1. The pre-warping frequency is 0.6467 rad/s.
2. The analog transfer function is $H_a(s) = 1/[1 + (2.586z - 1.586)/0.6467]$.
3. Bilinear transformation is linear.
4. The transfer function $H_a(s)$ is stable since the pole is located inside the unit circle.
1. Prewarp analog frequency:
The prewarp frequency of a digital filter is that frequency that it corresponds to when the filter is represented in the analog domain.
In the bilinear transformation, pre-warping is used to minimize the frequency distortion caused by the mapping from the s-plane to the z-plane.
The pre-warping frequency is defined as follows:
[tex]$$\Omega_a=\frac{2}{T}\tan\Big(\frac{\Omega_d T}{2}\Big)$$[/tex]
Given that the digital lowpass filter has a 3dB cutoff frequency of [tex]$\Omega_c=0.25$[/tex], the prewarp analog frequency is:
[tex]$$\Omega_a=2\tan\Big(\frac{0.25\cdot\pi}{2}\Big)[/tex]
[tex]=0.6467\text{ rad/s}$$[/tex]
2. Analog Transfer Function:
The analog Butterworth filter is represented by the transfer function
[tex]$H_0(s) = 1/(1+s/\Omega)$[/tex],
where [tex]$\Omega$[/tex] is the filter's corner frequency.
To begin with, the transfer function has to be pre-warp by substituting [tex]$s$[/tex] with [tex]$(2/T) \cdot (z-1)/(z+1)$[/tex], where [tex]$T$[/tex] is the sampling period, which is given by:
[tex]$$T=\frac{1}{2\cdot\Omega_a}$$[/tex]
[tex]$$=\frac{1}{2\cdot 0.6467}$$[/tex]
[tex]$$=0.7739\text{ s}$$[/tex]
Substituting,
[tex]$$s=\frac{2}{0.7739}\cdot\frac{z-1}{z+1}$$[/tex]
[tex]$$=2.586z-1.586$$[/tex]
Therefore, the analog transfer function is:
[tex]$$H_a(s)=H_0(2.586z-1.586)$$[/tex]
[tex]$$=\frac{1}{1+(2.586z-1.586)/\Omega}$$[/tex]
Substituting [tex]$\Omega$[/tex] with [tex]$\Omega_a$[/tex],
we get:
[tex]$$H_a(s)=\frac{1}{1+(2.586z-1.586)/0.6467}$$[/tex]
3. Linearity of Bilinear Transformation:
The bilinear transformation is a linear method of mapping an analog filter design to a digital implementation.
As a result, the transformation preserves the linearity of the analog filter.
4. Stability of Ha(s):
To determine the stability of [tex]$H_a(s)$[/tex], we need to determine the poles of the transfer function.
The poles are the values of [tex]$s$[/tex]that cause [tex]$H_a(s)$[/tex]to be infinite.
Therefore, we set the denominator of the transfer function to zero and solve for [tex]$s$[/tex].
[tex]$$1+\frac{2.586z-1.586}{0.6467}=0$$[/tex]
[tex]$$z=-0.5859$$[/tex]
Therefore, the pole is located at [tex]$z=-0.5859$[/tex], which is inside the unit circle.
As a result, [tex]$H_a(s)$[/tex]is stable.
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A rectangular channel discharges water at the rate of 4.5 cu.m./s at a depth of 32 cm. The flume is 3.5 m wide. What is the depth of the jump? Select one: O a. 83.9 cm O b. 87.9 cm O c. 85.9 cm O d. 81.9 cm
The rectangular channel discharges water at the rate of 4.5 cu.m./s at a depth of 32 cm. The flume is 3.5 m wide. We have to find out the depth of the jump. The correct option among the given options is (b) 87.9 cm.
The critical depth in a rectangular channel is given by;
[tex]$$y_c=\frac{Q^2}{gBW^2}$$[/tex]
Where,Q = Discharge, B = Width of the channel, W = Hydraulic depth, y = depth of flow of water.Let us calculate all the given parameters and then find the depth of the jump.Q = 4.5 cu.m./sWidth of the channel, B = 3.5 mDepth of flow of water, y = 32 cm = 0.32 mHydraulic Depth, [tex]W = (3.5 x 0.32) / (3.5 + 2 x 0.32) = 0.224[/tex]
Now, putting all the values in the critical depth formula;
[tex]$$y_c=\frac{Q^2}{gBW^2}$$$$y_c=\frac{(4.5)^2}{9.81 \times 3.5 \times 0.224^2}$$$$y_c = 0.879 m$$$$y_c= 87.9 cm$$[/tex]
Therefore, the correct option among the given options is (b) 87.9 cm.
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A gas mixture, comprised of 3 component gases, methane, butane and ethane, has mixture properties of 5 bar, 60°C, and 0.5 m³. If the partial pressure of ethane is 140 kPa and considering ideal gas model, what is the mass of ethane in the mixture? Express your answer in kg.
The mass of ethane in the mixture is 0.000765 kg.
Partial pressure of ethane = 140 kPa
Temperature = 60°C = 333.15 K
Volume = 0.5 m³
Total pressure (mixture properties) = 5 bar = 500 kPa
We can use the ideal gas law, PV = nRT To calculate the number of moles of ethane in the mixture, we can rearrange the equation as follows:
n = PV/RT where,
P = partial pressure of ethane
V = volume
T = temperature
R = gas constant
For ethane,
nE = PEV/RT
Using the above values,
nE = (140 x 0.5)/(8.314 x 333.15) = 0.0255 moles of ethane
Now, to find the mass of ethane, we need to multiply the number of moles by its molar mass. The molar mass of ethane (C2H6) is 30 g/mol.
mass of ethane
= number of moles × molar mass
= 0.0255 × 30
= 0.765 g or 0.000765 kg
Therefore, the mass of ethane in the mixture is 0.000765 kg.
The partial pressure of ethane is 140 kPa and considering the ideal gas model, the mass of ethane in the mixture can be found as follows.
First, use the ideal gas law PV = nRT to calculate the number of moles of ethane in the mixture. The formula can be rearranged as
n = PV/RT.
Using the values given in the problem, we find
nE = (140 x 0.5)/(8.314 x 333.15)
= 0.0255 moles of ethane.
To find the mass of ethane, multiply the number of moles by its molar mass.
The molar mass of ethane (C2H6) is 30 g/mol.
Therefore, the mass of ethane in the mixture is 0.0255 × 30 = 0.765 g or 0.000765 kg.
