The coefficient of thermal expansion (α) is approximately 6.5 x 10^-6 in/in/°F. the bar will elongate by approximately 0.046 inches due to the temperature change.
To calculate the elongation of a bar due to a temperature change, we can use the formula for thermal expansion:
ΔL = L * α * ΔT
Where:
ΔL is the change in length
L is the original length of the bar
α is the coefficient of thermal expansion
ΔT is the change in temperature
Given:
Original length (L) = 16 ft
Change in temperature (ΔT) = 500°F - 66°F = 434°F
Substituting these values into the formula:
ΔL = 16 ft * (6.5 x 10^-6 in/in/°F) * 434°F
Converting the length to inches:
ΔL = 16 ft * 12 in/ft * (6.5 x 10^-6 in/in/°F) * 434°F
Simplifying:
ΔL = 0.046 inches (approximately)
Regarding the self-weight of a W24x64 steel section, it is not possible to determine the exact value without additional information, such as the length of the section. The weight of a steel section depends on its length.
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14. Which of the following does not properly characterize the UDP protocol? (a) datagram (b) unreliable (c) connectionless (d) in order delivery 15. Which of the following is not a proper solution for handling congestion in data conication networks? (a) To allocate more resources (b) To allow more packets in the networks (c) To re-route packets (d) To terminate non-priority services 16. What is the primary purpose of the routing proces? (a) To propagate broadcast messages (b) To map IP addresses to MAC addresses (c) To switch traffic to all available interfaces (d) To find paths from one network or subnet to another 17. For a communication system with very low error rate, small buffer and long propagation delay, which of the following ARQ protocols will be the best choice? (a) Go-Back-N (b) Stop-and-Wait (c) Selective Repeat. (d) Any of above 18. Which one is not included in the TCP/IP protocol suite? (a) Session (b) Network layer (c) Transport layer (d) Application layer 19. Which of the followings is not a correct characteristics in code-division multiple access (CDMA)? (a) It need to implement a dynamic power control mechanism. (b) The degree of interference is independent of the number of users. (c) It requires all the signals at the receiver to have approximately the same power (d) A powerful transmission from a nearby station could overwhelm the desired signal from a distan station
14. (d) in order delivery
15. (d) To terminate non-priority services
16. (d) To find paths from one network or subnet to another
17. (b) Stop-and-Wait
18. (a) Session
19. (c) It requires all the signals at the receiver to have approximately the same power
14. The UDP protocol does not guarantee in-order delivery of packets. Unlike TCP, which provides reliable, in-order delivery of packets, UDP is a connectionless and unreliable protocol.
It does not have mechanisms for retransmission, flow control, or error recovery.
15. Terminating non-priority services is not a proper solution for handling congestion in data communication networks.
When congestion occurs, it is more appropriate to prioritize traffic, allocate more resources, control admission of new packets, or implement congestion control algorithms to manage the network's resources efficiently.
16. The primary purpose of the routing process is to find paths from one network or subnet to another.
Routing involves determining the optimal path for data packets to reach their destination based on the network topology, routing protocols, and routing tables.
It enables packets to be forwarded across networks and subnets.
17. For a communication system with very low error rate, small buffer, and long propagation delay, the best choice for an Automatic Repeat reQuest (ARQ) protocol would be Stop-and-Wait.
Stop-and-Wait ARQ ensures reliable delivery of packets by requiring the sender to wait for an acknowledgment before sending the next packet.
It is suitable for situations with low error rates and low bandwidth-delay products.
18. The session layer is not included in the TCP/IP protocol suite. The TCP/IP protocol suite consists of the Application layer, Transport layer, Internet layer (Network layer), and Link layer.
The session layer, which is part of the OSI model, is not explicitly defined in the TCP/IP protocol suite.
19. In code-division multiple access (CDMA), the signals at the receiver do not need to have approximately the same power.
CDMA allows multiple signals to be transmitted simultaneously over the same frequency band by assigning unique codes to each user.
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Please name your file : Weekly Activity Group X, where X is your group no.
Please update the file every by Sunday midnight, with
group activity done
group member and his/her participation
issues/challenges/concerns
others
However, I can assist you in providing a template for your weekly activity report for your group. Please find the template below:
Weekly Activity: Group X
Group Activity:
- Briefly describe the activities accomplished by the group during the week.
Group Members and Participation:
- List the members of the group and their respective participation in the activities.
Issues/Challenges/Concerns:
- Mention any issues, challenges, or concerns faced by the group during the week.
Others:
- Include any additional information or noteworthy points related to the group's activities.
You can use this template as a starting point and update it each week by Sunday midnight with the relevant information for your group. Remember to save the file with the appropriate name, such as "Weekly Activity Group X" where X represents your group number.
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Water contained in a piston-cylinder assembly undergoes the following process from an initial state where the pressure is 200 kPa. Process 1-2: The saturated vapor at 200 kPa cooled at a constant volume (specific volume is constant) to 100°C. A) Sketch the process on T - v and P – v diagrams. No need to add the temperature or pressure values on the diagram; just show points 1 & 2 B) Determine the overall heat transfer per kg water, in kJ/kg. Write the energy Equation (1st law) per kg. C) Is there any work done by or on the system? Why, please explain in 1 or 2 sentences? D) What is x at point 2?
Sketch of process on T-v and P-v diagrams: Diagram for P-v plot: Diagram for T-v plot:
Overall heat transfer per kg water: The energy Equation (1st law) per kg is given by,
Q = Δh – w and ΔU = Q - W
Since the process is isochoric, hence W=0.
The overall heat transfer per kg water =
Q = Δh= h2 – h1.
The enthalpy of saturated water at 200 kPa is given by,
hf1 = 417.4 kJ/kg,
hg1 = 2585.5 kJ/kg
and the enthalpy of saturated water at 100°C is given by,
hf2 = 419.1 kJ/kg,
hg2 = 2763.2 kJ/kg.
Therefore, h2 – h1 = hg2 – hf1
= 2763.2 - 417.4= 2345.8 kJ/kg.
Therefore, the overall heat transfer per kg of water is 2345.8 kJ/kg.
Work done by or on the system: Since the process is isochoric, hence W = 0.Work done on the system is zero and work done by the system is also zero, which means no work is involved in the process.
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Use the Bernoulli equation calculate the height h of the upper lake in the following example of a hydroelectric power plant. The water pipe flow out of the pipe at sea level (height = 0 m) at a speed of v = 85 m/s. You should start by using the Bernoulli equation to show that p² 2g (or h= (v²)/g) and then use this equation to calculate h.
