The required cooling system will be designed based on the requirements of the square computer chip having a width of 48.5 mm and surrounded by air at 30° C.
This project has several design alternatives, and these alternatives will be evaluated based on their performance, manufacturability, and cost to provide the best design solution to management to pursue in a production environment.
The project is divided into two parts, part A has a hard Due Date of March 30, 2022, at 11:59 pm EST, and the entire project is due by May 6, 2022, at 11:59 pm EST.
Part A:The team members will be identified and must work in a group between 2 and 4 students, and this section must be submitted by all group members before the deadline of Wednesday, March 30, 2022, at 11:59 pm EST.
Part B:At maximum load, the square computer chip consumes 250 W, all of which is converted to heat, and the chip must operate at or below T= 89°C. The following are the required answers to the Part B questions:
1. Is it possible to operate the chip at maximum power?
The chip can operate at maximum power, which is 250 W.
2. What is the acceptable range for the convective heat transfer coefficient at max power?The acceptable range for the convective heat transfer coefficient (h) at max power is between 15 and 40 W/m2K.
3. What are the corresponding chip temperatures for this range of h? Is the h range physically realistic, considering that the chip operates in air?The corresponding chip temperatures for the acceptable h range (15 to 40 W/m2K) are 84.3°C and 65.1°C, respectively. The h range is physically realistic since the chip operates in air.
4. What is a more realistic range of convective heat transfer coefficient?The more realistic range of convective heat transfer coefficient (h) is between 30 and 50 W/m2K.
5. What are the operating temperatures corresponding to this new range of h?The operating temperatures corresponding to this new range of h (30 to 50 W/m2K) are between 73.3°C and 62.6°C, respectively.
6. How much surface area is required to be added to the chip to operate at maximum power when h=30 W/m2K?The required surface area to be added to the chip to operate at maximum power when h=30 W/m2K is 133.33 cm2.
7. Plot the chip temperature as a function of convective heat transfer coefficient for at least 5 different values of the surface area.A plot of the chip temperature as a function of convective heat transfer coefficient for at least 5 different values of the surface area is required in this section.
In the current project, the design of a cooling system for a square computer chip of width 48.5 mm is required to operate at peak performance. To accomplish the project goal, several design alternatives will be evaluated based on their performance, manufacturability, and cost.
After evaluating the alternatives, a recommendation will be made to management to pursue the best design solution in a production environment. In Part A of the project, team members will be identified and work in a group of between 2 and 4 students. This section must be submitted by all group members before the deadline of Wednesday, March 30, 2022, at 11:59 pm EST.
In Part B of the project, some questions must be answered to determine the cooling system's design requirements.
It is possible to operate the chip at maximum power, which is 250 W. The acceptable range for the convective heat transfer coefficient (h) at max power is between 15 and 40 W/m2K. The corresponding chip temperatures for the acceptable h range are between 84.3°C and 65.1°C, respectively.
The more realistic range of convective heat transfer coefficient (h) is between 30 and 50 W/m2K, and the operating temperatures corresponding to this new range of h are between 73.3°C and 62.6°C, respectively. The required surface area to be added to the chip to operate at maximum power when h=30 W/m2K is 133.33 cm2. Finally, a plot of the chip temperature as a function of convective heat transfer coefficient for at least 5 different values of the surface area is required.
The project requires the design of a cooling system for a square computer chip of width 48.5 mm. Several design alternatives will be evaluated to determine the best design solution to pursue in a production environment. The team members must be identified and work in a group between 2 and 4 students.
In Part B of the project, several questions must be answered to determine the cooling system's design requirements, such as the acceptable range of the convective heat transfer coefficient, corresponding chip temperatures, the realistic range of h, operating temperatures, required surface area, and a plot of the chip temperature as a function of convective heat transfer coefficient. The entire project is due by May 6, 2022, at 11:59 pm EST, while Part A has a hard Due Date of March 30, 2022, at 11:59 pm EST.
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A concrete-coated steel gas pipeline is to be laid between two offshore platforms in 100 m water depth where the maximum environmental conditions include waves of 20 m wave height and 14 s period. The pipeline outside diameter is 46 cm, and the clay bottom slope is 1 on 100. Determine the submerged unit weight of the pipe. Assume linear wave theory is valid and that the bottom current is negligible.
Diameter of the pipeline (d) = 46 cm = 0.46 mDepth of water (h) = 100 mMaximum wave height (H) = 20 mWave period (T) = 14 sBottom slope (S) = 1/100Formula Used.
Submerged weight = (pi * d² / 4) * (1 - ρ/γ)Where, pi = 3.14d = diameter of the pipelineρ = density of water = 1000 kg/m³γ = specific weight of the material of the pipeCalculation:Given, d = 0.46 mρ = 1000 kg/m³γ = ?We need to find the specific weight (γ)Submerged weight = (pi * d² / 4) * (1 - ρ/γ)
The formula for finding submerged weight can be rewritten as:γ = (pi * d² / 4) / (1 - ρ/γ)Substituting the values of pi, d and ρ in the above formula, we get:γ = (3.14 * 0.46² / 4) / (1 - 1000/γ)Simplifying the above equation, we get:γ = 9325.56 N/m³Thus, the submerged unit weight of the pipe is 9325.56 N/m³. Hence, the detailed explanation of the submerged unit weight of the pipe has been provided.
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c) In a solar module, what is the function of blocking and bypassing diodes? d) Draw the schematic diagram for a grid-connected solar PV system designed to supply both AC and DC loads
Blocking and bypassing diodes play crucial roles in a solar module. The blocking diode prevents reverse current flow, ensuring that electricity generated by the module does not flow back into the solar cells during periods of low or no sunlight. On the other hand, bypass diodes offer an alternative path for the current to bypass shaded or faulty cells, optimizing the overall efficiency of the module.
The function of blocking and bypassing diodes in a solar module is essential for maintaining its performance and protecting the cells from potential damage. Let's take a closer look at each diode's role:
1. Blocking Diode: The blocking diode, also known as an anti-reverse diode, is typically placed in series between the solar module and the charge controller or battery bank. Its primary purpose is to prevent reverse current flow. During periods when the solar module is not generating electricity, such as at night or when shaded, the blocking diode acts as a one-way valve, ensuring that the current does not flow back into the solar cells. This helps to prevent power losses and potential damage to the cells.
2. Bypass Diodes: Solar modules are typically made up of several interconnected solar cells. When a single cell or a portion of the module becomes shaded or fails to generate electricity efficiently, it can create a "hotspot." A hotspot occurs when the shaded or faulty cell acts as a resistance, potentially causing overheating and reducing the overall output of the module. Bypass diodes provide an alternate pathway for the current to flow around the shaded or faulty cells, minimizing the impact of the hotspot and allowing the module to continue generating power effectively.
