The correct option is(B). The height of the transverse metacentre above the centre of buoyancy is called metacentric height.
The height of the transverse metacentre above the centre of buoyancy is called metacentric height (GM).
The distance between the centre of gravity of a floating body and its metacentre is referred to as the metacentric radius.
GM determines the stability of a floating body.
Metacentric height (GM) is the term used to describe the distance between the center of gravity of a floating body and its metacentre.
It is the height of the transverse metacentre above the centre of buoyancy. When a ship is floating in calm water, it will come to a stable position after it is heeled because of the change in its centre of gravity and buoyancy.
The stability of the floating body is determined by the metacentric height (GM).When the GM value is high, the ship will have more stability, and it will be challenging to overturn it.
When the GM value is low, the ship will be less stable, and it will be easier to overturn. If the value of GM is zero, the ship will become unstable and will capsize.
Therefore, the value of GM is critical in the design of a ship and is calculated by naval architects to ensure the safety and stability of the vessel.
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Heat exchanger will be used in reflux to utilize the sensible heat of flow of organic supply fluid mh = 0.16 kg / s and temperature Th = 85οC (Cph = 2 kJ / kg οC). This current will heat water supply mc = 0.10 kg / s temperature Tc = 22οC (Cpc = 4.2 kJ / kg οC). The total heat transfer coefficient is estimated as U = 400 W / m2 οC.
(a) Find the theoretical output temperatures of the two currents corresponding to an infinite surface heat exchanger.
(b) If a temperature approach equal to 5οC is selected, calculate the actual outlet temperatures and the required surface.
Theoretical output temperature for hot fluid at inlet is 85ºC. Therefore the theoretical output temperature for hot fluid at outlet is also 85ºC.Theoretical output temperature for cold fluid at inlet is 22ºC.
Therefore the theoretical output temperature for cold fluid at outlet is also 22ºC. b)Given, Mass flow rate of hot fluid, mh = 0.16 kg/s
Temperature of hot fluid at inlet, Th = 85 °C
Temperature of cold fluid at inlet, Tc = 22 °C
Mass flow rate of cold fluid, mc = 0.10 kg/s
The specific heat capacity of hot fluid, Cph = 2 kJ/kg.
K The specific heat capacity of cold fluid, Cpc = 4.2 kJ/kgK
The total heat transfer coefficient,
U = 400 W/m².K
temperature) = Q / (m_c * C_pc)U A
The specific heat capacity of hot fluid, Cph = 2 kJ/kg.
K The specific heat capacity of cold fluid, Cpc = 4.2 kJ/kgK
Therefore the actual outlet temperatures are T_h,out = 93.5°C and T_c,out = 10.57°C.
The theoretical output temperature for hot fluid at outlet is also 85ºC.Theoretical output temperature for cold fluid at inlet is 22ºC.
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A shaft and hole fit is 20 H9/d9 inch, what is the
allowance of the fit (in this case, the minimum clearance of the
fit)
A.
0.065 in
B.
0.104 in
C.
0.208 in
D.
0.052 in
Shaft and hole fits are the fit types between a shaft (external cylinder) and a hole (internal cylinder). The fit types are classified according to the tolerance or clearance.
There are four types of shaft and hole fits: clearance fit, transition fit, interference fit, and shrink fit. Here, the given fit is 20 H9/d9 inch. Therefore, the allowance of the fit (minimum clearance of the fit) can be found as follows: Allowance = [(Upper deviation of hole size) − (Lower deviation of shaft size)] .
where Upper deviation of hole size = IT9 = 25 microns Lower deviation of shaft size = IT7 = 50 microns the allowance = [(25 + 50) / 2] / 1000 inches= 0.0375 inches the option A, 0.065 in is the closest value to the calculated allowance.
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The return air from a space is mixed with the outside air in the ratio of (4:1) by mass. The mixed air is then entering the heating coil. The following data refer to the space: Inside design conditions (t-25°C; = 50%), outdoor air conditions (t= 5°C; = 60%), and the room Sensible Heat Ratio SHR is -0.5, Determine: (a) the supply air dry-bulb and wet-bulb temperature (b) the supply mass flow rate for 1 m³/min supply air; (c) the sensible and latent heat in kW; (d) the fresh air volume flow rate, in m³/min; and (d) the total load of the heating coil.
Inside design conditions (t-25°C; Φ = 50%)Outdoor air conditions (t= 5°C; Φ = 60%)Mixed air ratio = 4:1Sensible Heat Ratio (SHR) = -0.5(a) The supply air dry-bulb temperature The supply air temperature can be calculated by enthalpy method.
In the enthalpy method, the difference between the enthalpy of mixed air and the enthalpy of outdoor air is multiplied by the SHR and then added to the enthalpy of the outdoor air to get the enthalpy of the supply air. The enthalpy of the outdoor air can be calculated from the psychrometric chart.
It is found to be 20.07 kJ/kg. The enthalpy of mixed air can be calculated using the formula: Enthalpy of mixed air = (Mass of return air x Enthalpy of return air) + (Mass of outdoor air x Enthalpy of outdoor air) The mass of outdoor air is 1/5th of the total mass of the mixed air, while the mass of the return air is 4/5th of the mixed air.
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A message signal with a frequency of 5 kHz is obtained with an FM signal with a frequency deviation of (+-)3 kHz. What is the bandwidth of the FM signal?
The message signal has a frequency of 5 kHz and the frequency deviation is (+-)3 kHz.
The formula used to determine the bandwidth of the FM signal is given as:B = 2(Δf + fm) where Δf is the frequency deviation and fm is the message signal frequency.
Substituting the values given above:B = 2(3 kHz + 5 kHz)B = 16 kHz the bandwidth of the FM signal is 16 kHz.
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21. A(n) ____. is a material that has a very high resistance and resists the flow of electrons a. Circuit breaker b. insulator c. fuse d. conductor e. none of the above 22. The process by which general contractors and electrical contractors obey during construction for safety purposes around electrical equipment is referred to as: a. Saf-T-tag b. Keep out watch out c. Lock out tag out d. Suns out guns out 23. Explain the difference between 12-2 and 10-3 Romex: 24. Which type of light bulb currently used in construction draws the least amount of power? 25. (A) What does GFCI stand for? (B) What does a GFCI do, and where does it belong?
