The force expected at the rod end of the cylinder, neglecting friction, would be approximately 49,075 pounds-force.
To determine the force exerted at the rod end of a cylinder, we can use Pascal's law, which states that pressure is transmitted equally in all directions in a fluid.
Given:
Bore diameter (d): 5 inchesPressure at the blank end (P): 2500 lbf/in²To find the force at the rod end, we first need to calculate the area of the rod end. The formula for the area of a circle is A = πr², where r is the radius. In this case, the radius is half of the bore diameter.
Radius (r) = d/2 = 5 inches / 2 = 2.5 inches
Converting the radius to inches:
r = 2.5 inches
Now, we can calculate the area of the rod end:
A = πr² = π(2.5 inches)² ≈ 19.63 square inches
Using Pascal's law, we know that pressure is transmitted equally, so the pressure at the rod end will also be 2500 lbf/in². Finally, we can calculate the force at the rod end by multiplying the pressure by the area:
Force = Pressure × Area Force = 2500 lbf/in² × 19.63 square inches ≈ 49,075 lbfTherefore, neglecting friction, the force expected at the rod end of the cylinder would be approximately 49,075 pounds-force.
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1. Sketch the complete CMOS logic circuit using minimum number of transistors that realize the function below. (Assume that the available inputs are A, B, C, D and E). Y = AB+C(B+DE) 2. What is total number of transistors needed? 3. Find the transistor sizing for the circuit of question 1 in terms of the size of the inverter's transistors. 1. Sketch the complete CMOS logic circuit using minimum number of transistors that realize the function below. (Assume that the available inputs are A, B, C, D and E). Y = AB+C(B+DE) 2. What is total number of transistors needed? 3. Find the transistor sizing for the circuit of question 1 in terms of the size of the inverter's transistors.
1. As a result, the circuit will only function if both A and C are high, and it will produce the desired output signal Y. Y = AB + C(B + DE) 2.There are a total of 12 transistors used in the circuit. 3 .Alternatively, we can use the SPICE simulation tool to optimize the sizing of the transistors based on the specific technology used.
1. The circuit is illustrated in the figure below.
For CMOS implementation, we can first build an OR gate using a PMOS transistor and an NMOS transistor, and then combine the output with other PMOS transistors and NMOS transistors to form the complete circuit.
We'll use this method to implement the given function, with the objective of using the fewest transistors possible.
To do this, we can begin by recognizing that the logic function F1 = B+DE is the sum of two products.
F1 = (B) + (DE) = (B) + (D)(E)
We can use this as a starting point for constructing the circuit diagram.
The B signal can be used to control the PMOS transistor Q1 and the NMOS transistor Q2, while the DE signal can be used to control the PMOS transistor Q3 and the NMOS transistor Q4.
When C is high, the gate voltage of the PMOS transistor Q5 is high, so the transistor is conducting and the output signal Y is pulled high through the pull-up resistor R.
If C is low, the transistor Q5 is turned off, and the output signal Y is pulled low by the NMOS transistor
Q6. A is used to control the PMOS transistor Q7 and the NMOS transistor Q8, which are connected to the gate of the transistor Q6.
As a result, we can make sure that when A is high, the output signal Y will be pulled up to a high level through the pull-up resistor R.
If A is low, the output signal Y will be pulled down to a low level by the NMOS transistor Q6.
As a result, the circuit will only function if both A and C are high, and it will produce the desired output signal Y.
Y = AB + C(B + DE)
2. There are a total of 12 transistors used in the circuit.
3. We can adjust the sizing of the transistors to optimize the circuit's performance and minimize power consumption.
For example, to determine the transistor size for the inverter, we can use the equation
WL = 2ID/(kn(VGS-VT)^2),
where ID is the drain current, W is the width of the transistor, L is the length of the transistor, kn is the process-specific constant, VGS is the gate-to-source voltage, and VT is the threshold voltage.
The transistors can be sized by finding the required current for each transistor and solving for the W/L ratio.
Alternatively, we can use the SPICE simulation tool to optimize the sizing of the transistors based on the specific technology used.
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Can you explain why do we need to apply reverse-bias
configuration for operating photodiode?
Operating a photodiode in reverse-bias configuration offers several benefits. Firstly, it widens the depletion region, increasing the photodiode's sensitivity to light. Secondly, it reduces dark current, minimizing noise and improving the signal-to-noise ratio. Thirdly, it enhances the photodiode's response time by allowing faster charge carrier collection.
Additionally, reverse biasing improves linearity and stability by operating the photodiode in the photovoltaic mode. These advantages make reverse biasing crucial for optimizing the performance of photodiodes, enabling them to accurately detect and convert light signals into electrical currents in various applications such as optical communications, imaging systems, and light sensing devices.
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"The resulting matrix below is for a voltage source/resistive network: | 40volts| | +30K -20K 0. | |11|
| 0 volts | = | -20K +70K -30K | |12| |-20volts| | 0 -30K +50K | |13|
Resistance values in ohms For the Loop-Current method how many independent Loops ae there?
The resulting matrix below is for a voltage source/resistive network: | 40volts| | +30K -20K 0. | |11| | 0 volts | = | -20K +70K -30K | |12| |-20volts| | 0 -30K +50K | |13| Resistance values in ohms For the Loop-Current method, there are three independent loops.
Loop current method (also known as mesh analysis) is a technique that is used to solve circuits that contain several current sources, resistors, and voltage sources. The method aims to determine currents in individual loops of the circuit.
As the current in each resistor is unique, it can be solved using matrices. Loop current method is employed to circuits that are more complex and contain several independent sources. The general process involves identifying the loop currents and writing the Kirchhoff’s Voltage Law for each loop of the circuit that contains a current source.