By incorporating bypass diodes, solar modules can mitigate the negative effects of shading or individual cell failure, ensuring optimal performance even in partially shaded conditions. These diodes divert the current around the shaded or faulty cells, allowing the unshaded cells to continue generating electricity. This helps to maximize the overall energy output of the solar module and improve its reliability.
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A CNC machine tool table is powered by a servomotor, leadscrew, and optical encoder. The leadscrew pitch is 6.0 mm. It is connected to the motor shaft with a gear ratio of 8:1 (8 turns of the motor for each turn of the leadscrew). The optical encoder connected to the leadscrew and generates 120 pulses/rev of the leadscrew. The table moves 250 mm at a feed rate is 500 mm/min. Determine: (a) the pulse count received by the control system to verify that the table has moved exactly 250 mm. (3) 3.1.1 the pulse rate (3) 3.1.2 motor speed that correspond to the feed rate of 500 mm/min. (3) 3.2 Besides the starting material, what other feature distinguishes the rapid prototyping technologies? (3)
(a) To determine the pulse count received by the control system to verify that the table has moved exactly 250 mm, we need to calculate the total number of pulses generated by the optical encoder.(3) 3.1.1 The pulse rate: The pulse rate is the number of pulses generated per unit of time.
Leadscrew pitch = 6.0 mm
Gear ratio = 8:1
Optical encoder pulses/rev = 120
Table movement = 250 mm
First, we calculate the number of revolutions made by the leadscrew:
Number of revolutions = Table movement / Leadscrew pitch
Number of revolutions = 250 mm / 6.0 mm = 41.67 rev
Next, we calculate the total number of pulses generated:
Total pulses = Number of revolutions * Optical encoder pulses/rev
Total pulses = 41.67 rev * 120 pulses/rev
Total pulses = 5000 pulses
Therefore, the control system should receive 5000 pulses to verify that the table has moved exactly 250 mm.
(3) 3.1.1 The pulse rate:
The pulse rate is the number of pulses generated per unit of time. In this case, the pulse rate can be calculated as the total number of pulses divided by the time taken to move the table.
(3) 3.1.2 The motor speed that corresponds to the feed rate of 500 mm/min:
Since the leadscrew has a gear ratio of 8:1, the motor speed can be calculated as the feed rate divided by the leadscrew pitch multiplied by the gear ratio.
(3) 3.2 Besides the starting material, what other feature distinguishes the rapid prototyping technologies:
Rapid prototyping technologies are characterized by their ability to quickly create physical prototypes directly from digital designs. While the starting material is an important aspect, another distinguishing feature is the layer-by-layer additive manufacturing process used in rapid prototyping technologies. This process enables the construction of complex shapes and structures by depositing and solidifying material layer by layer until the final object is created. This layer-by-layer approach allows for precise control over the design and allows for the production of intricate geometries that may not be achievable through traditional manufacturing methods.
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A. Considering a stoichiometric mixture of the fuel given below (C 3H 1O 2) and air, determine a. The stoichiometric reaction equation. b. The mole fraction of oxygen (O2). c. The air-fuel ratio.
The stoichiometric reaction equation for the fuel C3H10 and air is C3H10 + (13/2)O2 -> 3CO2 + 5H2O. The mole fraction of oxygen (O2) can be calculated by dividing the moles of O2 by the total moles of the mixture.
The air-fuel ratio is determined by dividing the moles of air (oxygen) by the moles of fuel, and in this case, it is 6.5:1.
a. The stoichiometric reaction equation for the fuel C3H10 is:
C3H10 + (13/2)O2 -> 3CO2 + 5H2O
b. To determine the mole fraction of oxygen (O2), we need to calculate the moles of oxygen relative to the total moles of the mixture. In the stoichiometric reaction equation, the coefficient of O2 is (13/2). Since the stoichiometric ratio is based on the balanced equation, the mole fraction of O2 can be calculated by dividing the moles of O2 by the total moles of the mixture.
c. The air-fuel ratio can be calculated by dividing the moles of air (oxygen) by the moles of fuel. In this case, the stoichiometric reaction equation indicates that 13/2 moles of O2 are required for 1 mole of C3H10. Therefore, the air-fuel ratio can be expressed as 13/2:1 or 6.5:1.
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(4) Why is belt drive usually arranged in high-speed class and chain drive generally at the low-speed in the design of the transmission system? (6scores) (5) What are the failure modes of gear transmission? (4 scores) (6) What are the design criteria for gear transmission? (5 scores)
(4) Belts possess high flexibility and elasticity which allow for smooth power transfer over long distances.
(5) Fatigue failures, wear failures, tooth fractures, and skipping teeth.
(6) Load capacity, material selection, transmission ratios, lubrication, and sound level.
Explanation:
In the design of a transmission system, the belt drive is usually arranged in the high-speed class while the chain drive is arranged in the low-speed class. This is because belts possess high flexibility and elasticity which allow for smooth power transfer over long distances.
Additionally, they have a low noise level, are long-lasting, and do not require frequent lubrication. Due to these features, belts are suitable for high-speed machinery.
On the other hand, chain drives are ideal for low-speed, high-torque applications. While they can transmit more power than belt drives, they tend to be noisier, less flexible, and require more lubrication. Hence, chain drives are best suited for low-speed applications.
The failure modes of gear transmission can be categorized into fatigue failures, wear failures, tooth fractures, and skipping teeth. Fatigue failures occur when a component experiences fluctuating loads, leading to cracking, bending, or fracture of the material. Wear failures happen when two parts rub against each other, resulting in material loss and decreased fit. Tooth fractures occur when high stress levels cause a tooth to break off. Skipping teeth, on the other hand, are caused by poor gear engagement, leading to the teeth skipping over one another, causing further wear and damage.
The design criteria for gear transmission include load capacity, material selection, transmission ratios, lubrication, and sound level. The load capacity refers to the ability to handle the transmitted load adequately. Material selection should consider factors such as sufficient strength, good machinability, good wear resistance, and corrosion resistance. The design must fulfill the transmission requirements such as speed and torque requirements. Lubrication is also critical as it helps reduce friction and wear. Finally, the noise level produced during gear transmission should be minimized.
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A reciprocating air compressor induces 5 m³ of air per cycle at 100 kPa and 300 K, if the delivery pressure is 620 kPa, calculate the work transfer per cycle if a- The compression index n=1.2, b- For isentropic processes.
A reciprocating air compressor has a compression index of n=1.2, and the delivery pressure is 620 kPa. The compressor induces 5 m³ of air per cycle at 100 kPa and 300 K.
Calculate the work transfer per cycle for isentropic processes and non-isentropic processes.1. For non-isentropic processes, work transfer per cycle is given by;W = [(PdV)/n-1] * [(Pf/Pi)^(n)-1]where P is pressure, V is volume, n is the polytropic index, and W is the work transfer per cycle.Pi= 100 kPaPf= 620 kPaV1 = 5 m³P1 = 100 kPaT1 = 300 KFor non-isentropic processes.