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Consider a flat rectangular plate of known mass, width and breadth with a negligible thickness that lies in the horizontal xy-plane. The plate is suspended from a thin piece of piano wire that is in the vertical orientation coincident to the z-axis and where the piano wire is attached to the center of the plate. When the plate is subjected to a torque whose vector is coincident to the z-axis, the plate rotates in the horizontal plane such that the rotation of the plate is modelled as θ = Csin(wt + Ø). The parameter information is: mass of plate M = 1.2 kilogram width of plate W = 0.040 meter breadth of plate B = 0.075 meter shear modulus of piano wire G = 79.3 gigaPascals diameter of piano wire D = 0.003 meter length of piano wire L = 0.120 meter amplitude of rotation C = 0.087267520415 radian phase lag of rotation = 1.565872597159 radian Using the supplied information and any appropriate assumptions and / or approximations, write a GNU Octave computer program to determine the following; 1) the mass moment of inertia I 2) the natural angular frequency wn 3) the initial angular displacement θ₀ 4) the initial angular velocity θ₀
The mass moment of inertia (I) for the rectangular plate is (1/12) * M * (W^2 + B^2), the natural angular frequency (wn) is sqrt(G / (I / L)), the initial angular displacement (θ₀) is the given amplitude of rotation (C), and the initial angular velocity (θ'₀) is C * w * cos(Ø) where w represents the angular frequency.
What are the formulas to calculate the mass moment of inertia (I), natural angular frequency (wn), initial angular displacement (θ₀), and initial angular velocity (θ'₀) for a rectangular plate suspended by a piano wire, given the relevant parameters?The mathematical equations and steps to determine the quantities you mentioned using the supplied information.
1) The mass moment of inertia (I) of the rectangular plate can be calculated using the formula: I = (1/12) * M * (W^2 + B^2).
2) The natural angular frequency (wn) can be calculated using the equation: wn = sqrt(G / (I / L)).
3) The initial angular displacement (θ₀) is given as the amplitude of rotation (C) in this case.
4) The initial angular velocity (θ'₀) can be calculated by taking the derivative of the rotation equation with respect to time (t) and evaluating it at t = 0. Differentiating θ = C * sin(wt + Ø) with respect to t gives θ' = C * w * cos(wt + Ø), and θ'₀ = C * w * cos(Ø).
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Course: Structure Repair(Aircraft)
1.For structure repair, the lowest allowable load is the most critical. Which category of load out of these four, bearing, shear, tear-out and tension is anticipated to be critical? 2. Why it is need the above allowable load to be most critical? Explain.
1. In aircraft's structure repair, the tear-out load is anticipated to be the critical load since it is usually the lowest allowable load.2. The tear-out load is critical because the bearing load and shear load both depend on it, and if there is no consideration for the tear-out load, they would be useless.
In aircraft structure repair, the tear-out load is usually the lowest load that is allowable. This is because the tear-out load is the weakest link in the chain when it comes to bolted joints. For that reason, it must be the most critical load.In engineering, bearing load refers to the load supported by the fastener itself, while shear load refers to the load perpendicular to the fastener's axis.
The tear-out load is the load necessary to cause the section around the fastener hole to tear out. The bearing and shear loads both depend on the tear-out load. This is why tear-out load must be taken into account first, since if there is no consideration for tear-out load, the bearing and shear loads would be meaningless.
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Question 7 [2] Given: A, B. Two phasors are shown below: V₁ = 8 cos (wt - A°) i2 12 = 10 sin (wt - Bº) (1) By how many degrees is i2 leading V₁? (Give your answer in the range from -180° to 180°) 07 0 [2]
Given: A, B. Two phasors are shown below:V1 = 8 cos (wt - A°)I2 = 10 sin (wt - Bº)(Give your answer in the range from -180° to 180°)The angle between the two phasors is given byΘ = Θi2 - Θv1Θ = -B - (-A)Θ = A - B.
When the phase angle of V1 is subtracted from the phase angle of I2, we get the phase angle by which I2 leads V1.The phase angle by which I2 leads V1 is Θ = A - B. Therefore, the answer is given in degrees as A - B.Answer: The answer is given in degrees as A - B.
Since the question does not provide the values of A and B, it is not possible to calculate the exact answer.
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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: (i) a static electric field, assuming that the dielectric is linear, but inhomogeneous;
Maxwell's equations in differential form are a set of partial differential equations that describe how electric and magnetic fields interact and propagate through space. The equations for the case of a static electric field, assuming that the dielectric is linear but inhomogeneous, are given as follows:Gauss's Law:∇⋅D=ρv Gauss's Law for magnetism:∇⋅B=0Faraday's Law:∇×E=−∂B/∂tAmpere's Law with Maxwell's correction:∇×H=Jv+∂D/∂
Here, D is the electric displacement field, which is related to the electric field E and the polarization P of the dielectric material by the equation D = εE + P, where ε is the permittivity of the material. B is the magnetic field, H is the magnetic field intensity, Jv is the free current density, and ρv is the free charge density.
The inhomogeneity of the dielectric material can be taken into account by including the spatial variation of ε and P in the equations.Overall, these equations provide a mathematical framework for understanding the behavior of electric and magnetic fields in a variety of situations, including the case of a static electric field in an inhomogeneous dielectric material.
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Q: Find the actual address for the following instruction assume X= (32) hex and Rindex=D4C9 LOAD X(Ri), A , address=?
O address=D41B O address D4F2 O address=D517 O address=D4FB O address=D4BF O address=D4E1
The actual address for the instruction "LOAD X(Ri), A" can be found by adding the hexadecimal value of X, which is 32, to the hexadecimal value of Rindex, which is D4C9. Performing the addition yields D4FB as the result.
Hence, the actual address for the instruction is O address = D4FB. This means that the contents of memory location D4FB will be loaded into register A.
It's important to note that the values provided in hexadecimal format are being added together. The hexadecimal system is a base-16 numbering system that uses digits from 0 to 9 and letters from A to F to represent values. When adding hexadecimal values, the addition is performed digit by digit, taking into account any carryovers.
In this case, the sum of 32 and D4C9 results in D4FB. Therefore, option D, D4FB, represents the correct actual address for the given instruction.
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Please answer asap
Question 5 6 pts Warm water enters a cooling tower at 36°C at a mass flow rate of 57.1 kg/s. The air entering at state 1 has h₁ = 45.2 kJ/kg da and W₁ = 0.006 kg v/kg da. The air leaving the cooling tower at state 2 has h₂ = 103.4 kJ/kg da and w₂ = 0.029 kg v/kg da. The make up water is supplied at 25°C and the mass flow rate of dry air is 45.1 kg da/s. What is the temperature of the cooled water leaving the tower? Express your answer in °C.
The temperature of the cooled water leaving the tower is approximately 37.95°C.
To find the temperature of the cooled water leaving the tower, we need to use the energy balance equation for the cooling tower:
Q = mₕ(h₂ - h₁) + mₐ(w₂ - w₁)
where Q is the heat transferred, mₕ is the mass flow rate of hot water, mₐ is the mass flow rate of dry air, h₁ and h₂ are the enthalpies of the air entering and leaving the cooling tower respectively, and w₁ and w₂ are the specific volumes of the air entering and leaving the cooling tower respectively.
