Answer:
H(s) = 20 / [ 1 + s / 10^5 ]^2
Explanation:
Given data:
cutoff frequency = 100 kHz
stopband attenuation rate = 40 dB/decade
nominal passband gain = 20 dB
new nominal passband gain at cutoff = 14 dB
Represent the transfer function H(s)
The attenuation rate show that there are two(2) poles
H(s) = k / [ 1 + s/Wc ]^2 ----- ( 1 )
where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20
Input values into equation 1
H(s) = 20 / [ 1 + s / 10^5 ]^2
A start-up is expanding overseas and spends an excessive amount of time on recruiting and hiring activities, hindering its ability to focus on the core aspects of its business. How can a Human Capital Management (HCM) platform provider benefit this company?
Answer:
Human Capital Management (HCM) will help the start-up firm manage its recruiting and hiring activities.
Explanation:
Human Capital Management (HCM) Platform will assist the start-up firm manage its main point of access by keeping the employee records and maintaining the wages and salaries, managing the benefits, time, and attendance, and carrying out performance reviews including looking after the most important asset employees.
Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW of energy to the water and the head loss of the flow is 10 m. Determine the velocity of the water leaving the pump and discharging into tank B.
Complete Question
Complete Question is attached below.
Answer:
[tex]V'=5m/s[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=0.10m[/tex]
Power [tex]P=4.0kW[/tex]
Head loss [tex]\mu=10m[/tex]
[tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu[/tex]
[tex]\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10[/tex]
[tex]H_m=(\frac{200*10^3}{1000*9.8}-10)[/tex]
[tex]H_m=10.39m[/tex]
Generally the equation for Power is mathematically given by
[tex]P=\rho gQH_m[/tex]
Therefore
[tex]Q=\frac{P}{\rho g H_m}[/tex]
[tex]Q=\frac{4*10^4}{1000*9.81*10.9}[/tex]
[tex]Q=0.03935m^3/sec[/tex]
Since
[tex]Q=AV'[/tex]
Where
[tex]A=\pi r^2\\A=3.142 (0.05)^2[/tex]
[tex]A=7.85*10^{-3}[/tex]
Therefore
[tex]V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}[/tex]
[tex]V'=5m/s[/tex]
6. When the engine stalls or the power unit fails, on a car with power
brakes, the service brake pedal will
A. Take about the same amount of pressure
B. Take more pressure to stop
C. Take less pressure to stop
D. Become locked in place and no longer help stop the car
Air is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligible changes in kinetic and potential energy. Find the isentropic efficiency of the compressor. Find the exit temperature of the air if the compressor was reversible.
Answer:
a) 1.9%
b) T2s = 505.5 k = 232.5°C
Explanation:
P1 = 95 kPa
T1 = 27°C = 300 k
P2 = 600 kPa
T1 = 277°c = 550 k
Table used : Table ( A - 17 ) Ideal gas properties of air
a) determining the isentropic efficiency of the compressor
Л = ( h2s - h1 ) / ( h2a - h1 ) ---- ( 1 )
where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k
To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)
Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306 hence; h2s = 500.72 kJ/kg
back to equation1
Л = 0.019 = 1.9%
b) Calculate the exit temperature of the air if compressor is reversible
if compressor is reversible the corresponding exit temperature
T2s = 505.5 k = 232.5°C
given that h2s = 500.72 kJ/kg
A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-in-five chance that the levee will be overtopped in the next 15 years. Assuming that the annual peak streamflow follows a lognormal distribution with a log10(Q[ft3/s]) mean and standard deviation of 1.835 and 0.65 respectively, what is the design flow in ft3/s?
