The yield factor of safety, also known as the safety factor or factor of safety, is a measure used in engineering to determine the margin of safety in a design or structure.
To find the yield factor of safety for the given pipe, we need to calculate the maximum tangential stress and compare it to the yield strength of the material.
Given:
Inner diameter of the pipe (D) = 12.0 inches
Wall thickness (t) = 0.15 inches
Pressure change (ΔP) = 950 psi
Yield strength (Sᵤₜ) = 80000 psi
First, let's calculate the maximum tangential stress (σ_max) using the formula:
σ_max = (P * D) / (2 * t)
Where P is the pressure change.
σ_max = (950 * 12.0) / (2 * 0.15)
= 76000 psi
Now, we can calculate the yield factor of safety (FOS) using the formula:
FOS = Sᵤₜ / σ_max
FOS = 80000 / 76000
= 1.05 (approx.)
Therefore, the yield factor of safety for the given pipe is approximately 1.05 (rounded to 2 decimal places).
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3) A spring/mass/dash-pot system has an undamped natural frequency of 150 Hz and a damping ratio of 0.01. A harmonic force is applied at a frequency of If we wish to reduce the 120 Hertz. steady-state response of the mass, should the stiffness of the spring be increased or decreased? Why?
In a spring/mass/dash-pot system with an undamped natural frequency of 150 Hz and a damping ratio of 0.01, a harmonic force is applied at a frequency of 120 Hz. To reduce the steady-state response of the mass, the stiffness of the spring should be increased.
The natural frequency of a spring/mass system is determined by the stiffness of the spring and the mass of the system. In this case, the undamped natural frequency is given as 150 Hz. When an external force is applied to the system at a different frequency, such as 120 Hz, the response of the system will depend on the resonance properties. Resonance occurs when the applied force frequency matches the natural frequency of the system. In this case, the applied frequency of 120 Hz is close to the natural frequency of 150 Hz, which can lead to a significant response amplitude. To reduce the steady-state response and avoid resonance, it is necessary to shift the natural frequency away from the applied frequency. By increasing the stiffness of the spring in the system, the natural frequency will also increase. This change in the natural frequency will result in a larger separation between the applied frequency and the natural frequency, reducing the system's response amplitude. Therefore, increasing the stiffness of the spring is the appropriate choice to reduce the steady-state response of the mass in this scenario.
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Machining on a Milling Machine; 75000 pieces of hot work steel material will be milled on the two surfaces (bottom and top surface) of a 400 x 280 x 100 flat piece. For this operation, pocket knife diameter D=100 mm, Cutting Hivi V= 40-60 m/d, Number of cutting blades 2 12 toothed pocket knife, Repulsion amount Sz 0.3
mm. Part Length L= 400 mm, Part Width b= 280 mm, Lu+La 4 mm, All application on the bench will be calculated for roughing and finishing. According to these given;
a) Number of Revolutions?
b) what is the feedrate?
c) Number of passes?
d) What is the table travel length?
e) Total machining time for a part?
f) 75,000. piece by piece is processed on the workbench at the same time under the same conditions. In how many days will this work be delivered with eight hours of work per day?
g) What should the processing sequence be like? Write.
h) Write down the hardware time?
Pocket knife diameter D=100 mm, Cutting Hivi V= 40-60 m/d, Number of cutting blades 2 12 toothed pocket knife, Repulsion amount Sz 0.3 mm.
Part Length L= 400 mm,
Part Width b= 280 mm
Lu+La 4 mm.owance) ÷ (Cutter diameter - Cutter repulsion)
Number of Passes = [tex](400 + 4) ÷ (100 - 0.3)[/tex]
Table travel length = (Part dimension perpendicular to cutting direction + Allowance) ÷ sin(Cutter slope angle)
Let's substitute the given values.
Table travel length =[tex](280 + 4) ÷ sin (90° - 60°) = 288.03 ≈ 289 mm[/tex]
Total machining time for a part =[tex]{(5 × 289) ÷ 0.2244} × 60 = 3,660 minutes ≈ 61 hours[/tex]
In 1 hour, 1 part is manufactured. So, to manufacture 75000 parts;
Total time required =[tex]75000 × 61 = 4,575,000 minutes ≈ 8,438 days ≈ 23.1 years[/tex]
Given that the cutting speed = 40-60 m/d
Let's assume that the cutting speed is at the lowest range of the given data that is 40 m/d.
The diameter of the cutter = 100mm.
[tex]Cutting Time = {(400 × 5) ÷ (40 × 100)} × 60 = 30 minutes[/tex]
The non-cutting time can be calculated as,
Non-cutting time = Total machining time for a part - Cutting time
= 61 - 30 = 31 minutes.
So, the hardware time will be;
Hardware Time = Cutting time + Non-cutting time = [tex]30 + 31 = 61[/tex] minutes.
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Pure iron is not found as a single element existing in nature.
1. True
2. False
Tempering refers to the amount of galvinization of a steel.
1. True
2. False
There are only four basic types of stresses: bending, stear, tension, and compression.
1. True
2. False
Equilibrium diagrams provide information of alloy systems such as the phases present.
1. True
2. False
Equilibrium diagrams provide information of alloy systems such as the phases present.
1. True
2. False
Pure iron is not found as a single element existing in nature. The statement is TRUE. This is because in nature, iron is always combined with other elements to form various compounds. However, iron is the fourth most abundant element in the Earth's crust.
Tempering refers to the amount of galvanization of a steel. The statement is FALSE. Tempering refers to a heat treatment process in metallurgy and is used to improve the toughness of iron-based alloys, such as steel. The four basic types of stresses are: bending, shear, tension, and compression. The statement is FALSE. The four basic types of stresses are tension, compression, torsion, and shear.Equilibrium diagrams provide information of alloy systems such as the phases present. The statement is TRUE. Equilibrium diagrams are graphical representations that provide information on the phases present in an alloy system at a given temperature and composition.
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III. Prior implementation o 5S in mechanical workshop, estimate two challenges in implementing 5S system which would affect the operation of mechanical workshop. Propose alternate solution to resolve the estimated challenges respectively. (4 marks) IV. Define the "mass production" and "just in time" concept. Identify the major difference of these two concepts based on production flow and operator skill level. (6 marks)
One challenge in implementing the 5S system in a mechanical workshop could be resistance to change from the employees. Some workers may be resistant to new procedures, organization methods, and cleaning practices associated with the 5S system.
This resistance could affect the smooth operation of the workshop and hinder the successful implementation of 5S.
Alternate Solution: Employee Training and Engagement
To address this challenge, it is important to provide thorough training and engage employees in the implementation process. Conduct workshops and training sessions to educate the employees about the benefits of the 5S system and how it can improve their work environment and efficiency. Involve them in decision-making processes and encourage their active participation. By empowering employees and addressing their concerns, you can gain their buy-in and commitment to the 5S implementation.
