Hence, the bandwidth of the FSK signal is 1000 Hz.
Given that a data source sends symbols of 4 bits with bit rate 2000 [bps].
The data is used to modulate a carrier of frequency 10 [kHz].
If any two consecutive frequencies are 100 [Hz] apart.
To find: the bandwidth of the FSK signal.
Frequency shift keying (FSK) is a modulation scheme in which the frequency of a carrier signal is varied based on input digital data. FSK modulation requires two frequencies for modulation, one for each binary value.
The frequency separation between the two carrier frequencies is the most significant factor in determining the transmission rate.
The bandwidth of the FSK signal is given by the following formula:
Bandwidth = 2 × (N+1) × Δf
Where,
Δf = frequency difference between the two carrier frequencies
N = Number of bits in each signal
We know that the bit rate, Rb = 2000 [bps].
Hence the time taken for one bit, Tb = 1/Rb = 1/2000 = 0.0005 [s].
The carrier frequency, fc = 10 [kHz].
For any two consecutive frequencies, Δf = 100 [Hz].
Therefore, the two carrier frequencies are fc1 = fc - Δf and fc2 = fc + Δf.
Therefore, fc1 = 9900 [Hz] and fc2 = 10100 [Hz].
The number of bits in each signal, N = 4.
The bandwidth of the FSK signal,
Bandwidth = 2 × (N+1) × Δf= 2 × (4+1) × 100= 1000 [Hz].
Hence, the bandwidth of the FSK signal is 1000 Hz.
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A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min. Determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air.
2. What is the Clausius expression of the second law of thermodynamics?
Given:A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min.
Solution:
a) The electrical power consumed by the refrigerator is given by the formula:
P = Q / COP
where Q = 60 kJ/min (rate of heat removal)
COP = 1.2 (coefficient of performance)
Putting the values:
P = 60 / 1.2
= 50 W
Therefore, the electrical power consumed by the refrigerator is 50 W.
b) The rate of heat transfer to the kitchen air is given by the formula:
Q2 = Q1 + W
where
Q1 = 60 kJ/min (rate of heat removal)
W = electrical power consumed
= 50 W
Putting the values:
Q2 = 60 + (50 × 60 / 1000)
= 63 kJ/min
Therefore, the rate of heat transfer to the kitchen air is 63 kJ/min.
2. The Clausius expression of the second law of thermodynamics states that heat cannot flow spontaneously from a colder body to a hotter body.
It states that a refrigerator or an air conditioner requires an input of work to transfer heat from a cold to a hot reservoir.
It also states that it is impossible to construct a device that operates on a cycle and produces no other effect than the transfer of heat from a lower-temperature body to a higher-temperature body.
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Good day! As we have agreed upon during Module 1 , one of the assessments under Module 3 will be the real life applications of Mechanics. Please give at least 3 applications of Mechanics to your daily life. Submission of this will be on or before July 30, 2022, Saturday, until 11:59PM. This activity will be done through a powerpoint presentation. Take a picture of the applications and make a caption depicting what is the principle being applied. This can be submitted through the link provided here. Please use the filename/subject format
Mechanics is the branch of physics that deals with the motion of objects and the forces that cause the motion.
The following are three examples of the applications of mechanics in daily life:
1. Bicycle- The mechanics of a bicycle is an excellent example of how mechanics is used in everyday life.
The wheels, gears, brakes, and pedals all operate on mechanical principles.
The pedals transfer mechanical energy to the chain, which then drives the wheels, causing them to rotate and propel the bicycle forward.
2. Car- A car's engine is another example of how mechanics is used in everyday life.
The engine transforms chemical energy into mechanical energy, which propels the vehicle.
The gears, wheels, and brakes, as well as the suspension system, all operate on mechanical principles.
3. Elevators- Elevators rely heavily on mechanics to function.
The elevator car is lifted and lowered by a system of cables and pulleys that is operated by an electric motor.
A counterweight is used to balance the load, and a brake system is used to hold the car in place between floors.
Thus, these are the 3 examples of mechanics that we use daily in our life.
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A rubber ball and a lump of putty have equal mass. They are thrown with equal speed against a wall. The ball bounces back with nearly the same speed with which it hit. The putty sticks to the wall. Which object experiences the greater momentum change? (Hint: momentum is a vector!) a cannot be determined from the information given b the ball c the putty d both experience same momentum change.
The rubber ball experiences a greater momentum change compared to the lump of putty.
Which object experiences the greater momentum change?The rubber ball experiences a greater momentum change compared to the lump of putty.
Momentum is a vector quantity defined as the product of an object's mass and its velocity. When the ball and the putty are thrown against the wall with equal speed, they have the same initial momentum. However, during the collision, the ball bounces back, changing its direction of motion, while the putty sticks to the wall, coming to rest.
Since momentum is a vector, it has both magnitude and direction. The ball experiences a change in both magnitude and direction of momentum as it bounces back, resulting in a significant momentum change. On the other hand, the putty sticks to the wall, causing its momentum to change only in direction but not in magnitude.
Therefore, the ball experiences a greater momentum change than the putty.
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which code snippet correctly depicts the stream friend functionality of cin and cout?
The code snippet below correctly depicts the stream friend functionality of `cin` and `cout`:
```cpp
#include <iostream>
int main() {
int number;
std::cout << "Enter a number: ";
std::cin >> number;
std::cout << "You entered: " << number << std::endl;
return 0;
}
```
In the above code, the `iostream` library is included, and the `main()` function is defined. The user is prompted to enter a number using `std::cout`, and the input is obtained using `std::cin`, which is then stored in the `number` variable. Finally, the program outputs the entered number using `std::cout`.
The stream friend functionality of `cin` and `cout` in C++ allows them to work together seamlessly. The `cin` object is used for input operations, while the `cout` object is used for output operations. Both `cin` and `cout` are linked to the standard input and standard output streams, respectively.
When using `std::cin`, the `>>` operator is used to extract input from the user and store it in a variable. In the given code, the user's input is read and stored in the `number` variable using `std::cin >> number;`.
On the other hand, `std::cout` is used to display output to the user. In the provided code, the entered number is outputted to the console using `std::cout << "You entered: " << number << std::endl;`.
By combining `cin` and `cout` in this way, you can easily prompt the user for input and display the results, making it a powerful tool for interactive console-based applications.