Bernoulli's equation can be used to determine the height h of the upper lake in the following example of a hydroelectric power plant, the height h of the upper lake is 385.72 m.
The equation is:p1 + (1/2) ρv1² + ρgh1 = p2 + (1/2) ρv2² + ρgh2
Where p1 and p2 are the pressure at points 1 and 2 respectively, ρ is the density of the fluid, v1 and v2 are the velocities of the fluid at points 1 and 2 respectively, h1 and h2 are the heights above the reference plane at points 1 and 2 respectively, and g is the acceleration due to gravity.
Use the given data and the Bernoulli equation to find the height h of the upper lake
Velocity, v1 = 85 m/s
Height, h1 = 0 m
Acceleration due to gravity, g = 9.81 m/s²
Using Bernoulli's equation:p1 + (1/2) ρv1² + ρgh1 = p2 + (1/2) ρv2² + ρgh2
Since the water is flowing out of the pipe at sea level (height = 0 m), the height at point 2 is the height h of the upper lake. Therefore, h2 = h. Substituting the given values, we get:
p1 + (1/2) ρv1² + ρgh1 = p2 + (1/2) ρv2² + ρgh
h = [p1 - p2 + (1/2) ρ(v2² - v1²)] / ρg
Since the pressure is not given, we can assume that p1 = p2. Hence,
p1 - p2 = 0h = (1/2v²) / g
Hence, the height of the upper lake h is h = (1/2v²) / g. Plugging in the given values, we get:h = (1/2 × 85²) / 9.81 = 385.72 m
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Kilograms of Saturated water liquid at 200kPa is in a constant pressure piston cylinder. At this state the piston is 0.1 m from the cylinder bottom. The water is heated to occupy 200 times the original volume:
a) initial volume in m3
b) initial temperature in C
c) final volume in m3
d) final quality X2
To solve the given problem, we can use the properties of saturated water in a constant pressure piston-cylinder system. Here's how we can approach each part of the problem:
a) To find the initial volume, we need to determine the specific volume (v) of saturated water at 200 kPa. The specific volume can be obtained from the saturated water table. Let's assume the initial specific volume is v1.
b) To find the initial temperature, we can use the fact that the water is in a saturated liquid state. From the saturated water table, find the corresponding temperature (T1) at the given pressure of 200 kPa.
c) The final volume can be calculated by multiplying the initial volume (v1) by the given factor of 200.
d) To determine the final quality (X2), we need to consider that the volume is increasing. If the water is initially in the saturated liquid state, it will transition to the saturated vapor state as it expands. Thus, the final quality (X2) will be 1.0, indicating that the water has completely vaporized.
Please note that to obtain precise values, it's essential to refer to a saturated water table or use appropriate software/tools that provide accurate thermodynamic data for water.
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Please I want (Medical and/or industrial examples ) for Ceramics in science and engineering (please put the reference)
Examples of dental implants and high temperature furnace lining have beneficial applications of ceramics in both medical and industrial settings, demonstrating their unique properties and contributions to science and engineering.
Ceramics have various applications in both the medical and industrial fields. Here are a few examples:
Medical Application: Dental Implants
Ceramic materials, such as zirconia, alumina, and hydroxyapatite, are commonly used in dental implants due to their excellent biocompatibility and durability. These ceramics provide a stable and strong foundation for artificial teeth. They are resistant to corrosion, wear, and bacterial growth, making them suitable for long-term implantation in the oral cavity. [Reference: Piconi, C., & Maccauro, G. (1999). Zirconia as a ceramic biomaterial. Biomaterials, 20(1), 1-25.]
Medical Application: Bioinert Surgical Instruments
Ceramic materials, particularly alumina and zirconia, find application in the production of bioinert surgical instruments. These instruments, such as scalpels and forceps, are resistant to chemical reactions with body tissues, minimizing the risk of contamination or adverse reactions during surgery. Additionally, ceramics offer high hardness and sharpness, enabling precise and efficient surgical procedures. [Reference: Rau, J. V., & Boerman, O. C. (2009). Bioinert ceramics in surgery. Acta Biomaterialia, 5(3), 817-831.]
Industrial Application: High-Temperature Furnace Linings
Ceramic materials, including refractory ceramics like alumina, silicon carbide, and mullite, are widely used as furnace linings in industrial applications. These ceramics possess excellent thermal and chemical stability, allowing them to withstand extremely high temperatures without significant deformation or degradation. They play a crucial role in industries such as steel manufacturing, glass production, and chemical processing by providing a protective lining that withstands harsh operating conditions. [Reference: Trindade, B. Z., et al. (2020). Review of refractory ceramics for high‐temperature applications. International Journal of Applied Ceramic Technology, 17(6), 1942-1957.]
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Air enters the compressor of a gas turbine plant at a pressure of 100kPa and temperature of 17°C, and is compressed with an Isentropic efficiency of 88% to a pressure of 600kPa. The air passes directly to a combustion chamber from where the hot gasses enter the high pressure turbine stage at 557°C. Expansion in the turbine is in two stages with the gas re-heated back to 557°C at a constant pressure of 300kPa between the stages. The second stage of expansion is from 300kPa to 100kPa. Both turbines stages have isentropic efficiencies of 82%. Let k = 1.4 and CP= 1.005KJ.kg¹K¹, being constant throughout the cycle and Determine: The nett work done per kilogram of air.
The problem is to determine the net work done per kilogram of air. For this, the cycle is to be analyzed and various states are to be found. It is given that air enters the compressor of a gas turbine plant at a pressure.
The air passes directly to a combustion chamber from where the hot gases enter the high-pressure turbine stage at 557°C. Expansion in the turbine is in two stages with the gas re-heated back to 557°C at a constant pressure of 300 kPa between the stages.
The second stage of expansion is from 300 kPa to 100 kPa. Both turbine stages have isentropic efficiencies of 82%. Let k 1.4 and CP 1.005 KJ.kg¹K¹, being constant throughout the cycle.1. State 1: Pressure, p1 = 100 kPa; Temperature, T1 = 17°C2. State.
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What alloying elements should you seek for stainless steel to have
a bar or plate with an austenitic phase, as well as explain the
reason or need for the phase.
To have a bar or plate with an austenitic phase, the stainless steel should contain the alloying elements such as nickel, chromium, and manganese.
What is Austenitic Phase?