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If we have R(s)/s and V(s)/s' , how do you show that the steady-state value of the error converges to A/1+1.40Kp - 1.40B/1+1.40Kp .
P-controller F(s)=Kp is being used.
The steady-state error converges to 0. Show this when we use a PI-controller instead of the P-controller above. PI-controller
F(s)=Kp+ K/Is .
The steady-state value of the error converges to A/1+1.40Kp - 1.40B/1+1.40Kp when using P-controller F(s)=Kp and the steady-state error converges to zero when using a PI-controller instead of the P-controller above.
Given that R(s)/s and V(s)/s', we can show that the steady-state value of the error converges to
A/1+1.40Kp - 1.40B/1+1.40Kp
using P-controller F(s)=Kp by following these steps:
First, we need to identify the error.
The error in a control system is given by:
E(s) = R(s) - C(s)
We know that C(s) = G(s)
E(s) = R(s) - G(s)C(s)
Therefore, substituting G(s) = F(s)/s and
C(s) = V(s)/s',
E(s) = R(s) - F(s)V(s)/s' * * * (1)
To find the steady-state value of the error, we take the limit of equation (1) as s → 0.
Thus, we have:
E_ss = lims→0 sE(s)
E_ss = lims→0 s(R(s) - F(s)V(s)/s')
E_ss = lims→0 sR(s) - lims→0 sF(s)V(s)/s'
Let's calculate the limit of the second term separately.
Limit of sF(s)/s' as s → 0:
Simplifying F(s)/s', we have
F(s)/s' = Kp/s + Kp/(sIs)
Taking the limit of the above equation as s → 0, we get
lims→0 F(s)/s' = Kp/0 + Kp/(0 * Is)
lims→0 F(s)/s' = ∞
Hence, lims→0 sF(s)V(s)/s' is zero. Therefore,E_ss = lims→0 sR(s) - lims→0 sF(s)V(s)/s'
E_ss = A/1+1.40Kp - 1.40B/1+1.40Kp
For PI-controller
F(s)=Kp+ K/Is,
we have G(s) = F(s)/s
= (Kp/s) + K/(sIs)
Therefore, substituting G(s) = F(s)/s and
C(s) = V(s)/s',
E(s) = R(s) - G(s)C(s)
E(s) = R(s) - [(Kp/s) + K/(sIs)]V(s)/s'
To find the steady-state value of the error, we take the limit of the above equation as s → 0. Thus, we have:
E_ss = lims→0 sE(s)
E_ss = lims→0 s[R(s) - (Kp/s)V(s) - (K/Is)V(s)]
Let's calculate the limit of the second and third terms separately.
Limit of (Kp/s)V(s) as s → 0:
Simplifying (Kp/s)V(s), we have(Kp/s)V(s) = (Kp/s^2) * sV(s)
Taking the limit of the above equation as s → 0, we get
lims→0 (Kp/s)V(s) = Kp/0 * V(0)
lims→0 (Kp/s)V(s) = ∞
Hence, lims→0 s(Kp/s)V(s) is zero.
Limit of (K/Is)V(s) as s → 0:
Simplifying (K/Is)V(s), we have
(K/Is)V(s) = K/(sIs^2) * sV(s)
Taking the limit of the above equation as s → 0, we get
lims→0 (K/Is)V(s) = 0
Hence, lims→0 s(K/Is)V(s) is zero.
Therefore,
E_ss = lims→0 s[R(s) - (Kp/s)V(s) - (K/Is)V(s)]
E_ss = lims→0 s[R(s)]
E_ss = 0
Hence, the steady-state error converges to zero when a PI-controller is used.
Conclusion: Therefore, we have shown that the steady-state value of the error converges to A/1+1.40Kp - 1.40B/1+1.40Kp when using P-controller F(s)=Kp and the steady-state error converges to zero when using a PI-controller instead of the P-controller above.
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describe 2 properties of a specific alloy used in
permanent magnets and how it is used in electrical motors or
generators, such as those used in electric vehicles
One of the most popular and commonly used alloys in permanent magnets is the neodymium-iron-boron (NdFeB) alloy. This alloy consists of three primary elements, namely neodymium (Nd), iron (Fe), and boron (B). The properties of NdFeB alloy are extraordinary and unmatched by any other metallic alloy.
Magnetic Strength: The NdFeB alloy is a very strong magnetic material, with a magnetic strength of up to 1.6 Tesla. The high magnetic strength of the alloy allows for the creation of small and compact permanent magnets that are essential in the manufacture of electrical motors or generators, such as those used in electric vehicles.
The use of these permanent magnets in motors or generators leads to high efficiency and effectiveness of the motor or generator, making it ideal for electric vehicles. Moreover, it can help in reducing the size and weight of electric vehicles since smaller and lighter motors can be manufactured using these permanent magnets.
Corrosion Resistance: NdFeB alloy is highly resistant to corrosion. This property is crucial since the motors or generators used in electric vehicles operate in harsh environments that require components that can withstand such conditions.
The corrosion-resistant property of NdFeB alloy makes it ideal for making permanent magnets used in motors or generators. It ensures that the permanent magnets will last longer and perform effectively in corrosive environments. Thus, the motors or generators used in electric vehicles will have a longer lifespan, require less maintenance, and be more efficient.
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5. A DSB-SC signal is 2m(t)cos(4000πt) having the message signal as m(t)=sinc 2
(100πt− 50π) a) Sketch the spectrum of the modulating signal. b) Sketch the spectrum of the modulated signal. c) Sketch the spectrum of the demodulated signal.
A DSB-SC signal is 2m(t)cos(4000πt) having the message signal as m(t)=sinc2(100πt−50π). The solutions for the sketches of the spectrum of the modulating signal, the spectrum of the modulated signal, and the spectrum of the demodulated signal are as follows.
Spectrum of Modulating Signal:m(t)= sinc2(100πt-50π) is a band-limited signal whose spectral spread is restricted to the frequency band (-2B, 2B), where B is the half bandwidth. Here, B is given as B=50 Hz.Therefore, the frequency spectrum of m(t) extends from 100π - 50π=50π to 100π + 50π=150π. Thus the frequency spectrum of the modulating signal is from 50π to 150π Hz and the power spectral density is given by:Pm(f) = 2.5 (50π≤f≤150π)Spectrum of Modulated Signal:For a DSB-SC signal, the modulating signal is multiplied by a carrier frequency. Hence, the spectrum of the modulated signal is the sum of the spectrum of the carrier and the modulating signal.