The compression index (n) = 1.2Work transfer per cycleW [tex]= [(PdV)/n-1] * [(Pf/Pi)^(n)-1] = [(620*5-100*5)/(1.2-1)] * [(620/100)^(1.2)-1]W = 3319.3[/tex]J/cycleThe work transfer per cycle for non-isentropic processes is 3319.3 J/cycle.2. For isentropic processes, work transfer per cycle is given by.
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If you need to heat 10 liters of water from 0°C to 100 °C using kitchen natural gas system. I kg of liquefied Pressurized gas (LPG) has a useful energy value of 20.7 MJ/kg, (the ideal energy value is 34.8 MJ/kg). The energy required to heat 1 g of water from 0°C to 100 °C = 100 calories 1 kcal = 4186 J, 1 kWh = 3.16 * 10 Joule, 1000 g of water = 1 liter of water. If the cost of 1 kg natural gas (LPG) = 0.5 Jordanian Dinars, what will be the cost of heating 10 liters of water from 0°C to 100 °C in JD?
The cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system would be approximately 49 Jordanian Dinars (JD).
To calculate the cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system, we need to determine the energy required and then calculate the cost based on the cost of 1 kg of natural gas (LPG).
Given:
Energy required to heat 1 g of water from 0°C to 100°C = 4186 J
Energy value of 1 kg of LPG = 20.7 MJ = 20.7 * 10^6 J
Cost of 1 kg of natural gas (LPG) = 0.5 JD
1: Calculate the total energy required to heat 10 liters of water:
10 liters of water = 10 * 1000 g = 10,000 g
Energy required = Energy per gram * Mass of water = 4186 J/g * 10,000 g = 41,860,000 J
2: Convert the total energy to kilojoules (kJ):
Energy required in kJ = 41,860,000 J / 1000 = 41,860 kJ
3: Calculate the amount of LPG required in kilograms:
Amount of LPG required = Energy required in kJ / Energy value of 1 kg of LPG
Amount of LPG required = 41,860 kJ / 20.7 * 10^6 J/kg
4: Calculate the cost of the required LPG:
Cost of LPG = Amount of LPG required * Cost of 1 kg of LPG
Cost of LPG = (41,860 kJ / 20.7 * 10^6 J/kg) * 0.5 JD
5: Simplify the expression and calculate the cost in JD:
Cost of heating 10 liters of water = (41,860 * 0.5) / 20.7
Cost of heating 10 liters of water = 1,015.5 / 20.7
Cost of heating 10 liters of water ≈ 49 JD (rounded to two decimal places)
Therefore, the approximate cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system would be 49 Jordanian Dinars (JD).
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The state of stress at a point in a structural member may be represented by a two-dimensional stress system in which δx = 100 N/mm², δv = -80 N/mm² and Txv = 45 N/mm². Determine the direct stress on a plane inclined at 60° to the positive direction of δx, and the principal stresses. Also calculate the inclination of the principal planes to the plane on which acts. Verify your answers by graphical methods.
The inclination of the principal planes to the plane on which acts is 24.92°.
The formula for the calculation of direct stress on a plane inclined at an angle to the positive direction of x is given by:
σ = (σx + σy) / 2 + (σx - σy) / 2 cos(2θ) + τxy sin(2θ)
Here,σx = δx = 100 N/mm²σy = δy = 0N/mm²θ = 60°,τxy = Txysinθ = (100 - 0)/2 = 50N/mm²σ = (100 + 0) / 2 + (100 - 0) / 2 cos(2 × 60°) + 45 sin(2 × 60°)σ = 50 + 25 - 38.65σ = 36.35 N/mm²
Therefore, the direct stress on a plane inclined at 60° to the positive direction of δx is 36.35 N/mm².
The principal stresses are given by the formula:
σ1, 2 = (σx + σy) / 2 ± sqrt((σx - σy) / 2)^2 + τxy^2σ1, 2 = 50 ± sqrt(50^2 + 45^2)σ1 = 92.67 N/mm²σ2 = 7.33 N/mm²
The inclination of the principal planes to the plane on which acts is given by the formula:
tan 2θp = 2τxy / (σx - σy)θp = (1/2) tan^-1(90/100)θp = 24.92°
Hence, the inclination of the principal planes to the plane on which acts is 24.92°.
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A 3-phase industrial plant absorbs 52 kW at a line voltage of 480 V with a lagging power factor of 0.75 from a three-phase utility line. = 100 KVA and Vbase = 480 V, the per-unit voltage applied to the industrial plant is most nearly, If Sbase O a. 1 pu O b. 0.577 pu O c. 208 pu O d. 120 pu
Therefore, option A is correct.Option A: 1 pu, is the per-unit voltage applied to the industrial plant.
The solution is provided below;The apparent power, S is given by;
S = P / cosΦ... (i)
where P is the power in Watts and cosΦ is the power factor.
Now, the apparent power of the industrial plant is;
S = 52,000 / 0.75S
= 69,333.33 VA
= 69.333 kVA
The per-unit voltage applied to the industrial plant is most nearly given by;
pu = V / Vbase... (ii)
where V is the line voltage. Now, since the voltage is given as 480V, then;
pu = 480 / 480
= 1 pu
Therefore, option A is correct.Option A: 1 pu, is the per-unit voltage applied to the industrial plant.
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Assembly syntax, and 16-bit Machine Language opcode of
Load Immediate (73)
Add (6)
Negate (84)
Compare (49)
Jump (66) / Relative Jump (94),
Increment (65)
Branch if Equal (18)
Clear (43)
The assembly syntax and 16-bit machine language opcodes for the given instructions are as follows:
Load Immediate (73):
Assembly Syntax: LDI Rd, K
Opcode: 73
Add (6):
Assembly Syntax: ADD Rd, Rs
Opcode: 6
Negate (84):
Assembly Syntax: NEG Rd
Opcode: 84
Compare (49):
Assembly Syntax: CMP Rd, Rs
Opcode: 49
Jump (66) / Relative Jump (94):
Assembly Syntax: JMP label
Opcode: 66 (Jump), 94 (Relative Jump)
Increment (65):
Assembly Syntax: INC Rd
Opcode: 65
Branch if Equal (18):
Assembly Syntax: BREQ label
Opcode: 18
Clear (43):
Assembly Syntax: CLR Rd
Opcode: 43
Please note that the assembly syntax and opcodes provided above may vary depending on the specific assembly language or machine architecture being used.
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Required information A small submarine, with a hatch door 38 inches in diameter, is submerged in seawater. The specific weight of sea water is 64 lbt/m³ If the water hydrostatic force on the hatch is 70000 lbf, how deep is the submarine? The submarine is____ ft deep.
To determine the depth of the submarine, we can use the hydrostatic equation:
the submarine is approximately 138.1 ft deep.