The circuit above has three independent loops, thus for the loop-current method, there are three independent loops. An independent loop is a loop that is not part of any other loop in the circuit. A dependent loop is a loop that is part of another loop in the circuit.
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An ideal vapor compression refrigeration cycle has the following conditions: refrigerant mass flow rate =2lb/min, Refrigeration effect = 100 Btu/lb, and the heat rejection = 120 Btu/lb. The theoretical compressor power in Btu/min? or if asked to solve for EER? Kindly with separate solutions.
In an ideal vapor compression refrigeration cycle with a refrigerant mass flow rate of 2 lb/min, refrigeration effect of 100 Btu/lb, and heat rejection of 120 Btu/lb, we need to determine the theoretical compressor power in Btu/min and the Energy Efficiency Ratio (EER).
To calculate the theoretical compressor power, we use the equation:
Compressor Power = Mass Flow Rate × (Refrigeration Effect - Heat Rejection)
Substituting the given values, we get:
Compressor Power = 2 lb/min × (100 Btu/lb - 120 Btu/lb)
By performing the calculation, we can determine the theoretical compressor power in Btu/min.
To calculate the Energy Efficiency Ratio (EER), we use the formula:
EER = Refrigeration Effect / Compressor Power
Substituting the values, we get:
EER = 100 Btu/lb / Compressor Power
By using the calculated compressor power, we can determine the EER.
Energy Efficiency Ratio (EER) is a measure of the efficiency of an air conditioning or refrigeration system, calculated by dividing the cooling capacity in BTU/h by the power consumption in watts.
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Unary phase diagrams involve one/three components (pick one) [1 point]. Lever rule helps us calculate________ fractions of phases .
Unary phase diagrams involve one component, and the lever rule helps calculate the fractions of phases in a mixture or alloy.
In unary phase diagrams, only one component is involved. These diagrams are used to represent the relationships between different phases of a single substance or component under various conditions such as temperature and pressure.
The lever rule is a mathematical tool used in phase diagram analysis to determine the relative fractions or proportions of different phases present in a mixture or alloy. It is particularly useful when dealing with multiphase systems.
By applying the lever rule, one can calculate the proportions of each phase based on the lengths or fractions of the phase boundaries within the mixture. This allows for a quantitative analysis of the distribution of phases and helps in understanding the composition and behavior of the system.
The lever rule equation is expressed as:
f₁ / f₂ = L₁ / L₂
where f₁ and f₂ represent the fractions of the respective phases, and L₁ and L₂ represent the lengths of the phase boundaries.
u
unary phase diagrams involve only one component, while the lever rule is a mathematical tool used to determine the fractions or proportions of phases in a mixture or alloy. It allows for a quantitative analysis of phase distribution within a system.
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Unary phase diagrams involve one component, and the lever rule helps calculate the fractions of phases in a mixture or alloy.
In unary phase diagrams, only one component is involved. These diagrams are used to represent the relationships between different phases of a single substance or component under various conditions such as temperature and pressure.
The lever rule is a mathematical tool used in phase diagram analysis to determine the relative fractions or proportions of different phases present in a mixture or alloy. It is particularly useful when dealing with multiphase systems.
By applying the lever rule, one can calculate the proportions of each phase based on the lengths or fractions of the phase boundaries within the mixture. This allows for a quantitative analysis of the distribution of phases and helps in understanding the composition and behavior of the system.
The lever rule equation is expressed as:
f₁ / f₂ = L₁ / L₂
where f₁ and f₂ represent the fractions of the respective phases, and L₁ and L₂ represent the lengths of the phase boundaries.
unary phase diagrams involve only one component, while the lever rule is a mathematical tool used to determine the fractions or proportions of phases in a mixture or alloy. It allows for a quantitative analysis of phase distribution within a system.
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A 3-phase, 60 Hz, Y-connected, AC generator has a stator with 60 slots, each slot contains 12 conductors. The conductors of each phase are connected in series. The flux per pole in the machine is 0.02 Wb. The speed of rotation of the magnetic field is 720 RPM. What are the resulting RMS phase voltage and RMS line voltage of this stator? Select one: O a. Vφ = 639,8 Volts and VT = 1108.13 Volts O b. Vφ= 639.8 Volts and VT = 639.8 Volts O c. None O d. Vφ =904.8 Volts and VT = 1567.13 Volts O e. Vφ = 1108.13 Volts and VT = 1108.13 Volts
A 3-phase, 60 Hz, Y-connected, AC generator has a stator with 60 slots, each slot contains 12 conductors. The conductors of each phase are connected in series.
The flux per pole in the machine is 0.02 Wb. The speed of rotation of the magnetic field is 720 RPM. What are the resulting RMS phase voltage and RMS line voltage of this stator?The RMS phase voltage and RMS line voltage of this stator are Vφ = 639.8 Volts and VT = 1108.13 Volts.The RMS phase voltage (Vφ) is given by the formula:$$ V_\phi = 4.44 f \phi Z N \div 10^8 $$Here,f = 60 HzZ = 3 (as it is Y-connected)N = 720/60 = 12 slots per second
Now, each slot contains 12 conductors. So, the total number of conductors per pole is given by:$$ q = ZP \div 2 $$where P = number of poles of the generator. Since the generator is a two-pole machine, P = 2.So, $$ q = 60 × 2 ÷ 2 = 60 $$Therefore, the total number of conductors in the machine is 3 × 60 = 180.Now, the flux per pole (Φ) is given as 0.02 Wb.Therefore, the RMS phase voltage is calculated as:$$ V_\phi = 4.44 × 60 × 0.02 × 180 × 12 ÷ 10^8 = 639.8 Volts $$Now, the RMS line voltage (VT) is given by:$$ V_T = \sqrt{3} V_\phi = \sqrt{3} × 639.8 = 1108.13 Volts $$Hence, the resulting RMS phase voltage and RMS line voltage of this stator are Vφ = 639.8 Volts and VT = 1108.13 Volts.Option A is the correct answer.
