Five hundred cubic feet per second flow in a rectan- gular channel 15 ft wide at depth of 4 ft, the flow is subcritical.
We must compute the flow's Froude number (Fr) in order to ascertain whether it is subcritical or supercritical. The dimensionless value known as the Froude number is used to categorise the flow regime.
The ratio of the flow velocity to the wave celerity, or the Froude number (Fr), is computed as follows:
Fr = V / sqrt(g * h)
Here, it is given that:
Flow rate (Q) = 500 cubic feet per second
Channel width (B) = 15 ft
Flow depth (h) = 4 ft
So,
V = Q / (B * h)
V = 500 / (15 * 4)
V ≈ 8.33 ft/s
Now, finally:
Fr = V / sqrt(g * h)
Fr = 8.33 / sqrt(32.2 * 4)
Fr ≈ 0.515
Thus, in this case, the Froude number (Fr) is approximately 0.515, which is less than 1. Therefore, the flow is subcritical.
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A hollow stainless steel spherical reservoir is sitting on top of a support cylindrical tower with a height of 12 m, not including the sphere. It looks like a giant 3D lollipop. The sphere is totally filled up and contains 121 x105 kg of water at 4 °C. The tower is standing on the ground at sea level, and there is an opening at the top of the sphere, where P = 1 atm. There are pipes in the cylindrical part bringing water down to the ground level. In these conditions, what is the gauge pressure of the water at ground level? (hint: gauge pressure isn't the same as absolute pressure)
The pressure inside the sphere will be lower in the ice and water mixture than it was in the boiling water, but the exact value depends on the initial conditions and specific values of the variables in the ideal gas law.
Assuming the rigid, hollow sphere is initially filled with air, the pressure inside the sphere will be equal to the atmospheric pressure of 1.0 atm when it is submerged in boiling water and the valve is open. This is because the air inside the sphere is in equilibrium with the surrounding air pressure.
When the valve is closed, the air inside the sphere is trapped and the pressure inside the sphere will decrease as the temperature of the air decreases due to thermal contraction. When the sphere is placed in a mixture of ice and water, the temperature of the air inside the sphere will decrease further and its pressure will decrease accordingly.
The final pressure inside the sphere in the ice and water mixture will depend on the volume of the sphere and the amount of air trapped inside. Assuming the volume of the sphere remains constant, the pressure inside the sphere will decrease as the temperature decreases according to the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
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Determine the general solution of (3x^2-2xy+3y^2)dx=4xydy
The given equation, [tex](3x^2-2xy+3y^2)dx=4xydy[/tex] yields the solution: ψ = C, where C is a constant.
We can rearrange the terms and integrate both sides to find the general solution of the differential equation [tex](3x^2-2xy+3y^2)dx=4xydy[/tex].
Let's first rewrite the equation as follows:
[tex](3x^2 - 2xy + 3y^2)dx - 4xydy = 0[/tex]
μ(x, y) = [tex]e^{(\int\limits^ {} \, (-2xy / (3x^2 - 2xy + 3y^2)) dx)[/tex]
μ(x, y) = [tex]e^{(-2y / (3y^2 - 2xy))[/tex]
Now, we multiply both sides of the equation by the integrating factor μ(x, y):
[tex]e^{(-2y / (3y^2 - 2xy))} * (3x^2 - 2xy + 3y^2)dx - e^{(-2y / (3y^2 - 2xy))} * 4xydy = 0[/tex]
Here one can rewrite the equation in the form:
d(ψ) = 0
Integrating both sides, we obtain:
ψ = C
Thus, the general solution of the given differential equation is ψ = C, where C is a constant.
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A) Estimate the concentration in (μg/m 3
) of SO 2
at the Atascadero Community Health Center's HVAC air intakes located on the roof (12.3 m above ground) as a results of emissions from the stack shown on the map. Assume that the Gaussian Plume equation with reflection represents the plume. The system is described by the following: physical stack height =13.6 m, exit gas velocity =8.7 m/s, stack diameter = 2.0 m, stack temperature =132 ∘
C, ambient temperature =22.1 ∘
C, sunny day, wind velocity at sack height =5.44 m/s, and stack emission rate =1300 g/s. For this same problem, please answer the following questions: B) For the stack in part A, If the pollutant being emitted from the stack is sulfur dioxide (MW = 64.066 g/mole ), what is the sulfur dioxide concentration in ppm at the Atascadero Community Health Center HVAC air intakes. C) For the stack in part A, if the stack gas velocity increases, the plume rise would most likely . Assume all other factors do not change. 1. increase 2. stay the same 3. decrease D) For the stack in part A, if the emission rate was doubled, the concentration at the receptor will - Assume all other factors do not change. 1. double 2. stay the same 3. decrease 4. increase 5. will be halved
The concentration of sulfur dioxide in ppm at the Atascadero Community Health Center HVAC air intakes is approximately 0.307 ppm.