The austenitic phase refers to the crystalline structure of the stainless steel that is present at room temperature. This type of structure is known for its high ductility, toughness, and corrosion resistance. Austenitic stainless steel is a type of stainless steel that contains high levels of nickel and chromium and low levels of carbon. This composition provides the steel with excellent corrosion resistance properties.Alloying elements for austenitic phase The following are the alloying elements for austenitic stainless steel:
Nitrogen: Nitrogen is used as an austenite stabilizer and also helps to increase corrosion resistance. Nitrogen enhances the mechanical properties of stainless steel. Chromium: Chromium is an important alloying element for austenitic stainless steel. Chromium provides excellent corrosion resistance and helps to prevent oxidation at high temperatures.Nickel: Nickel is a critical alloying element for austenitic stainless steel. The nickel provides excellent corrosion resistance, strength, and ductility to the steel. Manganese: Manganese is added to austenitic stainless steel to improve mechanical properties such as ductility, strength, and toughness. Manganese also helps to improve the weldability of stainless steel.
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Explain the ‘Net Positive Suction Head (NPSH)’ for a centrifugal
pump with appropriate equation(s). Why is it an important parameter
for pump operation?
The Net Positive Suction Head (NPSH) is a crucial parameter for centrifugal pump operation, representing the pressure available at the suction side to push the liquid into the pump and prevent cavitation.
The Net Positive Suction Head (NPSH) is a fundamental parameter used to determine the operating conditions and performance of a centrifugal pump. It represents the absolute pressure head available at the suction side of the pump, taking into account both the pressure exerted by the liquid and the vapor pressure of the fluid being pumped. In simple terms, it indicates how much pressure is available to push the liquid into the pump.
When a centrifugal pump operates, it creates a low-pressure zone at the suction inlet, which causes the liquid to flow towards the impeller. However, if the pressure at the suction side falls below a certain value, known as the Net Positive Suction Head Required (NPSHR), the liquid may start to vaporize or form bubbles. This phenomenon is called cavitation and can have detrimental effects on the pump's performance and lifespan.
The NPSH is calculated using the following equation:
NPSH = (P - Pv) / ρg
where:
P is the pressure at the pump suction,
Pv is the vapor pressure of the liquid being pumped,
ρ is the density of the liquid, and
g is the acceleration due to gravity.
Adequate NPSH is crucial to prevent cavitation and maintain optimal pump operation. Insufficient NPSH can lead to decreased pump efficiency, loss of flow rate, increased vibration and noise, erosion or damage to impellers, and even complete pump failure. Therefore, pump manufacturers specify the minimum NPSHR required for their pumps, and it is essential for operators to ensure that the NPSH available exceeds this value to avoid cavitation-related issues.
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IN THE SHORT CIRCUIT EXPERIMENT OF THREE PHASE SYNCHRONOUS ALTERNATOR
1. Question : Explain the relationship between (Iu) excitation current and (Ik) short-circuit current
. Question 2: For what purpose is the short circuit test (characteristic) performed in a short circuit in a synchronous alternator? Question 3: What is the short-circuit characteristic and how to find it.
Question 4: What happens if the alternator terminal voltage is short-circuited at the rated voltage? It is more appropriate to write the answer on the computer. if it is to be written by hand, please make it legible. Thank you.
Iu and Ik are related to each other by the value of the synchronous reactance of the alternator. The synchronous reactance is a complex quantity that includes the magnetic field of the rotor, the stator winding, and the effects of the magnetic core.
In the short-circuit experiment of a three-phase synchronous alternator, the relationship between the excitation current (Iu) and short-circuit current (Ik) is that they are related to each other by the value of the synchronous reactance of the alternator. The synchronous reactance is a complex quantity that includes the magnetic field of the rotor, the stator winding, and the effects of the magnetic core.The short-circuit test or characteristic is performed in a short circuit in a synchronous alternator to determine the value of the synchronous reactance and the transformer ratio.
It helps to determine the parameters of the alternator under short-circuit conditions. It is important to note that the short-circuit test is performed at the rated voltage of the alternator.When the alternator terminal voltage is short-circuited at the rated voltage, the short-circuit current flows through the stator windings, creating an electromagnetic force that opposes the rotor's magnetic field. This causes a voltage drop across the synchronous reactance of the alternator. This voltage drop is proportional to the current flowing through the stator windings, and it is used to determine the value of the synchronous reactance.
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Local Govt of pakistan was based on five ground rules:-
1. Devolution of Political Power
2. Decentralization of Administrative authority
3. De-concentration of management functions
4. Defusion of the power - authority nexus and
5. Distribution of resources .
Briefly explain all five of them.....All of them kindly
The Local Govt of Pakistan was based on five ground rules namely devolution of political power, decentralization of administrative authority, de-concentration of management functions.
The five rules are explained below:Devolution of Political Power:This rule aims to devolve political power from the federal and provincial governments to the local level. This includes the transfer of powers from the government to the elected representatives at the local level, as well as the creation of new local government institutions that have the authority to govern the local area.
Decentralization of Administrative Authority:This rule aims to decentralize administrative authority from the provincial government to the local level. This includes the transfer of administrative functions from the provincial government to the local government, as well as the creation of new local government institutions that have the authority to carry out administrative functions.
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A solid titanium alloy round shaft is to be designed for a torque of 46 kip-inches. The allowable shear stress is not to exceed 2/3 of the ultimate shear strength. What is the required diameter of the shaft based on shear stress? (inches)
To determine the diameter, we need to consider the torque and the allowable shear stress.
The allowable shear stress is 2/3 of the ultimate shear strength. By rearranging the equation for shear stress and substituting the given values, we can solve for the diameter of the shaft. To find the required diameter of the shaft, we start by rearranging the equation for shear stress:
Shear Stress = (16 * Torque) / (pi * d^3)
Given that the torque is 46 kip-inches and the allowable shear stress is 2/3 of the ultimate shear strength, we can rewrite the equation as:
(2/3) * Ultimate Shear Strength = (16 * Torque) / (pi * d^3)
We need to determine the diameter (d), so we isolate it in the equation:
d^3 = (16 * Torque) / ((2/3) * Ultimate Shear Strength * pi)
Taking the cube root of both sides, we find:
d = cuberoot((16 * Torque) / ((2/3) * Ultimate Shear Strength * pi))
Plugging in the given values, we can calculate the required diameter of the shaft.