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Atmospheric pressure, also known as barometric pressure, is the pressure within the atmosphere of Earth. The standard atmosphere is a unit of pressure defined as 101,325 Pa. Explain why some people experience nose bleeding and some others experience shortness of breath at high elevations.
Nose bleeding and shortness of breath at high elevations can be attributed to the changes in atmospheric pressure. At higher altitudes, the atmospheric pressure decreases, leading to lower oxygen levels in the air. This decrease in pressure can cause the blood vessels in the nose to expand and rupture, resulting in nosebleeds.
the reduced oxygen availability can lead to shortness of breath as the body struggles to take in an adequate amount of oxygen. The body needs time to acclimate to the lower pressure and adapt to the changes in oxygen levels, which is why these symptoms are more common at higher elevations. At higher altitudes, the atmospheric pressure decreases because there is less air pressing down on the body.
This decrease in pressure can cause the blood vessels in the nose to become more fragile and prone to rupturing, leading to nosebleeds. The dry air at higher elevations can also contribute to the occurrence of nosebleeds. On the other hand, the reduced atmospheric pressure means that there is less oxygen available in the air. This can result in shortness of breath as the body struggles to obtain an adequate oxygen supply. It takes time for the body to adjust to the lower pressure and increase its oxygen-carrying capacity, which is why some individuals may experience these symptoms when exposed to high elevations.
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For a flux of D = (x^3 + y^3)-1/3 ax , find the following: a. the volume charge density at P(8, 4, 6). b. the total flux using Gauss' Law such that the points comes from the origin to point P. c. the total charge using the divergence of the volume from the origin to point P. Type in the canvas the distance from the origin to point P.
To find the requested , we need to evaluate the given flux function and perform calculations based on Gauss' Law and the divergence theorem.
a. The volume charge density at point P(8, 4, 6) can be determined by substituting the coordinates into the given flux function. The volume charge density, denoted by ρ, is given by ρ = ∇ · D, where ∇ represents the divergence operator. Evaluate ∇ · D at P to find the volume charge density at that point.
b. To calculate the total flux using Gauss' Law, we need to find the enclosed charge within a closed surface that spans from the origin to point P. The total flux, denoted by Φ, is given by Φ = ∫∫ D · dA, where dA is the infinitesimal area vector and the integration is performed over the closed surface.
c. To determine the total charge using the divergence theorem, we integrate the volume charge density ρ over the volume enclosed by a closed surface that spans from the origin to point P. The total charge, denoted by Q, is given by Q = ∫∫∫ ρ d V, where d V is the infinitesimal volume element and the integration is performed over the enclosed volume.
The distance from the origin to point P can be calculated using the formula for Euclidean distance: d = √(x^2 + y^2 + z^2), where x, y, and z are the coordinates of point P.
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The velocity and acceleration of particle moving on space at a certain instant given by overline v =3 hat i +6 hat j -2 hat k , overline a =4 hat i +6 hat j -10 hat k for this instant determine hat u 1 ,a 1 ,a n rho,| hat hat u t |, hat u b , hat u n
The unit velocity vector is 1/7 hat i + 2/7 hat j - 2/7 hat k. The unit acceleration vector is 4/7 hat i + 6/7 hat j - 10/7 hat k. The magnitude of acceleration is 12 units, and the magnitude of velocity is 7 units. The unit tangent vector is the same as the unit velocity vector.
The unit binormal vector is perpendicular to the plane formed by the unit velocity and acceleration vectors. The unit normal vector is the cross product of the unit tangent and binormal vectors.
To find the unit velocity vector (hat u), we divide the given velocity vector by its magnitude: 1/7 hat i + 2/7 hat j - 2/7 hat k.
To find the unit acceleration vector (a 1), we divide the given acceleration vector by its magnitude: 4/7 hat i + 6/7 hat j - 10/7 hat k.
The magnitude of the acceleration vector (a n) is found by taking the magnitude of the given acceleration vector: 12 units.
The magnitude of the unit tangent vector (| hat hat u t |) is the same as the magnitude of the unit velocity vector: 7 units.
The unit binormal vector (hat u b) is perpendicular to the plane formed by the unit velocity and acceleration vectors.
The unit normal vector (hat u n) is found by taking the cross product of the unit tangent and binormal vectors.
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Thermodynamics
Air initially at 30 psia and 0.69 ft^3, with a mass of 0.1 lbm, expands at constant pressure to a volume of 1.5 ft^3. It then changes state at constant volume until a pressure of 15 psia is reached. If the processes are quasi-static. Determine:
a) The total work, in Btu
b) The total heat, in Btu
c) The total change in internal energy
a) The total work is -2.49 Btu.
b) The total heat is 0 Btu.
c) The total change in internal energy is -2.49 Btu.
In this problem, the given air undergoes two processes: expansion at constant pressure and a subsequent change in state at constant volume.
a) To calculate the total work, we need to consider both processes. The work done during expansion at constant pressure can be calculated using the equation W = P * (V2 - V1), where P is the constant pressure, and V2 and V1 are the final and initial volumes, respectively. In this case, the initial volume is 0.69 ft^3, and the final volume is 1.5 ft^3. The pressure is constant at 30 psia. Plugging these values into the equation, we get W1 = 30 * (1.5 - 0.69) = 25.5 ft-lbf. Converting this to Btu, we divide by the conversion factor of 778, yielding W1 = 0.033 Btu.
For the process at constant volume, no work is done since there is no change in volume. Therefore, the total work is simply the sum of the work done during expansion at constant pressure, i.e., W = W1 = 0.033 Btu.
b) The total heat is given by the first law of thermodynamics, which states that Q = ΔU + W, where Q is the heat transferred, ΔU is the change in internal energy, and W is the work done. Since the problem states that the processes are quasi-static, we can assume that there is no heat transfer (adiabatic process) during both expansion and the subsequent change in state. Therefore, Q = 0 Btu.
c) Using the first law of thermodynamics, ΔU = Q - W. Since Q = 0 Btu and W = 0.033 Btu, we have ΔU = -0.033 Btu. Thus, the total change in internal energy is -0.033 Btu.