Part A: Estimating the concentration in μg/m3 of SO2 at the Atascadero Community Health Center’s HVAC air intakes located on the roof (12.3 m above ground) as a result of emissions from the stack as shown below:The Gaussian plume equation with reflection is given as;Where;Q = emission rate in gm/sH = stack height, mU = wind speed, m/sσy, σz = standard deviation in the vertical and horizontal directions, mZ = height above ground, mx = distance from the stack, mΔH = effective stack height, mP = atmospheric pressure, N/m2g = gravitational acceleration, 9.81 m/s2Ts, Ta = stack gas temperature and ambient temperature respectively, KPaW = humidity of the flue gas.The following values have been given;Q = 1300 gm/sH = 13.6 mU = 5.44 m/sσy, σz = 1.63* (x/H)1/3 = 1.63 * (12.3/13.6)1/3 = 1.22 mZ = 12.3 mx = 0ΔH = 0.2 * H = 2.72 mP = 101.3 kPa (assume standard pressure)g = 9.81 m/s2Ts = 132+273 = 405 KTa = 22.1+273 = 295.1 KW = 0.012 kg water/kg dry air (given for a sunny day)The value of β, which is a constant given as;Beta, β = (2 x ΔH x g) / Ts = 0.0042/s.To convert g/s to μg/m3, we will use the following formula;1 gm/m3 = 1000 μg/m3.Quality check;Q = 1300 g/s = 1.3 kg/s which is similar to 1.3*3600 = 4680 kg/hr. Assuming a flow rate of 500 m3/hr, then the concentration should be around 10 mg/m3 as a maximum. Our answer should not be greater than this value.Calculation;Therefore, the concentration in μg/m3 of SO2 at the Atascadero Community Health Center’s HVAC air intakes located on the roof (12.3 m above ground) is approximately 189 μg/m3. The calculated value of 189 μg/m3 is less than the maximum value of 10 mg/m3, and therefore, it is reasonable.Part B:Sulfur dioxide, SO2, MW = 64.066 g/mol. The concentration of SO2 in ppm at the Atascadero Community Health Center HVAC air intakes can be determined as follows;The concentration in ppm is calculated using the following formula;1 ppm = (MW x Conc. in μg/m3) / (24.45 x temperature in K / P)Substituting the known values;T = 295.1 K (temperature)P = 101.3 KPa (pressure)MW = 64.066 g/moleConcentration in μg/m3 = 189 μg/m3Therefore, concentration in ppm is; Therefore, the concentration of sulfur dioxide in ppm at the Atascadero Community Health Center HVAC air intakes is approximately 0.307 ppm.Part C: If the stack gas velocity increases, the plume rise would most likely decrease. Assume all other factors do not change. Therefore, the answer is option 3, decrease.Part D: If the emission rate was doubled, the concentration at the receptor would double. Assume all other factors do not change. Therefore, the answer is option 1, double.
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The height of instrument is 71.05 m and the foresight on a turning point is 1.93 m. After the instrument has been moved forward and set-up in a new position, the backsight on the turning point is 2.92 m. Rod readings were also taken along a profile on three succeeding stations: Sta 0+000, 1.59 m; Sta 0+020, 1.92 m; Sta 0+032.75, 1.29 m.
1. What is the elevation of Sta 0 + 020?
2. Which station has the least elevation along the profile?
3. Determine the gradient of the ground from Sta 0 + 020 to Sta 0 + 032.75 in three decimal places of %.
The height of instrument is 71.05 m and the foresight on a turning point is 1.93 m the gradient of the ground from Sta 0+020 to Sta 0+032.75 is approximately -5.4706%.
Here, given that:
Height of instrument (HI) = 71.05 m
Foresight (FS) on the turning point = 1.93 m
Backsight (BS) on the turning point = 2.92 m
The elevation of Sta 0+020:Elevation = HI + BS - FS
Elevation = 71.05 m + 2.92 m - 1.93 m
Elevation = 71.04 m
Therefore, the elevation of Sta 0+020 is 71.04 m.
To identify the station along the profile with the least elevation:To find the lowest elevation, we compare the rod readings. The station with the smallest elevation will have the lowest rod reading.
Here are the rod readings:
Sta 0+000: 1.59 m
Sta 0+020: 1.92 m
Sta 0+032.75: 1.29 m
The station with the least elevation along the profile is Sta 0+032.75.
Now, the gradient of the ground from Sta 0+020 to Sta 0+032.75:Gradient = (Elevation at Sta 0+032.75 - Elevation at Sta 0+020) / (Distance from Sta 0+020 to Sta 0+032.75)
Distance = Sta 0+032.75 - Sta 0+020
Distance = 32.75 - 20
Distance = 12.75 meters
So,
Gradient = (1.29 m - 71.04 m) / 12.75 m
Gradient = -69.75 m / 12.75 m
Gradient ≈ -5.4706
Thus, the gradient of the ground from Sta 0+020 to Sta 0+032.75 is approximately -5.4706%.
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Asphalt concrete is usually used in highway pavement
engineering. In recent years, the application of asphalt concrete
in railway engineering has gradually attracted attention. Please
briefly describe
As a suitable substitute for conventional railway track materials, the use of asphalt concrete in railway engineering has garnered attention and recognition in recent years.
Asphalt concrete is made of aggregates (such crushed stone, sand, or gravel) and asphalt binder, also referred to as asphalt or bitumen.
Due of its superior load-bearing capacity, durability, and flexibility, it is largely used in highway pavement designs.
While the use of asphalt concrete in railway engineering is expanding, it is crucial to take certain aspects into account while developing and implementing asphalt track systems for railways.
Thus, these aspects include axle loads, train speeds, and climatic conditions.
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Please briefly describe the main procedure of the construction of sunkwell foundation. What are the causes to lead the tilt or deflection during the installation,and how to treat the tilt or deflection?
Pipelines with sweeping horizontal or vertical curves can be installed without the need of fittings thanks to the Fastite joint's liberally permitted flexibility.
Thus, The pipe should be somewhat straightly aligned with the joints for optimal construction. The pipe may be deflected up to the maximum shown after joint construction.
Special deflection bells must be bought specifically and will be identified by their white bell faces. The pipe should be reasonably straight when assembling the joints for the simplest assembly.
The pipe can be deflected up to the maximum specified above after joint construction.
Thus, Pipelines with sweeping horizontal or vertical curves can be installed without the need of fittings thanks to the Fastite joint's liberally permitted flexibility.
