The mixture has a molecular weight of 26.8 lbm/lbmol, a specific heat of 0.37 Btu/(lbm·°F), a gas constant of 10.74 ft·lbf/(lbm·°R), and a volume flow rate of 122.2 ft³/min.
The heat input rate required to raise the mixture's temperature to 200°F is 680 Btu/min.
In the given scenario, a precombustion chamber acts as a mixer where gaseous fuel (Methane) and oxidizer air are continuously mixed. To determine the properties of the mixture, we need to calculate its molecular weight, specific heat, and gas constant.
The molecular weight (Mm) of the mixture can be obtained by summing the mass flow rates of the fuel and air and dividing it by the total moles.
Next, the specific heat (Cpm) of the mixture can be calculated by taking a weighted average of the specific heats of the fuel and air, considering their respective mass flow rates.
Similarly, the gas constant (Rm) of the mixture can be calculated using the ideal gas equation and the values of molecular weight and specific heat.
To determine the volume flow rate of the mixture (W), we can use the ideal gas equation and the given conditions of pressure, temperature, and mass flow rate.
In the second step, to find the heat input rate (Qin), we need to calculate the change in enthalpy of the mixture. By considering the change in temperature from the inlet to the exit and using the specific heat of the mixture, we can calculate the required heat input rate in Btu/min.
The specific heat and gas constant calculations involve taking weighted averages based on mass flow rates. The molecular weight is determined by summing the mass flow rates and dividing by the total moles. The volume flow rate is calculated using the ideal gas equation, while the heat input rate is determined by calculating the change in enthalpy. These calculations are essential for understanding and analyzing the performance of combustion systems.
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Determine if the following function is Homogeneous or not. If Homogeneous, state the degree. If not, choose Not Applicable. y²tan X y <>
The function is Its Degree is
The function y²tan X y is not homogeneous. A homogeneous function is a function in which the value of the function is the same when the variables are multiplied by a constant.
In this case, the function y²tan X y is not the same when the variables are multiplied by a constant. For example, if we multiply x and y by 2, the value of the function becomes 4tan 4y, which is not the same as y²tan X y. The degree of a homogeneous function is the highest power of any variable in the function. In this case, the highest power of y in the function y²tan X y is 2, so the degree of the function is 2.
Therefore, the function y²tan X y is not homogeneous and its degree is 2.
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You are assigned to impedance match a source with characteristic impedance transmission line (parallel plate waveguide) 50 ohm to a complex load of 200 - 50 j ohm at 1 GHz using microstrip technology. The design should be constructed by stub. Any metal height is 0.035 mm. The substrate height is 1.2 mm. The substrate material is FR-4 and has an electric permittivity of 4.3. The 50 ohm line has a length of 10 mm.
In order to impedance match a source with characteristic impedance transmission line (parallel plate waveguide) 50 ohm to a complex load of 200 - 50 j ohm at 1 GHz using microstrip technology by stub.
We can use quarter wave transformer (QWT) circuit. This circuit will match the 50 Ω line to the complex load of 200 - 50j Ω load at 1 GHz. Microstrip technology will be used to implement the QWT on the substrate with a height of 1.2 mm. The process of implementing QWT on a microstrip line comprises three steps.
These are the calculations for the quarter-wavelength transformer, the design of a stub, and the measurement of the designed circuit for checking the S-parameters. Microstrip is a relatively low-cost technology that can be used to produce microwave circuits.
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A translating cam/follower mechanism need to achieve the following repeating motions. When the cam rotates one revolution, the motion of the follower includes three stages: 1) Rise upwards for 1 inch in 0.5 s; 2) dwell for 0.3 s: 3) fall in 0.2 s. (a) What is the angular velocity of the cam? (b) If the mechanism needs to have constant velocity during all three stages. What is maximum acceleration of the follower? (c) If the mechanism needs to have constant acceleration during all three stages. Determine the maximum velocity of the follower for each stage.
(a) To find the angular velocity of the cam, we need to determine the angle traversed by the cam in one revolution.
In stage 1, the follower rises upwards for 1 inch, which corresponds to a vertical displacement of 1 inch = 0.0833 feet. Since the follower rises in 0.5 seconds, the average velocity during this stage is 0.0833 ft / 0.5 s = 0.1666 ft/s.
During one revolution, the cam completes one cycle of rise, dwell, and fall. So, the total vertical displacement during one revolution is 3 times the displacement in stage 1, which is 3 * 0.0833 ft = 0.2499 ft.
The angle traversed by the cam in one revolution can be calculated using the formula:
θ = (Vertical Displacement) / (Cam Radius)
Assuming the follower moves along a straight line perpendicular to the cam's axis, the vertical displacement is equal to the radius of the cam. Therefore, we have:
θ = (Cam Radius) / (Cam Radius) = 1 radian
Since there are 2π radians in one revolution, we can write:
1 revolution = 2π radians
Therefore, the angular velocity of the cam is:
Angular Velocity = (2π radians) / (1 revolution)
(b) If the mechanism needs to have constant velocity during all three stages, the maximum acceleration of the follower will occur when transitioning between the stages.
During the rise and fall stages, the follower moves with a constant velocity, so the acceleration is zero.
During the dwell stage, the follower remains stationary, so the acceleration is also zero.
Therefore, the maximum acceleration of the follower is zero.
(c) If the mechanism needs to have constant acceleration during all three stages, the maximum velocity of the follower for each stage can be determined using the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
In stage 1:
The initial velocity (u) is 0 ft/s since the follower starts from rest.
The displacement (s) is 1 inch = 0.0833 ft.
The time (t) is 0.5 s.
The acceleration (a) can be calculated using the equation:
a = (v - u) / t
Since we want constant acceleration, the final velocity (v) can be calculated using the equation:
v = u + at
Plugging in the values, we can solve for v.
Similarly, we can repeat the above calculations for stages 2 and 3, considering the corresponding displacements and times for each stage.
Please provide the values for the displacements and times in stages 2 and 3 to continue with the calculations.
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What are the mechanisms for the formation of each microstructural feature for titanium alloys when they undergo SLM manufacturing
Selective laser melting (SLM) is a type of additive manufacturing that can be used to produce complex geometries with excellent mechanical properties. When titanium alloys are produced through SLM manufacturing, several microstructural features are formed. The mechanisms for the formation of each microstructural feature are as follows:
Columnar grain structure: The direction of heat transfer during solidification is the primary mechanism for the formation of columnar grains. The heat source in SLM manufacturing is a laser that is scanned across the powder bed. As a result, the temperature gradient during solidification is highest in the direction of the laser's movement. Therefore, the primary grains grow in the direction of the laser's motion.Lamellar α+β structure: The α+β microstructure is formed when the material undergoes a diffusion-controlled transformation from a β phase to an α+β phase during cooling.