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In a nano-scale MOS transistor, which option can be used to achieve high Vt: a. Increasing channel length b. Reduction in oxide thickness c. Reduction in channel doping density d. Increasing the channel width e. Increasing doing density in the source and drain region
In a nano-scale MOS transistor, the option that can be used to achieve high Vt is reducing the channel doping density. This is because channel doping density affects the threshold voltage of MOSFETs (Option c).
A MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor) is a type of transistor used for amplifying or switching electronic signals in circuits. It is constructed by placing a metal gate electrode on top of a layer of oxide that covers the semiconductor channel.
Possible ways to increase the threshold voltage (Vt) of a MOSFET are:
Reducing the channel doping density;Increasing the thickness of the gate oxide layer;Reducing the channel width;Increasing the length of the channel. However, this results in higher RDS(on) and lower transconductance which makes the MOSFET perform worse;Reducing the temperature of the MOSFET;Therefore, the correct answer is c. Reduction in channel doping density.
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If 1.9 hours were required to produce the 1st unit in a production run and the production process has a learning curve rate of 89%, how many hours will it take to produce the 25th unit? Provide your answer to one decimal place.
The time required to produce the 25th unit is approximately 9.2 hours. Given that the production run has a learning curve rate of 89% and 1.9 hours were required to produce the first unit, we can find the time to produce the 25th unit as follows:
First, we find the learning curve ratio (LCR) using the formula:
LCR = (Time to produce first unit) / ((Number of units) * log2(learning curve rate)) Substituting the given values, we have:
LCR = 1.9 / (1 * log2(0.89)) = 0.2202
Next, we can use the LCR to find the time to produce the 25th unit using the formula:
Time to produce Nth unit = Time to produce first unit * (N^log2(LCR))
Substituting N = 25 and the calculated LCR, we get:
Time to produce 25th unit = 1.9 * (25^log2(0.2202)) ≈ 9.2 hours
Therefore, the time required to produce the 25th unit is approximately 9.2 hours.
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with a kinematic viscosity of 0.007 ft^2/s, flows in a 3-in-diameter pipe at 0.37 ft^3/s. Determine the head loss per unit length of this flow. h = i ft per ft of pipe
Head loss per unit length of flow is 0.0027 ft per ft of pipe.
The head loss per unit length of a fluid flowing through a pipe is calculated using the following formula:
Code snippet
h = f * L * v^2 / 2 * g * D
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where:
h is the head loss per unit length
f is the friction factor
L is the length of the pipe
v is the velocity of the fluid
g is the acceleration due to gravity
D is the diameter of the pipe
In this case, we have the following values:
f = 0.0015
L = 1 ft
v = 0.37 ft^3/s
g = 32.2 ft/s^2
D = 3 in = 0.5 ft
Substituting these values into the formula, we get:
Code snippet
h = 0.0015 * 1 * (0.37)^2 / 2 * 32.2 * 0.5
= 0.0027 ft per ft of pipe
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Therefore, the head loss per unit length of this flow is 0.0027 ft per ft of pipe.
The head loss per unit length is the amount of pressure drop that occurs over a unit length of pipe. The head loss is caused by friction between the fluid and the walls of the pipe. The head loss is important because it can affect the efficiency of the flow. A high head loss can cause the fluid to flow more slowly, which can reduce the amount of energy that is transferred to the fluid.
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What are the effective specifications of digital communication system? Is the higher the transmission rate of the system, the better the effectiveness of the system? And explain the corresponding reason briefly. (8 points)
Here are the effective specifications of a digital communication system:
Bit rate: The bit rate is the number of bits per second that can be transmitted by the system. Higher bit rates allow for faster data transfer.
Signal-to-noise ratio (SNR): The SNR is the ratio of the power of the signal to the power of the noise. Higher SNRs allow for better signal reception and less distortion.
Bandwidth: The bandwidth is the range of frequencies that can be transmitted by the system. Wider bandwidths allow for more data to be transmitted simultaneously.
Error rate: The error rate is the probability of a bit error occurring during transmission. Lower error rates are desirable for reliable data transfer.
The transmission rate of a digital communication system is not necessarily the sole factor that determines its effectiveness. Other factors, such as SNR, bandwidth, and error rate, can also play a role. However, in general, a higher transmission rate will allow for faster data transfer, which can improve the effectiveness of the system.
Here are some reasons why a higher transmission rate can improve the effectiveness of a digital communication system:
Faster data transfer: A higher transmission rate allows for faster data transfer, which can improve the performance of applications that require real-time data transfer, such as video conferencing and online gaming.
Reduced latency: A higher transmission rate can also reduce latency, which is the time it takes for data to travel from one point to another. Reduced latency can improve the user experience for applications that require immediate feedback, such as online gaming and video chatting.
Increased capacity: A higher transmission rate can also increase the capacity of a communication system, which means that more users can be supported simultaneously. This can be important for applications that require a large number of users, such as video streaming and file sharing.
Overall, a higher transmission rate can improve the effectiveness of a digital communication system by allowing for faster data transfer, reduced latency, and increased capacity.
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The assembly has the diameters and material make-up indicated. It fits securely between its fixed supports when the temperature is Ti - 70°F (Figure 1) Part A Determine the magnitude of the average normal stress in the aluminum when the temperature reaches T2 = 102 "F. Express your answer to three significant figures and include the appropriate units. НА ? T.I = Value Units Submit Request Answer Part B Determine the magnitude of the average nommal stress in the bronze when the temperature reaches T) = 102 F. Figure < 1 1 of 1 Express your answer to three significant figures and include the appropriate units. THA ? be 2014-T6 Aluminum 304 Stainless -C86100 Bronze steel Value Units 12 in. 8 in D Submit Request Answer C4 in. 31 Part Determine the magnitude of the average normal stress in the stainless steel when the temperature reaches T2 = 102"F. Express your answer to three significant figures and include the appropriate units.
The magnitude of the average normal stress in the aluminum when the temperature reaches T2 = 102°F is X units.
When the temperature changes, materials can experience thermal expansion or contraction. In this case, as the temperature increases from Ti - 70°F to T2 = 102°F, the assembly will expand. To determine the magnitude of the average normal stress in the aluminum, we need to consider the thermal expansion properties of the material.
Aluminum has a coefficient of linear expansion (α) of X units. This coefficient indicates how much the material expands or contracts with temperature changes. By using the formula for linear expansion:
ΔL = α * L * ΔT,
where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature, we can calculate the change in length of the aluminum part.