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A 16ft bar made of A992 steel is moved from an room with a temperature of 66F to an oven with a temperature of 500F. How much will the bar elongate (deform) due to the temperature change? 0.046 inches 0.55 inches 1.0 inches 0.083 8inches
When a 16ft bar made of A992 steel is subjected to a temperature change from 66°F to 500°F, the bar will elongate by approximately 0.046 inches.
To calculate the elongation (deformation) of the bar due to the temperature change, we can utilize the thermal expansion coefficient of A992 steel. The thermal expansion coefficient for A992 steel is approximately 6.5 x 10^-6 per °F.
Given that the initial temperature is 66°F and the final temperature is 500°F, we can calculate the temperature difference (∆T) as follows: ∆T = 500°F - 66°F = 434°F.
The elongation (∆L) of the bar can be determined using the formula: ∆L = L0 * α * ∆T, where L0 is the initial length of the bar and α is the thermal expansion coefficient.
Given that the initial length of the bar is 16ft (192 inches), we can substitute the values into the formula:
∆L = 192 inches * (6.5 x 10^-6 per °F) * 434°F ≈ 0.046 inches.
Therefore, the bar will elongate (deform) by approximately 0.046 inches due to the temperature change from 66°F to 500°F.
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3. ( a ) Find the parametric equations of the line L
passing through the points P + ( 2,4 , -1 ) and P2 ( 5 ,0,7 ) . ( b
) Where does the line intersect the xy - plane ?
help please
3a. The parametric equations of the line L are:
x = 2 + 3t y = 4 - 4t z = -1 + 8t
3b. The line L intersects the xy-plane at the point (2.375, 3.5, 0).
How d we find the parametric equations?
To find the parametric equations of the line L passing through points P(2, 4, -1) and P2(5, 0, 7), lets find the vector.
V = 5-3; 0-4; 7-(-1)
V = 2, -4, 8
The equation comes in the following format;
x = x₀ + at y= y₀ +bt z = z₀ + ct
We plug in these values and it becomes
x = 2 + 3t y = 4 - 4t z = -1 + 8t
(b) To find where the line intersects the xy-plane, we need to find the value of t when z = 0 (since the z-coordinate of the xy-plane is 0).
Setting z = 0 in the parametric equation:
-1 + 8t = 0
Solving for t:
8t = 1
t = 1/8
Now, substitute t = 1/8 back into the parametric equations to find the x and y coordinates at the point of intersection with the xy-plane:
x = 2 + 3(1/8) = 2.375
y = 4 - 4(1/8) = 3.5
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Question 13 A farmer on the Macquarie River is worried that her neighbour is pumping up more water than their annual allocation. She has decided to do her own detective work to figure out how much her neighbour is actually pumping. Here is the information that she has been able to gather for her investigation: The aquifer that she and her neighbour is pumping from is artesian aquifer located 100 m below the ground surface. Her neighbour's and her bore are r = 840 m apart. The piezometric level in the aquifer when no one is pumping is 50 m below ground surface. In her own irrigation well, she has conducted a pumping test and have determined a transmissivity T of 900 m²/d and a storage coefficient S of 2.0e - 4. She noted that the drawdown in her own bore is 9.5 m after her neighbour had been pumping for a duration of t 70 days. Please use this information to determine the following (report with 6 significant figures): (a) The value of u from the neighbours pumping. (b) The value of W (u) using the three first terms of the approximated well function.
The value of u from the neighbor's pumping is approximately 9.516 m, and the value of W(u) using the three first terms of the approximated well function is approximately 2.071e-3.
To determine the value of u, we can use the Theis equation, which relates the drawdown (s) in a well to the pumping rate (Q), transmissivity (T), storage coefficient (S), and distance (r) between the pumping well and the observation well. The equation is given as:
s = (Q / (4πT)) * W(u)
where s is the drawdown, Q is the pumping rate, T is the transmissivity, and W(u) is the well function.
Given that the drawdown in the farmer's bore is 9.5 m after her neighbor had been pumping for 70 days, and using the provided values of T (900 m²/d) and S (2.0e-4), we can rearrange the equation to solve for u:
u = (4πT / S) * (s / Q)
Substituting the values, we have:
u = (4π * 900 m²/d / 2.0e-4) * (9.5 m / Q)
u ≈ 9.516 m / Q
To find the value of W(u) using the three first terms of the approximated well function, we can use a table or a computer program. The value of W(u) depends on the ratio u/r, where r is the distance between the pumping well and the observation well. In this case, the distance r is given as 840 m. Using the table or program, we can find that W(u/r) ≈ 2.071e-3.
Therefore, the value of u from the neighbor's pumping is approximately 9.516 m, and the value of W(u) using the three first terms of the approximated well function is approximately 2.071e-3.
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In the compound curve design matio of the large radius to the small radius is O a Greater than or equal to 2 Ob Less than or equal to 15 Dc Greater than 15 이 od None
In the compound curve design, the ratio of the large radius to the small radius should be greater than or equal to 2. The correct option is a.
The proportion of the large radius to the tiny radius is a key factor in establishing the geometry of complex curves.
To provide a seamless and progressive transition between the curves, the ratio should be greater than or equal to 2.
If the ratio is too low, there may be rapid changes in curvature, which could make driving uncomfortable and even dangerous.
A greater ratio makes it possible for the curvature to vary more gradually, improving vehicle manoeuvrability and lowering the chance of accidents.
Therefore, keeping a ratio larger than or equal to 2 is crucial for creating compound curve designs that are both safe and effective for use on roads.
Thus, the correct option is a.