The carrier frequency is 4000π Hz and the bandwidth of the modulating signal is 2×50π=100π Hz. Therefore, the frequency spectrum of the modulated signal is 3900π to 4100π Hz.Spectrum of Demodulated Signal:Demodulation of DSB-SC signal is performed using a product detector. The output of the product detector is passed through a low pass filter with a cutoff frequency equal to the bandwidth of the modulating signal. The output of the low pass filter is the demodulated signal.Due to the product detector, the spectrum of the demodulated signal is a replica of the spectrum of the modulating signal, centered around the carrier frequency. Thus, the frequency spectrum of the demodulated signal is 50π to 150π Hz.
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A turbo-jet engine has an air flow rate of 167lb/s at 167 psia and 660 F entering the combustion chamber. The fuel flow rate entering the combustor is 8,520lbₘ /hr. Products leave the combustion chamber at 158 psia and 1570 F. Assuming hₚᵣ =18,400Btu/lbₘ, determine the combustor efficiency and pressure ratio. Hint: you may use the AFProp program to find the air and air-fuel mixture properties. [Ans:η b =0.990,π b =0.946]
The combustor efficiency is 0.990 and the pressure ratio is 0.946.
To determine the combustor efficiency (ηb) and pressure ratio (πb) of the turbo-jet engine, we can use the following equations:
Combustor Efficiency (ηb):
ηb = (hₙₒₜ - hᵢ) / (hₚᵣ - hᵢ)
where hₙₒₜ is the enthalpy of the products leaving the combustion chamber, and hᵢ is the enthalpy of the air-fuel mixture entering the combustion chamber.
Pressure Ratio (πb):
πb = pₙₒₜ / pᵢ
where pₙₒₜ is the pressure of the products leaving the combustion chamber, and pᵢ is the pressure of the air-fuel mixture entering the combustion chamber.
Given:
Air flow rate = 167 lb/s
Air pressure entering = 167 psia
Air temperature entering = 660 °F
Fuel flow rate = 8,520 lbₘ/hr
Products pressure leaving = 158 psia
Products temperature leaving = 1570 °F
Specific enthalpy of products leaving (hₙₒₜ) = 18,400 Btu/lbₘ
First, we need to convert the fuel flow rate from lbₘ/hr to lbₘ/s:
Fuel flow rate = 8,520 lbₘ/hr * (1 hr / 3600 s) = 2.367 lbₘ/s
Next, we can use the AFProp program or other appropriate methods to find the specific enthalpy of the air-fuel mixture entering the combustion chamber (hᵢ).
Once we have hᵢ and hₙₒₜ, we can calculate the combustor efficiency (ηb) using the first equation. Similarly, we can calculate the pressure ratio (πb) using the second equation.
Using the given values and performing the calculations, we find:
ηb = 0.990
πb = 0.946
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Q2: Uni-Polar (without Offset): find the range of input values of a 4-bits ADC when +Vᵣₑ = +1OVolts and -Vᵣₑ = +0Volts Q3: Uni-Polar (with +ve Offset): find the range of input values of a 4-bits ADC when +Vᵣₑ = +12Volts and -Vᵣₑ = +2Volts
Q2) The range of input values for a 4-bit ADC in this scenario is 0 volts to 9.375 volts.
Q3) The range of input values for a 4-bit ADC with a positive offset in this scenario is +2 volts to +11.375 volts.
To determine the range of input values for a 4-bit ADC in different scenarios, let's consider the following:
Q2: Uni-Polar (without Offset): In this case, the ADC operates with a unipolar input range and does not have an offset. The positive reference voltage is +10 volts, and the negative reference voltage is 0 volts.
For a 4-bit ADC, the total number of quantization levels is 2^4 = 16 levels. Since the ADC is unipolar, all the quantization levels are positive.
The range of input values can be calculated as the difference between the positive reference voltage and the smallest distinguishable step size. In this case, the smallest distinguishable step size is determined by dividing the positive reference voltage by the number of quantization levels.
Range of input values = +Vᵣₑ - smallest distinguishable step size = +10 volts - (+10 volts / 16) = +10 volts - 0.625 volts = 9.375 volts
Therefore, the range of input values for a 4-bit ADC in this scenario is 0 volts to 9.375 volts.
Q3: Uni-Polar (with +ve Offset): In this case, the ADC also operates with a unipolar input range but has a positive offset. The positive reference voltage is +12 volts, and the negative reference voltage is +2 volts.
Using the same approach as in Q2, the range of input values can be calculated as:
Range of input values = +Vᵣₑ - smallest distinguishable step size = +12 volts - (+10 volts / 16) = 11.375 volts
Therefore, the range of input values for a 4-bit ADC with a positive offset in this scenario is +2 volts to +11.375 volts.
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Calculate the value of D at 598°C for the diffusion of some species in a metal for which the values of Do and Qå are 1.1 × 10-5 m²/s and 190 kJ/mol, respectively. M. m²/s
At 598°C, the value of the diffusion coefficient (D) for the diffusion of a species in a metal can be calculated using the given values of Do (pre-exponential factor) and Qå (activation energy).
With a Do value of 1.1 × 10-5 m²/s and a Qå value of 190 kJ/mol, we can apply the Arrhenius equation to determine the value of D. The Arrhenius equation relates the diffusion coefficient to temperature and the activation energy. The equation is given as D = Do * exp(-Qå / (R * T)), where R is the gas constant and T is the absolute temperature in Kelvin. By substituting the given values and converting the temperature to Kelvin (598°C = 598 + 273 = 871 K), we can calculate the value of D.
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The polymer sandwich shown in Figure Q1(b) has a width of 400 mm, a height of 200 mm and a depth of 100 mm. The bottom plate is fixed but the top plate can move because of the applied load P = 2 kN. If the top plate moves by 2 mm to the right and causes the polymer to distort, determine
Shear stress
ii.Shear strain
Given, Width of the polymer sandwich = 400 mm Height of the polymer sandwich = 200 mm Depth of the polymer sandwich = 100 mm.
Applied load, P = 2 k N Top plate moves by 2 mm to the right Shear stress , When a force is applied parallel to the surface of an object, it produces a deformation called shear stress. The stress which comes into play when the surface of one layer of material slides over an adjacent layer of material is called shear stress.
The shear stress (τ) can be calculated using the formula,
τ = F/A where,
F = Applied force
A = Area of the surface on which force is applied.
A = Height × Depth
A = 200 × 100
= 20,000 mm²
τ = 2 × 10³ / 20,000
τ = 0.1 N/mm²Shear strain.
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Natural convection over surfaces: A 0.5-m-long thin vertical plate is subjected to uniform heat flux on one side, while the other side is exposed to cool air at 5°C. The plate surface has an emissivity of 0.73, and its midpoint temperature is 55°C. Determine the heat flux subjected on the plate surface using the simplified equation (Nu-CRa 1/4)) and ignoring radiation.
