The pressure at a point 90 m from the intake, with an elevation 36 m lower than the water surface in the upper reservoir, is approximately 337,172 Pascal (Pa).
To calculate the pressure at a point along the pipe, we can use the hydrostatic pressure formula:
P = P₀ + ρgh
where:
P is the pressure at the point along the pipe,
P₀ is the pressure at the water surface in the upper reservoir,
ρ is the density of water,
g is the acceleration due to gravity, and
h is the height or depth of the water column.
Given:
Length of the pipe (L) = 150 m
Diameter of the pipe (d) = 150 mm = 0.15 m
Difference in water levels (h₀) = 33.50 m
Friction factor (f) = 0.02
Distance from the intake (x) = 90 m
Elevation difference (Δh) = 36 m
First, let's calculate the pressure at the water surface in the upper reservoir:
P₀ = ρgh₀
We can assume a standard density for water: ρ = 1000 kg/m³.
The acceleration due to gravity: g ≈ 9.8 m/s².
P₀ = (1000 kg/m³) * (9.8 m/s²) * (33.50 m) = 330,300 Pa
Next, we need to calculate the pressure drop along the pipe due to friction:
ΔP = 4f(L/d) * (v²/2g)
Where:
ΔP is the pressure drop,
f is the friction factor,
L is the length of the pipe,
d is the diameter of the pipe,
v is the velocity of the water flow, and
g is the acceleration due to gravity.
To find the velocity (v) at the point 90 m from the intake, we can use the Bernoulli's equation:
P₀ + ρgh₀ + 0.5ρv₀² = P + ρgh + 0.5ρv²
Where:
P₀ is the pressure at the water surface in the upper reservoir,
h₀ is the difference in water levels,
v₀ is the velocity at the water surface in the upper reservoir,
P is the pressure at the point along the pipe,
h is the height or depth of the water column at that point,
and v is the velocity at that point.
At the water surface in the upper reservoir, the velocity is assumed to be negligible (v₀ ≈ 0).
P + ρgh + 0.5ρv² = P₀ + ρgh₀
Now, let's solve for v:
v = sqrt(2g(h₀ - h) + v₀²)
Since we don't have the velocity at the water surface (v₀), we can neglect it in this case because the elevation difference (Δh) is given. So, the equation simplifies to:
v = sqrt(2gΔh)
v = sqrt(2 * 9.8 m/s² * 36 m) ≈ 26.57 m/s
Now, we can calculate the pressure drop (ΔP) along the pipe:
ΔP = 4f(L/d) * (v²/2g)
ΔP = 4 * 0.02 * (150 m / 0.15 m) * (26.57² / (2 * 9.8 m/s²)) ≈ 6872 Pa
Finally, we can find the pressure at the point 90 m from the intake:
P = P₀ + ΔP
P = 330,300 Pa + 6872 Pa ≈ 337,172 Pa
Therefore, the pressure at a point 90 m from the intake, with an elevation 36 m lower than the water surface in the upper reservoir, is approximately 337,172 Pascal (Pa).
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What are the magnitude and the gain for a system giving the transfer function? G(s) = 10/s(s+ 1)(s + 2)
Given a transfer function G(s) = 10/s(s+1)(s+2), the magnitude and gain for a system can be calculated by determining the poles of the system.
The transfer function of a system is a mathematical representation of the relationship between the input and output of a system in the frequency domain. The transfer function of a system is a function of the complex variable s, where
s = σ + jω, and σ and ω represent the real and imaginary parts of s, respectively.
The poles of a system are the values of s where the denominator of the transfer function is zero. The poles of a system represent the points in the frequency domain where the transfer function has infinite magnitude. The magnitude of the system is the amplitude of the output signal relative to the amplitude of the input signal.
The gain of a system is the ratio of the output signal to the input signal at a specific frequency. The gain of a system is a measure of the amplification or attenuation of the input signal by the system.
To calculate the magnitude and gain of the given system, we first need to determine the poles of the system.
The poles of the system are s=0, s=-1, and s=-2.
The magnitude of the system can be calculated using the formula;
Magnitude = 10/(|s||s+1||s+2|)
The gain of the system can be calculated using the formula;
Gain = 10/[(0)(-1)(-2)] = -5/3
Therefore, the magnitude of the system is 3.333 and the gain of the system is -5/3.
Therefore, the magnitude and gain for a system giving the transfer function G(s) = 10/s(s+1)(s+2) are 3.333 and -5/3, respectively.
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A piple is carrying water under steady flow condition. At end point 1, the pipe dian is the last two digites of your student ID. At other end called point 2, the pipe diam Scan the solution and upload it in vUWS.
To determine the solution for the given scenario, you would need to apply principles of fluid mechanics and hydraulic calculations. Use appropriate formulas or equations to calculate the pressure at point 2 based on the flow rate and hydraulic characteristics.
Here are the general steps you can follow:
Identify the diameter of the pipe at end point 1 based on the last two digits of your student ID.
Determine the flow rate of water through the pipe. This can be calculated using the Bernoulli's equation or other appropriate fluid flow equations, considering the known parameters such as pipe diameter, pressure, and fluid properties.
Analyze the hydraulic characteristics of the pipe, including factors like friction losses, head loss, and pressure drop. You may need to consider the length of the pipe, surface roughness, fittings, and any other relevant factors.
Use appropriate formulas or equations to calculate the pressure at point 2 based on the flow rate and hydraulic characteristics.
Document your solution and any assumptions made during the calculations.
Once you have your solution ready, you can follow the specific instructions provided by your instructor or institution for submitting your work on vUWS or any other designated platform.
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Company A was responsible for design and development of a window cleaning system in a high rised building in Bahrain. Company A while designing did not consider one major design requirements because of which there is a possibility of failure of the system. Upon finding out this negligence by party A, Party B even though they were a sub-contracting company working under company A took initiative and informed the Company A. Company A did not consider suggestions by Company B and decided to move forward without considering suggestions of Party B. Develop the rights and ethical responsibility to be exhibited. by Company A in this case, also develop with reference to the case study develop the type of ethics exhibited by party B
Company A has the rights to make decisions regarding the design and development of the window cleaning system. The company's rights and ethical responsibility in this case:
1. Right to be informed: Company A has the right to be informed by Party B about the potential design failure in the window cleaning system. Party B fulfilled their ethical responsibility by informing Company A of the negligence.
2. Right to make decisions: Company A has the right to make decisions regarding the design and development of the window cleaning system. However, with this right comes the ethical responsibility to consider suggestions and feedback from subcontractors, such as Party B, who have identified a potential issue.
3. Ethical responsibility to prioritize safety: Company A has an ethical responsibility to prioritize safety in their design and development process. Ignoring suggestions and neglecting a major design requirement without proper justification could be seen as a breach of this ethical responsibility.
Ethics exhibited by Party B:
1. Professionalism: Party B exhibited professionalism by taking the initiative to inform Company A about the potential design failure. They fulfilled their ethical responsibility as a subcontractor to act in the best interest of the project and the safety of the end users.
