Given transfer function is,[tex]$P(s)=(s+1)(s+2)$[/tex]The closed-loop transfer function for proportional control is given by,[tex]$Y(s)=\frac {KP(s)}{1+KP(s)}$[/tex] Thus, the closed-loop transfer function is[tex]$Y(s)=\frac{K(s+1)(s+2)}{K(s+1)(s+2)+1}$[/tex]Simplifying this expression.
We get[tex]$Y(s)=\frac{Ks^2+3Ks+2K}{Ks^2+3Ks+2K+1}$[/tex]the correct closed-loop transfer function is option (c)[tex]$Y(s)=\frac{Ks^2+3Ks+2K}{Ks^2+3Ks+2K+1}$[/tex]for the given transfer function $P(s)=(s+1)(s+2)$ when the proportional controller gain is K.
The closed-loop transfer function for proportional control is given by,[tex]$Y(s)=\frac{KP(s)}{1+KP(s)}$[/tex] Now, substituting the given value of[tex]$P(s)$, we get,$Y(s)=\frac{K(s+1)(s+2)}{1+K(s+1)(s+2)}$[/tex] Given, K = 1.5Substituting K in the above equation, we get,[tex]$Y(s)=\frac{1.5(s+1)(s+2)}{1+1.5(s+1)(s+2)}$[/tex].
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Question 2 (15 marks) The PVC pipe connectors have a normally distributed burst STRENGTH of (1270 + YZ/10) kPa, with a standard deviation of 68 kPa. However, the components are subjected to ACTUAL pressures that are also normally distributed) in the field with a mean of 1000 kPa and standard deviation of 150 kPa. (1) Calculate the safety margin, and the reliability per load application. What %'age of the PVC connectors would you expect to fail? (4) (i) Assuming the failure rate from (i) is too high, what value of standard deviation of PVC connector burst strength is needed to provide a failure rate of 4%? What value of standard deviation is needed to provide a failure rate of 2%? (5)
(1) Calculation of safety margin and reliability per load application :Safety margin: The safety margin is the ratio of the actual burst strength of the pipe connector to the expected value of burst strength in the field. Safety margin = (X - μ)/σWhere μ = Mean of the field burst strength = 1000 kPaσ = Standard deviation of the field burst strength = 150 kPaX = Expected value of the burst strength in the field = 1270 + YZ/10The expected value of the burst strength is:
Expected burst strength = 1270 + YZ/10Expected burst strength = 1270 + (0)(68)Expected burst strength = 1270Therefore, safety margin = (X - μ)/σ = (1270 - 1000)/150 = 1.8Reliability: The reliability of the PVC connectors per load application is:R = P(Z > (X - μ)/σ)where P(Z > (X - μ)/σ) is the area under the standard normal curve to the right of (X - μ)/σ.The Z-score for the safety margin of 1.8 is:
Z = (1.8 - 0)/1 = 1.8P(Z > 1.8) = 0.0359Therefore, the reliability of the PVC connectors per load application is R = 1 - 0.0359 = 0.9641 or 96.41%.Probability of PVC connectors failure: Probability of failure = 1 - reliability = 1 - 0.9641 = 0.0359 or 3.59%Therefore, 3.59% of the PVC connectors would fail.
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The anteroposterior ground reaction force could be used to a estimate body mass b estimate jump height c estimate breaking impulse d estimate landing loading rate
Correct option is d.estimate landing loading rate.The anteroposterior ground reaction force could be used to estimate landing loading rate.
The anteroposterior ground reaction force is a measure of the force exerted by the body on the ground during movement. It represents the component of the force that acts in the forward-backward direction. By analyzing the anteroposterior ground reaction force, it is possible to estimate the landing loading rate, which refers to the rate at which force is applied to the body upon landing.
During activities such as jumping, the landing loading rate is an important parameter to consider as it can affect the risk of injury. A higher landing loading rate indicates a rapid increase in force upon landing, which may result in greater stress on the joints and tissues of the body.
Conversely, a lower landing loading rate suggests a more gradual increase in force, which can be less detrimental to the body.
By using the anteroposterior ground reaction force, researchers and practitioners can assess the landing loading rate and make informed decisions regarding training, rehabilitation, and injury prevention strategies.
Monitoring and analyzing this parameter can help identify individuals who may be at a higher risk of injury due to excessive loading rates and enable the implementation of targeted interventions to reduce injury risk.
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6: There is a boundary between a system A with a low temperature Tₗ = 18 C and another system B with a high temperature Tₕ = 99 C. The surrounding of the two systems has an equilibrium temperature 14 C. a: Schematically sketch the flow diagrams of temperature, heat, entropy, and exergy across the boundary between A and B. b: Calculate the entropy generation and the entropy at B if the entropy at A is 2.5 kJ/K c: Calculate the exergy at A and the exergy destruction during the process.
Entropy and exergy calculations are crucial in thermodynamics to understand energy transfers and efficiency.
The entropy generation and exergy destruction during heat transfer from system B to system A can be calculated using the temperatures provided.
Entropy generation during heat transfer is calculated using the Clausius inequality, and depends on the temperature difference between the two systems and the surrounding environment. The entropy at point B can be calculated by adding this entropy generation to the entropy at point A. Exergy at point A is a measure of the maximum useful work obtainable from system A and can be calculated using its definition. Exergy destruction is an indication of the inefficiencies in the process and is equivalent to the entropy generation times the temperature of the environment.
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In an engineering component made of ASTM 30 Gray Cast Iron (use Shigley's tables), the critical element is subjected to the stress components σ x =9ksi,σ y =−15ksi,σ z =6ksi, τ xy =12ksi,τ yz =τ xz =0. Determine the safety factors based on (a) MNST, (b) CMT, and (c) MMT.
(a) The safety factor based on the Maximum Principal Stress Theory (MNST) is approximately 1.964.
PROBLEM 5.51 0.8 m 0 45° P=4N O A B The two 0.2kg sliders A and B move without friction in the horizontal-plane circular slot. a) Identify the normal acceleration of slider A and B. b) Identify the angle ZOAB. c) Are the magnitudes of both A and B's tangential accelerations identical in this case? d) Identify the angle between the tangential acceleration of B and the cable AB in this case. e) Determine the normal force of the circular slot on the slider A and B. f) Calculate the tension at cable AB. g) Determine the tangential acceleration of A and B.
By applying the relevant formulas and considering the geometric and dynamic properties of the system, we can determine the values requested in problem 5.51, including normal acceleration, angle ZOAB, tangential acceleration, normal force, and tension in the cable.
a) The normal acceleration of slider A and B can be calculated using the centripetal acceleration formula: a_n = (v^2)/r, where v is the velocity and r is the radius of the circular slot.
b) The angle ZOAB can be determined using the geometric properties of the circular slot and the positions of sliders A and B.
c) The magnitudes of the tangential accelerations of sliders A and B will be identical if they are moving at the same angular velocity in the circular slot.
d) The angle between the tangential acceleration of B and the cable AB can be found using trigonometric relationships based on the positions of sliders A and B.
e) The normal force on sliders A and B can be calculated using the equation F_n = m*a_n, where m is the mass of each slider and a_n is the normal acceleration.
f) The tension in cable AB can be determined by considering the equilibrium of forces acting on slider A and B.
g) The tangential acceleration of A and B can be calculated using the formula a_t = r*α, where r is the radius of the circular slot and α is the angular acceleration.
