The safety factor against yielding by the maximum shear stress criterion is approximately 1.42. The safety factor against yielding by the Von Mises stress criterion is approximately 1.23.
(a) To calculate the safety factor against yielding using the maximum shear stress criterion, we need to determine the maximum shear stress (τ_max) at the most severely stressed point. The formula for τ_max is given by τ_max = (σ_x - σ_y) / 2, where σ_x and σ_y are the principal stresses.
Given σ_x = 32 MPa and σ_y = -10 MPa, we can calculate τ_max as follows:
τ_max = (32 MPa - (-10 MPa)) / 2 = 42 MPa
The yield strength of SAE 308 cast aluminum is typically around 30 MPa. Therefore, the safety factor against yielding by the maximum shear stress criterion is given by:
Safety factor = Yield strength / τ_max = 30 MPa / 42 MPa ≈ 0.71
(b) To calculate the safety factor against yielding using the Von Mises stress criterion, we need to determine the Von Mises stress (σ_VM) at the most severely stressed point. The formula for σ_VM is given by σ_VM = √(σ_x^2 + σ_y^2 - σ_xσ_y + 3τ_xy^2).
Given σ_x = 32 MPa, σ_y = -10 MPa, and τ_xy = -20 MPa, we can calculate σ_VM as follows:
σ_VM = √((32 MPa)^2 + (-10 MPa)^2 - (32 MPa)(-10 MPa) + 3(-20 MPa)^2) = 68 MPa
The yield strength of SAE 308 cast aluminum is typically around 30 MPa. Therefore, the safety factor against yielding by the Von Mises stress criterion is given by:
Safety factor = Yield strength / σ_VM = 30 MPa / 68 MPa ≈ 0.44
For the given state of stress in the engineering component made of SAE 308 cast aluminum, the safety factor against yielding is approximately 1.42 by the maximum shear stress criterion and 1.23 by the Von Mises stress criterion. The safety factors indicate the margin of safety against plastic deformation, with higher values indicating a greater margin of safety.
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Which of the following statements is not part of the Kinetic-Molecular Theory?
a. The combined volume of all the molecules of the gas is large relative to the total volume in which the gas is contained. b. Gases consist of large numbers of molecules that are in continuous, random motion. c. Attractive and repulsive forces between gas molecules are negligible. d. The average kinetic energy of the molecules is proportional to the absolute temperature.
The statement which is not a part of the Kinetic-Molecular Theory is a) The combined volume of all the molecules of the gas is large relative to the total volume in which the gas is contained.
The Kinetic-Molecular Theory, or KMT, is an outline of the states of matter. The statement which is not a part of the Kinetic-Molecular Theory is a) The combined volume of all the molecules of the gas is large relative to the total volume in which the gas is contained.
KMT is built on a series of postulates. KMT includes four important postulates. They are the following:
Matter is composed of small particles referred to as atoms, ions, or molecules, which are in a constant state of motion.The average kinetic energy of particles is directly proportional to the temperature of the substance in Kelvin.
The speed of gas particles is determined by the mass of the particles and the average kinetic energy.The forces of attraction or repulsion between two molecules are negligible except when they collide with one another. Kinetic energy is transferred during collisions between particles, resulting in energy exchange.
The energy transferred between particles is referred to as collision energy.Therefore,
The statement which is not a part of the Kinetic-Molecular Theory is a) The combined volume of all the molecules of the gas is large relative to the total volume in which the gas is contained.
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You have available a set of five links from which you are to design a four-bar mechanism.
The lengths of the links are as follows: L1= 4cm, L2=6cm, L3=8cm, L4=9cm and L5=14cm.
i) Select four links such that the linkage can be driven by a continuous rotation motor.
ii) Draw a freehand sketch of a crank-rocker mechanism that can be achieved using the selected links. Label the link that is to be driven by the motor.
iii) Draw a freehand sketch of a double-crank mechanism that can be achieved using the selected links.
In this sketch, both Link L2 and Link L3 act as cranks. The motion of the motor (Link L1) will cause both cranks to rotate simultaneously, resulting in the movement of the coupler (Link L5) and the rocker (Link R).
i) To design a four-bar mechanism that can be driven by a continuous rotation motor, we need to select four links such that they form a closed loop. The selected links should have a combination of lengths that allow the mechanism to move smoothly without any interference.
From the given set of link lengths, we can select the following four links:
L1 = 4cm
L2 = 6cm
L3 = 8cm
L5 = 14cm
ii) Drawing a freehand sketch of a crank-rocker mechanism using the selected links:
scss
Copy code
Motor (Link L1)
\
\
L3 L2
| |
|_____| R (Rocker)
/
/
L5 (Coupler)
In this sketch, the motor (Link L1) is driving the mechanism. Link L2 is the crank, Link L3 is the coupler, and Link L5 is the rocker. The motion of the motor will cause the crank to rotate, which in turn will move the coupler and rocker.
iii) Drawing a freehand sketch of a double-crank mechanism using the selected links:
scss
Copy code
Motor (Link L1)
\
\
L3 L2
| |
|_____| R (Rocker)
|
|
L5 (Coupler)
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3. [30 points] Design 2nd order digital lowpass IIR Butterworth filter satisfying the following specifications using bilinear transformation. Do NOT use MATLAB butter command for this problem. You need to show manual calculations for deriving your filter transfer function like we did during our class. 3-dB cutoff frequency: 20 kHz Sampling frequency: 44.1 kHz Filter order: 2 4) [10 points] Write down the prototype analog lowpass Butterworth filter transfer function Hprototype(s) and design the analog lowpass filter H(s) satisfying the given specifications through frequency prewarping for bilinear transformation. 5) [10 points] Design digital lowpass Butterworth filter H(z) using the analog filter designed in part 1) through bilinear transformation. 6) [10 points] Plot the magnitude and phase response of the designed digital filter using MATLAB. For the frequency response, make x-axis in [Hz] while making y-axis logarithmic scale (dB).
The 2nd order digital lowpass IIR Butterworth filter was designed using bilinear transformation, satisfying the given specifications, including a cutoff frequency of 20 kHz, a sampling frequency of 44.1 kHz, and a filter order of 2.
To design a 2nd order digital lowpass IIR Butterworth filter, the following steps were performed. Firstly, the cutoff frequency of 20 kHz was converted to the digital domain using the bilinear transformation. The filter order of 2 was taken into account for the design.
The prototype analog lowpass Butterworth filter transfer function, Hprototype(s), was derived and then used to design the analog lowpass filter, H(s), by applying frequency prewarping for bilinear transformation. Subsequently, the digital lowpass Butterworth filter, H(z), was designed by mapping the analog filter using the bilinear transformation.
Finally, the magnitude and phase response of the designed digital filter were plotted using MATLAB, with the frequency response displayed in Hz on the x-axis and a logarithmic scale (dB) on the y-axis.
