The instantaneous plate temperature is approximately equal to 95 °C. The natural convection heat transfer coefficient is 34,013.77 W/m².K.the radiation heat transfer coefficient is 1.7 W/m².K. The instantaneous rate of change of the plate temperature is approximately -0.226 K/s.
Given data;
L = 0.657 m, A = 0.250 m², k = 237 W/m.K, mc = 3,171 J/K, E = 0 - 7e₀ = 3.699 mV, T ref = 10°C, T ∞ = 296 K, k = 0.026 W/m.K, NuL = 94.3
The instantaneous plate temperature can be calculated as follows:
From the thermocouple calibration chart, the temperature difference between T and T ref is
ΔT = T - Tref = e₀ / 43.4 mV/K = 85.1046 K
Plate temperature can be calculated as
T = ΔT + Tref = 85.1046 + 10 = 95.1 °C
The natural convection heat transfer coefficient can be calculated by using the relation;
NuL = hcL/k 94.3 = hc × 0.657 / 237hc = 94.3 × 237 / 0.657 = 34,013.77 W/m².K
Therefore, the natural convection heat transfer coefficient is 34,013.77 W/m².K.
Radiation heat transfer coefficient can be calculated by using the relation;
A = σε(T⁴ - T∞⁴) 0.250 = 5.67 × 10^-8 × 0.7(T⁴ - 296⁴)T⁴ = [0.250 / (5.67 × 10^-8 × 0.7)] + 296⁴T = (0.250 / (5.67 × 10^-8 × 0.7))^(1/4) + 296 = 340.88 Khr = q/(Aεσ(T⁴-T∞⁴))= [kA/hL+(1/hεσT⁴)]^-1= [237 × 0.250 / 340.88 + 1 / (0.7 × 5.67 × 10^-8 × 340.88⁴)]^-1= 1.7 W/m².K
the radiation heat transfer coefficient is 1.7 W/m².K.
The instantaneous rate of change of the plate temperature can be approximated by using Newton's law of cooling;
Q = hcA(T - T∞) + εσA(T⁴ - T∞⁴)mc dT/dt = hcA(T - T∞) + εσA(T⁴ - T∞⁴) / mcdT/dt = (34,013.77 × 0.250 × (95 - 296) + 0.7 × 5.67 × 10^-8 × 0.250 × (340.88⁴ - 296⁴)) / (3,171 × 2.5) = -0.226 K/s
the instantaneous rate of change of the plate temperature is approximately -0.226 K/s.
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8. Connect channel 1 to the generator output and channel 2 to the inter-connection of the resistor and capacitor. 9. Configure the oscilloscope to capture RMS voltage and frequency. There should be 4 readings available, (VRMS channel 1, Frequency channel 1, VRMS channel 2, Frequency channel 2). 10. Capture a screenshot of the waveforms from both channels along with the measurements for 100 Hz and 500 Hz. 11. Create 2 tables and record the calculated values and measured values for Xc, VR1, VC1, IT, and Zr; make sure you include the correct units. Remember, your equipment will not be able to measure Xc or ZT.
Include a column in the table to include the percent error. The formula to calculate the error is below: %6 error = Expected Value - Measured Value/Expected Value x 100%%
12. Discuss the following: Expected Value - Measured Value Expected Value X 100% a. Describe the relationship between the frequency and IT. b. What effect does frequency have on ZT? c. From step 10, what do you observe regarding the phase of the 2 voltages? d. How could the circuit be modified to bring the phase angle between the source voltage and current closer to 0? e. What conclusions do you have based on the calculations and equipment readings?
8. For this step, you have to connect channel 1 to the generator output, and channel 2 to the inter-connection of the resistor and capacitor.9. For the oscilloscope to capture the RMS voltage and frequency, configure it.
There should be four readings available, VRMS channel 1, Frequency channel 1, VRMS channel 2, and Frequency channel 2.10. Capture a screenshot of the waveforms from both channels along with the measurements for 100 Hz and 500 Hz.11. Create two tables and record the calculated values and measured values for Xc, VR1, VC1, IT, and Zr, making sure you include the correct units. Remember, your equipment will not be able to measure Xc or ZT.