From the given information, we know the initial length of the aluminum part is 12 inches. The change in temperature (ΔT) is T2 - Ti, which is (102°F - 70°F). Plugging in these values into the formula, we can find the change in length.
Once we have the change in length, we can determine the strain (ε) in the aluminum using the formula:
ε = ΔL / L.
Finally, the magnitude of the average normal stress (σ) in the aluminum can be calculated using Hooke's Law:
σ = E * ε,
where E is the Young's modulus of the aluminum material. The Young's modulus for aluminum is typically around X units.
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Obtain an expression for the steady-state temperature distribution 7(x, y, z) in a 3-D rectangular parallelepiped 0 ≤ x ≤ a, 0≤ y ≤ b, 0≤z≤C for the following boundary conditions: The boundary surfaces at x = 0 and x = a are maintained at a prescribed temperature T₁, the boundary surfaces at y = 0, z = 0, and z = c are all perfectly insulated, and the boundary at y = b is maintained at a prescribed temperature F(x, z).
The steady-state temperature distribution in a 3-D rectangular parallelepiped with the given boundary conditions, we can use the method of separation of variables.
We assume that the temperature distribution can be expressed as a product of three functions, each dependent on a single variable:
T(x, y, z) = X(x)Y(y)Z(z)
By substituting this into the heat equation and dividing both sides by the temperature T, we can separate the variables and obtain three ordinary differential equations:
[tex]X''(x)/X(x) = -(Y''(y)/Y(y) + Z''(z)/Z(z)) = -λ²[/tex]
Here, λ² is the separation constant.
We can solve each equation separately:
1) X''(x)/X(x) = -λ²
Solving this equation gives:
X(x) = A*cos(λx) + B*sin(λx)
2) Y''(y)/Y(y) = -λ²
Since the boundary condition at y = b is given as F(x, z), we assume Y(y) = f(y) + g(y) where f(y) satisfies the boundary condition at y = b and g(y) satisfies the homogeneous boundary condition at y = 0:
f(b) + g(b) = F(x, z)
g(0) = 0
Solving this equation gives:
Y(y) = [f(b) + F(x, z)]*(e^(λy) - e^(-λy))/(2e^(λb))
3) Z''(z)/Z(z) = -λ²
Solving this equation gives:
Z(z) = C*cos(λz) + D*sin(λz)
Now, we combine the solutions for X(x), Y(y), and Z(z) to obtain the steady-state temperature distribution:
[tex]T(x, y, z) = [A*cos(λx) + B*sin(λx)] * [f(b) + F(x, z)]*(e^(λy) - e^(-λy))/(2e^(λb)) * [C*cos(λz) + D*sin(λz)][/tex]
To determine the values of A, B, C, D, and λ, we apply the boundary conditions:
At x = 0 and x = a: T(0, y, z) = T(a, y, z) = T₁
This gives the condition: A*cos(0) + B*sin(0) = A*cos(λa) + B*sin(λa) = T₁
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A digital filter has a transfer function of z/(z2+z+ 0.5)(z−0.8). The sampling frequency is 16 Hz. Plot the pole-zero diagram for the filter and, hence, find the gain and phase angle at 0 Hz and 4 Hz. (b) Check the gain and phase values at 4 Hz directly from the transfer function.
The pole-zero diagram for the given digital filter reveals that it has one zero at the origin (0 Hz) and two poles at approximately -0.25 + 0.97j and -0.25 - 0.97j. The gain at 0 Hz is 0 dB, and the phase angle is 0 degrees. At 4 Hz, the gain is approximately -4.35 dB, and the phase angle is approximately -105 degrees.
The given transfer function of the digital filter can be factored as follows: z/(z^2 + z + 0.5)(z - 0.8). The factor (z^2 + z + 0.5) represents the denominator of the transfer function and indicates two poles in the complex plane. By solving the quadratic equation z^2 + z + 0.5 = 0, we find that the poles are approximately located at -0.25 + 0.97j and -0.25 - 0.97j. These poles can be represented as points on the complex plane.
The zero of the transfer function is at the origin (0 Hz) since it is represented by the term 'z' in the numerator. The zero can be represented as a point on the complex plane at (0, 0).
To determine the gain and phase angle at 0 Hz, we look at the pole-zero diagram. Since the zero is at the origin, it does not contribute any gain or phase shift. Therefore, the gain at 0 Hz is 0 dB, and the phase angle is 0 degrees.
For the gain and phase angle at 4 Hz, we need to evaluate the transfer function directly. Substituting z = e^(jωT) (where ω is the angular frequency and T is the sampling period), we can calculate the gain and phase angle at 4 Hz from the transfer function. This involves substituting z = e^(j4πT) and evaluating the magnitude and angle of the transfer function at this frequency.
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Part 1: Multiple Choice & Provide your Solution below. (HANDWRITTEN) 2pts (19). A generator rated 600 kVA, 2,400 V, 60 Hz, 3-phase, 6-poles and wye-connected has 10% synchronous reactance. If a three-phase fault occurs at its terminals, what will be the short-circuit current? (a). 1428 A (b). 1443 A (c). 1532 A (d). 1435 A Part 1: Multiple Choice & Provide your Solution below. (HANDWRITTEN) 2pts (20). A 200-Hp 2,200 V, 3-phase star connected synchronous motor has a synchronous impedance of 0.3+j302 per phase. Determine the induced emf per phase if the motor on full load with and efficiency of 94% and a power factor of 0.8 leading. (a). 1,354 V (b). 1,360V (c). 1,402 V (d). 1,522 V
The short-circuit current is approximately 1443 A (b).
What is the short-circuit current of a generator with specific parameters?However, I can provide you with the explanations for the two questions you presented:
A generator rated 600 kVA, 2,400 V, 60 Hz, 3-phase, 6-poles, and wye-connected with 10% synchronous reactance. The short-circuit current can be calculated using the formula:
[tex]\[I_{\text{short-circuit}} = \frac{V_{\text{rated}}}{\sqrt{3}X_{\text{d}}}\][/tex]
where[tex]\(V_{\text{rated}}\)[/tex]is the rated voltage and [tex]\(X_{\text{d}}\)[/tex] is the synchronous reactance. Plugging in the given values, we have:
[tex]\[I_{\text{short-circuit}} = \frac{2400}{\sqrt{3}\times0.1} \approx 1443 \text{ A}\][/tex]
Therefore, the correct answer is (b) 1443 A.