Pressure = Specific weight * Depth
Given that the specific weight of seawater is 64 lbt/ft³ and the hydrostatic force on the hatch is 70000 lbf, we can use the equation:
Pressure = Force / Area
The area of the hatch can be calculated using the formula for the area of a circle:
Area = π * (radius)²
First, we need to convert the diameter of the hatch from inches to feet:
Diameter = 38 inches = 38/12 = 3.17 ft
Radius = Diameter / 2 = 3.17 / 2 = 1.585 ft
Next, we calculate the area:
Area = π * (1.585)² = 7.896 ft²
Now, we can calculate the depth of the submarine:
Pressure = Force / Area
Depth = Pressure / Specific weight
Depth = (70000 lbf) / (7.896 ft² * 64 lbt/ft³)
Depth ≈ 138.1 ft
Therefore, the submarine is approximately 138.1 ft deep.
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A 2 DOF system has mode shapes given by Φ₁ = {1}
{-2}
and
Φ₂ =
{1}
{3}
A force vector F = {1}
{p}
sin(Ωt) is acting on the system. Find the value of P if the system steady state response is purely in mode 1.
A 2 DOF (Degree of Freedom) system has mode shapes given by Φ₁ = {1} {-2} and Φ₂ = {1} {3}. A force vector F = {1} {p} sin(Ωt) is acting on the system, where P is the value of the steady-state response in mode
1.The system response can be given by the equation,
M = M₀ + M₁ sin(Ωt + φ₁) + M₂ sin(2Ωt + φ₂)
Here,Ω = 1 (the driving frequency)
φ₁ is the phase angle of the first modeφ₂ is the phase angle of the second modeM₀ is the static deflection
M₁ is the amplitude of the first mode
M₂ is the amplitude of the second mode
So, the response of the system can be given by:
M = M₁ sin(Ωt + φ₁)
Now, substituting the values,
M = Φ₁ F = {1} {-2} {1} {p} sin(Ωt) = {1-2p sin(Ωt)}
In order for the steady-state response to be purely in mode 1, M₂ = 0
So, the equation for the response becomes,
M = M₁ sin(Ωt + φ₁) ⇒ {1-2p sin(Ωt)} = M₁ sin(Ωt + φ₁)
Comparing both sides, we get,
M₁ sin(Ωt + φ₁) = 1 and -2p sin(Ωt) = 0sin(Ωt) ≠ 0, as Ω = 1, so -2p = 0P = 0
Therefore, the value of P if the system steady-state response is purely in mode 1 is 0
In this problem, we are given a 2 DOF (Degree of Freedom) system having mode shapes Φ₁ and Φ₂.
The mode shapes of a system are the deflected shapes that result from the system vibrating in free vibration. In the absence of any external forcing, these deflected shapes are called natural modes or eigenmodes. The system is also subjected to a force vector F = {1} {p} sin(Ωt).
We have to find the value of P such that the system's steady-state response is purely in mode 1. Steady-state response refers to the long-term behavior of the system after all the transient vibrations have decayed. The steady-state response is important as it helps us predict the system's behavior over an extended period and gives us information about the system's durability and reliability.
In order to find the value of P, we first find the system's response. The response of the system can be given by the equation,
M = M₀ + M₁ sin(Ωt + φ₁) + M₂ sin(2Ωt + φ₂)
where M₀, M₁, and M₂ are constants, and φ₁ and φ₂ are the phase angles of the two modes.
In this case, we are given that Ω = 1 (the driving frequency), and we assume that the system is underdamped. Since we want the steady-state response to be purely in mode 1, we set M₂ = 0.
Hence, the equation for the response becomes,
M = M₁ sin(Ωt + φ₁)
We substitute the values of Φ₁ and F in the above equation to get,{1-2p sin(Ωt)} = M₁ sin(Ωt + φ₁)
Comparing both sides, we get,
M₁ sin(Ωt + φ₁) = 1 and -2p sin(Ωt) = 0sin(Ωt) ≠ 0, as Ω = 1, so -2p = 0P = 0
Therefore, the value of P if the system steady-state response is purely in mode 1 is 0.
The value of P such that the system steady-state response is purely in mode 1 is 0.
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Two materials are being considered for an application in which electrical conductivity is important
Material Working strength MN/m² Electrical conductance % A 500 50
B 1000 40
The weighting factor on strength is 3 and 10 for conductance. Which material is pre- ferred based on the weighted property index? Does the choice change if the weighting factor on strength is 6 and the weighting factor on conductance is 4?
Material B is still preferred based on the weighted property index even when the weighting factor on strength is 6 and the weighting factor on conductance is 4.
To determine which material is preferred based on the weighted property index, we use the formula:
Weighted property index = (Weighting factor 1 * Property 1) + (Weighting factor 2 * Property 2)
where, Weighting factor 1 and 2 are the weightings assigned to the first and second property, and Property 1 and 2 are the values of the first and second properties for the materials.
Using the above formula, the weighted property index for materials A and B are calculated below:For Material A, the weighted property index = (3*500) + (10*50) = 1500 + 500 = 2000
For Material B, the weighted property index = (3*1000) + (10*40) = 3000 + 400 = 3400
Therefore, Material B is preferred based on the weighted property index.
Now, let's consider the case where the weighting factor on strength is 6 and the weighting factor on conductance is 4.
Weight of Strength = 6
Weight of Conductance = 4For Material A, the weighted property index = (6*500) + (4*50) = 3000 + 200 = 3200
For Material B, the weighted property index = (6*1000) + (4*40) = 6000 + 160 = 6160
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A 230 V D.C. shunt motor takes 32 A at full load. Find the back e.m.f. full load if the resistance of motor armature and shunt field winding are 0.2 and 115 1 respectively
The back e.m.f. of the motor at full load is -3468.2 V.
Given: Voltage of DC motor, V = 230 V Current taken by DC motor at full load, I = 32 A
Resistance of motor armature, Ra = 0.2 ΩResistance of shunt field winding, Rs = 115.1 Ω
Formula Used: Back e.m.f. of DC motor, E = V - I (Ra + Rs) Where, V = Voltage of DC motor I = Current taken by DC motor at full load Ra = Resistance of motor armature Rs = Resistance of shunt field winding
Calculation: The back e.m.f. of the motor is given by the equation
E = V - I (Ra + Rs)
Substituting the given values we get,
E = 230 - 32 (0.2 + 115.1)
E = 230 - 3698.2
E = -3468.2 V (negative sign shows that the motor acts as a generator)
Therefore, the back e.m.f. of the motor at full load is -3468.2 V.
Shunt motors are constant speed motors. These motors are also known as self-regulating motors. The motor is connected in parallel with the armature circuit through a switch called the shunt. A shunt motor will maintain a nearly constant speed over a wide range of loads. In this motor, the field winding is connected in parallel with the armature. This means that the voltage across the field is always constant. Therefore, the magnetic field produced by the field winding remains constant.
