Answer:
a) Solid solution strengthening and alloying, Precipitation hardening, work hardening
b) Absence of enough crystallographic misalignment in the grain boundary region for a small-angle
Explanation:
A) strengthening mechanism
i) Solid solution strengthening and alloying:
In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal
also the shear stress required can be represented as: Δz = Gb√Ce^3/2
where : C = solute concentration , e = strain on material
ii) Precipitation hardening:
During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments
particle bowing cab be written as : Δz = Gb / L-2x
iii) work hardening :
Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism
dislocation can be represented as ; Gb √ p
where : G = shear modulus , b = Burgess vector, p = dislocation density
B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle
What Is Soil Tunneling?
Answer:
A tunnel built in soft ground—such as clay, silt, sand, gravel or mud—requires specialized techniques compared to hard rock, to compensate for the shifting nature of the soil.
It's from web...
Soft ground tunneling describes the additional measures needed when Microtunneling through soil conditions that are vulnerable to collapse. ... This process ensures tunneling can happen effectively in soft grounds.
It's from me...
A test bar of nonferrous material has a diameter of 0.253 inches. Upon applying a tensile load, the sample exhibited 0.002 plastic strain at 3400 lb and the maximum load during testing was 6200 lb and occurred at an engineering strain of 0.65; and breaking occurred at 4400 lb. The sample diameter at fracture was measured to be 0.15 inches.
Required:
a. The yield strength of the material is :________
b. The UTS of the material is:________
Answer:
a. 67607.9psi
b. 123278.33
Explanation:
to get the yield strength of the material
= load/ cross sectional area
cross sectional area = π * 0.253²/4
= 0.0502927
The yield strenght
= 3400/0.0502927
= 67609.9 psi
b. the uts of the material
= maximum load/cross sectional area
= 6200/0.0502927
= 123278.33
A heat pump heats the air in a rigid, insulated cuboid room of size 25m x 10m x 4m. The heat pump consumes 15 kW of power. The initial temperature and pressure in this room are 12°C and 1 bar, respectively. With an average coefficient of performance of COPHP= 3.0 over the range of air temperature in this room.
Requried:
How long will it take to raise the temperature in the room to 27 °C?
Answer:
Time required = 287.2 secs
Explanation:
Volume of room = 25 * 10 * 4 = 1000 m^3
power consumed by pump = 15 kW
T1 ( initial temperature ) = 12°C
P1 ( Initial pressure ) = 1 bar
COPhp = 3
Calculate time taken to raise room Temp to 27°C
average heat supplied ( ∅ ) = COPhp * power consumed by pump
= 3 * 15 = 45 kW
Time required can be calculated using the relation below
∅t = P*V*Cv ( T2 - T1 ) [ p = 1.2 kg/m^3 , Cv = 0.718 KJ/kg ( air properties ) ]
45 * 10^3 ( t ) = 1.2*1000* 718 ( 27 - 12 )
∴ solving for t
t = 287.2 secs ≈ 4.79 mins
The AGC control voltage: ___________
a. varies as the signal strength of the received signal varies.
b. a negative feedback voltage.
c. is actually the dc voltage component produced by the mixing action in the AM demodulator stage.
d. is produced by an RC circuit having a much larger time constant than that of the detector.
e. all of the above
Answer:
The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.
Explanation:
The sample calculation for iron oxide in the IDEAS section of this experiment used known atomic weights to calculate an empirical formula. However, early chemists did not have any references in which they could look up atomic weights. Instead, they guessed at the formulas of compounds and measured the percent compositions of elements in compounds in order to calculate atomic weights. Calculate an atomic weight for iron using the hypothetical formula Fe101 and the composition data given in the example in the IDEAS section. You may assume the atomic weight of oxygen is known from other sources to be 16 amu.
Answer:
37.33 grams
Explanation:
The missing information embedded in the idea section is attached in the image below:
The aim of this question is to determine the atomic wt. of Iron (Fe) from the hypothetic formula:
Fe₁O₁
Here, we know that the mole ratio can be written as:
[tex]\dfrac{O}{Fe}=\dfrac{1}{1}[/tex]
Suppose we assume that the atomic wt. of Fe = β(unknown)???