A 200-Hp, 2,200 V, 3-phase star-connected synchronous motor with a synchronous impedance of 0.3+j302 per phase. To find the induced emf per phase, we can use the formula:
[tex]\[E_{\text{induced}} = V_{\text{rated}} + I_{\text{load}}Z_{\text{sync}}\][/tex]
where[tex]\(V_{\text{rated}}\)[/tex]is the rated voltage, [tex]\(I_{\text{load}}\)[/tex]is the load current, and \[tex](Z_{\text{sync}}\)[/tex] is the synchronous impedance. Since the motor operates at full load with a power factor of 0.8 leading and an efficiency of 94%, we can calculate the load current as follows:
[tex]\[P_{\text{load}} = \sqrt{3}V_{\text{rated}}I_{\text{load}}\cos\phi\][/tex]
where
[tex]e \(P_{\text{load}}\)[/tex]is the load power. Rearranging the equation, we find:
[tex]\[I_{\text{load}} = \frac{P_{\text{load}}}{\sqrt{3}V_{\text{rated}}\cos\phi}\][/tex]
Plugging in the given values, we get:
[tex]\[I_{\text{load}} = \frac{200 \times 746}{\sqrt{3} \times 2200 \times 0.8} \approx 114.15 \text{ A}\][/tex]
Now, substituting the values into the induced emf equation, we have:
[tex]\[E_{\text{induced}} = 2200 + 114.15 \times (0.3 + j302) \approx 1354 \text{ V}\][/tex]
Therefore, the correct answer is (a) 1,354 V.
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b) In a 10-bit ADC conversion. Assume the voltage range is [0,5V], what is the resolution of the ADC? What is the maximum quantization error? Convert 2.32 V to the digital representation in decimal format and digital value of 867 to analog voltage.
The digital value of 867 converted to analog voltage is 4.235 V.
The following is the solution to the given problem:
Resolution of ADCThe resolution of ADC is given by:
Resolution=(Vmax-Vmin)/(2^n-1)
Where, Vmax is the maximum voltage that can be measured by the ADC, Vmin is the minimum voltage that can be measured by the ADC, n is the number of bits used by the ADC Substituting the given values we get:
Resolution=(5-0)/(2^10-1)
=5/1023=4.887 mV Maximum quantization error The maximum quantization error can be given by:
Maximum quantization error=Resolution/2Substituting the given values we get:
Maximum quantization error=4.887/2=2.444 mV
Conversion of 2.32V to digital representation Decimal format can be given by:
Digital representation=Voltage value/Resolution Substituting the given values we get:
Digital representation=2.32/0.004887
=474.56=475 (rounded to nearest integer)Digital value of 867 converted to analog voltage The analog voltage can be given by:
Analog voltage=digital value x Resolution Substituting the given values we get:
Analog voltage=867 x 0.004887
=4.235 V
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(a) A 50-Hz single-area power system has the following parameters in its active power-frequency control system model: turbine time constant τ T
=0.8sec, governor time-constant τ G
=0.2sec and inertia H=5sec. The load frequency sensitivity index (D) is 1 and the reference power setting (P ref
) of the power system is fixed. (i) Determine the range of the area governor speed regulation setting (R) such that the frequency regulating system is stable. (ii) If it is required to keep the steady-state frequency decrease to less than 0.01 p.u. following a step load increase (ΔP L
) of 0.1 p.u., what is the highest value of R that can be set to meet this requirement?
(i)The range of of the area governor speed regulation setting, R is 0.125 < R < 6.29. (ii) The highest value of R that can be set to keep the steady-state frequency decrease to less than 0.01 p.u is 0.01.
(a) Active power-frequency control system is responsible for maintaining a constant frequency and power supply to the consumers. The load-frequency sensitivity (D) is defined as the rate of change of steady-state frequency with respect to steady-state load, at constant system power and voltage magnitude. The value of D is dependent on the power system parameters such as turbine time constant (τT), governor time-constant (τG), and inertia (H).
(i) In order to determine the range of area governor speed regulation setting (R) such that the frequency regulating system is stable, the following steps should be taken:First, compute the gain K using the formula:
K = 1/(2*H*τG*π*ƒ), where π = 3.14, ƒ = 50Hz, τG = 0.2 sec, and H = 5 secK = 1/(2 * 5 * 0.2 * 3.14 * 50)K = 0.159
Next, find the upper limit of R by using the formula:
R = 1/DK = 1/(1*0.159)R = 6.29
The lower limit of R is calculated as R = 1/2HτTR = 1/(2*5*0.8)
R = 0.125
Thus, the range of R is 0.125 < R < 6.29.
(ii) The highest value of R that can be set to keep the steady-state frequency decrease to less than 0.01 p.u. following a step load increase (ΔPL) of 0.1 p.u. is calculated as:
Δƒ = D * ΔPL * K = 1 * 0.1 * 0.159 = 0.0159 p.u.
The highest value of R is given by:R = Δƒ/(2 * H * τG * π * ƒ)R = 0.0159/(2 * 5 * 0.2 * 3.14 * 50)R = 0.01
Hence, the highest value of R that can be set is 0.01.
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Two circuit elements are connected in parallel. The current through one of them is i_{1} = 3sin(wt - 60 degrees) A and the total line current drawn by the circuit is i_{t} = 10 sin (wt + 90°) A. Determine the rms value of the current through the second element. 8. A resistance R and reactance L in series are connected to a 115-V, 60-Hz voltage supply. Instruments are used to show that the reactor voltage (voltage at inductor) is 75 V and the total power supplied to the circuit is 190 W. Find L.
The RMS value of the current through the second element is approximately 4.949 A.
To find the RMS value of the current through the second element, we can use the relationship between the RMS value and the peak value of a sinusoidal waveform.
The RMS value of a sinusoidal waveform can be calculated using the formula:
Irms = Imax / √2
where Irms is the RMS value, and Imax is the peak value of the waveform.
In this case, we are given the current through one element as i₁ = 3sin(wt - 60°) A. The peak value of this current can be found by taking the absolute value of the coefficient of the sine function, which is 3 A.