21 A(n) insulator. is a material that has a very high resistance and resists the flow of electrons
b. insulatorWhat contractors and electrical contractors must adhere to22. During construction, general contractors and electrical contractors must adhere to the lock out tag out process for safety purposes around electrical equipment.
c. Lock out tag out23. The numbers in 12-2 and 10-3 Romex refer to the gauge of the wire and the number of conductors.
12-2 Romex has a 12-gauge wire, which is thicker than 10-gauge wire. It contains two conductors, typically a black (hot) wire and a white (neutral) wire.
10-3 Romex has a 10-gauge wire, which is thicker than 12-gauge wire. It contains three conductors, typically a black (hot) wire, a red (hot) wire, and a white (neutral) wire.
The difference in gauge affects the current-carrying capacity of the wire, with lower gauge numbers being able to handle higher currents.
24. LED (Light Emitting Diode) light bulbs currently used in construction draw the least amount of power compared to traditional incandescent or fluorescent bulbs. LEDs are highly efficient and provide significant energy savings.
25. (A) GFCI stands for Ground Fault Circuit Interrupter.
(B) A GFCI is a safety device designed to protect against electrical shocks caused by ground faults. It constantly monitors the electrical current flowing through a circuit and quickly shuts off power if it detects any imbalance between the hot and neutral wires. It helps prevent electric shock hazards, particularly in areas with water such as bathrooms, kitchens, or outdoor outlets. GFCIs are typically installed in electrical outlets or incorporated into circuit breakers.
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Merits and Demerits between HRC/Drop-Out Fuses and other
different types of
fuses
HRC and drop-out fuses have both merits and demerits when compared to other types of fuses. It is up to the user to decide which type of fuse is best suited for their specific needs.
HRC (High Rupturing Capacity) and drop-out fuses are some of the types of fuses that have both merits and demerits as compared to other types of fuses.
The demerits and merits of each type of fuse are discussed in detail as follows:
Demerits of HRC and Drop-Out Fuses:
The following are the demerits of the HRC and drop-out fuses:
They are more expensive than other types of fuses. Due to their complexity, they require more maintenance, which adds to their cost.
They are unsuitable for low voltages because they require a lot of current to trigger, which can be dangerous.
They have a higher tripping time than other types of fuses, which can cause damage to equipment.
Merits of HRC and Drop-Out Fuses:
The following are the merits of the HRC and drop-out fuses:
They can handle a larger amount of current than other types of fuses, which means they can protect larger electrical systems.
They have a higher breaking capacity, which means they can handle large current surges without breaking down.
They have a longer lifespan than other types of fuses, which makes them more reliable.
They are safer because they have a lower risk of causing a fire or explosion due to their design. Other types of fuses have a higher risk of failure due to their design, which can lead to a fire or explosion.
Overall, HRC and drop-out fuses have both merits and demerits when compared to other types of fuses. It is up to the user to decide which type of fuse is best suited for their specific needs.
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Develop a project with simulation data of a DC-DC converter: Buck Boost a) 12V output and output current between (1.5 A-3A) b) Load will be two 12 V lamps in parallel/Other equivalent loads correction criteria c) Simulation: Waveforms (input, conversion, output) of voltage and current in general. Empty and with load. d) Converter efficiency: no-load and with load e) Frequency must be specified f) Development of the high frequency transformer, if necessary g) Smallest size and smallest possible mass. Reduce the use of large transformers. >>> Simulation can be done in Multisim or in another software of your choice.
Project Description:In this project, we will simulate a DC-DC converter known as a Buck-Boost converter. The objective is to design a converter that produces a 12V output with an output current ranging between 1.5A and 3A.
The load for the converter will consist of two 12V lamps connected in parallel or other equivalent loads as per the correction criteria.
The simulation will involve analyzing the waveforms of the input voltage and current, conversion voltage and current, and output voltage and current. The simulation will be conducted for both empty (no-load) conditions and with the specified load.
Efficiency analysis will be performed to determine the converter's efficiency under both no-load and loaded conditions. The efficiency will be calculated as the ratio of the output power to the input power.
The frequency of operation for the converter needs to be specified. Generally, a high-frequency operation is preferred to reduce the size and mass of the components. The specific frequency will depend on the requirements and constraints of the project.
If necessary, the design will involve the development of a high-frequency transformer. The transformer will be designed to meet the size and mass requirements while ensuring efficient power transfer.
The main objective of the project is to achieve the smallest possible size and mass for the converter while reducing the reliance on large transformers. The design will prioritize compactness and efficiency.
Simulation software such as Multisim or any other suitable software of your choice can be used to perform the simulation and analysis of the DC-DC converter.
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In a machining operation that approximates orthogonal cutting, the cutting tool has a rake angle = 8º. The chip thickness before the cut to=0.50mm and the chip thickness after the cut tc=1.25mm. The width of the orthogonal cutting operation w = 3.0 mm. It is known the shear strength of the work is 250MPa. Please determine the magnitude of the cutting force.
In orthogonal cutting operations, the chip thickness ratio is defined as the ratio of the thickness of the chip before the cut to the thickness of the chip after the cut. It is denoted by r.
Therefore, $r = \frac{t_c}{t_0}$Where, $t_c$ = Chip thickness after the cut$ t_0$ = Chip thickness before the cut. The shear angle and the shear plane angle can be calculated by using the rake angle and the friction angle. Shear angle φ is given as$\tan \phi = \frac{\tan \alpha - \mu}{1 + \tan \alpha \mu}$.
Where, α is the rake angle, and μ is the coefficient of friction at the shear plane. The shear plane angle $\phi_ s $ is equal to 90° - φ.The magnitude of the cutting force F can be calculated using the equation, F = \frac{T}{r} Where T is the cutting force per unit width of cut.
