Answer:
intrinsic semiconductors
Explanation:
An intrinsic semiconductor is also known as a pure conductor. In such a semiconductor there are no impurities, that is why it is said to be pure.
It has some of these properties:
1. Electrical conductivity is only based on temperature
2. The quantity of electrons is the same as the number of holes in the valence bond
3. Electrical conductivity is not on the high side
4. These materials exist in their pure forms.
how many crankshaft is in V8 engine?
Answer:
Explanation:
hola friend!!!!!
there is only one crankshaft in V8 engine
Hope this helps
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help out with that question
Answer:
(523.74 lb)∠-48.59°
Explanation:
a) We can find the sum of the force vectors several ways, but first we need to know the direction of vector BT. The angle PBT is identical to the angle shown as α, which we can find from the lengths BD and AD.
α = arctan(BD/AD) = arctan((3√3)/6)) ≈ 40.893°
Using the Law of Cosines, we can find the magnitude of PT from ...
PT^2 = PB^2 +BT^2 -2·PB·BT·cos(α)
PT^2 = 800^2 +600^2 -2·800·600·2/√7 = 1000000 -1920000/√7
PT ≈ 523.74 . . . . lb
The direction of PT can be found from the Law of Sines.
∠BPT = arcsin(BT/PT·sin(α)) ≈ 1.145597sin(40.893°) ≈ 48.59°
Relative to the +x axis, the resultant force (R) is ...
R = (523.74 lb)∠-48.59°
__
b) The graphical solution is shown in the attachment. The graphing tool measures the segment lengths and angles. Those measures confirm the above result. (The pound values shown are scaled up by a factor of 100 from the segment lengths on the diagram. They are "captions" for the respective vectors.)
In each situation, write a recurrence relation, including base case(s), that describes the recursive structure of the problem. You do not need to solve the recurrence.
a) Let B(n) be the number of length n bit sequences that have no three consecutive 0s (i.e., do not contain the substring "000"). Write a recurrence for B(n).
b) Let S(n) be the number of subsets of {1, 2, ..., n} having the following property: no two elements in the subset are consecutive integers. The empty set with no elements should be included in your count. Write a recurrence for S(n).
c) Say you are tiling a 2 times n rectangle with L-shaped tiles of area 3 (trominoes). To tile the rectangle is to cover it with tiles so that no tiles overlap and every cell is covered by some tile. Let T(n) denote the number of ways to tile the rectangle. Write a recurrence for T(n).
d) A ternary string is like a binary string except it uses three symbols, 0, 1, and 2. For example, 12210021 is a ternary string of length 8. Let T(n) be the number of ternary strings of length n with the property that there is never a 2 appearing anywhere after a 0. For example, 12120110 has this property but 10120012 does not. Write a recurrence for T(n).
Explanation:
a) Given B(n) is the number of the length 'n' bit sequences which have no three consecutive 0s(i.e., they does not contain substring “000”)
Any bit string which has no 000 should have a 1 in at least one of the 1st three positions. Then, the n we will break all the bit strings by avoiding the 000 by when the 1st 1 occurs. i.e., each of the bit of string of the length of n will avoid 000 falls into the exactly any one of these cases:
1 is followed by the bit string of the length of (n-1) avoiding the 000.
01 is followed by some bit string of the length of (n-2) avoiding a 000.
001 which is followed by any bit string of the length(n-3) avoiding the 000.
Therefore, recurrence is
B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(0)=1, B(1)=2, B(2)=4
Or
B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(1)=2, B(2)=4, B(3)=7
b) Let the S(n)={1,2,3,…,n}. We will say that the subset A of S(n) is very good if A does not have any of the two integers that are consecutive.
For any k, let a(k) be a number of any good subsets of a S(k).
There are two types of good subsets of S(n).
Type 1 of good subsets of S(n) which contain the element n,
Type 2 of good subsets of S(n) which do not contain n.