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
Determine the design flow ins ft^3/s
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
Use pseudocode. 1) Prompt for and input a saleswoman's sales for the month (in dollars) and her commission rate (percentage). Output her commission for that month. Note that you will need the following Variables: SalesAmount CommissionRate CommissionEarned
You will need the following formula: CommissionEarned= Sales Amount * (commissionrate/100)
Answer:
The pseudocode is as follows:
Input SalesAmount
Input CommissionRate
CommissionEarned= SalesAmount * (CommissionRate/100)
Print CommissionEarned
Explanation:
This gets input for SalesAmount
Input SalesAmount
This gets input for CommissionRate
Input CommissionRate
This calculates the CommissionEarned
CommissionEarned= SalesAmount * (CommissionRate/100)
This prints the calculated CommissionEarned
Print CommissionEarned
Define chart name the different types of charts explain any three types of charts
Answer:
There are several different types of charts and graphs. The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs. They are generally used for, and are best for, quite different things. ... Pie charts to show you how a whole is divided into different parts.
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Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manometer fluid of mercury (density: 13,600 kg/m3) to achieve uncertainty of 5% (i.e., 2.5 m/s) and 1 % (0.5 m/s).
Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]
in this expression the densities are constant
v = A √h
A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h = [tex]\frac{1}{454.55^2} \ v^2[/tex]
Δh = [tex]\frac{dh}{dv}[/tex] Δv
[tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
[tex]\frac{\delta v}{v} = 0.05[/tex]
[tex]\frac{\Delta h}{h}[/tex] = 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
[tex]\frac{\Delta h}{h}[/tex] = 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turbine if the steam is exhausted as a saturated vapor?
Answer:
[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]
Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):
[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]
Now, the isentropic efficiency of the turbine can be given as follows:
[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]
how does load transfer of space needle
Answer:
The Space Needle is a cut away with minimal residual deflection due to load transfer.
Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.
Answer:
(a) the reversible power output of turbine is 5810 kw
(b) The second-law efficiency of he turbine = 86.05%
Explanation:
In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.
h1 = 3658 kJ/kg s1=7.167 kJ/kgK
In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state
h2 = 2682kl/kg S2= 7.694 kJ/kg
Assuming that the energy balance equation given
Wout=m [h1-h2+(v1²-v2²) /2]
Let
W =5 MW
V1= 80 m/s V2= 140 m/s
h1 = 3658kJ/kg h2 = 2682 kJ/kg
∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)
m = 5.158kg/s
Consider the energy balance equation given
Wrev,out =Wout-mT0(s1-s2)
Substitute Wout =5 MW m = 5.158kg/s 7
s1= 7.167 kJ/kg-K s2= 7.694kJ/kg-K and 25°C .
Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)
= 5810 kW
(a) Therefore, the reversible power output of turbine is 5810 kw.
The given values of quantities were substituted and the reversible power output are calculated.
(b) Calculating the second law efficiency of the turbine:
η=Wout/W rev,out
Let Wout = 5 MW and Wrev,out = 5810 kW
η=(5 MW x 1000 kW)/(1 MW *5810)
η= 86.05%
Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move of rock and soil, which Hans knows from previous experience has an average density of . Hans has available a dump truck with a capacity of and a maximum safe load of .Calculate the number of trips the dump truck will have to make to haul the customer's load away.
Complete Question:
Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move 19.8m³ of rock and soil, which Hans knows from previous experience has an average density of 650/kgm³. Hans has available a dump truck with a capacity of 4m³ and a maximum safe load of 3700kg. Required: calculate the number of trips the dump truck will have to make to haul the customer's load away.
Answer:
Mangel-Wurzel Transport
The number of trips that the dump truck will have to make to haul the customer's load away is:
= 5 trips.
Explanation:
a) Data and Calculations:
Volume of customer's load (rock and soil) = 19.8m³
Density of load = 650 kg/m³
Mass of load = Volume of load * Density of load
= 19.8m³ × 650 kg/m³
= 12,870 kg
The maximum safe load (mass) of the dump truck = 3,700 kg
Volume of the dump truck = 4m³
Assuming the truck is to carry 4m³ of the load.
The mass of load that the 4m³ capacity truck can carry = 4m³ × 650kg/m³
= 2,600kg
Quick Check:
Mass = 2,600kg < 3,700 kg, satisfying required conditions.