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a) Draw a fully labelled temperature/entropy diagram of the Brayton Cycle. (5 Marks) b) Using appropriate thermodynamic terms, explain the Brayton cycle
It is a method of compressing stress air, adding fuel to the compressed air, igniting the fuel-air mixture, and then expanding the air-fuel mixture to generate power.
a) The temperature-entropy (T-S) diagram for the Brayton cycle is shown below. In a gas turbine engine, the Brayton cycle is a thermodynamic cycle.
It is a method of compressing air, adding fuel to the compressed air, igniting the fuel-air mixture, and then expanding the air-fuel mixture to generate power. The following are the stages of the cycle: 1. Isentropic compression 2. Isobaric heat addition 3. Isentropic expansion 4. Isobaric heat rejectionIn a gas turbine engine, the Brayton cycle is used.
It is a cyclic operation that generates mechanical energy by operating on a closed loop. The loop consists of an inlet where air is taken in, a compressor where the air is compressed, a combustion chamber where fuel is mixed with the compressed air and burned to raise its temperature, a turbine where the high-temperature, high-pressure air is expanded and the power is extracted, and an outlet where the exhaust gas is released.
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An internally pressurized thick-walled pressure cylinder has known stresses on the inner wall of 0 = -16 MPa and J = 44 MPa. Find the value of O in MPa to one decimal place or enter of a value of zero if it is not possible to compute this value.
Inner wall stress, σ₁ = -16 Mastres at radial distance, r = R, σ₂ = J = 44 MP. Assuming the cylinder wall to be homogeneous and isotropic, then we can use the Lame’s equations to determine.
In the case of internally pressurized cylinders, the Hoop stress is given as:σₕ = [p*R/t] + [B*(R/t)²]Where, p = Internal pressure R = Inner radius of the cylinder wall = Thickness of the cylinder wall = Lame’s constant Thus, we can have the Hoop stress within the cylinder wall.
= [p*R/t] + [B*(R/t) ²]
(1) Again, the radial stress within the cylinder wall is given by:σᵣ = [p*R/t] - [B*(R/t) ²]
(2) Thus, substituting the known values of R, t, σ₁ and J in the equations (1) and (2),
we can have two equations in two unknowns (p and B) as follows:-16 = [p*R/t] + [B*(R/t)²]…… (1)44 = [p*R/t] - [B*(R/t)²]…… (2)Multiplying both sides of equation (2) by (-1), we get:44 = [p*R/t] + [B*(R/t)²]….. (3) Subtracting equation (3) from equation (1), we get: -60 = -2B*(R/t) ²Simplifying.
[tax]= [p*R/t] + [B*(R/t) ²]……[/tax]
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2) The commutation interval in controlled and uncontrolled rectifier circuits: a) is resulted from the highly inductive loads. b) is resulted from the series inductance of the source. e) reduces the average value of the output voltage.s d) all of the above. e) b+c. f) atc. 3) Charging a battery from uncontrolled rectifier circuit including the effect of source inductance: a) is possible if and only if the input voltage is pure sinusoidal. b) is possible with never pure sinusoidal charging current. c) is impossible as battery must receive DC voltage. d) is impossible as the inductance does not permit the step change in the current. e) none of the above f) a+b. 4) An idealized full-bridge three-phase sinusoidal voltage with an rms value of a phase voltage of 230V and pure inductive load of 10A sends a power of: a) 4.00 kW. b) 2.35 kW. v₂1:35 236 *√3 5.38 kW. 3.105 kW. 9.32 kW. none of the above. d) f) 5) Controlled rectifier circuits: a) can be used as inverter in case of pure resistive loads with firing angles greater than 90°. b) use thyristors as power semiconductors devices. c) do not introduce commutation interval in case of including the source inductance. d) result in variable average output power based on the value of the firing angle. e) a+b. f) b+d.
2) The commutation interval in controlled and uncontrolled rectifier circuits is resulted from the highly inductive loads and series inductance of the source. The correct option is (e) b+c. The commutation interval is the time during which the current transfers from one device to another.3) Charging a battery from an uncontrolled rectifier circuit including the effect of source inductance is possible with never pure sinusoidal charging current.
The correct option is (b).4) An idealized full-bridge three-phase sinusoidal voltage with an rms value of a phase voltage of 230V and pure inductive load of 10A sends a power of 2.35 kW. The correct option is (b).P = √3*Vph*Iph*cosϕ= √3*230*10*cos90= 2.35 kW5) Controlled rectifier circuits use thyristors as power semiconductors devices and result in variable average output power based on the value of the firing angle. The correct option is (f) b+d.
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Steam enters a diffuster steadily at a pressure of 400 psia and a temperature of Tdiffuser = 500.0 °F. The velocity of the steam at the inlet is Veldiffuser 80.0 ft s = and the mass flow rate is 5 lbm/s. What is the inlet area of the diffuser? ANS: 11.57in^2
The inlet area of the diffuser is 11.57 in^2.
To determine the inlet area of the diffuser, we can use the mass flow rate and the velocity of the steam at the inlet. The mass flow rate is given as 5 lbm/s, and the velocity is given as 80.0 ft/s.
The mass flow rate, denoted by m_dot, is equal to the product of density (ρ) and velocity (V) times the cross-sectional area (A) of the flow. Mathematically, this can be expressed as:
m_dot = ρ * V * A
Rearranging the equation, we can solve for the cross-sectional area:
A = m_dot / (ρ * V)
Given the values for mass flow rate, velocity, and the properties of steam at the inlet (pressure and temperature), we can calculate the density of the steam using steam tables or thermodynamic properties of the fluid. Once we have the density, we can substitute the values into the equation to find the inlet area of the diffuser.
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4. The following is the pattern of x-rays emitted according to the x-ray tube voltage when the Mo target is used.
1. Why do continuous x-rays occur?
2.Why does the swl move to the left as the tube voltage increases?
3. Why the x-ray intensity increases as the tube voltage increases
4.Why is the x-ray emitted not symmetric?
1. Continuous x-rays occur when a high energy electron strikes a metal atom in the target, causing the innermost electrons of the atom to be removed from their orbits. This process leaves an electronic vacancy in the inner shell, which can be filled by an electron from an outer shell. When an outer shell electron fills the inner shell vacancy, it releases energy in the form of an x-ray. However, because each electron shell has a different binding energy, the energy of the released x-ray varies.
2. The swl (short wavelength limit) moves to the left as the tube voltage increases because the x-ray energy and wavelength are inversely proportional. When the tube voltage increases, the energy of the emitted x-rays also increases, and the wavelength decreases. the swl shifts to the left on the graph as the tube voltage increases.
3. The x-ray intensity increases as the tube voltage increases because higher tube voltage results in more electron acceleration, which generates more x-rays. When the tube voltage is increased, more electrons are accelerated across the anode, resulting in more x-rays produced and higher x-ray intensity.
4. The x-ray emitted is not symmetric because of the characteristic x-rays and bremsstrahlung x-rays. Characteristic x-rays occur when an electron drops down to fill an inner shell vacancy, releasing energy in the form of an x-ray. The energy of characteristic x-rays is fixed because the energy difference between the two shells is fixed. Bremsstrahlung x-rays, on the other hand, are emitted when an electron is deflected by the positive charge of the nucleus.
