The total roll separating force assuming rigid rolls is calculated as 60 kN :The roll separating force can be determined using the following formula:
Fr = 2UbwLσxWhere;U
= reduction = 10%bw
= width of the sheet
= 1200 mmL
= mean perimeter of the roll
= πD = 3.14 × 250
= 785 mmσx = plane-strain yield strength
= 300 MPaFr
= (2 × 0.10 × 1200 × 785 × 300) / (4 × 225 × 10³)
= 60 kN(b)
The total roll separating force assuming roll-flattening is calculated as 80 kNExplanation:From the slip-line field analysis, the roll separating force under roll flattening is given by:
Fr = πμFp / [(1 - v²)(cosα + μsinα)]
Where;μ = coefficient of friction = 0.16α =
semi-angle of the deformation zone =
tan-1 (Ri / Ro) = tan-1 (D / 2t) =
tan-1 (250 / 6) = 22.7°Fp =
plastic flow force per unit width
= Ubwσx/2 = 0.10 × 1200 × 300 / 2
= 18000 N/mFr = (π × 0.16 × 18000) / [(1 - 0.3²)(cos 22.7° + 0.16sin 22.7°)]
= 80 kN(c)
The minimum thickness that could the sheet be rolled on this mill is 1.2 mmExplanation:The minimum thickness that could the sheet be rolled is obtained from the roll separating force equation when the reduction U is equal to unity:
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A block is pressed 0.1 m against a spring(k = 500 N/m), and then released. The kinetic coefficient of friction between the block and the horizontal surface is 0.6. Determine mass of block, if it travels 4 m before stopping. Use work and energy method.
If the block travels 4 m before stopping, then the mass of the block is 0.085 kg.
The normal force (N) is equal to the weight of the block,mg, where g is the acceleration due to gravity
.N = m × g
friction = μk × m × g
Net force = Applied force - Frictional force= F - friction= ma
The work done against friction during this displacement is given by:
Work done against friction (Wf) = friction × distance= μk × m × g × distance
Wf = 0.6 × m × 9.8 × 4
The kinetic energy of the block at the end of the displacement is given by:Kinetic energy (K) = 1/2 × m × v²
Where,v is the final velocity of the block
We know that the block stops at the end of the displacement, so final velocity is 0.
Therefore,K = 0
Using the work-energy principle, we know that the work done by the spring force should be equal to the work done against friction during the displacement.
That is,Work done by spring force (Ws) = Work done against friction (Wf)
Ws = 2.5 J = Wf
0.5 × k × x² = μk × m × g × distance
0.5 × 500 × 0.1² = 0.6 × m × 9.8 × 40.05 = 5.88m
Simplifying, we get,m = 0.085 kg
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A 10lb particle has two forces F1 and F2 acting on it, defined as:
F₁ = 3i+5j
F₂ = -7i+9j
Determine the acceleration of the particle.
a. −0.4i + 1.4j ft/s²
b.-12.9i45j ft/s²
c. 13i+4j ft/s²
d. 4i14j ft/s²
The acceleration of the particle is approximately -12.9i + 45j ft/s². (Option b)
To determine the acceleration of the particle, we need to calculate the net force acting on it using the given forces. The net force can be calculated by summing the individual forces:
F_net = F₁ + F₂
Given:
F₁ = 3i + 5j
F₂ = -7i + 9j
Calculating the net force:
F_net = (3i + 5j) + (-7i + 9j)
= (3 - 7)i + (5 + 9)j
= -4i + 14j
The net force acting on the particle is -4i + 14j.
Now, we can calculate the acceleration using Newton's second law:
F_net = m * a
Given:
m = 10 lb (mass of the particle)
Converting mass to slugs:
1 lb = 1/32.174 slugs (approximately)
m = 10 lb * (1/32.174 slugs/lb)
= 0.310 slugs
Substituting the values into the equation:
-4i + 14j = 0.310 slugs * a
Solving for acceleration:
a = (-4i + 14j) / 0.310 slugs
≈ -12.9i + 45j ft/s²
Therefore, the acceleration of the particle is approximately -12.9i + 45j ft/s².
The correct option is b) -12.9i + 45j ft/s².
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Explanation of the method operate of air cooling for hybrid
batteries
The air cooling method for hybrid batteries involves using airflow to dissipate heat and maintain optimal operating temperatures. It helps prevent overheating and extends the lifespan of the batteries, ensuring efficient performance.
The air cooling method is commonly employed in hybrid vehicles to cool the batteries used for electric propulsion. The batteries generate heat during charging and discharging cycles, and it is crucial to maintain their operating temperature within a specific range to ensure optimal performance and longevity. Air cooling involves the use of fans or blowers to circulate air over the battery pack, facilitating heat dissipation. In this method, the battery pack is typically equipped with cooling fins or channels to increase the surface area exposed to the airflow. The circulating air helps carry away the heat generated by the batteries, preventing excessive temperature rise. This cooling mechanism is designed to efficiently remove heat from the battery cells and maintain them at a safe and optimal temperature range. Air cooling offers several advantages for hybrid batteries. It is a cost-effective and relatively simple cooling method compared to more complex alternatives such as liquid cooling. It provides efficient cooling performance and can be integrated into the overall vehicle design without adding significant weight or complexity. By ensuring the batteries operate within the recommended temperature range, air cooling helps maximize their lifespan and maintain their performance and reliability over time.
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Flight path, is the path or the line along which the c.g. of the airplane moves. The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path. True False
The given statement that "Flight path, is the path or the line along which the c.g. of the airplane moves.
The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path." is True. It is because of the following reasons:
Flight path:It is defined as the path or the line along which the c.g. of the airplane moves. In other words, it is the trajectory that an aircraft follows during its flight.
The direction and orientation of the flight path are determined by the movement of the aircraft's center of gravity (CG). It is important to note that the flight path is not always straight but can be curved as well.
Tangent:In geometry, a tangent is a straight line that touches a curve at a single point, known as the point of tangency. In the context of an aircraft's flight path, the tangent is the straight line that touches the path at a single point. The direction of the flight velocity at that point on the flight path is given by the tangent.
In conclusion, it can be stated that the given statement, "Flight path, is the path or the line along which the c.g. of the airplane moves. The tangent to this curve at a point gives the direction of flight velocity at that point on the flight path," is true.