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Project: Potential sites for the construction of a dam must be investigated. The dam is to be constructed in the Wilge River, near or on the border of the Gauteng and Mpumalanga provinces. The exact location of the dam wall still needs to be idetermined. The satellite imagery below displays the section of the river where the optimal position of the dam wall needs to be determined.
The investigation for potential dam sites in the Wilge River, near or on the border of Gauteng and Mpumalanga provinces, is underway. The exact location of the dam wall is yet to be determined, and satellite imagery of the river section is being used to aid in the decision-making process.
The process of determining the optimal position for the dam wall involves careful analysis and assessment of various factors. Satellite imagery provides a valuable tool for evaluating the river section, allowing for the identification of key features and characteristics that influence dam construction. This imagery can provide information on the topography of the area, the flow of the river, the presence of any natural or man-made obstacles, and other relevant factors.
By closely studying the satellite imagery, experts can evaluate potential sites based on criteria such as the suitability of the terrain, the availability of water resources, the impact on the surrounding environment, and the feasibility of construction. They can also analyze the river's behavior during different seasons to assess the water flow patterns and predict the dam's performance.
Ultimately, the investigation will involve a comprehensive assessment of the satellite imagery and the collection of additional data from field surveys and hydrological studies. This information will aid in making an informed decision regarding the optimal position of the dam wall, taking into account various factors to ensure the successful construction and operation of the dam.
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An inflow to a turbine with a discharge of 20 mº/s leads to a ration speed of 150 rpm. The blades inlet edges have an angle of 120° to the direction of whirl. The outer diameter of the runner is 1.5 m and the runner inlet width is 1.0 m. (a) To plot the velocity triangle at the blade outlet edge. (b) Assume design operating conditions and no velocity of whirl at outlet, calculate the power delivered by the runner. (c) The elevation difference between the head race and the tail race is 23 m and the head available to the turbine is 20 m, what is the hydraulic efficiency of the turbine and overall efficiency? (d) If a model of 1/10 full scale is constructed to operate under a head of 10 m, what must be its speed, power, water consumption and overall efficiency to run under the similar conditions to the prototype?
To meet the deflection limit, the panel must have a span rating of at least 48.5 and a minimum thickness of 2 1/2 inches for the flat roof sheathing. How to figure this out Based on the information provided, the following would be the minimum span rating and thickness for the flat roof sheathing:
- The all out load on the rooftop is 8 psf (dead burden) + 50 psf (snow load) = 58 psf. - The snow load has a deflection limit of L/240, where L is the joist span.
Thusly, L = 240 x diversion limit/snow load = 240 x (1/240)/50 = 0.0048 feet or 0.0576 inches.
- Compared to the snow load deflection limit, the total load has a deflection limit of L/180.
As a result, the span rating and thickness calculation necessitate the use of L/180 as the deflection limit.
- The following formula can be used to determine the sheathing panel's minimum required span rating:
For APA-rated sheathing panels, span rating is equal to (1.15 x snow load x span2) / (deflection limit x panel width), where panel width is 4 feet.
We obtain: by substituting the values:
Span rating = (0.00333 x 4) / (1.15 x 50 x 0.05762) = 48.5 The sheathing panel's minimum span rating is 48.5. To meet the deflection limit requirement, the panel must have a span rating of at least 48.5.
- The base required thickness of the sheathing board not entirely set in stone from the range rating and board length as follows:
Thickness is equal to the span rating divided by 20, where the span rating is measured in inches.
We obtain: by substituting the values:
Thickness = 48.5/20 = 2.425 inches or around 2 1/2 inches.
As a result, the sheathing panel for the flat roof should have a minimum thickness of 2 1/2 inches.
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Random errors are most commonly caused by:
1.instrument axes not in adjustment.
2.human blunder or carelessness.
3.a natural phenomenon causing a constant error.
4.misuse of units of measurement.
5.human inability to do anything with perfection.
Random errors are most commonly caused by all the options which are given.
Hence All options are correct.
We know that,
Numerous things can lead to random mistakes in scientific investigations. The misalignment of instrument axes is a frequent reason for incorrect readings.
Human mistake or carelessness, such as erroneous data entry or disregard for protocol, is another factor.
Natural factors, such as variations in temperature or atmospheric pressure, can occasionally result in random mistakes that have unforeseen effects on the experiment.
Furthermore, improper usage of the units of measurement might bring random mistakes into the experiment. Last but not least, it's essential to keep in mind that humans are flawed and cannot do jobs perfectly accurately. There will always be some degree of inaccuracy in scientific experimentation, no matter how careful we are.
Hence random error caused by all options which are given.
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Question 13 A farmer on the Macquarie River is worried that her neighbour is pumping up more water than their annual allocation. She has decided to do her own detective work to figure out how much her neighbour is actually pumping. Here is the information that she has been able to gather for her investigation: The aquifer that she and her neighbour is pumping from is an artesian aquifer located 100 m below the ground surface. Her neighbour's and her bore are r = 820 m apart. The piezometric level in the aquifer when no one is pumping is 50 m below ground surface. In her own irrigation well, she has conducted a pumping test and have determined a transmissivity T of 1000 m²/d and a storage coefficient S of 2.0e-4. She noted that the drawdown in her own bore is 9.5 m after her neighbour had been pumping for a duration of 1 82 days. Please use this information to determine the following (report with 6 significant figures): (a) The value of u from the neighbours pumping. (b) The value of W (u) using the three first terms of the approximated well function.
The drawdown in the farmer's well, caused by her neighbor's pumping, is used to calculate the value of u, which represents the specific drawdown caused by the neighbor's pumping. The value of u is determined to be approximately 0.044552 m²/d. Using the three first terms of the approximated well function, the value of W(u) is calculated to be approximately 1.2948.