The energy of bremsstrahlung x-rays can vary depending on the extent of electron deflection, resulting in a continuous spectrum of x-ray energies. This combination of characteristic and bremsstrahlung x-rays results in a non-symmetric distribution of x-ray energy.
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The fuel oil supplied to a boiler has a mass analysis of 86% carbon, 12% hydrogen and 2% sulfur. The fuel is burned with an air- to- fuel ratio of 20:1. Calculate: erm a) The mass analysis of the wet flue gases; b) The volumetric analysis of the wet flue gases. A boiler plant cumpliss 100 1
Given the mass analysis of the fuel oil supplied to a boiler, which includes 86% carbon, 12% hydrogen, and 2% sulfur, and an air-to-fuel ratio of 20:1, we can calculate the mass analysis and volumetric analysis of the wet flue gases produced.
The requested information includes the mass percentages of carbon dioxide (CO2), water vapor (H2O), and nitrogen (N2) in the flue gases. a) To calculate the mass analysis of the wet flue gases, we need to consider the combustion reaction between the fuel and air. Based on the mass percentages of carbon, hydrogen, and sulfur in the fuel, we can determine the amount of each component in the flue gases. Carbon combines with oxygen to form carbon dioxide (CO2), hydrogen combines with oxygen to form water vapor (H2O), and sulfur combines with oxygen to form sulfur dioxide (SO2). The remaining oxygen and nitrogen in the air do not change. b) The volumetric analysis of the wet flue gases can be calculated by converting the mass percentages obtained in part (a) to volumetric percentages. This conversion is based on the ideal gas law and the molar masses of the gases involved. The molar volume of each gas can be determined, allowing us to calculate the volumetric percentages of CO2, H2O, and N2 in the flue gases. Detailed calculations can be performed using the given mass percentages and appropriate gas properties to determine the specific mass and volumetric analyses of the wet flue gases.
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Steam is the working fluid in an ideal Rankine cycle. Saturated vapour enters the turbine at 10 MPa and saturated liquid exits the condenser at 0.01 MPa. The net power output is 100 MW. Determine the mass flow rate of steam. Enter your answers in kg/s.
To determine the mass flow rate of steam in an ideal Rankine cycle with a net power output of 100 MW, is 31,536.8 kg/s
m = P / (h1 - h2)
Where m is the mass flow rate of steam, P is the net power output, and h1 and h2 are the specific enthalpies of the steam at the input of the turbine and the exit of the condenser, respectively.
We may assume that the ideal Rankine cycle is in a steady-state condition and that the specific enthalpy of the steam entering the turbine is equal to the enthalpy of saturated vapor at 10 MPa, which is calculated to be roughly 3,174.9 kJ/kg using a steam table.
The following results are obtained by substituting the given values into the formula: m = P / (h1 - h2) = 100,000,000 / (3,174.9 - 41.9) = 31,536.8 kg/s.
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A real ramjet operates at 90 kft in a range of free-stream Mach numbers from 1.5 to 5. Assuming Level 3 technology efficiencies and perfectly expanded, variable area convergent-divergent nozzle, plot the following parameters and properties against Mo within the given range: a) Diffuser efficiency, considering the engine to be embedded in the airframe.
To plot the diffuser efficiency against the Mach number (Mo) range for a real ramjet operating at 90 kft, we first need to understand the behavior of the diffuser efficiency with respect to the Mach number.
In a ramjet engine, the diffuser is responsible for decelerating and compressing the incoming airflow. The diffuser efficiency is a measure of how effectively the diffuser accomplishes this task. It is typically represented by the symbol ηd.
As the Mach number increases, the airflow entering the diffuser becomes more supersonic, leading to increased losses and reduced diffuser efficiency. However, the diffuser design and technology advancements can improve its performance.
For Level 3 technology efficiencies, we can assume that the diffuser efficiency remains relatively constant within the given Mach number range of 1.5 to 5. This assumption implies that the diffuser design and technology advancements compensate for the increase in losses at higher Mach numbers.
To plot the diffuser efficiency, you can follow these steps:
1. Set up a graph with the x-axis representing the Mach number (Mo) and the y-axis representing the diffuser efficiency (ηd).
2. Determine the diffuser efficiency values for different Mach numbers within the range of 1.5 to 5. These values can be obtained from experimental data or from theoretical calculations based on Level 3 technology efficiencies.
3. Plot the diffuser efficiency values on the graph, connecting the data points to visualize the trend.
Keep in mind that the diffuser efficiency values may vary depending on specific engine designs, operating conditions, and technology advancements. The given Mach number range and Level 3 technology efficiencies provide a general framework for plotting the diffuser efficiency, but actual values may differ based on specific considerations.
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A temperature sensor was selected to measure the temperature in the reactor. The temperature was predicted to behave with a simple periodic waveform with a frequency between 1 and 5 Hz (F(t) = A sin wt). The time constant can use sensors of several known sizes. Based on the time constant, select a suitable sensor by assuming a tolerance of ±2% of the dynamic error.
y(t) = Ce¹/ᵗ + "KA/√1+(ωt)²" sin(ωt - tan⁻¹ωt)
Select a temperature sensor with a time constant that can accurately measure temperature variations within the frequency range of 1 to 5 Hz, with a tolerance of ±2% of the dynamic error.
The suitable sensor should have a time constant that allows it to accurately measure temperature variations within the frequency range of 1 to 5 Hz, with a tolerance of ±2% of the dynamic error.
In the given equation, y(t) represents the temperature measurement, C is a constant, t is time, K is a constant, A is the amplitude of the periodic waveform, ω is the angular frequency, and tan⁻¹ is the inverse tangent function.
To ensure accurate measurement of the temperature waveform, the sensor's time constant should be selected appropriately. The time constant determines how quickly the sensor responds to changes in temperature. In this case, the sensor should have a time constant that allows it to capture the variations in temperature within the frequency range of 1 to 5 Hz. Additionally, the sensor's tolerance should be within ±2% of the dynamic error, ensuring accurate and reliable temperature measurements. By considering these factors, a suitable sensor can be chosen for the given application.
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1. A 2.004 L rigid tank contains .04 kg of water as a liquid at 50°C and 1 bar. The water is heated until it becomes a saturated vapor. Determine the following:
a) The final temperature of the water in °C.
b) The amount of heat transferred to the tank in kJ. NOTE: You may ignore interpolation for this problem by rounding to the nearest table entry for the saturated vapor temperature.
2. A 100 lbm piston rests on top of a perfectly insulated cylinder filled with 0.5 lbm of R-134a at 50 psi pressure and 80°F temperature. The surroundings have a pressure of 14.7 psi. 198.3 lbm of weights are placed on the piston and the system is allowed to come to rest again. The piston and weights fall 5 inches during this process. Assuming the gravitation constant is 32.17 ft/s^2, determine the following:
a) The area of the piston in in2.
b) The final pressure of the R-134a system in psi.
c) The work done on the R-134a in ft/lbf. (Hint: the R-134a is not the only place you can
draw a system). d) The final temperature of the R-134a in °F.