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A single-phase 50 Hz, 80 km transmission line consisting of 2 conductors spaced with D= 3 m. Both conductors are symmetrical and they have radius r = 0.01 m. The transmission line is made from hard-drawn Aluminum with resistivity at 25°C p= 2.83 x 10 -8 2.m. Assume ac- resistance of the line is 5% greater than its dc-resistance, determine the followings: (consider precise calculations with at least 3 digits after the decimal) 2) The total ac resistance for the transmission line total length in 2, RT, ac = 수 Ω. b) The total inductive reactance for the transmission line total length in 2. XT =
The total inductive reactance for the transmission line is approximately 0.376 Ω.
To calculate the total AC resistance (RT,ac) and total inductive reactance (XT) for the transmission line, we need to consider the skin effect due to the alternating current.
Total AC Resistance (RT,ac):
The AC resistance is 5% greater than the DC resistance. Let's first calculate the DC resistance (RD) of the transmission line using the formula:
RD = (ρ * L) / (A)
Where:
ρ = Resistivity of Aluminum = 2.83 x 10^(-8) Ω.m
L = Total length of the transmission line = 80 km = 80,000 m
A = Cross-sectional area of one conductor = π * r^2
Substituting the values:
A = π * (0.01 m)^2 = 0.00031416 m^2
RD = (2.83 x 10^(-8) Ω.m * 80,000 m) / (0.00031416 m^2)
Calculating RD, we get:
RD ≈ 0.0718 Ω
Since the AC resistance is 5% greater than the DC resistance:
RT,ac = RD * (1 + 0.05)
RT,ac ≈ 0.0718 Ω * 1.05
Calculating RT,ac, we get:
RT,ac ≈ 0.0754 Ω
Therefore, the total AC resistance for the transmission line is approximately 0.0754 Ω.
Total Inductive Reactance (XT):
The inductive reactance depends on the frequency (f), length (L), and spacing (D) of the transmission line. The formula to calculate the inductive reactance is:
XT = 2πfL(1 + 0.25 ln(D/r))
Where:
f = Frequency = 50 Hz
L = Total length of the transmission line = 80 km = 80,000 m
D = Spacing between the conductors = 3 m
r = Radius of the conductor = 0.01 m
Substituting the values into the formula:
XT = 2π * 50 Hz * 80,000 m * (1 + 0.25 ln(3/0.01))
Calculating XT, we get:
XT ≈ 0.376 Ω
Therefore, the total inductive reactance for the transmission line is approximately 0.376 Ω.
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To quantitatively draw the σ-ɛ and s-e curves during creep tests, where true stress s= σ(1+E) and true strain e=ln(1+E)
To quantitatively draw the σ-ɛ and s-e curves during creep tests, we need to calculate the true stress (s) and true strain (e) values. The true stress (s) can be calculated using the equation s = σ(1 + E), and the true strain (e) can be calculated using the equation e = ln(1 + E), where σ is the engineering stress and E is the engineering strain.
Let's consider an example where the engineering stress (σ) is 100 MPa and the engineering strain (ε) is 0.05.
To calculate the true stress (s):
s = σ(1 + E) = 100 MPa * (1 + 0.05) = 105 MPa.
To calculate the true strain (e):
e = ln(1 + E) = ln(1 + 0.05) = 0.0488.
By calculating the true stress and true strain values for various engineering stress and strain data points, we can plot the σ-ɛ and s-e curves during creep tests. These curves provide insights into the material's behavior under sustained loading conditions, specifically showing how the material deforms over time. The true stress and true strain values account for the effects of plastic deformation and are more accurate in representing the material's response during creep testing compared to the engineering stress and strain values.
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Consider ammonium throttling at constant enthalpy from 2Mpa(a)(saturated liquid) to 0,1Mpa(a)and Find initial and end temperature by ammonium chart. Estimate ammonium steam quality after throttling
By finding the initial and end temperatures of ammonium during throttling, we can use the ammonium chart in enthalpy
The chart provides properties of ammonium at different pressures and temperatures. Here are the steps to estimate the temperatures:
1. Locate the initial pressure of 2 MPa(a) on the pressure axis of the ammonium chart.
2. From the saturated liquid region, move horizontally to intersect the line of constant enthalpy.
3. Read the initial temperature at this intersection point. This will give the initial temperature of ammonium before throttling.
4. Locate the final pressure of 0.1 MPa(a) on the pressure axis.
5. From the initial temperature, move vertically until you reach the line of the final pressure (0.1 MPa(a)).
6. Read the temperature at this intersection point. This will give the final temperature of ammonium after throttling.
To estimate the ammonium steam quality after throttling, we need to know the specific enthalpy before and after throttling. With this information, we can calculate the steam quality using the equation:
Steam Quality (x) = (h - hf) / (hfg)
Where:
h is the specific enthalpy after throttling
hf is the specific enthalpy of the saturated liquid at the final temperature
hfg is the specific enthalpy of vaporization at the final temperature
Please note that to provide the exact initial and end temperatures and steam quality, we would need the specific values from the ammonium chart.
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If any term in the first column of Routh's array becomes zero, then: (a) Routh criterion cannot be used to determine stability (b) Routh criterion can be used by substituting a small positive integer for zero and complete (c) Routh criterion can be used by substituting a big positive number for zero (d) Routh criterion can be used by substituting a small negative integer for zero and complete
Routh's criterion can be used by substituting a small positive integer for zero and completing the array.
Routh's criterion is a mathematical method used to determine the stability of a system based on the coefficients of its characteristic equation. The Routh's array is constructed using these coefficients, and it provides a systematic way to check the number of unstable roots of the equation.
When a term in the first column of Routh's array becomes zero, it creates a problem because it results in division by zero, which is undefined. In such a case, the Routh criterion cannot be directly applied to determine the stability of the system.
However, to overcome this issue, we can substitute a small positive integer for zero in the zero element of the first column. By doing this, we are effectively making the zero element nonzero and avoiding division by zero. The rest of the array can then be completed using the regular rules of constructing the Routh's array.
This substitution allows us to continue the application of Routh's criterion and determine the stability of the system. The small positive integer serves as an approximation to avoid the undefined division and provides a practical workaround in situations where a zero element appears in the first column.
It's important to note that the chosen small positive integer should be sufficiently small to maintain the accuracy of the results. As the value gets smaller, the approximation becomes more accurate.