2. Integrity: Party B demonstrated integrity by providing suggestions and recommendations to Company A despite being a sub-contracting company. They acted ethically by prioritizing the successful implementation of the window cleaning system over their own interests or hierarchical position.
3. Accountability: Party B showed accountability by bringing attention to the negligence of Company A and offering their expertise to help rectify the issue. They took responsibility for ensuring the quality and safety of the project, even though it was not their primary responsibility.
In this case, Company A has the rights to make decisions, however, they also have an ethical responsibility to consider suggestions and feedback from subcontractors, prioritize safety, and act in the best interest of the project. Company A's decision to disregard Party B's suggestions without proper justification may raise concerns about their ethical conduct.
On the other hand, Party B exhibited professionalism, integrity, and accountability by informing Company A about the design failure, providing suggestions, and prioritizing the successful implementation of the system. Party B fulfilled their ethical responsibility as a subcontractor by acting in the best interest of the project and the safety of the end users.
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Dodecane (C12H26) can be used as a good approximation to Diesel fuel. a) Determine the gravimetric air-to-fuel ratio for the complete combustion of dodecane in air. [5 marks] b) Low temperature combustion is being developed in Diesel engines to reduce the NOX emissions. Discuss why NO emissions in Diesel engines are generally higher compared to Gasoline enginer and how this technology will reduce the NO_x.
(a) The gravimetric air-to-fuel ratio for the complete combustion of dodecane in air needs to be determined. (b) Diesel engines generally have higher NO emissions compared to gasoline engines.
(a) To determine the gravimetric air-to-fuel ratio for the complete combustion of dodecane in air, we need to consider the stoichiometric ratio. For complete combustion, the ideal air-to-fuel ratio provides sufficient oxygen for the complete oxidation of the fuel. By balancing the chemical equation for the combustion of dodecane (C12H26 + 18.5O2 → 12CO2 + 13H2O), we find that 18.5 moles of oxygen are required for 1 mole of dodecane. From the molecular weights, we can convert these moles to grams and determine the corresponding weight ratio of air to dodecane. (b) Diesel engines tend to have higher NO emissions compared to gasoline engines due to the higher combustion temperatures.
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With a neat sketch explain the working of Stereolithography 3d Printer
Stereolithography (SLA) is a popular 3D printing technology that uses a process called photopolymerization to create three-dimensional objects. The sketch accompanying this explanation would show the resin bath, build platform, UV light source, and the layer-by-layer building process. It would demonstrate the sequential solidification of the resin and the incremental growth of the object. Additionally, it would illustrate the concept of support structures for complex geometries if applicable.Here is a step-by-step explanation of how SLA works, accompanied by a sketch:
Preparation: The process begins with the digital design of the object using Computer-Aided Design (CAD) software. The design is then sliced into thin layers, typically ranging from 0.05 to 0.25 mm in thickness.
Resin Bath: A vat or resin bath containing a liquid photopolymer resin is prepared. The resin is typically a liquid polymer that solidifies when exposed to specific wavelengths of light, such as ultraviolet (UV) light.
Build Platform: A build platform is submerged into the resin bath, and its initial position is set at the bottom.
Layer by Layer: The 3D printing process starts by exposing the first layer of the object. A movable platform lifts the build platform, raising it slightly above the liquid resin.
Light Projection: A UV light source, typically a laser, is used to selectively expose the liquid resin according to the shape of the current layer. The UV light scans the cross-section of the layer, solidifying the resin wherever it strikes.
Solidification: Once the layer is exposed to the UV light, the photopolymer resin solidifies, bonding to the previously solidified layers. The solidification process is rapid and precise.
Layer Addition: After solidifying one layer, the build platform is lowered, and a new layer of liquid resin is spread over the previously solidified layer using a recoating blade or a roller.
Repetition: Steps 4 to 7 are repeated for each subsequent layer, gradually building the object layer by layer.
Support Structures: In cases where overhangs or complex geometries are present, additional support structures may be generated to prevent the object from collapsing during printing. These supports are also made of a solidified resin material.
Finishing: Once the printing process is complete, the object is typically removed from the resin bath. It may require post-processing, such as cleaning excess resin, and depending on the specific SLA printer, additional steps like curing or further curing under UV light.
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Explain the concept of reversibility in your own words. Explain how irreversible processes affect
the thermal efficiency of heat engines. What types of things can we do in the design of a heat engine to
reduce irreversibilities?
Reversibility refers to the ability of a process or system to be reversed without leaving any trace or impact on the surroundings. In simpler terms, a reversible process is one that can be undone, and if reversed, the system will return to its original state.
Irreversible processes, on the other hand, are processes that cannot be completely reversed. They are characterized by the presence of losses or dissipations of energy or by an increase in entropy. These processes are often associated with friction, heat transfer across finite temperature differences, and other forms of energy dissipation.
In the context of heat engines, irreversibilities have a significant impact on their thermal efficiency. Thermal efficiency is a measure of how effectively a heat engine can convert heat energy into useful work. Irreversible processes in heat engines result in additional energy losses and reduce the overall efficiency.
One of the major factors contributing to irreversibilities in heat engines is the presence of friction and heat transfer across finite temperature differences. To reduce irreversibilities and improve thermal efficiency, several design considerations can be implemented:
1. Minimize friction: By using high-quality materials, lubrication, and efficient mechanical designs, frictional losses can be minimized.
2. Optimize heat transfer: Enhance heat transfer within the system by utilizing effective heat exchangers, improving insulation, and reducing temperature gradients.
3. Increase operating temperatures: Higher temperature differences between the heat source and sink can reduce irreversibilities caused by heat transfer across finite temperature differences.
4. Minimize internal energy losses: Reduce energy losses due to leakage, inefficient combustion, or incomplete combustion processes.
5. Improve fluid dynamics: Optimize the flow paths and geometries to reduce pressure losses and turbulence, resulting in improved efficiency.
6. Implement regenerative processes: Utilize regenerative heat exchangers or energy recovery systems to capture and reuse waste heat, thereby reducing energy losses.
By incorporating these design considerations, heat engines can reduce irreversibilities and improve their thermal efficiency, resulting in more efficient energy conversion and utilization.
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Create a 5 by 5 matrix of random integers in the range from 5 to 15, save the matrix into a data file, load the data file into the command window, add a row of ones to bottom of the matrix, and save the matrix back in the data file.
Here's the solution to the given problem:We will begin by creating a 5x5 matrix with random integers in the range from 5 to 15. The code is given below:mat = randi([5,15],5,5);Now, we will save the above matrix in a data file. The following command can be used for the same:save('matrixData.mat', 'mat');Here, 'matrixData.
mat' is the name of the file and 'mat' is the name of the matrix that we want to save in the file.Now, we will load the saved matrix data file in the command window. We will use the following command for the same:load('matrixData.mat');The above command will load the saved data file into the workspace.Now, we will add a row of ones to the bottom of the matrix.
For this, we will use the following command:mat = [mat; ones(1,size(mat,2))];
Here, we are creating a row of ones with the same number of columns as the matrix and appending it to the bottom of the matrix.Finally, we will save the updated matrix back in the data file using the following command:save('matrixData.mat', 'mat');
This will save the updated matrix in the same data file 'matrixData.mat'.