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When filled to capacity, the unpressurized storage tank contains water to a height of h = 34 ft. The outside diameter of the tank is 7.3 ft and the wall thickness is 0.646 in. Determine the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base. (Weight density of water = 62.4 lb/ft3.)
The given data:Height of the storage tank, h = 34 ftOutside diameter of the tank, D = 7.3 ftWall thickness, t = 0.646 inWeight density of water, w = 62.4 lb/ft³.
We need to determine the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base.So, the following formulae are used:Volume of the tank = [tex]πD²h/4 = π(7.3)²(34)/4 = 1988.29 ft³.[/tex]
Weight of the water = Volume of the tank × weight density of water = 1988.29 × 62.4 = 124236.1 lb.
The water in the tank is trying to expand and the tank is resisting this expansion. Thus, there will be a radial stress on the tank at the bottom.The maximum normal stress at the base of the tank,
σmax = wH/2t + P/4t
Where P = Weight of the water in the tank = 124236.1 lbH = Height of the water in the tank = 34 ft
[tex]σmax = (62.4 × 34)/(2 × 0.646) + 124236.1/(4 × 0.646) = 23618.2 + 48325.6 = 71943.8 lb/ft²= 71943.8/144 = 499.6 psi[/tex].
The absolute maximum shear stress on the outer surface of the tank at its base, τmax = P/2At the base, the direction of the normal stress is radial and the direction of the shear stress is tangential.
Therefore, τmax = 124236.1/2 = 62118.05 lb/ft²= 62118.05/144 = 431.4 psi
In this question, the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base is to be determined. The formulae used to solve this problem are as follows:
The maximum normal stress at the base of the tank, σmax = wH/2t + P/4tThe absolute maximum shear stress on the outer surface of the tank at its base, τmax = P/2When the water is filled in the tank, it tries to expand and the tank resists this expansion.
Therefore, there is a radial stress on the tank at the bottom. The maximum normal stress at the base of the tank is calculated by using the formula σmax = wH/2t + P/4t. Here, w is the weight density of water, H is the height of the water in the tank, t is the thickness of the wall, and P is the weight of the water in the tank.
Substituting the given values, we get
[tex]σmax = (62.4 × 34)/(2 × 0.646) + 124236.1/(4 × 0.646) = 23618.2 + 48325.6 = 71943.8 lb/ft².[/tex]
The absolute maximum shear stress on the outer surface of the tank at its base is calculated by using the formula τmax = P/2. Here, P is the weight of the water in the tank. Substituting the given values, we get
τmax = 124236.1/2 = 62118.05 lb/ft².
Therefore, the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base are 499.6 psi and 431.4 psi, respectively.
Thus, we can conclude that the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base are 499.6 psi and 431.4 psi, respectively.
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In the manufacturing of self-lubricating bearings by powder metallurgy, an important secondary operation that is out after sintering is. a) Infiltration b) impregnation c) Cold isostatic pressing d) Hot isostatic pressing
The correct option is b) Impregnation is an important secondary operation that is carried out after sintering in the manufacturing of self-lubricating bearings by powder metallurgy.
Impregnation involves filling the interconnected porosity of the sintered bearing with a lubricant or resin. This process helps to enhance the self-lubricating properties of the bearing by providing a continuous lubricating film within the bearing structure. The lubricant or resin infiltrates the pores of the sintered material, improving its ability to reduce friction and wear.
In contrast, infiltration (a) refers to the process of filling the porosity of a sintered part with a material different from the base material, such as a metal or alloy. Cold isostatic pressing (c) involves subjecting the sintered part to high-pressure isostatic compression at room temperature. Hot isostatic pressing (d) is a similar process but performed at elevated temperatures.
While these processes may be used in powder metallurgy, impregnation specifically addresses the enhancement of self-lubricating properties in bearings.
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There is an air flow with a temperature of 32.0℃, and it is humidified by making it flow over a container filled with water and whose length is 1.2 m. The temperature at the air-water interface is 20.0 ℃. If the initial humidity of the air is 25.0% and its speed is 0.15 m/s.
You are asked to determine:
a. The mass transfer coefficient.
b. The rate of evaporation of water per unit width of the container.
For this purpose, you must use the following empirical correlation:
Sℎ = 0.664Re^0.5Sc^0.333
- Sherwood number (Sh)
- Schmidt number (Sc)
Psat(20.0℃) = 0.02308 atm
Psat(32.0℃) = 0.04696 atm
R= 0.082 atm l/Kmol
Dwater in air = 2.77 ∙ 10−5 m^2⁄s
NH2O: it is expressed in mol/m^2s
The rate of evaporation of water per unit width of the container is 5.45 × 10^-6 mol/(m.s).
Given data:
Temperature of air, T_1 = 32.0 ℃
Length of the container, L = 1.2 m
Temperature at the air-water interface, T2 = 20.0 ℃
Initial humidity of air, H_1 = 25.0%
Speed of air, V = 0.15 m/s
Water vapour pressure at T2,
Psat = 0.02308 atm
Water vapour pressure at T1,
P = 0.04696 atm
Gas constant, R = 0.082 atm l/Kmol
Diffusion coefficient of water in air, Dwater = 2.77 × 10^-5 m^2⁄s
Using the Sherwood Number equation:
Sℎ = 0.664Re^0.5Sc^0.333
Where Re is Reynolds's Number and Sc is Schmidt's Number.