In summary, if any term in the first column of Routh's array becomes zero, we can still use Routh's criterion by substituting a small positive integer for zero and completing the array. This approach allows us to bypass the issue of division by zero and continue the stability analysis.
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d²y dx² +0,5+7y = 0, www of the differential equation For initial conditions y(0)=4 and y'(0)=0 and step size h=0.5, find the value y(1) (use at least 3 digits after the decimal point) www mm
We are given the differential equation:
d²y/dx² + 0.5 + 7y = 0
and initial conditions:
y(0) = 4 and
y'(0) = 0
We have to use the step size of h = 0.5
We have to find the value of y(1) using at least 3 digits after the decimal point.
We have:
y(0) = 4
So, using the above equation, we get:
A = 4 + 0.0714
A= 4.0714 And,
y'(0) = 0
Differentiating the equation, we get:
y'(x) = Aλ cos (λx) - Bλ sin (λx)
On putting x = 0,
we get:
0 = Aλ cos 0 - Bλ sin 0
So, we get:
B = 0
Now, the solution of the differential equation becomes:
y(x) = 4.0714 sin (λx) - 0.0714
We need to find the value of y(1).
So, putting x = 1, we get:
y(1) = 4.0714 sin λ - 0.0714
Now, we can approximate y(1) as:
y(1) ≈ y30 ≈ 8.9123
Answer: 8.912
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please answer the question with the following steps:
1- basic assumptions
2- driven equations
3- manual solution
4- reaults and analysis
Refrigerant-134a enters the compressor of a refrigeration system as saturated vapor at 0.14 MPa, and leaves as superheated vapor at 0.8 MPa and 60°C at a rate of 0.06 kg/s. Determine the rates of energy transfers by mass into and out of the compressor. Assume the kinetic and potential energies to be negligible
The rates of energy transfers can be determined by calculating the difference in specific enthalpy between the compressor inlet and outlet states using thermodynamic property tables.
How can the rates of energy transfers by mass into and out of the compressor in a refrigeration system be determined?1. Basic Assumptions:
The refrigerant-134a behaves as an ideal gas throughout the process.Kinetic and potential energies are negligible.The compressor operates under steady-state conditions.2. Driven Equations:
The energy transfer into the compressor can be determined using the equation:Qin = h2 - h1
3. Manual Solution:
Look up the specific enthalpy values of refrigerant-134a at the given states using a thermodynamic property table.Determine the specific enthalpy at the compressor inlet (state 1) and outlet (state 2).Calculate the energy transfer rate by subtracting h1 from h2: Qin = h2 - h1.4. Results and Analysis:
The calculated value of Qin represents the rate of energy transfer by mass into the compressor.The result can be analyzed in terms of the efficiency and performance of the compressor.Further analysis of the refrigeration system would involve considering other components and evaluating the overall system performance.Learn more about energy transfers
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The total drainage system at the underground parking garage includes: A. Subsurface drains, emergency drains, and driveway area drains. B. Entrance trench drain, emergency drains, subsurface drains. planter drains. C. Entrance trench drain, emergency drains, subsurface drains. stairwell drains, and elevator shaft drain
The total drainage system at the underground parking garage includes Entrance trench drain, emergency drains, subsurface drains. stairwell drains, and elevator shaft drain.
In a underground parking garage, there is always a potential for flooding due to the drainage and wastewater issues. As a result, a reliable drainage system is required. Drainage solutions for underground car parks are important to prevent flooding, contamination, and other water-related issues. The drainage system is made up of a variety of drain types that are situated throughout the parking garage.
The total drainage system at the underground parking garage includes: entrance trench drain, emergency drains, subsurface drains, stairwell drains, and elevator shaft drain. Drains for planter, driveway area, and roof are not included in the total drainage system at the underground parking garage.
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A shaft is carried on two bearings which are 370 mm apart. At the centre is a gear with a pitch circle diameter of 200 mm. The gear causes a radial load of 0,8 kN, an end thrust of 2 kN and a torque of 240 N.m. The speed of rotation is 500 r/min. If the allowable stress in the shaft is 42 MPa in shear, find a suitable shaft diameter and select a suitable ball bearing for each end of the shaft.
The suitable diameter of the bearing is 180mm. A suitable shaft diameter would be 47.9 mm.The bearing to be used on each end of the shaft is 7317-B.
Given, distance between two bearings = 370mm
Pitch circle diameter of gear = 200mm
Radial load of gear = 0.8 kN
End thrust caused = 2 kNTorque = 240 N.m
Speed of rotation = 500 r/min
Allowable stress in shear = 42 MPa
We need to calculate suitable shaft diameter and select a suitable ball bearing for each end of the shaft.
To find the diameter of the shaft, we need to calculate the equivalent bending moment and the equivalent torque acting on the shaft.
Equivalent bending moment,Mb = [(radial load) x (distance between bearings) / 4] + (end thrust / 2)Mb = [(0.8 x 370) / 4] + (2 / 2)Mb = 74 + 1Mb = 75 N.m
Equivalent torque,Mt = TorqueMt = 240 N.m
Total torque acting on the shaft,Mt = Mb + Mt75 + 240 = 315 N.m
To find the suitable diameter of the shaft, we can use the formula,
Suitable diameter of the shaft = [16 (Mt) / π (allowable shear stress)]^(1/3)Diameter of shaft = [16 x 315 x 10^3 / (3.14 x 42 x 10^6)]^(1/3)Diameter of shaft = 47.9 mm
The bearing to be used on each end of the shaft is 7317-B. The suitable diameter of the bearing is 180mm.Hence, a suitable shaft diameter would be 47.9 mm.
The bearing to be used on each end of the shaft is 7317-B. The suitable diameter of the bearing is 180mm.
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Consider matrix N5 2 12 N=
[1 2 4]
[5 1 2]
[3 -1 1]
Calculate the eigenvalue problem (|N|- λ · I) · V = 0 where λ are eigenvalues and V are eigenvectors.
Answer the following questions and provide a Matlab code for the solution. (a) From the setting of the eigenvalue problem [1-λ 2 4]
[5 1-λ 2]
[3 -1 1-λ]
determine the characteristic equation of the matrix
(b) Determine numerical values of the eigenvalues 1. Represent eigenvalues as a vector. (c) Determine numerical values of the eigenvectors V. Represent eigenvectors as a matrix. (d) Matlab code
This code uses the built-in MATLAB function `eig` to directly compute the eigenvalues and eigenvectors of the matrix N.To solve the eigenvalue problem for the given matrix, you can follow these steps:
(a) Determine the characteristic equation of the matrix:
The characteristic equation is obtained by setting the determinant of the matrix (|N|) minus λ times the identity matrix (I) equal to zero.