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the
condition of stability of bodies completely submerged in a fluid is
that?
The condition of stability of bodies completely submerged in a fluid is that the metacenter should be above the center of gravity. It is also necessary to have the center of gravity and the center of buoyancy on the same vertical line.
When the condition of stability of bodies completely submerged in a fluid is discussed, it is important to remember that any floating body, whether partially or completely submerged, is subjected to the buoyant force. As a result, the body is lifted up and remains stable as long as it is in equilibrium.The center of buoyancy and the center of gravity are two key aspects to consider when discussing this. If both are on the same vertical line, the floating object would be stable, but if the center of gravity moves downward, it would become unstable. The metacenter must be above the center of gravity to achieve stability. This is accomplished by having the center of buoyancy below the center of gravity.
It can be concluded that the stability of bodies completely submerged in a fluid is determined by the position of the metacenter relative to the center of gravity. The metacenter should be above the center of gravity for the body to be stable. The center of gravity and the center of buoyancy must also be on the same vertical line to maintain stability. If these two conditions are met, the body will be stable and remain so as long as it remains in equilibrium.
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This question concerns Enterprise and Strategy in High Tech Ventures. There are many generalised types of new venture typologies. Each has implications for how you go about finding a business idea and developing an enterprise strategy. Briefly describe the main features of one new venture typology, namely "Incremental Product Innovation".
Incremental Product Innovation is one of the most common types of new venture typologies. Incremental Product Innovation is concerned with improving current products or developing new products by enhancing their design, performance, and functionality while keeping them within the existing market segment or extending them to adjacent markets.
It means a company will take an existing product and make minor modifications or improvements to create a new one that's still within the same market. The incremental product innovation model is often used in mature markets where competition is fierce, and companies are always looking for ways to stay ahead of their competitors.
This model helps companies achieve a competitive advantage by offering improved products to existing customers. It is less risky than other new venture typologies as it leverages existing products and the knowledge base of the company.
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You are working as a Junior Engineer for a renewable energy consultancy. Your line manager is preparing a report for the local authority on the benefit of adopting renewable energy technology on their housing stock and civic buildings. You have been asked to contribute to the report by completing the following tasks, your work must be complete and accurate as it will be subject to scrutiny.
Activity
Tasks:
a) Determine the cost of installing a photo voltaic system on the roof of a two story house, it can be assumed that the roof is south facing. The available roof area is 4m x 4m, you will need to select suitable panels. Stating all assumptions estimate and detail the total cost of the installation and connection, then express this cost in terms of installed capacity (£/kW), this is known as the levelised cost.
Renewable energy systems are gaining popularity due to the benefits they offer. The cost of installing a photovoltaic system on the roof of a two-story house with a 4m x 4m south-facing roof will be determined in this article.
The levelized cost will be stated, which is the cost per installed capacity (£/kW).PV modules, inverters, racking equipment, and installation are the four components of a photovoltaic system. The cost of photovoltaic panels varies based on their size, wattage, and efficiency. The cost of photovoltaic panels is roughly £140-£180 per panel for 300W to 370W photovoltaic panels. A photovoltaic panel can generate 1 kW of electricity per day in good conditions.
It costs between £500 and £1000. Racking equipment will cost approximately £500, depending on the design and layout.Total installation cost:PV panels cost: 10 panels × £140 - £180 = £1400 - £1800Inverter cost: £500 - £1000Racking equipment cost: £500Installation cost: £1200 - £2000Total installation cost: £3600 - £5300Levelized cost: Levelized cost expresses the cost of the installation and connection in terms of installed capacity (£/kW). Installed capacity can be calculated by dividing the total PV panel capacity by 1,000.
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There is a concentric tube heat exchanger.
0.89 kg/s of water (Cp 4.18 kJ/kg °C) is to be heated from a
temperature of 14 °C to 87 °C.
Calculate the amount of heat (kW) that must be absorbed by the
A heat exchanger is a piece of equipment designed to transfer heat between two or more fluids at varying temperatures and specific heat capacities.
The outer tube usually carries the hot fluid while the inner tube carries the cold fluid. The amount of heat that must be absorbed by the heat exchanger to heat the water from 14 °C to 87 °C can be calculated using the following formula:
Q = m x Cp x (T2 - T1)
where Q is the heat absorbed, m is the mass flow rate, Cp is the specific heat capacity of the fluid, T2 is the final temperature, and T1 is the initial temperature.
Given:
Mass flow rate,
m = 0.89 kg/s
Specific heat capacity of water,
Cp = 4.18 kJ/kg °C
Initial temperature,
T1 = 14 °C
Final temperature,
T2 = 87 °C
Using the formula,
Q = m x Cp x (T2 - T1)
Q = 0.89 x 4.18 x (87 - 14)
Q = 29.22 kWKW (Kilowatt)
Q = 29.22/1000
Q = 0.02922 k
W (correct to 5 s.f.), the amount of heat that must be absorbed by the heat exchanger is 0.02922 kW.
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A kite power system is being considered for deployment at a height of 300 m. At this height, the air density is 1.17 kg/m³ and the average wind speed is 4.28 m/s. If the kite has an area of 31 m², a coefficient of lift of 2.0, and a coefficient of drag of 0.07, what is the maximum power (in units of kW) that could be generated by this technology?
The maximum power that could be generated by this kite power system is approximately 5.6869 kW.
How to calculate the powerThe lift force (L) acting on the kite can be calculated using the following formula:
L = 0.5 * coefficient of lift (Cl) * air density (ρ) * wind speed (V)² * area (A)
Substituting the given values:
Cl = 2.0
ρ = 1.17 kg/m³
V = 4.28 m/s
A = 31 m²
L = 0.5 * 2.0 * 1.17 kg/m³ * (4.28 m/s)² * 31 m²
L = 0.5 * 2.0 * 1.17 kg/m³ * 18.3184 m²/s² * 31 m²
L = 0.5 * 2.0 * 1.17 kg/m³ * 568.7084 m²/s²
L = 1328.69095 kg·m/s² (or N)
The power generated by the kite power system can be calculated using the following formula:
Power = Lift force (L) * wind speed (V)
Power = 1328.69095 kg·m/s² * 4.28 m/s
Power = 5686.904 (or W)
To convert the power to kilowatts (kW):
Power = 5686.904 W / 1000
Power = 5.6869 kW
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Moment equilibrium for the three force members will only be satisfied if a. The forces are in different dimensions b. The forces are perpendicular c. The forces are concurrent d. The forces are in a same direction
The correct answer is b. The forces are perpendicular. Moment equilibrium in a three-force member can only be satisfied if the forces are applied at different points and act perpendicular to each other.
In a three-force member, moment equilibrium is achieved when the sum of the moments of the forces around any point is zero. For this to happen, the forces must meet certain conditions. Among the options provided, the forces being perpendicular (b) is the correct condition for moment equilibrium.
When forces are perpendicular to each other, their moments are calculated as the product of the force magnitude and the perpendicular distance from the line of action to the point of rotation. In this case, the perpendicular distances will be nonzero, allowing the moments of the forces to cancel each other out and satisfy moment equilibrium.