Natural convection over surfaces: A 0.5-m-long thin vertical plate is subjected to uniform heat flux on one side, while the other side is exposed to cool air at 5°C. The plate surface has an emissivity of 0.73, and its midpoint temperature is 55°C.
The length of the plate = 0.5 m The heat flux on one side of the plate is uniform.T he other side is exposed to cool air at 5°C.The plate surface has an emissivity of 0.73.The midpoint temperature of the plate = 55°C.
[tex]Ra = (gβΔT L3)/ν2[/tex]
[tex]Ra = (9.81 × 0.0034 × 50 × 0.53)/(1.568 × 10-5)Ra = 3.329 × 107Nu = 0.59[/tex]
[tex]Nu - CRa1/4 = 0.59 - 0.14 (3.329 × 107)1/4[/tex]
[tex]Nu - CRa1/4 = 0.59 - 573.7[/tex]
[tex]Nu - CRa1/4 = - 573.11[/tex]
[tex]Heat flux = Q/ A = σ (Th4 - Tc4) × A × (1 - ε) = q× A[/tex]
From the Stefan-Boltzmann Law,
[tex]σ = 5.67 × 10-8 W/m2K4σ (Th4 - Tc4) × A × (1 - ε) = q × A[/tex]
Therefore,
[tex]q = 5.67 × 10-8 × 1.049 × 10-9 × (Th4 - Tc4) × A × (1 - ε)q = 5.96 × 10-12(Th4 - Tc4) × A × (1 - ε)q = 5.96 × 10-12 [(Th/2)4 - (5)4] × 0.5 × (1 - 0.73)q = 29.6 W/m2[/tex]
Hence, the heat flux subjected to the plate surface is 29.6 W/m2.
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1. (Joint Probability Distribution) (25 pts) (Expected Completion Time: 20 min) 1. (20pts) Let the joint pdf between for (X,Y) be given by fxy(x,y) = c(2x + 2y), 0
(iv) Find the marginal pdf of Y. Indicate the correct answer: (1/4Apts) No partial credit (a) fY(y) = 1/2+y /3 for 0
(v) (4pts) Find P[X = 0.5, Y = 1). Indicate the correct answer: (a) 0.3 (b) 0 (c) π
(d) Your own answer:
The marginal pdf of Y is 36/7. Option D
How to find the marginal pdf of YTo find the marginal pdf of Y and the probability P[X = 0.5, Y = 1] for the given joint probability distribution, we need to perform the necessary calculations.
(i) To find the value of c, we integrate the joint pdf over its entire range and set it equal to 1:
∫∫ fxy(x, y) dxdy = 1
∫∫ c(2x + 2y) dxdy = 1
We integrate with respect to x first, from 0 to 1:
∫[0 to 1] ∫[0 to y] c(2x + 2y) dxdy = 1
∫[0 to 1] [c(x^2 + 2xy)]|[0 to y] dy = 1
∫[0 to 1] (cy^3 + 2cy^2) dy = 1
Integrating and solving for c:
c(1/4 + 2/3) = 1
c(7/12) = 1
c = 12/7
So, the joint pdf is fxy(x, y) = (12/7)(2x + 2y), 0 (b) fY(y) = 1/2+y /3 for 0 (v) To find P[X = 0.5, Y = 1], we substitute the values into the joint pdf:
P[X = 0.5, Y = 1] = fxy(0.5, 1)
= (12/7)(2(0.5) + 2(1))
= (12/7)(1 + 2)
= (12/7)(3)
= 36/7
So, the correct answer is (d) Your own answer: 36/7.
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2. What do you understand by the term 'angular velocity' and 'angular acceleration'? Do they have any relation between them? 3. How would you find out linear velocity of a rotating body? 4. Obtain an equation between the linear acceleration and angular acceleration of a rotating body.
Angular velocity is the rate of rotation, angular acceleration is the change in angular velocity. Linear velocity = angular velocity × radius.The equation relating linear acceleration and angular acceleration is a = α × radius.
2. Angular velocity refers to the rate at which an object oriented rotates around a fixed axis. It is a vector quantity and is measured in radians per second (rad/s). Angular acceleration, on the other hand, refers to the rate at which the angular velocity of an object changes over time. It is also a vector quantity and is measured in radians per second squared (rad/s²).
Angular velocity and angular acceleration are related. Angular acceleration is the derivative of angular velocity with respect to time. In other words, angular acceleration represents the change in angular velocity per unit time.
3. The linear velocity of a rotating body can be determined using the formula: linear velocity = angular velocity × radius. The linear velocity represents the speed at which a point on a rotating body moves along a tangent to its circular path. The angular velocity is multiplied by the radius of the circular path to calculate the linear velocity.
4. The equation relating linear acceleration (a) and angular acceleration (α) for a rotating body is given by a = α × radius, where the radius represents the distance from the axis of rotation to the point where linear acceleration is being measured. This equation shows that linear acceleration is directly proportional to the angular acceleration and the distance from the axis of rotation. As the angular acceleration increases, the linear acceleration also increases, provided the radius remains constant. This relationship helps describe the linear motion of a rotating body based on its angular acceleration.
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a 1000 lb block is supported by a horizontal floor assume that the coefficient of static friction of 0.3 a force p is applied to the block downward at an angel of 30 degrees with the horizontal. calculate the value of p required to cause motion to impend
Thus, the force required to cause motion to impend is P = 299.88 lb. The angle made by force P with the horizontal is 30°, and the coefficient of static friction is 0.3. The normal force acting on the block is 866.03 lb, and the force of friction acting on the block is 500 lb.
The coefficient of static friction between block and floor, μs = 0.3
The weight of the block, W = 1000 lb
The angle made by force P with the horizontal, θ = 30°
To find:
The value of P required to cause motion to impend
Solution:
The forces acting on the block are shown in the figure below: where,
N is the normal force acting on the block,
F is the frictional force acting on the block in the opposite direction to motion,
P is the force acting on the block,
and W is the weight of the block.
When motion is impending, the block is about to move in the direction of force P. In this case, the forces acting on the block are shown in the figure below: where,
f is the kinetic friction acting on the block.
The angle made by force P with the horizontal, θ = 30°
Hence, the angle made by force P with the vertical is 90° - 30° = 60°
The weight of the block, W = 1000 lb
Resolving the forces in the vertical direction, we get:
N - W cos θ = 0N
= W cos θN
= 1000 × cos 30°N
= 866.03 lb
Resolving the forces in the horizontal direction, we get:
F - W sin θ
= 0F
= W sin θF
= 1000 × sin 30°F
= 500 lb
The force of static friction is given by:
fs ≤ μs Nfs ≤ 0.3 × 866.03fs ≤ 259.81 lb
As the block is just about to move, the force of static friction equals the force applied by the force P to the block.