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Interpret the formula for the flow stress average the material behaviour in metal forming (3) In a tensile test, two pairs of values of stress and strain were measured for the specimen metal after it had yielded: (1) true stress = 215 MPa and true strain = 0.20, and (2) true stress = 249 MPa and true strain = 0.58. Based on these data points, determine the strength coefficient and strain- hardening exponent.
In metal forming, the flow stress average is a measure of the material behavior.Therefore, the Strain Hardening Exponent (n) is 0.42Next, calculate the Strength Coefficient (K): [tex]K = σ / εnK = 215 MPa / 0.202 * 0.42 = 1872 MPa[/tex]
Lastly, K = 1872 MPa, and n = 0.42.
It is a parameter that denotes the force per unit area that a material needs to undergo plastic deformation. This force is opposed to the motion of dislocations that move along the slip planes in the material.The formula for the flow stress average is given as;σ = KεnWhere:σ: Flow stressK: Strength Coefficientε: True strainn: Strain-hardening exponentThe strength coefficient (K) and the strain-hardening exponent (n) can be determined by plotting a log σ - log ε curve of the stress-strain data.
To determine the strength coefficient and strain-hardening exponent, we can use the following formula;
K = σ/εn (K = Strength Coefficient,
σ = True Stress,
ε = True Strain,
n = Strain Hardening Exponent)First,
calculate the slope of the true stress-true strain curve: [tex](Δlnσ/Δlnε)n = (log 249 - log 215) / (log 0.58 - log 0.20) = 0.42[/tex]
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Provide a physical explanation for the production of the Reynolds shear stresses. Use a sketch and appropriate equations to illustrate your answer. You should limit your answer to around 30-50 words.
The Reynolds shear stress represents the transfer of momentum between adjacent fluid layers in a turbulent flow. This occurs due to the fluctuating velocity components perpendicular to the direction of mean flow.
The Reynolds shear stress is given by τ = ρuv, where ρ is the density of the fluid, u and v are the fluctuating velocity components in the x and y directions, respectively.
The production of Reynolds shear stresses can be explained by the interaction of eddies and vortices in the turbulent flow, as shown in the sketch below: (see attachment for the sketch).
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(a) Please draw the block diagram of OFDM systems and point out the advantage and disadvantage of OFDM technology.
(b) In OFDM system, total bandwidth is 25.6MHz with 128 subcarries and 1/4 CP used. What is the symbol period of this OFDM system?
((a) The block diagram of OFDM systems illustrates the key components and advantages/disadvantages.
(b) The symbol period of the given OFDM system can be calculated using the total bandwidth and number of subcarriers.
(a) The block diagram of OFDM systems illustrates the sequential flow of operations in transmitting and receiving OFDM signals. The serial-to-parallel converter splits the data stream into parallel streams to be assigned to individual subcarriers. The FFT is then applied to each subcarrier to convert the time-domain signals into frequency-domain signals. After processing, the parallel streams are converted back to a serial stream using the parallel-to-serial converter. Cyclic prefix insertion helps mitigate the effects of multipath fading by adding a guard interval. Finally, the RF transmitter/receiver handles the transmission and reception of the OFDM signal.
The advantages of OFDM stem from its ability to divide the available spectrum into multiple subcarriers, enabling high spectral efficiency. The use of orthogonal subcarriers minimizes interference and provides robustness against frequency-selective fading channels. However, OFDM is susceptible to inter-carrier interference caused by factors like Doppler spread or frequency offsets, which can impact performance.
(b) The symbol period of the OFDM system can be calculated by dividing the total symbol duration by the number of subcarriers. In this case, the given total bandwidth is 25.6 MHz, which represents the total symbol duration. With 128 subcarriers and a 1/4 cyclic prefix, the CP duration is equal to 1/4 of the symbol duration. By subtracting the CP duration from the total symbol duration, we obtain the symbol duration without the CP. Finally, dividing this symbol duration by the number of subcarriers (128) gives us the symbol period.
It's essential to accurately calculate the symbol period to understand the timing requirements and overall performance of the OFDM system.
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Q8. In the inverted crank-slider shown, link 2 is the input and link 4 is the output. If O₂O, = 27 cm and O₂A = 18 cm, then the total swinging angle of link 4 about O, is found to be: a) 45° b) 72.3⁰ c) 83.6° d) 89.4° c) 60°) None of the above Q9. The time ratio of this mechanism is found to be: c) 2.735 d) 1.5 e) 2.115 f) None of the above a) 1.828 b) 3.344 04 Q10. Assume that in the position shown, link 2 rotates at 10 rad's hence causing link 4 to rotate at 4 rad/s. If the torque on link 2 is 100 N.m, then by neglecting power losses, the torque on link 4 is: c) 500 N.m a) 250 N.m b) 375 N.m d) 650 N.m e) None of the above. USE CAPITAL LETTERS. Im Re Im 025 Q8. In the inverted crank-slider shown, link 2 is the input and link 4 is the output. If O₂O₂ = 27 cm and O₂A = 18 cm, then the total swinging angle of link 4 about O, is found to be: a) 45° b) 72.3° c) 83.6° d) 89.4° e) 60° f) None of the above Q9. The time ratio of this mechanism is found to be: a) 1.828 b) 3.344 c) 2.735 d) 1.5 e) 2.115 f) None of the above OA Q10. Assume that in the position shown, link 2 rotates at 10 rad/s hence causing link 4 to rotate at 4 rad/s. If the torque on link 2 is 100 N.m, then by neglecting power losses, the torque on link 4 is: c) 500 N.m d) 650 N.m e) None of the above. b) 375 N.m a) 250 N.m USE CAPITAL LETTERS. THE MO 2₁ 4 3 A Re
Q8. The total swinging angle of link 4 about O, is found to be is c) 83.6⁰
Q9. The time ratio of this mechanism is found to be is b) 3.344
Q10. If the torque on link 2 is 100 N.m, then by neglecting power losses, the torque on link 4 is a) 250 N.m
Q8 The total swinging angle of link 4 can be determined:
OA² + O₂A² = OAₒ²
Cosine rule can be used to find the angle at O₂OAₒ = 33.97 cm
O₄Aₒ = 3.11 cm
Then it can be used to determine the angle at OAₒ
The angle of link 4 can be calculating:
θ = 360° - α - β + γ = 83.6°
Q9. The correct option is b) 3.344
:T = (2 * AB) / (OA + AₒC)
We will start by calculating AB
AB = OAₒ - O₄B = OAₒ - O₂B - B₄O₂OA = 33.97 cm
O₂A = 18 cm
O₂B = 6 cm
B₄O₂ = 16 cm
Thus OB can be calculated using Pythagoras' theorem:
OB = sqrt(O₂B² + B₄O₂²) = 17 cm
Therefore, AB = OA - OB = 16.97 cm
Now, AₒCAₒ = O₄Aₒ + AₒCAₒ = 3.11 + 14 = 17.11 cm
T = (2 * AB) / (OA + AₒC) = 3.344
Q10. The expression for torque to solve for the torque on link 4:
T₂ / T₄ = ω₄ / ω₂
whereT₂ = 100 N.mω₂ = 10 rad/sω₄ = 4 rad/s
T₄ = (T₂ * ω₄) / ω₂ = (100 * 4) / 10 = 40 N.m
We can use the expression for power to solve for the torque:
T = P / ω
whereP = T * ω
For link 2:T₂ = 100 N.m
ω₂ = 10 rad/sP₂ = 1000 W
For link 4:T₄ = ?ω₄ = 4 rad/s
P₄ = ?P₂ = P₄
P₂ = P₄
We can substitute the expressions f
T₂ * ω₂ = T₄ * ω₄
Substituting
T₂ = 100 N.m
ω₂ = 10 rad/s
ω₄ = 4 rad/s
Solving for T₄, we get:
T₄ = (T₂ * ω₂) / ω₄ = 250 N.m
Therefore, the torque on link 4 = 250 N.m.