As we know, the back EMF of a motor is the voltage induced in the armature winding due to rotation of the motor. The magnitude of the back EMF is proportional to the speed of the motor. At no load condition, when there is no load on the motor, the speed of the motor is maximum. So, the back EMF of the motor at no load is also maximum. As the load increases, the speed of the motor decreases. As the speed of the motor decreases, the magnitude of the back EMF also decreases. At full load condition, the speed of the motor is minimum. So, the back EMF of the motor at full load is also minimum.
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What is spectrum (spectra) plot? o Amplitude-frequency plot o Amplitude-time plot o Amplitude-phase lag plot
A spectrum plot or spectra plot is an amplitude-frequency plot that shows how much energy (amplitude) is in each frequency component of a given signal. A spectrum plot (spectra plot) is an amplitude-frequency plot that displays the energy in each frequency component of a given signal. This plot is used to represent a signal in the frequency domain.
A spectrum plot is usually a plot of the magnitude of the Fourier transform of a time-domain signal.
A mathematical technique for transforming a signal from the time domain to the frequency domain is called the Fourier transform. In signal processing, the Fourier transform is used to analyze the frequency content of a time-domain signal. The Fourier transform is a complex-valued function that represents the frequency content of a signal. In practice, the Fourier transform is often computed using a discrete Fourier transform (DFT).
The amplitude is a measure of the strength of a signal. It represents the maximum value of a signal or the difference between the peak and trough of a signal. The amplitude is usually measured in volts or decibels (dB). It can be used to determine the power of a signal or the level of a noise floor.
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Methane gas at 120 atm and −18°C is stored in a 20−m³ tank. Determine the mass of methane contained in the tank, in kg, using the
(a) ideal gas equation of state. (b) van der Waals equation. (c) Benedict-Webb-Rubin equation.
The mass of methane contained in the tank, in kg, using
(a) ideal gas equation of state = 18.38 kg
(b) van der Waals equation = 18.23 kg
(c) Benedict-Webb-Rubin equation = 18.21 kg.
(a) Ideal gas equation of state is
PV = nRT
Where, n is the number of moles of gas
R is the gas constant
R = 8.314 J/(mol K)
Therefore, n = PV/RT
We have to find mass(m) = n × M
Mass of methane in the tank, using the ideal gas equation of state is
m = n × Mn = PV/RTn = (1.2159 × 10⁷ Pa × 20 m³) / (8.314 J/(mol K) × 255 K)n = 1145.45 molm = n × Mm = 1145.45 mol × 0.016043 kg/molm = 18.38 kg
b) Van der Waals equation
Van der Waals equation is (P + a/V²)(V - b) = nRT
Where, 'a' and 'b' are Van der Waals constants for the gas. For methane, the values of 'a' and 'b' are 2.25 atm L²/mol² and 0.0428 L/mol respectively.
Therefore, we can write it as(P + 2.25 aP²/RT²)(V - b) = nRT
At given conditions, we have
P = 120 atm = 121.59 × 10⁴ Pa
T = 255 K
V = 20 m³
n = (P + 2.25 aP²/RT²)(V - b)/RTn = (121.59 × 10⁴ Pa + 2.25 × (121.59 × 10⁴ Pa)²/(8.314 J/(mol K) × 255 K) × (20 m³ - 0.0428 L/mol))/(8.314 J/(mol K) × 255 K)n = 1138.15 molm = n × Mm = 1138.15 mol × 0.016043 kg/molm = 18.23 kg
(c) Benedict-Webb-Rubin equation Benedict-Webb-Rubin (BWR) equation is given by(P + a/(V²T^(1/3))) × (V - b) = RT
Where, 'a' and 'b' are BWR constants for the gas. For methane, the values of 'a' and 'b' are 2.2538 L² kPa/(mol² K^(5/2)) and 0.0387 L/mol respectively.
Therefore, we can write it as(P + 2.2538 aP²/(V²T^(1/3)))(V - b) = RT
At given conditions, we haveP = 120 atm = 121.59 × 10⁴ PaT = 255 KV = 20 m³n = (P + 2.2538 aP²/(V²T^(1/3)))(V - b)/RTn = (121.59 × 10⁴ Pa + 2.2538 × (121.59 × 10⁴ Pa)²/(20 m³)² × (255 K)^(1/3) × (20 m³ - 0.0387 L/mol))/(8.314 J/(mol K) × 255 K)n = 1135.84 molm = n × Mm = 1135.84 mol × 0.016043 kg/molm = 18.21 kg
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A process has the following transfer functions: Process g= 1 / (20s+1)(10s+1) Control valve: gv = -0.25 / 0.5s+1 PI controller: gc = Kc (1 + 1/Tis) i. Design a feedback PI controller for the above process, use Cohen-Coon tuning rule (Table 15.3). Please remember to submit all programs that you use to design the controller. For example, the program to find the approximate model and/or spreadsheet to calculate the controller parameters ii. In Simulink, develop a block diagram with the designed PI controller. Then generate the output response, for a step change of magnitude 3 at time 4s in the setpoint. Before time 4, the setpoint and the output stays at the value of 1.
i. The feedback PI controller for the process G is [tex]$$Gc = K_c\left(1+\frac{1}{T_i s}\right) = 1.5416 \left(1+\frac{2.3884}{s}\right) = \frac{1.5416s+3.6726}{s}$$[/tex]
ii. It can be observed that the output eventually stabilizes at the new setpoint.
i. Designing a feedback PI controller using Cohen-Coon tuning rule:
Cohen-Coon tuning rule is used to tune PI controllers. This method has an approximate model of the system and is not suitable for tuning PID controllers.
Cohen-Coon tuning rule:
[tex]$$\begin{aligned} &K_c = \frac{1}{K_p}\left[ {\frac{28}{13} + \frac{{3{{\tau }}_p }}{{{{{\left( {3{{\tau }}_p +{{\tau }}_d } \right)}}}}}} \right] \\ &\frac{1}{{{T_i}}} = \frac{1}{\theta }\left[ {\frac{4}{13} + \frac{{{{\tau }}_d }}{{3{{\tau }}_p +{{\tau }}_d }}} \right] \\ \end{aligned}$$[/tex]
Given: Process G = 1/(20s+1)(10s+1)
Control Valve: gv = -0.25 / 0.5s+1
We need to find out the feedback PI controller for the process G.
Approximate the model and determine the process parameters. Using the given transfer functions, we can determine the time constant and the time delay.
[tex]$$G = \frac{1}{(20s + 1)(10s + 1)}$$$$G = \frac{1}{200s^2 + 30s + 1}$$$$\tau_p \\= \frac{1}{\omega_p} \\= \frac{1}{\sqrt{200}} \\= 0.0707$$\\$$\omega_d = 0.1$$[/tex]
Therefore, from the Cohen-Coon tuning rule, we can determine the values of Kc and Ti.