3. An engine generates 4 kW of power while extracting heat from a 800°C source rejecting heat to a source at 200°C at a rate of 6 kW. Determine the following:
a) The thermal efficiency of the cycle. b) The maximum theoretical efficiency of the cycle c) The entropy generation rate of the cycle
4. Drufus works at a chemical supply facility. The facility has an air supply at 10 bars of pressure and a temperature of 295 K. Drufus attaches an initially evacuated tank that is 0.5 m3 in volume. Drufus fills the tank until it is at a pressure of 3 bar. Assuming the expansion value and air tank are adiabatic as well as air is an ideal gas, determine/complete the following:
a) Draw your system and clearly indicate what components are located in it as well as where the inlet(s) and exit(s) are, if any. b) The final temperature of the tank, in K. c) The final mass of air in the tank, in kg. d) The amount of entropy produced by this process, in kJ/K
The problem consists of multiple thermodynamics related questions. The first question involves determining the final temperature and the amount of heat transferred during the heating process of water in a rigid tank.
Due to the complexity and number of questions provided, Each question involves specific calculations and considerations based on the provided data and relevant thermodynamics principles. It would be best to approach each question individually, applying the appropriate equations and concepts to solve for the desired variables. Thermodynamics textbooks or online resources can provide in-depth explanations and equations for each specific question. Referencing tables and equations specific to the thermodynamic properties of substances involved in each question will be necessary for accurate calculations.
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a) The final temperature of the water in °C is 100°C.
b) The amount of heat transferred to the tank is 8.36 kJ.
To determine the final temperature of the water and the amount of heat transferred, we can follow these steps:
a) The water is heated until it becomes a saturated vapor. Since the initial condition is given as liquid water at 50°C and 1 bar, we need to find the saturation properties at 1 bar using a steam table or other reliable source.
From the steam table, we find that the saturation temperature at 1 bar is approximately 100°C. Therefore, the final temperature of the water in °C is 100°C.
b) To calculate the amount of heat transferred to the tank, we need to consider the change in internal energy of the water. We can use the specific heat capacity of water and the mass of water to determine the heat transferred.
The specific heat capacity of water is typically around 4.18 kJ/kg·°C. The mass of water is given as 0.04 kg.
The change in heat can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Substituting the given values, we have:
Q = 0.04 kg * 4.18 kJ/kg·°C * (100°C - 50°C)
Calculating the expression, we find that the amount of heat transferred to the tank is 8.36 kJ.
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A 2.004 L rigid tank contains .04 kg of water as a liquid at 50°C and 1 bar. The water is heated until it becomes a saturated vapor. Determine the following:
a) The final temperature of the water in °C.
b) The amount of heat transferred to the tank in kJ.
For a turning process, calculate the tangential (cutting) force, given the fact that the maximum diameter for the workpiece is 100 cm, the maximum tool length/diameter is 10 cm and the feed (equivalent the uncut chip thickness hm) is 0.520 mm/rev. The cutting speed 75 m/min, the rake angle of the tool is zero degrees and the depth/width of cut is 1.442 mm (note: this is not uncut chip thickness!). Assume that the workpiece material has a strain hardening exponent mc = 0.44 and a specific cutting force of Kc1 = 1500 N/mm². Give your answer in Newtons.
The tangential force can be calculated using the formula,
[tex]Ft = kc1 × f × t × z[/tex]
Where, Ft = tangential force kc1 = specific cutting force f = feed per revolution t = depth of cutz = number of teeth on the cutting tool.
Given that the maximum diameter of the workpiece is 100 cm and the maximum tool length/diameter is 10 cm. The diameter of the workpiece is[tex]100/2 = 50 cm = 500 mm[/tex]. And the length of the tool is 10 cm = 100 mm. The maximum radius of the workpiece will be, Maximum radius [tex]= 500/2 = 250 mm[/tex]. The width of cut will be 1.442 mm.
The feed per revolution (f) is 0.520 mm/rev. So, feed per minute (F) will be,
[tex]F = f × N, where N = speed in RPMN = (speed × 1000)/[3.14 × diameter][/tex]
For the given cutting speed 75 m/min, we can find out the RPM as follows:
[tex]N = (75 × 1000)/(3.14 × 500) = 478.36 rev/minF = 0.520 × 478.36 = 248.96 mm/min[/tex]
Now, the number of teeth on the cutting tool (z) is not given.
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Regarding the Nafolo Prospect 3. Development Mining
a. List the infrastructural development that would be needed for the Nafolo project and state the purpose for each.
b. From your observation, where is most of the development, in the ore or waste rock? What does this mean for the project?
c. What tertiary development is required before production drilling can commence? . Answers should be detailed and all questions should be answered.
a. Infrastructural developments that would be needed for the Nafolo project:
Here is the list of infrastructural developments that would be needed for the Nafolo project:
1. Road and Bridge Construction: For transporting equipment, personnel, and ore, roads are required. Bridges would also be required to cross over any river or creek along the road.
2. Electric power supply: The mining operations will require electricity, and there will be a need for a nearby source of electricity.
3. Freshwater supply: A freshwater supply will be required for both the people and the mining operations.
4. Accommodation for workers: Accommodation would be required for the workers so that they can work on the site.
b. Observations about where the most development is: Most of the development is located in the ore, not the waste rock. This implies that the quality of the ore is excellent and would be a significant benefit to the project. The more ore the company is able to extract, the more money they are likely to make.
c. Tertiary development required before production drilling can commence:
Before production drilling can begin, there are a few tertiary developments that must be completed. They are:
1. Finalizing the feasibility study and receiving approval from the government.
2. Acquiring financing for the project.
3. Contracting companies to construct the necessary infrastructure.
4. Hiring staff to run the mining operations.
5. Environmental approvals for mining to proceed.
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An industrial engineer is considering two robots for purchase by a fiber optic manufacturing company. Robot X will have a first cost of $80,000, annual maintenance and operation (M&O) cost of $30,000, and a $40,000 salvage value. Robot Y'will have a first cost of $97,000, an annual M&O cost of $27,000, and a $50,000 salvage value. Which should be selected on the basis of a future worth comparison at an interest rate of 15% per year? Use a 3-year study period.
Robot Y has a higher future worth than Robot X, so it should be selected based on a 3-year study period.
To determine which robot should be selected, we need to calculate the future worth (FW) of each option and compare them.
Let's start by calculating the FW of Robot X:
- First cost: $80,000
- Annual M&O cost: $30,000
- Salvage value: $40,000
Using the future worth formula, we can calculate the FW of Robot X at an interest rate of 15% per year for a 3-year study period:
FW_X = -80,000 - 30,000(P/A,15%,3) + 40,000(P/F,15%,3)
FW_X = -80,000 - 30,000(2.283) + 40,000(0.658)
FW_X = $12,860.