Mass transfer coefficient = Dwater / L ShSc= 0.7
for air-water interface at 25°CSc = 2.14 × 10^-5 / 0.0343 = 6.23 × 10^-4 (calculated from Sc = v/D)
Re = ρvd/μ = 1092.8 (calculated from Re = VDwater/ν, where ν = viscosity of air = 1.81 × 10^-5 kg/m.s)
Therefore, Sh = 2.0 (calculated from Sherwood Number equation)
Mass transfer coefficient = Dwater / L Sh
= 2.77 × 10^-5 / (1.2 × 2) = 1.15 × 10^-5 m/s
Calculating the rate of evaporation of water per unit width of the container:
RH1 = H1 Psat / P - Psat
= 6.85% (Relative humidity)
Mass transfer rate = KH2O A RH = KH2O L RH1
W= 1.15 × 10^-5 × 1.2 × 6.85 / 18
= 5.45 × 10^-6 mol/(m.s)
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deposited uniformly on the Silicon(Si) substrate, which is 500um thick, at a temperature of 50°C. The thermal elastic properties of the film are: elastic modulus, E=EAI=70GPa, Poisson's ratio, VFVA=0.33, and coefficient of thermal expansion, a FaA=23*10-6°C. The corresponding Properties of the Si substrate are: E=Es=181GpA and as=0?i=3*10-6°C. The film-substrate is stress free at the deposition temperature. Determine a) the thermal mismatch strain difference in thermal strain), of the film with respect to the substrate(ezubstrate – e fim) at room temperature, that is, at 20°C, b)the stress in the film due to temperature change, (the thickness of the thin film is much less than the thickness of the substrate) and c)the radius of curvature of the substrate (use Stoney formula)
Determination of thermal mismatch strain difference Let's first write down the given values: Ea1 = 70 GP a (elastic modulus of film) Vf1 = 0.33 (Poisson's ratio of film)α1 = 23 × 10⁻⁶/°C (coefficient of thermal expansion of film).
Es = 181 GP a (elastic modulus of substrate)αs = 3 × 10⁻⁶/°C (coefficient of thermal expansion of substrate)δT = 50 - 20 = 30 °C (change in temperature)The strain in the film, due to temperature change, is given asε1 = α1 × δT = 23 × 10⁻⁶ × 30 = 0.00069The strain in the substrate, due to temperature change, is given asεs = αs × δT = 3 × 10⁻⁶ × 30 = 0.00009.
Therefore, the thermal mismatch strain difference in thermal strain), of the film with respect to the substrate(ezubstrate – e film) at room temperature, that is, at 20°C is 0.0006. Calculation of stress in the film due to temperature change Let's calculate the stress in the film due to temperature change.
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Explain why semiconducting materials and the behaviour
of semiconductor junctions play an important role in the working
principle and performance of Light-emitting diode
(LED).
Semiconducting materials and the behaviour of semiconductor junctions play a crucial role in the working principle and performance of Light-emitting diode (LED).Explanation: LEDs work on the principle of electroluminescence, in which a material emits light in response to an electric current passing through it. This property is exhibited by certain semiconducting materials that have a bandgap, which is the difference in energy levels between the valence and conduction bands.
When an LED is connected to a power source, an electric current flows through the device and causes electrons to move from the negative (n-type) to the positive (p-type) region of the semiconductor material. The electrons release energy as they move from the conduction band to the valence band, which produces photons of light.The behaviour of the semiconductor junctions is also essential to the performance of LEDs. A junction is formed by the contact between the n-type and p-type regions of the semiconductor material, which creates a depletion region that acts as a barrier to the flow of electrons and holes. This region is crucial because it helps to confine the charge carriers to the active region of the device, which maximizes the efficiency of the electroluminescent process.The construction of the p-n junction is also critical in ensuring the proper functioning of LEDs. The junction must be carefully engineered to ensure that it has the correct doping levels, thickness, and quality of the interface, among other factors. This helps to ensure that the device has the correct electrical and optical properties to emit light efficiently.
Finally, the choice of semiconducting materials used in LEDs is critical to their performance. Different materials have different bandgap energies, which determine the color of light that is emitted when the device is activated. Materials such as gallium arsenide, indium gallium nitride, and silicon carbide are commonly used in the construction of LEDs because they exhibit excellent electroluminescent properties.
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Question 2 The RCM3 process entails asking eight questions about the asset or the system under review. 2.1 Which is the first question would you consider as part of the initial steps in the RCM process? (1) 2.2 With an aid of an example, explain the difference between a primary and a secondary function. Please note: examples taken from the textbook/study guide will not be considered. (4) 2.3 With an aid of an example, describe the multiple performance standards of an equipment of your choice. Please note: examples taken from the textbook/study guide will not be considered. (4) 2.4 With an aid of an example, explain the difference between partial failure and total failure of an equipment of your choice. Please note: examples taken from the textbook/study guide will not be considered. (4)
2.5 What is meant by the operating context of a physical asset in RCM? Provide an example of an asset with different operating contexts (2) [15]
The first question to consider as part of the initial steps in the RCM (Reliability Centered Maintenance) process is "What are the functions and performance standards of the asset or system?".
Why "what are the functions and performance standards of the asset or system"?
When initiating the RCM process, it is crucial to clearly identify and understand the functions and performance standards of the asset or system under review. This involves determining the primary purpose and objectives of the asset or system as well as the specific performance requirements it needs to meet.
By establishing a solid understanding of the functions and performance standards, the subsequent steps in the RCM process such as identifying failure modes and consequences can be carried out effectively. This initial question sets the foundation for conducting a comprehensive analysis of the asset or system and ensures that maintenance strategies align with the desired performance objectives.
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Explain Metrology and types of Metrology, Deatiled the standard caliberation procedure for Calibration of an Instrument? Define Limits, Fit and Deviations used in Precision Metrology. What are the factors affecting accuracy of the Instruments .
Metro logy is the study of the measurement of physical quantities, their standards, and equipment used to measure them.
It involves various aspects like measurement science, engineering, and statistical methods, etc. Metro logy is the science of measurement. The objective of metro logy is to establish traceability of measurements to recognized standards and ensuring measurement accuracy and precision.
In other words, it is a science of measurement which deals with the establishment, maintenance, and development of national measurement standards.Types of Metro logy:Scientific metro logy: It deals with the establishment of the primary measurement standards that can be traced to the fundamental or natural quantities.
