Answer:
V2= 21m³/h
Explanation:
According to Boyle's law, pressure and flow rate of gas can be calculated from the following equation:
P1V1=P2V2
3 × 35 = 5 × V2
V2= 21m³/h
A wheel tractor is operating in it is fourth gear range with full rated revolution per minute. Tractors speed is 7.00 miles per hour. Ambien air temperature is 60 Fahrenheit, and operating attitude is sea level. This tractor is towing a fill material loaded pneumatic trailer while climbing with 5 % slop. Rolling resistance is 55lb/ton. Tractor is single axle and it is operating weight is 74,946 lb. The loaded trailer weighs 55,000 lb. The weight distribution is for the combined tractor- trailer unit is 53% to the drive axle and 47% to the rear axle.
The manufacturer, for the environmental conditions as given above, rates the tractive effort of the new tractor at 330 rimpull HP. What percentage of this manufacturers rated rimpul HP actually develop?
(20 points) (Assessment of Outcome 1): A plant has two identical standby generator units for emergency use. In the area of the generators, the normal noise level registers 82 dBA on the sound-level meter with the generators turned off. When one generator switches on, the SLM needle jumps to 85.8 dBA. What will the dBA reading be when the second generator also turns on (so that both generators are on)
Answer:
It wouldn't get any louder then maybe 3db more
Explanation:
There's even a equation if you wanted to check this out but, if they are the same generator same model and all and made the same precise noise it wouldn't increase more then 3db.
A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The refrigerant enters the evaporator at 120 kPa with a quality of 34 percent and leaves the compressor at 70°C. If the compressor consumes 450 W of power, determine (a) the mass flow rate of the refrigerant, (b) the condenser pressure, and (c) the COP of the refrigerator
Answer:
(a) 0.0064 kg/s
(b) 800 KPa
(c) 2.03
Explanation:
The ideal vapor compression cycle consists of following processes:
Process 1-2 Isentropic compression in a compressor
Process 2-3 Constant-pressure heat rejection in a condenser
Process 3-4 Throttling in an expansion device
Process 4-1 Constant-pressure heat absorption in an evaporator
For state 4 (while entering compressor):
x₄ = 34% = 0.34
P₄ = 120 KPa
from saturated table:
h₄ = hf + x hfg = 22.4 KJ/kg + (0.34)(214.52 KJ/kg)
h₄ = 95.34 KJ/kg
For State 1 (Entering Compressor):
h₁ = hg at 120 KPa
h₁ = 236.99 KJ/kg
s₁ = sg at 120 KPa = 0.94789 KJ/kg.k
For State 3 (Entering Expansion Valve)
Since 3 - 4 is an isenthalpic process.
Therefore,
h₃ = h₄ = 95.34 KJ/kg
Since this state lies at liquid side of saturation line, therefore, h₃ must be hf. Hence from saturation table we find the pressure by interpolation.
P₃ = 800 KPa
For State 2 (Leaving Compressor)
Since, process 2-3 is at constant pressure. Therefore,
P₂ = P₃ = 800 KPa
T₂ = 70°C (given)
Saturation temperature at 800 KPa is 31.31°C, which is less than T₂. Thus, this is super heated state. From super heated property table:
h₂ = 306.9 KJ/kg
(a)
Compressor Power = m(h₂ - h₁)
where,
m = mass flow rate of refrigerant.
m = Compressor Power/(h₂ - h₁)
m = (0.450 KJ/s)/(306.9 KJ/kg - 236.99 KJ/kg)
m = 0.0064 kg/s
(b)
Condenser Pressure = P₂ = P₃ = 800 KPa
(c)
The COP of ideal vapor compression cycle is given as:
COP = (h₁ - h₄)/(h₂ - h₁)
COP = (236.99 - 95.34)/(306.9 - 236.99)
COP = 2.03
The Ph diagram is attached
A deep drawing operation is performed on a sheet-metal blank that is 1/8 in thick. The height of the cup = 3.8 in and its diameter = 5.0 in (both inside dimensions). (a) Assuming the punch radius = 0, compute the starting diameter of the blank to complete the operation with no material left in the flange. (b) Is the operation feasible (ignoring the fact that the punch radius is too small)?