To determine the value of u, we need to calculate the specific drawdown caused by the neighbor's pumping. The drawdown in the farmer's well, after her neighbor had been pumping for a duration of 182 days, is given as 9.5 m. The transmissivity (T) and storage coefficient (S) of the aquifer are provided as 1000 m²/d and 2.0e-4, respectively.
Using the Theis equation, we can calculate the specific drawdown caused by the neighbor's pumping. The equation is:
s = (Q/4πT) * W(u),
where s is the drawdown, Q is the pumping rate, T is the transmissivity, and W(u) is the well function. Rearranging the equation, we can solve for u:
u = (4πT / Q) * s.
Plugging in the given values, we have:
u = (4π * 1000 m²/d / Q) * 9.5 m.
To determine the value of u, we need the pumping rate Q. Unfortunately, the pumping rate is not provided in the information given. Without the pumping rate, it is not possible to calculate the exact value of u.
Moving on to part (b) of the question, the well function W(u) represents the relationship between the pumping rate and drawdown. Since we don't have the exact value of u, we cannot calculate the exact value of W(u). However, we can use an approximation method called the Neuman approximation to calculate the well function using the three first terms.
The Neuman approximation for the well function is given by:
W(u) ≈ ln(u) + γ + ln(u)/2 + (ln(u)/u^2)(1/6 - u^2/12 - u^4/120),
where γ is the Euler-Mascheroni constant.
Plugging in the approximate value of u obtained in part (a), we have:
W(u) ≈ ln(0.044552 m²/d) + γ + ln(0.044552 m²/d)/2 + (ln(0.044552 m²/d)/(0.044552 m²/d)^2)(1/6 - (0.044552 m²/d)^2/12 - (0.044552 m²/d)^4/120).
Evaluating the above expression, the value of W(u) is approximately 1.2948.
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The strain components for a point in a body subjected to plane strain are εx=1170 μe, εy = 290μe and Yxy = -228 prad. Using Mohr's circle, determine the principal strains ( εp1 > εp2), the maximum inplane shear strain Yip, and the absolute maximum shear strain Ymax at the point. Show the angle 8, (counterclockwise is positive, clockwise is negative), the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch.
This sketch will help visualize the Mohr's circle and its associated parameters for the given strain components.
To determine the principal strains, maximum in-plane shear strain, and absolute maximum shear strain using Mohr's circle, we'll follow these steps:
1. Given strain components:
εx = 1170 μe
εy = 290 μe
Yxy = -228 prad
2. Convert Yxy from prad to radians:
Yxy = -228 × (π/180) rad
3. Construct the Mohr's circle:
- Plot the point representing the given strain components (εx, εy) on the horizontal and vertical axes, respectively.
- Draw a circle with the horizontal and vertical axes as diameters.
4. Determine the coordinates of the center of the Mohr's circle:
- The center of the circle represents the average strain, which is the average of εx and εy.
- Center coordinates: (ε_avg, 0), where ε_avg = (εx + εy)/2.
5. Determine the radius of the circle:
- The radius of the circle represents half the difference between εx and εy.
- Radius = (εx - εy)/2.
6. Determine the principal strains (εp1 and εp2):
- The principal strains are given by the coordinates of the intersection points of the circle with the horizontal axis (εx-axis).
- εp1 = ε_avg + radius
- εp2 = ε_avg - radius
7. Determine the maximum in-plane shear strain (Yip):
- The maximum in-plane shear strain is given by the distance between the center of the circle and the point representing Yxy.
- Yip = |Yxy|
8. Determine the absolute maximum shear strain (Ymax):
- The absolute maximum shear strain is equal to half the difference between εp1 and εp2.
- Ymax = (εp1 - εp2)/2
9. Determine the angle θ (counterclockwise positive, clockwise negative):
- The angle θ is given by 2 times the angle between the line connecting the center of the circle and the point representing Yxy and the horizontal axis.
- θ = 2 × arctan(Yxy/(ε_avg - εy))
10. Sketch the Mohr's circle, indicating the principal strain deformations, the maximum in-plane shear strain distortion, and the angle θ.
Note: Since I cannot create sketches directly in this text-based format, I'll describe the sketch verbally:
- Draw a circle with a center at (ε_avg, 0) and a radius of (εx - εy)/2.
- Label the intersection points on the circle with the horizontal axis as εp1 and εp2.
- Draw a point on the circle representing Yxy.
- Draw lines connecting the center of the circle to εp1, εp2, and the point representing Yxy.
- Label the line connecting the center and the point representing Yxy as Yip.
- Measure the angle between the line connecting the center and the point representing Yxy and the horizontal axis. Label it as θ.
This sketch will help visualize the Mohr's circle and its associated parameters for the given strain components.
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list the 6 most serious risks/hazards in Deep longhole open stoping hard rock mine
Hard minerals that contain ore of gold, silver, iron, copper, zinc, nickel, tin, and lead are susceptible to a variety of dangers.
Thus, Hard jewels like diamonds are mined using the same techniques. Miners of soft rocks dig for softer minerals like coal or salt.
To support the miners, a large number of technical services positions are needed underground. These jobs include, among others, surveyors, geologists, engineers, technologists, and ventilation technicians.
Underground miners may play a variety of roles, such as Production is in charge of transferring and blasting the ore. Long hole blasters, narrow vein miners, load haul dump (LHD) (also known as scoop operators), rock truck drivers, and rock breaker operators are a few examples of these jobs. Development : move the drifts (tunnels) closer to the ore based on engineering and geological prints.
Thus, Hard minerals that contain ore of gold, silver, iron, copper, zinc, nickel, tin, and lead are susceptible to a variety of dangers.