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For a 3-bus power system, the real and reactive powers are specified at all buses except the swing bus. The Newton Rephson method is chosen to solve the lood flow problem 1- What is the order of the Jacobian matrix ? 2- Determine the element in the Jacobson matrix, representing the variation of the real power at bus 2 with respect to the variation of the magnitude of the voltage at bus 2 3- Determine the element in the Jacobian matrix, representing the variation of the reactive power at bus 3 with respect to the variation of the angle of the voltage at bus 2
1. The order of the Jacobian matrix is equal to the number of unknowns in the power flow problem. In a 3-bus power system, the unknowns typically include the voltage magnitudes and voltage angles at each bus except the swing bus. Therefore, the order of the Jacobian matrix would be (2n - 1), where n is the number of buses in the system. In this case, since there are three buses, the order of the Jacobian matrix would be (2 * 3 - 1) = 5.
2. To determine the element in the Jacobian matrix representing the variation of the real power at bus 2 with respect to the variation of the magnitude of the voltage at bus 2, we need to compute the partial derivative of the real power at bus 2 with respect to the voltage magnitude at bus 2 (∂P2/∂|V2|).
The Jacobian matrix for the power flow problem consists of partial derivatives of the power injections at each bus with respect to the voltage magnitudes and voltage angles at all buses. Let's denote the Jacobian matrix as J.
The element representing ∂P2/∂|V2| in the Jacobian matrix can be denoted as J(2, 2), indicating the second row and second column of the matrix.
To determine the element in the Jacobian matrix representing the variation of the reactive power at bus 3 with respect to the variation of the angle of the voltage at bus 2, we need to compute the partial derivative of the reactive power at bus 3 with respect to the voltage angle at bus 2 (∂Q3/∂θ2).
Similarly to the previous question, the element representing ∂Q3/∂θ2 in the Jacobian matrix can be denoted as J(3, 2), indicating the third row and second column of the matrix.
1. The order of the Jacobian matrix for a 3-bus power system is 5.
2. The element in the Jacobian matrix representing the variation of the real power at bus 2 with respect to the variation of the magnitude of the voltage at bus 2 is J(2, 2).
3. The element in the Jacobian matrix representing the variation of the reactive power at bus 3 with respect to the variation of the angle of the voltage at bus 2 is J(3, 2).
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E Briefly explain how we take into account the variability of materials properties in: (i) Laboratory situations (ii) Real-world applications (3 marks)
In the laboratory and real-world applications, we take into account the variability of material properties.
In laboratory situations:In laboratory situations, material properties are assessed by carrying out experiments on a specimen of the material. In this situation, the variability of the material's properties is taken into account. In the laboratory, the variability of material properties is reduced by controlling environmental variables like temperature, humidity, and pressure.
Real-world applications:In real-world applications, materials are exposed to environmental factors that can affect their properties. The variability of material properties is taken into account by designing products that take into account the expected range of variability. Engineers will use the highest possible values of the material properties in their design calculations to account for the worst-case scenario
. Furthermore, manufacturers use statistical techniques to test the materials to ensure that the properties fall within an acceptable range. In addition, there are also safety factors that are built into designs to account for the variability of material properties.
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Q5) Given the denominator of a closed loop transfer function as expressed by the following expression: S²+85-5Kₚ + 20 The symbol Kₚ denotes the proportional controller gain. You are required to work out the following: 5.1) Find the boundaries of Kₚ for the control system to be stable.
5.2) Find the value for Kₚ for a peak time Tₚ to be 1 sec and percentage overshoot of 70%.
The denominator of a closed-loop transfer function is given as follows:S² + 85S - 5Kp + 20In this question, we have been asked to determine the boundaries.
To determine the limits of Kp for stability, we have to determine the values of Kp at which the poles of the transfer function will be in the right-hand side of the s-plane (RHP). This is also referred to as the instability criterion. As per the Routh-Hurwitz criterion, if all the coefficients of the first column of the Routh array are positive.
So let us form the Routh array for the given transfer function. Routh array:S² 1 -5Kp85 20The first column of the Routh array is [1, 85]. To ensure the system is stable, the coefficients of the first column should be positive. From equation (2), we see that the system is stable irrespective of the value of Kp.
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Angle of loll (10 marks) (a) A vessel is experiencing an Angle of Loll. What is the value of the righting lever GZ in this situation? (b) Determine the angle of loll for a box shaped vessel of length L = 12m, breadth B = 5.45m when floating on an even-keel at a draft of d = 1.75m. The KG is 2.32m.
(a) The value of the righting lever GZ in a vessel experiencing an Angle of Loll can be determined based on the vessel's stability characteristics.
The righting lever, GZ, represents the moment arm between the center of buoyancy (B) and the center of gravity (G), indicating the vessel's stability. To calculate GZ, the metacentric height (GM) and the heeling arm (GZh) must be considered. GM is the vertical distance between the center of gravity and the metacenter, while GZh is the distance between the center of gravity and the center of buoyancy at a given heel angle. GZ is then determined by subtracting GZh from GM.
(b) To determine the angle of loll for a box-shaped vessel, several factors need to be considered. The angle of loll occurs when a vessel has a negative metacentric height (GM) and is in an unstable condition. The formula to calculate the angle of loll is:
Angle of Loll = arctan(GM / KG)
In this case, the vessel has a length (L) of 12m, breadth (B) of 5.45m, and draft (d) of 1.75m. The KG, which represents the distance from the keel to the center of gravity, is given as 2.32m. By substituting these values into the formula, the angle of loll can be determined.
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In the following problems, the given loads are superimposed service loads; that is, they do not include the weights of the beams (unless noted otherwise). For structural steel beams (unless otherwise noted), assume a yield stress of 50 ksi (345 MPa). For timber beams, all beams are solid, rectangular shapes and Appendices E and F are applicable. Consider only moment and shear (unless otherwise noted). Select the lightest W shape to support a uniformly distrib- uted load of 2.1 kips/ft on a simple span of 24 ft.
The lightest W shape that can support a uniformly distributed load of 2.1 kips/ft on a simple span of 24 ft is [insert the W shape designation].
To determine the lightest W shape, we need to consider the maximum moment and shear forces generated by the given load. Given a uniformly distributed load of 2.1 kips/ft and a span of 24 ft, the total load on the beam can be calculated as (2.1 kips/ft) x (24 ft) = 50.4 kips.