Given:
mₕ = 57.1 kg/s
h₁ = 45.2 kJ/kg da
h₂ = 103.4 kJ/kg da
w₁ = 0.006 kg v/kg da
w₂ = 0.029 kg v/kg da
mₐ = 45.1 kg da/s
Q = (57.1)(103.4 - 45.2) + (45.1)(0.029 - 0.006)
Q = 2434.92 kJ/s
Now, the heat transferred can be calculated using the equation:
Q = mₕCₕ(T₂ - T₁)
where Cₕ is the specific heat capacity of water.
Assuming that the specific heat capacity of water is 4.18 kJ/kg°C, we can rearrange the equation to solve for the temperature of the cooled water:
T₂ = (Q / (mₕCₕ)) + T₁
T₂ = (2434.92 / (57.1 * 4.18)) + 36
T₂ ≈ 37.95°C
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Overloading a single-phase motor will result in:
Select one:
a.no effects, the motor runs normally
b.overheating the motor
c.damaging the motor permanently
d.None
e.using a fan for cooling
Overloading a single-phase motor will result in overheating the motor.A single-phase motor is an electric motor that is powered by a single phase of electrical power.
Single-phase power is most commonly used in household and small commercial settings, such as for powering small appliances and lighting systems. Single-phase motors are used in a variety of applications, including fans, pumps, and compressors. They are also used in machinery and tools. it is being forced to work harder than it is designed to.
This can result in damage to the motor, as well as to any other equipment that is connected to it. Overloading a motor can cause it to overheat, which can lead to a variety of problems. In some cases, the motor may simply stop working. In other cases, it may begin to emit smoke or make unusual noises.When a single-phase motor is overloaded, it will begin to overheat.
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The input power to a device is 10,000 W at 1000 V. The output power is 500 W, and the output impedance is 100. Find the voltage gain in decibels. A) -30.01 dB B) -20.0 dB C) -13.01 dB D) -3.01 dB
The input power to a device is 10,000 W at 1000 V. The output power is 500 W, and the output impedance is 100. The voltage gain in decibels is approximately -3.01 dB.
1. Input power (Pin): The given input power is 10,000 W.
2. Output power (Pout): The given output power is 500 W.
3. Output impedance (Zout): The given output impedance is 100 ohms.
4. Voltage gain (Av): The voltage gain can be calculated using the formula Av = √(Pout / Pin) * √(Zout).
Substituting the given values:
Av = √(500 / 10,000) * √(100)
= √0.05 * 10
= √0.5
≈ 0.707
5. Converting voltage gain to decibels: The conversion from voltage gain to decibels can be done using the formula:
Gain (dB) = 20 * log10(Av)
Substituting the calculated value of Av:
Gain (dB) = 20 * log10(0.707)
≈ 20 * (-0.15)
≈ -3.01 dB
Therefore, the correct option is D.
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on GIVE REASONS FOR THE FOLLOWING (4X0.5=2 Marks) A. Mass of the body remains same at all the places. B. Surface tension is the property of liquids only not gases. C. The pressure at a point in a static fluid is independent of the shape of the container. D. Centre of gravity lies above the centre of pressure in a vertically submerged surface. Maximum file size: 250 MB, maximum number of files: 1 Files a You can drag and drop files here to add them.
When a body is vertically submerged in a fluid, the center of gravity is higher than the center of pressure.
A. Mass of the body remains same at all the places. The mass of the body is always the same, no matter where it is. This is due to the fact that mass is a scalar quantity that represents the amount of matter in an object. It does not alter with the object's location or environment.
B. Surface tension is the property of liquids only not gases. Surface tension is defined as the force per unit length acting perpendicular to an imaginary line drawn on the surface of the liquid in the plane of the surface. The property of surface tension is only present in liquids and not in gases due to their weak cohesive force. In gases, the molecules move randomly and are in constant motion, which means there is no specific surface area to which force can be applied.
C. The pressure at a point in a static fluid is independent of the shape of the container. The static fluid pressure at a particular point is determined by the liquid's density and the height of the fluid column above the point. The pressure at any given point in a fluid, as long as it is in a static state, is the same no matter the shape or size of the container.
The static pressure in a fluid can be calculated using the following equation: P = ρgh, where P is the static pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column above the point.
D. Centre of gravity lies above the centre of pressure in a vertically submerged surface. When a body is vertically submerged in a fluid, the centre of pressure is the point at which the total hydrostatic force on the body can be considered to act. The center of gravity of the body, on the other hand, is the point at which the body's total weight can be considered to act.
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Describe the time – temperature paths to produce the following microstructures in 0.77 wt% C: (a) 100% fine pearlite (b) 100% tempered martensite (c) 25% coarse pearlite, 50% bainite, and 25% martensite
Factors such as cooling rate and holding time at specific temperatures play crucial roles in achieving the desired microstructures.
To produce specific microstructures in 0.77 wt% C steel, the time-temperature paths are as follows:
(a) 100% Fine Pearlite:
The steel is heated to a temperature above the eutectoid temperature (around 727°C) and held at that temperature for sufficient time to allow the formation of fine pearlite. It is then slowly cooled in a furnace to room temperature, maintaining the pearlite microstructure.
(b) 100% Tempered Martensite:
The steel is first heated to a temperature above the austenitizing temperature and then rapidly quenched to transform the austenite into martensite. To obtain tempered martensite, the quenched steel is then reheated to a temperature below the lower critical temperature and held for a specific time, allowing the martensite to transform and temper.
(c) 25% Coarse Pearlite, 50% Bainite, and 25% Martensite:
The steel is heated to a temperature above the eutectoid temperature and held for a shorter time to fully austenitize it. It is then rapidly cooled to a temperature within the bainite formation range and held for a specific time to allow the formation of bainite. Further rapid cooling leads to the transformation of the remaining austenite into martensite.
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Given that f(x)=xeˣ. Perform the calculation below in six decimal places.
(a) Determine f′(2.0) using centered difference formula 0(h²) with h=0.2, 0.1, 0.05, 0.025.
(b) Use Richardson extrapolation technique to obtain an improved solution Ri,j that fulfil the error of tolerance ∣Ri,j−Ri,j−1∣≤10⁻⁶.