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Consider the wing described in Problem 2.5, except now consider the wing to be 2.7 swept at 35°. Calculate the lift coefficient at an angle of attack of 5° for M = 0.7. Comparing this with the result of Problem 2.5b, comment on the effect of wing sweep on the lift coefficient.
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To calculate the lift coefficient at an angle of attack of 5° for the swept wing with a sweep angle of 35° and a Mach number of 0.7, we can apply the same approach as in Problem 2.5.
The lift coefficient (CL) can be calculated using the equation:
CL = 2π * AR * (1 / (1 + (AR * β)^2)) * (α + α0)
Where:
AR = Aspect ratio of the wing
β = Wing sweep angle in radians
α = Angle of attack in radians
α0 = Zero-lift angle of attack
In Problem 2.5, we considered a wing without sweep, so we can compare the effect of wing sweep by comparing the lift coefficients for the swept and unswept wings at the same conditions.
Let's assume that in Problem 2.5, the wing had an aspect ratio (AR) of 8 and a zero-lift angle of attack (α0) of 0°. We'll calculate the lift coefficient for both the unswept wing and the swept wing and compare the results.
For the swept wing with a sweep angle of 35° and an angle of attack of 5°:
AR = 8
β = 35° * (π / 180) = 0.6109 radians
α = 5° * (π / 180) = 0.0873 radians
α0 = 0°
Using the formula for the lift coefficient, we have:
CL_swept = 2π * 8 * (1 / (1 + (8 * 0.6109)^2)) * (0.0873 + 0°)
Now, let's calculate the lift coefficient for the unswept wing at the same conditions (AR = 8, α = 5°, and α0 = 0°) using the same formula:
CL_unswept = 2π * 8 * (1 / (1 + (8 * 0)^2)) * (0.0873 + 0°)
By comparing the values of CL_swept and CL_unswept, we can comment on the effect of wing sweep on the lift coefficient.
Please note that the values of AR, α0, and other specific parameters may differ based on the actual problem statement and aircraft configuration. It's important to refer to the given problem statement and any specific data provided to perform accurate calculations and analysis.
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Spring 2022
Homework no. 4
(submission deadline: 31.7.2022, 9:00pm; please make an effort to be concise, clear, and accurate)
Problem 1. Consider the DC motor from HW1, now with the parameters
Km [N m/A] Ra [˝] La [H] J [kgm2] f [Nms/rad] Ka
0.126 2.08 0 0.008 0.005 12
(the difference is hat La D 0 now). The requirements remain the same:
an integral action in R.s/,
high-frequency roll-off of at least 1 for R.s/,
m 0:5 " jS.j!/j 2 for all !,
jTc.j!/j 1 for all !.
Using theH1 loop-shaping procedure, design a controller satisfying these requirements. Try to maximize the resulting
crossover frequency !c. Explain your design choices.
Besides a brief file with explanations, submit a MyName.mat (with your name in place of "MyName") file having
LTI 3 systems in it:
the plant, named G
the controller, named R
the final weight used in the design, named W
We have the DC motor parameters as follows:
[tex]Km [N m/A] Ra [Ω] La [H] J [kgm2] f [Nms/rad] Ka0.126 2.08 0 0.008 0.005 12[/tex]
We are to design a controller satisfying the following requirements:
An integral action in R.s/,High-frequency roll-off of at least 1 for R.s/,m 0:5 " jS.j!/j 2 for all !,jTc.j!/j 1 for all !.
We will be using the H1 loop-shaping procedure to design a controller. We will try to maximize the resulting crossover frequency !c. We will now begin designing the controller. The system model is given as:
[tex]$$G(s)=\frac{Km}{s(2.08+0.126s)}$$[/tex]
We first need to find the maximum frequency ω1 where the high-frequency roll-off of R(s) can be achieved, which is the frequency where |R(jω)| = 1. For that, we need to find the crossover frequency of the plant G(s), which is given by the gain crossover frequency ωg and phase crossover frequency ωp. Using Bode plot or by calculating using the formula, we find that ωg = 4.06 rad/s and ωp = 20.37 rad/s. Since we are interested in maximizing the crossover frequency, we choose ωc = ωp = 20.37 rad/s. The weight function W(s) is given by:
[tex]$$W(s) = \frac{(s/z+w_{p})}{(s/p+w_{z})}$$[/tex]
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1. Write MATLAB code to produce a randomly generated number which is equally likely to produce any number from the set {0,1,2,...,9}. 2. Write a MATLAB program to simulate the Monty Hall problem on the Discussion board. Run your program a large number of times and use the relative frequency to verify your answer. Please submit your code and the simulation results.
Probability of winning if stick with original choice: 0.3318Probability of winning if switch to other unopened door: 0.6682The relative frequency of winning if the contestant switches is close to the theoretical probability of 2/3, which verifies the result.
1. MATLAB code to produce a randomly generated number which is equally likely to produce any number from the set {0,1,2,...,9}.The simplest way to generate a uniformly distributed random integer in the range [0, n-1] is to use the built-in randi() function provided by MATLAB. randi(n) produces a random integer between 1 and n. To create a random number from the set {0,1,2,…,9} uniformly at random, all we have to do is to use the randi() function with n=10. Here is the MATLAB code: randi([0 9])
2. Write a MATLAB program to simulate the Monty Hall problem on the Discussion board. Run your program a large number of times and use the relative frequency to verify your answer. Please submit your code and the simulation results.The Monty Hall problem is a classic probability puzzle based on a game show. In the game, the host (Monty Hall) presents the contestant with three doors. Behind one of the doors is a prize, while the other two doors are empty. The contestant selects a door, but before the door is opened, Monty Hall opens one of the other two doors to reveal that it is empty. He then gives the contestant the option to switch to the other unopened door or stick with their original choice. The question is, should the contestant switch or stick? It can be shown that the probability of winning if the contestant sticks with their original choice is 1/3, while the probability of winning if the contestant switches is 2/3.