Next, we need to calculate the maximum moment and shear values at the critical sections of the beam. For a simply supported beam under a uniformly distributed load, the maximum moment occurs at the center of the beam, and the maximum shear occurs at the supports.
Using standard beam formulas, we can determine the maximum moment (M) as (wL[tex]^2[/tex])/8, where w is the load per unit length and L is the span length. Substituting the values, we get M = (2.1 kips/ft) x (24 ft)[tex]^2[/tex] / 8 = 151.2 kip-ft.
The maximum shear (V) can be calculated as wL/2, which gives V = (2.1 kips/ft) x (24 ft) / 2 = 50.4 kips.
With the maximum moment and shear values, we can refer to the load tables for W shapes to find the lightest beam that can support these loads. The selection should consider the yield stress of the structural steel beams, which is given as 50 ksi.
By comparing the load capacity of different W shapes, we can identify the lightest shape that can safely support the given load. The specific W shape designation will depend on the load tables provided, and it should be chosen to ensure the beam's capacity is greater than or equal to the calculated maximum moment and shear values.
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Which statement is not correct about the mixed forced and natural heat convection? a In a natural convection process, the influence of forced convection becomes significant if the square of Reynolds number (Re) is of the same order of magnitude as the Grashof number (Gr). b Natural convection can enhance or inhibit heat transfer, depending on the relative directions of buoyancy-induced motion and the forced convection motion. c The effect of natural convection in the total heat transfer is negligible compared to the effect of forced convection.
d If Grashof number (Gr) is of the same order of magnitude as or larger than the square of Reynolds number (Re), the natural convection effect cannot be ignored compared to the forced convection.
Natural convection can enhance or inhibit heat transfer, depending on the relative directions of buoyancy-induced motion and the forced convection motion.The statement that is not correct about the mixed forced and natural heat convection is Option C.
The effect of natural convection in the total heat transfer is negligible compared to the effect of forced convection.
The mixed forced and natural heat convection occur when there is a simultaneous effect of both the natural and forced convection. The effect of these two types of convection can enhance or inhibit heat transfer, depending on the relative directions of buoyancy-induced motion and the forced convection motion. Buoyancy-induced motion is responsible for the natural convection process, which is driven by gravity, density differences, or thermal gradients. Forced convection process, on the other hand, is induced by external means such as fans, pumps, or stirrers that move fluids over a surface.Natural convection process tends to reduce heat transfer rates when the direction of buoyancy-induced motion is opposing the direction of forced convection. Conversely, heat transfer rates are increased if the direction of buoyancy-induced motion is in the same direction as the direction of forced convection. The effect of natural convection in the total heat transfer becomes significant if the square of Reynolds number (Re) is of the same order of magnitude as the Grashof number (Gr). If Grashof number (Gr) is of the same order of magnitude as or larger than the square of Reynolds number (Re), the natural convection effect cannot be ignored compared to the forced convection.
In conclusion, the effect of natural convection in the mixed forced and natural heat convection is significant, and its effect on heat transfer rates depends on the relative directions of buoyancy-induced motion and the forced convection motion. Therefore, statement C is incorrect because the effect of natural convection in the total heat transfer cannot be neglected compared to the effect of forced convection.
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If the number of turns in the coil is increased, the induced electromotive force in the coil will A. Increase B. Decrease C. Remains same D. None of the above
If the number of turns in the coil is increased, the induced electromotive force in the coil will A. Increase.
According to Faraday's law of electromagnetic induction, the magnitude of the induced electromotive force (EMF) in a coil is directly proportional to the rate of change of magnetic flux passing through the coil. The magnetic flux is influenced by factors such as the strength of the magnetic field and the number of turns in the coil.
When the number of turns in the coil is increased, more individual loops are present, resulting in a larger surface area for magnetic flux to pass through. As a result, a greater amount of magnetic flux is linked with the coil, leading to a higher rate of change of flux and an increased induced EMF.
Therefore, increasing the number of turns in the coil enhances the effectiveness of electromagnetic induction, resulting in a greater induced electromotive force.
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SAE30 oil flows through a 1.5 cm diameter at 5m/s. Determine the length of entrance region, le? 8. Gasoline flows through a 1.5 cm diameter at 1 m/s. Determine the length of entrance region, le?
To determine the length of the entrance region (le) for SAE30 oil and gasoline flowing through pipes, calculate the Reynolds number and use empirical correlations to estimate le based on flow conditions and pipe geometry.
To determine the length of the entrance region (le) for SAE30 oil and gasoline flowing through pipes, calculate the Reynolds number and use empirical correlations to estimate le based on flow conditions and pipe geometry.
For SAE30 oil:
- Calculate the Reynolds number using the formula Re = (ρvd) / μ, where ρ is the density of the oil, v is the velocity, d is the diameter, and μ is the dynamic viscosity of the oil.
- Use empirical correlations or charts to estimate the length of the entrance region (le) based on the Reynolds number and pipe geometry.
For gasoline:
- Follow the same process as for SAE30 oil, but use the properties specific to gasoline (density and dynamic viscosity) to calculate the Reynolds number and estimate the length of the entrance region (le).
The specific values and calculations can be obtained from relevant fluid property tables and empirical correlations for entrance region length.
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The base current is 34.23 mA and current gain, Boc is 100. The collector current in A is
In the given problem, we are given the value of base current, which is I_{B}=34.23 mAand current gain, which is B_{oc}=100. To find the collector current, we can use the relation, I_{C}=B_{oc}*I_{B} Putting the given values in the above relation,
we getI_{C}=100*34.23*10^{-3}I_{C}=3.423 A Therefore, the collector current is 3.423 A (Amperes). The current gain of a transistor is the ratio of the output current to the input current. It is a dimensionless quantity. The collector current of a transistor is controlled by the base current through the current gain of the transistor. When the base current flows, the transistor is switched on and it allows the current to flow through the collector-emitter junction.The current gain of a transistor is usually denoted by the symbol B. The value of B can be in the range of a few tens to several hundred. It is usually given by the manufacturer and is one of the key parameters of the transistor.