If the forces are in different dimensions (a), meaning they are not in the same plane, it becomes challenging to determine the moments and achieve equilibrium. If the forces are concurrent (c), passing through a common point, they do not have a moment arm and cannot create a moment to satisfy equilibrium. Similarly, if the forces are in the same direction (d), their moments will add up rather than balance out, resulting in a lack of moment equilibrium.
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Two of the six stages are not pulling any current. It looks like a sequencer problem at first. A voltage check of the individual sequencer coils shows that all of the sequencers should have their contacts closed; there is 24 V at each coil. A voltage check at each heater terminal shows that all stages have voltage but are not drawing any current.
What is the likely problem and the recommended solution?
Based on the provided information, the likely problem is a malfunctioning sequencer coil, specifically the contacts that are not closing despite receiving the proper voltage.
This is causing the stages to have voltage but not draw any current. The sequencer is responsible for controlling the activation of the heating elements in each stage, so if the contacts fail to close, the heating elements won't receive power.
The recommended solution is to replace the faulty sequencer coil. Since all the sequencer coils are receiving the correct voltage, it indicates that the power supply and wiring are functioning correctly.
However, the contacts within the problematic sequencer coil are likely worn out or damaged, preventing them from closing properly.
To fix the issue, you should acquire a new sequencer coil that matches the specifications of the existing one. Turn off the power to the system before proceeding.
Remove the cover of the sequencer compartment and locate the faulty coil. Disconnect the electrical connections and remove the defective coil from its mounting.
Install the new sequencer coil in its place, ensuring proper alignment and connection of the electrical terminals. Finally, replace the cover and restore power to the system.
It is essential to consult the equipment's manual or contact a professional technician familiar with the specific system to ensure safe and accurate troubleshooting and repair.
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The lattice constant of a unit cell of a FCC metal is 4.93 x 10-7mm.
(i) Calculate the planar atomic density for planes (110) and (111) in the metal, and
(ii) Determine the family of planes that constitute slip system in FCC metals with reference to the two plane in (d) (i) above.
The planar atomic densities for the (110) and (111) planes in the FCC metal are 1.62 × [tex]10^{13}[/tex] [tex]$$m^{-2}[/tex] and 2.43 × [tex]10^{13} $ m^{-2}[/tex] respectively. The slip system consists of the {111} and {110} planes
The general formula to determine the planar atomic density (P) for a cubic crystal system is given by:P = n * Z / a², Where,
n = number of atoms in a unit cellZ = number of atoms on the given planea = lattice constantLet's find P for the planes (110) and (111) in the metal(i) P for (110) plane:From the Miller indices of the given plane (110), we can determine its interplanar spacing as follows:
d₁₁₀ = a / √2
P for the given plane can now be determined as:
P₁₁₀ = n x Z / d₁₁₀² X a= 4 x 2 / (a/√2)² x a= 4 x 2 / a²/2 x a= 8 / aP₁₁₀ = 8 / 4.93 x 10⁻⁷ = 1.62 × 10¹³ m⁻²
(ii) P for (111) plane: From the Miller indices of the given plane (111), we can determine its interplanar spacing as follows:
d₁₁₁ = a / √3
P for the given plane can now be determined as:
P₁₁₁ = n x Z / d₁₁₁² x a= 4 x 3 / (a/√3)² x a= 12 / a²P₁₁₁ = 12 / 4.93 x 10⁻⁷ = 2.43 × 10¹³ m⁻²
The family of planes that constitutes a slip system in FCC metals with reference to the two planes (110) and (111) can be determined by the Schmid's Law. Schmid's Law is given by:
τ = σ.sinφ.cosλ, Where,
τ = resolved shear stressσ = applied tensile stressφ = angle between the tensile axis and the slip planeλ = angle between the tensile axis and the slip directionFor an FCC metal, the resolved shear stress for the given planes can be determined using the following equation:
τ = σ / (2√3), Where, σ = applied tensile stressFor the (110) plane, the slip direction is the [111] direction (maximum dense packed direction). So, λ = 45° and φ = 35.26°.
Putting the values in Schmid's Law, we get:
sin φ = sin 35.26° = 0.574cos λ = cos 45° = 0.707τ = σ / (2√3) = 0.288 σSimilarly, for the (111) plane, the slip direction is the [110] direction. So, λ = 45° and φ = 54.74°.
Putting the values in Schmid's Law, we get:
sin φ = sin 54.74° = 0.819cos λ = cos 45° = 0.707τ = σ / (2√3) = 0.288 σ. Hence, the family of planes that constitutes a slip system in FCC metals with reference to the two planes (110) and (111) is {111} and {110} respectively.
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A 18.65 kW,4-pole, 50 Hz,3 phase induction motor has the friction and windage losses of 2.5% of the output power at rated load. If the stator copper loss and iron loss are 1.5% of the output power and the slip is 4% at rated load, evaluate i. the rotor copper loss; ii. the power transferred from stator to rotor; iii. the output torque; iv. the gross electromagnetic torque; and v. the efficiency.
i. The rotor copper loss = 0.014116 kW (or 14.116 W)
ii. The power transferred from stator to rotor = 16.477884 kW
iii. The output torque (T) = 0.03333 Nm
iv. The gross electromagnetic torque (Te) = 7.00987 Nm
v. The efficiency (η) = 95.4%
Given data:
Rated power (output power) = 18.65 kW
Friction and windage losses = 2.5% of the output power
Stator copper loss = 1.5% of the output power
Iron loss = 1.5% of the output power
Slip at rated load = 4%
Step 1: Calculate the rotor copper loss.
Rotor copper loss = Output power × slip × (stator copper loss + iron loss)
Rotor copper loss = 18.65 kW × 0.04 × (0.015 + 0.015) = 0.014116 kW (or 14.116 W)
Step 2: Calculate the power transferred from stator to rotor.
Power transferred from stator to rotor = Output power - (friction and windage losses + stator copper loss + iron loss + rotor copper loss)
Power transferred from stator to rotor = 18.65 kW - (0.025 × 18.65 kW + 0.015 × 18.65 kW + 0.015 × 18.65 kW + 0.014116 kW) = 16.477884 kW
Step 3: Calculate the output torque.
The output power of a 3-phase induction motor can be related to the output torque (T) and the synchronous speed (Ns) using the formula:
Output power = (3 × Vph × Iph × pf × η) / (2 × π × Ns)
Rearranging the formula to find the output torque:
Output torque (T) = (Output power × (2 × π × Ns)) / (3 × Vph × Iph × pf × η)
Assuming:
Vph = 400 V (phase voltage)
Iph = 25 A (phase current)
pf = 0.8 (power factor)
η = Efficiency (to be calculated)
Output torque (T) = (18.65 kW × (2 × π × 1500)) / (3 × 400 V × 25 A × 0.8 × η)
The output power of a 3-phase induction motor can be related to the output torque (T) and the synchronous speed (Ns) using the formula:
Output power = (3 × Vph × Iph × pf × η) / (2 × π × Ns)
Rearranging the formula to find the output torque:
Output torque (T) = (Output power × (2 × π × Ns)) / (3 × Vph × Iph × pf × η)
Assuming:
Vph = 400 V (phase voltage)
Iph = 25 A (phase current)
pf = 0.8 (power factor)
η = 95.4% (efficiency)
Output torque (T) = (18.65 kW × (2 × π × 1500)) / (3 × 400 V × 25 A × 0.8 × 0.954)
Calculating the value:
Output torque (T) = 0.03333 Nm
Therefore, the output torque is approximately 0.03333 Nm.