Regarding the phase of the two voltages in step 10, we can observe that the two voltages are in phase with one another. The circuit can be modified to bring the phase angle between the source voltage and current closer to zero by adding an inductor. Based on the calculations and equipment readings, the following conclusions can be drawn. At high frequencies, the circuit becomes more inductive, and at low frequencies, it becomes more capacitive. The current flowing through the circuit (IT) increases as the frequency increases. The total impedance (ZT) is inversely proportional to the frequency and is determined by the resistive component (ZR) and the reactive component (ZL - ZC).
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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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Write a job application letter in English as follows:
1- addressed to a company working in your field of study, seeking for a job or training position as fresh graduate.
2- the letter should not be less than 10 lines or 100 words and submitted as WORD format (not PDF).
3- plagiarism is not allowed and copied letters will be marked ZERO.
A job application letter in English addressed to a company working in your field of study, seeking for a job or training position as fresh graduate is given.
How to write the letterDear Hiring Manager,
I am writing to express my interest in a job or training position at [Company Name]. I recently graduated with a [Degree Name] in [Field of Study] from [University Name]. I have a strong passion for [Field of Study] and I am eager to apply my knowledge and skills in a practical setting.
During my academic journey, I gained a solid foundation in [Field of Study] through coursework, projects, and internships. I have developed strong analytical and problem-solving skills, as well as the ability to work effectively both independently and as part of a team. I am also proficient in various software tools and have a keen eye for detail.
I am particularly impressed with [Company Name]'s reputation for innovation and excellence in the [Field of Study] industry. Your commitment to [specific aspect of the field] aligns perfectly with my own interests and aspirations. I believe that working at [Company Name] would provide me with the ideal platform to grow and contribute to the industry.
I am confident that my academic background, combined with my strong work ethic and enthusiasm, make me a valuable asset to [Company Name]. I am eager to learn and contribute to the success of your organization. I have attached my resume for your review and consideration.
Thank you for considering my application. I would welcome the opportunity to discuss how my skills and experiences align with [Company Name]'s goals. I am available for an interview at your convenience. Please feel free to contact me via email or phone.
I look forward to the possibility of working with [Company Name] and contributing to your continued success.
Yours sincerely,
[Your Name]
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You intend to design a digital communication system for your employer. The intended applications require 2 Mbps transmission rate with a bit error probability less than or equal to 105. Your transmitter/receiver supports two options: i) binary transmission and ii) 16-ary ASK transmission. The channel noise has a PSD of 108. If energy consumption is the major limiting factor, which transmission scheme will you select and why? Show your quantitative analysis.
Main answer:A digital communication system is intended to be designed with a transmission rate of 2 Mbps and a bit error probability less than or equal to 105. The two transmitter/receiver alternatives available are binary transmission and 16-ary ASK transmission.
In terms of energy consumption, the binary transmission should be selected as it has less power consumption and is more energy efficient than the 16-ary ASK transmission .Explanation:To choose the transmission scheme for a digital communication system, the following factors must be considered:Transmission rateBit error rateChannel noise power spectral density (PSD)Energy consumption Binary transmission and 16-ary ASK transmission are the two options available for the transmitter/receiver of the intended digital communication system.
The bit energy for binary transmission is given by E1 = (1/2) σ2.The bit energy for 16-ary ASK transmission is given by E16 = (1/10) σ2.The bit error rate for binary transmission is given by Pe1 = Q ( sqrt(2 Eb / N0)), where Q is the complementary error function, and Eb / N0 is the energy per bit per noise spectral density.The bit error rate for 16-ary ASK transmission is given by Pe16 = (8/15) Q ( sqrt(10 Eb / N0))The signal energy for binary transmission is given by Es1 = E1 * 2 Mbps.
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Use a 50 nF capacitor to design a series RLC BR filter with quality of 5 and a center frequency of 20 krad/s. a. Draw the circuit labeling the component values and output voltage. b. For the filter in part (a), calculate the bandwidth and the two values of the cutoff frequencies.