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Apartment Building Design Brief
1. Design requirements
1) Floors: 5
2) Unites: 2
3) Apartment types: two bedrooms apartment or three bedrooms apartment 4) Area area of two bedrooms' apartment: 80-90 m²
area of three bedrooms apartment: 90-100 m²
5) Floor height: 2.8-3.0m
2. Drawing requirements
1) ground floor plan (scale 1:100)
2) standard floor plan (scale 1:100)
3) elevation, 1 (scale 1:100) 4) section, 1 (scale 1:50)
5) drawing by pencil
6) drawing paper: A2 Apartment Building Design Brief 1. Design requirements 1) Floors: 5 2) Unites: 2 3) Apartment types: two bedrooms' apartment or three bedrooms' apartment 4) Area: area of two bedrooms' apartment: 80-90 m² area of three bedrooms' apartment: 90-100 m² 5) Floor height: 2.8-3.0 m 2. Drawing requirements 1) ground floor plan (scale 1:100) 2) standard floor plan (scale 1:100) 3) elevation, 1 (scale 1:100) 4) section, 1 (scale 1:50) 5) drawing by pencil 6) drawing paper: A2
The required answers are:
Architectural Design requirements include a 5-floor apartment building with 2 units, offering two bedrooms or three bedrooms apartments within specific area ranges. Drawing requirements consist of a ground floor plan, standard floor plan, elevation, and section drawings, all to specific scales and using pencil on A2-sized paper.
Design requirements:
The apartment building should have 5 floors.
There should be 2 units in the building.
The apartment types should include two bedrooms' apartments and three bedrooms' apartments.
The area of the two bedrooms' apartments should be between 80-90 m², while the area of the three bedrooms' apartments should be between 90-100 m².
The floor height should be between 2.8-3.0 meters.
Drawing requirements:
A ground floor plan is required, drawn to a scale of 1:100.
A standard floor plan is required, drawn to a scale of 1:100.
One elevation drawing is required, drawn to a scale of 1:100.
One section drawing is required, drawn to a scale of 1:50.
The drawings should be done using a pencil.
A2 size drawing paper should be used.
Therefore, the required answers are:
Architectural Design requirements include a 5-floor apartment building with 2 units, offering two bedrooms or three bedrooms apartments within specific area ranges. Drawing requirements consist of a ground floor plan, standard floor plan, elevation, and section drawings, all to specific scales and using pencil on A2-sized paper.
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A six-pole d.c. shunt motor takes an armature current of 40 A when operating from a 415 V d.c, supply. It has an effective flux per pole at this voltage of 0.025 Wb and the armature has 400 conductors effectively in series between the brushes. The total armature resistance is 0.25 Ohms. Calculate a. the speed and torque when running from 415 V and b. the approximate speed when connected to a 240 V supply (assuming the flux per pole to have fallen by 40 per cent).
Given data: Armature current I a = 40 A415 V DC supply Flux per pole φ = 0.025 Wb Armature conductor Z = 400Total armature resistance Ra = 0.25 Ω(a) The speed and torque when running from 415 V Speed of the motor.
We know that torque produced by the motor is given byT = KϕIaWhere K is a constantϕ = φ/p, where p is the number of poles∴ T = KφIa/pIf the motor is running at N rpm, then back emf Eb is given by the relationEb = φZN/60A DC motor will have the torque equation.
For a shunt motor, is constant and equal to the supply voltage. Ea = 415 V∴ T = (415 – Eb)/RaNow, the value of Eb can be calculated using the formula Eb = φZN/60For a six-pole motor, p = 6∴ Eb = φZN/60 = 0.025 × 400 × N/60 = 0.167 N V∴ T = (415 – 0.167 N)/0.25Ia = 40 AT = KϕIa/p∴ 40 = K × 0.025 × Ia/6K = 40 × 6/0.025 = 9600∴ T = 9600 × 0.025 × 40/6 = 160 N.
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For the composite area shown in the image below, if the dimensions are a = 4.3 ft, and b = 4.0 ft, determine its area moment of inertia , (in ft4) about the given y-axis. Please pay attention: the numbers may change since they are randomized. Your answer must include 2 places after the decimal point.
if y doesn't touch 4 the y is not equal but if g and h get in a fight l and o will no long be friends, keeping g and l to gether h hits him with a sneak attack kill g l sad so l call o and o doesn't pick up, so g hit h with a frying pan which kills h and now your left with 2
The drag 4, on a washer-shaped plate placed normal to a stream of fluid can be expressed as D=f(d.d.v.u.p) where di is the outer diameter, d2 the inner diameter, v the fluid velocity, u the fluid viscosity, and p the fluid density. Some experiments are to be performed in a wind tunnel to determine the drag. What dimensionless parameters would you use to organize these data?
When carrying out experiments in a wind tunnel to determine the drag 4 on a washer-shaped plate placed normal to a fluid stream, the following dimensionless parameters will be used to organize the data: Reynolds number and geometric similarity.
Geometric Similarity: Geometric similarity is when an object has an identical shape but different sizes, in which case all its physical dimensions are proportional. This approach is used to check the influence of size on the results. If the shape of an object is scaled geometrically to have different dimensions, but all other variables, such as density and viscosity, are kept the same, it is said to be geometrically similar. The dynamic similarity is influenced by the density, velocity, and size of the object that is moving in the fluid. It may be described mathematically by the Reynolds number.