Answer:
a) Db = 10.05 in
b) The operation is not feasible
Explanation:
Let's begin by listing out the given variables:
Thickness = 1/8 in, Height (h) = 3.8 in,
Diameter (Dp) = 5 in ⇒ rp = Dp ÷ 2 = 5 ÷ 2 = 2.5 in
a) Area of cup = Area of wall + Area of base
A = 2πh(rp) + π(rp)²
A = (2π * 2.5 * 3.8) + (π * 2.5²)
A = 59.69 + 19.635 = 79.325 in²
A ≈ 79.33 in²
But π(rb)² = 79.33 ⇒ rb² = 79.33 ÷ π
rb² = 25.25
rb = 5.025 ⇒ Db = 2 * rb = 2 * 5.025 = 10.05 in
Db = 10.05 in
b) To calculate for feasibility, we use the formula, draw ratio equals diameter of the blank divided by diameter of the punch
Mathematically,
DR = Db ÷ Dp
DR = 10.05 ÷ 5 = 2.01
DR = 2.01
DR > 2 ⇒ the operation is not feasible
For an operation to be feasible, it must have a drawing ratio limit of 2 or lesser
8. When supplying heated air for a building, one often chooses to mix in some fresh outside air with air that has been heated from the building as it passes through the furnace. An insulated mixing chamber is used to combine two streams of air to be used in a building. One stream of air, brought from the outside, enters at 2 kg/s, at a pressure of 120 kPa, and a temperature of 5oC. The second stream of air, coming from the building’s furnace, has a mass flow rate of 8 kg/s, a pressure of 120 kPa, and a temperature of 35oC. The combined stream is then delivered to the warm space at 120 kPa. Determine the rate of entropy generation for this mixing chamber.
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Air at 100°F, 1 atm, and 10% relative humidity enters an evaporative cooler operating at steady state. The volumetric flow rate of the incoming air is 1765 ft3/min. Liquid water at 68°F enters the cooler and fully evaporates. Moist air exits the cooler at 70°F, 1 atm. There is no significant heat transfer between the device and its surroundings and kinetic and potential energy effects can be neglected. Determine the mass flow rate at which liquid enters, in lb(water)/min.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
A thin‐walled tube with a diameter of 12 mm and length of 25 m is used to carry exhaust gas from a smoke stack to the laboratory in a nearby building for analysis. The gas enters the tube at 200°C and with a mass flow rate of 0.006 kg/s. Autumn winds at a temperature of 15°C blow directly across the tube at a velocity of 2.5 m/s. Assume the thermophysical properties of the exhaust gas are those of air
Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube.
Answer:
The average heat transfer coefficient for the exhaust gas flowing inside the tube, h = 204.41 W/m^2 - K
Explanation:
The detailed solution is attached as files below.
However, the steps followed are highlighted:
1) The average temperature was calculated as 380.5 K
2) The properties of air at 380.5 K was highlighted
3) The Prandti number was calculated. Pr = 0.693
4) The Reynold number was calculated, Re = 28716.77
5) The Nusselt umber was calculated, Nu = 75.94
6) From Nu = (hD)/k , the average heat transfer coefficient, h, was calculated and a value of 204.41 W/m^2 - K was gotten.
Your company has been retained to start steel erection for a project. The steel will be delivered on an 18-wheel flat-bed truck and then lifted into place. You will have a crane on site, a Grove GMK 5175 all-terrain crane and you will be lifting with ½ steel chains. The crane is a 175 Ton telescoping crane and the crane will be sitting on Type B soil. The steel delivery consists of 48 pieces of W18x65 structural steel, Grade A 36, broken down as follows and delivered in 12 bundles of 4 pieces each.