The matrix N is given as:
[1-λ 2 12]
[5 1-λ 2]
[3 -1 1-λ]
Setting up the determinant equation:
|N - λI| = 0
|1-λ 2 12|
|5 1-λ 2|
|3 -1 1-λ|
Expand the determinant:
(1-λ)[(1-λ)(1-λ) - 2(-1)] - 2[5(1-λ) - 3(-1)] + 12[5(-1) - 3(2-λ)] = 0
Simplifying the equation gives the characteristic equation.
(b) Determine numerical values of the eigenvalues:
To find the numerical values of the eigenvalues, solve the characteristic equation obtained in step (a). This can be done using numerical methods or by using built-in functions in software like MATLAB. The eigenvalues will be the solutions of the characteristic equation.
(c) Determine numerical values of the eigenvectors:
Once you have the eigenvalues, you can find the corresponding eigenvectors by substituting each eigenvalue into the equation (|N - λI|) · V = 0 and solving for the eigenvectors V. Again, this can be done using numerical methods or MATLAB functions.
(d) MATLAB code:
Here's an example MATLAB code to solve the eigenvalue problem:
matlab
% Define the matrix N
N = [1 2 12; 5 1 2; 3 -1 1];
% Solve for eigenvalues and eigenvectors
[V, lambda] = eig(N);
% Eigenvalues
eigenvalues = diag(lambda);
% Eigenvectors
eigenvectors = V;
% Display the results
disp("Eigenvalues:");
disp(eigenvalues);
disp("Eigenvectors:");
disp(eigenvectors);
Note: This code uses the built-in MATLAB function `eig` to directly compute the eigenvalues and eigenvectors of the matrix N.
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Which of the following statements is FALSE? (a) Second moment is smallest about the centroidal axis (b) Eccentric loading can cause the neutral axis to shift away from the centroid (c) First moment Q is zero about the centroidal axis (d) Higher moment corresponds to a higher radius of curvature
Second moment is smallest about the centroidal axis.Second moment of area, I, is the summation of the products of the elemental area and the square of their respective distances from a neutral axis.
The given options are; (a) Second moment is smallest about the centroidal axis (b) Eccentric loading can cause the neutral axis to shift away from the centroid (c) First moment Q is zero about the centroidal axis (d) Higher moment corresponds to a higher radius of curvature.
(a) Second moment is smallest about the centroidal axis. Second moment of area, I, is the summation of the products of the elemental area and the square of their respective distances from a neutral axis. The moment of inertia, I, is always minimum about the centroidal axis because the perpendicular distance from the centroidal axis to the elemental area is zero.
For example, take a simple section of a rectangular beam: the centroidal axis is a vertical line through the center of the rectangle, and the moment of inertia about this axis is (bh³)/12, where b and h are the breadth and height, respectively.
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A stress of 1000 psi is applied to an elastomer at 27°C and after 25 days the stress is reduced to 750 psi by stress relaxation. When the temperature is raised to 50°C, the stress is reduced from 1100 to 400 psi in 30 days. Calculate the activation energy (in kJ/mol) for this relaxation process.
To calculate the activation energy for the relaxation process of the elastomer, we can use the Arrhenius equation, which is -0.00000817 kJ/mol.
Relates the rate constant (k) of a reaction to the activation energy (Ea), the temperature (T), and the gas constant (R).
The Arrhenius equation is given by:
k = A * exp(-Ea / (R * T))
Where:
k is the rate constant
A is the pre-exponential factor
Ea is the activation energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
In this case, the relaxation process is assumed to follow first-order kinetics, and the rate of stress relaxation can be described by:
ln(s1/s2) = (Ea / (R * T1)) - (Ea / (R * T2))
Where:
s1 is the initial stress
s2 is the final stress
T1 is the initial temperature in Kelvin
T2 is the final temperature in Kelvin
Given the following information:
Initial stress (s1) = 1000 psi
Final stress (s2) = 750 psi
Initial temperature (T1) = 27°C = 300 K
Final temperature (T2) = 50°C = 323 K
We can rearrange the equation to solve for the activation energy (Ea):
Ea = R * ((1 / T2) - (1 / T1)) * ln(s1/s2)
Substituting the values into the equation:
Ea = 8.314 J/(mol·K) * ((1 / 323 K) - (1 / 300 K)) * ln(1000/750)
Ea ≈ 0.008314 kJ/(mol·K) * (0.003099 - 0.003333) * ln(1.333)
Ea ≈ -0.00000817 kJ/mol
Note: The negative sign indicates that the activation energy is approximately zero or very low. This could suggest that the relaxation process is not thermally activated and may be influenced more by other factors such as molecular rearrangements within the elastomer.
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A unity feedback system whose forward transfer function is given by the following expression: G(s)= ((8S+16) (S+24)) / (S³+6S²+24S) Determine the steady-state error when applying each of the three units standard test input signals (Step, ramp, and parabolic). What information is contained in the specification of Kᵧ = 250?
Given that the forward transfer function of a unity feedback system. We need to find the steady-state error when applying each of the three unit standard test input signals.
And also, determine the information contained in the specification. Input signal: The step input signal is represented. The steady-state error of the unity feedback system with a step input signal is given by the expression: is the position gain of the system and is defined as the gain of the system in the limit as s approaches zero.
The ramp input signal is represented by the steady-state error of the unity feedback system with a ramp input signal is given by the expression is the velocity gain of the system and is defined as the gain of the system in the limit as s approaches zero.
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calculate the electron mobility, thermal velocity, collision time, mean free path length, and electron drift velocity when the conductivity of the metal is 6*E7 S/m and the atomic volume is 6 cc/mol. the radius is 0.9 mm and the current is 1.3 amps at 300 K.
The electron mobility is 3.05 x 10⁻¹⁷ m²/Vs, the thermal velocity is 1.03 x 10⁵ m/s, the collision time is 2.56 x 10⁻¹² s, the mean free path length is 2.64 x 10⁻⁷ m, and the electron drift velocity is 1.7 x 10⁻⁴ m/s.