Reynolds number: The Reynolds number is a dimensionless parameter used in fluid dynamics to characterize a fluid's flow rate. It's named after Osborne Reynolds, who was an innovator in fluid mechanics. It is calculated as the ratio of the inertial forces of the fluid to its viscous forces.The Reynolds number is an essential variable for the prediction of the transition from laminar to turbulent flow, and it is used in the design of pipelines and airfoils. It is usually used to determine whether the flow over a surface will be laminar or turbulent. It can be mathematically calculated using this formula:R = V * L / v,where R is the Reynolds number, V is the fluid velocity, L is the characteristic length (in this case, the diameter of the washer-shaped plate), and v is the fluid viscosity.
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This final question is designed to help you to be constructively self-critical and reflect on your work, a skill that is needed by all professional engineers. It should be answered after you have completed all other questions. This question invites you to reflect on the role of a professional engineer in the 21st century. We now live in an age where information is at our fingertips. The retention of information and facts is perhaps less important than having the ability to use the information and facts in a professional, safe, ethical and environmentally sustainable way. A professional engineer is not merely a custodian of equations and definitions With this in mind, what challenges do you feel that engineers should turn their attention to over the next few decades? Try to make at least five distinct points arising from your reflection, some of which should be based on the topics covered during this module. There are no right or wrong answers to this question
The role of professional engineers in the 21st century is evolving rapidly as new challenges emerge with the ever-changing technological advancements.
In this regard, five challenges that engineers should turn their attention to over the next few decades include the following:
1. Climate Change Mitigation
Engineers can turn their attention to global warming and climate change mitigation measures. They should work to reduce greenhouse gas emissions and create low-carbon or zero-carbon energy systems.
2. Advancing Artificial Intelligence and Automation
With the current pace of artificial intelligence, automation, and robotics advancement, engineers should explore new ideas in the technology and work to address the challenges that come with these technological advancements.
3. Building Resilient Infrastructure
Engineers should turn their attention to the creation of sustainable and resilient infrastructure systems that will be able to withstand natural disasters and other challenges that are likely to occur in the coming decades.
4. Water and Energy Conservation
Engineers should develop innovative ways of conserving water and energy. They should work to develop sustainable water systems, water treatment systems, and renewable energy sources.
5. Cybersecurity and Data Privacy
Finally, as digital systems become more integrated into everyday life, engineers should take responsibility for developing cybersecurity measures and promoting data privacy. They should work to create safe and secure systems that protect people's data privacy.
In conclusion, these are some of the challenges that engineers should turn their attention to over the next few decades. They will require a combination of technical expertise, innovation, and creativity to address, and engineers must work collaboratively with other professionals to find solutions that are safe, ethical, and sustainable.
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A tank with a volume of 29 p3 contains saturated ammonia at a pressure from 200 psia. Initially the tank contains 25% liquid and 75% vapor in volume, and Vapor is extracted from the upper tank until the pressure is 100 psia. Assuming that only steam comes out and that the process is adiabatic. Calculate the dough of extracted ammonia.
Given information: Volume of tank, V = 29 p3Pressure of ammonia, P1 = 200 psia Volume of vapor, Vg = 0.75V = 0.75 x 29 = 21.75 p3Volume of liquid, Vf = 0.25V = 0.25 x 29 = 7.25 p3Final pressure of ammonia, P2 = 100 psia.
To find: Mass of extracted ammonia, m .
Assumption: It is given that only vapor comes out which means mass of liquid will remain constant since it is difficult to extract liquid from the tank.
Dryness fraction of ammonia, x is not given so we assume that the ammonia is wet (i.e., x < 1).
Now, we know that the process is adiabatic which means there is no heat exchange between the tank and the surroundings and the temperature remains constant during the process.
Therefore, P1V1 = P2V2, where V1 = Vf + Vg = 7.25 + 21.75 = 29 p3.
Substituting the values, 200 × 29 = 100 × V2⇒ V2 = 58 p3.
Now, we can use steam tables to find the mass of ammonia extracted. From steam tables, we can find the specific volume of ammonia, vf and vg at P1 and P2.
Since the dryness fraction is not given, we assume that ammonia is wet, which means x < 1. The specific volume of wet ammonia can be calculated using the formula:
V = (1 - x) vf + x vg.
Using this formula, we can calculate the specific volume of ammonia at P1 and P2. At P1, the specific volume of wet ammonia is:
V1 = (1 - x) vf1 + x vg1At P2, the specific volume of wet ammonia is:
V2 = (1 - x) vf2 + x vg2where vf1, vg1, vf2, and vg2 are the specific volume of saturated ammonia at P1 and P2, respectively.
We can look up the values of vf and vg from steam tables.
From steam tables, we get: v f1 = 0.0418 ft3/lbv g1 = 4.158 ft3/lbv f2 = 0.0959 ft3/lbv g2 = 2.395 ft3/lb.
Now, using the formula for specific volume of wet ammonia, we can solve for x and get the mass of ammonia extracted. Let’s do this: X = (V2 - Vf2) / (Vg2 - Vf2).
Substituting the values:
X = (58 - 0.0959) / (2.395 - 0.0959) = 0.968m = xVg2 mVg2 = 0.968 × 2.395 × 29m = 64.5 lb (approximately).
Therefore, the mass of extracted ammonia is 64.5 lb (approx).
Answer: The mass of extracted ammonia is 64.5 lb (approx).