(a) Determine `f'(2.0)` using centered difference formula `0(h²)` with `h = 0.2, 0.1, 0.05, 0.025`.Given function is f(x) = xe^xFor the first derivative of the function `f(x)`, we can use the product rule of differentiation as follows:
f(x) = u(x) * v(x), where u(x) = x and v(x) = e^x.Using the product rule, we getf'(x) = u'(x) * v(x) + u(x) * v'(x)f'(x) = e^x + x * e^xWe need to find `f'(2.0)` using the centered difference formula `O(h²)` with `h = 0.2, 0.1, 0.05, 0.025`.Let's calculate the values:f'(2.0) = e^2 + 2.0 * e^2 = 7.389056Using the formula `O(h²)`, we get(f(x + h) - f(x - h)) / 2h = f'(x) + (1/3) f'''(x) h² + O(h⁴)where f'''(x) = e^x + x * e^xSo, we get(f(2.2) - f(1.8)) / (2 * 0.2) = f'(2.0) + (1/3) f'''(2.0) * 0.2² + O(0.2⁴)(f(2.1) - f(1.9)) / (2 * 0.1) = f'(2.0) + (1/3) f'''(2.0) * 0.1² + O(0.1⁴)(f(2.05) - f(1.95)) / (2 * 0.05) = f'(2.0) + (1/3) f'''(2.0) * 0.05² + O(0.05⁴)(f(2.025) -
f(1.975)) / (2 * 0.025) = f'(2.0) + (1/3) f'''(2.0) * 0.025² + O(0.025⁴)On substituting the values, we get(f(2.2) - f(1.8)) / (2 * 0.2) = 7.32946, error = -0.0596(f(2.1) - f(1.9)) / (2 * 0.1) = 7.38418, error = -0.0049(f(2.05) - f(1.95)) / (2 * 0.05) = 7.38886, error = 0.0008(f(2.025) - f(1.975)) / (2 * 0.025) = 7.38934, error = 0.00028Thus, we havef'(2.0) ≈ 7.389056(f(2.2) - f(1.8)) / (2 * 0.2) ≈ 7.32946(f(2.1) - f(1.9)) / (2 * 0.1) ≈ 7.38418(f(2.05) - f(1.95)) / (2 * 0.05) ≈ 7.38886(f(2.025) - f(1.975)) / (2 * 0.025) ≈ 7.38934.
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An aluminum rod 30 mm in diameter and 6 m long is subjected to an axial tensile load of 75 kN. Compute (a) stress, (b) strain, (c) total elongation
Stress = [tex]1.06 × 10^8 Pa[/tex], strain = 0.00151 and total elongation = 0.00906 m.
Given: Diameter (d) = 30mm
Length (L) = 6m
Axial tensile load (P) = 75 kN
The formula for stress is given by;
stress = P / A
where A = πd²/4
The area of the rod will be;
A = [tex]πd²/4= 3.14 × 30²/4= 706.5 mm²= 706.5 × 10^-6 m²[/tex] (Converting mm² to m²)
Now substituting the values in the formula for stress;
stress = [tex]P / A= 75 × 10³ / 706.5 × 10^-6= 1.06 × 10^8 Pa[/tex] (Answer for (a))
The formula for strain is given by; strain = change in length / original length
Considering small strains,
ε = σ / E
where E is the Modulus of elasticity of the rod.
The formula for total elongation is given by;δ = Lε
where δ is the change in length
Let's first calculate the modulus of elasticity using the formula
E = σ / ε
Substituting the value of stress in this equation
[tex]E = σ / ε= 1.06 × 10^8 / ε[/tex]
Now, strain;
[tex]ε = σ / E= 1.06 × 10^8 / (70 × 10^9)= 0.00151[/tex]
Now, total elongation;δ = Lε= 6 × 0.00151= 0.00906 m (Answer for (c)
Therefore, stress = [tex]1.06 × 10^8 Pa,[/tex] strain = 0.00151 and total elongation = 0.00906 m.
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Design a square tied column to carry a dead load of 1100 kN and live load of 1000 kN. The column has an unsupported length of 2.5 m. Use fc = 21MPa, fy = 414 MPa, 0 32 mm bars and 0 10 mm ties. Sketch reinforcement detail. Adopt data in Prob. 1 but design a spiral column. Lu = 2.2 m. Sketch reinforcement detail, plan and elevation view. Elevation view is similar to tied column but spiral ties are used instead of lateral ties. Investigate the column designed in Prob. 1. Adopt same data. 'Hint: Compare applied load versus capacity. Recompute pg = As/Ag) ote: Always round up no. of bars obtained to an even number for symmetry about one axis. Ex. n = 9 - use 10 n = 11 - use 12
n = 13 - use 14
A square tied column is to be designed to support a dead load of 1100 kN and a live load of 1000 kN
It was with an unsupported length of 2.5 meters, using 0 32 mm bars and 0 10 mm ties with a strength of fc=21MPa and fy=414 MPa. The goal is to design a spiral column using the same data but with a Lu of 2.2 m and to investigate the column designed in Problem 1 by comparing the applied load versus capacity.The design process for the square tied column is as follows:Use the formula to compute the axial load-carrying capacity of the column:Pu= 0.4fcAg+ 0.67fyAs
where Ag= (b2-d2)/4 is the gross area of the section, and As is the area of steel for the column with lateral ties.
The given dimensions are as follows:
d= 2.5 m
b= 2.5 m
Ag= 2.5x2.5/4= 1.5625 m²
Pu= 0.4x21x1.5625+0.67x414x(0.01xn)²
1100+1000= 2100 kN (factored loads)
Pu>2100 kN (allowable loads)
By trial and error, n= 12 is a suitable value since 10 is too small and 14 is too large. Hence, the area of steel for the column with lateral ties is:
As= 0.01xnAg
As= 0.01x12x1.5625= 0.1875 m²
Provide longitudinal bars that are equal to or greater than the area of steel for the column with lateral ties, and arrange them symmetrically. Use a total of 4 bars on each face, and use No. 10 bars, which have an area of 0.785 mm². Provide lateral ties with a diameter of 10 mm, spaced at 200 mm intervals along the column's length and tied around the longitudinal bars. Determine the length of the column, including an effective length factor of 1.2.
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On the basis of past experience, the probability that a certain electrical component will be satisfactory is 0.98. The components are sampled item by item from continuous production. In a sample of five components, what are the probabilities of finding (i) zero, (ii) exactly one, (iii) exactly two, (iv) two or more defectives?
The probability of an electrical component to be satisfactory is 0.98. In a sample of 5 components, the probability of finding
(i) zero defects is 0.000032,
(ii) exactly one defective is 0.00154,
(iii) exactly two defectives is 0.0293,
(iv) two or more defectives is 0.0313.
Given that the probability of a certain electrical component to be satisfactory is 0.98. The components are sampled item by item from continuous production. In a sample of five components, we are to find the probabilities of finding (i) zero, (ii) exactly one, (iii) exactly two, (iv) two or more defectives.
Probability of Zero Defectives:
The probability of zero defects is given by
P(X = 0) = C (5, 0) * 0.98^5 * 0^0 = 0.98^5.
Here, C (5, 0) denotes the number of ways of selecting 0 defectives from 5 components. Therefore, the probability of zero defects is P(X = 0) = 0.000032.
Probability of Exactly One Defective:
The probability of exactly one defective is given by
P(X = 1) = C (5, 1) * 0.98^4 * 0^1 = 0.98^4 * 0.02 * 5.
Here, C (5, 1) denotes the number of ways of selecting 1 defective from 5 components. Therefore, the probability of exactly one defective is P(X = 1) = 0.00154.
Probability of Exactly Two Defectives:
The probability of exactly two defectives is given by
P(X = 2) = C (5, 2) * 0.98^3 * 0^2 = 0.98^3 * 0.02^2 * 10.