13 pieces - 25’ long
17 pieces -50’ long
18 pieces – 15’ long
Based on the above, answer the following questions:
a. Neglecting the weight of the truck, what is the total weight of the steel load?
b. Assuming that the steel will be bundled and lifted as follows, calculate the weight of each of the lifts
4 pieces of 15 feet long, W18x65 Weight_______________________
4 pieces of 50 feet long, W18x65 Weight_______________________
4 pieces of 25 feet long, W18x65 Weight_______________________
c. With the soil conditions noted above,
Will the crane be able to safely lift the heaviest lift?
i. Yes
ii. No
d. What will be the maximum amount or load that can be lifted by this crane using the boom length of 155 feet?
How does a car batteray NOT die?
Answer:
bye hooking plugs up to it to amp it up
If the symbol is on the top and bottom of the reference line, we call this __________ sides.
A. arrow
B. other
C. both
The amusement park ride consists of a fixed support near O, the 6-m arm OA, which rotates about the pivot at O, and the compartment, which remains horizontal by means of a mechanism at A. At a certain instant, β=30ο, 2 0.75 rad/s, and 0.5 rad/s , all clockwise. Determine the horizontal and vertical forces (F and N) exerted by the bench on the 75-kg rider at P. Compare your results with the static values of these forces. (Use x-y coordinate system and vector equations
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values to explain your result. Assume the water pressure is 30 psistr
Answer:
hoop stresslongitudinal stressmaterial usedall this could led to the failure of the garden hose and the tear along the length
Explanation:
For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :
hoop stress ( which acts along the circumference of the pipe):
αh = [tex]\frac{PD}{2T}[/tex] EQUATION 1
and Longitudinal stress ( acting along the length of the pipe )
αl = [tex]\frac{PD}{4T}[/tex] EQUATION 2
where p = water pressure inside the hose
d = diameter of hose, T = thickness of hose
we can as well attribute the failure of the hose to the material used in making the hose .
assume for a thin cylindrical pipe material used to be
[tex]\frac{D}{T}[/tex] ≥ 20
insert this value into equation 1
αh = [tex]\frac{20 *30}{2}[/tex] = 60/2 = 30 psi
the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose
What is Postflow used to protect?
Answer:
The idea is to protect the puddle while it cools
Explanation:
A steady green traffic light means
Answer:
Its C. you may proceed, but only if the path is clear
Explanation:
I just gave Quiz and its correct
The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for time (second) and the 2nd stands for force (Volt), respectively. This data were retrieved during an impact event. In this test, an impactor strikes to a sample. A force-ring sensor that attached to the impactor generates voltage during collision. A data acquisition card gathers the generated signals.
• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:
• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:
b) Compute the impulse of this event.
c) Obtain acceleration, velocity and displacement histories by using Newton’s second law of motion.
d) Compute the absorbed energy during collision.
Answer:
A.) 1mv = 2000N
B.) Impulse = 60Ns
C.) Acceleration = 66.67 m/s^2
Velocity = 4 m/s
Displacement = 0.075 metre
Absorbed energy = 60 J
Explanation:
A.) Using a mathematical linear equation,
Y = MX + C
Where M = (2000 - 0)/( 898 - 0 )
M = 2000/898
M = 2.23
Let Y = 2000 and X = 898
2000 = 2.23(898) + C
2000 = 2000 + C
C = 0
We can therefore conclude that
1 mV = 2000N
B.) Impulse is the product of force and time.
Also, impulse = momentum
Given that
Mass M = 30kg
Velocity V = 2 m/s
Impulse = M × V = momentum
Impulse = 30 × 2 = 60 Ns
C.) Force = mass × acceleration
F = ma
Substitute force and mass into the formula
2000 = 30a
Make a the subject of formula
a = 2000/30
acceleration a = 66.67 m/s^2
Since impulse = 60 Ns
From Newton 2nd law,
Force = rate of change in momentum
Where
change in momentum = -MV - (- MU)
Impulse = -MV + MU
Where U = initial velocity
60 = -60 + MU
30U = 120
U = 120/30
U = 4 m/s
Force = 2000N
Impulse = Ft
Substitute force and impulse to get time
60 = 2000t
t = 60/2000
t = 0.03 second
Using third equation of motion
V^2 = U^2 + 2as
Where S = displacement
4^2 = 2^2 + 2 × 66.67S
16 = 4 + 133.4S
133.4S = 10
S = 10/133.4
S = 0.075 metre
D.) Energy = 1/2 mV^2
Energy = 0.5 × 30 × 2^2
Energy = 15 × 4 = 60J
Determine the drag on a small circular disk of 0.02-ft diameter moving 0.01 ft/s through oil with a specific gravity of 0.89 and a viscosity 10000 times that of water. The disk is oriented normal to the upstream velocity. By what percent is the drag reduced if the disk is oriented parallel to the flow?