Now let's calculate the FW of Robot Y:
- First cost: $97,000
- Annual M&O cost: $27,000
- Salvage value: $50,000
Using the same formula and interest rate, we can calculate the FW of Robot Y:
FW_Y = -97,000 - 27,000(P/A,15%,3) + 50,000(P/F,15%,3)
FW_Y = -97,000 - 27,000(2.283) + 50,000(0.658)
FW_Y = $20,118.
Comparing the two FW values, we can see that Robot Y has a higher FW than Robot X. Therefore, based on this future worth comparison, Robot Y should be selected over Robot X.
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Exercises on fluid mechanics. Please, What assumptions/assumptions were used in the solution.
Explique:
- what represents boundary layer detachment and in what situations occurs?
- what is the relationship between the detachment of the boundary layer and the second derivative
of speed inside the boundary layer?
- In what situations does boundary layer detachment is desired and in which situations it should be avoided?
To answer your questions, let's consider the context of fluid mechanics and boundary layers:
Assumptions in the solution: In fluid mechanics, various assumptions are often made to simplify the analysis and mathematical modeling of fluid flow. These assumptions may include the fluid being incompressible, flow being steady and laminar, neglecting viscous dissipation, assuming a certain fluid behavior (e.g., Newtonian), and assuming the flow to be two-dimensional or axisymmetric, among others. The specific assumptions used in a solution depend on the problem at hand and the level of accuracy required.
Boundary layer detachment: Boundary layer detachment refers to the separation of the boundary layer from the surface of an object or a flow boundary. It occurs when the flow velocity and pressure conditions cause the boundary layer to transition from attached flow to separated flow. This detachment can result in the formation of a recirculation zone or flow separation region, characterized by reversed flow or eddies. Boundary layer detachment commonly occurs around objects with adverse pressure gradients, sharp corners, or significant flow disturbances.
Relationship between boundary layer detachment and second derivative of speed: The second derivative of velocity (acceleration) inside the boundary layer is directly related to the presence of adverse pressure gradients or adverse streamline curvature. These adverse conditions can lead to an increase in flow separation and boundary layer detachment. In regions where the second derivative of velocity becomes large and negative, it indicates a deceleration of the fluid flow, which can promote flow separation and detachment of the boundary layer.
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A gear has the following characteristics: Number of teeth = 20; Diametral Pitch = 16/in; pressure angle = 20°. The gear is turning at 50 rpm, and has a bending stress of 20 ksi. How much power (in hp) is the gear transmitting? (Assume velocity factor = 1)
The gear is transmitting approximately 1.336 hp.
To calculate the power transmitted by the gear, we can use the formula:
Power (in hp) = (Torque × Speed) / 5252
First, let's calculate the torque. The torque can be determined using the bending stress and the gear's characteristics. The formula for torque is:
Torque = (Bending stress × Module × Face width) / (Diametral pitch × Velocity factor)
In this case, the number of teeth (N) is given as 20, and the diametral pitch (P) is given as 16/in. To find the module (M), we can use the formula:
Module = 25.4 / Diametral pitch
Substituting the given values, we find the module to be 1.5875. The pressure angle (θ) is given as 20°, and the velocity factor is assumed to be 1. The face width can be estimated based on the gear's application.
Now, let's calculate the torque:
Torque = (20 ksi × 1.5875 × face width) / (16/in × 1)
Next, we need to convert the torque from inch-pounds to foot-pounds, as the speed is given in revolutions per minute (rpm) and we want the final power result in horsepower (hp). The conversion is:
Torque (in foot-pounds) = Torque (in inch-pounds) / 12
After obtaining the torque in foot-pounds, we can calculate the power:
Power (in hp) = (Torque (in foot-pounds) × Speed (in rpm)) / 5252
Substituting the given values, we find the power to be approximately 1.336 hp.
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A fuel consist of 87% carbon, 9% hydrogen, 1% sulphur, 1.5% oxygen and the remainder incombustibles. the actual air/fuel ratio is 18,5: 1.calculate mass of oxygen, theoretical mass of air required , mass of excess air , mass of excess air
1. Theoretical mass of air required is 9.484375 units
2. Actual air/fuel ratio is 0.0948
3. Mass of excess air is 18.4052
How to calculate the value1. Theoretical mass of air required = Mass of carbon/12 + Mass of hydrogen/4 + Mass of sulphur/32 - Mass of oxygen/32
Theoretical mass of air required = (87/12) + (9/4) + (1/32) - (1.5/32)
Theoretical mass of air required = 7.25 + 2.25 + 0.03125 - 0.046875
Theoretical mass of air required = 9.484375 units
2 Actual air/fuel ratio = Theoretical mass of air required / Total fuel mass
Actual air/fuel ratio = 9.484375 / 100
Actual air/fuel ratio ≈ 0.0948
3 Mass of excess air = Actual air/fuel ratio - Stoichiometric air/fuel ratio (assuming stoichiometric ratio of 18.5)
Mass of excess air = 18.5 - 0.0948
Mass of excess air ≈ 18.4052
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This is the distance between the parallel axes of spur gears or parallel helical gears, or the distance between the crossed axes of helical gears and worm gears. It can be defined also as the distance between the centers of pitch circles. What is this distance? A) Clearance B) Addendum C) Center distance D) Space width
The distance between the parallel axes of gears or the crossed axes of helical gears and worm gears is known as the "Center distance" (C).
The distance between the parallel axes of spur gears or parallel helical gears, or the distance between the crossed axes of helical gears and worm gears is known as the "Center distance" (C).
The center distance is an important parameter in gear design and is defined as the distance between the centers of the pitch circles of two meshing gears. The pitch circle is an imaginary circle that represents the theoretical contact point between the gears. It is determined based on the gear module (or tooth size) and the number of teeth on the gear.
The center distance is crucial in determining the proper alignment and engagement of the gears. It affects the gear meshing characteristics, such as the transmission ratio, gear tooth contact, backlash, and overall performance of the gear system.
In spur gears or parallel helical gears, the center distance is measured along a line parallel to the gear axes. It determines the spacing between the gears and affects the gear ratio. Proper center distance selection ensures smooth and efficient power transmission between the gears.
In helical gears and worm gears, where the gear axes are crossed, the center distance refers to the distance between the lines that are perpendicular to the gear axes and pass through the point of intersection. This distance determines the axial positioning of the gears and affects the gear meshing angle and efficiency.
The center distance is calculated based on the gear parameters, such as the module, gear tooth size, and gear diameters. It is essential to ensure proper center distance selection to avoid gear tooth interference, premature wear, and to optimize the gear system's performance.
In summary, the center distance is the distance between the centers of the pitch circles or the axes of meshing gears. It plays a critical role in gear design and influences gear meshing characteristics, transmission ratio, and overall performance of the gear system.
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Three kg of air at 150 kPa and 77°C temperature at state 1 is compressed polytropically to state 2 at pressure 750 kPa, index of compression being 1.2. It is then cooled to the original temperature by a constant pressure process to state 3. Find the net work done. (Show the processes on the P-V diagram). [Take Rair =0.287 kJ/kg K]
Given dataThree kg of air at 150 kPa and 77°C temperature at state 1 is compressed polytropically to state 2 at pressure 750 kPa, index of compression being 1.2.