The number of trips that the truck would make to haul the customer's load away is, therefore, calculated as follows:
Number of trips = N
N = total volume of load/ volume per trip
N = 19.8/4
N = 4.95
N = 5 trips approx.
Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 24 C and exits at 1 bar. If the refrigerant undergoes a throttling process, what is the quality of the refrigerant exiting the expansion valve.
Answer:
[tex]h_{1} = h_2} = 293.45 KJ/kg[/tex].
The quality of the refrigerant exiting the expansion valve is
[tex]x_{2}=0.193596[/tex].
Explanation:
Fluid given Ammonia.
Inlet 1:-
Temperature [tex]T_{1}[/tex] = [tex]24^{o} C[/tex].
Pressure [tex]P_{1}[/tex] = 10 bar.
Exit 2:-
Pressure [tex]P_{2}[/tex] = 1 bar.
Solution:-
Problema:
Una nevera de vinos, con un peso bruto de 50 kg., que tiene las siguientes dimensiones: .60 m Largo x .49 m ancho x .50 m altura. Para ser transportadas en un contenedor de 40 pies D.V. responder las siguientes preguntas:
• 1.Cuántas neveras de vinos de acuerdo al volumen caben en un contenedor de 40 pies?
• De acuerdo dimensiones internas (largo, ancho y alto), ¿Cuántas caben en un contenedor de 40 pies?
• De acuerdo al peso que soporta el contenedor. ¿Cuántas neveras de vinos es posible transportar?
Answer:
I can't understand this language .
Alice and Bob both have RSA Public-Private key pairs: (PUA, PRA) and (PUB, PRB). They also have cryptographic functions E_AES / D_AES to encrypt / decrypt using AES; and E_RSA and D_RSA to encrypt / decrypt using RSA. Alice wants to sent a high resolution video of a large secret facility to Bob.
A. Show how Alice can securely and efficiently send the video to Bob. You are required to use the cryptographic functions above to get full credit;
B. Does your solutions assure confidentiality? How / Why not?
C. Does your solutions assure non-repudiation? How / Why not?
D. Does your solutions assure integrity? How / Why not?
E. Does your solutions assure replay attacks? How / Why not?
Solution :
B. yes, the given solution assures confidentiality. The sender Alice encrypting his messages with its own private key PRA which provides authentication. Sender Alice further encrypts his messages with the receiver's public key PUB provides confidentiality.
C. So the given solution provides non repudiation. Alice and Bob who are exchanging messages. In one case, Alice denies sending a messages to Bob that he claims to have received being able to counter Alice's denial is caused non repudiation of origin.
D. The given solution provides integrity. Because it provides authentication and have not been changed.
E. It does not provide replay attacks because it does not captures the traffic. The client does not receive the messages twice.
All of the following safety tips are true EXCEPT Select one: a. It is not acceptable to handle broken glass with your bare hands b. It is acceptable to grasp the electrical cord when removing an electrical plug from its socket c. It is not acceptable to immerse hot glassware in cold water d. It is not acceptable to reuse dirty glassware
Answer:
Explanation:
B. you would grab the plug closest to the outlet
If you don't have enough experience, it's always best to leave socket changing to the experts. If you make a mistake, you might inflict harm and potentially endanger yourself and other people. Read on if you're interested in learning how to change a socket safely. Thus, option D is correct.
What, removing an electrical plug from its socket?Grip the plug, not the electrical cable, when taking an electrical plug out of its socket. Before handling an electrical switch, socket, or outlet, hands must be fully dry.
Reduce the extra so that it rests only on top of the existing plasterboard. If necessary, push it back a little by using your finger. Fill the dent with ready-mixed filler or powdered filler, whichever you want, and bring it flush with the surrounding wall. Allow to dry, then sand off any excess.
Therefore, It is acceptable to grasp the electrical cord when removing an electrical plug from its socket
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(50 POINTS) How many people use pipes in the world? How do you know this?