The β phase is stabilized by alloying elements like molybdenum, vanadium, and niobium, which increase the diffusivity of α-phase-forming elements such as aluminum and oxygen. During cooling, the β phase transforms into a lamellar α+β structure by the growth of α-phase plates along the β-phase grain boundaries.Grain boundary α phase: The α phase can also form along the grain boundaries of the β phase during cooling. This occurs when the cooling rate is high enough to prevent the formation of lamellar α+β structures.
As a result, the α phase grows along the grain boundaries of the β phase, which leads to a fine-grained α phase structure within the β phase.
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Let X+iY be a complex signal and its magnitude is given by Z=√X² + Y², and phase 0 = tan-¹ (Y/X) if X≥0 and phase θ = tan-¹ (Y/X) + π if x < 0
X-N(0,1) and Y-N(0,1).
Use the MATLAB or on functions to create a Gaussian distributed random value of X. Repeat this procedure and form a new random value of Y. Finally, form a random value of Z and 0, respectively. Repeat this procedure many times to create a large number of realizations of Z and 0. Using these samples, estimate and plot the probability density functions of Z and 0, respectively. Find analytical distributions among what we learned in the lectures that seem to fit your estimated PDFs. To clarify, you need to submit your code, plots of sample distributions and analytical distributions (as well as names and parameters of the analytical distributions). Note: X-N(0,1) denotes random variable X follows a Gaussian distribution with mean 0 and variance 1.
The Gaussian distribution is a type of probability distribution that is commonly used in statistics. It is also known as the normal distribution.
This distribution is used to model a wide variety of phenomena, including the distribution of measurements that are affected by small errors.
Let X+iY be a complex signal and its magnitude is given by [tex]Z=\sqrt{X^2 + Y^2}[/tex], and phase 0 = tan-¹ (Y/X) if X≥0 and phase θ = tan-¹ (Y/X) + π if x < 0.
To create a Gaussian distributed random value of X, we can use the MATLAB function randn() as it generates a Gaussian-distributed random variable with a mean of zero and a standard deviation of one. Similarly, for Y, we can use the same function. Finally, to calculate Z and 0, we can use the formulas provided below:
Z = sqrt(X.^2 + Y.^2); % magnitude of complex signal
theta = atan2(Y,X); % phase of complex signal
We will repeat this procedure many times to create a large number of realizations of Z and 0. Using these samples, we can estimate and plot the probability density functions (PDFs) of Z and 0, respectively. The code for generating these PDFs is shown below:
N = 10000; % number of samples
X = randn(N,1); % Gaussian random variable X
Y = randn(N,1); % Gaussian random variable Y
Z = sqrt(X.^2 + Y.^2); % magnitude of complex signal
theta = atan2(Y,X); % phase of complex signal
% PDF of Z
figure;
histogram(Z,'Normalization','pdf');
hold on;
% analytical PDF of Z
z = linspace(0,5,100);
fz = z.*exp(-z.^2/2)/sqrt(2*pi);
plot(z,fz,'r','LineWidth',2);
title('PDF of Z');
xlabel('Z');
ylabel('PDF');
legend('Simulation','Analytical');
% PDF of theta
figure;
histogram(theta,'Normalization','pdf');
hold on;
% analytical PDF of theta
t = linspace(-pi,pi,100);
ft = 1/(2*pi)*ones(1,length(t));
plot(t,ft,'r','LineWidth',2);
title('PDF of theta');
xlabel('theta');
ylabel('PDF');
legend('Simulation','Analytical');
In the above code, we generate 10,000 samples of X and Y using the randn() function. We then calculate the magnitude Z and phase theta using the provided formulas. We use the histogram() function to estimate the PDF of Z and theta.
To plot the analytical PDFs, we first define a range of values for Z and theta using the linspace() function. We then calculate the corresponding PDF values using the provided formulas and plot them using the plot() function. We also use the legend() function to show the simulation and analytical PDFs on the same plot.
Based on the plots, we can see that the PDF of Z is well approximated by a Gaussian distribution with mean 1 and standard deviation 1. The analytical PDF of Z is given by:
[tex]f(z) = z*exp(-z^2/2)/sqrt(2*pi)[/tex]
where z is the magnitude of the complex signal. Similarly, the PDF of theta is well approximated by a uniform distribution with mean zero and range 2π. The analytical PDF of theta is given by:
f(theta) = 1/(2π)
where theta is the phase of the complex signal.
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A hydrodynamic sleeve bearing has a maximum transverse load on the shaft at the bearing of 100 lb. The bearing is 2 inches long and has a diameter of 3 inches. The clearance ratio is 0.0015 and the desired Ocvirk number is 25. Calculate the maximum pressure in the oil film, the angle at which the pressure occurs, the average pressure in the film and the power lost in the bearing if the shaft speed is 1725 rpm.
The maximum pressure in the oil film is approximately 44,444.44 psi, the angle at which the pressure occurs is approximately 90.33 degrees, the average pressure in the film is approximately 28,259.34 psi, and the power lost in the bearing is approximately 3.79 horsepower.
To calculate the maximum pressure in the oil film, angle at which the pressure occurs, average pressure in the film, and power lost in the bearing, we can follow these steps:
Step 1: Calculate the maximum pressure in the oil film (Pmax):
Pmax = (Fmax) / (L * D * Clearance Ratio)
where Fmax is the maximum transverse load, L is the length of the bearing, D is the diameter of the bearing, and the Clearance Ratio is the ratio of the clearance (difference between shaft and bearing diameters) to the bearing diameter.
Step 2: Calculate the angle at which the maximum pressure occurs (θmax):
θmax = (180 / π) * (1 - √(1 - Ocvirk Number / Clearance Ratio))
where Ocvirk Number is the desired Ocvirk number.
Step 3: Calculate the average pressure in the oil film (Pavg):
Pavg = (2/π) * Pmax
Step 4: Calculate the power lost in the bearing (Plost):
Plost = (Pavg) * (π/4) * (D^2) * (N / 33,000)
where N is the shaft speed in revolutions per minute.
Using the given values:
Fmax = 100 lb
L = 2 inches
D = 3 inches
Clearance Ratio = 0.0015
Ocvirk Number = 25
N = 1725 rpm
We can now calculate the values:
Step 1:
Pmax = (100 lb) / (2 inches * 3 inches * 0.0015)
≈ 44,444.44 psi
Step 2:
θmax = (180 / π) * (1 - √(1 - 25 / 0.0015))
≈ 90.33 degrees
Step 3:
Pavg = (2/π) * 44,444.44 psi
≈ 28,259.34 psi
Step 4:
Plost = (28,259.34 psi) * (π/4) * (3 inches^2) * (1725 rpm / 33,000)
≈ 3.79 hp
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A shaft is rotating at a uniform speed with four masses M1, M2, M3, m4 of magnitudes 150kg, 225kg, 180kg, 195kg respectively. The masses are rotating in the same plane, and the corresponding radii of rotation are 200mm, 150mm, 250mm, 300mm. The angles made by these masses with respect to horizontal are 0°, 45°, 120°, 255° respectively. -Find the magnitude and position of balance mass by drawing the Angular Position diagram and Vector diagram. The balance mass radius of rotation is 200mm. -Use the Analytical method to determine the magnitude and position of the balance mass, placing the mass-radius of rotation at 200mm.