It is then cooled to the original temperature by a constant pressure process to state 3.We have to find out the net work done.
Conversion of temperature from Celsius to Kelvin
K = 273 + CK = 273 + 77K = 350 K
Specific gas constant of air is given as
Rair = 0.287 kJ/kg K
Weight of the air is given as 3 kg.
Work Done is given as,
The work done in polytropic process is given as:Work done in a constant pressure process is given as:
In order to find the specific volume and temperature of air in state 2, we will use the polytropic process formula as given below:For process 1-2From the polytropic process formula,
P1V1n = P2V2nV1/V2
= (P2/P1)^(1/n)V1/V2
= (750/150)^(1/1.2)V1/V2
= 4.187
We know that,The process 1-2 is polytropic so
PV^n = Constant
From state 1 to 2, n = 1.2
Therefore;P1V1^n = P2V2^nV2
= V1*(P1/P2)^(1/n)
Putting values,We get;
V2 = 0.00887 m^3/kg
We can now use the ideal gas law equation to find the temperature in state 2:We know that PV = mRTWhere m = mass, R = gas constant, T = Temperature, and P = pressure
Therefore;T2 = (P2V2)/(mR)T2
= (750*0.00887)/(3*0.287)T2
= 60.2 K
For process 2-3, the temperature is constant and is equal to
T3 = T1 = 350 K
For process 1-2, n = 1.2
The work done in process 1-2 is given by:
For process 2-3, P3 = P2 = 750 kPaV3
= mRT3/P3
= 3*287*350/750*10^3
= 0.351 m^3
The work done in process 2-3 is given by:
Therefore, the net work done isAnswer:
The work done in process 1-2 is 4.29 kJ
The work done in process 2-3 is -0.858 kJ
Therefore, the net work done is 3.432 kJ.
This is a thermodynamics problem in which we are given the initial state of a gas and we are required to find its final state. We are given the temperature and pressure of air in state 1 and are asked to compress it polytropically to state 2 at pressure 750 kPa and an index of compression being 1.2. It is then cooled to the original temperature by a constant pressure process to state 3. We are required to find the net work done in this process.In order to solve this problem, we first converted the temperature from Celsius to Kelvin and then found the weight of the air given. We used the polytropic process formula to find the specific volume and temperature of air in state 2. We then used the ideal gas law equation to find the temperature in state 2. Finally, we used the work done formula for process 1-2 and process 2-3 to find the net work done. The main answer for this question is 3.432 kJ.
In conclusion, we can say that this was a simple thermodynamics problem in which we were required to find the net work done in a process. We solved this problem by using the polytropic process formula and the ideal gas law equation. We found that the net work done in this process is 3.432 kJ.
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Roughening the faying surfaces tends to ___ the strength of an adhesively bonded joint \
A. Increase
B. Increase or decrease C. have no effect on D. decrease
Roughening the faying surfaces tends to increase the strength of an adhesively bonded joint. When two surfaces are bonded using an adhesive, the contact surfaces of the two materials are called faying surfaces.
These are the surfaces that are meant to be bonded by the adhesive. Roughening the faying surfaces means increasing the roughness of the surface texture. Roughening of faying surfaces of the adhesive improves the adhesive bonding strength.
Roughening the faying surfaces enhances the mechanical interlocking of the adhesive and the surfaces to be bonded. By increasing the surface area and surface energy of the faying surfaces, it increases the strength of an adhesively bonded joint.
The increased roughness increases the surface area of the faying surfaces, allowing more surface area for bonding to take place. This provides a stronger bond. Moreover, the increased surface area promotes better adhesive wetting of the faying surfaces.
This reduces the possibility of entrapped air between the faying surfaces.
Overall, roughening the faying surfaces tends to increase the strength of an adhesively bonded joint.
Therefore, the correct answer is option A, which states that roughening the faying surfaces tends to increase the strength of an adhesively bonded joint.
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1) Proof the back work ratio of an ideal air-standard Brayton cycle is the same as the ratio of compressor inlet (T1) and turbine outlet (T4) temperatures in Kelvin. Use cold-air standard analysis. (5
The back work ratio of an ideal air-standard Brayton cycle is the same as the ratio of compressor inlet (T1) and turbine outlet (T4) temperatures in Kelvin. Use a cold-air standard analysis.
Given data T1 = More than 100 in KelvinT4 = More than 100 in Kelvin Formula, Back Work Ratio (BWR) = Wc / Q_ in (or) W_ t / Q_ in, Where Wc = Work of compressor, W_ t = Work of turbine, and Q_ in = Heat Supplied to the cycle. Proof: The Brayton cycle is a closed-cycle in which the working fluid receives and rejects heat in the same manner.
Rankine cycle, but the working fluid is not water but air. The cycle comprises four basic components: compressor, heat exchanger, turbine, and heat exchanger, with two adiabatic expansion and compression processes. The first process is compression by the compressor.
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a) (10 pts). Using a decoder and external gates, design the combinational circuit defined by the following three Boolean functions: F1 (x, y, z) = (y'+ x) z F2 (x, y, z) = y'z' + xy + yz' F3 (x, y, z) = x' z' + xy
Given Boolean functions are:F1 (x, y, z) = (y'+ x) z F2 (x, y, z) = y'z' + xy + yz' F3 (x, y, z) = x' z' + xyThe Boolean function F1 can be represented using the decoder as shown below: The diagram of the decoder is shown below:
As shown in the above figure, y'x is the input and z is the output for this circuit.The Boolean function F2 can be represented using the external gates as shown below: From the Boolean expression F2, F2(x, y, z) = y'z' + xy + yz', taking minterms of F2: 1) m0: xy + yz' 2) m1: y'z' From the above minterms, we can form a sum of product expression, F2(x, y, z) = m0 + m1Using AND and OR gates.
The above sum of product expression can be implemented as shown below: The Boolean function F3 can be represented using the external gates as shown below: From the Boolean expression F3, F3(x, y, z) = x' z' + xy, taking minterms of F3: 1) m0: x'z' 2) m1: xy From the above minterms.
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A particle P has velocity:
v(t) = 5 + 3t a) Find the acceleration of the particle
b) Express position (x) as a function of time given the initial condition given the initial condition x(0) = 3m (4) c) Find the distance traversed by the particle in the first 5 seconds of its motion
The particle has an acceleration of 3 m/s^2. Its position as a function of time is x = 5t^2 + 3 m, given the initial condition x(0) = 3 m. The distance traversed by the particle in the first 5 seconds is 75 m.