Then the grams of O and Fe that is contained in Fe₁O₁ can be expressed as:
For O:
1 × 16 grams of Oxygen = 16 grams of O
For Fe:
1 × β grams of Fe = β grams of Fe
Now, let's take a look at the idea experiment, the mole solution can be computed as:
[tex]\dfrac{O}{Fe} = \dfrac{3}{2} \\ \\ \text{It implies that} \implies \dfrac{(3\times 16) \text{grams of O}}{(2 \times 56 ) \ \text{grams of Fe}}[/tex]
Equating both expressions above, we have:
[tex]\implies \dfrac{16}{ \beta} = \dfrac{3\times 16}{2\times 56}[/tex]
[tex]{ \beta} = \dfrac{(2\times 56)\times 16}{ 3\times 16}[/tex]
[tex]\mathbf{{ \beta} = 37.33 \ grams}[/tex]
From what year did 3.5G enter Vietnam?
Answer:
From what year did 3.5G enter Vietnam?
Explanation:
Vietnam does not have 3.5G network. MobiFone's first trial with 3.5G technology on this band was able to cover the whole waters of Vung Tau and Con Dao in August 2014.
Problem
In the clevis shown in Fig. find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P= 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi
Answer:
In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi.
127-clevis-double-shear-bolt.gif
Solution 127
Hide Click here to show or hide the solution
127-fbd-clevis-double-shear-bolt.gifFor shearing of rivets (double shear)
P=τA
14=12[2(14πd2)]
d=0.8618in → diameter of bolt answer
For bearing of yoke:
P=σbAb
14=20[2(0.8618t)]
t=0.4061in → thickness of yoke answer
A stream of ethylene glycol vapor at its normal boiling point and 1atm flowing at a rate of 175 kg/min is to be condensed at constant pressure. The product stream from the condenser is liquid g lycol at the condensation temperature.
a. Calculate the rate at which heat must be transferred from the condenser (kW).
b. If heat were transferred at a lower rate than that calculated in part (A), what would the state of the product stream be? (Dedu ce as much as you can about the phase and the temperature of the stream.)
c. If heat were transferred at a higher rate than that calculated in part (A), what could you deduce about the state of the product stream?
Answer: hello attached below is the question properly written
a) 2670 Kw
b) product will be made up of vapor and liquid
c) Product will be a super cooled liquid
Explanation:
mass Flow rate ( m ) = 175 kg/min
pressure = 1 atm
molecular weight of ethylene glycol ( mw ) = 62.07 g/mol
enthalpy of vaporization ( ΔHv ) = 56.9 KJ/mol
Using values from the table 8.1 related to the question
a) Determine the rate at which heat must be transferred from condenser
Using values from the table 8.1 related to the question
ΔH = 2670 Kw
b) If heat is transferred at a lower temperature the product will be made up of vapor and liquid
c) If heat was transferred at a higher temperature the product will be a super cooled liquid
Which of the following conditions would completely shut down a circuit
Here are the city gas mileages for 13 different midsized cars in 2008. 16, 15, 22, 21, 24, 19, 20, 20, 21, 27 , 18 , 21 , 48 What is the minimum ?
Answer:
Minimum city gas mileage is 15
Explanation:
Minimum city gas mileage among 13 different car sizes in 2008 is 15.
A circuit diagram for a lighting circuit is shown in Figure 6.
Figure 6
230 V AC
A
RL1
+
B T 12 V
04
4
Explain the function of the relay RL1 in the lighting circuit shown in Figure 6.
[2 marks)
Answer:
is there a picture of the figure?
The calculated value of the thermal conductivity of the carbon nano tube was found as: KCN = 3113 W/m-K, however, the theoretical value of the thermal conductivity of the wire is actually: K = 4500 W/m-K and the island separation is 5 μm (this is the actual spacing between the two islands). The difference between the measured and theoretical values is due to the contact resistance between the nano tube and the islands in the experiment.
Required:
a. Calculate the thermal contact resistance (Rtd) that exists between the carbon nano tube and the top surfaces of the heated and sensing islands.
b. Using the value of thermal contact resistance calculated in part A, calculate the fraction of the total resistance between the heated and sensing islands that is due to the thermal contact resistances for island separation distance of 5, 10, 15, and 20 μm.