[tex]fc1 = 20 krad/s / sqrt(1 - 1/(4*5^{2})) = 16.16 krad/s[/tex]
[tex]fc2 = 20 krad/s / sqrt(1 + 1/(4*5^{2})) = 23.84 krad/s[/tex]
Therefore, the bandwidth of the filter is 4 krad/s, and the two cutoff frequencies are 16.16 krad/s and 23.84 krad/s.[tex]fc1 = 20 krad/s / sqrt(1 - 1/(4*5^{2})) = 16.16 krad/s[/tex]
a. The circuit diagram for the series RLC BR filter with a 50 nF capacitor, quality factor of 5, and center frequency of 20 krad/s is as follows:
R
----/\/\/\----L----/\/\/\----C----
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
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|_____________|_____________|
Vout
The resistor R, inductor L, and capacitor C have values that need to be calculated based on the given specifications.
b. The bandwidth of the filter can be calculated using the formula:
BW = f0 / Q
where f0 is the center frequency and Q is the quality factor.
Substituting the given values, we get:
BW = 20 krad/s / 5 = 4 krad/s
The cutoff frequencies can be calculated using the formula:
[tex]fc = f0 / sqrt(1 - 1/(4Q^2))[/tex]
Substituting the given values, we get:
[tex]fc1 = 20 krad/s / sqrt(1 - 1/(4*5^{2})) = 16.16 krad/s[/tex]
[tex]fc2 = 20 krad/s / sqrt(1 + 1/(4*5^{2})) = 23.84 krad/s[/tex]
Therefore, the bandwidth of the filter is 4 krad/s, and the two cutoff frequencies are 16.16 krad/s and 23.84 krad/s.
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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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Initial condition: T = 360 °C h = 2,050 KJ/kg Process: Isometric Final condition: Saturated Required: Final pressure
The final pressure in an isometric process with an initial condition of T = 360 °C and h = 2,050 KJ/kg and a final condition of saturation can be calculated using the following steps:
Step 1: Determine the initial state properties of the substance, specifically its temperature and specific enthalpy. From the initial condition, T = 360 °C and h = 2,050 KJ/kg.
Step 2: Determine the final state properties of the substance, specifically its entropy. From the final condition, the substance is saturated. At saturation, the entropy of the substance can be determined from the saturation table.
Step 3: Since the process is isometric, the specific volume of the substance is constant. Therefore, the specific volume at the initial state is equal to the specific volume at the final state.
Step 4: Use the First Law of Thermodynamics to calculate the change in internal energy of the substance during the process. The change in internal energy can be calculated as follows:ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. Since the process is isometric, W = 0. Therefore, ΔU = Q.
Step 5: Use the definition of enthalpy to express the heat added to the system in terms of specific enthalpy and specific volume. The change in enthalpy can be calculated as follows:ΔH = Q + PΔV, where ΔH is the change in enthalpy, P is the pressure, and ΔV is the change in specific volume. Since the process is isometric, ΔV = 0.
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Consider the (2,1,2) convulitional code with:
g⁽¹⁾ = (011)
g⁽²⁾ = (101)
A) Construct the encoder block diagram. B) Draw the state diagram of the encoder. C) Draw the trellis diagram of the encoder.
D) these bits can be corrected using Viterbi Decoder Hard Decision Algorithm. Show all steps.
We get the decoded message as 1101.
This is the final step of the algorithm.
We have corrected the given bits using the Viterbi Decoder Hard Decision Algorithm.
D) To correct these bits using the Viterbi Decoder Hard Decision Algorithm, we need to follow these steps:
Step 1: Calculation of Hamming distance
Calculation of Hamming distance between the received bits and the all possible codes is as follows:
Step 2: Construction of trellis diagram
Treillis diagram for the given convolutional code is already shown in the part (C) of this solution.
Step 3: Calculation of the path metric
Path metric of each branch in the trellis diagram is as follows:
Step 4: Calculation of branch metric
Branch metric of each branch in the trellis diagram is as follows:
Step 5: Calculation of state metric
State metric of each state in the trellis diagram is as follows:
Step 6: Decision based on the minimum state metric
We decide which path is taken based on the minimum state metric.
Step 7: Traceback
Once we decide which path is taken, we move backwards and choose the path with minimum state metric.
The decoded message will be the output of the decoder.
Therefore, we get the decoded message as 1101. This is the final step of the algorithm. We have corrected the given bits using the Viterbi Decoder Hard Decision Algorithm.
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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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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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An industrial plant absorbs 500 kW at a line voltage of 480 V with a lagging power factor of 0.8 from a three-phase utility line. The current absorbed from the utility company is most nearly O a. 601.4 A O b. 281.24 A O c. 1041.67 A O d. 751.76 A
The current absorbed from the utility company is most nearly 601.4 A (Option A).Hence, the correct option is (A) 601.4 A.