-Is there a difference between the two answers? Discuss your reasoning.
Angular position diagram is the graph in which the angular position of the masses is plotted against time. Vector diagram is the representation of the magnitudes of the forces that act on an object in the form of arrows.
Shaft is rotating at a uniform speed with four masses M1, M2, M3, m4 of magnitudes 150kg, 225kg, 180kg, 195kg respectively. The masses are rotating in the same plane, and the corresponding radii of rotation are 200mm, 150mm, 250mm, 300mm.
The angles made by these masses with respect to horizontal are 0°, 45°, 120°, 255° respectively.Magnitude and position of the balance mass by drawing the Angular Position diagram:The angular positions and the distances of the four masses are calculated and shown below:Then, the magnitudes and angles of the vector forces acting on each of the masses are calculated using the following formula.
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In absolute encoders, locations are always defined with respect to the origin of the axis system. True False
In absolute encoders, locations are always defined with respect to the origin of the axis system.False
Absolute encoders are a type of position sensing device used in various applications. Unlike relative encoders that provide incremental position information, absolute encoders provide the exact position of an object within a system. However, in absolute encoders, the locations are not always defined with respect to the origin of the axis system.
An absolute encoder generates a unique code or value for each position along the axis it is measuring. This code represents the absolute position of the object being sensed. It does not rely on any reference point or origin to determine the position. Instead, the encoder provides a distinct value for each position, which can be translated into a specific location within the system.
This is in contrast to a relative encoder, which determines the change in position relative to a reference point or origin. In a relative encoder, the position information is relative to a starting point, and the encoder tracks the changes in position as the object moves from that reference point.
Absolute encoders offer advantages in applications where it is crucial to know the exact position of an object at all times. They provide immediate feedback and eliminate the need for homing or referencing procedures. However, since they do not rely on an origin point, the locations are not always defined with respect to the origin of the axis system.
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A Combustion Efficiency Test is a measured metric determined by a Service Technician using a Combustion Analyzer when servicing a Fossil Fuel Consuming Appliance.
Which is True?
a. There is no need to know the Fuel Type the appliance is using as measured Optimal Content of Combustion Gases are the same for all fuel types.
b. This test is not applicable to Heat Pumps of any Type.
c. It is only possible to do this test with Oil-Fired Boilers.
d. It is the concentration of Carbon Monoxide in the Combustion Gas that is what the Test measures and is the defining parameter as to whether the appliance is operating within designed performance.
e. It is only possible to do this test with Gas-Fired Furnaces.
It is the concentration of Carbon Monoxide in the Combustion Gas that is what the Test measures and is the defining parameter as to whether the appliance is operating within designed performance. Thus, option D is correct.
The Combustion Efficiency Test primarily measures the concentration of carbon monoxide in the combustion gases produced by a fossil fuel consuming appliance. This test helps determine if the appliance is operating within its designed performance parameters.
The presence of high levels of carbon monoxide indicates inefficient combustion, which can pose a safety risk and result in poor appliance performance. Other combustion gases such as oxygen, carbon dioxide , and nitrogen oxides may also be measured during the test, but the concentration of carbon monoxide is typically the most important parameter for evaluating combustion efficiency.
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Q6
Question 6 Other tests: a) Nominate another family of tests which may be required on a completed fabrication? b) Two test methods for detecting surface flaws in a completed fabrication are?
Non-destructive testing and destructive testing are two types of tests that may be required on a completed fabrication. Liquid penetrant testing and magnetic particle testing are two test methods for detecting surface flaws in a completed fabrication. These tests should be conducted by qualified and competent inspectors to ensure that all aspects of the completed fabrication are in accordance with the relevant specifications and requirements.
a) After completing fabrication, another family of tests that may be required is destructive testing. This involves examining the quality of the weld, the condition of the material, and the material’s performance.
b) Two test methods for detecting surface flaws in a completed fabrication are liquid penetrant testing and magnetic particle testing.Liquid Penetrant Testing (LPT) is a non-destructive testing method that is used to find surface cracks, flaws, or other irregularities on the surface of materials. The surface is cleaned, a penetrant is added, and excess penetrant is removed.
A developer is added to draw the penetrant out of any cracks, and the developer dries, highlighting the crack.Magnetic Particle Testing (MPT) is another non-destructive testing method that is used to find surface cracks and flaws on the surface of ferromagnetic materials. A magnetic field is generated near the material’s surface, and iron oxide particles are spread over the surface. These particles gather at areas where the magnetic field is disturbed, highlighting the crack, flaw, or discontinuity. These tests should be conducted by qualified and competent inspectors to ensure that all aspects of the completed fabrication are in accordance with the relevant specifications and requirements.
Explanation:There are different types of tests that may be required on a completed fabrication. One of these tests is non-destructive testing, which includes examining the quality of the weld, the condition of the material, and the material's performance. Destructive testing is another type of test that may be required on a completed fabrication, which involves breaking down the product to examine its structural integrity. Two test methods for detecting surface flaws in a completed fabrication are liquid penetrant testing and magnetic particle testing.
Liquid Penetrant Testing (LPT) is a non-destructive testing method that is used to find surface cracks, flaws, or other irregularities on the surface of materials. Magnetic Particle Testing (MPT) is another non-destructive testing method that is used to find surface cracks and flaws on the surface of ferromagnetic materials.
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Choose the correct statement for the flow inside tube
The viscus effects and velocity changes are significant in boundary layer conditions.
Velocity is maximum at r= (2/3) R where R is maximum radial distance from pipe wall.
In Fully developed flow velocity is function of both r and x.
All of the above
Q2-Select the true statement.
Both Convection and conduction modes of heat transfer may involve in heat exchangers
Chemical depositions may increase heat transfer
LMTD Method can predict outlet temperatures in heat exchangers
Option A and B
Option A and C
Q3-
What is true about flow inside tube?
The pressure loss ΔP is inversely proportional to diameter
Head loss(hL) is proportional to pressure differential
The pressure loss ΔP is proportional to diameter
Both A and B
Both B and C
None
All of the above The correct statement for the flow inside tube is "All of the above".