We will first get an expression for a number of Type 1 of good subsets of the S(n) , where n≥2. This a subset does not contain n-1. So any Type 1 of the good subset of the S(n) is obtainable when adding n to good subset of the S(n-2) . Also, on adding n to a good subset of the S(n-2) , we always get a Type 1 good subset of the S(n). Thus there are exactly as many good Type 1 of subsets of S(n) as there is good subsets of S(n-2) . By the definition there are a(n-2) good subsets of S(n-2) .
A good subset of a S(n) is either Type 1 or Type 2. Thus the number of the good subsets of S(n) is a(n-2)+a(n-1)
We have therefore shown that a(n) = a(n-2)+a(n-1)
c). Here firstly see that if the n is not the multiple of a 3, then there will be no chance to tile the rectangle. And also if n is a multiple of a 3, then there may be two ways to tile the 1st three columns:
And the rest of tilling is the tilling of 2x(n-3) rectangle for which there are T(n-3).
So, the recurrence is
T(n )= {2T(n-3), with T(0)=1 if n=0(mod 3) or 0 else
We could use the base case T(3)=2
So, the following recurrence can be aslo equivalent to T(n)= 2T(n-3), with T(0)=1, T(1)=0,T(2)=0
Or
T(n)= 2T(n-3), with T(1)=0, T(2)=0,T(3)=2
d)A ternary string may be defined as a sequence of some digits, where each digit has either 0,1,or 2.
According to the problem given, we do not have a 2 anywhere after 0, and the dot which represents the binary string of length(n-1) with the property which we cannot use 2 anywhere after we use the 0.
Now for base case, we should note that any of the ternary string which has a length of 1 satisfies the given required property. Hence the recurrence is
T(1)=3
T(n)=2T(n-1)+2^(n-1)
If the equation for the velocity profile is given by: 3/2 = 4yv . Assuming v is in ft/s, what is the velocity gradient at the boundary and at y = 0.25 ft and 0.5 ft from boundary?
Answer:
When y = 0.25 ft, velocity gradient = -6 ft/s
When y = 0.5 f, velocity gradient = -1.5 ft/s
Explanation:
Given;
equation for the velocity profile, 3/2 = 4yv
Rearrange this equation, you will get;
[tex]4yv = \frac{3}{2}\\\\v = \frac{3}{2} *\frac{1}{4y} \\\\v = \frac{3}{2}(\frac{1}{4} )(\frac{1}{y} )\\\\v = \frac{3}{8}y^{-1}\\\\ gradient \ of \ velocity \ = \frac{dv}{dy} \\\\\frac{dv}{dy} = -1(\frac{3}{8})y^{-2}[/tex]
When y = 0.25 ft
[tex]\frac{dv}{dy} = -1(\frac{3}{8})(0.25)^{-2}\\\\\frac{dv}{dy} = -6 \ ft/s[/tex]
When y = 0.5 ft
[tex]\frac{dv}{dy} = -1(\frac{3}{8})(0.5)^{-2}\\\\\frac{dv}{dy} = -1.5 \ ft/s[/tex]
A hollow shaft of diameter ratio 3/8 (internal dia to outer dia) is to transmit 375 kW power at 100 rpm. The maximum torque being 20% greater than the mean torque. The shear stress is not to exceed 60 N/mm2 and twist in a length of 4m not to exceed 2o. Calculate its external and internal diameters which would satisfy both the above conditions. (G= 0.85 X 105 N/mm2)
Answer:
External diameter = 158.15 mm mm
Internal diameter = 59.31 mm
Explanation:
We are given;
Diameter ratio; d_i = ⅜d_o
Where d_i is internal diameter and d_o is external diameter
Power;P = 375 KW = 375000 W
Rotational speed;N = 100 rpm
Max torque is 20% greater than mean torque; T_max = 1.2T_avg
Shear stress;τ = 60 N/mm²
Length; L = 4m = 4000 mm
Angle of twist; θ = 2° = 2π/180 radians
Modulus of rigidity;G = 0.85 X 10^(5) N/mm²
Formula for the power transmitted by the shaft is;
P = 2πNT_avg/60
Plugging in the relevant values, we have ;
375000 = 2π × 100T_avg/60