Answer:
33.3%
Explanation:
Given that:
specific gravity (SG) = 0.89
Diameter (D) = 0.01 ft/s
Density of oil [tex]\rho= SG\rho _{h20} = 0.89 * 1.94=1.7266\frac{sl}{ft^3}[/tex]
Since the viscosity 10000 times that of water, The reynold number [tex]R_E=\frac{\rho VD}{\mu} =\frac{1.7266*0.01*0.01}{0.234}=7.38*10^{-4}[/tex]
Since RE < 1, the drag coefficient for normal flow is given as: [tex]C_{D1}=\frac{24.4}{R_E}= \frac{20.4}{7.38*10^{-4}}=2.76*10^4[/tex]
the drag coefficient for parallel flow is given as: [tex]C_{D2}=\frac{13.6}{R_E}= \frac{13.6}{7.38*10^{-4}}=1.84*10^4[/tex]
Percent reduced = [tex]\frac{D_1-D_2}{D_2} *100=\frac{2.76-1.84}{3.3}=33.3[/tex] = 33.3%
Hot oil is to be cooled by water in a 1-shell-pass and 8-tube-passes heat exchanger. The tubes are thin-walled and are made of copper with an internal diameter of 1.4 cm. The length of each tube pass in the heat exchanger is 5 m, and the overall heat transfer coefficient is 310 W/m2.K. Water flows through the tubes at a rate of 0.2 kg/s, and the oil through the shell at a rate of 0.3 kg/is. The water and the oil enter at temperatures of 20 C and 150 C, respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.
Answer:
Rate of heat transfer is 66.8°C
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
g In the above water treatment facility, chemical concentration (mg/gal) within the tank can be considered uniform. The initial chemical concentration inside the tank was 0 mg/gal, the concentration of effluent coming in is 10 mg/gal. The volume of the tank is 10,000 gallons. The fluid coming in rate is equal to fluid going out is equal to 50 gal/min. Establish a dynamic model of how the concentration of the chemical inside the tank increases over time.
Answer:
0.05 mg / gallon
Explanation:
mass of chemecila coming in per minute = 50*10 = 500 mg/min
at a time t min , M = mass of chemical = 500*t mg
conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon
(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each? (c) In bullet format compare and contrast the expected mechanical behavior of hypoeutectoid and hypereutectoid steels in terms of: (i) Yield strength (ii) Ductility (iii) Hardness (iv) Tensile strength (d) If you want to choose an alloy to make a knife or ax blade would you recommend a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your recommendation in 1-3 bullet points. (e) If you wanted a steel that was easy to machine to make a die to press powders or stamp a softer metal, would you choose a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your choice in 1-3 bullets.
Answer:
See explanation below
Explanation:
Hypo-eutectoid steel has less than 0,8% of C in its composition.
It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.
Ferrite has a higher tensile strength than cementite but cementite is harder.
Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:
Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel
Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel
Hyper-eutectoid steel is harder than Hyper-eutectoid steel
Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.
When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because
1. It is harder
2. It has low cost
3. It is lighter
When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because
1. It is ductile
2. It has high tensile strength
3. It is durable
Answer:
(a)
Steels having carbon within 0.02% – 0.8% which consist of ferrite and pearlite are known as hypoeutectoid steel.
Steels having greater than 0.8% carbon but less than 2.0% are known as hypereutectoid steel.
(b)
The proeutectoid ferrite formed at a range of temperatures from austenite in the austenite+ferrite region above 726°C. The eutectoid ferrite formed during the eutectoid transformation as it cools below 726°C. It is a part of the pearlite microconstiutents . Note that both hypereutectoid and hypoeutectoid steels have proeutectoid phases, while in eutectoid steel, no proeutectoid phase is present.