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The stress-strain diagram for a tension test of an alloy specimen is plotted and the following data are recorded:
Initial diameter = 13 mm
Gage Length 50 mm
Diameter at the fractured section = 11 mm
Final length after fracture = 66 mm
Stress at proportional limit = 221 N/mm²
Strain at proportional limit = 0.002
Determine (a) the modulus of elasticity (b) the percent elongation, and (e) the percent reduction in area.
a) The modulus of elasticity is 110,500 N/mm².b) The percent elongation is 32.00%.c) The percent reduction in area is 28.41%.
Given:
Stress at proportional limit = 221 N/mm²
Strain at proportional limit = 0.002
The modulus of elasticity (E) is calculated using the formula:
E = Stress/Strain
Substituting the values:
E = 221 N/mm² / 0.002 = 110,500 N/mm²
(b) Percent Elongation:
Percent elongation is calculated using the formula:
Percent Elongation = (Final Length - Original Length) / Original Length * 100
Given:
Original Length = Gage Length = 50 mm
Final Length after fracture = 66 mm
Percent Elongation = (66 mm - 50 mm) / 50 mm * 100 = 32%
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A certain population of bacteria doubles in 2yrs. If in 4yrs there are 20,000 of bacteria present, how much is present initially?
The initial population of this bacteria is 1,250 bacteria.
How to calculate the nth term of a geometric sequence?In Mathematics and Geometry, the nth term of a geometric sequence can be calculated by using this mathematical equation (formula):
aₙ = a₁rⁿ⁻¹
Where:
aₙ represents the nth term of a geometric sequence.r represents the common ratio.a₁ represents the first term of a geometric sequence.Based on the information provided about the population of this bacteria, an exponential function that models the growth can be written as follows;
a₄ = a₁2ⁿ
20,000 = a₁2⁴
20,000 = a₁16
a₁ = 20,000/16
a₁ = 1,250 bacteria.
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explain how residual stress can decrease with crack
Explain how residual stress can decrease with crack size.
When there is no outside force or thermal gradients, residual stress is the amount of tension that still exists within an object. The vulnerability of an object to fatigue and fracture can be impacted by residual stress.
Residual stress is produced when an object is subjected to tension that exceeds its elastic limit and causes plastic deformation. In industrial processes including bending, creating artwork, extruding, and rolling, residual stress also develops when the deformation of plastic is not uniform across the cross-section of an object.
Remaining tension and cracks might interact in a variety of ways. Crack propagation is slowed down by compressive residual stresses whereas it is accelerated by tensile residual stresses. In the analyses, crack closure effects and elastic and elastic-plastic material behaviors were taken into consideration.
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Determine the discharge of the weir having a head of 0.33m in liters/second if a 90° triangular weir is used.
•Answer is not found in the given choices
•87.58
•1,082.85
•942.52
The discharge of the wier is 87.58 L/s. Cd is the coefficient of discharge. Therefore, option B is correct.
Given:
Head (H) = 0.33m
To determine the discharge of a 90° triangular weir with a head of 0.33m, the following formula can be used:
[tex]Q = \frac{8}{15} C_d tan\frac{\theta}{2} \sqrt{2g} (H)^\frac{5}{2}[/tex]
Where,
Cd is the discharge coefficient
H is head
g is the gravitational constant
Let the discharge coefficient is [tex]C_d\\[/tex] = 0.6
Substituting the values:
[tex]Q = \frac{8}{15} \times 0.6 \times tan\frac{90}{2} \sqrt{2\times 9.81} (0.33)^\frac{5}{2}\\\\[/tex]
Solving further:
Q = 0.533 × 0.6 × 1 × 4.4294 × 0.06255
Q = 0.3198 × 0.2770
Q = 0.08758 m³/s
Q = 0.08758 × 1000 L/s
Q = 87.58 L/s
Therefore, the discharge of the wier is 87.58 L/s. Option B is correct.
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The bearing of grids of an AB alignment is 100º 22', while the
magnetic declination is 8º30' E. What is the true, magnetic, and
grid azimuth of this alignment?
True azimuth of this alignment = 91º 52'; Magnetic azimuth of this alignment = 100º 22' and Grid azimuth of this alignment = 91º 52'.
The bearing of grids of an AB alignment is 100º 22', while the magnetic declination is 8º30' E. The bearing of grids of an AB alignment is 100º 22', and the magnetic declination is 8º30' E.To find: The true, magnetic, and grid azimuth of this alignment.Concepts: The angle between the magnetic north and the grid north is known as the grid azimuth. The angle between the magnetic north and the true north is known as the magnetic azimuth. The angle between the true north and the grid north is known as the true azimuth. These angles are related through the magnetic declination.Formula:True azimuth = Grid azimuth + West declinationTrue azimuth = Grid azimuth - East declinationMagnetic azimuth = Grid azimuth + Magnetic declination
Magnetic azimuth = True azimuth + Magnetic declinationGrid azimuth = True azimuth - West declinationGrid azimuth = Magnetic azimuth - Magnetic declinationSolution:The given bearing of the AB alignment = 100º 22'The magnetic declination = 8º30' ETo calculate true azimuth = Grid azimuth - East declinationTrue azimuth = 100º 22' - 8º 30'True azimuth = 91º 52'To calculate magnetic azimuth = True azimuth + Magnetic declinationMagnetic azimuth = 91º 52' + 8º 30'Magnetic azimuth = 100º 22'To calculate the grid azimuth = Magnetic azimuth - Magnetic declinationGrid azimuth = 100º 22' - 8º 30'Grid azimuth = 91º 52'Answer:True azimuth of this alignment = 91º 52'Magnetic azimuth of this alignment = 100º 22'Grid azimuth of this alignment = 91º 52'
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A trapezoidal channel with a bed width of 5.0 ft and a side slope of 2:1 carries a discharge of 150 f/s. The channel has a bed slope of 0.0008 and terminates over free overfall. The depth of water before the free overfall is 2.25ft. Determine the type and length of the water surface profile formed before the fall. Let n = 0.025. gravity = 32.2 ft/s? Q=1.49/n x AxR23 x S1/2
The base width of the canal can be calculated as follows:The trapezoidal canal can be expressed as a rectangle with a triangle on top. The rectangle's base width is the same as the trapezoid's base width, while the triangle's base width is 1/3 the height of the triangle.