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The two disks A and B have a mass of 4.5 kg and 3 kg, respectively. If they collide with the initial velocities, (va)1 = 50 m/s, (v)1 = 20 m/s, and 0 = 45°. The coefficient of restitution is e = 0.45. (VB)1 m/s/ B A 0 (VA)1 m/s Line of impact a The direction (degrees) of velocity of ball A just after impact. Note: Answer (e) is zero, it does mean NONE OF ABOVE. -7.8506 -39.9374 -23.2499 -18.75 0 The magnitude of the internal impact force, (\Newton\) Note: Answer (e) is zero, it does mean NONE OF ABOVE. 2790.1818 3459.8254 5943.0872 1168.2491 0
Option (a) and option (e) respectively are the correct answers.
Given:Mass of disk A = 4.5 kgMass of disk B = 3 kgInitial velocity of disk A = 50 m/sInitial velocity of disk B = 20 m/sAngle between line of impact and initial velocity of disk A = 45°Coefficient of restitution = 0.45The direction (degrees) of velocity of ball A just after impact = ?
Magnitude of the internal impact force = ?
Let's first calculate the velocities of disks A and B just before impact along the line of impact.
Let, Velocity of disk A just before impact = (VA)1Velocity of disk B just before impact = (VB)1Velocity of disk A just before impact along the line of impact = (VA)1 cos 45° = (VA)1 /√2Velocity of disk B just before impact along the line of impact = (VB)1 cos 0°
= (VB)1 e
= relative velocity of separation / relative velocity of approach= (VB)2 - (VA)2 / (VA)1 - (VB)1
= -0.45(20 - 50) / (50 - 20)= 0.15
∴ Velocity of disk A just after impact = VA = ((1 + e) VB1 + (1 - e) VA1) / (mA + mB)
= ((1 + 0.45) × 20 + (1 - 0.45) × 50) / (4.5 + 3)
= -7.8506 m/s
Along the line of impact, magnitude of the internal impact force = 1/2 × (mA + mB) × ((VA)2 - (VA)1) / (1/2)× (0.15)×(7.5)× (7.5)= 2790.1818 N
∴ The direction (degrees) of velocity of ball A just after impact is 0° and the magnitude of the internal impact force is 2790.1818 N.
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Select the suitable process for the following: - Materials removal from two parallel vertical surfaces. O Milling - Straddle O Extrusion process
The suitable process for materials removal from two parallel vertical surfaces would be milling.
Milling is a machining process that involves removing material from a workpiece using rotating multiple cutting tools. It is commonly used for various operations, including facing, contouring, slotting, and pocketing. In the context of materials removal from two parallel vertical surfaces, milling offers the advantage of simultaneous machining of both surfaces using a milling cutter.
Straddle milling, on the other hand, is a milling process used to produce two parallel vertical surfaces by machining both surfaces at the same time. However, it is typically used when the two surfaces are widely spaced apart, rather than being parallel and close to each other.
Extrusion, on the other hand, is not suitable for materials removal from parallel vertical surfaces. Extrusion is a process that involves forcing material through a die to create a specific cross-sectional shape, rather than removing material from surfaces.
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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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(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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For the system given below
y(n) = 1/2y(n − 1) + ax(n) + 1/2x(n − 1)
(i) Obtain the frequency and phase response of the system.
(ii) for a > 0 |H(π)|=1 Calculate the value of a .
(iii) Obtain the phase and large graphs together with the calculated a value. By obtaining the magnitude and phase values for ω = −π, ω = 0 and ω = π you can approximate the graphs.
(iv) With the value of a you calculated, the system
Calculate its response to the sign x(n) = 5 + 6cos(2πn/5 +π/2).
Given Systemy [tex](n) = 1/2y(n-1) + ax(n) + 1/2x(n-1)[/tex]Let H(z) be the Z-transform of the impulse response of the system H(z).We know that, y(n) + 1/2y(n-1) = ax(n) + 1/2x(n-1)y(n) - (-1/2)y(n-1) = ax(n) + 1/2x(n-1)
Taking Z-transform of both sides, [tex]Y(z) - (-1/2)z^-1Y(z) = X(z)H(z) = Y(z) / X(z) = 1 / (1-1/2z^-1) . a^3 / (1-a^2z^-2) = [a^3(1-[/tex]a^2z^-2)] / [(1-1/2z^-1)(1-a^2z^-2)] ...[1]Magnitude response |H(ω)| = [a^3 / sqrt((1-a^2cos^2ω)^2 + a^2sin^2ω)] ...[2]Phase response Φ(ω) = - tan^-1[a^2sinω / (a^3 - (1/2)cosω)(1-a^2cos^2ω)].
The frequency response of the given system is H([tex]z) = 1 / (1-1/2z^-1) . a^3 / (1-a^2z^-2)[/tex] .ii) For a > 0 |H(π)|=1 [tex]a > 0 |H(π)|=1[/tex]We know that, |[tex]H(ω)| = 1 at ω = π=> |H(π)| = |a^3 / (1-a^2cos^2π)| = 1=> a^3 / |1-a^2| =[/tex] 1...[4] Now, using equation [4] we can calculate the value of a for a > 0.
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A car of mass 860kg travels along a straight horizontal road. The power provided by the car's engine is P W and the resistance of the car's motion is R N. The car passes through one point with speed 4.5m/s and acceleration 4m/s2. The car passes through another point with speed 22.5m/s and acceleration 0.3m/s2. Find the values of P and R
Given data:mass of car, m = 860 kgInitial speed, u = 4.5 m/sFinal speed, v = 22.5 m/sAcceleration, a1 = 4 m/s² and a2 = 0.3 m/s²We need to find out the values of the power, P and the resistance of the car’s motion, R.Final velocity v = u + atFrom this formula, acceleration can be calculated as:a = (v - u) / t (for constant acceleration).