Proeutectoid signifies is a phase that forms (on cooling) before the eutectoid austenite decomposes. It has a parallel with primary solids in that it is the first phase to crystallize out of the austenite phase. If the steel is hypoeutectoid it will produce proeutectoid ferrite and if it is hypereutectoid it will produce proeutectoid cementite. The carbon concentration for both ferrites is 0.022 wt% C.
(c)
(i) Yield strength: The hypoeutectoid steel have good yield strength and hypereutectoid steels have little higher yield strengh.
(ii) Ductility: The hypoeutectoid steel is more ductile and the ductility has decreased by a factor of three for the eutectoid alloy. In hypereutectoid alloys the additional, brittle cementite on the pearlite grain boundaries further decreases the ductility of the alloy. The proeutectoid cementite restricts plastic deformation to the ferrite lamellae in the pearlite.
(iii) Hardness: hypoeutectoid steels are softer and hypereutectoid steel contains low strength cementite at grain boundary region which makes it harder than hypoeutectoids.
(iv) Tensile strength: Grain boundary regions of hypereutectoid steel are high energy regions prone to cracking because of cementite in the grain boundaries, its tensile strength decreases drastically even though pearlite is present. Hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains, so grain boundaries being the high energy state region, it has a higher tensile strength.
(d)
I would recommend hypereutectoid steel alloy to make a knife or ax blade
1- Hardness is required at the surface of the blades.
2- Ductility is not needed for such application.
3- Due to constant impact, the material will not easily yield to stress.
(e)
I would choose a hypoeutectoid steel alloy to make a steel that was easy to machine.
1- hypoeutectoid steel alloys have high machinability, hence better productivity
2- It will be used on softer metals, hence its fitness for the application
3- Certain amount of ductility is required which hypoeutectoid steel alloys possess.
Explanation:
See all together above
Suppose you have a coworker who is a high Mach in your workplace. What could you do to counter the behavior of that individual? Put the high Mach individual in charge of a project by himself, and don’t let others work with him. Set up work projects for teams, rather than working one on one with the high Mach person. Work with the high Mach individual one on one, rather than in a team setting. Explain to the high Mach individual what is expected of him and ask him to agree to your terms.
Answer:
To counter the behavior of a high Mach individual in my workplace, I could put the individual in charge of a project by himself, and don't let others work with him.
Explanation:
A high Mach individual is one who exhibits a manipulative and self-centered behavior. The personality trait is characterized by the use of manipulation and persuasion to achieve power and results. But, such individuals are hard to be persuaded. They do not function well in team settings and asking them to agree to terms is very difficult. "The presence of Machiavellianism in an organisation has been positively correlated with counterproductive workplace behaviour and workplace deviance," according to wikipedia.com.
Mach is an abbreviation for Machiavellianism. Machiavellianism is referred to in psychology as a personality trait which sees a person so focused on their own interests that they will manipulate, deceive, and exploit others to achieve their selfish goals. It is one of the Dark Triad traits. The others are narcissism and psychopathy, which are very dangerous behaviors.
A sheet of steel 4.4 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 5.9 × 10^(-11) m^2/s, and the diffusion flux is found to be 4.7 × 10^(-7) kg/m^2.s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4.5 kg/m^3.
How far into the sheet from this high-pressure side will the concentration be 2.7 kg/m^3? Assume a linear concentration profile.
Answer:
0.544×10–³
Explanation:
Please see the attached file for the solution
Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rate of 0.264 kg/s. The maximum absolute temperature in the cycle is 1.15 times the minimum absolute temperature, and the net power input to the cycle is 5 kW. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the ratio of the maximum to minimum pressures in the cycle.
Answer:
7.15
Explanation:
Firstly, the COP of such heat pump must be measured that is,
[tex]COP_{HP}=\frac{T_H}{T_H-T_L}[/tex]
Therefore, the temperature relationship, [tex]T_H=1.15\;T_L[/tex]
Then, we should apply the values in the COP.