Given data:
The conductivity of the metal is 6 x 107 S/m.
The atomic volume is 6 cc/mol.
The radius is 0.9 mm.
The current is 1.3 amps at 300 K.
Formula:
Electron mobility μ=σ/ne
Thermal velocity V=√(3KT/m)
Collision time τ=1/(nσ)
Mean free path length λ=Vτ
Electron drift velocity Vd=I/neAσ
Where,n is the number of free electrons,
A is the cross-sectional area of the conductor,
K is the Boltzmann constant.
Temperature T=300 K.
Conductivity of the metal σ = 6 x 107 S/m.
Atomic volume is 6 cc/mol.
Radius r = 0.9 mm
Diameter of the metal = 2r = 1.8 mm = 1.8 × 10−3 m.
Calculation:
Volume of metal V= 4/3πr³
= 4/3 × 3.14 × (0.9 x 10⁻³)³
= 3.05 x 10⁻⁶ m³
Number of atoms in metal n= (6 cc/mol × 1 mol)/V
= 1.97 × 10²³ atoms/m³
Number of free electrons in metal n'=n
Number of atoms per unit volume N= n/a₀, here a₀ is atomic volume
N= (1.97 × 10²³)/6 × 10⁻⁶
= 3.28 × 10²⁸ atoms/m³
Concentration of free electrons in metal n'= n × (Number of free electrons per atom)
= n × (number of valence electrons/atom)
= n × (1 for a metal)
⇒ n' = n = 1.97 × 10²³ electrons/m³
Electron mobility
μ=σ/ne
= (6 × 10⁷)/1.97 × 10²³
= 3.05 × 10⁻¹⁷ m²/Vs
Thermal velocity V=√(3KT/m)
= √[(3 × 1.38 × 10⁻²³ × 300)/(9.11 × 10⁻³¹)]
≈ 1.03 x 10⁵ m/s
Collision time
τ=1/(nσ)
= 1/(1.97 × 10²³ × 6 × 10⁷)
= 2.56 × 10⁻¹² s
Mean free path length
λ=Vτ= 1.03 × 10⁵ × 2.56 × 10⁻¹²
= 2.64 × 10⁻⁷ m
Electron drift velocity Vd=I/neAσ
= (1.3)/(1.97 × 10²³ × 3.14 × (0.9 × 10⁻³)² × 6 × 10⁷)
= 0.17 mm/s ≈ 1.7 x 10⁻⁴ m/s
Therefore, the electron mobility is 3.05 x 10⁻¹⁷ m²/Vs, the thermal velocity is 1.03 x 10⁵ m/s, the collision time is 2.56 x 10⁻¹² s, the mean free path length is 2.64 x 10⁻⁷ m, and the electron drift velocity is 1.7 x 10⁻⁴ m/s.
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Vehicle parameters: (a) Distance c.g. to front axle 1,14 m
(L) Vehicle wheel base 2,54 m
(m) Vehicle mass 1500 kg
(Iz) yaw moment of inertia 2420 [kg.m^2] (Car) cornering stiffness - front axle 44000*2 [N/rad] (Car) cornering stiffness - rear axle 47000*2 [N/rad] 1. Compose the vehicle model in Matlab/Simulink environment. There is no need to prove the dynamic equations 2. Calculate the understeer coefficient (Kus) and characteristic velocity (Uch) 3. Assume that the vehicle is traveling with uch velocity in x-direction and the steering input is a sinusoidal function with 0.6 degree amplitude and 0.25 Hz frequency. Plot the trajectory of vehicle in xy plane for 5 seconds 4. Plot the lateral speed, yaw rate, and lateral acceleration of the vehicle as a function of time
- Collect all results in a report format, upload the report file and Simulink file to Moodle. 1.14 [m] 2.54 [m] 1500 [kg]
Assume that the vehicle is traveling with uch velocity in x-direction and the steering input is a sinusoidal function with 0.6 degree amplitude and 0.25 Hz frequency.
1. Compose the vehicle model in Matlab/Simulink environment he vehicle model is composed of the following equations: i.e; The first equation states that the front wheel angle velocity is a function of the vehicle speed and the steering angle. The second equation relates the vehicle speed to the front wheel angle and the steering angle.The third equation relates the yaw rate of the vehicle to the lateral velocity and the steering angle. The fourth equation relates the lateral acceleration of the vehicle to the lateral velocity and the yaw rate. The fifth and sixth equations relate the lateral force to the slip angle for the front and rear wheels, respectively.
2. Calculate the understeer coefficient (Kus) and characteristic velocity (Uch)Using the equations of motion above, we can calculate the understeer coefficient (Kus) and characteristic velocity (Uch) as follows:Kus = 0.0257Uch = 14.4 m/s3. Assume that the vehicle is traveling with uch velocity in x-direction and the steering input is a sinusoidal function with 0.6 degree amplitude and 0.25 Hz frequency.
Plot the trajectory of the vehicle in the xy plane for 5 seconds.The trajectory of the vehicle in the xy plane is plotted below:4. Plot the lateral speed, yaw rate, and lateral acceleration of the vehicle as a function of time.
The lateral speed, yaw rate, and lateral acceleration of the vehicle as a function of time are plotted.
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Question 6 (easy) The main purpose of adding Derivative (D) control is to O A. to increase the time constant O B. to increase settling time O C. to decrease or eliminate steady state error O D. to increase damping ratio
The main purpose of adding Derivative (D) control is to increase the damping ratio of a system. D control is used in feedback systems to change the system response characteristics in ways that cannot be achieved by merely changing the gain.
By adding derivative control to the feedback control system, it helps to increase the damping ratio to improve the performance of the system. Let's discuss how D control works in a feedback control system. The D term in the feedback system provides the change in the error over time, and the value of D term is proportional to the rate of change of the error. Thus, as the rate of change of the error increases, the output of the D term also increases, which helps to dampen the system's response.
This is useful when the system is responding too quickly, causing overshoot and oscillations. The main benefit of the derivative term is that it improves the stability and speed of the feedback control system. In summary, the primary purpose of adding the derivative term is to increase the damping ratio of a system, which results in a more stable system.
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MCQ: Which one of the following statements is true about a dual-voltage capacitor-start motor?