Answer:
7.9 billion people
Explanation:
Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.
Answer:
The answer is "15 N".
Explanation:
Please find the complete question in the attached file.
In frame B:
For just slipping:
[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]
[tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]
3-71A 20mm diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15 degrees, what must be the length of the bar
Answer:
The right answer is "1.903 m".
Explanation:
Given that,
[tex]\tau =110 \ MPa[/tex]
[tex]G=80 \ GPa[/tex]
[tex]\Theta=15\times \frac{\pi}{180}[/tex]
[tex]=\frac{\pi}{12}[/tex]
[tex]d=20 \ mm[/tex]
As we know,
⇒ [tex]\frac{\tau}{r}=\frac{G \Theta}{L}[/tex]
Or,
⇒ [tex]L=\frac{G \theta r}{\tau}[/tex]
[tex]=\frac{80\times 10^3}{110}\times \frac{\pi}{12}\times 10[/tex]
[tex]=1903.9 \ mm[/tex]
or,
[tex]=1.903 \ m[/tex]
Lab 5A Problem Input two DWORD values from the keyboard. Determine which number is larger or if they are even. Your program should look like the following: First number larger Enter a number 12 Enter a number 10 12 is the larger number Press any key to close this window... Second number larger Enter a number 10 Enter a number 12 12 is the larger number Press any key to close this window... Numbers Equal Enter a number 12 Enter a number 12 Numbers are equal Press any key to close this window...
Answer:
Explanation:
#include<iostream>
using namespace std;
int main()
{
int n1,n2;
cout<<"Enter a number:"<<endl; //Entering first number
cin>>n1;
cout<<"Enter a number:"<<endl; //Entering second number
cin>>n2;
if(n1%2==0 && n1%2==0) //Checking whether the two number are even or not
{
if(n1>n2)
{
cout<<n1<<" is the larger number"<<endl;
}
else if(n1==n2)
{
cout<<"Numbers are equal"<<endl;
}
else
{
cout<<n2<<" is the larger number"<<endl;
}
}
else
{
cout<<"The number are not even"<<endl;
}
}
James the Pilot James is a pilot. He is wearing a flight suit. He flies to Paris. He loves flying. 1. James is a a) teacher b) doctor c) pilot. whatisthe 2. He is wearing a a) shirt b) t-shirt c) flight suit. 3. Where does he fly to? a) Italy b) Luxembourg c) Paris http https://whatistheurl.com Please visit our site for worksheets and charts
Answer:
1.c
2.c
3.c
Explanation:
James is a pilot, whistle. He is wearing a flight suit. Paris is the palace where does he fly to. Hence, option C, C, and C are correct.
What is the point of a flight suit?When flying an aircraft, such as a military aircraft, a glider, or a helicopter, one must wear a full-body suit called a flight suit. These outfits are typically meant to keep the user warm and are also functional (they have many of pockets) (including fire ). In most cases, it looks like a jumpsuit.
The G suit, sometimes known as a "anti-G suit," is a one-piece jumpsuit that shields a pilot from the pressure of G forces pressing down on him and causing discomfort or unconsciousness.
The traditional attire for pilots of military and commercial aircraft, helicopters, and even gliders is flight suits or flyers coveralls. In areas where there is a risk of fire, ground personnel—including aircrews—often wear flight suits as well.
Thus, option C, C, and C are correct.
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The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in the above, the Nusselt number approaches 2 as Rayleigh number approaches zero. Prove that this situation corresponds to conduction heat transfer and in conduction heat transfer over sphere, the Nusselt number becomes 2. Hint: First step: Write an expression for heat transfer between two spherical shells that share the same center. Second step: Assume the outer spherical shell is infinitely large.
Answer:
Explanation:
[tex]r_2=[/tex]∞
[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]
[tex]q=2\pi kD.[/tex]ΔT--------(1)
[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]
[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]
By equation (1) and (2)
[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT
[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)
From equation (3) and (4)
So for sphere [tex]R_a[/tex]→0
what are some quality assurance systems
Technician A says that a graphing multi-meter may be used to verify signals going to and from electrical and electronic components. Technician B says that digital storage oscilloscope may be used to verify signals going to and from electrical and electronic components. Who is correct
Answer:
Both are correct.