T_avg = (375000 × 60)/(2π × 100) = 35809.862 N.m = 35809862 N.mm
Since T_max = 1.20T_avg
Thus, T_max = 1.20(35809862) = 42971834.4 N.mm
Checking for strength, we'll use;
τ = Tr/J_p
Or since r = d/2
It can be written as;
τ = T(d_o)/2J_p - - - (1)
Where T is T_max
But Polar moment of inertia of hollow shaft is;
J_p = [π(d_o)⁴ - π(d_i)⁴]/32
Now, we are told that d_i = ⅜d_o
Thus;
J_p = [π(d_o)⁴ - π(⅜d_o)⁴]/32
J_p = (π/32) × d_o⁴(1 - 3⁴/8⁴)
J_p = 0.0926 d_o⁴
Plugging this for J_p in eq 1,we have;
τ = T(d_o)/2(0.0926d_o⁴)
Making d_o the subject gives;
d_o³ = T/(2 × 0.0926τ)
Plugging in the relevant values to give;
d_o³ = 42971834.4/(2 × 0.0926 × 60)
d_o³ = 3867155.7235421166
d_o = ∛3867155.7235421166
d_o = 156.96 mm
Thus, d_i = ⅜ × 156.96 = 58.86 mm
Checking for stiffness, we'll use;
T/J_p = Gθ/L
Again T is T_max
Plugging in the relevant values, we have;
42971834.4/0.0926 d_o⁴ = (0.85 × 10^(5) × 2π/180)/4000
464058686.825054/d_o⁴ = 0.7417649321
d_o⁴ = 464058686.825054/0.7417649321
d_o⁴ = 625614216.5028806
d_o = ∜625614216.5028806
d_o = 158.15 mm
d_i = ⅜ × 158.15 = 59.31 mm
So we will pick the highest valies.
Thus;
d_o = 158.15 mm
d_i = 59.31 mm
The spring-held follower AB has weight W and moves back and forth as its end rolls on the contoured surface of the cam, where the radius is r and z = asin(2θ). If the cam is rotating at a constant rate θ', determine the force at the end A of the follower when θ = θ 1. In this position the spring is compressed δ1. Neglect friction at the bearing C.
Answer:
some parts of your question is missing attached below is the missing part
Answer : Fa = 4.46 Ib
Explanation:
use the equation
Z = 0.1 sin2∅
next we will differentiate the equation to get the locus of the velocity
z = 0.2 cos2∅∅
differentiate the equation furthermore to get the locus of acceleration in the horizontal axis
z = -0.4sin2∅(∅)^2 + 0.2cos2∅∅
note : express ∅ as 6 rad.[tex]s^{-1}[/tex] for angular velocity and ∅ = 0 for angular acceleration
equation above becomes :
Z = - 0.4 sin 2∅ ( 6)^2 + 0.2 cos 2∅(0)
= - 14.4 sin 2∅ ( acceleration of the follower in horizontal direction )
next calculate The force at the end of A of the follower
Fa - Kx = mz
note: m = w / g hence : Fa - Kx = w/g z ------- (2)
w = weight of the spring-held follower = 0.75 Ib
x = compression of the spring = 0.4
k = spring stiffness = 12 Ib/ft
∅ = 45⁰
g = 32.2 ft/s^2
input these values into equation 2
hence : Fa = 4.46 Ib ( force at the end A of the follower )
Can I get a hand? Thanks y’all much luv
Answer:
C
Explanation:
As the impurity concentration in solid solution is increased, the tensile and yield strengths ____________________.
increase
decrease
Answer:
As the impurity concentration in solid solution is increased, the tensile and yield strengths increases.
Explanation:
The addition of impurities in solid solutions shows an improved tensile and yield strength due to the grain refinement and obstacles to the motion of dislocation.
Example, the addition of carbon as impurity into iron, which forms steel shows a significant increase in the tensile and yield strengths of iron.
Blacksmiths also use work hardening to introduce dislocation into solid solutions in order to increase their tensile and yield strengths.
Therefore, as the impurity concentration in solid solution is increased, the tensile and yield strengths increases.
A solid solution is a mixture of crystalline solids and is soluble over the partial or evenly complete range.