[tex]=\frac{1.15\;T_L}{1.15-1}[/tex]
[tex]=7.67[/tex]
The number of heat rejected by the heat pump must then be calculated.
[tex]Q_H=COP_{HP}\times W_{nst}[/tex]
[tex]=7.67\times5=38.35[/tex]
We must then calculate the refrigerant mass flow rate.
[tex]m=0.264\;kg/s[/tex]
[tex]q_H=\frac{Q_H}{m}[/tex]
[tex]=\frac{38.35}{0.264}=145.27[/tex]
The [tex]h_g[/tex] value is 145.27 and therefore the hot reservoir temperature is 64° C.
The pressure at 64 ° C is thus 1849.36 kPa by interpolation.
And, the lowest reservoir temperature must be calculated.
[tex]T_L=\frac{T_H}{1.15}[/tex]
[tex]=\frac{64+273}{1.15}=293.04[/tex]
[tex]=19.89\°C[/tex]
the lowest reservoir temperature = 258.703 kpa
So, the pressure ratio should be = 7.15
The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est
Question:
The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.
Answer:
See explanation below
Explanation:
Given:
d = 2m = 2*10³ = 2000
thickness, t = 10 mm
Length of strain guage = 20 mm
i) Let's calculate d/t
[tex] \frac{d}{t} = \frac{2000}{10} = 200 [/tex]
Since [tex] \frac{d}{t}[/tex] is greater than length of strain guage, the pressure vessel is thin.
For the minimum normal stress, we have:
[tex] \sigma max= \frac{pd}{4t} [/tex]
[tex] \sigma max= \frac{2000p}{4 * 20} [/tex]
= 50p
For the minimum normal strain due to pressure, we have:
[tex] E_max= \frac{change in L}{L_g} [/tex]
[tex] = \frac{0.012}{20} = 0.60*10^-^3[/tex]
The minimum normal stress for a thin pressure vessel is 0.
[tex] \sigma _min = 0 [/tex]
i) Let's use Hookes law to calculate the pressure causing this deformation.
[tex] E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)] [/tex]
Substituting figures, we have:
[tex] 0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)] [/tex]
[tex] 120 * 10^6 = 35p [/tex]
[tex] p = \frac{120*10^6}{35}[/tex]
[tex] p = 3.429 * 10^6 [/tex]
p = 3.4 MPa
ii) Calculating the maximum in-plane shear stress, we have:
[tex] \frac{\sigma _max - \sigma _int}{2}[/tex]
[tex] = \frac{50p - 50p}{2} = 0 [/tex]
Max in plane shear stress = 0
iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:
[tex] \frac{\sigma _max - \sigma _min}{2}[/tex]
[tex] = \frac{50p - 0}{2} = 25p [/tex]
since p = 3.429 MPa
25p = 25 * 3.4 MPa
= 85.71 ≈ 85.7 MPa
The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa
(a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2 (b) Show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs. (c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs
Answer:
Explanation:
a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2
We are to show how Two-2-to -1 multiplexers could be connected to form 3-to-1 MUX
If AB = 00 select [tex]I_o[/tex]
If AB = 01 select [tex]I_1[/tex]
If AB = 1_(B is don't care), select [tex]I_2[/tex]
However, the truth table is attached and shown in the first file below.
Also, the free- body diagram for 2- to - 1 MUX is shown in the second diagram attached below.
b) We are show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
The perfect illustration showing how they are connected in displayed in the third free-body diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
For four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs, we have a perfect illustration of the diagram in the last( which is the fourth) diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output
This question is a multiplexer which is a topic in digital circuit.
Multiplexer is a type of combination circuit that consist of a maximum of [tex]2^n[/tex] data inputs 'n' selection lines and single output line. One of these data inputs will be connected to the output based on the values of selection lines. Another name for multiplexers is MUX.
If we have 'n' selection lines, we will get [tex]2^n[/tex] possible combinations zero and ones. Each combination will select a maximum of only one data input.
a)
Two 2-to-1 multiplexers to form a 3-to-1 MUX.