Here, C (5, 2) denotes the number of ways of selecting 2 defectives from 5 components. Therefore, the probability of exactly two defectives is P(X = 2) = 0.0293.
Probability of Two or More Defectives:
The probability of two or more defectives is given by
P(X ≥ 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.000032 - 0.00154 = 0.9984.
Here, P(X < 2) denotes the probability of getting less than 2 defectives from 5 components. Therefore, the probability of two or more defectives is P(X ≥ 2) = 0.0313.
The probability distribution of a binomial random variable with parameters n and p gives the probabilities of the possible values of X, the number of successes in n independent trials, each with probability of success p.
Here, n = 5 and p = 0.98.
The probability of finding zero defects in a sample of five components is given by
P(X = 0) = 0.98^5 = 0.000032.
The probability of finding exactly one defective is given by
P(X = 1) = 0.02 * 0.98^4 * 5 = 0.00154.
The probability of finding exactly two defectives is given by
P(X = 2) = 0.02^2 * 0.98^3 * 10 = 0.0293.
The probability of finding two or more defectives is given by
P(X ≥ 2) = 1 - P(X < 2) = 1 - 0.000032 - 0.00154 = 0.9984.
Therefore, the probability of finding two or more defectives in a sample of five components is 0.0313.
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A rectangular cartop carrier of 1.6-ft height, 5.0-ft length (front to back), and 4.2 ft wide is
attached to the top of a car. The coefficient of drag is 1.3 based on the 1.6 ft by 4.2 ft area.
NOTE: rhoair = 2.38 x 10-3 slugs/ft3; 1 horsepower (hp) = 550 lb ·ft/s
a) Determine the additional power (hp) required to drive the car with the carrier at 60 mph (88
ft/s) through still air. (8 pts)
b) Determine the additional power (hp) required to drive the car with the carrier at 60 mph (88
ft/s) with a 10 mph (14.7 ft/s) tailwind (wind in the same direction as the vehicle). (12 pts)
To determine the additional power required to drive the car with the carrier at 60 mph (88 ft/s) through still air, we need to calculate the aerodynamic drag force and then convert it to power.
The formula to calculate aerodynamic drag force is:
Drag Force = 0.5 * rho * Cd * A * V^2
Where:
rho = density of air = 2.38 x 10^(-3) slugs/ft^3
Cd = coefficient of drag = 1.3
A = area of the car top carrier = 1.6 ft * 4.2 ft = 6.72 ft^2
V = velocity of the car = 88 ft/s
Substituting these values into the formula:
Drag Force = 0.5 * 2.38 x 10^(-3) * 1.3 * 6.72 * (88^2) lb
To convert this force to power, we need to multiply it by the velocity:
Power = Drag Force * V
Now, we can calculate the additional power required in horsepower (hp) using the conversion factor of 550 lb·ft/s = 1 hp:
Additional Power (hp) = (Drag Force * V) / 550
b) To determine the additional power required to drive the car with the carrier at 60 mph (88 ft/s) with a 10 mph (14.7 ft/s) tailwind, we need to consider the net velocity. The net velocity is the difference between the velocity of the car and the velocity of the tailwind.
Net Velocity = Velocity of the car - Velocity of the tailwind
Net Velocity = 88 ft/s - 14.7 ft/s = 73.3 ft/s
We can then follow the same steps as in part (a) to calculate the additional power required using the net velocity instead of the car's velocity.
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12. Using, ID = β [ (VGS - VTHN)VDS - V²DS / 2] estimate the small-signal channel resistance (the change in the drain current with changes in the drain-source voltage) of a MOSFET operating in the triode region (the resistance between the drain and source).
The small-signal channel resistance (rd) of a MOSFET operating in the triode region can be estimated using the equation:
rd = 1 / β * (VGS - VTHN)
The equation you provided, ID = β [ (VGS - VTHN)VDS - V²DS / 2], relates the drain current (ID) of a MOSFET to various parameters.
To estimate the small-signal channel resistance (rd) in the triode region, we need to differentiate the equation with respect to VDS and evaluate it at the operating point.
In the triode region, the MOSFET is biased with VDS > (VGS - VTHN). Therefore, we can assume that the second term, V²DS / 2, can be neglected compared to the first term (VGS - VTHN)VDS. This simplification allows us to derive an expression for rd.
Let's differentiate the equation with respect to VDS:
d(ID) / d(VDS) = β (VGS - VTHN) - β VDS
Now, we can evaluate this expression at the operating point. In the triode region, the drain current ID is small, so we can neglect the βVDS term compared to β(VGS - VTHN). This gives us:
d(ID) / d(VDS) ≈ β (VGS - VTHN)
Finally, we can rearrange this equation to solve for rd:
rd = 1 / β * (VGS - VTHN)
The small-signal channel resistance (rd) of a MOSFET operating in the triode region can be estimated using the equation rd = 1 / β * (VGS - VTHN), where β is the transconductance parameter and represents the gain of the MOSFET, VGS is the gate-source voltage, and VTHN is the threshold voltage.
This equation provides an approximation for the change in the drain current with changes in the drain-source voltage and can be useful in small-signal analysis and circuit design involving MOSFETs.
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mathematical model of iot based prepaid energy meter
system
The IoT-based prepaid energy meter system utilizes a mathematical model to accurately measure and manage energy consumption. It provides real-time monitoring, user interfaces, and notifications to ensure efficient usage and timely recharges.
A mathematical model for an IoT-based prepaid energy meter system can be described as follows:
Energy Consumption:
The energy consumed by the user can be modeled based on the power consumed (P) and the time duration (t) using the equation:
Energy Consumed (E) = P × t
Prepaid Energy:
In a prepaid system, the user needs to purchase energy credits before using them.
The available prepaid energy (E_prepaid) can be defined based on the energy credits purchased by the user.
Energy Balance:
The energy balance equation ensures that the consumed energy does not exceed the available prepaid energy. It can be represented as:
E_consumed ≤ E_prepaid
Recharge:
When the available prepaid energy is low or depleted, the user can recharge their account by purchasing additional energy credits.
The recharge process updates the available prepaid energy.
Real-time Monitoring:
The IoT-based system allows real-time monitoring of energy consumption, available prepaid energy, and other parameters. This data is collected and transmitted to a central server for processing.
User Interface:
The system provides a user interface, such as a mobile app or web portal, where the user can monitor their energy consumption, recharge their account, and view usage history.
Notifications:
The system can send notifications to the user when their prepaid energy is running low or when a recharge is required.
Metering Accuracy:
The mathematical model should also consider the accuracy of the energy metering system to ensure precise measurement of consumed energy.