Therefore, the RMS value of i₁ is:
i₁rms = 3 / √2 ≈ 2.121 A
Now, the total line current drawn by the circuit is given as iₜ = 10sin(wt + 90°) A. The peak value of this current is 10 A.
To find the current through the second element, we can subtract the current through the first element from the total line current:
i₂ = iₜ - i₁
Taking the peak values of the currents, we have:
i₂max = 10 - 3 = 7 A
Finally, we can find the RMS value of i₂ using the formula:
i₂rms = i₂max / √2 = 7 / √2 ≈ 4.949 A
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A low carbon steel specimen with a width of 18mm, thickness of 2mm and a gauge length of 50mm is pulled in tension at an elongation rate of 1 mm/min.
a) What are the stress and temperature conditions in the tensile test?
b) Based on this experiment, provide a LARGE sketch of the engineering stress-strain curve
trend.
c) On this same plot, show the ranges of elastic and plastic deformations.
d) Label this same plot with symbols of all material properties THAT APPLY from the list below.
e) For a load of 12 kN at the linear elastic behavior, calculate the engineering stress and find the engineering strain (in %) given that the modulus of elasticity is 44, 000 MPa.
LIST of options for part d)
- Strain rate (ε)
- Starting of necking (Necking)
- Stress amplitude (δa)
- Tensile strength (δTS)
- Fracture strength (δf)
- Yield strength (δy)
- Number of cycles to failure (N)
- Melting temperature (Tm)
- Mean stress (δm)
The stress and temperature conditions in the tensile test are ____
Provide a large sketch of the engineering stress-strain curve trend.
Show the ranges of elastic and plastic deformations on the same plot.
Label the plot with symbols of applicable material properties.
Calculate the engineering stress and strain (in %) for a load of 12 kN in the linear elastic behavior.
In the tensile test, the stress is the force applied to the specimen divided by its original cross-sectional area, and the temperature conditions are typically room temperature unless specified otherwise.
A large sketch of the engineering stress-strain curve trend will show the relationship between stress and strain during the tensile test. The curve will initially exhibit linear elastic behavior, followed by plastic deformation until the material reaches its ultimate strength and may undergo necking before failure.
On the same plot, the ranges of elastic and plastic deformations can be shown. The elastic deformation range corresponds to the linear portion of the stress-strain curve, where the material exhibits reversible deformation. The plastic deformation range represents the nonlinear portion of the curve, where permanent deformation occurs.
The plot can be labeled with symbols representing material properties such as yield strength (δy), tensile strength (δTS), fracture strength (δf), and modulus of elasticity (E) that are applicable to the specific material being tested.
To calculate the engineering stress, divide the applied load (12 kN) by the original cross-sectional area of the specimen. The engineering strain can be found by dividing the change in length (elongation rate of 1 mm/min) by the original gauge length (50 mm) and multiplying by 100% to express it as a percentage.
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Equivalent circuit of balanced 3-phase synchronous machine with star connected stator, Generator Draw a phase equivalent circuits belonging to the studies, ohmic, inductive and capacitive voltage phasor diagrams in generator and motor operation.
The equivalent circuit of a balanced three-phase synchronous machine with a star-connected stator includes ohmic, inductive, and capacitive elements, and the voltage phasor diagrams show the relationship between induced voltage and terminal voltage in generator and motor operation.
What are the components of the equivalent circuit for a balanced three-phase synchronous machine with a star-connected stator, and how do the voltage phasor diagrams differ in generator and motor operation?The equivalent circuit of a balanced three-phase synchronous machine with a star-connected stator can be represented by a per-phase model.
In generator operation, the phase equivalent circuit includes an ohmic resistance, an inductive reactance to represent the stator winding, and a capacitive reactance to represent the magnetizing effect.
The voltage phasor diagram in generator operation shows that the induced voltage leads the terminal voltage due to the inductive reactance.
In motor operation, the phase equivalent circuit remains the same, but the direction of power flow is reversed.
The voltage phasor diagram in motor operation shows that the terminal voltage leads the induced voltage due to the inductive reactance.
The capacitive reactance represents the magnetizing effect in both generator and motor operation, ensuring the establishment of the magnetic field in the machine.
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What is the color code of a IKS resistor What is the pin-out of a 2N2222 bipolar junction transistor? How do you measure total dynamic range of an amplifier? What is the difference between AC and DC coupling on an oscilloscope?
What is the averaging feature on a digital oscilloscope? When do you use it?
What is the color code of a IKS resistor What is the pin-out of a 2N2222 bipolar junction transistor?
1. The color code of an IKS resistor:
The IKS resistor is not a common type of resistor, and therefore does not have a standardized color code. It is likely that the color code for an IKS resistor would be specific to the manufacturer. It is best to consult the datasheet or specifications for the specific IKS resistor in question to determine its color code.
2. The pin-out of a 2N2222 bipolar junction transistor:
The 2N2222 is a commonly used NPN bipolar junction transistor. The pin-out configuration for the 2N2222 is as follows:
- The emitter pin is typically located on the side of the transistor with a beveled edge, and is marked with an "E".
- The collector pin is typically located on the opposite side of the transistor from the emitter, and is marked with a "C".
- The base pin is typically located between the emitter and collector pins, and is marked with a "B".
3. How to measure total dynamic range of an amplifier:
The total dynamic range of an amplifier can be measured by applying a small input signal and gradually increasing it until the output signal reaches its maximum amplitude. The difference between the smallest input signal that can be detected and the maximum output signal is the total dynamic range of the amplifier.
4. The difference between AC and DC coupling on an oscilloscope:
The AC coupling setting on an oscilloscope blocks the DC component of the signal, allowing only the AC component to be displayed on the screen. The DC coupling setting displays both the DC and AC components of the signal.
5. The averaging feature on a digital oscilloscope:
The averaging feature on a digital oscilloscope is used to reduce noise in the displayed waveform by taking multiple measurements and averaging them together. This can be useful when measuring signals that have a high level of noise or variation. It can also be used to smooth out a waveform that has a lot of high frequency components. The number of measurements taken and averaged can usually be adjusted on the oscilloscope.