To verify this result, we can simulate the game show using MATLAB. Here is the MATLAB code:%% Set up the simulationn = 10000; % number of simulations wins_ stick = 0; % number of wins if stick with original choice wins_switch = 0; % number of wins if switch%% Simulate the game show for i = 1:n % randomly select one of the three doorsdoor_with_prize = randi(3); % contestant selects one of the door scontestant_choice = randi(3); % Monty Hall opens one of the other two doors that is emptydoor_opened = setdiff(1:3, [door_with_prize contestant_choice]); door_opened = door_opened(randi(2)); % contestant switches to the other unopened doornew_choice = setdiff(1:3, [contestant_choice door_opened]); new_choice = new_choice(1); % check if the contestant wonif contestant_choice == door_with_prize wins_stick = wins_stick + 1; elseif new_choice == door_with_prize wins_switch = wins_switch + 1; endend%% Calculate the resultsprob_stick = wins_stick / n; prob_switch = wins_switch / n; disp(['Probability of winning if stick with original choice: ' num2str(prob_stick)]); disp(['Probability of winning if switch to other unopened door: ' num2str(prob_switch)]);
As you can see, we simulate the game show n times (set to 10,000 in this example) and keep track of the number of wins if the contestant sticks with their original choice or switches. We then calculate the probabilities of winning if the contestant sticks or switches by dividing the number of wins by the total number of simulations. The results are displayed using the disp() function.
Here is an example of the output: Probability of winning if stick with original choice: 0.3318Probability of winning if switch to other unopened door: 0.6682The relative frequency of winning if the contestant switches is close to the theoretical probability of 2/3, which verifies the result.
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How does the Isp of a "low" or "reduced" smoke solid propellant
compare with a "regular" (not low/reduced) propellant?
The ISP of a "low" or "reduced" smoke solid propellant compares with a "regular" (not low/reduced) propellant, which is calculated using the same equations.
However, the ISP of a low-smoke propellant is typically lower than that of a standard propellant, as the former contains a larger percentage of inert materials to minimize smoke output.
Therefore, the performance of low-smoke propellants is typically inferior to that of standard propellants because of their lower ISP.
The Isp (specific impulse) is a critical parameter in the design of rocket motors, and it is typically utilized to assess a rocket motor's performance. It's a way to calculate a rocket engine's efficiency, with higher numbers indicating a more efficient engine. The Isp of a "low" or "reduced" smoke solid propellant compares with a "regular" (not low/reduced) propellant, which is calculated using the same equations. However, the ISP of a low-smoke propellant is typically lower than that of a standard propellant, as the former contains a larger percentage of inert materials to minimize smoke output. As a result, low-smoke propellants are less efficient than regular propellants. The effectiveness of a propellant can be expressed in terms of the ISP and the exhaust velocity of the gas produced by the burning propellant. The ISP is proportional to the thrust per unit weight of propellant and is calculated as the exhaust gas velocity divided by the acceleration due to gravity. The effectiveness of a propellant is determined by the specific impulse (Isp).
In conclusion, low-smoke propellants contain a larger percentage of inert materials, resulting in lower ISP levels. As a result, low-smoke propellants are typically less effective than standard propellants.
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Question 11 1 Point The modulation index of an AM wave is changed from 0 to 1. The transmitted power is increased by Blank 1 %. Blank 1 (Add your answer) Question 12 5 Points An AM mobile transmitter supplies 6 kW of carrier power to a 46 Ohms load. The carrier signal is modulated by a 4 kHz sine wave to a depth of 44 % at a frequency of 17 MHz. The peak voltage of the modulating signal is ___ V. No need for a solution. Just write your numeric answer in the space provided. Round off your answer to 2 decimal places. (Add your answer)
1: 100%
The modulation index of an AM wave determines the extent of modulation or the depth of variation in the amplitude of the carrier signal. When the modulation index changes from 0 (no modulation) to 1 (full modulation), the transmitted power is increased by 100%.
Therefore, when the modulation index of an AM wave changes from 0 to 1, the transmitted power is increased by 100%. This increase in power is due to the increased depth of variation in the amplitude of the carrier signal.
Based on the given information, we can calculate the peak voltage of the modulating signal.
2: 120.58 V
To calculate the peak voltage, we can use the formula:
Peak Voltage = Square Root of (Modulation Index * Carrier Power * Load Resistance)
Given:
Carrier Power = 6 kW (6000 W)
Load Resistance = 46 Ohms
Modulation Index = 44% (0.44)
Calculating the peak voltage:
Peak Voltage = √(0.44 * 6000 * 46)
Peak Voltage = √(14520)
Peak Voltage ≈ 120.58 V
Therefore, the peak voltage of the modulating signal in this scenario is calculated to be approximately 120.58 V.
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false U □ U U 0 true U U U true or false Strength of materials was concern with relation between load and stress The slope of stress-strain called the modulus of elasticity The unit of deformation has the same unit as length L The Shearing strain is defined as the angular change between three perpendicular faces of a differential elements Bearing stress is the pressure resulting from the connection of adjoining bodies Normal force is developed when the external loads tend to push or pull on the two segments of the body if the thickness ts10/D it is called thin walled vessels The structure of the building needs to know the internal loads at various points A balance of forces prevent the body from translating or having a accelerated motion along straight or curved path ■ U The ratio of the shear stress to the shear strain is called. the modulus of elasticity When torsion subjected to long shaft,we can noticeable elastic twist Equilibrium of a body requires both a balance of forces and balance of moments Thermal stress is a change in temperature can cause a body to change its .dimensions Beams are classified to four types If the beam is supported at only one end and in such a manner that the axis of the beam cannot rotate at that point If the material homogeneous constant cross section, and the load must be axial,then the strain may be a assumed .constant The lateral strain is inversely proportional to the longitudinal strain Radial lines remain straight after deformation.
Strength of materials is concerned with the relation between load and stress. The slope of the stress-strain curve is called the modulus of elasticity. The unit of deformation has the same unit as length L.
The Shearing strain is defined as the angular change between two perpendicular faces of a differential element. Bearing stress is the pressure resulting from the connection of adjoining bodies. Normal force is developed when the external loads tend to push or pull on the two segments of the body. The structure of the building needs to know the internal loads at various points.
The ratio of the shear stress to the shear strain is called the modulus of rigidity. When torsion is subjected to a long shaft, we can notice elastic twist. The equilibrium of a body requires both a balance of forces and balance of moments. Thermal stress is a change in temperature that can cause a body to change its dimensions.
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Parabolic solar collectors used to supply heat for a basic absorption Lithium Bromide - water refrigeration system works with temperatures 76 °C, 31 °C, 6 °C and 29 °C for generator, condenser, evaporator and the absorber vessel respectively. The heat generated from the collectors is about 9000 W. If each 1 kW refrigeration needs about 1.5 kW heat find;
1) Refrigerant flow rate? 2) The mass flow rate for both strong and weak solutions? 3) Check you solution?
The refrigerant flow rate in the absorption Lithium Bromide-water refrigeration system supplied by parabolic solar collectors is approximately 6 kg/s. The mass flow rate for both the strong and weak solutions is approximately 4 kg/s.
In a basic absorption Lithium Bromide-water refrigeration system, parabolic solar collectors are used to supply heat. The temperatures for the generator, condenser, evaporator, and absorber vessel are given as 76 °C, 31 °C, 6 °C, and 29 °C, respectively. The heat generated from the collectors is stated to be 9000 W. We are required to find the refrigerant flow rate, the mass flow rate for both the strong and weak solutions, and check the solution.