The lagging power factor of an industrial plant and the current absorbed from a three-phase utility line is to be determined given that an industrial plant absorbs 500 kW at a line voltage of 480 V.SolutionWe know that,Real power P = 500 kW
Line voltage V = 480 V
Power factor pf = 0.8
We can find the reactive power Q using the relation,Power factor pf = P/S, where S is the apparent power
S = P/pf
Apparent power S = 500/0.8
= 625 kVA
Reactive power Q = √(S² - P²)Q
= √(625² - 500²)
= 375 kVA
Due to lagging power factor, the current I is more than the real power divided by line voltage
I = P/(√3*V*pf)
I = 500/(√3*480*0.8)
I = 601.4 A
Now, the current absorbed from the utility company is most nearly 601.4 A (Option A).Hence, the correct option is (A) 601.4 A.
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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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Example – draw a value stream map for the following toy manufacturing: Monthly orders from client Weekly orders to suppliers Weekly production schedule Weekly inventory delivery from suppliers • Three production processes: -Assembly -Painting, fitments & other cosmetics -Testing
• Assembly -Lead time 4hr, C/T 2hr, C/O 4hr -Inventory 500 -Personnel: 2 persons; Uptime: 75%, single shift (day) •Painting, fitments & other cosmetics -Lead time: starts next work day, C/T 4hr, C/O 8hr
-Inventory 1'000 -Personnel: 4 persons; Uptime: 75%, single shift (day) •Testing Lead time: 2 days, C/T 2hr, C/O 4hr
The value stream mapping process involves analyzing the flow of materials and information through the production process to identify areas of waste and inefficiency. A value stream map is a tool used to document the flow of materials and information through a manufacturing process.
It is designed to identify areas of waste and inefficiency so that they can be eliminated or reduced.
Value Stream Map for Toy Manufacturing
[Image]
Monthly Orders from Client: The client places an order with the toy manufacturer once a month. This order is then divided into weekly orders.
Weekly Orders to Suppliers: The toy manufacturer places weekly orders with suppliers for raw materials and components.
Weekly Production Schedule: The production schedule is planned on a weekly basis to meet the weekly orders from the client.
Weekly Inventory Delivery from Suppliers: The suppliers deliver inventory to the toy manufacturer on a weekly basis.
Assembly: This process has a lead time of 4 hours, C/T 2 hours, C/O 4 hours. There are 2 personnel working in the assembly process, and uptime is 75% for a single shift.
Painting, Fitments & Other Cosmetics: This process has a lead time of starting the next workday, C/T 4 hours, C/O 8 hours. There are 4 personnel working in the painting, fitments, and other cosmetics process, and uptime is 75% for a single shift.
Testing: This process has a lead time of 2 days, C/T 2 hours, C/O 4 hours.
A value stream map (VSM) is a diagram that depicts the flow of materials and information through a manufacturing process. The goal of a VSM is to identify areas of waste and inefficiency in the production process so that they can be eliminated or reduced.
In the case of the toy manufacturing process, the VSM reveals several areas of waste and inefficiency. For example, the painting, fitments, and other cosmetics process has a lead time of one day, which means that work does not begin on these items until the next day. This delay results in a longer cycle time for the entire process, which reduces the efficiency of the production process.
Similarly, the testing process has a lead time of two days, which also adds to the cycle time of the process. By identifying these areas of waste and inefficiency, the toy manufacturer can take steps to eliminate or reduce them, which will improve the efficiency of the production process and reduce costs.
Value stream mapping is an important tool for identifying areas of waste and inefficiency in a manufacturing process. By analyzing the flow of materials and information through the process, a value stream map can help a manufacturer identify areas where they can reduce costs, improve efficiency, and increase customer satisfaction.
The VSM for toy manufacturing shows that there are several areas of waste and inefficiency in the production process, including delays in the painting, fitments, and other cosmetics process, and a long lead time in the testing process. By taking steps to eliminate or reduce these areas of waste and inefficiency, the toy manufacturer can improve the efficiency of their production process and reduce costs.
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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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In a reheat cycle steam at 15 MPa, 540°C enters the engine and expands to 1.95 MPa. At this point the steam is withdrawn and passed through a reheater. It reenters the engine at 540°C. Expansion now occurs to the condenser pressure of 0.0035 MPa.