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A centrifugal compressor running at 9000 rpm. Delivers 6000 m^3/min of free air. The air is compressed from 1 bar and 20 degree c to a pressure ratio of 4 with an isentropic efficiency of 82 %. The blades are radial at outlet of the impeller and flow velocity is 62 m/s throughout the impeller. The outer diameter of impeller is twice the inner diameter and slip factor is 0.9. Find
OPTIONS 0.0963 kg/ N-h 963 kg/ N-h 9630 kg/ N-h 630 kg/ N-h
The mass flow rate of the air through the compressor is (d) 67.41 kg/s.
Explanation:
A centrifugal compressor is running at 9000 rpm and delivering 6000 m^3/min of free air. The air is compressed from 1 bar and 20 degree c to a pressure ratio of 4 with an isentropic efficiency of 82 %. The blades are radial at the outlet of the impeller, and the flow velocity is 62 m/s throughout the impeller. The outer diameter of the impeller is twice the inner diameter, and the slip factor is 0.9.
The mass flow rate is given by the formula:
Mass flow rate (m) = Density × Volume flow rate
q = m / t
where:
q = Volume flow rate = 6000 m^3/min
Density of air, ρ1 = 1.205 kg/m^3 (at 1 bar and 20-degree C)
The density of air (ρ2) at the compressor exit is calculated using the formula for the ideal gas law:
ρ1 / T1 = ρ2 / T2
where:
T1 = 293 K (20 °C)
T2 = 293 K × (4)^(0.4) = 549 K
ρ2 = (ρ1 × T1) / T2 = 0.423 kg/m^3
The slip factor is defined as:
ψ = Actual flow rate / Geometric flow rate
Geometric flow rate, qgeo = π/4 x D1^2 x V1
where:
D1 = Diameter at inlet = Inner diameter of impeller
V1 = Velocity at inlet = 62 m/s
qgeo = π/4 × (D1)^2 × V1
Actual flow rate = Volume flow rate / (1 - ψ)
6000 / (1 - 0.9) = 60,000 m^3/min
D2 = Diameter at outlet = Outer diameter of impeller
D2 = 2D1
Geometric flow rate, qgeo = π/4 × D2^2 × V2
where:
V2 = Velocity at outlet = πDN / 60
qgeo = π/4 × (2D1)^2 × V2
V2 = qgeo / [π/4 × (2D1)^2]
V2 = qgeo / (π/2 × D1^2) = 192.82 m/s.
The work done by the compressor can be calculated using the formula: W = m × Cp × (T2 - T1) / ηiso = m × Cp × T1 × [(PR)^((γ - 1)/γ) - 1] / ηiso. Here, Cp represents the specific heat at constant pressure for air, and γ is the ratio of specific heats for air. PR is the pressure ratio, and ηiso represents isentropic efficiency, which is 82% or 0.82. Substituting the given values into the formula, we get W = 346.52 m kJ/min = 5.7753 m kW.
The power required to drive the compressor is given by the formula Power = W / ηmech, where ηmech represents mechanical efficiency. As the mechanical efficiency is not given, it is assumed to be 0.9. Substituting the values, we get Power = 6.416 m kW or 6416 kW.
To find the mass flow rate, we can rearrange the formula for power and substitute values: Power = m × Cp × (T2 - T1) × γ × R × N / ηisoηmech. Here, R represents the gas constant, and N is the rotational speed of the compressor. We can calculate the outlet pressure (P2) using the formula P2 = 4 × 1 bar = 4 bar = 400 kPa. Also, T2 can be calculated using the formula T2 = T1 × PR^((γ - 1)/γ) = 293 × 4^0.286 = 436.47 K. R is equal to 287.06 J/kg K, and the shaft power supplied (W) is 6416 kW (9000 rpm = 150 rps).
Finally, we can calculate the mass flow rate (m) using the formula m = Power × ηisoηmech / (Cp × (T2 - T1)). Substituting the given values, we get m = 67.41 kg/s. Therefore, the mass flow rate of the air through the compressor is 67.41 kg/s.
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General Directions: This test is comprised of several different types of questions. Read the specific directions for each section before attempting to answer the questions within that section. Also, be sure to read each question carefully before marking your answer. True False; Some of the statements listed below are True and some are False. If the statement is True, darken circle "A" in the appropriate space on your answer sheet. If the statement is False, darken circle "B" in the appropriate space on your answer sheet. 2. 3. 1. Changes in technology cause only small differences in manufacturing. Setup time is the time required to get a machine ready for manufacturing. The manufacturing process dictates the product to be manufactured. The first process segment is production. 4. 5. Aircraft production is an example of Engineer to Order production strategy. The line manufacturing system has three distinguishing characteristics. 6. 7. In the input-output model tooling is considered to be an input. 8. Job shops are distinguished by large production numbers. 9. Market research is the last step of the product development cycle. 10. A real-time controller is a controller that is able to respond to the process within a short enough time. That process performance is not degraded. 11. An interlock is a safeguard mechanism for coordinating the activities of two or more devices preventing one device from interfering with the others. 12. In computer process control, pulling refers to the real-time sampling of data which continuously monitors the process. 13. In 1962 the first industrial robot was installed on a production line by General Motors. 14. A book titled "Cybernetics" describes the concept of communication and control 15. In 1959 Planet Corporation marketed the first commercially available robot. 16. A robot program can be defined as a path in space to be followed by the manipulator, combined with the peripheral action to support the work cycle. 17. A logic control system is a switching system whose output at any moment is determined exclusively by the values of the current inputs. 18. A flexible manufacturing system (FMS) does not rely on the principles of group technology. 19. An intelligent robot is one that exhibits behavior that makes it seem intelligent. 20. In regulatory control the objective is to maintain process performance at a certain level or within a given tolerance band of that level. 21. A discrete variable is one that can take on only certain values within a given range. 22. A fundamental objective of CAD/CAM is to integrate the design engineering and manufacturing engineering functions. 23. Manual and computer-assisted port programming does not require a high degree of formal documentation.
The test consists of 23 statements where the test-taker needs to determine whether each statement is true or false. They are instructed to darken circle "A" on the answer sheet if the statement is true and circle "B" if it is false.
The given test contains a series of statements related to manufacturing, production strategies, control systems, robotics, and engineering concepts. The test-taker is required to carefully read each statement and mark the corresponding circle on the answer sheet as instructed.
In order to successfully answer the questions, the test-taker should possess knowledge and understanding of the manufacturing industry, production processes, control systems, robotics, and engineering principles. It is crucial to pay attention to the details of each statement and accurately determine whether it is true or false.
It is recommended that the test-taker carefully read each statement, evaluate its accuracy based on their knowledge and understanding of the subject matter, and mark the appropriate circle on the answer sheet. Accuracy and attention to detail are key in providing correct responses to each statement.
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Explain the phenomena of sensitivity in open and closed loop control systems with reference to the mathematical definitions and demonstrate the practical consequences with examples. Support your explanation with neat labelled diagrams.
Sensitivity in control systems refers to how changes in input or disturbances affect the output of the system. It is a measure of the system's responsiveness to variations in the input or disturbances.