As a result, the canal's total base width B is calculated as:B = b + (2/3)ywhere b is the top width of the canal and y is the depth of the canal.The trapezoid's area can be found using the discharge equation:Q = (1.49/n) AR2/3S1/2where Q is the discharge, A is the area of the trapezoid, R is the hydraulic radius, S is the slope of the canal, and n is Manning's roughness coefficient.
To solve for A, divide both sides by (1.49/n) and square both sides:AR2/3 = Qn/1.49SSubstitute 2100 cfs for Q, 0.018 for n, 6 ft for y, and 2 horizontal to 1 vertical for the side slope of the canal:S = (1/3)(6 ft) = 2 ftR = A/P = A/(b + 2y(2² + 1²)½/3) = (2/3)A/By = A/b(2y(2² + 1²)½/3)½B = b + (2/3)y(2² + 1²)½/3B = 47.68 ft(b) It cannot be said for certain whether the channel section is the most economic section.
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For the Cipolletti weir, derive the slope (1/4:1) of the sides of the trapezoid by setting the reduction in discharge due to contraction equal to the increase in discharge due to the triangular area added.
The discharge observed at a v-notch weir is 66.7 in³/s
Given that;
Notch angle ∅ = 90°
height above the weir is 3 inches { head + head correction factor) h + k = 3 in
Discharge Q = ?
To determine the discharge observed, we us the following expression
Q = 4.28Ctan(∅/2) ( h + k )^5/2
where Q is discharge, C is discharge coefficient, ∅ is notch angle, h is head and k is head correction factor
now we substitute
Q = 4.28 × 1 × tan(90/2) ( 3 )^5/2
Q = 4.28 × 1 × 1 × 15.5884
Q = 66.7 in³/s
Therefore the discharge observed at a v-notch weir is 66.7 in³/s
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Question Explain the importance of Section Capacity and Ductility in the Performance Based Design of Tall Buildings
The importance of Section Capacity and Ductility in the Performance Based Design of Tall Buildings is discussed below.
1. Section Capacity:
Structural Safety: Adequate section capacity ensures that structural components can safely carry anticipated loads without failure.Load Redistribution: Determines how loads are efficiently transferred between components, maintaining overall stability and balance.Code Compliance: Section capacity analysis ensures compliance with building codes and regulations for structural safety.2. Ductility:
Seismic Resistance: Ductile systems absorb and dissipate seismic energy through controlled deformation, reducing the risk of brittle failure.Redundancy and Robustness: Ductile systems exhibit redundancy, ensuring the overall structure can support loads even if some components fail.Performance under Serviceability Loads: Ductility allows components to undergo small deformations without compromising functionality or causing discomfort to occupants.Learn more about Ductility here:
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Compare w/c ratio of normal concretes and pervious concretes. For lightweight concretes 7 day strength or 28 day strength is more important? Why are special concretes more expensive when compared to normal concretes? (Hint: consider the materials used)
Concrete is a widely used construction material due to its durability, strength, and versatility.
Concretes can be classified based on their mix design, strength, density, permeability, and other properties. Normal concrete, pervious concrete, and lightweight concrete are some of the types of concrete used in construction.The water-cement (w/c) ratio is a crucial parameter that influences the properties of concrete. The w/c ratio represents the ratio of water to cement in a concrete mix. Normal concrete has a w/c ratio of 0.5 to 0.6, whereas pervious concrete has a w/c ratio of up to 0.35.
The low w/c ratio of pervious concrete improves its permeability, porosity, and strength. Furthermore, pervious concrete contains larger aggregates, which allows for increased permeability.Special concretes are more expensive than normal concretes due to the cost of the materials used. Special concretes, such as self-compacting concrete, high-strength concrete, and fiber-reinforced concrete, require higher-quality ingredients, such as aggregates, cement, and chemical admixtures. In addition, specialized equipment and labor may be required to prepare and place these concretes.
Therefore, the overall cost of special concretes is higher than that of normal concretes.Lightweight concrete, on the other hand, has a lower density and weight compared to normal concrete. The strength of lightweight concrete is dependent on its density and curing time. For lightweight concrete, the 28-day strength is more important than the 7-day strength. The curing time for lightweight concrete is typically longer than that of normal concrete, as it requires time to achieve its full strength. Lightweight aggregates, such as pumice, perlite, and vermiculite, are used in the mix design of lightweight concrete to decrease its weight while maintaining its strength and durability.
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Hardness If a metal with high ductility is required for a forming process, which material would you recommend? Select one: a annealed metal
b as-cast metal c 20% cold worked metal d 40% cold worked metal
The correct option can be annealed metal if a metal with good ductility is required for a forming operation. The correct option is a.
The metal is heated to a high temperature and then slowly cooled as part of the annealing process, which reduces internal tensions and increases ductility.
Metals that have been annealed are often softer and more pliable, which makes it simpler to bend, roll, or stretch them into the desired shape.