Answer:
a) 1,607,973.9 K/W
b)
i) 0.3082 = 30.82%
ii) 0.1821 = 18.21%
iii) 0.1293 = 12.93%
iv) 0.1002 = 10.02%
Explanation:
Value of thermal conductivity ( calculated value ) KCN = 3113 W/m-k
Thermal conductivity ( theoretical value ) K = 4500 W/m-k
Island separation = 5 μm
a) Determine the thermal contact resistance
Resistance due to contact between carbon nano tube and top surfaces can be determined using the relation below
( I / A*K ) + 2Rc = ( l / A*KCN ) ------- ( 1 )
where ; I = 5 * 10^-6 m
A = π * ( 14 * 10^-9 )^2 m^2 = 153.93 * 10^-18 , K = 4500 , KCN = 3113
input values into equation 1 above
hence Rc = 1,607,973.9 K/W
b) Determine fraction of total resistance between heated and sensing
fraction of total resistance ; f1 = [tex]\frac{2 Rc}{I/KA + 2Rc}[/tex]
where : Rc = 1607973.9, K = 4500, A = 153.93 * 10^-18 ,
i) for I = 5 * 10^-6 m
fraction = 0.3082 = 30.82%
ii) for I = 10 * 10^-6 m
fraction = 0.1821 = 18.21%
iii) for I = 15 * 10^-6 m
fraction = 0.1293 = 12.93%
iv) for I = 20*10^-6
fraction = 0.1002 = 10.02%
In the figure below, this “double” nozzle discharges water (at 10°C, density= 1000 kg/m3) into the atmosphere at a rate of 0.50 m3/s. The pressure at the inlet is to be 315612 Pa. If the nozzle is lying in a horizontal plane. Jet A is 10 cm in diameter, jet B is 12 cm in diameter, and the pipe (1) is 30 cm in diameter. The x-component of force (Rx) acting through the flange bolts is required to hold the nozzle in place is:
Solution :
Given data :
p = 315612 Pa
[tex]$V_1=7.07 \ m/sec$[/tex]
At exit of B,
p = [tex]$P_{atm}$[/tex]
[tex]$V_B = 26.1 \ m/sec$[/tex]
At exit of A,
[tex]p=P_{atm}[/tex]
[tex]$V_{A} = 26.1 \ m/s$[/tex]
We need to determine X component of force ([tex]$R_x$[/tex]) to hold in its place.
From figure,
[tex]$\sum F_x = m_0'V_{0x} - m_iV_{ix} $[/tex]
[tex]$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$[/tex]
[tex]$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$[/tex]
Substitute all the values,
[tex]$=F_x=[-315612 \times \frac{\pi}{4}(0.3)^2 \sin 30]-[26.1 \times 1000 \times 26.1 \frac{\pi}{4}(0.1)^2]-[7.07 \times 1000\times 0.5 \sin 30]$[/tex][tex]$=F_x = -11154.64-5350.21-1767.28$[/tex]
[tex]$F_x = -18.2733 \ kN$[/tex]
Therefore, the force required to hold the nozzle in its place along horizontal direction.
[tex]$F_x = -18.2733 \ kN$[/tex]
Ma puteti ajuta cu un argument de 2 pagini despre inlocuirea garniturii de etansare de pe pistonul etrierului de franare la un autoturism ?
Answer:
can you translate
Explanation:
what Is that?
The steps for proper studing for the exam. -1 (use: First, Then, Next, After that, Finally)
Answer:
here is your answer
Explanation:
1. First observe the syllabus for all subjects.
2. Then gather all your books
3. For a particular exam read the book related to it ( if possible read early in the morning.
4. After that write the things you learnt by reading.
5. Finally give your exams by giving it all
HOPE IT HELPS......
OM NAMO SHIVAYE
An assembly line has 3 fail safe sensors and one emergency shutdown switch.The line should keep moving unless any of the following conditions arise:
(1) If the emergency switch is pressed
(2) If the senor1 and sensor2 are activated at the same time.
(3) If sensor 2 and sensor3 are activated at the same time.
(4) If all the sensors are activated at the same time
Suppose a combinational circuit for above case is to be implemented only with NAND Gates. How many minimum number of 2 input NAND gates are required.
Answer:
1 NAND gate
Explanation:
The minimum number of 2 input NAND gates that can be used to implement the combinational circuit = 1
The only true combinations conditions that can produce a false result ( i.e. condition/result different from the expected result as stated in the question )
Sensor 2 activated + Emergency switch pressed = False ( Line will keep moving )
To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 297(106)ft2. Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.
Required:
What is the area measurement, 293 (106) ft^2, in SI units?
This question is incomplete, the complete question is;
To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 293 × 10⁶ ft². Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.
Required:
What is the area measurement, 293 × 10⁶ ft², in SI units?