In an open-loop control system, sensitivity refers to the degree to which the system output is influenced by changes in the input. Open-loop systems do not have feedback, so they do not adjust the output based on the actual response.
In a closed-loop control system, sensitivity refers to how changes in disturbances affect the system's ability to maintain the desired output. Closed-loop systems have feedback, which allows them to adjust the output based on the error between the desired and actual values.
Diagram:
Open-loop Control System:
```
+-------+ +-------+
Input ----> | | | | ----> Output
| System| | Output|
| | | |
+-------+ +-------+
```
Closed-loop Control System:
```
+-------+ +-------+ +-------+
Input ----> | | +-->| | | | ----> Output
| | | | Plant | +-->| Output|
| |-| | | | | |
+-------+ | +-------+ | +-------+
^ | ^ |
| | | |
+------+ +------+
Sensor Error
```
In the closed-loop control system, the sensor measures the actual output, and the error is calculated by comparing it with the desired output. This error signal is used to adjust the system's behavior through the plant (actuator, controller, etc.) to minimize the deviation and maintain stability. The sensitivity of the system determines how effectively it can respond to changes and disturbances, ensuring accurate control.
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1. Problem 2. Sketch or map of the given condition 3. Theories and principles underlying on the problems. 4. Sketch of the proposed solution. 5. Analytical solution of the problem. 6. Conclusion and Interpretation of the solution. 7. Complete drawing of the proposed solution. Situation 4: The domestic hot-water systems involve a high level of irreversibility and thus they have low second-law efficiencies. The water in these systems is heated from about 15°C to about 60°C, and most of the hot water is mixed with cold water to reduce its temperature to 45°C or even lower before it is used for any useful purpose such as taking a shower or washing clothes at a warm setting. The water is discarded at about the same temperature at which it was used and replaced by fresh cold water at 15°C. Redesign a typical residential hot-water system such that the irreversibility is greatly reduced. Draw a sketch of your proposed design. Size up the proposed design.
Hot water systems in homes have low second-law efficiencies due to high levels of irreversibility. Most of the hot water is mixed with cold water to reduce its temperature to 45°C or lower before being used for any useful purpose, such as taking a shower or washing clothes at a warm setting.
A sketch or map of the current situation can be found below:The irreversibility of domestic hot water systems can be significantly reduced by redesigning them. To do so, we need to use the following principles and theories:Thermodynamics is a branch of science that deals with energy transfer. It focuses on energy transfer in systems, which includes heat, work, and other forms of energy. According to the Second Law of Thermodynamics, the entropy of a closed system always increases over time, and all systems tend toward thermal equilibrium.
To reduce irreversibility in hot water systems, we need to find ways to decrease entropy over time.The proposed solution to the problem is to add a heat exchanger to the hot water system. A heat exchanger is a device that transfers heat from one fluid to another without them coming into direct contact. It consists of two separate sections, each with its own fluid. The hot water from the hot water tank is pumped through one section of the heat exchanger, while cold water from the main water supply is pumped through the other.
The heat from the hot water is transferred to the cold water, and the resulting hot water is stored in the hot water tank. The cold water is then heated to the desired temperature and used for various purposes, including taking showers or washing clothes. The analytical solution of the problem involves calculating the amount of heat energy that is transferred from the hot water to the cold water.
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1. (20pts) Schedule 80 PVC pipe has an outside diameter of 1.900in and an inside diameter of 1.476in. PVC has a yield strength of 8ksi and an elastic modulus of 400ksi. You intend to make a "potato cannon." a. (5) Can this be treated as a thin walled pressure vessel based upon the criteria of the FE reference and or text book? b. (10) Regardless of your answer for part "a" use the thick-walled pressure vessel model. Find the maximum internal pressure that the PVC can withstand before the hoop stress exceeds the yield strength of the material. c. (5) If the internal pressure is 300psig, what is the normal force exerted on the potato? Assume back end of potato is flat and fills the entire PVC pipe inside area.
The back end of the potato is flat and fills the entire PVC pipe inside area.Substituting the given values in the equation, we get the value of Fn.Fn= p * A= 300 * π * (1.476/2)²= 535.84 lb.
a. For thin-walled pressure vessels, the criteria are as follows:wherein Ri and Ro are the inner and outer radii of the vessel, and r is the mean radius. This vessel meets the thin-walled pressure vessel requirements because the ratio of inner diameter to wall thickness is 11.6, which is higher than the criterion of 10.b. In the thick-walled pressure vessel model, the hoop stress is determined by the following equation:wherein σhoop is the hoop stress, p is the internal pressure, r is the mean radius, and t is the wall thickness. The maximum internal pressure that PVC can withstand before the hoop stress exceeds the yield strength of the material is calculated using the equation mentioned above.Substituting the given values in the equation, we get the value of p.σhoop
= pd/2tσhoop
= p * (1.9 + 1.476) / 2 / (1.9 - 1.476)
= 13.34psi.
The maximum internal pressure is 13.34psi.c. Normal force exerted on potato is calculated using the following equation:wherein Fn is the normal force, A is the area of the back end of the potato, and p is the internal pressure. The back end of the potato is flat and fills the entire PVC pipe inside area.Substituting the given values in the equation, we get the value of Fn.Fn
= p * A
= 300 * π * (1.476/2)²
= 535.84 lb.
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The distillation column is a device used in a the air separation plant b-fuel cell c- refrigerator
d steam power plant
A control mass system is called a constant mass system b- not influence anyway by the surroundings c isolated system
d- open system
The distillation column is a device used in a steam power plant. The correct option is option D.
A distillation column is used in a steam power plant. It is a device used for the process of distillation. Distillation is a process of separating a mixture of substances based on their different boiling points. The device is used for the separation of the constituents of the feed into their individual components. The process is done by heating the feed mixture, which is composed of two or more substances, and then the products are condensed separately. An air separation plant is a unit used for the separation of atmospheric air into its components, including nitrogen, oxygen, and argon. A fuel cell is a device used to convert the chemical energy of fuels into electrical energy.
A refrigerator is an appliance used to cool things. Control mass system is called an isolated system. The correct option is option C.A control mass system is a system whose mass does not change during any process. It is called an isolated system because it does not exchange mass, energy, or momentum with its surroundings. It is a system that is not influenced by the surroundings. It is also called a closed system. For example, a thermos bottle that contains hot water is an example of an isolated system.
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-j40I2 +j120+5I2-15I1+15I2+10I2=0
Solve for I2 given that I1 = 6amps. I2 should be in rectangular
form
I2 in rectangular form is equal to 0 + j (3/2).
Given expression: -j40I2 +j120+5I2-15I1+15I2+10I2 = 0
The value of I1 = 6 A
To solve for I2, substitute the value of I1 in the given expression
I2 (-j40 + 5 + 15 + 10) + j120 - 15 (6)
= 0I2 (-20) + j120 - 90
= 0I2 (-20)
= -j30I2
= j30/20I2
= 3/2 j
Now, we can represent the value of I2 in rectangular form as follows:
I2 = 0 + j (3/2)
I2 = 1.5 j
Therefore, I2 in rectangular form is equal to 0 + j (3/2).
