm from this line, draw another parallel line to represent the front face of the prism. From the corners of this line, draw lines to the midpoints of the opposite sides of the hexagon.
Step 5: Draw the remaining edges of the prism Complete the projections by joining the corners of the front and top views with lines. Mark the dimensions and label the points as required.
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Activity Definition is typically performed by which of the following:
a) Project Team Members responsible for the work package
b) Project Officer
c) Project Stakeholder
d) Project Manager who created the WBS
Activity Definition is typically performed by the Project Manager who created the WBS.What is Activity Definition?Activity definition is the process of dividing the project deliverables into smaller, more manageable components, known as activities.
It involves identifying and documenting the specific required to complete the work package.How Activity Definition is performed?Activity Definition is typically performed by the Project Manager who created the WBS. The WBS (Work Breakdown Structure) outlines the major deliverables and their component tasks.Activity Definition is an essential component of project planning as it gives the project team a clear understanding of the work to be performed and how it will be completed.
When activities are defined accurately, project managers can estimate project timelines more precisely, identify resource needs, and develop a more robust project schedule for all project stakeholders.What is the importance of Activity Definition?Activity definition is a critical component of project management as it provides the necessary foundation for project planning, scheduling, budgeting, and performance measurement. By defining activities accurately, project managers can ensure that project timelines are more precisely estimated, project risks are identified and mitigated, and project progress is tracked against the schedule. When the activities are defined and the project schedule is created, it becomes easier to understand the project's timeline.
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Boeing, Inc. is attempting to determine whether an existing
machine is capable of milling an engine part that has a design
specification of 5.0mm+- 0.10 mm. The standard deviation of the
process is
The goal of Boeing, Inc is to determine whether an existing machine can mill an engine part that has a design specification of 5.0mm +- 0.10 mm. The process has a standard deviation of σ = 0.02 mm.
Given that the standard deviation of the process is σ = 0.02 mm. Now, we need to determine if the existing machine can produce a part that meets the design specification using the given standard deviation (σ).The tolerance limit (T.L) can be calculated as;
T.L = (Upper Limit – Lower Limit) / 2 where;
Upper Limit = 5.0 + 0.10 = 5.10 mm
Lower Limit = 5.0 – 0.10 = 4.90 mm
T.L = (5.10 – 4.90) / 2 = 0.10 / 2 = 0.05 mm
The value of T.L = 0.05 mm
The process standard deviation (σ) = 0.02 mm
Using the formula;
P = 0.6827, approximately 68.27% of the parts produced by the existing machine fall within the tolerance limit (5.0 ± 0.10)
With a process standard deviation of σ = 0.02 mm and a tolerance limit of 0.05 mm, approximately 68.27% of the parts produced by the existing machine fall within the tolerance limit (5.0 ± 0.10). It can, therefore, be concluded that the existing machine can produce parts that meet the design specification.
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Two gray bodies: Body 1 at 200oC and Body 2 at 100oC with emissivities of 0.6 and 0.8 respectively.
Required: The net heat transfer rate by radiation from Body 1 to Body 2 per m2, if: Bodies 1 and 2 are parallel disks each with 4 cm radius and 5 cm apart.
The net heat transfer rate by radiation from Body 1 to Body 2 per square meter is approximately 10.35 W/m^2.
The problem concerns the radiation heat transfer between two parallel disks. The disks have radii of 4 cm and emissivities of 0.6 and 0.8. They are maintained at temperatures of 200°C and 100°C respectively, with a distance of 5 cm between them. The objective is to determine the net heat transfer rate per square meter from Body 1 to Body 2, assuming no convective heat transfer occurs between the disks.
To calculate the heat transfer rate by radiation, the Stefan-Boltzmann law is applied:
q'' = σA1A2 (T1^4 - T2^4) / πD^2
Where:
q'' is the heat transfer rate by radiation
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2-K^4)
A1 and A2 are the areas of the two disks
T1 and T2 are the temperatures of the two disks
D is the distance between the two disks
From the equation, it is evident that the heat transfer rate by radiation is directly proportional to the area of the disks and the temperature difference between them, while inversely proportional to the square of the distance between them.
Calculating the area of each disk, A, we find A = πr^2 = 3.14 x 4^2 / 10000 = 0.000502 m^2.
Substituting the given values into the Stefan-Boltzmann law, we obtain:
q'' = 5.67 x 10^-8 x 0.000502 x 0.000502 x (473.15^4 - 373.15^4) / (π x 0.05^2)
≈ 10.35 W/m^2
Therefore, the net heat transfer rate by radiation from Body 1 to Body 2 per square meter is approximately 10.35 W/m^2.
Answer: 10.35 W/m^2.
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