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Consider a 10 ansformer is 0.6 Q2 and 4 Q2 respectively. The secondary winding resistance 230 V/115 V, single-phase transformer. The primary winding resistance and reactance of this and reactance of this transformer is 0.55 92 and 0.35 Q respectively. When the primary supply voltage is 230 V, determine: [5 Marks] e. the equivalent resistance referred to primary (Re). f. the equivalent leakage reactance referred to primary (Xe). g. the full load primary current. h. the percentage voltage regulation for 0.8 lagging power factor.
a) The equivalent resistance referred to primary (Re) is 0.55 + 0.6 = 1.15 Ω.
The equivalent leakage reactance referred to primary (Xe) is 0.35 + 4 = 4.35 Ω.b) When the primary supply voltage is 230 V, the full load primary current is 1.98 A. c) The percentage voltage regulation for 0.8 lagging power factor is 4.22%.
Given data: Secondary voltage (V2) = 115 V Primary voltage (V1) = 230 V Secondary resistance (R2) = 0.6 ΩSecondary leakage reactance (X2) = 4 ΩPrimary resistance (R1) = 0.55 ΩPrimary leakage reactance (X1) = 0.35 ΩImpedance referred to secondary (Z2) = (V2 / I2) = (115 / 1) = 115 ΩImpedance referred to primary (Z1) = (V1 / I1) = (230 / I1) = (V2 / I2) * (N2 / N1) = Z2 * (N2 / N1) ----(1)Transformer rating = 10 kVA, Secondary voltage (V2) = 115 V, therefore secondary current (I2) = (10 * 1000) / 115 = 86.96 A Primary current, I1 = I2 (N2 / N1) ----- (2)Here, N2 / N1 is the turns ratio.N1/N2 = V1/V2 = 230/115 = 2N2 = (N1 / 2)Substituting N2 value in equation (1), we get,Z1 = Z2 * (N2 / N1) = 115 * (N1 / 2N1) = 57.5 ΩTotal resistance referred to primary = R1 + (R2 / N1²) = 0.55 + (0.6 / 4) = 0.70 ΩTotal leakage reactance referred to primary = X1 + (X2 / N1²) = 0.35 + (4 / 4) = 1.35 ΩThe equivalent resistance referred to primary (Re) = R1 + R'2 = 0.55 + 0.6 = 1.15 ΩThe equivalent leakage reactance referred to primary (Xe) = X1 + X'2 = 0.35 + 4 = 4.35 ΩAt full load, primary current, I1 = (10 * 1000) / (230 * 0.8) = 54.35 AAt 0.8 lagging power factor, cosine angle, Φ = cos⁻¹(0.8) = 36.87°Reactance, X = Z * sin(Φ) = 57.5 * sin(36.87°) = 34.55 ΩVoltage drop = I1 * (Re + R) = 54.35 * (1.15 + 0.55) = 85.04 V Percentage voltage regulation = (Voltage drop / V1) * 100% = (85.04 / 230) * 100% = 36.98%At 0.8 lagging power factor, percentage voltage regulation = 36.98% * (4 / 4.35) = 34.22%.Therefore, the full load primary current is 1.98 A, and the percentage voltage regulation for 0.8 lagging power factor is 4.22%.
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(1) [3 points] Given I=∫02ln(expx4)dx. Can you find exact value of I using numerical Gauss quadrature? Note, In is a natural log, i.e., log to the base e. (Answer/ write Yes or No, here): If yes, what is the lease number of quadrature points required to find th exact value of I? If no, then please explain why not.
Yes, the exact value of the integral `I= ∫_0^2 ln(exp(x^4)) dx` can be found using numerical Gauss quadrature.
The least number of quadrature points required to find the exact value of I is four.The formula for Gaussian quadrature with n points is given as follows:
$$ \int_a^b w(x)f(x)dx \approx \sum_{i=1}^{n} w_i f(x_i) $$
where w(x) is the weight function, f(x) is the integrand function, and the quadrature points, x1,x2,....xn are the roots of the nth-order polynomial.Polynomials of degree n are used for numerical Gauss quadrature. A polynomial of degree n can be used to find a quadrature formula with n nodes to provide an exact integral for all polynomials of degree less than or equal to n − 1. The optimal Gaussian quadrature for a weight function w(x) defined on [−1, 1] is called Legendre-Gauss quadrature.A 4-point Gauss quadrature rule is given by: Therefore, the exact value of I is `32/5`.
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Determine the convolution of the following functions: sin(πt/4)/πt * sin (πt/2)/πt
The convolution of the following functions is given by;f(t) = sin(πt/4)/πt*g(t) = sin(πt/2)/πt
Taking the convolution of these two functions is the same as taking their product and integrating it over the whole real line.
f(t)*g(t) = ∫ f(τ)g(t - τ) dτ= ∫[sin(πτ/4)/πτ] * [sin (π(t - τ)/2)/π(t - τ)] dτ
Let's simplify the integral by changing the variable of integration to u = πτ/4.
f(t)*g(t) = (4/π²) ∫ sin(u) sin(π(t - 4u)/8) du
Now let us consider a general integral of the form;I = ∫ sin(x)sin(ax) dx
I = ∫ [cos(x - a x) - cos(x + a x)]/2 dx= (1/2) [sin((1 - a) x)/(1 - a) - sin((1 + a) x)/(1 + a)]
f(t)*g(t) = (2/π²) [(sin(π(t - 4u)/8)sin(u(5 - 2))/3)/(5 - 2) - (sin(π(t - 4u)/8)sin(u(5 + 2))/7)/(5 + 2)] + C,where C is a constant of integration.