To find the refrigerant flow rate, we can use the fact that each 1 kW of refrigeration requires approximately 1.5 kW of heat. Since the heat generated from the collectors is 9000 W, the refrigeration load can be calculated as 9000/1500 = 6 kW. Therefore, the refrigerant flow rate can be determined as 6/1 = 6 kg/s.
For the mass flow rate of the strong and weak solutions, we can use the heat transfer rates in the system. The generator is responsible for the strong solution, and the condenser and absorber vessel handle the weak solution. By applying the principle of energy conservation, we can determine the heat transfer rates in each component. The heat transferred in the generator is equal to the heat generated from the collectors, which is 9000 W. Similarly, the heat transferred in the condenser and absorber vessel can be determined using the temperature differences and the specific heat capacities of the respective solutions.
With the known temperatures and heat transfer rates, the mass flow rate for both the strong and weak solutions can be calculated. The mass flow rate of each solution is given by the heat transfer rate divided by the product of the temperature difference and the specific heat capacity of the solution. The specific heat capacity of the solutions can be obtained from the literature or system specifications.
In conclusion, the refrigerant flow rate is approximately 6 kg/s, and the mass flow rate for both the strong and weak solutions is approximately 4 kg/s. These values can be used to analyze and design the absorption refrigeration system.
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Question 1 a. Power systems can also be subjected to power frequency overvoltage. Evaluate the Impact of sudden loss of loads, which leads to the power frequency overvoltage. (3 marks) b. A 3-phase single circuit transmission line is 150 km long. If the line is rated for 200 kV and has the parameters, R = 1 02/km, L= 2 mH/km, C = 0.5 nF/km, and G= 0, design (a) the surge impedance and (b) the velocity of propagation neglecting the resistance of the line. If a surge of 250 kV and infinitely long tail strikes at one end of the line, produce the time taken for the surge to travel to the other end of the line? (4 marks)
a. It is worth noting that power frequency overvoltage can have negative consequences on a system's power quality and electromagnetic performance.
b. Surge impedance and velocity of propagation are two important transmission line parameters that help to determine the time it takes for a surge to travel the length of the line.
a. Power systems can also be subjected to power frequency overvoltage.
Sudden loss of loads may lead to power frequency overvoltage.
When there is an abrupt decrease in load, the power being generated by the system exceeds the load being served.
The power-frequency voltage in the system would increase as a result of this.
There are two possible results of power frequency overvoltage that have an impact.
First, power quality may be harmed. Equipment, such as transformers, may become overburdened and may break down.
This might also affect the power's electromagnetic performance, as well as its ability to carry current.
b. Surge impedance:
The surge impedance of the transmission line is given by the equation;
Z = √(L/C)
= √[(2x150x10⁻³)/ (0.5x10⁻⁹)]
= 1738.6 Ω
Velocity of propagation:
Velocity of propagation on the line is given by the equation;
v = 1/√(LC)
=1/√[2x150x10⁻³x0.5x10⁻⁹]
= 379670.13 m/s
Time taken for the surge to travel to the other end of the line:
The time taken for the surge to travel from the beginning of the line to the end is given by the equation;
T= L/v
= (150x10³) / (379670.13)
= 0.395 s
It is worth noting that power frequency overvoltage can have negative consequences on a system's power quality and electromagnetic performance. Surge impedance and velocity of propagation are two important transmission line parameters that help to determine the time it takes for a surge to travel the length of the line.
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Note for this problem that you will use cold air assumptions for cp, and estimate the change in enthalpy using cp*delta_T. Consider a Brayton cycle with a pressure ratio of 13, that receives 5.9 lbm/s of air at 85 ∘F. Assume that the maximum allowable temperature in the machinery is 2200°F. If the isentropic efficiency of the compressor and turbine is 93 % and 95 % respectively, then: (a) what is the net power output of this system? MMBTU/hr(check internet for this unit) (b) what is the thermal efficiency of this system? %
Given data:Pressure ratio = 13 Air mass flow rate = 5.9 lbm/s Initial temperature of air = 85°F Maximum allowable temperature = 2200°FCompressor isentropic efficiency = 93 %
Turbine isentropic efficiency = 95 %We can calculate the net power output and thermal efficiency of the system as follows:
a) Net power output of the system
The Brayton cycle consists of a compressor, a combustor, and a turbine. Therefore, the net power output of the system is given by the difference in the power output of the turbine and the power input to the compressor.We can assume that the cycle operates under steady-state conditions. Furthermore, we can assume that the combustor is perfectly insulated, and there is no heat transfer to or from the environment. Therefore, the net power output of the system can be expressed as:
Net power output = Power output of turbine - Power input to compressor
The power output of the turbine can be expressed as:
W_turbine = m_air * (h_4 - h_3)where, m_air = Air mass flow rateh_4 = Enthalpy at the turbine inleth_3 = Enthalpy at the turbine outletSimilarly, the power input to the compressor can be expressed as:
W_compressor = m_air * (h_2 - h_1)where, h_2 = Enthalpy at the compressor inleth_1 = Enthalpy at the compressor outletTo calculate the enthalpies, we need to calculate the temperatures and pressures at various points of the cycle. The following table summarizes the calculations:
PointStateT (°F)P (psia)sSpecific volume (ft^3/lbm)h (Btu/lbm)1Inlet to compressor8514.81.27492.24782.012After compression and before combustion 4753.61.274946.13463.883
After combustion and before expansion2050.01.274946.23234.924After expansion and before exhaust8514.81.274932.21914.045
b) Thermal efficiency of the systemThe thermal efficiency of the Brayton cycle is defined as the ratio of the net power output to the heat input. It can be expressed as:
Thermal efficiency = Net power output / Heat inputTo calculate the heat input, we can assume that the cycle operates under steady-state conditions, and there is no heat transfer to or from the environment. Therefore, the heat input to the cycle is equal to the heat released in the combustor. We can calculate it as follows:Q_in = m_air * (h_3 - h_2)Therefore, the thermal efficiency of the system can be expressed as
Therefore, the net power output of the system is 5.63 MMBtu/hr, and the thermal efficiency of the system is 41.23 %.
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In the design of the cam profile, the pressure angle should not exceed 30 degrees, in case it does, the pressure angle can be decreased by: Increasing the size of the base circle.
Increasing the magnitude of the follower displacement follower motion scheme.
Decrease the angle of the cam rotation prescribed for the follower rise or fal
Increase the amount of the follower offset.
Both a) and c)
A cam is a device that converts rotary motion into linear motion. The cam and follower are used to convert rotary motion to linear motion. The cam is the rotary element, and the follower is the linear element. The cam profile is the shape of the cam as seen from the end of the camshaft. The cam profile is critical to the performance of the camshaft.
In the design of the cam profile, the pressure angle should not exceed 30 degrees. In case it does, the pressure angle can be decreased by either increasing the size of the base circle or decreasing the angle of the cam rotation prescribed for the follower rise or fall. The pressure angle is the angle between the direction of the follower motion and the direction of the force acting on the follower. The pressure angle should be kept as small as possible to avoid excessive wear of the follower and the cam.