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Write a Matlab code to plot the continuous time domain signal for the following spectrum:
X (jω) = 2sin(ω)/ω
Here is a MATLAB code to plot the continuous-time domain signal for the given spectrum: X(jω) = 2sin(ω)/ω.
% Define the frequency range
w = -10*pi:0.01*pi:10*pi;
% Compute the spectrum X(jω)
X = 2*sin(w)./w;
% Plot the signal in the time domain
plot(w, X)
xlabel('Frequency (rad)')
ylabel('Amplitude')
title('Continuous-Time Domain Signal')
grid on
The MATLAB code provided above allows us to plot the continuous-time domain signal for the given spectrum X(jω) = 2sin(ω)/ω.
First, we define the frequency range 'w' over which we want to evaluate the spectrum. In this case, we use a range of -10π to 10π with a step size of 0.01π.
Next, we compute the values of the spectrum X(jω) using the element-wise division operator './'. We calculate 2*sin(w)./w to obtain the values of X for each frequency 'w'.
Finally, we plot the signal in the time domain using the 'plot' function. The 'xlabel', 'ylabel', and 'title' functions are used to label the axes and title of the plot. The 'grid on' command adds a grid to the plot for better visualization.
By running this MATLAB code, we can obtain a plot that represents the continuous-time domain signal corresponding to the given spectrum.
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i).Briefly explain the differences between impact hammer method and shaker method in modal testing. When it is appropriate to use one method instead of the other?
ii).You are tasked to measure the vibration level of a diesel engine. Propose four (4) suitable measures to ensure validity in your measurement results.
The impact hammer method and shaker method are two different approaches used in modal testing to determine the dynamic characteristics of a structure or system.
1. Impact Hammer Method:
In the impact hammer method, an instrumented hammer is used to deliver a mechanical impact to the structure at specific points. The response of the structure to the impact is measured using accelerometers. This method is typically used for small and medium-sized structures, and it provides localized excitation and measurement. It is suitable for measuring high-frequency modes and for structures with limited accessibility.
2. Shaker Method:
In the shaker method, a shaker or electrodynamic exciter is used to apply controlled vibrations to the structure over a range of frequencies. Accelerometers are used to measure the response of the structure at various points. This method is commonly used for larger structures and allows for excitation over a wide frequency range. It provides a more controlled and uniform excitation compared to the impact hammer method.
When to use each method:
- Impact Hammer Method: The impact hammer method is appropriate when there is limited access to the structure or when localized excitation and measurement are needed. It is suitable for small and medium-sized structures and high-frequency modes. It can be used in situations where it is challenging to mount a shaker or apply controlled vibrations to the entire structure.
- Shaker Method: The shaker method is suitable for larger structures and when a wide frequency range of excitation is required. It provides more controlled and uniform excitation compared to the impact hammer method. It is often used in modal testing of aerospace, automotive, and large structural components.
ii). To ensure validity in measuring the vibration level of a diesel engine, the following measures can be considered:
1. Calibration: Calibrate the measuring instruments, including accelerometers and data acquisition systems, to ensure accurate and reliable measurements. Regular calibration checks should be performed to maintain measurement accuracy.
2. Sensor Placement: Carefully select and position the accelerometers on the engine to capture representative vibration data. Consider the critical points and components that experience significant vibrations and ensure proper mounting and orientation of the sensors.
3. Signal Conditioning: Use appropriate signal conditioning techniques to filter and amplify the measured vibration signals. This helps to eliminate noise and enhance the accuracy of the measurements.
4. Data Analysis: Employ advanced data analysis techniques such as frequency analysis, power spectral density estimation, and statistical analysis to extract meaningful information from the vibration data. Validate the results by comparing them with known standards or reference measurements, if available.
By implementing these measures, one can enhance the validity of the measurement results and ensure accurate assessment of the vibration levels in a diesel engine.
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Answer the below questions 1- What do we mean by stall angle of attack, and what happens to the air streams when we reach this angle (you may use some sketches)? 2- Explain the lifting principle of aircraft with the help of the Coandă effect. 3- Why we don't consider the equal time principle a correct explanation for lift force in aircraft? 4- Write a small paragraph that describes the wind tunnel (Lift force) experiment. Plot the graph for the lift and drag coefficient versus the angle of attack.
The air streams over the wings are disturbed when the angle of attack is reached. The air in the lower part of the wing is relatively undisturbed, whereas the air in the upper part is more disturbed. As a result of the separation, the wing produces less lift, and the drag increases.
1. Stall angle of attack: Stall angle of attack refers to the angle of attack where the wing's lift coefficient starts to decrease rapidly. At this angle of attack, the airflow over the wing's upper surface separates from the wing's surface, resulting in a decrease in lift and an increase in drag.
2. Lifting Principle: According to the Coanda effect, a fluid, when flowing over the curved surface of an object, tends to follow the surface rather than continue flowing in a straight line. The curvature of the wing's upper surface causes the airflow to follow the surface.
3. Equal time principle: According to the equal time principle, air flowing over the top of a wing and air flowing over the bottom of a wing must meet at the back of the wing at the same time. This theory is incorrect because it does not account for the wing's curvature and the Coanda effect.
4. Wind Tunnel Experiment: In a wind tunnel experiment to measure lift and drag coefficients versus the angle of attack, a model of the wing is mounted in the wind tunnel and subjected to varying airspeeds at different angles of attack. By measuring the forces generated on the wing, the lift and drag coefficients can be determined.
The plot of the lift coefficient versus the angle of attack is shaped like an elongated S curve, while the plot of the drag coefficient versus the angle of attack is shaped like a U curve.
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Denign a fiot-order digital lowpass niter with a 3 dB cutoff rieciency of =0.25+by applying the bilinear transformation to the analog Bultenuorth iner H₀(s) = 1/1+s/Ω
1. What is the prewarp analog frequency frequency 2. Find the analog transfer function 3. Is bilinear transformation linear? 4. Is Ha(s) stable?