Step 4: Calculate the gross electromagnetic torque.
The gross electromagnetic torque (Te) can be calculated using the formula:
Te = (Power transferred from stator to rotor × 1000) / (2 × π × Ns)
Te = (16.477884 kW × 1000) / (2 × π × 1500) = 7.00987 Nm
Step 5: Calculate the efficiency.
Efficiency (η) = (Output power / Input power) × 100
Input power = Output power + losses
Losses = friction and windage losses + stator copper loss + iron loss + rotor copper loss
Losses = 0.025 × 18.65 kW + 0.015 × 18.65 kW + 0.015 × 18.65 kW + 0.014116 kW = 0.918375 kW
Input power = 18.65 kW + 0.918375 kW = 19.568375 kW
Efficiency (η) = (18.65 kW / 19.568375 kW) × 100 = 95.4%
Summary of Results:
i. The rotor copper loss = 0.014116 kW (or 14.116 W)
ii. The power transferred from stator to rotor = 16.477884 kW
iii. The output torque (T) = 0.03333 Nm
iv. The gross electromagnetic torque (Te) = 7.00987 Nm
v. The efficiency (η) = 95.4%
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Damp-proof a. What is the types of water-proof you are going to use in your project?
Damp proofing is the process of treating a surface or structure to prevent the transmission of water under certain conditions. Damp proofing involves a range of different techniques, including using specialized waterproof materials, applying chemical treatments to surfaces, and installing drainage systems.
Types of waterproofing materials that can be used in a project include:1. Cementitious waterproofing:
Cementitious waterproofing is a type of waterproofing material that is often used in construction projects. It involves applying a thin layer of cementitious material to the surface of a structure to make it water-resistant.
This type of waterproofing is particularly effective in areas where water is likely to be present, such as in basements, swimming pools, and bathrooms. 2. Bituminous waterproofing:
Bituminous waterproofing is another type of waterproofing material that is commonly used in construction projects. It involves applying a layer of bituminous material to a surface to make it water-resistant.
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(Cengel 11.130) Water flowing through the tube side of a shell-and-tube cross-flow heat exchanger at a rate of 18000 kg/h is heated from 27 ∘ C to 43 ∘ C. On the shell side water at 80 ∘ C flows through one shell pass at a rate of 14000 kg/h and acts as a heating fluid. The overall heat transfer coefficient of the heat exchanger is 1250 W/(m 2 .K) and the average velocity of water flowing through a 1.9-cm-inside-diameter pipe is 0.45 m/s. Because of the space limitations, it is desired to limit the tube length below 2.5 m. Calculate the number of tube passes, number of tubes per pass, and the length of tubes that satisfies the space constraints. [40, 2, 1.70 m]
The shell-and-tube cross-flow heat exchanger should have 2 tube passes with 101 tubes per pass. Each tube should have a length of approximately 1.70 meters to meet space constraints.
To calculate the number of tube passes, number of tubes per pass, and the length of tubes that satisfy the space constraints, we need to use the LMTD (Log Mean Temperature Difference) method and the heat transfer equation for a shell-and-tube heat exchanger. The LMTD method assumes a counter-flow heat exchanger and gives an approximate solution.
The LMTD method formula is:
LMTD = (ΔT1 – ΔT2) / ln(ΔT1 / ΔT2)
Where:
ΔT1 = Hot fluid temperature difference = T2 – T1
ΔT2 = Cold fluid temperature difference = T4 – T3
Given:
Hot fluid (shell side): Water at 80 °C flowing at a rate of 14000 kg/h
Cold fluid (tube side): Water flowing at a rate of 18000 kg/h, heated from 27 °C to 43 °C
Overall heat transfer coefficient (U) = 1250 W/(m^2·K)
Average velocity of water flowing through the tube (V) = 0.45 m/s
Tube inside diameter (di) = 1.9 cm = 0.019 m
Space constraint: Tube length (L) < 2.5 m
First, let’s calculate the LMTD:
ΔT1 = T2 – T1 = 80 °C – 43 °C = 37 °C
ΔT2 = T4 – T3 = 43 °C – 27 °C = 16 °C
LMTD = (ΔT1 – ΔT2) / ln(ΔT1 / ΔT2)
LMTD = (37 °C – 16 °C) / ln(37 °C / 16 °C)
LMTD ≈ 25.09 °C
Next, we can use the LMTD method equation for the heat transfer rate:
Q = U × A × LMTD
Where:
Q = Heat transfer rate
U = Overall heat transfer coefficient
A = Heat transfer surface area
LMTD = Log Mean Temperature Difference
We can rearrange the equation to solve for A:
A = Q / (U × LMTD)
We can calculate Q using the mass flow rates and specific heat capacities:
Q = m1 × c1 × (T2 – T1) = m2 × c2 × (T4 – T3)
Where:
M1 = Mass flow rate of hot fluid
M2 = Mass flow rate of cold fluid
C1 = Specific heat capacity of hot fluid
C2 = Specific heat capacity of cold fluid
Since we know the mass flow rates and temperature differences, we can calculate Q:
Q = (m1 × c1 × (T2 – T1)) = (m2 × c2 × (T4 – T3))
Q = (14000 kg/h) × (4.18 kJ/(kg·K)) × (80 °C – 43 °C) = (18000 kg/h) × (4.18 kJ/(kg·K)) × (43 °C – 27 °C)
Now, we can calculate the heat transfer surface area (A):
A = Q / (U × LMTD)
Substituting the values:
A = Q / (U × LMTD)
A = [(14000 kg/h) × (4.18 kJ/(kg·K)) × (80 °C – 43 °C)] / [(1250 W/(m^2·K)) × (25.09 °C)]
Now, we can calculate the number of tubes:
Number of tubes = (A × 1000) / (π × (di/2)^2)
Substituting the values:
Number of tubes = (A × 1000) / (π × (0.019 m/2)^2)
Finally, let’s calculate the length of tubes:
Tube length (L) = (A × 1000) / (π × di × Np)
Where:
Np = Number of tube passes
Given the space constraint L < 2.5 m, we can solve for Np:
Np = (A × 1000) / (π × di × L)
Substituting the values, we can find Np.
Calculating these values, we get:
Q ≈ 2,272,727.27 kJ/h
A ≈ 3.04 m^2
Number of tubes ≈ 100.85 tubes
Np ≈ 2
Tube length (L) ≈ 1.70 m
Therefore, to satisfy the space constraints, you would need approximately 2 tube passes with 101 tubes per pass, and the length of each tube would be approximately 1.70 meters.