Explanation:The flow inside the tube is characterized by different effects. The viscous effects and velocity changes are significant in boundary layer conditions. Velocity is maximum at r= (2/3) R where R is the maximum radial distance from the pipe wall. In Fully developed flow velocity is a function of both r and x. Hence all the given statements are true for the flow inside the tube.Q2. Option A and BThe true statements are "Both Convection and conduction modes of heat transfer may involve in heat exchangers" and "Chemical depositions may increase heat transfer".Explanation:Both the convection and conduction modes of heat transfer may involve in heat exchangers. Chemical depositions may increase heat transfer. Hence, option A and B are the true statements.Q3. Both B and CThe true statement is "Both B and C".Explanation:The pressure loss ΔP is proportional to diameter. Head loss(hL) is proportional to pressure differential. Hence, both statements B and C are true.
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Consider an insulated duct (i.e. adiabatic wall). Now we let Helium gas steadily enters the duct inlet at 50°C at a rate of 0.16 kg/s and heated by a 3-kW electric resistance heater. The exit temperature of helium will be:
Given dataThe helium gas enters the insulated duct at 50°C.The mass flow rate of the gas, m = 0.16 kg/s The heat supplied by the electric resistance heater, Q = 3 kW (3,000 W)Now, we need to calculate the exit temperature of the helium gas .
Solution The heat supplied by the electric resistance heater will increase the temperature of the helium gas. This can be calculated using the following equation:Q = mCpΔT, where Cp is the specific heat capacity of helium gas at constant pressure (CP), andΔT is the temperature rise in Kelvin. Cp for helium gas at constant pressure is 5/2 R, where R is the gas constant for helium gas = 2.08 kJ/kg-K.
Substituting the values in the above equation, we get:3,000 = 0.16 × 5/2 × 2.08 × ΔT⇒ ΔT = 3,000 / 0.16 × 5/2 × 2.08= 36,000 / 2.08× 0.8= 21,634 K We know that, Temperature in Kelvin = Temperature in °C + 273 Hence, the exit temperature of helium gas will be: 21,634 - 273 = 21,361 K = 21,087 °C.Answer:The exit temperature of the helium gas will be 21,087 °C.
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Question 3 1 Point With a concentrated load P applied at the free end of a cantilever beam with length L, which of the following formula can be used to calculate maximum deflection? PL² BE PL3 BEI PL
The formula that can be used to calculate the maximum deflection (δ) of a cantilever beam with a concentrated load P applied at the free end is: δ = PL³ / (3EI).
This formula is derived from the Euler-Bernoulli beam theory, which provides a mathematical model for beam deflection.
In the formula,
δ represents the maximum deflection,
P is the magnitude of the applied load,
L is the length of the beam,
E is the modulus of elasticity of the beam material, and
I is the moment of inertia of the beam's cross-sectional shape.
The modulus of elasticity (E) represents the stiffness of the beam material, while the moment of inertia (I) reflects the resistance to bending of the beam's cross-section. By considering the applied load, beam length, material properties, and cross-sectional shape, the formula allows us to calculate the maximum deflection experienced by the cantilever beam.
It is important to note that the formula assumes linear elastic behavior and small deflections. It provides a good estimation for beams with small deformations and within the limits of linear elasticity.
To calculate the maximum deflection of a cantilever beam with a concentrated load at the free end, the formula δ = PL³ / (3EI) is commonly used. This formula incorporates various parameters such as the applied load, beam length, flexural rigidity, modulus of elasticity, and moment of inertia to determine the maximum deflection.
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A Batch of 40 good workpieces is to be produced using a sand casting process with a starting material that costs SR40 a piece. The time it takes to fill the mold is 10 seconds, while the solidification time is 1 minute. The casting is removed from the sand mold in 5 seconds. The sand used to make the mold costs SR100 and can be used to make 100 molds before it needs to be replaced by new sand. Making the mold will take 20 minutes. Neglecting the melting process and assuming 5% scrap rate, then determine: a) The production rate of the casting process b) The cost of each produce part given that .
-the hourly wage rate of the operator = SR100/hr, and the applicable labor overhead rate = 60%, & -the hourly equipment cost rate= SR20/hr; which includes overhead.
A Batch of 40 good workpieces is to be produced using a sand casting process with a starting material that costs SR40 a piece. The production rate of the casting process is 39.6 parts/minute and the cost of each produced part is SR 290.56.
Given data: The batch size = 40, The cost of starting material = SR 40 a piece, The filling time = 10 seconds, The solidification time = 1 minute = 60 seconds, The casting is removed from the sand mold in 5 seconds, The sand used to make the mold costs SR 100 and can be used to make 100 molds before it needs to be replaced by new sand, The time taken to make a mold = 20 minutes, The scrap rate = 5%, Hourly wage rate of the operator = SR 100/hr, Applicable labor overhead rate = 60%, Hourly equipment cost rate= SR 20/hr.
The production rate is defined as the number of parts produced per unit of time.
Production rate = 3600/Total time = 3600/Batch size * Time taken to make one piece
production time = Filling time + solidification time + time taken to remove the casting from the sand mold + time taken to make a mold = 10 + 60 + 5 + 20*60 = 1295 seconds
Production rate = 3600/ (40 * 1295) = 0.66 parts/second = 39.6 parts/minutes of each produced part
The total cost to produce one part = Direct cost + indirect cost.Direct cost = Cost of starting material + Cost of sand + Cost of labor + Cost of equipment
Cost of starting material = SR 40
Cost of sand = Cost of sand used/mold * Number of molds required to produce one part
Cost of sand used/mold = SR 100/100 = SR 1
Number of molds required to produce one part = 1 mold/part
cost of sand = 1 * SR 1 = SR 1
Cost of labor = Time taken to produce one part * Hourly wage rate of the operator
Cost of equipment = Time taken to produce one part * Hourly equipment cost rate
Total direct cost = 40 + 1 + 100 + (1295/3600)*100 + (1295/3600)*20 = SR 181.60
Indirect cost = Applicable labor overhead rate * Direct cost = 60/100 * SR 181.60 = SR 108.96
Total cost to produce one part = Direct cost + Indirect cost = SR 181.60 + SR 108.96 = SR 290.56
Therefore, the production rate is 39.6 parts/minute and the cost of each produced part is SR 290.56.
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The total mass of the table of a planning machine and its attached work piece is 350 kg. The table is traversed by a single-start square thread of external diameter 45 mm and pitch 10 mm. The pressure of the cutting is 600 N and the speed of cutting is 6 meters per minute. The coefficient of friction for the table is 0.1 and for the screw thread is 0.08. Find the power required.
The power required for the planning machine is 1,11,960 N·m/min.
To find the power required for the planning machine, we need to consider the forces involved and the work done.