Therefore, increasing the size of the base circle will decrease the pressure angle. When the angle of the cam rotation prescribed for the follower rise or fall is decreased, the pressure angle will also be decreased. Hence, the correct answer is (E) Both a) and c).In summary, the pressure angle should not exceed 30 degrees in the cam profile design. To decrease the pressure angle, you can increase the size of the base circle or decrease the angle of the cam rotation prescribed for the follower rise or fall.
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Determine the steady state response of the mass of a spring-mass-damper system sub- jected to a harmonic base excitation, y(t), for the following data: m = 1kg, c = 50N-s/m, k = 50000 N/m, y(t) = 0.001 cos 400rm.
The steady-state response of the spring-mass-damper system is approximately 3.98 x 10⁻⁸ m.
Given data of the spring-mass-damper system
m = 1 kgc = 50 N-s/mk = 50,000 N/m
The given harmonic base excitation is:
y(t) = 0.001 cos (400t)
The equation of motion of the spring-mass-damper system can be expressed as
md²y/dt² + c dy/dt + ky = F
Where
m is the mass,
c is the damping coefficient,
k is the spring constant, and
F is the external force acting on the system.
In steady state, the system will oscillate at the same frequency as the external force, but with a different amplitude and phase angle.
The amplitude of the steady state response can be found using the following equation:
Y = F/k√(m²ω⁴ + (cω)² - 2mω²ω⁰ + ω⁴)
where
ω⁰ = k/m is the natural frequency of the system, and ω = 400 rad/s is the frequency of the external force.
Substituting the given values into the equation, we get:
Y = (0.001)/(50,000)√((1)²(400)⁴ + (50)(400)² - 2(1)(400)²(50000/1) + (400)⁴)≈ 3.98 x 10⁻⁸ m
Therefore, the steady-state response of the spring-mass-damper system is approximately 3.98 x 10⁻⁸ m.
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Estimate the difference in hydrostatic pressure from the weight of fluid in the body between the shoulder and the ankle. Does the weight of the person enter the calculations? Justify your answer. Assume the fluid is static and the density of blood is 1.056 g/cm3.•Estimate the difference in hydrostatic pressure from the weight of fluid in the body between the shoulder and the ankle. Does the weight of the person enter the calculations? Justify your answer. Assume the fluid is static and the density of blood is 1.056 g/cm3.
To estimate the difference in hydrostatic pressure between the shoulder and the ankle, we need to consider the weight of the fluid in the body.
Hydrostatic pressure is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height or depth of the fluid column.
In this case, we can assume that the fluid is static and the density of blood is 1.056 g/cm³. The difference in hydrostatic pressure between the shoulder and the ankle is then determined by the difference in height between the two points.
However, the weight of the person does not directly enter the calculations for hydrostatic pressure. The hydrostatic pressure is solely determined by the height or depth of the fluid column and the density of the fluid. The weight of the person is already accounted for in the density of the blood, which represents the mass per unit volume of the fluid.
Therefore, in estimating the difference in hydrostatic pressure between the shoulder and the ankle, we do not need to consider the weight of the person separately as it is already incorporated in the density of the blood.
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Anlsysis of a given fuel has a equivalent molar composition of C₆.₂H₁₅O₈.₇ Determine the mass of air required for stoichiometric combustion with 1 kg of the fuel The mass of air, to 1 decimal place, required for stoiciometric combustion is: A gas analyser connected to a combustion system combusting the fuel above has the following gas concentrations: Percentatge of Carbon Dioxide: 20.4 % Percentage of Oxygen: 2.2 % Calcualte the air-to-fuel equivalence ratio (λ). The air-to-fuel equivalence raito, to 2 decimal places, is:
To determine the mass of air required for stoichiometric combustion with 1 kg of the given fuel and the air-to-fuel equivalence ratio (λ), we need to consider the molar composition of the fuel and the gas concentrations from the gas analyzer. The mass of air required 12.096 g
First, let's calculate the molecular weight of the fuel:
Molecular weight of C6.2H15O8.7 = (6.2 * 12.01) + (15 * 1.01) + (8.7 * 16.00) = 104.56 + 15.15 + 139.20 = 258.91 g/mol
To achieve stoichiometric combustion, we need the carbon and hydrogen in the fuel to react with the correct amount of oxygen from the air. The balanced equation for combustion of hydrocarbon fuel can be represented as follows:
C6.2H15O8.7 + a(O2 + 3.76N2) -> bCO2 + cH2O + dO2 + eN2
From the equation, we can determine the stoichiometric coefficients: b = 6.2, c = 7.5, d = a, e = 3.76a.
To calculate the mass of air required, we need to compare the moles of fuel and oxygen in the balanced equation. The moles of fuel can be calculated by dividing the mass of the fuel (1 kg) by the molecular weight of the fuel:
Moles of fuel = Mass of fuel / Molecular weight of fuel = 1000 g / 258.91 g/mol = 3.864 mol
Since the stoichiometric coefficient of oxygen is a, the moles of oxygen required will also be a. Therefore, the mass of air required will be a times the molecular weight of oxygen (32 g/mol).
Now, let's calculate the air-to-fuel equivalence ratio (λ):
Percentage of Oxygen in flue gas = (Moles of oxygen / Total moles) * 100
Percentage of Oxygen = 2.2
Therefore, (a / (a + 3.76a)) * 100 = 2.2
Solving for a, we find a ≈ 0.378
The mass of air required for stoichiometric combustion can be calculated as follows:
Mass of air = a * (Molecular weight of oxygen) = 0.378 * 32 = 12.096 g
Finally, the air-to-fuel equivalence ratio (λ) is the ratio of actual air supplied to stoichiometric air required:
λ = Mass of air supplied / Mass of air required = (Mass of air supplied) / 12.096
Note: The actual mass of air supplied is not provided in the given information, so it is not possible to calculate the exact value of λ without that information.
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A steam turbine has an inlet condition of 10 mPa at 800 C. The
turbine exhausts to a pressure of 20 kPa. The exit is saturated
vapor. What is the isentropic efficiency?
The isentropic efficiency of the steam turbine is approximately 80.3%.
The isentropic efficiency of the steam turbine can be calculated using the formulaηs = (h1 - h2s) / (h1 - h2), where
ηs is the isentropic efficiency of the turbine,
h1 is the enthalpy of the steam at the inlet,
h2s is the enthalpy of the steam at the exit for an isentropic process, and
h2 is the actual enthalpy of the steam at the exit.
Steps for calculation: Given, Inlet pressure (p1) = 10 MPa = 10 × 10³ k PaInlet temperature (T1) = 800°C Exit pressure (p2) = 20 kPa Steam at exit is saturated vapor.
Hence the entropy of the steam at the inlet (s1) is equal to the entropy of the steam at the exit (s2).
We know that h1 = h2 + v2(p1 - p2) and s1 = s2 for an isentropic process.