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A cylindrical specimen of some metal alloy 10 mm in diameter is stressed elastically in tension. A force of 10,000 N produces a reduction in specimen diameter of 2 × 10^-3 mm. The elastic modulus of this material is 100 GPa and its yield strength is 100 MPa. What is the Poisson's ratio of this material?
A cylindrical specimen of some metal alloy 10 mm in diameter is stressed elastically in tension.A force of 10,000 N produces a reduction in specimen diameter of 2 × 10^-3 mm.
The elastic modulus of this material is 100 GPa and its yield strength is 100 MPa.Poisson’s ratio (v) is equal to the negative ratio of the transverse strain to the axial strain. Mathematically,v = - (delta D/ D) / (delta L/ L)where delta D is the diameter reduction and D is the original diameter, and delta L is the length elongation and L is the original length We know that; Diameter reduction = 2 × 10^-3 mm = 2 × 10^-6 mL is the original length => L = πD = π × 10 = 31.42 mm.
The axial strain = delta L / L = 0.0032/31.42 = 0.000102 m= 102 μm Elastic modulus (E) = 100 GPa = 100 × 10^3 M PaYield strength (σy) = 100 MPaThe stress produced by the force is given byσ = F/A where F is the force and A is the cross-sectional area of the specimen. A = πD²/4 = π × 10²/4 = 78.54 mm²σ = 10,000/78.54 = 127.28 M PaSince the stress is less than the yield strength, the deformation is elastic. Poisson's ratio can now be calculated.v = - (delta D/ D) / (delta L/ L)= - 2 × 10^-6 / 10 / (102 × 10^-6) = - 0.196Therefore, the Poisson's ratio of this material is -0.196.
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A closed system contains an ideal gas, which molecular weight is W-60 kg/kmol, and its standard state entropy is so-0. The system undergoes the following cycle: at state 1 the temperature is 293.15 K, the pressure is 94 kPa, and the entropy is 45.428 J/(kg.K). The gas is compressed polytropically at n=1.45 until the specific volume is 10 times lower than that at state 1 (state 2). Then 84552.2 J/kg of heat is added at constant specific volume (state 3). After that heat is added at constant pressure until entropy is 333.333 J/(kg.K) (state 4). In the next process the system undergoes isentropic expansion (and reaches state 5). Finally there is a constant volume rejection of heat (until state 1). Determine a) the values of p, v, T and s, at each cycle point
The constant volume rejection of heat from state 5 to state 1 means that the pressure and maxium temperature change, but the volume remains constant.
a) The values of **p, v, T, and s at each cycle point are as follows:
State 1:
p1 = 94 kPa
v1 = Unknown
T1 = 293.15 K
s1 = 45.428 J/(kg·K)
State 2:
p2 = Unknown
v2 = 10 * v1
T2 = Unknown
s2 = Unknown
State 3:
p3 = p2 (constant specific volume)
v3 = v2
T3 = Unknown
s3 = Unknown
State 4:
p4 = Unknown
v4 = Unknown
T4 = Unknown
s4 = 333.333 J/(kg·K)
State 5:
p5 = p1
v5 = Unknown
T5 = Unknown
s5 = s1
To determine the values at each state, we need to use the appropriate thermodynamic relationships and equations. The polytropic process in state 2 can be described using the equation p2 * v2^n = constant. The heat added at constant volume in state 3 does not affect the pressure, but increases the temperature. The heat added at constant pressure in state 4 increases the temperature and entropy.
The isentropic expansion from state 4 to state 5 implies that entropy remains constant. Finally, the constant volume rejection of heat from state 5 to state 1 means that the pressure and temperature change, but the volume remains constant. By applying the relevant equations and conditions, the values of p, v, T, and s at each state can be determined
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List three (3) basic attributes required for the operation of PV Cells.
What technology is used to generate electricity from solar power?
Three basic attributes required for the operation of PV cells (Photovoltaic cells) are: Sunlight: PV cells require sunlight or solar radiation to generate electricity.
Semiconductor Material: PV cells are made of semiconductor materials, typically silicon-based, that have the ability to convert sunlight into electricity. Electric Field: PV cells have an internal electric field created by the junction between different types of semiconductor materials. This electric field helps separate the generated electron-hole pairs, allowing the flow of electric current.
The technology used to generate electricity from solar power is called solar photovoltaic technology or solar PV technology. Solar PV technology involves the use of PV cells to directly convert sunlight into electricity.This electric current can then be harnessed and used to power electrical devices or stored in batteries for later use.
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A regenerative steam turbine has a throttle pressure of 3.8 MPa at 380ºC and a condenser at 0.01 MPa. Steam are extracted at the following points: 2.0 MPa, 1.0 MPa, and at 0.2 MPa. For the ideal cycle, find (a) The amount of steam extracted (b) W, QA and e. (c) For an ideal engine and the same states, compute (d) W, QA and e and
The given values are, Throttle pressure (P1) = 3.8 MPaTemperature (T1) = 380°CCondenser pressure (P3) = 0.01 MPaSteam extraction points = 2.0 MPa, 1.0 MPa, and 0.2 MPa.
Regarding the Ideal Rankine cycle, we can write,
QN + W = Qout
where QN is the heat input, W is the work done, and Qout is the heat rejected.
Now, QA is the difference between QN and Qout, i.e.,
QA = QN - Qout
where QA = W + Q3 - Q2
For the Regenerative Rankine cycle, we can write,
QA = W + Q3 - Q2 - Qextracted
where Qextracted is the heat extracted through steam at the extraction points.
Using the table for steam properties, at 3.8 MPa, we get,
Tsat = 208.34°C, h1 = 3137.9 kJ/kg, and s1 = 6.8697 kJ/kg.K.
At 0.01 MPa, we get, h3 = 191.81 kJ/kg.
Now, we can find the heat input as, QN = h1 - h4
where we can assume h4 = h3 (because we have no other information about it).
Qout = h3 - h2
Where,we can assume that the extracted steam at 2 MPa, 1 MPa, and 0.2 MPa is dry saturated.
Using the steam table, we can get the enthalpy values of the extracted steam as,
h2a = 3053.7 kJ/kg,
h2b = 2987.2 kJ/kg,
h2c = 2834.9 kJ/kg.
As we are using the extracted steam for feedwater heating, we can assume that the feedwater enters the feedwater heater (FWH) at the condenser pressure and exits at the same pressure.
Using the above values, we can find the enthalpies at state 4 as,
h4a = 2873.2 kJ/kg,
h4b = 2728.6 kJ/kg,
h4c = 2335.5 kJ/kg.
Now we can find the heat input as,
QN = h1 - h4a = 3137.9 - 2873.2 = 264.7 kJ/kg.
(a) The amount of steam extracted =
m(flow rate of extracted steam) = m2a + m2b + m2c.
From the enthalpy values of the extracted steam, we can write,
m2a = (h2a - h3) / (h1 - h4a) = 0.0237 kg/kg,
m2b = (h2b - h3) / (h1 - h4b) = 0.0294 kg/kg,
m2c = (h2c - h3) / (h1 - h4c) = 0.0462 kg/kg,
Therefore, the flow rate of extracted steam is m = m2a + m2b + m2c = 0.0993 kg/kg.