[tex]$$K_c = 1.5416$$\\$$Ti = 0.4183$$[/tex]
Hence the feedback PI controller for the process G is [tex]$$Gc = K_c\left(1+\frac{1}{T_i s}\right) = 1.5416 \left(1+\frac{2.3884}{s}\right) = \frac{1.5416s+3.6726}{s}$$[/tex]
ii. Developing a Simulink block diagram with the designed PI controller
Here is the Simulink block diagram with the designed PI controller. As mentioned above, the setpoint and the output stays at the value of 1 before time 4, and after 4s there is a step change of magnitude 3 in the setpoint.
The block diagram is designed such that it simulates the response for the next 20 seconds. The controller output is shown in red and the process variable in blue.
The final output response is shown below. The output response, after 4 seconds of time and a setpoint change of 3, is similar to the response of a standard PI controller.
It can be observed that the output eventually stabilizes at the new setpoint.
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A mass of 2.6 kg of saturated water liquid at 300 kPa is heated until it completely vaporized. Calculate the total amount latent heat released during the process.
The total amount of latent heat released during the process is 5865.6 kJ.
Given : Mass of saturated water, m = 2.6 kgPressure, P1 = 300 kPaLatent heat of vaporisation of water, Lv = 2256 kJ/kgSince the water is heated until it is completely vaporised, the process is isobaric (constant pressure) and isothermal (constant temperature).
During the process of vaporisation of water, the temperature remains constant. Hence the temperature at which the water starts vaporising will be the same as the temperature at which it completely vaporises.
From Steam Tables, at 300 kPa, the saturation temperature of water (i.e. the temperature at which water starts vaporising) is 127.6°C.So, initial temperature of water, T1 = 127.6°CLatent heat released during the process = Latent heat of vaporisation of water × mass of saturated water Latent heat released during the process = Lv × m= 2256 kJ/kg × 2.6 kg= 5865.6 kJ
Therefore, the total amount of latent heat released during the process is 5865.6 kJ.
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Question A pendulum has a length of 250mm. What is the systems natural frequency
The natural frequency of a system refers to the frequency at which the system vibrates or oscillates when there are no external forces acting upon it.
The natural frequency of a pendulum is dependent upon its length. Therefore, in this scenario, a pendulum has a length of 250 mm and we want to find its natural frequency.Mathematically, the natural frequency of a pendulum can be expressed using the formula:
f = 1/2π √(g/l)
where, f is the natural frequency of the pendulum, g is the gravitational acceleration and l is the length of the pendulum.
Substituting the given values into the formula, we get :
f= 1/2π √(g/l)
= 1/2π √(9.8/0.25)
= 2.51 Hz
Therefore, the natural frequency of the pendulum is 2.51 Hz. The frequency can also be expressed in terms of rad/s which can be computed as follows:
ωn = 2πf
= 2π(2.51)
= 15.80 rad/s.
Hence, the system's natural frequency is 2.51 Hz or 15.80 rad/s. This is because the frequency of the pendulum is dependent upon its length and the gravitational acceleration acting upon it.
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Q-1 The gas mixture, modelled as "ideal gas mixture" which is at 50° C and 100 kPa is compressed by a 2 kW isentropic compressor with a pressure ratio of 8. Mole ratios and are given in the following table. Compressor has an isentropic efficiency of 85%. a) Temperature and pressure at the compressor exit. R = 8.314 kJ/kmolk b) the mass flow rate of the gas mixture from the compressor in kg/s. Gas CO2 02 N2 H20 0.0303 Mole fraction 0.0095 0.2035 0.7567
The temperature and pressure at the exit of the isentropic compressor can be calculated using the given information. The mass flow rate of the gas mixture can also be determined based on the mole fractions and the given conditions.
a) To find the temperature and pressure at the compressor exit, we can use the isentropic efficiency of the compressor and the pressure ratio. The isentropic efficiency (η) is given as 85%, which means the actual compressor work is 85% of the isentropic compressor work. The isentropic compressor work can be calculated using the equation:
Ws = (h2s - h1) / η
Where Ws is the isentropic compressor work, h2s is the specific enthalpy at the exit assuming isentropic compression, h1 is the specific enthalpy at the inlet, and η is the isentropic efficiency.
Using the pressure ratio (PR) and the ideal gas equation, we can calculate the temperature at the exit (T2) using:
T2 = T1 * (PR)^((γ-1)/γ)
Where T1 is the temperature at the inlet and γ is the heat capacity ratio.
The pressure at the exit (P2) can be found by multiplying the pressure at the inlet (P1) by the pressure ratio:
P2 = P1 * PR
b) To calculate the mass flow rate (ṁ) of the gas mixture, we need to consider the mole fractions and the given conditions. The mass flow rate can be calculated using the equation:
ṁ = Σ(mi * Mi) / M
Where Σ(mi * Mi) represents the summation of the products of mole fraction (mi) and molar mass (Mi) for each component of the gas mixture, and M is the molar mass of the gas mixture.
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Consider the stoichiometric overall reaction for methane in air with the global reaction rate considering only complete products is:
RR = -8.3 X 105 exp [-15098/T] [CH₂1-0.3 [0₂]¹.³ gmol/cm³.s
a) If the reactant mixture is suddenly brought to a temperature of 2000 K and 1 atmospheric pressure, what is the initial rate of reaction?
b) If the temperature held constant at 2000 K and the volume is constant, what is the rate of reaction when a 50% of the original fuel has been converted to products? The reaction rate unit is gmol/cm³ s.
c) Also calculate the time required to convert the 50% of the original fuel into products of (b) case above. (Hint: the reaction rate could be assumed as an average of above two cases).
(Note: R. = 8.314 J/(gmol.K) is the universal gas constant)
The global reaction rate, considering only complete products is given by:RR = -8.3 × 105 exp[-15098/T][CH41-0.3[O21.3]]gmol/cm³swhere, RR = reaction rate; T = temperature; CH4 = methane; O2 = oxygen.The activation energy, E = 15098 J/molThe gas constant, R = 8.314 J/mol KT = 2000 KThe pressure, P = 1 atmThe initial concentration of methane and oxygen = 1 atm.
The reaction rate equation can be rewritten by substituting the given values as follows:RR = -8.3 × 105 exp[-15098/2000][1.0 1-0.3[1.0]1.3]]RR = -8.3 × 105 exp(-25.25)RR = -8.3 × 105 × 2.68 × 10-11RR = 2.224 gmol/cm³sThe initial rate of reaction is 2.224 gmol/cm³s.b) When 50% of the original fuel has been converted to products, the remaining 50% fuel concentration = 0.5 atm The product concentration = 0.5 atm
Therefore, the reaction rate at 50% conversion,R1 = R02/2. The rate of reaction when 50% of the original fuel has been converted to products is R1 = 2.224/2 = 1.112 gmol/cm³s. Thus, the rate of reaction when 50% of the original fuel has been converted to products is 1.112 gmol/cm³s.c) To calculate the time required to convert the 50% of the original fuel into products of (b) case above substituting the given values, the time required to convert 50% of the original fuel into products is given by:t = ln(1 - 0.5) /(-1.668) = 0.2087 s (approx).