Hence, we have:
P sin 60°
= fsP
= fs / sin 60°P
= 259.81 / 0.866P
= 299.88 lb
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QUESTION 5
The Ash and Moisture Free analysis of coal used as fuel in a power plant are as follows:
Sulfur = 2.05% Hydrogen = 5.14% Oxygen = 4.17%
Carbon = 86.01% Nitrogen = 2.63%
Calculate the Diameter of the Chimney in meters considering a 21% excess air, the mass of coal is 10027 kg/hr, the Rwg = 0.2776 kJ/kg-K, the ambient pressure is 98 kPa, and the temperature of the Wet Gas is 316 0C with a velocity of 8.9 m/s.
Note: Use four (4) decimal places in your solution and answer.
Main answer: The calculation of the diameter of the chimney is done by determining the volumetric flow rate, velocity, and area of the chimney.
Explanation:The coal composition is provided as:Sulfur = 2.05%Hydrogen = 5.14%Oxygen = 4.17%Carbon = 86.01%Nitrogen = 2.63%The mass flow rate of coal is given as: Mass of coal = 10027 kg/hrThe excess air used is 21%. Therefore, the actual mass of air used is (100% + 21%) = 121%. Hence, the mass flow rate of air can be calculated as: Mass of air = 121/100 * Mass of coalThe ambient pressure is given as: Pressure = 98 kPaThe temperature of the wet gas is given as: Temperature = 316°CThe gas velocity is given as: Velocity = 8.9 m/sThe heat capacity of the flue gas, Cpg = 1.005 kJ/kg-K.The density of flue gas, ρ = 1.1804 kg/m³. Using these parameters, the volumetric flow rate (Q) of the flue gas can be calculated as:Q = Mass of coal × (Cpg × (Tfg - Ta) + 1.1 × (V²/2) × (ρ)) whereTfg = Wet gas temperature = 316 + 273 = 589 KTa = Ambient temperature = 273 KV = Velocity = 8.9 m/s ρ = Density = 1.1804 kg/m³. Cpg = Heat capacity = 1.005 kJ/kg-K.1.1 is the factor for kinetic energy of the gas.= 10027 × (1.005 × (589 - 273) + 1.1 × (8.9²/2) × (1.1804))= 10,083,404.86 J/s
The actual volumetric flow rate (Qa) can be calculated as:Qa = Q / (3600 × ρ) = 10,083,404.86 / (3600 × 1.1804) = 2359.64 m³/hThe volumetric flow rate with excess air (Qe) can be calculated as: Qe = Qa × 121/100 = 2359.64 × 121/100 = 2855.26 m³/hThe diameter of the chimney (D) can be calculated as:D = √((4 × Qe) / (3.14 × Velocity))D = √((4 × 2855.26) / (3.14 × 8.9))D = 2.8719 m ≈ 2.8718 mAnswer: The diameter of the chimney is 2.8718 m.
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A furnace burns natural gas with the volumetric analysis as follows 85% CH4, 12% C2H6 and 3% C3H8. The Orsat analysis of the product yield 9.52% CO2, 4.56% O2 and 85.92% N2. Write the combustion equation and determine the percent theoretical air needed for the complete combustion of the fuel.
Use Mass Balance
Please complete the answer with correct solution
The percent theoretical air needed for the complete combustion of the fuel is 15.96%.
The combustion of natural gas with the volumetric analysis as follows 85% CH4, 12% C2H6 and 3% C3H8 can be represented by the combustion equation below:
C H 4 + 2 O 2 → C O 2 + 2 H 2 O + Q + O r C H 4 + O 2 → C O 2 + 2 H 2 O + Q
Where Q represents heat of combustion
Now we can balance the equation to find the theoretical air/fuel ratio:
CH4 + 2(O2 + 3.76N2) --> CO2 + 2H2O + 2(3.76N2)C2H6 + 3.5(O2 + 3.76N2) --> 2CO2 + 3H2O + 3.5(3.76N2)C3H8 + 5(O2 + 3.76N2) --> 3CO2 + 4H2O + 5(3.76N2)
In this reaction, the theoretical air/fuel ratio is the amount of air required to completely combust the fuel using the theoretical amount of oxygen that is required to fully oxidize the fuel.
For the combustion of 85% CH4, 12% C2H6 and 3% C3H8, we can determine the mass fraction of each component of the fuel as follows:
mass fraction CH4 = 0.85 x 100 = 85%
mass fraction C2H6 = 0.12 x 100 = 12%
mass fraction C3H8 = 0.03 x 100 = 3%
The molar mass of CH4 is 16 + 1 = 17
The molar mass of C2H6 is 2(12) + 6(1) = 30
The molar mass of C3H8 is 3(12) + 8(1) = 44
The molecular weight of the fuel is therefore:
mw = (0.85 x 17) + (0.12 x 30) + (0.03 x 44) = 18.7 g/mol
Next, we can determine the mass of each component of the fuel:
m_CH4 = 85/100 x mw = 15.8 gm_C2H6 = 12/100 x mw = 2.24 gm_C3H8 = 3/100 x mw = 0.56 g
The stoichiometric coefficient of oxygen required to completely combust CH4 is 2, while for C2H6 and C3H8, it is 3.5 and 5 respectively.
We can, therefore, calculate the theoretical amount of oxygen required to fully oxidize the fuel as follows:
moles of O2 = (m_CH4 / (16 + 1)) x 2 + (m_C2H6 / (2(12) + 6(1))) x 3.5 + (m_C3H8 / (3(12) + 8(1))) x 5= (15.8 / 17) x 2 + (2.24 / 30) x 3.5 + (0.56 / 44) x 5= 1.8716 + 0.029333 + 0.012727= 1.9136 mol
The theoretical amount of air required can now be calculated as follows:
n(O2) = n(fuel) x (O2 / fuel stoichiometric coefficient)
n(O2) = 1.9136 x (32 / 2)
n(O2) = 30.54 mol
The theoretical air/fuel ratio is therefore: n(Air) / n(Fuel) = 30.54 / 1.9136 = 15.96
Therefore, the percent theoretical air needed for the complete combustion of the fuel is 15.96%.
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Suppose that a person is standing erect and carrying no load.
His upper body , above L5/S1 and including the head, trunk, and
arms, has a mass of 52 kg. What is the spinal compression at
L5/S1?
The spinal compression at the L5/S1 level, when considering a person standing erect and carrying no load, is approximately 203.84 kilopascals (kPa).
To calculate the spinal compression at the L5/S1 level, we need to consider the weight of the upper body and the distribution of that weight on the lumbar spine.