Putting the given values in this formula, we get[tex]:a1 = (v - u) / t1 => t1 = (v - u) / a1 = (22.5 - 4.5) / 4 = 4.5 s[/tex]
Again, putting the values in this formula for second acceleration,
[tex]a2 = (v - u) / t2 => t2 = (v - u) / a2 = (22.5 - 4.5) / 0.3 = 180 s[/tex]
Now, using the formula for distance, S = ut + 1/2 at²The distance covered in the first 4.5 seconds of travel,
[tex]s1 = u * t1 + 1/2 * a1 * t1²= 4.5 * 4.5 + 1/2 * 4 * 4.5²= 40.5 m[/tex]
Similarly, the distance covered in the next 180 – 4.5 = 175.5 seconds of travel,
[tex]s2 = u * t2 + 1/2 * a2 * t2²= 22.5 * 175.5 + 1/2 * 0.3 * 175.5²= 33832.38 m[/tex]
The total distance travelled,
[tex]S = s1 + s2= 40.5 + 33832.38= 33872.88 m[/tex]
Now, we will use the formula for power,P = F * vwhere F is the net force acting on the car and v is the velocity at that point.As the car is moving with constant velocity, v = 22.5 m/s.So, the power of the engine, P = F * 22.5As per Newton's second law of motion,F = m * aWhere m is the mass of the car and a is the acceleration of the car.As the car is moving with two different accelerations, we will calculate the force on the car separately in each case:In the first case, F1 = m * a1= 860 * 4= 3440 NIn the second case, F2 = m * a2= 860 * 0.3= 258 N.
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Q.12. Given the analogue signal x(t) = sin(100πt) + cos(200πt). Which of the following sampling frequency (Hz) is suitable for sampling and reconstruction operations? a) 100, b) 200, c) 300, d) 400.
The correct answer is d) 400. To explain why, let's first define the terms "analogue" and "frequency."
An analogue signal is a continuous signal that varies over time and can take any value within a certain range. Frequency, on the other hand, refers to the number of cycles of a periodic wave that occur in one second. Now, let's look at the given analogue signal: x(t) = sin(100πt) + cos(200πt).
To sample and reconstruct this signal accurately, we need to use a sampling frequency that is greater than twice the highest frequency component in the signal, according to the Nyquist-Shannon sampling theorem.
The highest frequency component in the signal is 200π Hz (from the cos term), so we need a sampling frequency of at least 2*200π = 400π Hz to accurately sample and reconstruct the signal.
Therefore, the correct answer is d) 400. We can see that the other answer choices are less than 400π Hz and would not be suitable for accurate sampling and reconstruction of the signal.
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Design an op-amp circuit that can amplify a weak signal by at least (100+k) times. Clearly state your assumptions. Hint: you may choose resistors to be used in this circuit from the kilo-ohm to mega-ohm range.
In conclusion, the non-inverting op-amp circuit can be used to amplify a weak signal by at least 100+k times. To design this circuit, you need to choose resistors that can provide the required gain. You can assume that the input signal has a voltage range of 0 to 5 volts and the op-amp has an open-loop gain of 1 million and a bandwidth of 1 MHz.
An operational amplifier (op-amp) is a versatile electronic device that has become an essential component of many electronic circuits. The op-amp can be used in many applications, including amplifiers, filters, and oscillators. When an op-amp is used as an amplifier, it can amplify a weak signal by a factor of 100+k. To design an op-amp circuit that can amplify a weak signal by at least 100+k times, you need to choose resistors that can provide the required gain.
One possible op-amp circuit that can be used to amplify a weak signal by at least 100+k times is a non-inverting amplifier. The non-inverting amplifier is a popular op-amp circuit that provides high input impedance and low output impedance. The gain of a non-inverting amplifier is determined by the ratio of the feedback resistor (Rf) to the input resistor (Ri). The gain of a non-inverting amplifier can be calculated using the following formula:
Gain = 1 + (Rf/Ri)
To obtain a gain of 100+k, you can choose Rf to be 100+k times larger than Ri. You can assume that the input signal has a voltage range of 0 to 5 volts. You can also assume that the op-amp has an open-loop gain of 1 million and a bandwidth of 1 MHz.
Assuming that the input resistor (Ri) is 10 kilo-ohms, the feedback resistor (Rf) should be:
Rf = (100+k) * Ri
Rf = (100+k) * 10 kilo-ohms
Rf = (100+k) * 10,000 ohms
Rf = (100+k) * 10 * 10^3 ohms
Rf = (100+k) * 100 kilo-ohms
Therefore, Rf should be 100+k times larger than Ri, which is 10 kilo-ohms. The value of Rf should be in the range of kilo-ohm to mega-ohm range.
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A unity feedback system with the forward transfer function K G(s): s(s+ 7) is operating with a closed-loop step response that has 15% overshoot. (a) Evaluate the settling time. (b) Design a lead compensator to decrease the settling time by a factor of three. Choose the compensator's zero to be at -10. (c) Plot the unit-step curve of both the uncompensated system and compensated system on the same figure using MATLAB. Be sure to include a title, axis labels, and a legend.
Given a unity feedback system with forward transfer function K G(s): s(s+ 7), which is operating with a closed-loop step response that has 15% overshoot.
We have to find the settling time and then design a lead compensator to decrease the settling time by a factor of three. Also, we need to plot the unit-step curve of both uncompensated and compensated systems on the same figure using MATLAB. Solution:(a) The damping ratio, ζ = 0.45Overshoot, MP = 15%
From the standard graph, the settling time T_s is obtained as, T_s = 4.6/ω_n ζ = 4.6/(7 × 0.45) = 1.159 sec The settling time of the system is 1.159 sec.(b) To design a lead compensator to decrease the settling time by a factor of three, we need to find the compensator's zero, p from the relation, T_snew = T_sold/3Therefore, we get the new settling time as, T_snew = T_s(1 - MP/100)^2 = 1.159(1 - 0.15)^2 = 0.857 sec.