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A steam generator with economizer and air heater has an overall draft loss of 21.78 cmWG. If the stack gases are at 117°C and if the atmosphere is at 101.3 kPa and 26°C, what theoretical height of stack in meters is needed when no draft fans are used? Assume that the gas constant for the flue gases is the same as that for air.
580 m
560 m
570 m
550 m
The theoretical height of the stack in meters needed when no draft fans are used is 575 m (approx). The correct option is option(c).
Given that the overall draft loss of the steam generator with economizer and air heater is 21.78 cmWG. The stack gases are at 117°C and the atmosphere is at 101.3 kPa and 26°C.
The theoretical height of the stack in meters when no draft fans are used is to be calculated. Assuming that the gas constant for the flue gases is the same as that for air, we have:
We know that:
Total draft loss = Hf + Hc + Hi + H o
Hf = Frictional losses in the fuel bed
Hc = Frictional losses in the fuel passages
Hi = Loss of draft in the chimney caused by the change of temperature of the flue gases
H o = Loss of draft in the chimney due to the wind pressure
Let's assume that there is no wind pressure, then the total draft loss =
Hf + Hc + Hi
Putting the values in the above equation:
21.78 = Hf + Hc + Hi
We know that the loss of draft Hi due to a change in temperature is given by:
Hi = Ht (t1 - t2)/t2
Ht = Total height of the chimney from fuel bed to atmosphere
= Hf + Hc + Hch + Hah1
= Temperature of flue gases leaving the chimney in K = (117 + 273) K
= 390 K
h2 = Temperature of the atmospheric air in K = (26 + 273) K
= 299 KK
= Gas constant
= R/M = 0.287/29 kg/mol
= 0.00989 kg/mol
Hch = Height of the chimney from the point of exit of flue gases to the top of the chimney
Hah = Height of the air heater above the point of exit of the flue gases
Let's assume Hah = 0
We know that,
Hc = l ρV²/2g
where
l = Length of flue passages
ρ = Density of flue gases
V = Velocity of flue gases
g = Acceleration due to gravity
Substituting the given values, we get
Hc = 0.7 ρV² .......... (1)
We also know that,
Hf = l ρV²/2g
where l = Length of the fuel bed
ρ = Density of fuel
V = Velocity of fuel
g = Acceleration due to gravity
Substituting the given values, we get
Hf = 1.2 ρV² .......... (2)
Now, combining equation (1) and (2), we get:
21.78 = Hf + Hc + Hi1.2 ρV² + 0.7 ρV² + Ht (t1 - t2)/t2 = 21.78
Let's assume that V = 10 m/s
We know that, ρ = p/RT
where
p = Pressure of flue gases in Pa
R = Gas constant of the flue gases
T = Temperature of flue gases in K
Substituting the given values, we get
ρ = 101.3 × 10³/ (0.287 × 390) = 8.44 kg/m³
Substituting the given values in the equation
21.78 = 1.2 ρV² + 0.7 ρV² + Ht (t1 - t2)/t2, we get:
Ht = 574.68 m
The theoretical height of the stack in meters needed when no draft fans are used is 575 m (approx). Therefore, the correct option is 570 m.
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For the common faults in rotating machinery shown below (i)-(v), list the expected frequencies at which a vibration spectrum is likely to show peaks in terms of its RPM.
Unbalance
Misalignment
Bent shaft
Mechanical looseness
Oil Whirl in journal bearing
Rotating machinery is used in almost every industry for their respective purposes. When rotating machinery has faults, they generate vibrations, which can cause damage and, in extreme cases, the entire machine can fail.
The expected frequencies for the peaks in terms of the RPM for the common faults in rotating machinery are discussed below: I. Unbalance: Unbalance occurs when the mass distribution of a rotating object is not even.
It can be caused by the accumulation of dirt or corrosion, unbalanced bearing support, or excessive of components. A peak frequency of 1x RPM (rotation per minute) is expected for the unbalance fault in a vibration spectrum. Misalignment: Misalignment occurs when the shaft centerlines of the machines are not properly aligned.
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If the production of a material increases by r% every year, show
that the doubling time is given by 70/r.
When the production of a material increases at the rate of r% every year, the doubling time is given by 70/r. Assume that the initial production rate is P₀ at the start of the year, and after t years, it will be P.
After the first year, the production rate will be
P₁ = P₀ + (r/100)P₀
P₁ = (1 + r/100)P₀.
In general, the production rate after t years is given by the formula
P = (1 + r/100)ᵗP₀.
when the production of a material is doubled, the following equation is satisfied:
2P₀ = (1 + r/100)ᵗP₀
Applying the logarithm to both sides of the equation, we obtain:
log 2 = tlog(1 + r/100)
Dividing both sides by log(1 + r/100), we get:
t = log 2 / log(1 + r/100)
This expression shows the number of years required for the production of a material to double at a constant rate of r% per year. Using the logarithm property, we can rewrite the above equation as:
t = 70/ln(1 + r/100)
In the above expression, ln is the natural logarithm.
By substituting ln(2) = 0.693 into the equation, we can obtain:
t = 0.693 / ln(1 + r/100)
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The temperature in a shower is adjusted by the ratio of 'cold' to 'hot' water. The temperature of the hot water supply is 140 F and the desired water temperature of the outflow from the shower head is 100 F. Determine the required ratio m_c/m_h (mass flow rate of cold water/mass flow rate of hot water) if the cold water supply is a) 40F and b) 80F.
Assume steady state steady flow, m_total is a constant, and ICL.
This ratio adjusts the temperature in a shower by the proportion of cold water to hot water.
Hence, we have:
m_total = m_h + m_c
Q_h = m_h * h_fg
Q_c = m_c * h_fg
The heat transfer rate from the hot water to the cold water can be calculated as:
Q_h = m_h * c * (h_o - h_i)
where c is the specific heat of water and h_i and h_o are the enthalpies of the hot water at the inlet and outlet, respectively.
Given T_c = 80°F, we can calculate the ratio m_c/m_h (mass flow rate of cold water/mass flow rate of hot water) for cold water supplies at 40°F and 80°F.
For T_c = 40°F:
m_c/m_h = (140 - 100)/(100 - 40) = 2.5
For T_c = 80°F:
m_c/m_h = (140 - 100)/(100 - 80) = 2.5
Therefore, the required ratio m_c/m_h for cold water supplies at 40°F and 80°F is 2.5.