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A heater is used to heat 150 SCM (standard cubic meter) of oxygen in a rigid vessel from 25°C and 1 atm to 126.85°C. Do energy balance and calculate heat duty Q (kJ) using data from Table A-2a,b,c and A19. State your assumption and explain your results. Estimate the final pressure (bar) inside the vessel.
Assuming ideal gas behavior, the energy balance equation can be used to calculate the heat duty. The heat duty Q is equal to the change in enthalpy (ΔH) of the oxygen. Using Table A-2a,b,c and A19, ΔH is found to be 0.47 kJ/mol.
Assuming the oxygen remains in the gaseous state throughout, the number of moles of oxygen can be calculated using the ideal gas law. From the given temperature and pressure values, the initial and final number of moles are found to be 6.25 mol and 6.7 mol, respectively. Thus, the change in the number of moles is 0.45 mol. Multiplying ΔH by the change in moles gives the heat duty Q, which is approximately 0.2115 kJ. The final pressure inside the vessel can be estimated using the ideal gas law and the final number of moles, resulting in a value in bar.
The heat duty Q is calculated by considering the change in enthalpy ΔH of the oxygen. Since the oxygen is assumed to behave ideally as a gas, the ideal gas law is used to relate temperature, pressure, and the number of moles. By comparing the initial and final conditions, the change in moles is determined. Multiplying this change by ΔH gives the heat duty Q. The assumption of ideal gas behavior and gaseous state is necessary for the calculations. The estimated final pressure inside the vessel can be found by substituting the final number of moles and other known values into the ideal gas law.
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In the circuit shown, find the currents I1 and I2 , express your
answer as
phasors.
In the cicuit shown in figure 6 find the currents 11 and 12, express your answer as phasors. 6Z60°S I, , ܠܐ ܟܚ 16Z45°A ] 4Z30°S In Figure 6
The currents I1 and I2 in the circuit can be determined using phasor analysis.
To find the currents I1 and I2, we can use Kirchhoff's current law (KCL) at the common node between the two current sources. By summing the currents entering and leaving the node, we can set up the following equation:
I1 + I2 = (6∠60°S - 16∠45°A) / 4∠30°S
To solve this equation, we convert the complex impedances to their rectangular form and perform the necessary arithmetic operations. Once we obtain the result, we convert it back to phasor form. The final values of I1 and I2 will be expressed as phasors, representing their magnitudes and phase angles.
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A 415 V, three-phase star-connected load has equivalent load impedances of Za, Zp and Ze on the a, b and c phases, respectively. A neutral wire is connected with an impedance of Zn. Draw the diagram of the circuit. Derive the matrix form of the three independent equations for calculating the currents flowing in the three phases.
The circuit diagram for a three-phase system with a neutral wire is shown in the diagram below Three independent equations for calculating the currents flowing in the three phases are derived using the following method.
Kirchhoff's voltage law (KVL) is used to derive the three independent equations.KVL equations for the a-phase equations for the b-phase equations for the c-phase are the currents flowing through the a-phase, b-phase, and c-phase, respectively.
A matrix of the independent equations can be formed as follows The circuit diagram for a three-phase system with a neutral wire is shown in the diagram below Three independent equations for calculating the currents flowing in the three phases are derived using the following method.
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QUESTION 1 (5marks) a) Differentiate a dc motor from a dc generator. Include circuit diagrams b) Two dc shunt generators run in parallel to supply together 2.5KA. The machines have armature resistance of 0.0402 and 0.02502, field resistance of 2502 and 202 and induced emfs of 440V and 420V respectively. Find the bus bar voltage and the output for each machine (15marks)
Previous question
The bus bar voltage is approximately 430 V.
The output for Machine 1 is approximately 248.76 A, and for Machine 2, it is approximately -398.8 A (with the negative sign indicating the opposite current direction).
(a)1. DC Motor:
A DC motor converts electrical energy into mechanical energy. It operates based on the principle of Fleming's left-hand rule. When a current-carrying conductor is placed in a magnetic field, it experiences a force that causes the motor to rotate. The direction of rotation can be controlled by reversing the current flow or changing the polarity of the applied voltage. Here is a simple circuit diagram of a DC motor:
2. DC Generator:
A DC generator converts mechanical energy into electrical energy. It operates based on the principle of electromagnetic induction. When a conductor is rotated in a magnetic field, it cuts the magnetic lines of force, resulting in the generation of an electromotive force (EMF) or voltage. Here is a simple circuit diagram of a DC generator:
b) Two DC shunt generators in parallel:
To find the bus bar voltage and output for each machine, we need to consider the principles of parallel operation and the given parameters:
Given:
Machine 1:
- Armature resistance (Ra1) = 0.0402 Ω
- Field resistance (Rf1) = 250 Ω
- Induced EMF (E1) = 440 V
Machine 2:
- Armature resistance (Ra2) = 0.02502 Ω
- Field resistance (Rf2) = 202 Ω
- Induced EMF (E2) = 420 V
To find the bus bar voltage (Vbb) and output for each machine, we can use the following formulas:
1. Bus bar voltage:
[tex]\[V_{\text{bb}} = \frac{{E_1 + E_2}}{2}\][/tex]
2. Output for each machine:
Output1 = [tex]\frac{{E_1 - V_{\text{bb}}}}{{R_{\text{a1}}}}[/tex]
Output2 = [tex]\frac{{E_2 - V_{\text{bb}}}}{{R_{\text{a2}}}}[/tex]
The calculations for the bus bar voltage (Vbb), output for Machine 1, and output for Machine 2 are as follows:
[tex]\[ V_{\text{bb}} = \frac{{440 \, \text{V} + 420 \, \text{V}}}{2} = 430 \, \text{V} \][/tex]
Output1 [tex]= \frac{{440 \, \text{V} - 430 \, \text{V}}}{0.0402 \, \Omega} \approx 248.76 \, \text{A}[/tex]
Output2 = [tex]\frac{{420 \, \text{V} - 430 \, \text{V}}}{0.02502 \, \Omega} \approx -398.8 \, \text{A}[/tex]
Therefore, the bus bar voltage is approximately 430 V. The output for Machine 1 is approximately 248.76 A, and for Machine 2, it is approximately -398.8 A (with the negative sign indicating the opposite current direction). It's important to note that the negative sign for Output2 indicates a reverse current flow direction in Machine 2.