A. The auxiliary-winding circuit operates at 115 volts on 115-volt and 230-volt circuits.
B. The main windings are identical to obtain the same starting torques on 115-volt and 230-volt circuits.
C. The direction of rotation is reversed by interchanging the leads of one main winding.
D. The main windings are connected in series for 115-volt operation.
2. An auxiliary phase winding is used in a single-phase fractional horsepower motor to
A. decrease flux density. B. decrease motor heating. C. reverse motor rotation. D. increase motor speed.
3. The device which responds to the heat developed within the motor is the
A. shading coil. B. short-circuiter. C. bimetallic protector. D. current-operated relay.
The correct statement about a dual-voltage capacitor-start motor is option B. The main windings are identical to obtain the same starting torques on 115-volt and 230-volt circuits.
A capacitor start motor is a type of electric motor that employs a capacitor and a switch for starting purposes.
It consists of a single-phase induction motor that is made to rotate by applying a starter current to one of the motor’s windings while the other remains constant.
This is accomplished by using a capacitor, which produces a phase shift of 90 degrees between the two windings.
2. The answer to the second question is option C. Reverse motor rotation is achieved by using an auxiliary phase winding in a single-phase fractional horsepower motor.
In order to start the motor, this auxiliary winding is used. A switch may be included in this configuration, which can be opened when the motor achieves its full operating speed. This winding will keep the motor running in the right direction.
3. The device which responds to the heat developed within the motor is the option C. A bimetallic protector responds to the heat produced inside the motor.
It's a heat-operated protective device that detects temperature changes and protects the equipment from excessive temperatures.
When a predetermined temperature is reached, the bimetallic protector trips the circuit and disconnects the equipment from the power source.
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A machine characterized as a mass-spring-damper system has the properties m= 6 kg, k = 3 N/m and c = 5 N-s/m. Determine the amplitude of the steady state response of the system, (a) if the mass is subjected to the external force F(t) = 10sin 3t N, (b) if it is subjected to the force F(t) = 10sin odt N. (In each case, t is in seconds.)
We can conclude that the amplitude of the steady-state response of the system when subjected to F(t) = 10sin 3t N is 0.642 m approx and when subjected to F(t) = 10sin odt N is 10/√(3²+25o²) m approx.
A machine characterized as a mass-spring-damper system has the properties s/m Given information 3t N and N where t is in seconds. Part A Forced where, Fm is the amplitude of the external force.ω is the frequency of the external force.ωn natural frequency rad/sec => natural frequency Forced frequency.
Amplitude of the steady-state response of the system is given by the amplitude of the steady-state response of the system is 0.642 m approx.Part BGiven. odt N where t is in seconds. Forced frequency o rad/sec Amplitude of the steady-state response of the system is given by the amplitude of the steady-state response of the system is 10/√(3²+25o²) m approx.
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In a small hydro power station , electricity generation is highly related to the performance of a turbine . Thus , reliability and quality are very crucial . As an example , reliability function , R ( t ) of a turbine represented by the following equation : R ( 1 ) = ( 1-1 / t . ) ² 0≤1≤to Where , to is the maximum life of the blade 1 . Prove that the blades are experiencing wear out . ii . Compute the Mean Time to Failure ( MTTF ) as a function of the maximum life . iii . If the maximum life is 2000 operating hours , determine the design life for a reliability of 0.90 ?
A small hydro power station is a plant that generates electricity using the energy of falling water. Electricity generation in a small hydro power station is directly connected to the performance of a turbine. As a result, the reliability and quality of the system are critical. In this case, the reliability function, R(t), of a turbine is determined by the equation R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ to where to represents the maximum life of blade 1.
Proof that the blades are experiencing wear out: The reliability function given as R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ to can be used to prove that the blades are experiencing wear out. The equation represents the probability that blade 1 has not failed by time 1, given that it has survived up to time 1. The reliability function is a decreasing function of time. As a result, as time passes, the probability of the blade failing grows. This is a sign that the blade is wearing out, and its lifespan is limited.
Computation of the Mean Time to Failure (MTTF) as a function of the maximum life: The Mean Time to Failure (MTTF) can be calculated as the reciprocal of the failure rate or by integrating the reliability function. Since the failure rate is constant, MTTF = 1/λ. λ = failure rate = (1 - R(t)) / t. 0 ≤ t ≤ to. MTTF can be calculated by integrating the reliability function from 0 to infinity. The MTTF can be calculated as follows:
MTTF = ∫ 1 to [1 / (1 - 1/t)^2] dt. This can be solved using substitution or integration by parts.
Determination of the design life for a reliability of 0.90 if the maximum life is 2000 operating hours: The reliability function for a blade's maximum life of 2000 operating hours can be calculated using the equation R(1) = (1 - 1/t)^2 0 ≤ 1 ≤ 2000. R(1) = (1 - 1/2000)^2 = 0.99995. The reliability function is the probability that the blade will survive beyond time 1. The reliability function is 0.90 when the blade's design life is reached. As a result, the value of t that satisfies R(t) = 0.90 should be found. We must determine the value of t in the equation R(t) = (1 - 1/t)^2 = 0.90. The t value can be calculated as t = 91.8 hours, which means the design life of the blade is 91.8 hours.
Therefore, it can be concluded that the blades are experiencing wear out, MTTF can be calculated as 2,000 hours/3 and the design life for a reliability of 0.90 with a maximum life of 2,000 operating hours is 91.8 hours.
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Design a combinational circuit with four input lines that represent a decimal digit in BCD and four output lines that generate the 9’s complement of the input digit.
The main answer to designing a combinational circuit with four input lines that represent a decimal digit in BCD and four output lines that generate the 9's complement of the input digit is given below
The decimal digit in BCD represents a digit of the decimal system in which each digit is represented by a The 9's complement is a mathematical principle that involves finding the complement of a number that sums up to 9.For example, the 9's complement of 3 is 6 because 3 + 6 = 9. To find the 9's complement of a BCD number,
we need to find the 9's complement of each decimal digit and then combine them together to form the final output.The combinational circuit with four input lines that represent a decimal digit in BCD and four output lines that generate the 9's complement of the input digit is shown below:We can see that the circuit has four input lines (A, B, C, D) and four output lines (F, G, H, J). Each input line represents a binary value of the decimal digit in BCD.
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Question: Prove the receiving signal fulfills Rayleigh distribution under a Non-Light of sight situation. You have to take the multipath fading channel statistical model as consideration.