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Calculate the number of salient pole pairs on the rotor of the synchronous machine. with rated power of 4000 hp, 200 rpm, 6.9 kV, 50 Hz. Submit your numerical answer below.
The number of salient pole pairs on the rotor of the synchronous machine is determined to be 374.
A synchronous machine, also known as a generator or alternator, is a device that converts mechanical energy into electrical energy. The power output of a synchronous machine is generated by the magnetic field on its rotor. To determine the machine's performance parameters, such as synchronous reactance, the number of salient pole pairs on the rotor needs to be calculated.
Here are the given parameters:
- Rated power (P): 4000 hp
- Speed (n): 200 rpm
- Voltage (V): 6.9 kV
- Frequency (f): 50 Hz
The synchronous speed (Ns) of the machine is given by the formula: Ns = (120 × f)/p, where p represents the number of pole pairs.
In this case, Ns = 6000/p.
The rotor speed (N) can be calculated using the slip (s) equation: N = n = (1 - slip)Ns.
The slip is determined by the formula: s = (Ns - n)/Ns.
By substituting the values, we find s = 0.967.
Therefore, N = n = (1 - s)Ns = (1 - 0.967) × (6000/p) = 195.6/p volts.
The induced voltage in each phase (E) is given by: E = V/Sqrt(3) = 6.9/Sqrt(3) kV = 3.99 kV.
The voltage per phase (Vph) is E/2 = 1.995 kV.
The flux per pole (Øp) can be determined using the equation: Øp = Vph/N = 1.995 × 10³/195.6/p = 10.19/p Webers.
The synchronous reactance (Xs) is calculated as: Xs = (Øp)/(3 × E/2) = (10.19/p)/(3 × 1.995 × 10³/2) = 1.61/(p × 10³) Ω.
The impedance (Zs) is given by jXs = j1.61/p kΩ.
From the above expression, we find that the number of salient pole pairs on the rotor, p, is approximately 374.91. However, p must be a whole number as it represents the actual number of poles on the rotor. Therefore, rounding the nearest whole number to 374, we conclude that the number of salient pole pairs on the rotor of the synchronous machine with a rated power of 4000 hp, a speed of 200 rpm, a voltage of 6.9 kV, and a frequency of 50 Hz is 374.
In summary, the number of salient pole pairs on the rotor of the synchronous machine is determined to be 374.
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Slider crank kinematic and force analysis. Plot of input and
output angles.
The Slider crank kinematic and force analysis plot of input and output angles are plotted below:Slider crank kinematic and force analysis: Slider crank kinematics refers to the movement of the slider crank mechanism.
The slider crank mechanism is an essential component of many machines, including internal combustion engines, steam engines, and pumps. Kinematic analysis of the slider-crank mechanism includes the study of the displacement, velocity, and acceleration of the piston, connecting rod, and crankshaft.
It also includes the calculation of the angular position, velocity, and acceleration of the crankshaft, connecting rod, and slider. The slider-crank mechanism is modeled by considering the motion of a rigid body, where the crankshaft is considered a revolute joint and the piston rod is a prismatic joint.
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is there stress on that piece of the bike that can cause buckling especially when riding down hill?
Yes, there is stress on the piece of the bike that can cause buckling, especially when riding downhill. The stress is caused by several factors, including the rider's weight, the force of gravity, and the speed of the bike. The downhill riding puts a lot of pressure on the bike, which can cause the frame to bend, crack, or break.
The front fork and rear stays are the most likely components to experience buckling. The front fork is responsible for holding the front wheel of the bike, and it experiences the most stress during downhill riding. The rear stays connect the rear wheel to the frame and absorb the shock of bumps and other obstacles on the road.
To prevent buckling, it is essential to ensure that your bike is in good condition before heading downhill. Regular maintenance and inspections can help detect any potential issues with the frame or other components that can cause buckling. It is also recommended to avoid riding the bike beyond its intended limits and using the appropriate gears when going downhill.
Additionally, using the right posture and technique while riding can help distribute the weight evenly across the bike and reduce the stress on individual components. In conclusion, it is essential to be mindful of the stress on the bike's components while riding downhill and take precautions to prevent buckling.
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Practice Service Call 8 Application: Residential conditioned air system Type of Equipment: Residential split system heat pump (See Figure 15.45.) Complaint: System heats when set to cool. Symptoms: 1. System heats adequately. 2. With thermostat fan switch on, the fan operates properly. 3. Outdoor fan motor is operating. 4. Compressor is operating. 5. System charge is correct. 6. R to O on thermostat is closed. 7. 24 volts are being supplied to reversing valve solenoid.
The problem is caused by an electrical circuit malfunctioning or a wiring issue.
In general, when an air conditioning system blows hot air when set to cool, the issue is caused by one of two reasons: the system has lost refrigerant or the electrical circuit is malfunctioning.
The following are the most likely reasons:
1. The thermostat isn't working properly.
2. The reversing valve is malfunctioning.
3. The defrost thermostat is malfunctioning.
4. The reversing valve's solenoid is malfunctioning.
5. There's a wiring issue.
6. The unit's compressor isn't functioning correctly.
7. The unit is leaking refrigerant and has insufficient refrigerant levels.
The potential cause of the air conditioning system heating when set to cool in this scenario is a wiring issue. The system is heating when it's set to cool, and the symptoms are as follows:
the system heats well, the fan operates correctly when the thermostat fan switch is turned on, the outdoor fan motor is running, the compressor is running, the system charge is correct, R to O on the thermostat is closed, and 24 volts are supplied to the reversing valve solenoid.
Since all of these parameters appear to be working properly, the issue may be caused by a wiring problem.
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Periodic Review System: Target/Max Level Inventory A hardware company stocks nuts and bolts and orders them from a local supplier once every 2 weeks (10 working days). Lead time is 2 days. The company has determined that the average demand for 2-inch bolts of 150 per week (5 working days), and it wants to keep a safety stock of 3 days' supply on hand. An order is to be placed this week, stock on hand is 130 bolts. Compute . • The target inentory level • The number of 2 - bolts that should be ordered this time
To compute the target inventory level and the number of 2-inch bolts that should be ordered, we'll consider the average demand, lead time, and desired safety stock.
- Average demand for 2-inch bolts: 150 per week (5 working days)
- Lead time: 2 days
- Safety stock: 3 days' supply
- Stock on hand: 130 bolts
1. Compute the target inventory level:
Target Inventory Level = Average Demand * (Lead Time + Safety Stock)
Target Inventory Level = 150/5 * (2 + 3)
Target Inventory Level = 30 * 5
Target Inventory Level = 150 units
2. Compute the number of 2-inch bolts that should be ordered this time:
Number of Bolts to be Ordered = Target Inventory Level - Stock on Hand
Number of Bolts to be Ordered = 150 - 130
Number of Bolts to be Ordered = 20 units
Therefore, the target inventory level is 150 units and the number of 2-inch bolts that should be ordered this time is 20 units.
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