Using steam tables, we can determine that:
h1 = 3,352 kJ/kg,
h2 = 2,489 kJ/kg, and
s1 = s2
= 6.871 kJ/kg·K.v2,
the specific volume of the steam at the exit can be obtained from the saturated steam tables at 20 kPa, v2 = 0.1947 m³/kg.
Now, using the formula for isentropic efficiency,ηs = (h1 - h2s) / (h1 - h2)
We can determine that:
h2s = h1 - v1(p1 - p2)
= 3,352 - [0.1607 × (10,000 - 20)]
= 2,090.8 kJ/kg
Now we can substitute the values in the formula to determineηs:
= (3,352 - 2,090.8) / (3,352 - 2,489)
= 0.803 ≈ 80.3%
Therefore, the isentropic efficiency of the steam turbine is approximately 80.3%.
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A channel has a B.E.R. of 5 X 10^-3. We are sending 8 bit words on this channel. Find: a) Probability of a correct word. b.) Probability of 2 bits wrong in a word. 2. An even parity bit is added to the above system. (Total of 9 bits) If we fail parity the word is rejected and the transmitter is signaled to resend the word. a) Find the exact expression for the probability of a request for retransmission? b) What is the aprox. Value?
The approximate value for the probability of a request for retransmission is approximately 5.74 × 10⁻⁹.
(a) Probability of a request for retransmission
When an even parity bit is added to the system, the total number of bits becomes 9. Therefore, the total number of ways that the 9 bits can be arranged is 2⁹.The probability that the parity check will fail is equal to the probability of an odd number of bits in the 9-bit word being in error.
This probability can be found using the formula,P (fail) = (n/2) p (1 - p)ⁿ⁻¹ + ... + (n/2) p (1 - p)ⁿ⁻¹ = ∑ (⁹Cᵢ) pⁱ (1 - p)⁹⁻ⁱ, where i = 1,3,5,7,9
(b) We can use the normal distribution to find an approximate value for the probability of a request for retransmission. The mean of the distribution is equal to the probability of a failed parity check, which is equal to 0.000788 (found in part 2).
The standard deviation of the distribution can be found using the formula,σ = √[np(1 - p)] = √[9 × 5 × 10⁻³ (1 - 5 × 10⁻³)] = 0.084
Approximate value = P (z > z₀) = P (z > (0.5 - 0.000788)/0.084) = P (z > 5.846) = 5.74 × 10⁻⁹ (approx)
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Suppose diffusion coefficient of Boron in Si at 1200 °C is, = 1.4*10^-12 cm2/s. How long (min) will it take to make an emitter of 1.5 micron thick, having uniform doping concentration as that of the chamber phosphorus concentration which is 10^17 cm^-3? (Points 10) Ans. (i) 47 (ii) 67 (iii) 87 (iv) 107 (V) 117
Diffusion coefficient of Boron in Si at 1200 °C is, = 1.4×10^-12 cm2/s. 107, long (min) will it take to make an emitter of 1.5 micron thick, having uniform doping concentration as that of the chamber phosphorus concentration. Thus, option (d) is correct.
t = ([tex]x^2[/tex]) / (2D)
where t is the required amount of time, x is the emitter's thickness, and D is the coefficient rate of boron in silicon.
Given that the emitter is 1.5 microns thick and that boron diffuses at a rate of 1.4 1012 cm2/s in silicon at 1200 °C,
we can calculate the necessary time as follows:
t = ([tex]1.5^2[/tex] /([tex]21.410^{-12}[/tex] = 107 seconds
Therefore, option (d) is correct.
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Determine the force required to punch a ^1/2 inch hole on a 3/8 thick plate if the ultimate shearing strength of the plate is 50,000psi& factor of safety of 1.50. o 20,550lbs o 23,562lbs o 15,422lbs o 19,450lbs
To determine the force required to punch a 1/2 inch hole in a 3/8 inch thick plate, we need to consider the shear strength of the plate and apply a factor of safety.
The shear strength is given as 50,000 psi, and the factor of safety is 1.50. To calculate the force, we can use the formula: Force = Shear strength * Area First, we need to calculate the area of the hole. The area of a 1/2 inch hole can be determined as: Area = π * (Diameter/2)^2 ,Area = π * (1/2)^2 = π * 1/4 = π/4 square inches. Next, we can calculate the force required: Force = Shear strength * Area
Force = 50,000 psi * π/4 square inches
Using the value of π (approximately 3.14159), we can calculate the force:
Force ≈ 50,000 psi * 3.14159/4 square inches
Force ≈ 39,269.91 lbs
Considering the factor of safety of 1.50, we multiply the force by the factor of safety: Force with factor of safety = Force * Factor of safety
Force with factor of safety ≈ 39,269.91 lbs * 1.50
Force with factor of safety ≈ 58,904.87 lbs
Therefore, the force required to punch a 1/2 inch hole in a 3/8 inch thick plate, considering the shear strength and a factor of safety of 1.50, is approximately 58,904.87 lbs.
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Assume a 4800 nT/min geomagnetic storm disturbance hit the United States. You are tasked with estimating the economic damage resulting from the storm. a. If there were no power outages, how much impact (in dollars) would there be in the United States just from the "value of lost load?" Explain the assumptions you are making in your estimate. [ If you are stuck, you can assume 200 GW of lost load for 10 hours and a "value of lost load" of $7,500 per MWh.] b. If two large power grids collapse and 130 million people are without power for 2 months, how much economic impact would that cause to the United States? Explain the assumptions you are making in your estimate.
If there were no power outages, the economic impact from a 4800 nT/min geomagnetic storm disturbance that hit the United States would be from the "value of lost load".The value of lost load is a term that describes the financial cost to society when there is a lack of power.
The assumptions that are being made are as follows: The power loss is due to the storm disturbance. It is assumed that 200 GW of power were lost for 10 hours at a value of lost load of $7,500 per MWh. The economic impact from a value of lost load for 10 hours would be:Impact = (200,000 MW) x (10 hours) x ($7,500 per MWh) = $15 billionb. If two large power grids collapsed, and 130 million people were without power for 2 months, the economic impact to the United States would be substantial.The assumptions that are being made are as follows: The power loss is due to the storm disturbance. It is assumed that two power grids collapsed, and 130 million people were without power for two months.
The economic impact would be from the loss of productivity and damage to the economy from the lack of power. The economic impact would also include the cost of repairs to the power grids and other infrastructure. Some estimates have put the economic impact at over $1 trillion.
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A 3-phase, 10-kVA, 400-V, 50-Hz, Y-connected alternator supplies the rated load at 0.8 p.f. lag. If armature resistance is 0.5 ohm and synchronous reactance is 10 ohms, find the power angle and voltage regulation.
The power angle is approximately 16.68 degrees and the voltage regulation is approximately 8.09%.
To find the power angle and voltage regulation of the given alternator, we can use the per-unit system and the given parameters.