(b) We can calculate the work done as,
W = QN - Qout = 264.7 - 179.1 = 85.6 kJ/kg.
QA = W + Q3 - Q2
where Q3 = h3 and Q2 = (m2a * h2a + m2b * h2b + m2c * h2c)
Using these values, we get, QA = 85.6 + 191.81 - (0.0237 * 3053.7 + 0.0294 * 2987.2 + 0.0462 * 2834.9) = -56.5 kJ/kg.
(c) For an ideal engine and the same states, compute (d) W, QA, and e
The values for the ideal cycle can be calculated using the formulae,
e = 1 - (P3 / P1) ^ (γ - 1) / γ = 1 - (0.01 / 3.8) ^ 0.286 = 0.4821.
W = m (h1 - h3) = 0.0993 (3137.9 - 191.81) = 296.54 kJ/kg.
Qout = m (h3 - h4a) = 0.0993 (191.81 - 2873.2) = -266.96 kJ/kg.
QN = m (h1 - h4a) = 0.0993 (3137.9 - 2873.2) = 264.7 kJ/kg.
QA = W + Q3 - Q2
where Q3 = h3 and Q2 = 0,
Using these values, we get,QA = 296.54 + 191.81 = 488.35 kJ/kg
In conclusion, the given parameters were used to find the values for the amount of steam extracted, W, QA, and e for the ideal and regenerative Rankine cycle. The problem can be solved using the formulae provided and the enthalpy values from the steam table.
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Steam enters the high-pressure turbine of a steam power plant that operates on the ideal reheat Rankine cycle at 6 MPa and 500°C and leaves as saturated vapor. Steam is then reheated to 400°C before it expands to a pressure of 10 kPa. Heat is transferred to the steam in the boiler at a rate of 6 × 104 kW. Steam is cooled in the condenser by the cooling water from a nearby river, which enters the condenser at 7°C. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the pressure at which reheating takes place, (b) the net power output and thermal efficiency, and (c) the minimum mass flow rate of the cooling water required. mains the same
a) Pressure at which reheating takes place The given steam power plant operates on the ideal reheat Rankine cycle. Steam enters the high-pressure turbine at 6 MPa and 500°C and leaves as saturated vapor.
The cycle on a T-s diagram with respect to saturation lines can be represented as shown below :From the above diagram, it can be observed that the steam is reheated between 6 MPa and 10 kPa. Therefore, the pressure at which reheating takes place is 10 kPa .
b) Net power output and thermal efficiency The net power output of the steam power plant can be given as follows: Net Power output = Work done by the turbine – Work done by the pump Work done by the turbine = h3 - h4Work done by the pump = h2 - h1Net Power output = h3 - h4 - (h2 - h1)Thermal efficiency of the steam power plant can be given as follows: Thermal Efficiency = (Net Power Output / Heat Supplied) x 100Heat supplied =[tex]6 × 104 kW = Q1 + Q2 + Q3h1 = hf (7°C) = 5.204 kJ/kgh2 = hf (10 kPa) = 191.81 kJ/kgh3 = hg (6 MPa) = 3072.2 kJ/kgh4 = hf (400°C) = 2676.3 kJ/kgQ1 = m(h3 - h2) = m(3072.2 - 191.81) = 2880.39m kJ/kgQ2 = m(h4 - h1) = m(26762880.39m - 2671.09m = 209.3m x 100= [209.3m / (2880.39m + 2671.09m)] x 100= 6.4 %c)[/tex]
Minimum mass flow rate of the cooling water required Heat rejected by the steam to the cooling water can be given as follows: Q rejected = mCpΔTwhere m is the mass flow rate of cooling water, Cp is the specific heat capacity of water, and ΔT is the temperature difference .Qrejected = Q1 - Q2 - Q3 = 209.3 m kW Q rejected = m Cp (T2 - T1)where T2 = temperature of water leaving the condenser = 37°C, T1 = temperature of water entering the condenser = 7°C, and Cp = 4.18 kJ/kg K Therefore, m = Qrejected / (Cp (T2 - T1))= 209.3 x 103 / (4.18 x 30)= 1.59 x 103 kg/s = 1590 kg/s Thus, the minimum mass flow rate of cooling water required is 1590 kg/s.
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i (hydraulic gradient) = 0.0706
D= 3 mm v=0.2345 mis Find Friction factor ? Friction factor (non-dimensional): f = i 2gD/V²
To Find: Friction factor (f) Formula Used: Friction factor (non-dimensional) formula: f = i 2gD/V² Using the given values in the formula, we get the friction factor as 0.3184.
Hydraulic gradient (i) = 0.0706
Diameter of pipe (D) = 3 mm
Velocity of water (V) = 0.2345 m/s
Using the formula for friction factor, f = i 2gD/V²
= (0.0706)2 × 9.81 × 0.003 / (0.2345)²
= 0.01754 / 0.05501
= 0.3184 (approximately)
Therefore, the friction factor (f) is 0.3184. Friction factor is a dimensionless quantity used in fluid mechanics to calculate the frictional pressure loss or head loss in a fluid flowing through a pipe of known diameter, length, and roughness.
Where, i is the hydraulic gradient, D is the diameter of the pipe, V is the velocity of water, g is the acceleration due to gravity. To calculate the friction factor in this problem, we have given the hydraulic gradient, diameter of pipe, and velocity of water. Using the given values in the formula, we get the friction factor as 0.3184.
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A cantilever beam has length 24 in and a force of 2000 lbf at the free end. The material is A36/. For a factor of safety of 2, find the required cross section dimensions of the beam. The cross section can be assumed as square, rectangular, pipe or I-beam.
The formula for the shear stress in a cantilever beam subjected to a transverse force can be used to find the required cross-section dimensions for the beam.The formula is; τmax = VQ/ItWhere;V = the maximum force (2000 lbs.)Q = the first moment of the area around the neutral axis.
I = the moment of inertia.The maximum shear stress for A36 steel is 20,000 psi. For a factor of safety of 2, this value can be doubled to 40,000 psi.So,τmax = VQ/It = 40000 psi.The dimensions of the beam can be found using the shear stress equation and the bending moment equation.
Mmax = PL/4 = 2000 lbs. × 24 in./4 = 12000 in. lbs.τmax = Mmax*c/I = 40000 psiThe required cross-section dimensions of the beam can be found as follows;For a square beam;a = b ⇒ c = a / √6P = 12000 lbs.
[tex]Q = b × h × h / 2 = a × a × a / 2√3h = a/√3I = a^4/12c = I × τmax / b × h²a = (6 × P / (τmax × h²))^(1/4).[/tex]
For a rectangular beam;
[tex]a < b ⇒ c = a / √6P = 12000 lbs.Q = b × h × h / 2 = a × b × b / 2h = √(2a / 3)I = ab^3/12c = I × τmax / b × h²a = (6 × P / (τmax × h² × b))^(1/3) × b^2/3.[/tex]
For a pipe;a = b and D = 2rP = 12000 lbs.τavg = P/ (2A - a²) = 40000 psiThe diameter of the pipe can be found using the following equation;
[tex]r = (P/2τavg)(D² - d²)/D²d = D - 2ta = πr² - πr²/4A = πr²D = 2r(1 + (4a²/(πr^2))^(1/2)).[/tex]
For an I-beam;the required dimensions can be found by assuming that the beam is an equivalent rectangular beam and then using the above rectangular beam formula. In the equivalent rectangular beam, the width of the flanges is equal to the thickness of the web multiplied by a factor of 1.2 to 1.5. The thickness of the web is taken as the distance between the midpoints of the flanges.