293 × 10⁶ ft² = ?km²
Answer:
the area measurement is 27.221 km²
Explanation:
Given the data in the question;
What is the area measurement, 293 × 10⁶ ft², in SI units
we are to the result of the measured area from ft² to km²
we know that;
1 meter = 3.2808 ft
1 km = 1000 m
1 ft = (1 / 3.2808)m
1 m = ( 1/1000 ) km
since our measured are is 293 × 10⁶ ft²
hence
A = 293 × 10⁶ × [ (1 / 3.2808)m ]²
A = 27221252.74 m²
A = 27221252.74 × [ ( 1/1000 ) km ]²
A = 27.221 km²
Therefore, the area measurement is 27.221 km²
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
78950W the answer
Explanation:
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles of the gas are present in the tank? What is the molecular weight of the gas? Assuming that the gas to be a pure element can you identify it?
Answer:
The molecular weight will be "28.12 g/mol".
Explanation:
The given values are:
Pressure,
P = 10 atm
= [tex]10\times 101325 \ Pa[/tex]
= [tex]1013250 \ Pa[/tex]
Temperature,
T = 298 K
Mass,
m = 11.5 Kg
Volume,
V = 1000 r
= [tex]1 \ m^3[/tex]
R = 8.3145 J/mol K
Now,
By using the ideal gas law, we get
⇒ [tex]PV=nRT[/tex]
o,
⇒ [tex]n=\frac{PV}{RT}[/tex]
By substituting the values, we get
[tex]=\frac{1013250\times 1}{8.3145\times 298}[/tex]
[tex]=408.94 \ moles[/tex]
As we know,
⇒ [tex]Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}[/tex]
or,
⇒ [tex]MW=\frac{m}{n}[/tex]
[tex]=\frac{11.5}{408.94}[/tex]
[tex]=0.02812 \ Kg/mol[/tex]
[tex]=28.12 \ g/mol[/tex]
It is essential to wait until the end of the project to check if the sponsor/customer requirements and expectations have been met regarding the quality of the project deliverables.
T/F
Answer:
False.
Explanation:
Project management can be defined as the process of designing, planning, developing, leading and execution of a project plan or activities using a set of skills, tools, knowledge, techniques and experience to achieve the set goals and objectives of creating a unique product or service.
Generally, projects are considered to be temporary because they usually have a start-time and an end-time to complete, execute or implement the project plan.
The fundamentals of Project Management includes;
1. Project initiation.
2. Project planning.
3. Project execution.
4. Monitoring and controlling of the project.
5. Adapting and closure of project.
Basically, the specifications of a project or manufacturing process outlines the minimum requirements and quality that are acceptable. Thus, it must be adhered to strictly in order to achieve a successful and desired outcome.
As a rule, it is essential to check at every stage of a project if the sponsor or customer requirements and expectations have been met regarding the quality of the project deliverables.
The pressure gage on a 2.5-m3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank (mass in kg) if the temperature is 28°C and the atmospheric pressure is 97 kPa.
Answer:
[tex]n=5.36kg[/tex]
Explanation:
From the question we are told that:
Volume [tex]V=2.5m^3[/tex]
Pressure[tex]\rho=500Kpa[/tex]
Temperature [tex]T=28^o[/tex]
Atmospheric pressure [tex]\rho_{atm} =97 kPa.[/tex]
Generally the equation for an Ideal gas is mathematically given by
[tex]PV=nRT[/tex]
Therefore
[tex]n=\frac{500*2.5}{8.314*28}[/tex]
[tex]n=5.36kg[/tex]
Request for proposal (RFP) is a type of document that contains the information and proposals mostly through the bidding process. This document is regarding the valuable assets, services, entity, commodity, etc.
Answer:
Answer to the following is as follows;
Explanation:
A request for proposal is a documentation that invites prospective contractors to submit business opportunities to an agency or corporation interested in procuring a commodities, product, or valuable resource through a bid procedure.
A request for proposal (RFP) is a commercial document that introduces a project, defines it, and invites eligible contractors to compete on its completion.
A pressure transducer has the following specifications: Input rage: 0-100 psi and the corresponding Output Range: 0-5 Volts. Linearity Error: 0.10% of the Reading Hysteresis Error: 0.10% of the Reading Sensitivity Error: 0.15% of the Reading Zero Drift Error: 0.20% of the Reading Its output is read via a voltmeter with instrument error of 0.10% of the reading and resolution of 0.01 V. If the applied pressure on the transducer is 65 psi, what is the design stage uncertainty of this pressure measurement system?