If AB = 00, select [tex]I_o[/tex]
If AB = 01, select [tex]I_1[/tex]
If AB = 1- (B is don't care) select I
The truth table for the above scenario is in the attached document below.
Figure 1 and 2 represents the solution to this question.
b).
Two 4-to-1 multiplexers and one 2-to-1 multiplexers and one 2-to-1 multiplexers are used to form an 8-to-1 MUX.
In the attached diagram, figure 3 shows a comprehensive detail of how it is structured.
Where [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Four 2-to-1 multiplexers and one 4-to -1 multiplexer are used to form 8-to-1 MUX.
In the attached diagram, figure 4 shows how it is structured.
We would see that [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output in the system.
Learn more about multiplexers here;
https://brainly.com/question/25953942
Tech A says that when checking tire pressure, the tire should be " cold." Tech B says that the tires should be driven more than 3 miles before checking tire presure. Who is correct?
Answer: Technician A is correct.
Explanation:
Technician A is correct because temperature of a tire will affect its pressure reading.
Tires attract heat because of their dark colour and then motion on the road generates heat. A car owner or technician should know that tire pressure is most accurate when the tire is cold (especially when atmospheric temperature is cool too).
If the tires are driven three miles first, their temperature will be high (due to the rubbing of the tires on the surface of the road). This higher temperature will result in higher per square inch (psi) readings.
Temperature has a great influence on the tire pressure.
Even if the tire is driven up to or more than 3 miles, it should still be left to cool, before tire pressure is checked.
The tire manufacturer's rating should be the maximum possible tire pressure. If an abnormal reading is gotten, the gauge should be properly checked.
For what type of metal is high speed steel drill best suited?
Answer:
high speed steel I believe
a) The current that goes through a 100 mH inductor is given as
i(t) = 6 - 2e^-2t A t >= 0
Find the voltage v(t) across the inductor.
b) The voltage v(t) = 5sin(5t) V is applied across the terminals of a 200 mH inductor. The initial current through the inductor is i(0) = -10 A. Find the current i(t) through the inductor for t > 0.
Answer:
A) V(t) = 0.4e^-2t
B) i(t) = (25tsin5t+10) A for t>0
Explanation:
Formula for calculating voltage across an inductor is expressed as:
V = Ldi/dt
Given L = 100mH = 100×10^-3
If i(t) = 6 - 2e^-2t A t >= 0
di/dt = (-2)(-2)e^-2t
di/dt = 4e^-2t
If t ≥ 0
V(t) = 100×10^-3 × (4e^-2t)
V(t) = 0.1×4e^-2t
V(t) = 0.4e^-2t for t≥0
B) Applying the same formula as above
V = Ldi/dt
Vdt = Ldi
V/L dt = di
On integration
Vt/L = i + C
When t = 0, i = -10A
Substituting the values into the formula
V(0)/L = -10 + C
0 = -10+C
C = 10
To get the current i(t) through the inductor for t>0,
Since Vt/L = i + C
Given V(t) = 5sin5t Volts
L = 200mH = 200×10^-3H
C = 10
On substituting
(5sin5t)t/0.2 = i + 10
25tsin5t = i + 10
i(t) = (25tsin5t-10) A for t>0
The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for time (second) and the 2nd stands for force (Volt), respectively. This data were retrieved during an impact event. In this test, an impactor strikes to a sample. A force-ring sensor that attached to the impactor generates voltage during collision. A data acquisition card gathers the generated signals.
Answer:
A.) 1mv = 2000N
B.) Impulse = 60Ns
C.) Acceleration = 66.67 m/s^2
Velocity = 4 m/s
Displacement = 0.075 metre
Absorbed energy = 60 J
Explanation:
A.) Using a mathematical linear equation,
Y = MX + C
Where M = (2000 - 0)/( 898 - 0 )
M = 2000/898
M = 2.23
Let Y = 2000 and X = 898
2000 = 2.23(898) + C
2000 = 2000 + C
C = 0
We can therefore conclude that
1 mV = 2000N
B.) Impulse is the product of force and time.