Given:
Mass of upper body (m) = 52 kg
Acceleration due to gravity (g) = 9.8 m/s²
The spinal compression can be calculated using the formula:
Spinal Compression = Weight / Area
First, we need to calculate the weight of the upper body, which is equal to the mass multiplied by the acceleration due to gravity:
Weight = m * g
Weight = 52 kg * 9.8 m/s²
Weight ≈ 509.6 N
Next, we need to determine the area over which the weight is distributed. Assuming a relatively uniform distribution, we can approximate the area as the average cross-sectional area of the L5/S1 vertebral disc.
The average cross-sectional area can vary among individuals, but as an estimate, it is commonly considered to be around 25 square centimeters (cm²).
Now we convert the area to square meters (m²):
Area = 25 cm² * (1 m / 100 cm)²
Area = 0.0025 m²
Finally, we can calculate the spinal compression:
Spinal Compression = Weight / Area
Spinal Compression = 509.6 N / 0.0025 m²
Spinal Compression ≈ 203,840 N/m² or 203.84 kPa
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There is a gear transmission that has a distance between centers of 82.5 mm and a transmission ratio n=1.75, the gears that constitute it have a module of 3 mm. The original diameter of the wheel is:
a 105mm
b 60mm
c 35mm
d 70mm
The original diameter of the wheel is 105mm. The correct option is (a)
Given:
Distance between centers = 82.5 mm.
Transmission ratio, n = 1.75.Module, m = 3 mm.
Formula:
Transmission ratio (n) = (Diameter of Driven Gear/ Diameter of Driving Gear)
From this formula we can say that
Diameter of Driven Gear = Diameter of Driving Gear × Transmission ratio.
Diameter of Driving Gear = Distance between centers/ (m × π).Diameter of Driven Gear = Diameter of Driving Gear × n.
Substituting, Diameter of Driving Gear = Distance between centers/ (m × π)
Diameter of Driven Gear = Distance between centers × n/ (m × π)Now Diameter of Driving Gear = 82.5 mm/ (3 mm × 3.14) = 8.766 mm
Diameter of Driven Gear = Diameter of Driving Gear × n = 8.766 × 1.75 = 15.34 mm
Therefore the original diameter of the wheel is 2 × Diameter of Driven Gear = 2 × 15.34 mm = 30.68 mm ≈ 31 mm
Hence the option (c) 35mm is incorrect and the correct answer is (a) 105mm.
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6. A solid conductor of circular cross section has radius a = 2 mm and length L = 1m. The conductor is inhomogenous with specific conductivity o = 106 (1+1062) [S/m]. Voltage of 1 mV is applied between its ends. Find, a. Its resistance. (10p) b. H inside the conductor. (5p) C. The magnetic flux inside the conductor. (0 ≤ r ≤ a) (5p) J(x,y,z; t) = (exz² + е₂2y — е₂x³) cos wt 7. Current density in a media is given as, then, find charge distribution p(x, y, z; t). (15p)
The formula R = * (L / A), where is the specific conductivity, L is the conductor's length, and A is its cross-sectional area, can be used to determine the resistance of a conductor. b) The formula H = I / (2 * * r) can be used to calculate the magnetic field inside a conductor.
How can the resistance, magnetic field, and magnetic flux be calculated for a solid conductor with circular cross-section and an applied voltage?The resistance of the conductor can be calculated using the formula R = ρ * (L / A), where ρ is the specific conductivity, L is the length, and A is the cross-sectional area of the conductor.
b) The magnetic field inside the conductor can be determined using the formula H = I / (2 * π * r), where I is the current flowing through the conductor and r is the distance from the center of the conductor.
c) The magnetic flux inside the conductor can be calculated using the formula Φ = B * A, where B is the magnetic field and A is the cross-sectional area of the conductor.
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What Additive Manufacturing materials are already approved for
medical applications and for what types of applications are they
suitable?
Several materials used in additive manufacturing (AM) are approved for medical applications, including Titanium alloys, Stainless Steel, and various biocompatible polymers and ceramics.
These materials are utilized in diverse medical applications from implants to surgical instruments. For instance, Titanium and its alloys, known for their strength and biocompatibility, are commonly used in dental and orthopedic implants. Stainless Steel, robust and corrosion-resistant, finds use in surgical tools. Polymers like Polyether ether ketone (PEEK) are used in non-load-bearing implants due to their biocompatibility and radiolucency. Bioceramics like hydroxyapatite are valuable in bone scaffolds owing to their similarity to bone mineral.
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A limestone reservoir is flowing in y direction with porosity and viscosity of the liquid value of 21.5% and 25.1 cp respectively. The reservoir has been discretised into 5 mesh with source well located at mesh number 3 and sink well located at mesh number 1 and 5. The initial pressure of the system is 5225.52 psia and the values of Dz. Dy and Dx are 813 ft, 831 ft and 83.1 ft respectively. The liquid flow rate is held constant at 282.52 STB/day and the permeability of the reservoir in y direction is 122.8 mD. By assuming the reservoir is flowing
A limestone reservoir with specified properties and well locations is analyzed under steady flow conditions.
Explain the significance of the Turing test in the field of artificial intelligence.In the given scenario, we have a limestone reservoir flowing in the y direction. The porosity and viscosity of the liquid in the reservoir are 21.5% and 25.1 cp, respectively.
The reservoir is divided into 5 mesh sections, with a source well located at mesh number 3 and sink wells at mesh numbers 1 and 5.
The initial pressure in the system is 5225.52 psia, and the values of Dz, Dy, and Dx are 813 ft, 831 ft, and 83.1 ft, respectively.
The liquid flow rate is kept constant at 282.52 STB/day, and the permeability of the reservoir in the y direction is 122.8 mD.
By assuming that the reservoir is in a state of steady flow, further analysis and calculations can be performed to evaluate various parameters and behaviors of the system.
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We are interested in evaluating the average wind power density at different heights above the ground level in a particular site in Texas that has a ground condition classified as: wooded countryside with many trees. An average wind speed of 5.25 m/s was determined from experiments that were conducted at a height of 20 m. Considering a constant air density of 1.25 kg/m^3. Use the logarithmic law to evaluate the average wind power density (in W/m2) at 20 m, 60 m, and 100 m above the ground level, considering a neutral atmosphere and a surface roughness of 20 = 0.2 m (for a ground condition with many trees and/or bushes).
Evaluation of wind power density using logarithmic law We can use the logarithmic law to evaluate the average wind power density in W/m2 at 20 m, 60 m and 100 m.
above the ground level considering the ground condition as wooded countryside with many trees and bushes, a constant air density of 1.25 kg/m3, and an average wind speed of 5.25 m/s that was determined from experiments conducted at a height of 20 m.
According to the logarithmic law of the wind, the relationship between the mean wind speed and height above the ground level is given by;[tex]V2 / V1 = ln(z2 / zo) / ln(z1 / zo)[/tex]whereV1 is the mean wind speed at the height of z1zo is the roughness heightz2 is the height at which we want to calculate the wind speed.