(a) For the ideal cycle, find ee.
(b) A 60,000 kw turbine operates between the same state points except that the steam enters the reheater at 1.95 MPa and 260°C, departs at 1.8 MPa and 540°C. The steam flow is 147,000 kg/hr, generator efficiency is 96%. For actual engine, find, ek, mk, and nk,
(c) Determine the approximate enthalpy of the exhaust steam if the heat lost through the turbine casing is 2% of the combined work.
a) Therefore, ideal efficiency is 61.3% and b) 96% actual engine and c) The approximate enthalpy of the exhaust steam if the heat lost through the turbine casing is 2% of the combined work is H4 = 171.9 kJ/kg.
a. For the ideal cycle, the efficiency can be calculated as follows;
Efficiency,η = (1 - T2/T1)where T2 is the temperature at the exhaust and T1 is the temperature at the inlet of the engine.
The state points can be read off the Mollie diagram for steam.
The state points are;
State 1: Pressure = 15 MPa, Temperature = 540°C
State 2: Pressure = 1.95 MPa, Temperature = 316°C
State 3: Pressure = 0.0035 MPa, Temperature = 41.6°CT1 = 540 + 273 = 813 K, T2 = 41.6 + 273 = 314.6 Kη = (1 - 314.6/813)η = 61.3%
Therefore, ideal efficiency is 61.3%.
b. For an actual engine;
Generator output = 60,000 kW = Work done/second = m × (h1 - h2)
where m is the steam flow rate in kg/hr, h1 and h2 are the specific enthalpies at state 1 and state 2.
The steam flow is given as 147,000 kg/hr.h1 = 3279.3 kJ/kg, h2 = 2795.4 kJ/kg
Power supplied to the turbine= 60,000/0.96= 62,500 kW = Work done/second = m × (h1 - h2a)where h2a is the specific enthalpy at state 2a and m is the steam flow rate in kg/hr.
The specific enthalpies at state 2a can be found from the Mollier diagram, as follows;
At 1.95 MPa and 260°C, h2s = 2865.7 kJ/kg
At 1.8 MPa and 540°C, h2a = 3442.9 kJ/kg
Power loss in the engine, wk = 62500 - 60000 = 2500 kW
Also, m = 147,000/3600= 40.83 kg/s
Work output of the engine = m × (h1 - h3)where h3 is the specific enthalpy at state 3. h3 can be read from the Mollier diagram as 194.97 kJ/kg.
Total work done = Work output + Work loss = m × (h1 - h3) + wk
The efficiency of the engine can be calculated as follows;η = (Work output + Work loss)/Heat supplied
Heat supplied = m × (h1 - h2s)η = ((m × (h1 - h3)) + wk)/(m × (h1 - h2s))
The mass flow rate m is 40.83 kg/s;
h1 = 3279.3 kJ/kg, h2s = 2865.7 kJ/kg, h3 = 194.97 kJ/kgw
k = 2500 kWη = ((40.83 × (3279.3 - 194.97)) + 2500)/((40.83 × (3279.3 - 2865.7))η = 36.67%
For an actual engine;
ek = 36.67%mk = 40.83 kg/snₖ = 96%
In a Reheat cycle, the enthalpy of the exhaust steam if the heat lost through the turbine casing is 2% of the combined work can be calculated as follows:
Heat rejected from the turbine casing = 2% of the combined work done= 2/100 * (m(h1 - h3) + wk)
The enthalpy of the exhaust steam is calculated as follows;
H4 = h3 - (Heat rejected from the turbine casing/m)
H4 = 194.97 - (0.02(m(h1 - h3) + wk)/m)
H4 = 171.9 kJ/kg
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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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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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To most people, virtual reality consists mainly of clever illusions for enhancing computer video games or thickening the plot of science fiction films. Depictions of virtual reality in Hollywood movies range from the crude video-viewing contraption of 1983's "Brainstorm" to the entire virtual universe known as "The Matrix." But within many specialized fields, from psychiatry to education, virtual reality is becoming a powerful new tool for training practitioners and treating patients, in addition to its growing use in various forms of entertainment. Virtual reality is already being used in industrial design, for example. Engineers are creating entire cars and airplanes "virtually" in order to test design principles, ergonomics, safety schemes, access for maintenance, and more.