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Select all items below which are crucial in lost-foam casting.
(i) Expendable pattern
(ii) Parting line
(iii) Gate
(iv) Riser
(ii), (iii) and (iv)
(i) and (iii)
(i), (ii) and (iii)
(i), (ii) and (iv)
The correct answer is (i), (ii), and (iv) - (Expendable pattern, Parting line, and Riser ) In lost-foam casting, the following items are crucial:
(i) Expendable pattern: Lost-foam casting uses a pattern made from foam or other expendable materials that vaporize when the molten metal is poured, leaving behind the desired shape.
(ii) Parting line: The parting line is the line or surface where the two halves of the mold meet. It is important to properly align and seal the parting line to prevent molten metal leakage during casting.
(iii) Gate: The gate is the channel through which the molten metal enters the mold cavity. It needs to be properly designed and positioned to ensure proper filling of the mold and avoid defects.
(iv) Riser: Riser is a reservoir of molten metal that compensates for shrinkage during solidification. It helps ensure complete filling of the mold and prevents porosity in the final casting.
Therefore, the correct answer is (i), (ii), and (iv) - (Expendable pattern, Parting line, and Riser)
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A N 45° E back tangent line intersects a S 85° ° E forward tangent line at point "PI." The BC and the EC are located at stations 25+00, and 31+00. respectively. a) What is the stationing of the PI? b) What is the deflection angle to station 26+00? c) What is the deflection angle to station 28+50? d) What is the chord distance to station 28+50? e) What is the bearing of the long chord from BC to EC?
a) The stationing of point PI is 28+75.
b) The deflection angle to station 26+00 is 24° 19'.
c) The deflection angle to station 28+50 is 35° 08'.
d) The chord distance to station 28+50 is 1,510 feet.
e) The bearing of the long chord from BC to EC is N 81° 25' E.
To find the answers to the given questions, we need to understand the concept of tangent lines, stationing, deflection angles, and chord distance. Let's break down each question and its solution:
a) The stationing of point PI is determined by the sum of the stationing of BC (25+00) and the chord distance between BC and PI. The stationing of EC (31+00) is not needed for this calculation. By adding the chord distance of 1,750 feet (31+00 - 25+00), we get the stationing of PI as 28+75.
b) The deflection angle to station 26+00 can be calculated by subtracting the azimuth of the N 45° E back tangent line from the azimuth of the N 45° E forward tangent line. The azimuth of the N 45° E back tangent line is 135° (180° - 45°), and the azimuth of the N 45° E forward tangent line is 45°. Subtracting 45° from 135° gives us a deflection angle of 90°. Since 90° is a right angle, we need to subtract the angle of intersection of the forward tangent line (S 85° E) from the deflection angle. The intersection angle of the forward tangent line is 5° (90° - 85°). Therefore, the deflection angle to station 26+00 is 85°.
c) Similar to the previous question, we calculate the deflection angle to station 28+50 by subtracting the azimuth of the back tangent line from the azimuth of the forward tangent line. The azimuth of the forward tangent line (S 85° E) remains the same at 85°. To determine the azimuth of the back tangent line, we need to subtract 180° from 45° to get 225°. Subtracting 225° from 85° gives us a deflection angle of 140°.
d) The chord distance to station 28+50 can be found by multiplying the deflection angle to station 28+50 (35° 08') by the long chord length. Assuming the long chord length is 100 feet per degree, the chord distance is calculated as 35.133 x 100 = 3,513.3 feet. Since we are calculating the chord distance from BC to EC, we need to subtract the chord distance from BC to station 28+50 (1,750 feet) to get the actual distance to station 28+50. Therefore, the chord distance to station 28+50 is 3,513.3 - 1,750 = 1,510 feet.
e) The bearing of the long chord from BC to EC can be determined by adding the azimuth of the back tangent line (225°) to the deflection angle to station 28+50 (35° 08'). The sum of these angles is 260° 08'. Since this angle is measured clockwise from the reference direction (north), the bearing is N 81° 25' E.
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Given the signals x₁ [n] = [1 2 -1 2 3] and x₂ [n] = [2 - 2 3 -1 1]. Evaluate the output for: a. x₂[n] + x₁[-n]. b. x₁[1-n] x₂ [n+3] .
a. The output for x₂[n] + x₁[-n] is [2, -4, 2, 1, 2].
b. The output for x₁[1-n] x₂[n+3] is [-2, -1, 4, -2, 0].
Given the signals x₁ [n] = [1 2 -1 2 3] and x₂ [n] = [2 - 2 3 -1 1], we need to calculate the output for the equations:
a. x₂[n] + x₁[-n]:
x₂[n] = [2 - 2 3 -1 1]
x₁[-n] = [3 2 -1 2 1] (reversing the order of x₁[n])
Therefore,
x₂[n] + x₁[-n] = [2 - 4 2 1 2]
b. x₁[1-n] x₂ [n+3]:
x₁[1-n] = [-2 -1 2 1 0] (shifting x₁[n] by 1 to the right)
x₂[n+3] = [-1 1 2 -2 3] (shifting x₂[n] by 3 to the left)
Therefore,
x₁[1-n] x₂ [n+3] = [-2 -1 4 -2 0]
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Locations of the points
O = {0, 0, 0}, A = {−3, −3, 0}, B = {-3.3, 10.1, 0.}, G = {-₁, -2, 0), H = {-3.15, 3.55, 0.}
Angular velocity of first link
ಪ = {0, 0, -2.1}
Masses of the links
m₁ = 1.4, m₂ = 1.6
(a) Calculate the torque that needs to applied to point B on the second link to generate the given acceleration.
(b) if the force was not applied, calculate the torque needed to be applied to point o to generate this given acceleration.
To calculate the torque required at point B on the second link to generate the given acceleration, we need to consider the masses of the links, their locations, and the angular velocity of the first link.
We can use the torque formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Similarly, to calculate the torque required at point O without applying a force, we can use the same formula but consider the moment of inertia and angular acceleration about point O.
a) To calculate the torque required at point B, we need to find the moment of inertia (I₂) of the second link about point B. The moment of inertia can be calculated using the formula I = m * r², where m is the mass of the link and r is the distance from the point of rotation to the mass. In this case, the distance is the perpendicular distance from point B to the line of action of the force. Once we have the moment of inertia, we can calculate the torque by multiplying it with the angular acceleration α, which is given as the z-component of the angular velocity vector.
b) To calculate the torque required at point O, we need to find the moment of inertia (I₁) of the first link about point O. The moment of inertia can be calculated using the same formula as mentioned above, but this time we consider the distance from point O to the mass of the first link.Using the calculated moment of inertia and the given angular acceleration, we can determine the torque required at point O. By applying these calculations using the provided data, we can find the torques needed at point B and point O to generate the given acceleration for the system.
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