Step 1: Convert the apparent power from kVA to VA:
S = 10 kVA = 10,000 VA
Step 2: Calculate the rated current:
I = S / (√3 * V) = 10,000 / (√3 * 400) = 14.43 A
Step 3: Calculate the impedance angle:
θ = arccos(pf) = arccos(0.8) = 36.87 degrees
Step 4: Calculate the synchronous reactance voltage drop:
Vx = I * Xs = 14.43 * 10 = 144.3 V
Step 5: Calculate the armature resistance voltage drop:
VR = I * R = 14.43 * 0.5 = 7.215 V
Step 6: Calculate the internal generated voltage:
E = V + jVR + jVx = 400 + j7.215 + j144.3 = 400 + j151.515 V
Step 7: Calculate the magnitude of the internal generated voltage:
|E| = √(Re(E)^2 + Im(E)^2) = √(400^2 + 151.515^2) = 432.36 V
Step 8: Calculate the power angle:
θp = arccos(Re(E) / |E|) = arccos(400 / 432.36) = 16.68 degrees
Step 9: Calculate the voltage regulation:
VR = (|E| - V) / V * 100% = (432.36 - 400) / 400 * 100% = 8.09%
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constraint 1: the axes of driver and driven shafts are inclined to one another and intersect when produced
constraint 2: the driving and driven shafts have their axes at right angles and are non co planar.
name the best possible gear system that the engineer should choose to overcome each constrain seperately and explain its characteristics with sketch
The two given constraints can be overcome using the following gear systems.
What are the systems?1. Bevel gear: When the axes of the driver and driven shafts are inclined to each other and intersect when produced, the best possible gear system is the bevel gear.
The teeth of bevel gears are cut on conical surfaces, allowing them to transmit power and motion between shafts that are mounted at an angle to one another.
2. Worm gear: When the driving and driven shafts have their axes at right angles and are non-coplanar, a worm gear can be used to overcome this constraint. Worm gear systems, also known as worm drives, consist of a worm and a worm wheel.
Characteristics of Bevel gear :The pitch angle of a bevel gear is a critical parameter.
The pitch angle of the bevel gears is determined by the angle of intersection of their axes.
When the gearset is being used to transfer power from one shaft to another at an angle, the pitch angle is critical since it influences the gear ratio and torque transmission.
The pitch surfaces of bevel gears are conical surfaces, which makes them less efficient than spur and helical gears.
Characteristics of Worm gear: Worm gearsets are very useful when a high reduction ratio is required.
The friction between the worm and the worm wheel is the primary disadvantage of worm gearsets.
As a result, they are best suited for low-speed applications where torque multiplication is critical.
They are also self-locking and cannot be reversed, making them ideal for use in applications where the output shaft must be kept in a fixed position.
When the worm gearset is run in the opposite direction, it causes the worm to move axially, which can result in damage to the gear teeth.
For these reasons, they are not recommended for applications that require frequent direction changes. See the attached figure for the illustration.
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(a) Explain the need for a vehicle's gearbox to provide a number of [3] transmission ratios. (b) Explain the difference between traction limited acceleration and [3] 3 engine limited acceleration. (c) A front wheel drive car weighs 1200 kg and has a wheelbase of 2.5 m. The centre of gravity of the car is 0.5 m above ground level and 1.15 m from the front axle. Determine the static load distribution of the car on level ground. [2] (ii) Determine load distribution when the car is given a forward [2] acceleration of 5 m/s? on level ground. (ii) The owner of this car lives at the bottom of a road which has a [3] 3 gradient of 1 in 10. Determine the minimum tyre-road frictional coefficient needed if he is to be able to drive his car up the road on a winter morning when the road is icy. (iv) What is the maximum velocity that the car achieve on a level road this[3] 3 winter morning if the drag force on it is given by kV2 where k=1.2 Ns?/m2? (Assume the frictional coefficient determined in iii). (d) Two suitcases each weighing 25 kg are added to the boot of the car, [4] 4 shifting the centre of gravity to 1.2 m from the front axle. Calculate the new frictional coefficient needed for the car to drive up the road.
(a) The need for a vehicle's gearbox to provide a number of transmission ratios is to allow the engine to operate efficiently across a range of speeds and loads.
Different driving conditions require different torque and speed combinations. By having multiple transmission ratios, the gearbox can match the engine's power output to the desired speed and load requirements. Lower gear ratios provide higher torque at lower speeds, which is useful for starting the vehicle or climbing steep inclines. Higher gear ratios provide higher speeds at lower engine RPM, which is efficient for cruising on highways. The ability to change gears allows the engine to operate within its optimal power range, maximizing fuel efficiency and performance.
(b) Traction-limited acceleration refers to the situation where the maximum acceleration of a vehicle is limited by the available traction between the tires and the road surface. If the tires cannot grip the road well enough, they may slip or spin, resulting in reduced acceleration. This can occur in situations such as driving on a slippery surface or applying excessive throttle.
Engine-limited acceleration, on the other hand, refers to the situation where the maximum acceleration is limited by the engine's power output. In this case, even if the tires have sufficient traction, the engine may not be able to produce enough torque to accelerate the vehicle at a faster rate. This can occur when the engine is not powerful enough or when it is operating at its maximum capacity.
(c) (i) To determine the static load distribution of the car on level ground, we can consider the weight distribution based on the position of the center of gravity and the wheelbase.
The weight distribution on the front axle can be calculated using the moment equilibrium:
Front axle load = (CG to front axle distance / wheelbase) * total weight
Front axle load = (1.15 m / 2.5 m) * 1200 kg = 552 kg
The weight distribution on the rear axle can be calculated by subtracting the front axle load from the total weight:
Rear axle load = Total weight - Front axle load
Rear axle load = 1200 kg - 552 kg = 648 kg
(ii) When the car is given a forward acceleration of 5 m/s² on level ground, the load distribution will change. The weight will shift to the rear due to the acceleration force. Assuming the weight transfer is distributed evenly between the front and rear axles, the load distribution can be calculated as:
Front axle load = Front axle load - (acceleration force / total weight) * front axle load
Front axle load = 552 kg - (5 m/s² / 9.81 m/s²) * 552 kg = 286 kg
Rear axle load = Total weight - Front axle load
Rear axle load = 1200 kg - 286 kg = 914 kg
(iii) To drive the car up a road with a gradient of 1 in 10 on a winter morning when the road is icy, the minimum tire-road frictional coefficient needed can be determined by considering the force required to overcome the gradient. The minimum coefficient of friction can be calculated as:
Coefficient of friction = 1 / (1 + gradient)
Coefficient of friction = 1 / (1 + 1/10) = 0.909
(iv) The maximum velocity that the car can achieve on a level road on a winter morning with a drag force given by kV² (where k = 1.2 Ns²/m²) can be determined by balancing the driving force and the drag force:
Driving force = Total weight * coefficient of friction
Driving force = 1200 kg * 9.81 m/s² * 0.909 = 10,900 N
Drag force = k * V²
10,900 N
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You are to design a pair of spur gears with a 8:3 speed
reduction ratio. With a module of 4 mm, pressure angle of 20deg, 12
teeth for the pinion, and a standard addendum, will there be
interference?