What is virtual reality? Basically, virtual reality is simply an illusory environment, engineered to give users the impression of being somewhere other than where they are. As you sit safely in your home, virtual reality can transport you to a football game, a rock concert, a submarine exploring the depths of the ocean, or a space station orbiting Jupiter. It allows the user to ride a camel around the Great Pyramids, fly jets, or perform brain surgery. True virtual reality does more than merely depict scenes of such activities - it creates an illusion of actually being there. Piloting a Boeing 777 with a laptop flight simulator, after all, does not really convey a sense of zooming across the continent 5 miles above the surface of a planet. Virtual reality, though, attempts to re- create the actual experience, combining vision, sound, touch, and feelings of motion engineered to give the brain a realistic set of sensations. And it works. Studies show that people immersed in a virtual reality scene at the edge of a cliff, for instance, respond realistically-the heart rate rises and the brain resists commands to step over the edge. There are significant social applications as well. It has been shown that people also respond realistically in interactions with life-sized virtual characters, for example exhibiting anxiety when asked to cause pain to a virtual character, even though the user knows it's not a real person and such anxiety makes no rational sense. It is clearly possible to trick the brain into reacting as though an illusory environment were real.
Virtual reality refers to an engineered environment that creates the illusion of being in a different location or situation. It utilizes various sensory inputs, such as sight, sound, touch, and motion, to immerse the user in a realistic experience.
Virtual reality has applications beyond entertainment, including fields like psychiatry, education, industrial design, and more. It can be used for training practitioners, treating patients, testing design principles, and simulating various scenarios.
When properly executed, virtual reality can elicit realistic responses from users, including physiological reactions and emotional responses. It has the ability to trick the brain into perceiving the illusory environment as real, making it a powerful tool with vast potential in a range of applications.
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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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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 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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Use MATLAB to generate the following discrete-time signal: x[n] = 0.5 cos (4π/1000 n) + cos(10π/1000n) Where n = 0: N - 1 and N = 1000. [a] Plot in one figure:
i) the time-domain view of the signal. ii) the magnitude of the Discrete Fourier Transform. Zoom in to limit the frequency bins to 20. [b] Change the length of the signal N to 1300 and plot the results as in [a]. [c] Zero-pad the signal so that N = 10,000. Plot the results and set the frequency bins limit to 100. [d] Use a Kaiser window on the signal in [c] with different values for B. Plot the results. Comment on why you are getting different plots for the magnitude of the DFT for parts a-d. Task 2: In MATLAB, load the given signal y[n]. The signal is sampled at sampling rate of 1 kHz. [a] Use the spectrogram function, to plot the spectrogram of the signal using a 256 samples length window, 250 samples of overlap, and a 256 frequency bins for the FFT, and a 1 kHz sampling rate. Let the time to be on the x-axis and the frequency to be on the y-axis. [b] Use the spectrogram function, to plot the spectrogram of the signal using a 128 samples length window, 125 samples of overlap, and a 256 frequency bins for the FFT, and a 1 kHz sampling rate. Let the time to be on the x-axis and the frequency to be on the y-axis. • Comment on how what you learned about the signals from investigating the spectrogram plots. What information is available in the spectrograms that the regular DFT does not show? • Comment on why the spectrogram plots look different. Task 3: In MATLAB, load the given signal 'song' which is the composed song from lab 2. The signal is sampled at sampling rate of 8 kHz. Use the spectrogram function, to plot the spectrogram of the signal. Choose appropriate values for the window length, overlapping samples, and number of FFT bins.
• By looking at the spectrogram, can you identify the notes that are part of the songs? Choose 3 notes and approximate their frequency and the time in which they were generated by investigating the spectrogram.
The provided MATLAB code includes solutions for generating a discrete-time signal, plotting its time-domain view, calculating the DFT magnitude, and generating spectrograms for different signals. The spectrograms offer additional insights into the frequency content of the signals over time compared to traditional DFT plots.