1. The pre-warping frequency is 0.6467 rad/s.
2. The analog transfer function is $H_a(s) = 1/[1 + (2.586z - 1.586)/0.6467]$.
3. Bilinear transformation is linear.
4. The transfer function $H_a(s)$ is stable since the pole is located inside the unit circle.
1. Prewarp analog frequency:
The prewarp frequency of a digital filter is that frequency that it corresponds to when the filter is represented in the analog domain.
In the bilinear transformation, pre-warping is used to minimize the frequency distortion caused by the mapping from the s-plane to the z-plane.
The pre-warping frequency is defined as follows:
[tex]$$\Omega_a=\frac{2}{T}\tan\Big(\frac{\Omega_d T}{2}\Big)$$[/tex]
Given that the digital lowpass filter has a 3dB cutoff frequency of [tex]$\Omega_c=0.25$[/tex], the prewarp analog frequency is:
[tex]$$\Omega_a=2\tan\Big(\frac{0.25\cdot\pi}{2}\Big)[/tex]
[tex]=0.6467\text{ rad/s}$$[/tex]
2. Analog Transfer Function:
The analog Butterworth filter is represented by the transfer function
[tex]$H_0(s) = 1/(1+s/\Omega)$[/tex],
where [tex]$\Omega$[/tex] is the filter's corner frequency.
To begin with, the transfer function has to be pre-warp by substituting [tex]$s$[/tex] with [tex]$(2/T) \cdot (z-1)/(z+1)$[/tex], where [tex]$T$[/tex] is the sampling period, which is given by:
[tex]$$T=\frac{1}{2\cdot\Omega_a}$$[/tex]
[tex]$$=\frac{1}{2\cdot 0.6467}$$[/tex]
[tex]$$=0.7739\text{ s}$$[/tex]
Substituting,
[tex]$$s=\frac{2}{0.7739}\cdot\frac{z-1}{z+1}$$[/tex]
[tex]$$=2.586z-1.586$$[/tex]
Therefore, the analog transfer function is:
[tex]$$H_a(s)=H_0(2.586z-1.586)$$[/tex]
[tex]$$=\frac{1}{1+(2.586z-1.586)/\Omega}$$[/tex]
Substituting [tex]$\Omega$[/tex] with [tex]$\Omega_a$[/tex],
we get:
[tex]$$H_a(s)=\frac{1}{1+(2.586z-1.586)/0.6467}$$[/tex]
3. Linearity of Bilinear Transformation:
The bilinear transformation is a linear method of mapping an analog filter design to a digital implementation.
As a result, the transformation preserves the linearity of the analog filter.
4. Stability of Ha(s):
To determine the stability of [tex]$H_a(s)$[/tex], we need to determine the poles of the transfer function.
The poles are the values of [tex]$s$[/tex]that cause [tex]$H_a(s)$[/tex]to be infinite.
Therefore, we set the denominator of the transfer function to zero and solve for [tex]$s$[/tex].
[tex]$$1+\frac{2.586z-1.586}{0.6467}=0$$[/tex]
[tex]$$z=-0.5859$$[/tex]
Therefore, the pole is located at [tex]$z=-0.5859$[/tex], which is inside the unit circle.
As a result, [tex]$H_a(s)$[/tex]is stable.
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It is true that the continuity equation below is valid for viscous and inviscid flows, for Newtonian and Non-Newtonian fluids, compressible and incompressible? If yes, are there(are) limitation(s) for the use of this equation? Detail to the maximum, based on the book Muson.δt/δrho +∇⋅(rhoV)=0
The continuity equation given by Muson,
δt/δrho +∇⋅(rhoV) = 0
is true for viscous and inviscid flows, for Newtonian and Non-Newtonian fluids, compressible and incompressible. This is because the continuity equation is a fundamental equation of fluid dynamics that can be applied to different types of fluids and flow situations.
The continuity equation is a statement of the principle of conservation of mass, which means that mass can neither be created nor destroyed but can only change form. In fluid dynamics, the continuity equation expresses the fact that the mass flow rate through any given volume of fluid must remain constant over time. The equation states that the rate of change of mass density (ρ) with time (δt) plus the divergence of the mass flux density (ρV) must be zero.There are limitations to the use of the continuity equation, however. One limitation is that it assumes that the fluid is incompressible, which means that its density does not change with pressure. This is a reasonable assumption for many fluids, but it is not valid for all fluids.
For example, gases can be compressed and their density can change significantly with pressure.Another limitation of the continuity equation is that it assumes that the fluid is homogeneous and isotropic, which means that its properties are the same in all directions. This is not always the case, especially in complex flow situations such as turbulent flow. In these situations, the continuity equation may need to be modified or replaced with more complex equations to account for the effects of turbulence.
Furthermore, it is important to note that the continuity equation is a local equation, which means that it applies only to a small volume of fluid. To apply it to a larger volume of fluid, it must be integrated over the entire volume. Finally, it should be noted that the continuity equation is a linear equation, which means that it applies only to small changes in fluid density and velocity. For larger changes, nonlinear effects may need to be taken into account.
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A specimen of aluminum having a rectangular cross section 9.8 mm×12.8 mm(0.3858in×0.5039in.) is pulled in tension with 35300 N(7936Ibf) force, producing only elastic deformation. The elastic modulus for aluminum is 69GPa (or 10×10^6psi ). Calculate the resulting strain.
The resulting strain experienced by the aluminum specimen under a tensile force of 35300 N is approximately 0.00051, or 0.051%.
This value is obtained using the stress-strain relationship, which is derived from Hooke's law.
To explain further, the stress on the aluminum specimen is calculated first. Stress is the force applied divided by the area over which it is distributed. In this case, the cross-sectional area is 9.8 mm × 12.8 mm = 0.12544 cm². The stress thus equals the force (35300 N) divided by the area (0.12544 cm²), which gives 281300000 Pascal or 281.3 MPa. Using the formula for strain (which is stress divided by the modulus of elasticity), the strain equals 281.3 MPa divided by 69000 MPa (which is 69 GPa), resulting in a strain of approximately 0.00051, or 0.051%.
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