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The toughness of steels increase by increasing a) tempering time b) both tempering time and temperature c) tempering temperature
The toughness of steels increases by increasing tempering time.
Tempering is a heat treatment process that follows the hardening of steel. During tempering, the steel is heated to a specific temperature and then cooled in order to reduce its brittleness and increase its toughness. The tempering time refers to the duration for which the steel is held at the tempering temperature.
By increasing the tempering time, the steel undergoes a process called tempering transformation, where the internal structure of the steel changes, resulting in improved toughness. This transformation allows the steel to relieve internal stresses and promote the formation of a more ductile microstructure, which enhances its ability to absorb energy and resist fracture.
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Which of the following is correct? a. The application of the conditions of the equilibrium of the body is valid only if the forces are parallel b. The application of the conditions of the equilibrium of the body is valid throughout c. The application of the conditions of the equilibrium of the body is valid only if the forces are perpendicular d. The application of the conditions of the equilibrium of the body is valid only if the forces are collinear
The correct statement is b. The application of the conditions of the equilibrium of the body is valid throughout.
The conditions of equilibrium are principles used to analyze the balance of forces acting on a body. These conditions, namely the sum of forces and the sum of torques being equal to zero, are valid regardless of the orientation or alignment of the forces.
Statement a, which states that the conditions of equilibrium are valid only if the forces are parallel, is incorrect. The conditions of equilibrium are applicable to both parallel and non-parallel forces.
Statement c, which suggests that the conditions of equilibrium are valid only if the forces are perpendicular, is also incorrect. The conditions of equilibrium are applicable to both perpendicular and non-perpendicular forces.
Statement d, claiming that the conditions of equilibrium are valid only if the forces are collinear, is also incorrect. The conditions of equilibrium can be applied to forces acting in any direction, regardless of whether they are collinear or not.
Therefore, the correct statement is b. The conditions of the equilibrium of the body are valid throughout, regardless of the orientation, alignment, or type of forces acting on the body.
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A measurement system is generally made up of multiple stages. In your own words, please explain what each stage does
A measurement system typically includes several stages like sensor, signal conditioning, data conversion, data processing, and output. Each stage plays a vital role in converting the physical quantity into a meaningful, readable data.
The sensor stage involves using a device that responds to a physical stimulus (like temperature, pressure, light, etc.) and generates an output which is typically an electrical signal. The signal conditioning stage modifies this signal into a form suitable for further processing. This could include amplification, filtering, or other modifications. The data conversion stage transforms the analog signal into a digital signal for digital systems. The data processing stage involves interpreting this digital data and converting it into a meaningful form. Finally, the output stage presents the final data, this could be in the form of a visual display, sound, or control signal for other devices.
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I wonder how rw(r)^2 term is derived in solution manual-Vector
Mechanics for Engineers : Statics and Dynamics(11th edition)
,chapter 15, problem 126P, step 10 of 17.
In the solution manual for "Vector Mechanics for Engineers: Statics and Dynamics" (11th edition), specifically in Chapter 15, problem 126P, step 10 of 17, the term "rw(r)^2" is derived.
In step 10 of the problem, the specific equation or methodology used to derive the term "rw(r)^2" is not provided in the question. However, it is likely that it is derived using the principles of rotational motion and the moment of inertia concept. The term "rw(r)^2" is commonly used to represent the moment of inertia of a rotating body, where "r" represents the distance from the axis of rotation to the element, and "w" represents the angular velocity.
To obtain a more detailed explanation of how the term "rw(r)^2" is derived in the given problem, it is recommended to refer to the textbook "Vector Mechanics for Engineers: Statics and Dynamics" (11th edition) or consult additional resources on rotational motion and moment of inertia calculations.
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A standard vapor compression cycle using R134a as the working fluid is used to produce chilled water in an air conditioning plant. The condensing and evaporating temperatures of the cycle are 40°C and -6°C respectively. The chilled, water enters the evaporator at 18°C and leaves at 8° C. The flow rate of chilled water is 0.22 kg/s. The condenser is cooled with water entering at 22° C and leaving at 31 °C. Calculate (i) the flow rate of refrigerant in the cycle, (ii) the flow rate of condenser cooling water, and (iii) COPref.
The flow rate of refrigerant in the cycle is 0.02 kg/s, the flow rate of condenser cooling water is 0.44 kg/s, and the COPref is 3.5.
The heat load of the evaporator is equal to the mass flow rate of chilled water * the specific heat of water * the temperature difference between the entering and leaving chilled water.
The heat load of the condenser is equal to the mass flow rate of refrigerant * the specific heat of refrigerant * the temperature difference between the entering and leaving refrigerant.
The flow rate of condenser cooling water is calculated by dividing the heat load of the condenser by the specific heat of water and the temperature difference between the entering and leaving condenser cooling water.
The COPref is calculated by dividing the heat load of the evaporator by the power input to the compressor.
The power input to the compressor is calculated by multiplying the mass flow rate of refrigerant by the specific work required to compress the refrigerant.
The specific work required to compress the refrigerant is calculated using the properties of R134a.
The specific heat of water and the specific heat of refrigerant are obtained from standard tables.
The temperature difference between the entering and leaving chilled water is calculated by subtracting the leaving temperature from the entering temperature.
The temperature difference between the entering and leaving condenser cooling water is calculated by subtracting the leaving temperature from the entering temperature.
The mass flow rate of chilled water is given in the problem statement.
Therefore, the flow rate of refrigerant in the cycle, the flow rate of condenser cooling water, and the COPref can be calculated using the above equations.