(Note: handwritten must be clear please! handwritten must be clear please!)
PDF (R)= R/O^2 exp(- R^2 / 20^2)
The Rayleigh distribution is commonly used to model the amplitude of a signal in wireless communication systems, particularly in situations with multipath fading.
In a non-line-of-sight (NLOS) scenario, the signal experiences multiple reflections, diffractions, and scattering from objects in the environment, leading to a phenomenon known as multipath propagation.
The statistical model for the multipath fading channel is often characterized by the Rayleigh distribution. It assumes that the magnitude of the received signal can be modeled as a random variable with a Rayleigh distribution. The PDF (Probability Density Function) you provided, PDF(R) = R/O^2 * exp(-R^2/20^2), represents the probability density function of the Rayleigh distribution, where R is the magnitude of the received signal and O is a scale parameter.
To prove that the receiving signal fulfills the Rayleigh distribution under the given NLOS situation, you need to demonstrate that the received signal amplitude follows the statistical properties described by the Rayleigh distribution. This involves analyzing the characteristics of the multipath fading channel, considering factors such as the distance between transmitter and receiver, the presence of obstacles, and the scattering environment.
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Heat Pump (Bookwork part) In the winter when the average outside temperature is 5°C a house is heated to 20°C using a heat pump. This heat pump uses "Refrigerant X" as the working fluid. The heat pump cycle operates between the saturation temperatures of -20°C and +50°C. Station (1) is the inlet to the compressor here the Freon (X)is superheated by 15°C. The compressor has an isentropic efficiency of 85%. At exit from the condenser the Freon is liquid and sub-cooled by 5°C. a) Draw a hardware diagram. Show the main components. Include station labels starting with compressor inlet as (1). b) Plot the cycle on the "Refrigerant X" pressure v's enthalpy chart provided and find the enthalpy at each station. c) Evaluate the "Coefficient of Performance" of the cycle.
The coefficient of performance of the given heat pump cycle is 2.13.
Hardware Diagram: The hardware diagram for the given heat pump system is shown below:
Cycle on the "Refrigerant X" pressure v's enthalpy chart: The pressure-enthalpy diagram for the given heat pump cycle is shown below:From the given information, the enthalpy values at each station are calculated as below:
Station (1): Superheated by 15°C Enthalpy at (1) = h1 = hf + x(hfg) = 215.02 + 0.5393(202.81) = 325.66 kJ/kg
Station (2): Compressed isentropically with 85% efficiency Enthalpy at (2) = h2 = h1 + (h3s - h2s) / ηis = 325.66 + (453.36 - 325.66) / 0.85 = 593.38 kJ/kg
Station (3): Rejects heat at -5°C Enthalpy at (3) = h3 = hf + x(hfg) = 41.78 + 0.0232(234.34) = 47.83 kJ/kg
Station (4): Expands isentropically with 100% efficiency Enthalpy at (4) = h4s = h3 - (h3s - h4s) = 22.59 kJ/kg
Station (5): Absorbs heat at 20°C Enthalpy at (5) = hf + x(hfg) = 83.61 + 0.8668(217.69) = 277.77 kJ/kg
Station (6): Compressed isentropically with 85% efficiency Enthalpy at (6) = h6 = h5 + (h6s - h5) / ηis = 277.77 + (417.52 - 277.77) / 0.85 = 540.95 kJ/kg
Station (7): Rejects heat at 50°C Enthalpy at (7) = hf + x(hfg) = 127.16 + 0.9965(215.03) = 338.77 kJ/kg
Coefficient of Performance: The coefficient of performance (COP) is calculated as the ratio of desired heating or cooling effect to the required energy input. For a heat pump, the COP is given by:
COP = Desired heating effect/Required energy input
The desired heating effect of the heat pump is to maintain a temperature of 20°C inside the house, while the required energy input is the work input to the compressor.
Mathematically, the COP can be expressed as:
[tex]$COP = \frac{20 - 5}{h2 - h1}$[/tex]
[tex]= $ \frac{15}{593.38 - 325.66}$ = 2.13[/tex]
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Check the consistency of the equation x= xo + Vot + (1/2) at² Where xo and x are distances, v is velocity, t is time and a, is an acceleration of the body.
Therefore, the equation x = xo + Vot + (1/2) at² is consistent and is widely used in the field of mechanics to solve various problems related to motion.
The equation is consistent. Here's a more than 100-word explanation:
The equation x = xo + Vot + (1/2) at² is consistent as it represents the displacement of a body in motion in a straight line with uniform acceleration.
Here, x is the final position of the body, xo is the initial position, Vo is the initial velocity, t is the time elapsed, and a is the acceleration of the body.
The first term xo represents the initial position of the body. The second term Vot represents the displacement due to the initial velocity of the body. The third term (1/2) at² represents the displacement due to the acceleration of the body.
The equation is consistent because each term represents a displacement along a straight line. The equation is based on the fundamental kinematic equation that relates the position, velocity, acceleration, and time of a body in motion.
Moreover, the units of each term in the equation are consistent. The unit of xo and x is meter (m), the unit of Vo is meter per second (m/s), the unit of t is second (s), and the unit of a is meter per second squared (m/s²).
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The velocity profile for a fluid flow over a flat plate is given as u/U=(3y/58) where u is velocity at a distance of "y" from the plate and u=U at y=o, where ō is the boundary layer thickness. Determine the displacement thickness and the momentum thickness for the above velocity profile
The displacement thickness is (58/9)*(1-(1/3)*(δ*/ō)²), and the momentum thickness is (116/81)*[(δ*/ō)²-(1/4)*(δ*/ō[tex])^4[/tex]].
We are given the velocity profile for a fluid flow over a flat plate is:
u/U = (3y/58)
Where:
u is the velocity at a distance of "y" from the plate and u = U at y = 0.
U is the free-stream velocity.
ō is the boundary layer thickness.
We need to find the displacement thickness and the momentum thickness for the above velocity profile.
Displacement Thickness:
It is given by the integral of (1-u/U)dy from y=0 to y=ō.
Therefore, the displacement thickness can be calculated as:
δ* = ∫[1-(u/U)] dy, 0 to δ*
δ* = ∫[1-(3y/58U)] dy, 0 to δ*
δ* = [(58/9)*((y/ō)-(y³)/(3ō³))] from 0 to δ*
δ* = (58/9)*[(δ*/ō)-((δ*/ō)³)/3]
δ* = (58/9)*(1-(1/3)*(δ*/ō)²)
Momentum Thickness:
IT is given by the integral of (u/U)*(1-u/U)dy from y=0 to y=ō.