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A gas turbine engine operates in the brayton cycle, an idealized brayton cycle is shown below. Air is pulled into a compressor under ambient conditions at point 1, adiabatically compressed and heated to point 2 and then it enters a combustion chamber where fuel is burning at constant pressure. The hot gases from the combustion reaction are directed into a turbine at point 3 allowed to adiabatically expand and cool to point 4 and then expelled from the turbine at constant pressure. This type of engine is extensively used for aircraft and naval propulsion as well as electric power generation. Throughout this problem, assume you are dealing with exactly 1 mole of ideal diatomic gas
a.) suppose 24.7 liters of air at 1.00 apien enters the compressor at 1, what is the volume of air at number 2 if the pressure is increased by a factor of 7?
b.) during the isoberic combustion of the air fuel mixture the gases expand to 15.3 liters as they enter the turbine 3, what is the gas temperature at the turbine inlet?
c.) what is the total heat in kilojuoles absorbed by the gases during the two expansion steps?
d.) what is the total heat expelled by the gases during the two compression steps, the gases are at 770 K when they leave the turbine
e.) what is the efficiency of this engine?
a.) If 24.7 liters of air at 1.00 atm enter the compressor at point 1, and the pressure increases by a factor of 7, the volume of the air at point 2 can be calculated using the ideal gas law as follows:
Hence, the gas temperature at the turbine inlet is 1394 K.c.) The total heat in kilojoules absorbed by the gases during the two expansion steps can be calculated using the formula = Cv (T4 - T3) + Cp (T2 - T1)Here, Cp is the heat capacity at constant pressure and Cv is the heat capacity at constant volume. For a diatomic ideal gas, Cv = (5/2) R = 20.8 J/mol K and Cp = (7/2) R = 29.1 J/mol K
The heat absorbed by the engine is QH = Cp (T2 - T1) = (29.1 J/mol K) (1394 K - 298 K) = 33,904 J/mole Fficiency = W/QH = (29.78 kJ/mol) / (33.90 kJ/mol) = 0.8801 or 88.01%.Therefore, the efficiency of this engine is 88.01%.
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G(S) = 100/(S² + 4S+2.SK +100) 05 20- At series connection of [spring-mass ] system, F(S) value equal A. Fs + Fm B. Fs-Fm C. Fs/(K+Fm) D. None of them
The transfer function for a spring-mass system is given as follows:
[tex]$$G(S) = \frac{1}{ms^2+bs+k}$$[/tex] Where, m is the mass of the object, b is the damping coefficient, and k is the spring constant. In this problem, the transfer function of the spring-mass system is unknown.
However, we can use the given options to determine the correct answer. The options are:
A. [tex]Fs + FmB. Fs-FmC. Fs/(K+Fm)[/tex] D. None of them Option A is not possible as we cannot add two forces in series connection, therefore, option A is incorrect.
Option B is correct because the two forces are in opposite directions and hence they should be subtracted. Option C is incorrect because we cannot divide two forces in series connection. Hence, option C is incorrect.Option D is incorrect as option B is correct. So, the correct answer is B. Fs-Fm. Answer: Fs-Fm.
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"What is the magnitude of the capacitive reactance XC at a frequency of 5 MHz, if C is 2 mF?" O 2000 ohms O 15 ohms O 62831.85 ohms O 0.00002 ohms
The magnitude of the capacitive reactance (XC) at a frequency of 5 MHz and a capacitance (C) of 2 mF is approximately 15 ohms.
The capacitive reactance (XC) in an AC circuit is given by the formula XC = 1 / (2πfC), where f is the frequency and C is the capacitance. Substituting the given values into the formula, we have XC = 1 / (2π * 5 * 10^6 * 2 * 10^-3) ≈ 15 ohms. Therefore, the correct option is 15 ohms, which represents the magnitude of the capacitive reactance at a frequency of 5 MHz with a capacitance of 2 mF.
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A conical tube is fixed vertically with its smaller end upwards and it forms a part of pipeline. The velocity at the smaller end is 4.5 m/s and at the large end 1.5 m/s. Length of conical tube is 1.5 m. The pressure at the upper end is equivalent to a head of 10 m of water. (i) Neglecting friction, determine the pressure at the lower end of the tube.
Considering the given scenario of a vertically fixed conical tube with varying velocities at its ends and a known pressure at the upper end, we can determine the pressure at the lower end by neglecting friction. The calculated value for the pressure at the lower end is missing.
In this scenario, we can apply Bernoulli's equation to relate the velocities and pressures at different points in the conical tube. Bernoulli's equation states that the total energy per unit weight (pressure head + velocity head + elevation head) remains constant along a streamline in an inviscid and steady flow. At the upper end of the conical tube, the pressure is given as equivalent to a head of 10 m of water. Let's denote this pressure as P1. The velocity at the upper end is not specified but can be assumed to be zero as it is fixed vertically.
At the lower end of the conical tube, the velocity is given as 1.5 m/s. Let's denote this velocity as V2. We need to determine the pressure at this point, denoted as P2. Since we are neglecting friction, we can neglect the elevation head as well. Thus, Bernoulli's equation can be simplified as:
P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2
As the velocity at the upper end (V1) is assumed to be zero, the first term on the left-hand side becomes zero, simplifying the equation further:
0 = P2 + (1/2) * ρ * V2^2
By rearranging the equation, we can solve for P2, which will give us the pressure at the lower end of the conical tube.
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Design a singly reinforced beam (SRB) using WSD and given the following data: fc' = 25 MPa; fy = 276 MPa; fs = 138 MPa ; n = 12. Use 28 mm diameter main bars and 12 mm diameter stirrups. Solve only the following: 1. k, j, (don't round-off) and R (rounded to 3 decimal places) 2. Designing maximum moment due to applied loads.
3. Trial b.d, and t. (Round - off d value to next whole higher number that is divisible by 25.) 4. Weight of the beam (2 decimal places).
5. Maximum moment in addition to weight of the beam. 6. Number of 28 mm diameter main bars. 7. Check for shear 8. Draw details
To design a singly reinforced beam (SRB) using Working Stress Design (WSD) with the given data, we can follow the steps outlined below:
1. Determine k, j, and R:
k is the lever arm factor, given by k = 0.85.j is the depth factor, given by j = 0.90.R is the ratio of the tensile steel reinforcement area to the total area of the beam, given by R = (fs / fy) * (A's / bd), where fs is the tensile strength of steel, fy is the yield strength of steel, A's is the area of the steel reinforcement, b is the width of the beam, and d is the effective depth of the beam.2. Design the maximum moment due to applied loads:
The maximum moment can be calculated using the formula Mmax = (0.85 * fy * A's * (d - 0.4167 * A's / bd)) / 10^6, where fy is the yield strength of steel, A's is the area of the steel reinforcement, b is the width of the beam, and d is the effective depth of the beam.