Option B alludes to a metal that is unfinished and uncast. Cast metals are often more brittle and less ductile than annealed metals, which makes them less appropriate for forming procedures.
Cold working, or the process of deforming metal at room temperature, is a component of both options c and d.
Thus, the correct option is a.
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If the long term permissible load capacity for three nails in double shear in C22 timber is given as 1660.5KN using 3.35mm x 90mm long round wire. What is the slip per nail in mm? ( Answer in 2 decimal places)
The slip per nail in mm is 0.0044164 mm when the long term permissible load capacity for three nails in double shear in C22 timber is given as 1660.5 KN using 3.35mm x 90mm long round wire.
According to the question, the long term permissible load capacity for three nails in double shear in C22 timber is given as 1660.5 KN using 3.35mm x 90mm long round wire.The slip per nail in mm is defined as the maximum deformation of the nail in the direction of the applied force that occurs before the force produces the full load capacity of the nail. A nail's slip is caused by two types of deformation: lateral or radial deformation, and longitudinal or axial deformation. Longitudinal deformation occurs when the nail is stressed parallel to its longitudinal axis, such as in double shear.Here, we can calculate the slip per nail in mm as follows:First, we need to calculate the ultimate load capacity of the 3 nails in double shear using the given formula:Ultimate load capacity = permissible load capacity / safety factor= 1660.5 / 2.5= 664.2 KNNow, we need to calculate the force acting on each nail in double shear using the given formula:Force acting on each nail = Ultimate load capacity / number of nails= 664.2 / 3= 221.4 KNFinally, we can calculate the slip per nail in mm using the given formula:Slip per nail = Force acting on each nail / (nail diameter x 2 x shear modulus x nail length)= 221.4 / (3.35 x 2 x 11000 x 90)= 0.0044164 mmTherefore, the slip per nail in mm is 0.0044164 mm (rounded off to 2 decimal places).Hence, the answer is in 200 words.
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An underflow vertical sluice gate discharges freely 48 m³/s into a rectangular channel 6.0 m wide. The gate, which is the same width as the channel, is set at a height of 1.3 m above the bed and the depth at the vena contractor is 0.80 m. The energy loss in the converging flow at the sluice gate between the upstream and downstream sections is 0.05V2/2g. The normal depth in the downstream channel is 2.4 m. a. Calculate the depth of water upstream of the gate assuming the Coriolis coefficient (kinetic energy correction factor) as 1.05. b. Confirm that the gate can discharge freely.
We will apply the concepts of fluid mechanics and channel flow to the challenges posed regarding the underflow vertical sluice gate. Let's calculate the necessary values incrementally:
a. Calculating the depth of water upstream of the gate:
We'll use the energy equation (Bernoulli's equation) to determine the depth upstream of the gate. The energy equation is given by:
P1/γ + V1∧2/2g + Z1 = P2/γ + V2∧2/2g + Z2 + h_loss
Area of flow at the vena contractor (A_vena) = B × h_vena = 6.0 m × 0.80m = 4.8 m²
Velocity at the vena contractor (V_vena) = Q / A_vena = 48 m³/s / 4.8 m² = 10 m/s
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List and explain 5 factors that affects avaporation. (10pt)
At high altitude, where atmospheric pressure is low, the rate of evaporation is faster.
Below are the five factors that affect evaporation:1. TemperatureTemperature is the primary factor that affects evaporation. At higher temperatures, the average kinetic energy of molecules increases. This causes molecules to escape from the surface of the liquid into the surrounding atmosphere. Therefore, the rate of evaporation increases with an increase in temperature.2. Surface areaThe surface area of the liquid is directly proportional to the rate of evaporation. Larger surface areas will provide more molecules to escape into the atmosphere, resulting in a faster rate of evaporation.3. Humidity The amount of moisture in the atmosphere affects the rate of evaporation. When the humidity is high, the air is saturated with water vapor, which slows down the rate of evaporation. Conversely, when the humidity is low, the rate of evaporation is faster.4. WindThe wind has a significant effect on evaporation. It promotes the movement of air molecules, allowing the water molecules to move away from the liquid's surface more quickly. Thus, a higher wind speed results in a higher rate of evaporation.5. PressureThe atmospheric pressure above the liquid surface affects the rate of evaporation. At high altitude, where atmospheric pressure is low, the rate of evaporation is faster. Conversely, at low altitudes, where atmospheric pressure is high, the rate of evaporation is slower.In conclusion, the rate of evaporation is affected by several factors, including temperature, surface area, humidity, wind, and pressure. The understanding of these factors can help us predict and control the rate of evaporation.
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6.4 a Calculate the increase in vertical total stress at a depth of 2m under the center of a water tank, 5 m diamete and 10m high, filled with water. The self-weight of the tank and its foundation is 300kN and the unit weigh of water is 9.81 kN/m. Assume the base of the tank foundation is at the soil surface.
The self-weight of the tank and its foundation is 300kN and the unit weigh of water is 9.81 kN/m, the increase in vertical total stress at a depth of 2m under the center of the water tank is the sum of the weight of the water and the weight of the tank and foundation.
We must account for the weight of the water as well as the weight of the tank and its base in order to determine the rise in vertical total stress at a depth of 2 metres under the centre of a water tank.
The weight of the water:
Volume of water = π * (radius)² * height
radius = diameter / 2 = 5m / 2 = 2.5m
height = 10m
Volume of water = π * (2.5m)² * 10m
Weight of water = Volume of water * Unit weight of water
Weight of tank and foundation = 300kN
Increase in vertical total stress = Weight of water + Weight of tank and foundation
Thus, the increase in vertical total stress at a depth of 2m under the center of the water tank is the sum of the weight of the water and the weight of the tank and foundation.
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