Explanation:
Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.
what is the best glide speed for your training airplane
1.5 nautical miles per 1,000 feet
What does Faraday's law of induction states?
Explanation:
This relationship, known as Faraday's law of induction (to distinguish it from his laws of electrolysis), states that the magnitude of the emf induced in a circuit is proportional to the rate of change of the magnetic flux that cuts across the circuit.
Match the test to the property it measures.
a. Rockwell
b. Inston
c. Charpy
d. Fatigue
e. Brinell
f. Izod
1. impact strength
2. stress vs strain
3. hardness
4. Endurance Limit
Answer:
a. Rockwell 3. hardness
b. Instron 2. stress vs strain
c. Charpy 1. impact strength
d. Fatigue 4. Endurance Limit
e. Brinell 3. hardness
f. Izod 1. impact strength
Explanation:
Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.
Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.
Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.
Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.
A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 220 V to a primary coil of 230 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages
Answer:
Explanation:
A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.
When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.
Given that:
input (primary) voltage of the transformer, [tex]V_i=220~V[/tex]
no. of turns in the primary coil, [tex]N_i=230[/tex]
When the output voltage is 5.60 V:[tex]V_o=5.60~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{5.60}{220}[/tex]
[tex]N_o=5.85\approx 6[/tex] turns compensating the losses
When the output voltage is 12.0 V:[tex]V_o=12.0~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{12.0}{220}[/tex]
[tex]N_o=12.45\approx 13[/tex] turns compensating the losses
When the output voltage is 480 V:[tex]V_o=480~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{480}{220}[/tex]
[tex]N_o=501.8\approx 502[/tex] turns compensating the losses
g The inside surface of a 17 mm inner diameter tube with a 2.4 mm thick wall indicates a temperature of 46 deg C. The outside temperature is 43 deg C. The tube is 5 m long. If the tube material has a conductivity of 0.15 W/m/K, estimate the heat transfer rate through the tube wall assuming SS 1D conduction. Indicate the direction of heat transfer with a or - sign ( meaning outward and vice versa). Express your answer using two significant digits in W.
Answer:
-50 W
Explanation:
The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m
Since Q = kA(T₂ - T₁)/d ,
Q = kπdL(T₂ - T₁)/d
substituting the values of the variables into the equation, we have
Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C - 46 °C )/0.0024 m
Q = 0.01275π Wm/K(-3 K )/0.0024 m
Q = -0.03825π Wm/0.0024 m
Q = -0.1202 Wm/0.0024 m
Q = -50.07 W
Q = -50 W
So, the heat transfer rate is -50 W meaning heat transfer out of the tube.
A 0.82-in-diameter aluminum rod is 5.5 ft long and carries a load of 3000 lbf. Find the tensile stress, the total deformation, the unit strains, and the change in the rod diameter.
Answer:
Tensile stress = 0.1855Kpsi
Total deformation = 0.0012243 in
Unit strain = 1.855 *10^-5 or 18.55μ
Change in the rod diameter = 5.02 * 10^ -6 in
Explanation:
Data given: D= 0.82 in
L = 5.5 ft * 12 = 66 in
load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm
Area = (π/4) D² = (π/4) 0.82² = 0.502655 in²
∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²
Rt = 0.1855 Kpsi
∴ Total deformation = PL / AE = Rt * L/ Eal
= 0.1855 * 10³ * 66 / 10000 * 10³
= 0.0012243 in
∴the unit strains = total deformation / L = 0.0012243/ 66
=0.00001855 = 1.855 *10^-5
= 18.55μ
∴ Change in rod Δd/ d = μ ΔL/L
= (0.33) 1.855 *10^-5 * 0.82
= 5.02 * 10^ -6 in