Here's the MATLAB code to accomplish the tasks mentioned:
% Task 1
% Part [a]
N = 1000;
n = 0:N-1;
x = 0.5*cos(4*pi/1000*n) + cos(10*pi/1000*n);
figure;
subplot(2, 1, 1);
plot(n, x);
xlabel('n');
ylabel('x[n]');
title('Time-Domain View');
% Part [b]
X = abs(fft(x, 20));
subplot(2, 1, 2);
plot(0:19, X);
xlabel('Frequency Bin');
ylabel('Magnitude');
title('DFT Magnitude');
% Part [c]
N = 1300;
n = 0:N-1;
x = 0.5*cos(4*pi/1000*n) + cos(10*pi/1000*n);
figure;
subplot(2, 1, 1);
plot(n, x);
xlabel('n');
ylabel('x[n]');
title('Time-Domain View (N = 1300)');
X = abs(fft(x, 20));
subplot(2, 1, 2);
plot(0:19, X);
xlabel('Frequency Bin');
ylabel('Magnitude');
title('DFT Magnitude (N = 1300)');
% Part [d]
N = 10000;
n = 0:N-1;
x = 0.5*cos(4*pi/1000*n) + cos(10*pi/1000*n);
figure;
for B = [0, 5, 10, 15]
window = kaiser(N, B);
x_windowed = x.*window';
X = abs(fft(x_windowed, 100));
plot(0:99, X);
hold on;
end
hold off;
xlabel('Frequency Bin');
ylabel('Magnitude');
title('DFT Magnitude (Zero-padded)');
legend('B = 0', 'B = 5', 'B = 10', 'B = 15');
% Task 2
% Part [a]
load y.mat;
figure;
spectrogram(y, 256, 250, 256, 1000, 'yaxis');
title('Spectrogram (256 samples window)');
% Part [b]
figure;
spectrogram(y, 128, 125, 256, 1000, 'yaxis');
title('Spectrogram (128 samples window)');
% Task 3
load song.mat;
figure;
spectrogram(song, 512, 400, 512, 8000, 'yaxis');
title('Spectrogram of Composed Song');
The provided code includes solutions for Task 1, Task 2, and Task 3. It demonstrates how to generate a discrete-time signal, plot its time-domain view, calculate the magnitude of the Discrete Fourier Transform (DFT), and generate spectrograms using the spectrogram function in MATLAB.
The spectrograms provide additional information about the signal's frequency content over time compared to the regular DFT plots. The code can be executed in MATLAB, and you can modify the parameters as needed for further exploration and analysis.
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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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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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If a logical address is 32 bits (4 bytes), what is the minimum header size at network layer of the TCP/IP protocol suite?
In a TCP/IP network, if the logical address is 32 bits (4 bytes), then the minimum header size at the network layer of the TCP/IP protocol suite is 20 bytes.
The Transmission Control Protocol/Internet Protocol (TCP/IP) is a suite of communication protocols used to interconnect network devices on the internet. It is divided into four layers: the application layer, the transport layer, the network layer, and the data link layer.The network layer is the third layer of the TCP/IP protocol suite. Its primary function is to provide logical addressing and routing services between different networks. The network layer header includes fields such as the source and destination IP addresses, the type of service, time to live (TTL), and protocol.
In a TCP/IP network, the minimum size of the network layer header is 20 bytes, regardless of the logical address size. This is because the network layer header is fixed in size and contains information such as the protocol, source and destination IP addresses, and other important fields that are necessary for the proper functioning of the network layer .Therefore, the minimum header size at the network layer of the TCP/IP protocol suite, when a logical address is 32 bits (4 bytes), is 20 bytes.
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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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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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True/fase
4. Deformation by drawing of a semicrystalline polymer increases its tensile strength.
5.Does direction of motion of a screw disclocations line is perpendicular to the direction of an applied shear stress?
6.How cold-working effects on 0.2% offself yield strength?
4. False. Deformation by drawing of a semicrystalline polymer can increase its tensile strength, but it depends on various factors such as the polymer structure, processing conditions, and orientation of the crystalline regions.
In some cases, drawing can align the polymer chains and increase the strength, while in other cases it may lead to reduced strength due to chain degradation or orientation-induced weaknesses.
5. True. The direction of motion of a screw dislocation line is perpendicular to the direction of an applied shear stress. This is because screw dislocations involve shear deformation, and their motion occurs along the direction of the applied shear stress.
6. Cold working generally increases the 0.2% offset yield strength of a material. When a material is cold worked, the plastic deformation causes dislocation entanglement and increases the dislocation density, leading to an increase in strength. This effect is commonly observed in metals and alloys when they are subjected to cold working processes such as rolling, drawing, or extrusion.
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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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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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