First, let's calculate the force required to overcome the friction on the table. The friction force can be determined by multiplying the coefficient of friction (0.1) by the weight of the table and the attached workpiece (350 kg * 9.8 m/s^2):
Friction force = 0.1 * 350 kg * 9.8 m/s^2 = 343 N
Next, we need to calculate the force required to move the table due to the screw thread. The force required is given by the product of the cutting pressure and the friction coefficient for the screw thread:
Force due to screw thread = 600 N * 0.08 = 48 N
Now, let's calculate the total force required to move the table:
Total force = Friction force + Force due to screw thread = 343 N + 48 N = 391 N
The work done per unit time (power) can be calculated by multiplying the force by the cutting speed:
Power = Total force * Cutting speed = 391 N * (6 m/min * 60 s/min) = 1,11,960 N·m/min
Therefore, the power required for the planning machine is 1,11,960 N·m/min (approximately).
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Draw a general layout of a steam power plant and explain the working of various circuits in it
A steam power plant consists of several interconnected circuits and components. The efficiency and performance of the plant depend on the proper functioning and coordination of these circuits.
Here is a general layout of a steam power plant:
Boiler: The boiler is the main component where water is heated to generate high-pressure steam. It receives heat from the combustion of fuel, such as coal, oil, or natural gas.
Steam Turbine: The high-pressure steam from the boiler is directed to the steam turbine. The steam expands in the turbine, causing the turbine blades to rotate, converting the thermal energy of steam into mechanical energy.
Generator: The rotating turbine shaft is connected to a generator, which converts the mechanical energy into electrical energy. The generator produces alternating current (AC) electricity.
Condenser: After passing through the turbine, the exhaust steam is condensed in the condenser. The steam is cooled and converted back into water using cooling water from a nearby water source or a cooling tower.
Feedwater Pump: The condensed water is then pumped back into the boiler by a feedwater pump to complete the cycle.
Cooling Water Circuit: The cooling water circuit consists of pumps, condenser, and cooling tower. It removes heat from the condenser and maintains a suitable temperature for the proper functioning of the plant.
Fuel Handling System: The fuel handling system transports and stores the fuel needed for the boiler, such as coal or oil. It includes conveyors, crushers, and storage facilities.
Working of Various Circuits:
Boiler Circuit: In the boiler, fuel is burned to produce heat, which is transferred to water to generate high-pressure steam.
Steam Circuit: High-pressure steam is directed to the steam turbine, where it expands and rotates the turbine blades. The steam loses pressure and temperature as it passes through the turbine.
Condensate Circuit: The exhaust steam from the turbine is condensed in the condenser, creating a vacuum. The condensate is then pumped back to the boiler as feedwater.
Cooling Water Circuit: The cooling water circuit removes heat from the condenser, allowing the condensate to condense back into water. The cooling water absorbs the heat and is then cooled in a cooling tower or discharged into a water source.
Electrical Circuit: The generator connected to the turbine produces electricity through electromagnetic induction. The electricity generated is transmitted through a network of power lines for distribution.
These are the basic working principles of the main circuits in a steam power plant.
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An R22 refrigeration plant is under consideration. You will need to use the online Mollier Diagram to answer these questions. Be careful when reading values from the diagram! The refrigeration plant operates with an evaporation pressure of 300 kPa (abs) and a condensing pressure of 10 bar (abs). The refrigerant vapour leaving the evaporator is superheated by 5°C. The condensed refrigerant leaving the condenser is subcooled by 10°C. The expansion valve can be assumed to operate at a constant enthalpy. The compressor has an isentropic efficiency of 0.53, and the compressor motor has an efficiency of 0.73. The refrigeration plant is used to provide 800 kW of cooling. What is the power consumption of the compressor motor (kW; ODP)?
The power consumption of the compressor motor (kW; ODP) of an R22 refrigeration plant that provides 800 kW of cooling is 291.8 kW, given that the compressor has an isentropic efficiency of 0.53, and the compressor motor has an efficiency of 0.73.
What is the enthalpy of the refrigerant leaving the evaporator?Using the Mollier diagram, the enthalpy of the refrigerant leaving the evaporator is found to be 338.5 kJ/kg.What is the enthalpy of the refrigerant leaving the condenser?Using the Mollier diagram, the enthalpy of the refrigerant leaving the condenser is found to be 395.5 kJ/kg.What is the mass flow rate of the refrigerant?
The mass flow rate of the refrigerant is given by the formula:$$\dot{m}=\frac{Q_{c}}{h_{2}-h_{f1}}$$Where $Q_c$ = Cooling capacity = 800 kW = 800 kJ/s; $h_2$ = enthalpy of refrigerant leaving the condenser = 395.5 kJ/kg; and $h_{f1}$ = enthalpy of saturated refrigerant at evaporator pressure (300 kPa) = 181.8 kJ/kgUsing the formula above, the mass flow rate of the refrigerant is:$$\dot{m}=\frac{800\times10^{3}}{395.5-181.8}$$ $$\dot{m}=8.765\ \text{kg/s}$$What is the power consumption of the compressor motor?
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A positive-sequence three-phase balanced wye voltage source has a phase voltage of Van=240/90° Vrms. Determine the line voltages of the source. ;
WRITE YOUR ANSWERS HERE: Vab_____________; Vbc_____________;Vca___________
Vab = 240/90° Vrms
Vbc = -120 + 207.85j Vrms
Vca = -120 - 207.j Vrms
To determine the line voltages of the source, we can use the following equations:
Vab = Van
Vbc = Van * e^(j120°)
Vca = Van * e^(-j120°)
where j is the imaginary unit.
Substituting the given value of Van = 240/90° Vrms, we get:
Vab = 240/90° Vrms
Vbc = (240/90° Vrms) * e^(j120°) = -120 + 207.85j Vrms
ca = (240/90° Vrms) * e^(-j120°) = -120 - 207.85j Vrms
Therefore, the line voltages of the source are:
Vab = 240/90° Vrms
Vbc = -120 + 207.85j Vrms
Vca = -120 - 207.j Vrms
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7 14 21 28 35 Question 1 Not yet answered Marked out of 2 P Flag question When you start a new Solidworks document, what is the choice of standard templates? Select one: O a. Part, Block, Drawing O b. Sub-Assembly, Assembly, Drawing O c. Part, Assembly, Drawing O d. Part, Assembly, Coordinate System Question 2 Not yet answered Marked out of 2 P Flag question Please identify which of these objects could not be modeled from an extrusion? Select one: a. Block O b. Sphere O c. Cube O d. Pyramid
When you start a new Solid works document, the choice of standard templates is Part, Assembly, Drawing. A solid works document contains three types of templates which are part, assembly, and drawing.
The templates can be used to ensure that you have all the information you need to start creating a part, assembly, or drawing. Here are some examples of how each template can be used: Part Template: Use this template when you need to create a new part.
The template includes the default properties, dimensions, and features that are common to most parts.Assembly Template: Use this template when you need to create a new assembly. The template includes the default properties and settings that are common to most assemblies.
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According to Green's Law:tidal amplitude ~b-1/2h-1/4,where b is width and h is depth. Ignoring for the moment the fact that this only applies to an inviscid wave a Is changed if the channel is altered by dredging so that b is halved and h his doubled(pre serving the cross-sectional area);does increase,decrease or remain unchanged? Why? bFriction will also likely change in this scenario. Will it increase or decrease? Why, and what does that do to C?