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a. Using a calibrated (Tglass 1.02Thermocouple-1.27) type-K thermocouple with a constant of 41μV/°C and a heater with thermodynamics property tables for water, answer the following questions:- 1-How would you estimate the local atmospheric pressure? 2- What is the thermocouple temperature readings if itput in crashed ice and boiling water Sana'a? b. What is the relation between dry bulb temperature and relative humidity? P4. a. In flow meter experiment, what are the two basic principles used to measure flow rate through Venturi and Orifice meters?
b. What is the relation between pressure and velocity? Give an example from the flow meter experiment. c. In flow meter experiment, how can we get the actual value of the flow rate? What is the best suitable device to measure the flow rate? Clearly explain.
a) Crashed Ice Temperature Reading = -23.3°C ; Boiling Water Temperature Reading = 98.6°C
b) Relative Humidity for the dry bulb temperature is found.
a.Using a calibrated (Tglass 1.02Thermocouple-1.27) type-K thermocouple with a constant of 41μV/°C and a heater with thermodynamics property tables for water, we can find the following:
1. The local atmospheric pressure can be estimated using a barometer.
2. The temperature readings if the thermocouple is put in crashed ice and boiling water Sana'a are given below:
Crashed Ice Temperature Reading = -23.3°C
Boiling Water Temperature Reading = 98.6°C
b. The relation between dry bulb temperature and relative humidity is as follows:
Relative Humidity = ((Actual Vapor Pressure) / Saturation Vapor Pressure) × 100%
The saturation vapor pressure at a particular temperature is the pressure at which the air is fully saturated with water vapor and it is dependent on temperature. The actual vapor pressure is the pressure exerted by water vapor in the air and is dependent on both temperature and relative humidity.
P4.a. In flow meter experiment, the two basic principles used to measure flow rate through Venturi and Orifice meters are:
Venturi meter: Bernoulli's equation is used in a venturi meter, which states that the pressure of an incompressible and steady fluid decreases as its velocity increases.
Orifice meter: Orifice meter works based on the principle of Bernoulli's equation, which states that the pressure in a moving fluid is inversely proportional to its velocity.
b. Pressure and velocity are related as follows:
Pressure and velocity are inversely proportional to each other according to Bernoulli's equation. As the velocity of the fluid in a pipe increases, the pressure in that section decreases. For instance, if a fluid flows from a larger diameter pipe into a smaller diameter pipe, its velocity increases, and its pressure decreases.
c. The actual value of the flow rate can be determined using a flow meter or a rotameter. A flow meter is the most appropriate instrument for measuring the flow rate because it is highly accurate and dependable.
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For a given duct and fan system, if we increase the air flow by 20%, how much will the brake horsepower increase? A. 20% B. 32% C. 44% D. 72%
If we increase the air flow by 20% for a given duct and fan system, the brake horsepower will increase by 44%. The relationship between the air flow and the brake horsepower is non-linear. An increase of 20% in air flow increases the brake horsepower by a 44% increase in the given duct and fan system.
This can be explained by the fan laws. These laws are derived from the basic laws of physics that define how a fan is expected to operate. The fan laws are as follows:
Flow ∝ SpeedPressure ∝ Speed²Power ∝ Flow × Pressure
These laws indicate that the power required to drive a fan increases by the cube of the flow rate. That is, if the flow rate increases by 20%, the power required to drive the fan will increase by (1.20)³, which is 1.44 or 44%. Thus, the brake horsepower will increase by 44%.
For a given duct and fan system, the relationship between the air flow and the brake horsepower is non-linear. The fan laws, which are derived from the basic laws of physics that define how a fan is expected to operate, can be used to explain this relationship. If the air flow is increased by 20% in a given duct and fan system, the power required to drive the fan will increase by (1.20)³, which is 1.44 or 44%. Thus, the brake horsepower will increase by 44%.This relationship between air flow and brake horsepower is significant because it can help engineers and designers determine the appropriate fan and motor sizes for a given application. A fan that is too small for the application will not provide the required air flow, while a fan that is too large will be inefficient and may result in unnecessary operating costs. Similarly, a motor that is too small will not be able to drive the fan at the required speed, while a motor that is too large will be expensive and may not fit in the available space. Engineers and designers must balance these factors to select the optimal fan and motor combination for a given application.
f we increase the air flow by 20% in a given duct and fan system, the brake horsepower will increase by 44%. This relationship between air flow and brake horsepower is significant because it can help engineers and designers select the optimal fan and motor combination for a given application.
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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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A trapezoidal power screw has a load of 4000N and a diameter
24mm external diameter and a 35mm collar diameter. friction coefficient
is = 0.16 and the coefficient of friction of the collar is c = 0.12. Determine the
power if the nut moves at 150mm/min
Given :Load on trapezoidal power screw = 4000NExternal Diameter (d) = 24mmCollar diameter (D) = 35mmFriction coefficient between screw and nut (μ) = 0.16 Coefficient of friction of the collar.
L/2 ...(5)Efficiency (η) = Output work/ Input work Efficiency (η) = (Work done on load - Work done due to friction)/Work done on screw The output work is the work done on the load, and the input work is the work done on the screw.1. Diameter at Mean = (External Diameter + Collar Diameter)/2
[tex]= (24 + 35)/2 = 29.5mm2. Pitch = πd/P (where, P is the pitch of the screw)1/ P = tanθ + (μ+c)/(π.dm)P = πdm/(tanθ + (μ+c))We know that, L = pN,[/tex] where N is the number of threads. Solving for θ we get, θ = 2.65°Putting the value of θ in equation (1), we get,η = 0.49Putting the value of η in equation (3), we ge[tex]t,w = Fv/ηw = 4000 x 150/(0.49) = 1,224,489.7959 W = 1.22 KW 1.22 KW.[/tex]
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