From the above, we can conclude that the cross-section dimensions of a square beam, rectangular beam, pipe, and I-beam can be found.
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Load has a power factor of — (lagging) 1) is this lood inductive or capacitive 2) You connect a test voltage source of 20my (cosωt) and measure the current. Absolute value of the current is 5mA. Find its phase. and 3) Find average power, reactive power apparent power Draw the vectors.
The real power, or average power, is represented by the adjacent side of the triangle, while the reactive power is represented by the opposite side. The real power vector is horizontal, while the reactive power vector is vertical.
Load has a power factor of 1, which is lagging, indicating that the load is inductive. The load is inductive because the power factor is lagging and is between 0 and 1. A lagging power factor indicates that the current is not in phase with the voltage.
The test voltage source is 20mV (cosωt), and the absolute value of the current is 5mA. To determine the phase angle, we'll need to use Ohm's law.
Since the current and voltage are out of phase, we'll need to utilize complex arithmetic to determine the phase angle. We'll have to compute the product of the two complex numbers.
In this case, Z=V/I,
where V = 20mV,
I = 5mA.
Therefore, Z = (20 x 10^-3)/(5 x 10^-3) = 4.
The angle of this complex number is the same as the phase angle of the circuit.
Therefore, tan θ = 0.5, and θ = 26.56 degrees.
The following formulae were used to find the average power, reactive power, and apparent power:
Average power = Vrms * Irms * cosθ = 20mV * 5mA * cos 26.56 degrees
= 0.444mWReactive power
= Vrms * Irms * sinθ
= 20mV * 5mA * sin 26.56 degrees
= 0.208mWApparent power
= Vrms * Irms = 20mV * 5mA
= 0.1mW
The power vectors can be drawn to represent the power characteristics of the circuit. The apparent power is represented by the hypotenuse of the power triangle.
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Implementation of the quadcopter model in Matlab (for example a
state-space model or transfer matrix one), simulation results (step
responses).
The quadcopter is an aerial vehicle that has gained a lot of attention and interest in recent times due to its application in different fields. It has different flight controls, including lift, pitch, roll, and yaw, which make it versatile and efficient.
The implementation of a quadcopter model in Matlab involves the creation of a mathematical representation of the system that simulates the flight behavior of the quadcopter.The state-space model or transfer matrix is the common representation used to simulate the quadcopter's dynamics. The state-space model represents the quadcopter's states in the form of differential equations that describe how the system changes over time.
The quadcopter model's implementation involves the following steps:
1. Define the system inputs and outputs: The system inputs are the control signals, while the outputs are the states of the system.
2. Develop the mathematical model: This involves deriving the equations that represent the quadcopter's dynamics.
3. Linearize the system: The quadcopter model is a nonlinear system, and linearizing it simplifies its dynamics and makes it easier to simulate.
4. Create the state-space model or transfer matrix: Using the derived equations, the state-space model or transfer matrix is created.
5. Simulate the system: The created model is used to simulate the system's response to different inputs, including step responses. The simulation results help to analyze and evaluate the quadcopter's behavior and performance.
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Explain briefly the advantages" and "disadvantages of the "Non ferrous metals and alloys" in comparison with the "Ferrous alloys (15p). Explain briefly the compositions and the application areas of the "Brasses"
The advantages are : 1. Non-ferrous metals are generally more corrosion resistant than ferrous alloys. 2. They are also more lightweight and have a higher melting point. 3. Some non-ferrous metals, such as copper, are excellent conductors of electricity. The disadvantages are : 1. Non-ferrous metals are typically more expensive than ferrous alloys. 2. They are also more difficult to machine and weld. 3. Some non-ferrous metals, such as lead, are toxic.
Here is a brief explanation of the compositions and application areas of brasses:
1. Brasses are copper-based alloys that contain zinc.
2. The amount of zinc in a brass can vary, and this can affect the properties of the alloy.
3. For example, brasses with a high zinc content are more ductile and machinable, while brasses with a low zinc content are more resistant to corrosion.
4. Brasses are used in a wide variety of applications, including:
Electrical connectors
Plumbing fixtures
Musical instruments
Jewelry
Coins
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Polyethylene (PE), C2H4 has an average molecular weight of 25,000 amu. What is the degree of polymerization of the average PE molecule? Answer must be to 3 significant figures or will be marked wrong. Atomic mass of Carbon is 12.01 Synthesis is defined as a. The shaping of materials into components to cause changes in the properties of materials.
b. The making of a material from naturally occurring and/or man-made material. c. The arrangement and rearrangement of atoms to change the performance of materials. d. The chemical make-up of naturally occurring and/or engineered material.
The degree of polymerization (DP) of a polymer is defined as the average number of monomer units in a polymer chain.the degree of polymerization of the average PE molecule is approximately 890.
In the case of polyethylene (PE), which has an average molecular weight of 25,000 amu, we can calculate the DP using the formula:
DP = (Average molecular weight of polymer) / (Molecular weight of monomer)
The molecular weight of ethylene (C2H4) can be calculated as follows:
Molecular weight of C2H4 = (2 * Atomic mass of Carbon) + (4 * Atomic mass of Hydrogen)
= (2 * 12.01 amu) + (4 * 1.01 amu)
= 24.02 amu + 4.04 amu
= 28.06 amu
Now, we can calculate the DP:
DP = 25,000 amu / 28.06 amu
≈ 890.24
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A centrifugal pump may be viewed as a vortex, where the 0.4m diameter impeller, rotates within a 1m diameter casing at a speed of 200 rpm.
Determine
The circumferential velocity, in m/s at a radius of 0.45 m
A centrifugal pump may be viewed as a vortex.
It consists of an impeller that rotates within a casing.
The impeller's diameter is 0.4m and rotates within a 1m diameter casing at a speed of 200rpm.
To determine the circumferential velocity, use the formula provided below:
Formula:
Circumferential velocity (v) = 2π x Radius (r) x Rotational Speed (N) / 60
Given:
Radius (r) = 0.45 m
Rotational speed
(N) = 200 rpm
Diameter of impeller = 0.4m
Diameter of casing = 1m
Solution:
Circumference of the impeller= π
diameter= π x 0.4 m
= 1.2566 m
Therefore,
Circumferential velocity (v) = 2π x Radius (r) x Rotational Speed (N) / 60
= (2 x π x 0.45 m x 200 rpm) / 60
= (0.1414 x 200) m/s
= 28.28 m/s
Therefore, the circumferential velocity at a radius of 0.45 m is 28.28 m/s.
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