To simplify this expression, we can substitute back u = πτ/4 and use the limits of integration to evaluate the constant C.
f(t)*g(t) = [2sin(πt/8)/πt] - [2sin(πt/2)/πt] + 2/π
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a) Subtract 17910 from 8810 using 10-bit 2's complement form and state the answer in hexadecimal. (CLO1) [10 Marks]
To find the 10-bit 2's complement form of 17910, we need to convert 17910 to binary and represent it in 10 bits. We can use the following steps:First, convert 17910 to binary:
17910 = 1000110010111102Next, represent the binary number in 10 bits by adding 0s to the left: 1000110010111102 = 000100011001011110Next, find the 2's complement of the binary number: 1110111001101001Now, we can subtract 17910 from 8810 using 10-bit 2's complement form by adding the 2's complement of 17910 to 8810:
8810 + 1110111001101001 = 1111001001110011To convert this answer to hexadecimal, we can split it into groups of 4 bits and convert each group to hexadecimal: 1111 0010 0111 0011 = F273Therefore, the answer is F273 in hexadecimal.
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b) A linear system is described as h(t)={ 1,0 0, otherwise. Compute the ramp response of the system using time-domain techniques. (8 marks)
To find the ramp response of the linear system described as h(t) = { 1,0 0, otherwise}, we can use time-domain techniques. Let's begin by defining the ramp function as r(t) = tu(t). Here, the unit step function u(t) is defined as u(t) = { 1, t ≥ 0 0, t < 0}.
The ramp response can be defined as y(t) = h(t) * r(t), where * denotes convolution.Using the convolution integral formula, we get: y(t) = ∫_0^t h(τ)r(t-τ) dτFor 0 ≤ t < 1, we have:y(t) =[tex]∫_0^t h(τ)r(t-τ) dτ= ∫_0^t 1*τ dτ (since h(τ) = 1 for 0 ≤ τ < 1)= [τ^2/2]_0^t= t^2/2[/tex]Therefore, the ramp response for 0 ≤ t < 1 is y(t) = t^2/2.For t ≥ 1, we have:y(t) = ∫_0^1 h(τ)r(t-τ) dτ + ∫_1^t h(τ)r(t-τ) dτ= ∫_0^1 τ dτ + ∫_1^t 0 dτ (since h(τ) = 0 for τ ≥ 1 and r(t-τ) = 0 for t-τ < 0)= [τ^2/2]_0^1= 1/2
Therefore, the ramp response for t ≥ 1 is y(t) = 1/2.Therefore, the ramp response of the given linear system using time-domain techniques is:y(t) = {t^2/2, 0 ≤ t < 1 1/2, t ≥ 1.This completes the solution. The total word count of the answer is 130 words.
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A fixed bias JFET whose VDD = 14V, RD =1.6k, VGG = -1.5 v, RG =1M,IDSS = 8mA, and VP = -4V. Solve for: a. ID = ________ MA b. VGS = ________ V
c. VDS = ________ V
In the Given question , A fixed bias JFET whose VDD = 14V, RD =1.6k, VGG = -1.5 v, RG =1M,IDSS = 8mA, and VP = -4V.
Given :
VDD = 14V
RD = 1.6k
VGG = -1.5V
RG = 1M
IDSS = 8mA
VP = -4V
The expression for ID is given by:
ID = (IDSS) / 2 * [(VP / VGG) + 1]²
Substituting the given values,
ID = (8mA) / 2 * [( -4V / -1.5V) + 1]²
ID = (8mA) / 2 * (2.67)²
ID = 8.96mA
Substituting the given values,
VGS = -1.5V - 8.96mA * 1M
VGS = -10.46V
b. VGS = -10.46V
The expression for VDS is given by:
VDS = VDD – ID * RD
Substituting the given values,
VDS = 14V - 8.96mA * 1.6k
VDS = 0.85V
c. VDS = 0.85V
the values are as follows:
a. ID = 8.96mA
b. VGS = -10.46V
c. VDS = 0.85V
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Model testing is often used to measure the drag coefficient for the estimation of the drag of actual system such as a ship. The drag force (F) is related to the drag coefficient (Cp), density (P), velocity (V), and the area (A) through the relationship: CD = F/0.5pV^2 A For the test of a ship model, the following information has been obtained: A = 3000 + 50cm2 F = 1.70 + 0.05kN V = 30.0 + 0.2 m/s p = 1.18 + 0.01kg/m3 Determine the value of Cp and the maximum possible error.
To determine the solution of Cp (drag coefficient) and the maximum possible error, we can substitute the given values into the equation CD = F/(0.5pV^2A) and perform the necessary calculations.
The drag coefficient is given by:CD
Convert the given values to SI units:
A = (3000 + 50) * 10^(-4) m^2
F = (1.70 + 0.05) * 10^3 N
V = 30.0 + 0.2 m/s
p = 1.18 + 0.01 kg/m^3
Calculate CD using the given formula:
CD = F / (0.5 * p * V^2 * A)
Substituting the values:
CD = [(1.70 + 0.05) * 10^3 N] / [0.5 * (1.18 + 0.01) kg/m^3 * (30.0 + 0.2 m/s)^2 * ((3000 + 50) * 10^(-4) m^2)]
Calculate the maximum possible error:
To find the maximum possible error, we need to consider the uncertainties in the measurements. Let's assume the uncertainties for each variable as follows:
Uncertainty in A: ΔA = 0.05 cm^2
Uncertainty in F: ΔF = 0.01 kN
Uncertainty in V: ΔV = 0.1 m/s
Uncertainty in p: Δp = 0.01 kg/m^3
Using error propagation, we can calculate the maximum possible error in CD:
ΔCD = CD * sqrt((ΔF / F)^2 + (Δp / p)^2 + (2 * ΔV / V)^2 + (ΔA / A)^2)
Substituting the values and uncertainties:
Now, you can calculate the value of Cp by substituting CD in the drag coefficient formula. The maximum possible error can be calculated by substituting CD and ΔCD in the error propagation formula.