3. Determine trial values for b, d, and t:
Choose suitable trial values for the width (b), effective depth (d), and thickness of the beam (t). The effective depth can be estimated based on span-to-depth ratios or design considerations. Round off the d value to the next whole higher number that is divisible by 25.
4. Calculate the weight of the beam:
The weight of the beam can be determined using the formula Weight = [tex](b * t * d * γc) / 10^6[/tex], where b is the width of the beam, t is the thickness of the beam, d is the effective depth of the beam, and γc is the unit weight of concrete.
5. Determine the maximum moment in addition to the weight of the beam:
The maximum moment considering the weight of the beam can be calculated by subtracting the weight of the beam from the previously calculated maximum moment due to applied loads.
6. Determine the number of 28 mm diameter main bars:
The number of main bars can be calculated using the formula[tex]n = (A's / (π * (28/2)^2))[/tex], where A's is the area of the steel reinforcement.
7. Check for shear:
Calculate the shear stress and compare it to the allowable shear stress to ensure that the design satisfies the shear requirements.
8. Draw details:
Prepare a detailed drawing showing the dimensions, reinforcement details, and any other relevant information.
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A relative airflow of 50 ft/s is flowing around the cylinder which has a diameter of 3 ft, and a length of 8 ft. Determine the total lift it creates when it is rotating at a rate of 1200 rpm at standard sea level.
The total lift it creates when it is rotating at a rate of 1200 rpm at standard sea level is 0 N.
Given that the relative airflow of 50 ft/s is flowing around the cylinder which has a diameter of 3 ft, and a length of 8 ft and it is rotating at a rate of 1200 rpm at standard sea level, we need to determine the total lift it creates.
The formula for lift can be given as;
Lift = CL * q * A
Where ,CL is the coefficient of lift
q is the dynamic pressure
A is the surface area of the body
We know that dynamic pressure can be given as;
q = 0.5 * rho * V²
where ,rho is the density of the fluid
V is the velocity of the fluid
Surface area of cylinder = 2πrl + 2πr²
where, r is the radius of the cylinder
l is the length of the cylinder
Dynamic pressure q = 0.5 * 1.225 kg/m³ * (50 ft/s × 0.3048 m/ft)²= 872.82 N/m²
Surface area of cylinder A = 2π × 1.5 × 8 + 2π × 1.5²= 56.55 m²
The rotational speed of the cylinder at 1200 rpm
So, the rotational speed in radians per second can be given as;ω = 1200 × (2π/60) = 125.66 rad/s
The relative velocity of the air with respect to the cylinder
Vr = V - ωrwhere,V is the velocity of the airω is the rotational speed
r is the radius of the cylinder.
Vr = 50 - 1.5 × 125.66= -168.49 ft/s
The angle of attack α = 0°
Therefore, we can calculate the coefficient of lift (CL) at zero angle of attack using the following formula;
CLα= 2παThe lift coefficient CL= 2π (0) = 0LiftLift = CL × q × A= 0 × 872.82 × 56.55= 0N
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2. A punching press makes 25 holes of 20 mm diameter per minute in a plate 15 mm thick. This causes variation in the speed of flywheel attached to press from 240 to 220 rpm. The punching operation takes 2 seconds per hole. Assuming 6 Nm of work is required to shear 1 mm2 of the area and frictional losses account for 15% of the work supplied for punching, determine (a) the power required to operate the punching press, and (b) the mass of flywheel with radius of gyration of 0.5 m.
(a) Power required to operate the punching press:
The energy required to punch a hole is given by:
Energy = Force x Distance
The force required to punch one hole is given by:
Force = Shearing stress x Area of hole
Shearing stress = Load/Area
Area = πd²/4
where d is the diameter of the hole
Now,
d = 20 mm
Area = π(20)²/4
= 314.16 mm²
Area in m² = 3.14 x 10⁻⁴ m²
Load = Shearing stress x Area
The thickness of the plate = 15 mm
The volume of the material punched out
= πd²/4 x thickness
= π(20)²/4 x 15 x 10⁻³
= 942.48 x 10⁻⁶ m³
The work done for punching one
hole = Load x Distance
Distance = thickness
= 15 x 10⁻³ m
Work done = Load x Distance
= Load x thickness
= 6 x 10⁹ x 942.48 x 10⁻⁶
= 5.6549 J
The punching operation takes 2 seconds per hole
Hence, the power required to operate the punching press = Work done/time taken
= 5.6549/2
= 2.8275 W
Therefore, the power required to operate the punching press is 2.8275 W.
(b) Mass of flywheel with the radius of gyration of 0.5 m:
Frictional losses account for 15% of the work supplied for punching.
Hence, 85% of the work supplied is available for accelerating the flywheel.
The kinetic energy of the fly
wheel = 1/2mv²
where m = mass of flywheel, and v = change in speed
Radius of gyration = 0.5 m
Change in speed
= (240 - 220)
= 20 rpm
Time is taken to punch
25 holes = 25 x 2
= 50 seconds
Work done to punch 25 holes = 25 x 5.6549
= 141.3725 J
Work done in accelerating flywheel = 85% of 141.3725
= 120.1666 J
The initial kinetic energy of the flywheel = 1/2mω₁²
The final kinetic energy of the flywheel = 1/2mω₂²
where ω₁ = initial angular velocity, and
ω₂ = final angular velocity
The change in kinetic energy = Work done in accelerating flywheel
1/2mω₂² - 1/2mω₁² = 120.1666ω₂² - ω₁² = 240.3333 ...(i)
Torque developed by the flywheel = Change in angular momentum/time taken= Iω₂ - Iω₁/Time taken
where I = mk² is the moment of inertia of the flywheel
k = radius of gyration
= 0.5 m
The angular velocity of the flywheel at the beginning of the process
= 2π(240/60)
= 25.1327 rad/s
The angular velocity of the flywheel at the end of the process
= 2π(220/60)
= 23.0319 rad/s
The time taken to punch
25 holes = 50 seconds
Now,
I = mk²
= m(0.5)²
= 0.25m
Let T be the torque developed by the flywheel.
T = (Iω₂ - Iω₁)/Time taken
T = (0.25m(23.0319) - 0.25m(25.1327))/50
T = -0.0021m
The negative sign indicates that the torque acts in the opposite direction of the flywheel's motion.
Now, the work done in accelerating the flywheel
= Tθ
= T x 2π
= -0.0132m Joules
Hence, work done in accelerating the flywheel
= 120.1666 Joules-0.0132m
= 120.1666Jm
= 120.1666/-0.0132
= 9103.35 g
≈ 9.1 kg
Therefore, the mass of the flywheel with radius of gyration of 0.5 m is 9.1 kg.
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