Answer:
0.287
Explanation:
Design-stage uncertainty can be expressed as :
Ud = √ Uo^2 + Uc^2 ------ ( 1 )
where : Uo = 1/2( resolution value ) = 1/2 * 0.01 V = 0.005 V
Uc = √(0.10)^2 + (0.10)^2 + (0.15)^2 + (0.20)^2 = 0.287
back to equation 1
Ud = √ ( 0.005)^2 + ( 0.287 )^2 = 0.287
Answer:
0.287
Explanation:
In a certain pressing operation, the metallic powder fed into the open die has a packing factor of 0.5. The pressing operation reduces the powders to 70% of their starting volume. In the subsequent sintering operation, shrinkage amounts to 10% on a volume basis. Given that these are the only factors that affect the structure of the finished part, determine its final porosity.
Answer:
0.2063
Explanation:
Given data:
packing factor = 0.5
percentage of reduction of powders = 70%
Calculate the final porosity
after sintering Bulk specific volume = 0.9 * 0.7 = 0.63
assuming true specific volume = 1
packing factor = 0.5 , bulk specific volume = 2
packing factor after pressing and sintering
= 1 / ( 2 * 0.63 ) = 0.7937
hence : porosity = 1 - packing factor
= 1 - 0.7937 = 0.2063
find the volume of the pond with the following dimension length 40m breadth 10m height 1.2m depth 0.9m express in both meters and feet
Answer:
The volume for this is 29.7
Explanation:
Trust me on this I'm an expert
quy trình sản xuất bao bì plastic dạng túi
A copper block receives heat from two different sources: 5 kW from a source at 1500 K and 3 kW from a source at 1000 K. It loses heat to atmosphere at 300 K. Assuming the block to be at steady state, determine (a) the net rate of heat transfer in kW; (b) the rate of entropy generation in the system's universe
Answer:
a) Zero
b) the rate of entropy generation in the system's universe = ds/dt = 0.2603 KW/K
Explanation:
a) In steady state
Net rate of Heat transfer = net rate of heat gain - net rate of heat lost
Hence, the rate of heat transfer = 0
b) In steady state, entropy generated
ds/dt = - [ Qgain/Th1 + Qgain/Th2 - Qlost/300 K]
Substituting the given values, we get –
ds/dt = -[5/1500 + 3/1000 – (5+3)/300]
ds/dt = - [0.0033 + 0.003 -0.2666]
ds/dt = 0.2603 KW/K
1. What is the maximum value of the linear density in a crystalline solid (linear density defined as the fraction of the line length occupied by atoms, assumed as spheres and only counted it their center is on the line)?
2. What family of directions has the highest linear density in the FCC system?
3. What family of directions has the highest linear density in the BCC system?
4. What family of planes has the highest planar density in the FCC system?
5. What family of planes has the highest planar density in the BCC system?
6. What family of planes has the highest planar density in the HCP sytem?
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Answer:
ay man ima be real, i just need the points yo
A confined aquifer with a transmissivity of 300 m2/day and a storativity of 0.0005 and a well radius of 0.3 m. Find the drawdown in the well at 100 days if the following pumping schedule is followed after a long period of time of no pumping.
Period
1 2 3 4
Time (days) 0-20 20-50 50-90 90-100
Q (m3/day) 500 300 800 0
Answer:
8.4627 m
Explanation:
Transmissivity( T ) = 300 m^2/day
Storativity( S ) = 0.0005
well radius ( r ) = 0.3m
Determine the drawdown in well at 100 days
Drawdown at 100 days = ∑ Drawdown at various period
We will use the equation : S = Q / U*π*T [ -0.5772 - In U ] ----- ( 1 )
where : Q = discharge , T = transmissivity
S = drawdown ,
U = r^2*s / 4*T*t --- ( 2 )
r = well radius , S = Storativity, t = time period
i) During 0-20
U1 = r^2*s / u*π*t = 1.875 * 10^-9
Input values into equation 1
S1 = 2.5885
ii) During 20-50
U2 = r^2*s / 4*π*t = 0.3^2 * 30 / u * 300 * 30 = 1.25 * 10^-9
input values into equation 1
S2 = 1.5854 m
iii) During 50 -90
U3 = r^2*s / 4*π*t = 9.375 * 10^-10
input values into equation 1
S3 = 4.2888 m
iv) During 90-100
U4 = 0
s4 = 0
Drawdown at 100 days = ∑ Drawdowns at various period
= s1 + s2 + s3 + s4 = 2.5885 + 1.5854 + 4.2888 + 0
= 8.4627 m