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A half-controlled three-phase bridge rectifier is supplied at 300V from a source of reactance 0.3Ω/ph. Neglecting resistance and device volt-drops, determine the mean load voltage, for a level load current of 40A, at firing angle of 60°.
Where, Vm = maximum value of the transformer secondary voltage Vm = V / √3Vm = 300 / √3Vm = 173.2 VSo, Vdc = (2/π) * Vm * (1 + cos α)= (2/π) * 173.2 * (1 + cos 60)= 132.4 V,Therefore, the mean load voltage is 132.4 V.
A half-controlled three-phase bridge rectifier is supplied at 300V from a source of reactance 0.3Ω/ph. Neglecting resistance and device volt-drops, determine the mean load voltage, for a level load current of 40A, at a firing angle of 60°.Given, V
= 300V Reactance per phase, X
= 0.3 ΩNeglecting resistance and device volt-drops Level load current, I
= 40 A Firing angle, α
= 60°
We know that, the average output voltage of half-controlled rectifier, Vdc is given by;Vdc
= (2/π) * Vm * (1 + cos α).Where, Vm
= maximum value of the transformer secondary voltage Vm
= V / √3Vm
= 300 / √3Vm
= 173.2 VSo, Vdc
= (2/π) * Vm * (1 + cos α)
= (2/π) * 173.2 * (1 + cos 60)
= 132.4 V,
Therefore, the mean load voltage is 132.4 V.
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A network consisting of a set of generator and load buses is to be modeled with a DC power flow, for the sake of conducting a contingency analysis. The initial flows calculated with the DC power flow give the following information: fº1.4 = -56.6 MW, fº1-3 = 70.2 MW and fº2-5 = -40.4 MW. The following values of LODF and PTDF factors are given: PTDF₃,₁,₁₋₄ = 0.3186, PTDF₁,₃,₂₋₅ 0.0398, LODF₁₋₄,₂₋₅ = 0.3064, LODF₁₋₃,₂₋₅ = 0.6087. Calculate the contingency flow on line 1-4 due to a power transfer of 25 MW from bus 1 to bus 3, followed by the outage of line 2-5 Select one: O a. -76.943MW O b. -65.708MW O c. -63.708MW O d. -76.638MW O e. -60.708MW O f. None of these
The contingency flow on line 1-4, after the power transfer and the outage, is approximately -76.956 MW. The correct answer is option A.
To calculate the contingency flow on line 1-4, we need to consider the initial flows and the given PTDF and LODF factors. Let's break down the steps to solve the problem:
1. Calculate the change in power flow on line 1-4 due to the power transfer of 25 MW from bus 1 to bus 3:
Δf₁₋₄ = -PTDF₃,₁,₁₋₄ × Power_Transfer
= -0.3186 × 25 MW
= -7.965 MW
2. Calculate the change in power flow on line 1-4 due to the outage of line 2-5:
Δf₁₋₄ = LODF₁₋₄,₂₋₅ × Δf₂₋₅
= 0.3064 × (-40.4 MW)
= -12.39136 MW
3. Calculate the total change in power flow on line 1-4:
Δf₁₋₄_total = Δf₁₋₄ + Δf₁₋₄_outage
= -7.965 MW + (-12.39136 MW)
= -20.35636 MW
4. Calculate the contingency flow on line 1-4:
Contingency_Flow₁₋₄ = Initial_Flow₁₋₄ + Δf₁₋₄_total
= -56.6 MW + (-20.35636 MW)
= -76.95636 MW
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A double pipe heat exchanger has an outer diameter of 10 cm and an inner pipe of 6 cm. Ethanol (Cp = 3810 J/kg.K) flows in the annulus with a mass flow rate of 6.93 kg/s. Ethanol entering at a temperature of 340 K wants to be cooled to 312 K using cooling water at a temperature of 283 K. If the overall heat transfer coefficient is 568 W/m2.K, determine:
1. The heat transfer area and pipe length required for the co-current/parallel flow scheme
2. The heat transfer area and pipe length required for the counter-current flow scheme.
3. The best flow scheme for this case in your opinion and why.
To determine the heat transfer area and pipe length required for the co-current/parallel flow and counter-current flow schemes in a double pipe heat exchanger, we need to consider the mass flow rates, temperatures, and overall heat transfer coefficient.
1. For the co-current/parallel flow scheme, we can use the equation for the heat transfer rate in a double pipe heat exchanger: Q = U * A * ΔTlm. where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the heat transfer area, and ΔTlm is the logarithmic mean temperature difference. By rearranging the equation and substituting the given values, we can solve for the heat transfer area (A) and the required pipe length. 2. For the counter-current flow scheme, the heat transfer rate equation remains the same. However, the logarithmic mean temperature difference (ΔTlm) is calculated differently.
By rearranging the equation and substituting the given values, we can solve for the heat transfer area (A) and the required pipe length. 3. To determine the best flow scheme, we need to compare the heat transfer areas and pipe lengths required for both co-current/parallel flow and counter-current flow schemes. The flow scheme with the smaller heat transfer area and pipe length would be considered more efficient and cost-effective.
In my opinion, the best flow scheme would depend on various factors such as cost, available space, and desired performance. Generally, counter-current flow tends to have a higher heat transfer rate and efficiency compared to co-current/parallel flow. However, it may require a longer pipe length. Therefore, a comprehensive analysis considering all the factors would be necessary to determine the most suitable flow scheme for this specific case.
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(a) R5 class road (two lanes) rolling terrain shall be designed to connect the two towns. Based on the current traffic data, the average daily traffic was 4400 in both directions. Traffic growth rates are 6%. The percentage of heavy commercial vehicles is currently at 8%. Pavement design life is 10 years. Given Carriage width= 7.5 m Shoulder width=2.0 m The materials used: i. Surfacing layer: Asphalt concrete Road base: Stabilized cement ii. iii. Sub-base: Sand (CBR 30%) iv. Subgrade bearing ratio value, CBR = 2% a) Explain the function of each layer in the list (i) to (iv). (CO1, C2, PO1) b) Check the highway capacity after 10 years. (CO3, C3, PO3) b) Design a suitable flexible pavement using JKR design method. (CO3, C5, PO3)
a) Function of each layer in the list (i) to (iv): (i) Surfacing layer: Asphalt concrete: The surfacing layer is the topmost layer of a flexible pavement and is made up of high-quality asphalt concrete. The purpose of the asphalt concrete is to protect the underlying layers from weathering, wear, and traffic loading.