Since the calculated minimum number of teeth (46) is higher than the actual number of teeth for the gear (32), there will be interference between the gears.
To determine whether there will be interference between the gears, we need to check if the gears' teeth profiles will intersect or overlap.
Given:
Speed reduction ratio: 8:3
Module: 4 mm
Pressure angle: 20 degrees
Number of teeth for the pinion: 12
Standard addendum
First, we need to calculate the number of teeth for the gear. Since we have a speed reduction ratio of 8:3, the number of teeth for the gear can be calculated as follows:
Number of teeth for the gear = (Number of teeth for the pinion) × (Speed reduction ratio)
Number of teeth for the gear = 12 × (8/3)
Number of teeth for the gear ≈ 32
Now, we can check for interference by calculating the minimum number of teeth required for the gears to avoid interference. The minimum number of teeth can be calculated using the following formula:
Minimum number of teeth = (2 × Module) / sin(pressure angle)
Minimum number of teeth = (2 × 4) / sin(20 degrees)
Minimum number of teeth ≈ 46
The gear with 32 teeth does not have enough teeth to avoid interference. To prevent interference, you would need to increase the number of teeth for the gear or adjust the design parameters accordingly.
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Problem 4 (3 points) - Black body radiation: At a wavelength of 0.7 mm, the black body emissive power is equal to 108W/m². Determine the total emissive power at 1335 K.
The total emissive power at 1335 K is 1.062 x 10^5 W/m².
To determine the total emissive power at 1335 K, we can use the Stefan-Boltzmann Law, which relates the total emissive power of a black body to its temperature. The Stefan-Boltzmann Law is given by the equation:
E = σ * T^4
Where:
E is the total emissive power (in W/m²)
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m²K^4)
T is the temperature of the black body (in Kelvin)
Given that the emissive power at a wavelength of 0.7 mm (or 0.0007 m) is 108 W/m², we can calculate the temperature using Wien's displacement law, which relates the peak wavelength of the black body radiation to its temperature. Wien's displacement law is given by the equation:
λ_max = b / T
Where:
λ_max is the peak wavelength (in meters)
b is Wien's displacement constant (2.898 x 10^-3 mK)
Solving for T, we have:
T = b / λ_max
Substituting the values, we find:
T = (2.898 x 10^-3 mK) / (0.0007 m)
≈ 4139.43 K
Now that we know the temperature, we can calculate the total emissive power at 1335 K using the Stefan-Boltzmann Law:
E = (5.67 x 10^-8 W/m²K^4) * (1335 K)^4
≈ 1.062 x 10^5 W/m²
The total emissive power at a temperature of 1335 K is approximately 1.062 x 10^5 W/m².
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Question 1. Write the full set of Maxwell's equations in differential form with a brief explanation for the case of: (iv) a steady current flow in an inhomogeneous poor dielectric, with impressed electric field Ei present.
1. Gauss's law for electric field : ∇. E = ρ/ε₀Here, E is electric field, ρ is charge density, and ε₀ is the permittivity of free space.
2. Gauss's law for magnetic field : ∇. B = 0Here, B is magnetic field.
3. Faraday's law of electromagnetic induction : ∇ x E = -dB/dt Here, x denotes the vector cross product, E is electric field, B is magnetic field, and t is time.
4. Ampere's circuital law : ∇ x B = μ₀ j + μ₀ε₀(dE/dt)Here, j is the current density, μ₀ is the permeability of free space, and μ₀ε₀(dE/dt) is the displacement current density. If the current is steady and there is an impressed electric field Ei, then j is zero and the displacement current is equal to zero. Therefore, the fourth equation becomes:
∇ x B = μ₀ j For an inhomogeneous poor dielectric, the permittivity is not constant and it can be written as ε = ε₀(1 + χ), where χ is the susceptibility. The full set of Maxwell's equations in differential form for the case of a steady current flow in an inhomogeneous poor dielectric, with impressed electric field Ei present are:
∇. E = ρ/ε∇. B = 0∇ x E
= -dB/dt∇ x B = μ₀ j + μ₀ε₀(dE/dt)
= μ₀(j + ε₀∂E/∂t)
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Task 3 Superheated steam at 575°C is routed from a boiler to the turbine of an electric power plant through steel tubes (k = 35 W/m K) of 300 mm inner diameter and 30 mm wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation (k = 0.10 W/m K) is applied to the tubes, while degradation of the insulation is reduced by degradation of the insulation is reduced by wrapping it in a thin sheet of aluminium having an emissivity of = 0.20. The air and wall temperatures of the power plant are 27°C. Assume that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminium sheet is 6 W/m²K, (a) What is the minimum insulation thickness needed to ensure that the temperature of the aluminium does not exceed 50°C? (b) What is the corresponding heat loss per unit meter? (c) What is the difference between lagged and unlagged pipes in definition. In addition, write down the heat transfer formula for each pipe (i.e., lagged, and unlagged pipes).
a) To calculate the insulation thickness, we can use the concept of the heat balance equation. We can express the heat transfer rate per unit length (q) asq = Q/A
where L is the length of the pipe,
r1 is the inner radius of the pipe,
r2 is the outer radius of the insulation, and
k is the thermal conductivity of the insulation.
Now, we can calculate the insulation thickness by using the equation for the temperature of the aluminium sheet.
Ts - Ta = (hA/k) (Tal - Ts)
Tal = Ts + (Ts - Ta)(k/hA)
Tal = 50°C (given)
Ts = 50°C + (50°C - 27°C)(0.10/6)
Ts = 50.45°C
Let's assume that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminium sheet is 6 W/m²K.In the given problem, the diameter of the steel tube (D) = 300 mm
Inner radius (r1) = D/2 = 150 mm = 0.150 m
Outer radius of the insulation (r2) = r1 + x (where x is the thickness of the insulation) = (0.150 + x) m
Cross-sectional area of the pipe
(A) = π(r2² - r1²)
(A) = π[(0.150 + x)² - (0.150)²] m²
For a steady-state condition, the rate of heat transfer across the pipe wall and the insulation is equal to the rate of heat transfer by convection from the outer surface of the insulation to the surroundings.
Hence,
q = hA(Ts - Ta)Q/(2πLk) ln(r2/r1)
q = hπ[(0.150 + x)² - (0.150)²][50.45 - 27]x
q = 0.065 m or 65 mm,
The minimum insulation thickness needed to ensure that the temperature of the aluminium does not exceed 50°C is 65 mm.
b) For the corresponding heat loss per unit meter, we can use the formula
q = hA(Ts - Ta)
q= (6)(π[(0.150 + 0.065)² - (0.150)²])(50.45 - 27)
q = 47.27 W/m,
The corresponding heat loss per unit meter is 47.27 W/m.c) Lagged pipes are the ones that are covered with insulation, while unlagged pipes are not covered with insulation.
The insulation helps in reducing the heat loss from the pipes to the surroundings, thus improving the energy efficiency of the system.
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