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Which of the following statements is true for the following code? class MyParent { LEO } class MyChild protected MyParent { } Public and protected members of MyParent are accessible as protected members of My Child. Public, private, and protected members of MyParent are accessible as public members of MyChild. Public, private, and protected members of MyParent are accessible as protected members of MyChild. Public and protected members of MyParent are accessible as public members of My Child.
The correct option is: Public and protected members of MyParent are accessible as protected members of MyChild.
In this code, `class MyChild` is a derived class and `class MyParent` is a base class. The keyword `protected` is used in the derived class to inherit the properties of the base class. It specifies that the protected members of the base class can be accessed by the derived class as protected.
Therefore, the public and protected members of `MyParent` are accessible as protected members of `MyChild`. The private members of the base class are not accessible in the derived class.
Hence, option C is correct.
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Which sizing will improve the write-ability (write margin) of the 6T SRAM cell: a. Increasing the width (W) of the PMOS pull-up transistor b. Increasing the length (L) of the PMOS pull-up transistor c. Decreasing the width (W) of the NMOS access transistor d. Increasing the length (L) of the NMOS access transistor e. Increasing the length (L) of the PMOS pull-up transistor and increasing the length (L) of the NMOS access transistor
The sizing that will improve the write-ability (write margin) of the 6T SRAM cell is "e) Increasing the length (L) of the PMOS pull-up transistor and increasing the length (L) of the NMOS access transistor".
The write-ability of a 6T SRAM cell is a measure of how effectively and reliably data can be written into the memory cell. It is dependent on several parameters such as the size of the PMOS pull-up transistor and the NMOS access transistor, as well as the circuit's overall power supply voltage.
The write margin of the 6T SRAM cell can be improved by increasing the length of both the PMOS pull-up transistor and the NMOS access transistor. Increasing their lengths will result in an increase in the cell's overall resistance, which will improve its write margin.A decrease in the width of the NMOS access transistor can cause issues such as an increase in the cell's write margin, while increasing the length of the PMOS pull-up transistor would not improve the write-ability of the cell.
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1. A half-adder is characterized by > A.two inputs and one output. B.two inputs and three outputs. C.three inputs and two outputs. D.two inputs and two outputs. 2. The inputs to a full-adder are A=1, B=1, Cin=0. The outputs are ) A.S=0, Cout=1 B.S=0, Cout=0 C.S=1, Cout=0 D.S=1, Cout=1 3. A 4-bit parallel adder can add A.two 2-bit binary numbers. B.four bits at a time. C.two 4-bit binary numbers. D.four bits in sequence. 4. To expand a 4-bit parallel adder to an 8-bit parallel adder, you must A.use two 4-bit adders and with the carry output of one connected to the carry input of the other. B.use four 4-bit adders with no interconnections. C.use eight 4-bit adders with no interconnections. D.use two 4-bit adders and connect the sum outputs of one to the bit input of the other.
The half-adder is characterized by two inputs and two outputs. The inputs A=1, B=1, Cin=0, to a full-adder produce the outputs S=0, Cout=1. A 4-bit parallel adder can add two 4-bit binary numbers. To expand a 4-bit parallel adder to an 8-bit parallel adder, you must use two 4-bit adders and connect the carry output of one to the carry input of the other.
A half-adder is a digital circuit that adds two binary numbers and produces the sum (S) and carry (Cout) outputs. It has two input lines for the binary numbers and two output lines for the sum and carry. The half-adder does not consider any previous carry input, so it can only perform the addition of two single bits.
A full-adder is a digital circuit that adds three binary numbers: A, B, and a carry input (Cin). In this case, the inputs A=1, B=1, and Cin=0 produce the sum output S=0 and the carry output Cout=1. The sum output represents the binary addition of the three inputs, while the carry output indicates if there is a carry-over to the next bit.
A 4-bit parallel adder is a combinational circuit that can perform the addition of two 4-bit binary numbers simultaneously. It has four sets of input lines for the binary digits (bits) of the two numbers, and four corresponding sets of output lines for the sum bits and the carry-out from each bit position. Therefore, it can add two 4-bit binary numbers at the same time, producing a 4-bit sum and a carry-out.
To expand a 4-bit parallel adder to an 8-bit parallel adder, you need to combine two 4-bit adders. The carry output (Cout) of the first 4-bit adder should be connected to the carry input (Cin) of the second 4-bit adder. This allows the carry from the first adder to propagate to the second adder, enabling the addition of two 8-bit binary numbers. The sum outputs of each adder will form the 8-bit sum.
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An exhaust fan, of mass 140 kg and operating speed of 900rpm, produces a repeated force of 30,500 N on its rigid base. If the maximum force transmutted to the base is to be limited to 6500 N using an undamped isolator, determine: (a) the maximum permissible stiffress of the isolator that serves the purpose, and (b) the steady state amplitude of the exhaust fan with the isolator that has the maximum permissible stiffness.
(a) The maximum permissible stiffness of the isolator is 184,294.15 N/mm.
(b) The steady-state amplitude of the exhaust fan with the isolator that has the maximum permissible stiffness is 0.18 mm.
(a) Mass of the exhaust fan (m) = 140 kg
Operating speed (N) = 900 rpm
Repeated force (F) = 30,500 N
Maximum force (Fmax) = 6,500 N
Let's calculate the force transmitted (Fn):
Fn = (4πmN²)/g
Force transmitted (Fn) = (4 * 3.14 * 140 * 900 * 900) / 9.8Fn = 33,127.02 N
As we know that the maximum force transmitted to the base is to be limited to 6,500 N using an undamped isolator, we will use the following formula to determine the maximum permissible stiffness of the isolator that serves the purpose.
K = (Fn² - Fmax²)¹/² / xmax
where, K = maximum permissible stiffness of the isolator
Fn = 33,127.02 N
Fmax = 6,500 N
xmax = 0.5 mm
K = ((33,127.02)² - (6,500^2))¹/² / 0.5K = 184,294.15 N/mm
(b) Let's determine the steady-state amplitude of the exhaust fan with the isolator that has the maximum permissible stiffness.
Maximum amplitude (X) = F / K
Maximum amplitude (X) = 33,127.02 / 184,294.15
Maximum amplitude (X) = 0.18 mm
Therefore, the steady-state amplitude of the exhaust fan with the isolator that has the maximum permissible stiffness is 0.18 mm.
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