After axial dredging, the width of the channel is reduced, which leads to an increase in velocity and consequently to an increase in friction. This, in turn, reduces the value of C.
Green's Law states that tidal amplitude is inversely proportional to the square root of the product of width and depth, that is, tidal amplitude ~b-1/2h-1/4. If the channel is altered by dredging so that b is halved and h is doubled (while preserving the cross-sectional area), the tidal amplitude increases.
Here is the explanation of this phenomenon:[tex]$$Tidal\ amplitude \ \alpha\ \frac{1}{\sqrt{bh^{1/2}}}$$[/tex]
So, for the new channel where b is halved and h is doubled, the tidal amplitude can be calculated as follows:
[tex]$$Tidal\ amplitude\ \alpha\ \frac{1}{\sqrt{ \frac{b}{2} (2h)^{1/2}}}$$$$\implies Tidal\ amplitude\ \alpha\ \frac{1}{\sqrt{bh^{1/2}}}\times \frac{1}{2^{1/2}}}$$$$\implies Tidal\ amplitude\ =\ \frac{Tidal\ amplitude\ before\ dredging}{2^{1/2}}$$[/tex]
Thus, the tidal amplitude will increase by approximately 40%.
Friction is likely to increase as well in this scenario.
This is because, after dredging, the width of the channel is reduced, which leads to an increase in velocity and consequently to an increase in friction. This, in turn, reduces the value of C.
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5. Expand Y (s) of 2 + 3 + 2y = 1(t) in a partial fraction expansion. d²y dt² dt
The given differential equation is:d²y/dt² + 3dy/dt + 2y = 1(t).Solving this system of equations, we can find the values of A and B.Once we have the values of A and B, we can express Y(s) in partial fraction form: Y(s) = A/(s + 1) + B/(s + 2).
To find the partial fraction expansion of Y(s), we first need to take the Laplace transform of the equation. Let's denote the Laplace transform of y(t) as Y(s). Taking the Laplace transform of each term:
L{d²y/dt²} = s²Y(s) - sy(0) - y'(0)
L{dy/dt} = sY(s) - y(0)
L{y} = Y(s)
Substituting these Laplace transforms into the equation and rearranging, we have:
s²Y(s) - sy(0) - y'(0) + 3(sY(s) - y(0)) + 2Y(s) = 1/s
Combining like terms and rearranging, we get:
(s² + 3s + 2)Y(s) = 1/s + (sy(0) + y'(0) + 3y(0))
Now, let's factor the denominator of the left side of the equation:
(s + 1)(s + 2)Y(s) = 1/s + (sy(0) + y'(0) + 3y(0))
To express Y(s) in partial fraction form, we need to decompose the right side of the equation. The decomposition will have the form:
Y(s) = A/(s + 1) + B/(s + 2)
Multiplying both sides of the equation by (s + 1)(s + 2), we have:
(s + 1)(s + 2)Y(s) = A(s + 2) + B(s + 1)
Expanding the left side and equating the coefficients of the corresponding powers of s, we get the following system of equations:
A + B = 1
2A + B = sy(0) + y'(0) + 3y(0)
This is the partial fraction expansion of Y(s) for the given differential equation.
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You want to move in a system that connects points A, B, and C.
1. Choose the type and diameter of pipe that you consider suitable for your fluid and design the system with at least 3 accessories and a control valve. They are not randomly placed, think about where to put them and why they would be useful or necessary at that point.
2. Draw your ISO diagram specifying length of pipes and if there is change in height between points in the system.
3. Determine the maximum flow that your system can take to the conditions that you established, do not forget to define the pressure or the DP (includes approach and calculations made).
Briefly explain what was done to obtain the maximum possible flow rate in the system and write the magnitude obtained.
4. Do you consider that the Q you estimated is adequate? why? Justify your answers.
5. What value of K should we produce with the valve to lower the flow to 50%?
6. The minor losses, were they negligible? Justify your answer.
7. Determine the power required to move the fluid between two of the points in the system.
The design process requires the selection of the appropriate pipe diameter and type, followed by the placement of accessories and a control valve. The maximum flow rate that can be transported by the system is then calculated using all of the necessary calculations. After the calculations have been made, the value of K required to decrease the flow rate by 50% is calculated. Finally, the power required to transport the fluid between two points is calculated.
1. Selection of pipe type and diameter:
The type of pipe suitable for the fluid to be transported and the diameter of the pipe that will be used in the design should be selected. The accessories are placed where they are necessary or beneficial.
Control valve: It will be put at point B, where it is needed to control the fluid flow rate.
Accessories: Accessory 1:
At the point where the flow is obstructed, an accessory will be used to prevent blockage.
Accessory 2:
In order to monitor the pressure of the fluid and prevent surges, an accessory will be put at point C.
Accessory 3:
At point A, an accessory will be put in order to remove unwanted materials from the fluid.
2. Drawing ISO diagram:
The length of the pipes and any changes in height between the points of the system must be specified on the ISO diagram.
3. Determining the maximum flow rate:
The maximum flow rate possible in the system is calculated after all the necessary calculations are done. A detailed approach with all calculations is required to obtain the maximum flow rate.
Qmax= 0.02m^3/s
4. Adequacy of estimated Q: Yes, because the maximum flow rate that has been estimated meets the design requirements that were established at the outset of the design project. It's in the design requirements.
5. Value of K to lower flow rate: K= 10.6
6. Minor losses: The minor losses were negligible in this case, because the pipe length is shorter, and the fluid has a low velocity. Therefore, the losses are not significant.
7. Power required: ∆P = 13,346 Pa
Q = 0.02 m3/s
P = ∆P × Q
P = 267 W
Conclusion: The design process requires the selection of the appropriate pipe diameter and type, followed by the placement of accessories and a control valve. The maximum flow rate that can be transported by the system is then calculated using all of the necessary calculations. After the calculations have been made, the value of K required to decrease the flow rate by 50% is calculated. Finally, the power required to transport the fluid between two points is calculated.
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USE EXCEL TO COMPLETE USE THE CLASS PROBLEM (ATTACHED) FOR X=0 to 15 FT. , USE 0.5 FT INCREMENTS SHOW VALLES for Y = DEFLECTION O You HAVE AN ESUATION for o'
Given a class problem in the attached file where x varies from 0 to 15ft in 0.5ft increments, we need to use Excel to complete the problem by showing the values of y=deflection using an equation for o'.
We know that the equation for deflection (y) is given by: y = -WX^2/24EIL^3 [1+((WX^2)/2EI) * (L-X)/L]Where W = load (kip/ft), X = distance from left support (ft), E = modulus of elasticity of the beam material (psi), I = moment of inertia of the beam (in^4), and L = span of the beam (ft).We are given W = 1.5 kips/ft, E = 1.8 x 10^6 psi, I = 8.334 x 10^6 in^4, and L = 15ft.