Also, impulse = momentum
Given that
Mass M = 30kg
Velocity V = 2 m/s
Impulse = M × V = momentum
Impulse = 30 × 2 = 60 Ns
C.) Force = mass × acceleration
F = ma
Substitute force and mass into the formula
2000 = 30a
Make a the subject of formula
a = 2000/30
acceleration a = 66.67 m/s^2
Since impulse = 60 Ns
From Newton 2nd law,
Force = rate of change in momentum
Where
change in momentum = -MV - (- MU)
Impulse = -MV + MU
Where U = initial velocity
60 = -60 + MU
30U = 120
U = 120/30
U = 4 m/s
Force = 2000N
Impulse = Ft
Substitute force and impulse to get time
60 = 2000t
t = 60/2000
t = 0.03 second
Using third equation of motion
V^2 = U^2 + 2as
Where S = displacement
4^2 = 2^2 + 2 × 66.67S
16 = 4 + 133.4S
133.4S = 10
S = 10/133.4
S = 0.075 metre
D.) Energy = 1/2 mV^2
Energy = 0.5 × 30 × 2^2
Energy = 15 × 4 = 60J
Water enters the tubes of a cold plate at 70°F with an average velocity of 40 ft/min and leaves at 105°F. The diameter of the tubes is 0.25 in. Assuming 14 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation and the remaining 86 percent is removed by the cooling water, determine the amount of heat generated by the electronic devices mounted on the cold plate. The properties of water at room temperature are rho = 62.1 lbm/ft3 and cp = 1.00 Btu/lbm·°F.
Answer:
The total amount of heat generated;Q' = 2067 Btu/h
Explanation:
We are given;
Water entering temperature;T1 = 70°F
Water leaving temperature;T2 = 105°F
average velocity of water;V = 40 ft/min
Diameter of tube;D = 0.25 in = 0.25/12 ft = 0.02083 ft
Water density;ρ = 62.1 lbm/ft³
cp = 1.00 Btu/lbm·°F.
Now, the mass flow rate of the water is calculated from;
m' = ρAV
Where ρ is density, A is area and V is velocity
Area = πD²/4 = π*0.02083²/4 = 0.00034077555 ft²
m' = 62.1 * 0.00034077555 * 40
m' = 0.8465 lbm/min
Converting to lbm/hr = 0.8465 * 60 = 50.79 lbm/hr
From energy balance equation, we have;
E_in = E_out
So,
Q_in,w + m'h1 = m'h2
Q_in,w = m'h2 - m'h1
Q_in,w = m'(h2 - h1)
Now, m'(h2 - h1) can be written as;
m'cp(T2 - T1).
Thus ;
Q_in,w = m'cp(T2 - T1)
Plugging in the relevant values, we have;
Q_in,w = (50.79*1)(105 - 70)
Q_in,w = 1777.65 Btu/h
We are told that remaining 86 percent of heat generaged is removed by the cooling water. Thus;
The total amount of heat generated could be defined as;
Q' = Q_in,w/0.86
Q' = 1777.65/0.86
Q' = 2067 Btu/h
Discuss the ethics of the circumstances that resulted in the Columbia shuttle disaster. Considering the predictions that were made years before the disaster, as well as the reliability of the Binomial distribution and its implications, what could or should the engineers associated with the program have done differently
Explanation:
This is not so much a mathematical issue as a case study, because the response will inevitably require us to test the special Columbic shuttle disaster scenario. I would suggest that you read this in detail and present the points accordingly. Here I give as many points as I think are relevant.
The failure of a space program is definitely a complex situation, more than a simple binomial distribution. It's definitely not as simple as repeating the flip of a coin. There are several coherent factors and situations that govern the overall coordination and execution of such an event. The problem is, those who are running a project like this are still making a trade off,It is never the case that they sealed the lid on any chance of failure between multiple parameters. You try to do something, but often, as is the case above, the potentially dangerous situation is impossible or uncontrollable. Since the root cause of failure, which is dried out tiles that can not withstand heat and water, it appears that owing to the constant use of the shuttle the head architects have not foreseen this.