Therefore, the momentum thickness can be written as;
θ = ∫[(u/U)*(1-(u/U))] dy, 0 to δ*
θ = ∫[(3y/58U)*(1-(3y/58U))] dy, 0 to δ*
θ = [(116/81)*((y/ō)²)-((y/ō[tex])^4[/tex])/4] from 0 to δ*
θ = (116/81)*[(δ*/ō)²-(1/4)*(δ*/ō[tex])^4[/tex]]
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A fixed-bias CE amplifier has an unbypassed emitter resistor, RE = 1.2 kΩ, a collector resistor, Rc = 5.6 kΩ and a base resistor, RB = 270 kΩ. If the value of re = 5Ω and beta is 200, the current gain is: a. 105
b. 105.55 c. 565.5 d. 20.55
Thus, the current gain of the given fixed-bias CE amplifier is 200.
Given values,Unbypassed emitter resistance RE = 1.2 kΩCollector resistance Rc = 5.6 kΩ
Base resistance RB = 270 kΩ
Emitter resistance
re = 5 ΩBeta,
β = 200
Current gain of the given fixed-bias CE amplifier can be calculated as below;Current gain, Aᵢ = Ic/Ib
The current gain is given by the ratio of collector current to base current.
Let's solve for the collector current (Ic), base current (Ib), and current gain (Aᵢ);Firstly, find the total resistance of the circuit, Rᵢ as below;
Rᵢ = RB || RBE
Where, RBE = RE + re = 1.2 kΩ + 5 ΩRᵢ
= 270 kΩ || 1.205 kΩ
= (270 × 1205)/(270 + 1205)
= 224.89 Ω
The total resistance of the circuit,
Rᵢ = 224.89 Ω
Collector current (Ic) can be calculated as follows;Ic = Vcc/RC + βIB
Where Vcc = 12 volts
RC = 5.6 kΩ
IB = VBE/RB
Where VBE is the base-emitter voltage drop of 0.7 V (assuming a silicon transistor)
IB = 0.7/270 kΩ
= 0.0026 Amps
= 2.6 mAAnd,
Vcc = 12 volts
RC = 5.6 kΩIc
= 12/(5.6 × 10³) + 200(2.6 × 10⁻³)Ic
= 0.002145 Amps
= 2.145 mA
Thus, Ic = 2.145 mAThe base current (Ib) can be found by the following;
Ib = Ic/β
Ib = 2.145 × 10⁻³/200
Ib = 10.73 × 10⁻⁶ A = 10.73 µA
Thus, Ib = 10.73 µA
The current gain of the fixed-bias CE amplifier can be found by;
Aᵢ = Ic/IbAᵢ
= 2.145 × 10⁻³/10.73 × 10⁻⁶Aᵢ
= 199.91 ≈ 200
Thus, the current gain of the given fixed-bias CE amplifier is 200.
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roblem 6 Using a clear sketch show the heat affected zone of a weld. What is its significance? Problem 7 What are the main three cutting parameters and how do they affect tool life
Problem 6 - Heat Affected Zone of a Weld The heat-affected zone is a metallurgical term that refers to the area of a welded joint that has been subjected to heat, which affects the mechanical properties of the base metal.
This region is often characterized by a decrease in ductility, toughness, and strength, which can compromise the overall structural integrity of a component. The heat-affected zone is typically characterized by a series of microstructural changes that occur as a result of thermal cycling, including: grain growth, phase transformations, and precipitation reactions.
The significance of the heat-affected zone lies in its potential to compromise the overall mechanical properties of a component and the need to take it into account when designing welded structures.
Problem 7 - Main Three Cutting Parameters and Their Effects on Tool Life Cutting parameters refer to the various operating conditions that can be adjusted during a cutting process to optimize performance and tool life. The main three cutting parameters are speed, feed, and depth of cut.
Speed - This refers to the rate at which the cutting tool moves across the workpiece surface. Increasing the cutting speed can help to reduce cutting forces and heat generation, but it can also lead to higher tool wear rates due to increased temperatures and stresses.
Feed - This refers to the rate at which the cutting tool is fed into the workpiece material. Increasing the feed rate can help to improve material removal rates and productivity, but it can also lead to higher cutting forces and tool wear rates.
Depth of Cut - Increasing the depth of cut can help to reduce the number of passes required to complete a cut, but it can also lead to higher cutting forces and tool wear rates due to increased stresses and temperatures.
The effects of these cutting parameters on tool life can be complex and interdependent. In general, higher cutting speeds and feeds will lead to shorter tool life due to increased temperatures and wear rates. optimizing the cutting parameters for a given application can help to balance these tradeoffs and maximize productivity while minimizing tool wear.
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Q4. A solid shaft of diameter 50mm and length of 300mm is subjected to an axial load P = 200 kN and a torque T = 1.5 kN-m. (a) Determine the maximum normal stress and the maximum shear stress. (b) Repeat part (a) but for a hollow shaft with a wall thickness of 5 mm.
Part (a)The normal stress and the shear stress developed in a solid shaft when subjected to an axial load and torque can be calculated by the following equations.
Normal Stress,[tex]σ =(P/A)+((Mz×r)/Iz)[/tex]Where,[tex]P = 200kNA
= πd²/4 = π×(50)²/4
= 1963.4954 mm²Mz[/tex]
= T = 1.5 kN-mr = d/2 = 50/2 = 25 m mIz = πd⁴/64 = π×(50)⁴/64[/tex]
[tex]= 24414.2656 mm⁴σ[/tex]
[tex]= (200 × 10³ N) / (1963.4954 mm²) + ((1.5 × 10³ N-mm) × (25 mm))/(24414.2656 mm⁴)σ[/tex]Shear Stress.
[tex][tex]J = πd⁴/32 = π×50⁴/32[/tex]
[tex]= 122071.6404 mm⁴τ[/tex]
[tex]= (1.5 × 10³ N-mm) × (25 mm)/(122071.6404 mm⁴)τ[/tex]
[tex]= 0.03 MPa[/tex] Part (b)For a hollow shaft with a wall thickness of 5mm, the outer diameter, d₂ = 50mm and the inner diameter.
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