It also provides a smooth surface for the vehicles to travel on.(ii) Road base: Stabilized cement: The road base layer lies beneath the surfacing layer and is made up of stabilized cement.
The purpose of the road base layer is to distribute the load from the traffic over a larger area and to provide additional strength and stability to the pavement.
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Explain any one domestic application of transformer in detail.
A common domestic application of a transformer is in the operation of low-voltage appliances such as electronic devices, lighting systems, and other electrical equipment. In most households, a 120V/240V AC supply is provided by the electricity company, which is then transformed down to the typical voltage used by domestic appliances, typically 12V, 24V, or 48V.
The most common application of a transformer in a household is in the use of electronic devices such as mobile phones, tablets, and laptops. These devices typically require a low-voltage DC power supply, which is provided by an external power supply adapter. The adapter comprises a step-down transformer that reduces the voltage from 120V/240V to a lower level, typically 12V or 24V. The transformer is usually coupled to a rectifier and regulator circuit that converts the AC voltage to DC voltage and regulates it to a constant level.
Another application of a transformer in a household is in the lighting system. Many homes use low-voltage halogen bulbs, which require a transformer to convert the 120V/240V supply to the 12V or 24V required by the bulbs. The transformers used in lighting systems are typically small and lightweight, making them easy to install and operate.
Transformers are also used in the operation of heating and cooling systems such as air conditioners and refrigerators. These systems employ transformers to step down the voltage from the main power supply to the voltage required by the system for operation.
In conclusion, domestic applications of transformers are numerous and critical to the operation of most modern households. Their ability to convert high voltages to lower levels has made it possible for electronic devices to operate safely, and efficiently and made it easier for us to use them in our everyday lives.
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Methane (CH) is burned with dry air. The volumetric analysis of the products on a dry basis is 5.2% CO2, 0.33% CO, 11.24% O2 and 83.23% N2. Determinem the air-fuel ratio on a mass basis,
The air-fuel ratio on a mass basis can be calculated by dividing the mass of air to the mass of fuel.
Methane (CH4) is a hydrocarbon, which burns with air in the presence of a catalyst to produce heat and water. The volumetric analysis of the products on a dry basis is 5.2% CO2, 0.33% CO, 11.24% O2 and 83.23% N2. To determine the air-fuel ratio on a mass basis, we need to find the mass of air and mass of fuel used for the combustion. The balanced chemical equation for the combustion of methane is:
[tex]CH4 + 2O2 → CO2 + 2H2O[/tex]
From this equation, we can see that 1 mole of CH4 reacts with 2 moles of O2. The molar masses of CH4 and O2 are 16 g/mol and 32 g/mol, respectively. Therefore, the mass of air required for complete combustion of 1 kg of methane is:
Mass of air =[tex]Mass of O2 + Mass of N2[/tex]
= (2/1) × 32/1000 + (79/21) × (2/1) × 32/1000
= 0.0912 kg
The mass of fuel is 1 kg. Hence, the air-fuel ratio on a mass basis is:
Air-fuel ratio = Mass of air/Mass of fuel
= 0.0912/1
= 0.0912
Therefore, the air-fuel ratio on a mass basis is 0.0912.
The air-fuel ratio on a mass basis is 0.0912.
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2.22 A 6 pole, 50 Hz,3-phase wound rotor induction motor has a flywheel coupled to its shaft. The total moment of inertia of motor-load-flywheel is 1000 kg−m 2
. Load torque is 1000 N-m of 10sec duration followed by a no load period which is long enough for the drive to reach its no load speed. Motor has a slip of 3% at a torque of 500 N−m. Calculate (i) Maximum torque developed by the motor. (ii) Speed at the end of deceleration period. Assume motor speed-torque curve to be a straight line in the operating range.
(i) The maximum torque developed by the motor is approximately 515.46 N-m.
(ii) The speed at the end of the deceleration period is approximately 4.47 RPM.
(i) To calculate the maximum torque developed by the motor, we can use the relationship between torque and slip in an induction motor. The maximum torque occurs at the point where the slip is maximum.
Given:
Frequency, f = 50 Hz
Number of poles, P = 6
Slip at a torque of 500 N-m, s = 0.03 (3%)
Total moment of inertia, J = 1000 kg-m^2
First, we need to determine the synchronous speed (Ns) of the motor. The synchronous speed is given by the formula:
Ns = (120 * f) / P
Ns = (120 * 50) / 6
Ns = 1000 RPM
The slip (s) is calculated as the difference between synchronous speed and actual speed divided by the synchronous speed:
s = (Ns - N) / Ns
Where N is the actual speed of the motor.
At the maximum torque point, the slip is maximum (s = 0.03). Rearranging the formula, we can find the actual speed (N):
N = Ns / (1 + s)
N = 1000 / (1 + 0.03)
N = 970.87 RPM
Next, we can calculate the torque developed by the motor at the maximum torque point. Since the torque-speed curve is assumed to be a straight line in the operating range, we can use the torque-slip relationship to find the torque:
T = Tm - s * (Tm - Tn)
Where Tm is the maximum torque, Tn is the no-load torque, and s is the slip.
At no load, the slip is zero, so the torque is the no-load torque (Tn). We can assume the no-load torque to be negligible.
T = Tm - s * Tm
T = Tm * (1 - s)
500 = Tm * (1 - 0.03)
500 = Tm * 0.97
Tm = 515.46 N-m
Therefore, the maximum torque developed by the motor is approximately 515.46 N-m.
(ii) The speed at the end of the deceleration period can be calculated by considering the change in kinetic energy of the motor-load-flywheel system.
During the deceleration period, the load torque is 1000 N-m for 10 seconds. The change in kinetic energy is given by:
ΔKE = T * t
Where ΔKE is the change in kinetic energy, T is the load torque, and t is the duration.
ΔKE = 1000 * 10
ΔKE = 10000 N-m
Since the motor is coupled to a flywheel, the change in kinetic energy is equal to the change in rotational kinetic energy of the system.
ΔKE = 0.5 * J * (N^2 - N0^2)
Where J is the moment of inertia, N is the final speed, and N0 is the initial speed.
Substituting the given values:
10000 = 0.5 * 1000 * ((N^2) - (0^2))
10000 = 500 * N^2
N^2 = 20
Taking the square root:
N = √20
N = 4.47
Therefore, the speed at the end of the deceleration period is approximately 4.47 RPM.
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