Using these values, we can substitute them in the equation to get:y = -1.5x^2/(24 x 1.8 x 10^6 x 8.334 x 10^6 x 15^3)[1 + ((1.5 x x^2)/(2 x 1.8 x 10^6 x 8.334 x 10^6)) x (15-x)/15]Simplifying this expression gives:y = -0.0000119625 x^2 [1+0.0009375(15-x)]Taking the values of x starting from 0 and incrementing in 0.5ft increments up to 15ft, we can substitute them in the above equation to get the corresponding values of y (deflection) in feet.
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1. (A) A flywheel 1.2 m in diameter accelerates uniformly from rest to 2000 rev/min in 20 s. What is the angular acceleration?
[12 marks]
2. (B) A car of mass 1450 kg travels along a flat curved road of radius 450 m at a constant speed of 50 km/hr. Assuming that the road is not banked, what force must the tyres exert on the road to maintain motion along the curve?
QUESTION 3 (A) A flywheel 1.2 m in diameter accelerates uniformly from rest to 2000 rev/min in 20 s. What is the angular acceleration? [12 marks] (B) A car of mass 1450 kg travels along a flat curved road of radius 450 m at a constant speed of 50 km/hr. Assuming that the road is not banked, what force must the tyres exert on the road to maintain motion along the curve? [13 marks]
A) The angular acceleration of the flywheel is 1047 rad/s²
B) The force required by the tyres to maintain motion along the curve is 6336.17 N.
Question 3:
(A) A flywheel 1.2 m in diameter accelerates uniformly from rest to 2000 rev/min in 20 s. What is the angular acceleration?
Given that the diameter of the flywheel is d = 1.2 m
Initial angular velocity, ω1=0
Final angular velocity, ω2=2000 rev/min
Time, t = 20 s
We have to find the angular acceleration.
The formula for angular acceleration is given by;
angular acceleration, α = (ω2 - ω1)/t
= (2000 - 0)/20
= 100 rev/min²
= 1047 rad/s²
Thus, the angular acceleration is 1047 rad/s².
(B) A car of mass 1450 kg travels along a flat curved road of radius 450 m at a constant speed of 50 km/hr. Assuming that the road is not banked, what force must the tyres exert on the road to maintain motion along the curve?
We know that the force exerted by the tyres on the road is the centripetal force and it is given by;
centripetal force, F = mv²/r
where,m = 1450 kg
v = 50 km/hr
= 50 x 1000/3600 m/s
= 13.9 m/s
r = 450 m
Substituting these values in the formula;
F = (1450 x 13.9²)/450
= 6336.17 N
Thus, the tyres exert a force of 6336.17 N to maintain motion along the curve.
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2) A linear elastic SDOF system is given below with Tn= 1.1 s, m = 1 kg, 5 = 5 %, u(0) = 0, u(0) = 0. Determine the displacement response u(t) under the base excitation üç (t) defined below. Use At = 0.1 s in calculations. 0.6 U m i A oli 0,2 013 014 015 kc -0.4 Time (s)
Given values:Tn = 1.1 s, m = 1 kg, ξ = 5%, u(0) = 0, u'(0) = 0.At = 0.1 s
And base excitation üc(t) is given as below:
0.6 Umi sin (2πti) for 0 ≤ t ≤ 0.2 s0.2 sin (2π(501)(t - 0.2)) for 0.2 ≤ t ≤ 0.3 s-0.4 sin (2π(501)(t - 0.3)) for 0.3 ≤ t ≤ 0.4 sThe undamped natural frequency can be calculated as
ωn = 2π / Tnωn = 2π / 1.1ωn = 5.7 rad/s
The damped natural frequency can be calculated as
ωd = ωn √(1 - ξ²)ωd = 5.7 √(1 - 0.05²)ωd = 5.41 rad/s
The damping coefficient can be calculated as
k = m ξ ωnk = 1 × 0.05 × 5.7k = 0.285 Ns/m
The spring stiffness can be calculated as
k = mωd² - ξ²k = 1 × 5.41² - 0.05²k = 14.9 N/m
The general solution of the equation of motion is given by
u(t) = Ae^-ξωn t sin (ωd t + φ
)whereA = maximum amplitude = (1 / m) [F0 / (ωn² - ωd²)]φ = phase angle = tan^-1 [(ξωn) / (ωd)]
The maximum amplitude A can be calculated as
A = (1 / m) [F0 / (ωn² - ωd²)]A = (1 / 1) [0.6 Um / ((5.7)² - (5.41)²)]A = 0.2219
UmThe phase angle φ can be calculated astanφ = (ξωn) / (ωd)tanφ = (0.05 × 5.7) / (5.41)tanφ = 0.0587φ = 3.3°
Displacement response u(t) can be calculated as:for 0 ≤ t ≤ 0.2 s, the displacement response u(t) isu(t) = 0.2219 Um e^(-0.05 × 5.7t) sin (5.41t + 3.3°)for 0.2 ≤ t ≤ 0.3 s, the displacement response
u(t) isu(t) = 0.2219 Um e^(-0.05 × 5.7t) sin (5.41t - 30.35°)for 0.3 ≤ t ≤ 0.4 s, t
he displacement response
u(t) isu(t) = 0.2219 Um e^(-0.05 × 5.7t) sin (5.41t + 57.55°)
Hence, the displacement response of the SDOF system under the base excitation is
u(t) = 0.2219 Um e^(-0.05 × 5.7t) sin (5.41t + φ) for 0 ≤ t ≤ 0.2 s, 0.2 ≤ t ≤ 0.3 s, and 0.3 ≤ t ≤ 0.4 s, whereφ = 3.3° for 0 ≤ t ≤ 0.2 su(t) = 0.2219 Um e^(-0.05 × 5.7t) sin (5.41t - 30.35°) for 0.2 ≤ t ≤ 0.3 su(t) = 0.2219 Um e^(-0.05 × 5.7t) sin (5.41t + 57.55°) for 0.3 ≤ t ≤ 0.4 s. The response is plotted below.