The acceleration of the particle is found by differentiating the velocity function v(t) = 5 + 3t to get a(t) = 3 m/s^2. The position of the particle as a function of time is found by integrating the velocity function v(t) = 5 + 3t to get x(t) = 5t^2 + 3 m, given the initial condition x(0) = 3 m. The distance traversed by the particle in the first 5 seconds is found by evaluating x(5) - x(0) = 5(5)^2 + 3 - 3 = 75 m.
a) Find the acceleration of the particle
a(t) = v'(t) = 3
b) Express position (x) as a function of time given the initial condition given the initial condition x(0) = 3m
x(t) = ∫ v(t) dt = ∫ (5 + 3t) dt = 5t^2 + 3 m
The initial condition x(0) = 3 m is used to evaluate the constant of integration.
c) Find the distance traversed by the particle in the first 5 seconds of its motion
x(5) - x(0) = 5(5)^2 + 3 - 3 = 75 m
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QUESTION 1 (10 MARKS) As a Graduate Metallurgist you are working in a plant to produce a product as fast as possible. The process uses a liquid with specific heat capacity of 75.44 J.K-1mol-1 . In the first furnace you start with a temperature of 45°C and your aim temperature is 155°C. You use 2 kg of fuel with a mole mass of 22 gmol-1 . In your second furnace you start at a temperature of 82°C and also aim for 155°C. You have here 150 kg of fuel, the same type as in furnace 1. In furnace 1 you use an element of 1200W and for furnace 2 you use 3650W. Do metallurgical calculations to determine which furnace you will opt for, with special reference to time
QUESTION 2 (10 MARKS) Calculate the heat required for 2.4 kg of steel to be heated from 130°C to 920°C if the specific heat capacity is 54.6 + 2.1 x 10-3T – 6.5 x 105T -2 . Mole mass of steel is 56 gmol-1 . If the heat price is R36.50 per 2000J and your budget account is R2420.00 will you be able to buy the energy? Prove by thermodynamic calculations. QUESTION 3 (5 MARKS) A company producing aluminium is using coke as a fuel. The furnace work temperature is 1100°C. Is it possible to produce aluminium in this furnace, comment on your answer? Do metallurgical calculations to prove your answer. You are told that when the enthalpy is zero chemical equilibrium is reached. Given: 2Al + 3 2 O2 = Al2O3 ∆H = −216510 + 169.4T Joules C + 1 2 O2 = CO ∆H = −165550 + 170.2T Joules 3
QUESTION 4 (15 MARKS) The transformation of manganese is as follows: Mn(α) at 720°C → Mn(β) at 1100°C → Mn(γ) at 1136°C → Mn(δ) Calculate the heat of this process when Mn is oxidised by pure oxygen to form MnO at 1250°C. Given: Mn(α) + 1 2 O2 → MnO ∆H298 = −384900 Jmol−1 Mn(α) → Mn(β) ∆Hf = 2100 Jmol−1 Mn(β) → Mn(γ) ∆Hf = 2380 Jmol−1 Mn(γ) → Mn(δ) ∆Hf = 1840 Jmol−1 Cp(αMn) = 21.55 + 15.56 × 10−3T JK −1mol−1 Cp(βMn) = 34.85 + 2.76 × 10−3T JK −1mol−1 Cp(γMn) = 45.55 JK −1mol−1 Cp(δMn) = 47.28 JK −1mol−1 Cp(MnO) = 46.44 + 8.12 × 10−3T − 3.68 × 105T −2 JK −1mol−1 Cp(Oxygen) = 29.96 + 4.184 × 10−3T − 1.67 × 105T −2 JK −1mol−1
The specific heat capacity of the steel is given as a function of temperature, and we are required to use thermodynamic calculations to determine the heat energy required. Thus, we need to integrate the given equation with respect to T over the given temperature range, and obtain the average value of the integrand. In th
e first furnace, the heat energy required is obtained using the formula, Q = mcΔT.
1. Hence, Q = 2 x 75.44 x (155 - 45) = 22632.32 J. In the second furnace, the heat energy required is obtained using the same formula, but we are not given the mass of the liquid.
Thus,t = Q/P
For furnace 2, the time required is given by,t = (150 x 1000) / 3650 = 41.1 sThus, furnace 2 will produce the product faster.
2. The heat energy required to heat the steel from 130°C to 920°C is obtained using the formula,
Q = mcΔT
Hence, Q = (2.4 x 10^3) x (1/790) x (54.6(790-130) + 2.1 x 10^-3 (790^2 - 130^2)/2 - 6.5 x 10^-5 (790^-1 - 130^-1)) = 5.63 x 10^5 J
The cost of the heat energy is given by,
C = (Q/2000) x R36.50 = (5.63 x 10^5 / 2000) x R36.50 = R101.31
Since R101.31 is less than R2420.00, the budget is sufficient to buy the energy.
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A cam follower mechanism with a displacement diagram that has the following sequence, rise 2 mm in 1.2 seconds, dwell for 0.3 seconds, fall 1 r in 0.9 seconds, dwell again for 0.6 seconds and then continue falling for 1 E in 0.9 seconds.
a) The cam rotation angle during the rise is 120.5 degrees.
b) The rotational speed of the cam is 14.38 rpm.
c) The cam rotation angle during the second fall is 82.9 degrees.
d) Both b) and c).
e) None of the above.
The cam follower mechanism with a displacement diagram that has the following sequence, rise 2 mm in 1.2 seconds, dwell for 0.3 seconds, fall 1 r in 0.9 seconds, dwell again for 0.6 seconds and then continue falling for 1 E in 0.9 seconds can be analyzed as follows:a) To determine the cam rotation angle during the rise, we should know that it took 1.2 seconds to rise 2 mm.
We must first compute the cam's linear velocity during the rise:Linear velocity = (Displacement during the rise) / (Time for the rise)= 2 / 1.2 = 1.67 mm/s Then we can calculate the angle:Cam rotation angle = (Linear velocity * Time) / (Base circle radius)= (1.67 * 1.2) / 10 = 0.2 radian= (0.2 * 180) / π = 11.47 degrees Therefore, the cam rotation angle during the rise is 11.47 degrees. Therefore, option a) is incorrect.b) The rotational speed of the cam can be calculated as follows:Linear velocity = (Displacement during the second fall) / (Time for the second fall)= 1 / 0.9 = 1.11 mm/s
Therefore, the rotational speed of the cam is 71.95 rpm. Therefore, option b) is incorrect.c) To determine the cam rotation angle during the second fall, we should know that it took 0.9 seconds to fall 1 E. We must first compute the cam's linear velocity during the fall:Linear velocity = (Displacement during the fall) / (Time for the fall)= 1 / 0.9 = 1.11 mm/s Then we can calculate the angle:Cam rotation angle = (Linear velocity * Time) / (Base circle radius)= (1.11 * 0.9) / 10 = 0.0999 radians= (0.0999 * 180) / π = 5.73 degrees
Therefore, the cam rotation angle during the second fall is 5.73 degrees. Therefore, option c) is incorrect.Therefore, the answer is option e) None of the above.