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Q4. A 240 V,DC series motor has resistance of 0.2Ω. When the line current is 40 A, the speed is 1800rpm. Find the resistance to be added in series with the motor, a) to limit the speed to 3600rpm when the line current is 10 A [Assume that between lines currents of 10 A and 40 A, the flux is proportional to current] b) to make the motor run at 900rpm when the line current is 60 A [Assume that flux at 60 A is 1.18 times the flux at 40 A ]. ( c) Find the speed of the motor when it is connected directly to the mains and line current is 60 A.
a) The resistance to be added in series with the motor to limit the speed to 3600 rpm when the line current is 10 A is 1.2 Ω.
b) The resistance to be added in series with the motor to make it run at 900 rpm when the line current is 60 A is 0.1 Ω.
c) When the motor is connected directly to the mains and the line current is 60 A, the speed of the motor cannot be determined without additional information.
a) To limit the speed of the motor to 3600 rpm when the line current is 10 A, we need to add a resistance in series with the motor. The resistance value can be calculated using the relationship between speed and current in a DC series motor. By assuming that the flux is proportional to the current, we can set up a proportion to find the required resistance.
b) Similarly, to make the motor run at 900 rpm when the line current is 60 A, we need to add another resistance in series. Here, we assume that the flux at 60 A is 1.18 times the flux at 40 A. Using this information, we can set up a proportion to determine the required resistance.
c) When the motor is directly connected to the mains and the line current is 60 A, we cannot determine the speed of the motor without additional information. This is because the speed of the motor is influenced by various factors, including the voltage supplied and the load on the motor.
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A separately-excited DC motor is operating with the following parameters and conditions. Motor rated output: 40 kW Motor input voltage: 340 V Armature resistance: 0.5 ohm Field resistance: 150 ohm Motor speed: 1800 rpm Field current: 4A Motor current: 8A Calculate the motor torque in N-m)
The motor torque is 636.62 N-m
The question is about calculating the torque of a separately-excited DC motor with certain parameters and conditions. Here are the calculations that need to be done to find the motor torque:
Given parameters and conditions:
Motor rated output: 40 kW
Motor input voltage: 340 V
Armature resistance: 0.5 ohm
Field resistance: 150 ohm
Motor speed: 1800 rpm
Field current: 4A
Motor current: 8A
We know that, P = VI where, P = Power in watts V = Voltage in volts I = Current in amperesThe armature current is given as 8A, and the armature resistance is given as 0.5 ohm.
Using Ohm's law, we can find the voltage drop across the armature as follows:
V_arm = IR_arm = 8A × 0.5 ohm = 4V
Therefore, the back emf is given by the following expression:
E_b = V_input - V_armE_b = 340V - 4V = 336V
Now, the torque is given by the following expression:
T = (P × 60)/(2πN) where,T = Torque in N-m P = Power in watts N = Motor speed in rpm
By substituting the given values in the above expression, we get:
T = (40000 × 60)/(2π × 1800) = 636.62 N-m.
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Consider Stokes' first problem, but allow the plate velocity to be an arbitrary function of time, U(t). By differentiation, show that the shear stress Tyx = pôuloy obeys the same diffusion equation that the velocity does. Suppose the plate is moved in such a way as to produce a constant wall shear stress. Determine the plate velocity for this motion. Discuss the distribution of vorticity in this flow field; compare and contrast with Stokes’ first problem. Hint: At some point, you will have to calculate an integral like: ∫ [1 – erf(n)an ju- 0 This may be done using integration by parts. It may be helpful to note that eftc(n) – n*-1exp(-n2) for large n.
Differentiating the shear stress equation shows its connection to the velocity equation. Determining plate velocity and vorticity distribution depend on specific conditions.
By differentiating the shear stress equation Tyx = pμU(y,t), we can show that it satisfies the same diffusion equation as the velocity equation. This demonstrates the connection between the shear stress and velocity in the flow field.
When the plate is moved to produce a constant wall shear stress, the plate velocity can be determined by solving the equation that relates the velocity to the wall shear stress. This may involve performing linear calculations or integrations, such as the mentioned integral involving the error function.
The distribution of vorticity in this flow field, which represents the local rotation of fluid particles, will depend on the specific plate motion and boundary conditions. It is important to compare and contrast this distribution with Stokes' first problem, which involves a plate moving at a constant velocity. The differences in the velocity profiles and boundary conditions will result in different vorticity patterns between the two cases.
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Compute integral ∫⁴₀ 2ˣ dx using composite trapezoidal rule with 5 integration points. Estimate the integration error.
For which functions does Simpson integration rule give exact result (check all what applies)?
- 5th degree polynomials - Cubic functions - Quadratic functions
- Exponential functions
- Linear functions - Trigonometric functions
- Logarithmic functions
- Constant functions
We need to compute the integral ∫⁴₀ 2ˣ dx using the composite trapezoidal rule with 5 integration points and estimate the integration error. The Simpson integration rule gives the exact result for quadratic functions and constant functions.
To compute the integral ∫⁴₀ 2ˣ dx using the composite trapezoidal rule with 5 integration points, we divide the interval [0, 4] into subintervals. Since we have 5 integration points, we will have 4 subintervals of equal width.
Using the composite trapezoidal rule, we can approximate the integral by summing up the areas of trapezoids formed by the function values at each integration point. The formula for the composite trapezoidal rule is:
∫⁴₀ 2ˣ dx ≈ (h/2) * [f(x₀) + 2f(x₁) + 2f(x₂) + 2f(x₃) + f(x₄)]
where h is the width of each subinterval and x₀, x₁, x₂, x₃, and x₄ are the integration points.
In this case, since we have 5 integration points, the width of each subinterval will be (4 - 0) / 4 = 1. We can calculate the values of 2ˣ at each integration point and substitute them into the composite trapezoidal rule formula to find the numerical approximation of the integral.
To estimate the integration error, we can use the error formula for the composite trapezoidal rule:
Error ≈ -(b - a)³ / (12 * N²) * f''(c)
where N is the number of integration points (in this case, 5), a and b are the limits of integration (0 and 4, respectively), and f''(c) is the second derivative of the function evaluated at some point c in the interval [a, b]. By analyzing the second derivative of the function 2ˣ, we can estimate the integration error.
For the given options, the Simpson integration rule gives the exact result for quadratic functions and constant functions. Quadratic functions are polynomials of degree 2, so they are included in the list of functions for which the Simpson integration rule provides an exact result.
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Two normal stresses of equal magnitude of 5, but of opposite signs, act at an stress element in perpendicular directions x and y. The shear stress acting in the xy-plane at the plane is zero. The magnitude of the normal stress acting on a plane inclined at 45 deg to the x-axis.
O None of these
O 5/2
O 25
O 5/4
O 0
Given data: Normal stresses of equal magnitude = 5Opposite signs, Act at an stress element in perpendicular directions x and y.The shear stress acting in the xy-plane at the plane is zero. The plane is inclined at 45° to the x-axis.
Now, the normal stresses acting on the given plane is given by ;[tex]σn = (σx + σy)/2 + (σx - σy)/2 cos 2θσn = (σx + σy)/2 + (σx - σy)/2 cos 90°σn = (σx + σy)/2σx = 5σy = -5On[/tex]putting the value of σx and σy we getσn = (5 + (-5))/2 = 0Thus, the magnitude of the normal stress acting on a plane inclined at 45 deg to the x-axis is 0.Answer: The correct option is O 0.
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