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a) A company that manufactures different components of bike such as brake lever, cranks pins, hubs, clutch lever and wants to expand their product line by also producing tire rims. Begin the development process of designing by first listing the customer requirements or "WHAT" the customer needs or expects then lists the technical descriptors or "HOW" the company will design a rim. Furthermore, it is necessary to break down the technical descriptors and customer requirements to the tertiary level. Develop the Basic House of Quality Matrix using all the techniques including technical competitive assessment, Customer competitive assessment, absolute weight, and relative weights. Make reasonable assumptions where required. b) Prioritization matrices prioritize issues, tasks, characteristics, and so forth, based on weighted criteria using a combination of tree and matrix diagram techniques. Once prioritized, effective decisions can be made. A construction company was not able to complete the construction of bridge in planned time. The main causes of failure may include the people, machines, or systems. An audit company was given contract to conduct detailed analysis for this failure and provide feedback to avoid it in future. As a manager of this audit company, identify six implementation options and four implementation criteria, construct the tree diagram, and prioritize the criteria using nominal group techniques. Rank order the options in terms of importance by each criterion. Compute the option importance score under each criterion by multiplying the rank with the criteria weight. Develop the prioritization matrices.
15+15=30
a) Customer Requirements:The customer expects the following features in the bike tire rim:Durability: Tire rim must be strong enough to withstand rough terrain and last long.Aesthetics: Rim should look attractive and appealing to the eye.Corrosion resistance: Rim should not corrode and should be rust-resistant.Weighting Factors:The relative weight of durability is 0.35, aesthetics is 0.30 and corrosion resistance is 0.35. Technical Descriptors:The following technical descriptors will be used to design the rim:Diameter:
The diameter of the rim should be between 26-29 inches to fit standard bike tires.Material: Rim should be made of high-quality and lightweight material to ensure durability and strength.Weight: Weight of the rim should not be too high or too low.Spokes: Rim should have adequate spokes for strength and durability.Braking: Rim should have a braking system that provides good stopping power.Rim tape:
Rim tape should be strong enough to handle the high pressure of the tire.Weight allocation: The weight of each technical descriptor is diameter 0.10, material 0.30, weight 0.20, spokes 0.15, braking 0.10, and rim tape 0.15. Quality Matrix: The quality matrix is based on the given customer requirements and technical descriptors, with quality ranking from 1 to 5, and the corresponding weight is allocated to each parameter. The formula used to calculate the values in the matrix is given below: (Weight of customer requirements) * (Weight of technical descriptors) * Quality rankingFor instance, if the quality ranking of the diameter is 4 and the relative weight of the diameter is 0.1, the value of the quality matrix is (0.35) * (0.10) * 4 = 0.14.
The House of Quality Matrix is as follows:Technical Competitive Assessment: The company can research other manufacturers to see how they design and develop bike tire rims and determine the technical competitive assessment.Customer Competitive Assessment: The company can also conduct surveys or collect data on what customers require in terms of tire rim quality and design. Absolute weight: The weights that are not dependent on other factors are absolute weight.Relative weight: The weights that are dependent on other factors are relative weight.b)Implementation Options:Organizational structure, training, and development strategies.Resource allocation strategies, procurement strategies, financial strategies.Risk management strategies, conflict resolution strategies, and communication strategies.Process improvement strategies, quality management strategies, and compliance strategies. Implementation Criteria: Cost,
Time, Effectiveness, and Customer satisfaction. Tree Diagram: Prioritization Matrix:Nominal Group Technique:Ranking based on the Criteria and Weight:Organizational structure and Training: 22Resource allocation strategies and Financial strategies: 20Process improvement strategies and Quality management strategies: 19Risk management strategies and Conflict resolution strategies: 17Procurement strategies and Communication strategies: 16Therefore, Organizational structure and Training are the highest-ranked implementation options based on the criteria and weight.
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Consider a 210-MW steam power plant that operates on a simple ideal Rankine cycle. Steam enters the turbine at 10MPa and 5008C and is cooled in the condenser at a pressure of 10kPa. Sketch the cycle on a T-s diagram with respect to saturation lines, and determine: (a) the quality of the steam at the turbine exit, (b) the thermal efficiency of the cycle, (c) the mass flow rate of the steam. (d) Repeat Prob. (a)-(c) assuming an isentropic efficiency of 85 percent for both the turbine and the pump.
Given data:Pressure of steam entering turbine (P1) = 10 MPaTemperature of steam entering turbine (T1) = 500 degree CPressure of steam at the condenser (P2) = 10 kPaPower generated (W) = 210 MWNow, let's draw the T-s diagram with respect to saturation lines below:
1. The quality of steam at the turbine exit:From the T-s diagram, we can see that at the turbine exit, the state point lies somewhere between the two saturation lines.Using the steam tables, we can find the saturation temperature and pressure at the exit state:Pressure at the exit (P3) = 10 kPaSaturated temperature corresponding to P3 = 46.9 degree CEnthalpy of saturated liquid corresponding to P3 (h_f) = 191.81 kJ/kgEnthalpy of saturated vapor corresponding to P3 (h_g) = 2676.5 kJ/kgThe quality of steam (x) at the exit state is given by:x = (h - h_f)/(h_g - h_f)Where, h is the specific enthalpy at the exit state.
h = 191.81 + x(2676.5 - 191.81)h = 191.81 + 2421.69x= (h - h_f)/(h_g - h_f)x = (191.81 + 2421.69 - 191.81)/(2676.5 - 191.81)x = 0.91The quality of steam at the turbine exit is 0.91.2. Thermal efficiency of the cycle:For an ideal Rankine cycle, thermal efficiency is given by:eta_th = 1 - (T2/T1)Where, T2 and T1 are the temperatures of the steam at the condenser and the turbine inlet respectively.
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Heat recovery steam boiler (HRSB) was designed to produce 4600 kg/h saturated steam at pressure 20 atm with exhaust gas flow mg = 34000 kg / h and temperatures Tgin = 540οC, Tgout = 260οC. During its operation with reduced load (mg = 22800 kg / h, Tgi = 510οC) the exhaust temperature of the exhaust gas Tgο = 271οC is measured. Can you comment on the possibility of deterioration of the boiler operation due to the formation of deposits?
The lower exhaust gas temperature observed during reduced load operation suggests a potential improvement in heat transfer efficiency, but a thorough assessment of the specific operating conditions and potential deposit formation is necessary to evaluate the overall impact on boiler performance.
The formation of deposits in a boiler can have negative effects on its operation. Deposits are usually formed by the condensation of impurities contained in the exhaust gas onto the heat transfer surfaces. These deposits can reduce heat transfer efficiency, increase pressure drop, and potentially lead to corrosion or blockage. In this case, the decrease in exhaust gas temperature (Tgο) from the designed operating conditions could suggest improved heat transfer due to reduced fouling or deposit formation. The lower exhaust gas temperature indicates that more heat is being transferred to the steam, resulting in a higher steam production temperature. However, it is important to consider other factors such as the composition of the exhaust gas and the properties of the deposits. Different impurities and operating conditions can lead to varying degrees of deposit formation. A comprehensive analysis, including a study of the exhaust gas composition, flue gas analysis, and inspection of the boiler surfaces, would be required to make a definitive conclusion about the possibility of boiler operation deterioration due to deposits.
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