A solute may be replaced by a solvent particle. It may be used for heating the is related to the melting point. The tensile and strength of the solution form a solid increase as the concentration of the impurities increase.Hence the option Increases is correct.
Learn more about the concentration in solid solution is increased.
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Thermoplastics that exhibit a definite TM point have this kind of crystalline structure:
A. semi-crystalline
B. isotactic
C.amorphous
D. syndiotactic
Answer:
The correct option is A
Explanation:
Thermoplastics are polymers that can be recycled. They can be melted with heat and moulded into any shape which becomes hard when cooled. The heat applied to it in the process of recycling does not affect its chemical composition.
There are two main classes of thermoplastics; semi-crystalline and amorphous.
An amorphous thermoplastic has a disorderly arranged molecular structure which makes it difficult to have a definite melting temperature (Tm). However, a semi-crystalline thermoplastic has a well arranged molecular structure which makes it melt after a certain/definite amount of heat is absolved. Hence, the correct answer is A
In which of the following branches of engineering is the practice not restricted?
a) civil engineering.
b) mechanical engineering.
c) nuclear engineering.
Answer:
a) civil engineering.
Explanation:
Civil engineering is a professional engineering program that deals with the construction, design, and maintenance of all the natural and man-made environments including dams, buildings, railways, and roads.
Civil engineering is the branch of engineering that is the practice not restricted because civil engineer is not restricted to academic profession but practice in designing and construction can make someone a professional civil engineer.
Hence, the correct answer is "a)."
The branch of engineering in which the practice is not restricted is; Civil Engineering
Civil Engineering is a branch of engineering that deals with the design, construction and maintenance of the physical and naturally built environment.
These physically and naturally built environment includes; houses, roads, bridges, airports, railways, canals, dams, sewage systems, pipelines e.t.c
Now, mechanical engineering involves design, production, operation and maintenance of mechanical systems or machineries.
In conclusion, we see that the mechanical branch of engineering is restricted to machinery unlike civil engineering that is not restricted to only buildings but also includes pipelines, bridges, roads, railways, dams, sewage systems e.t.c
Finally, the nuclear engineering branch is also restricted to only nuclear fission and fusion applications.
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Students would like to sell cold drinks to raise money for a field trip. They need to keep the drinks cold for 3 hours at the ball game. Which step of the engineering design process are the students at?
Answer:
Problem definition
Explanation:
The engineering design is a series of step that the engineer gradually take in other to create a functioning product or process. There are many different models of the engineering design process , but they can be shrunk into four basic steps which are
problem definition conceptual designpreliminary designdetailed designdesign communication.The students are within the first step, and the problem here is 'an appropriate means for the students to keep their drinks cold, for 3 hours for the ball game.'
A piston-cylinder device contains Xenon gas. During a reversible, adiabatic process, the entropy of the gas will (never, sometimes, always) increase.
Answer:
Never
Explanation:
In a reversible adiabatic process, there is not transfer of heat or matter between the system and its environment. An adiabatic reversible process is a process with constant entropy, i.e ΔQ=0. The internal energy is solely dependent on the work done either due to compression or expansion. So the entropy of the gas will never increase.
The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on conventional gasoline vehicles.
Answer:
Batteries are safe when handled properly.
Explanation:
Just like the battery in your phone, the battery in some variant of an electric car is just as safe. If you puncture/smash just about any common kind of charged battery, it will combust. As long as you don't plan on doing anything extreme with the battery (or messing with high voltage) you should be fine.
The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on conventional gasoline vehicles: False.
Safety risks can be defined as an assessment of the risks and occupational hazards associated with the use, operation or maintenance of an equipment or automobile vehicle that is capable of leading to the;
Harm of a worker (technician).Injury of a worker (technician).Illness of a worker (technician).Death of a worker (technician).Hybrid electric vehicles (HEVs) or EVs are typically designed and developed with parts or components that operates through the use of high voltage electrical systems ranging from 100 Volts to 600 Volts. Also, these type of vehicles have an in-built HEV batteries which are typically encased in sealed shells so as to mitigate potential hazards to a technician.
On the other hand, conventional gasoline vehicles are typically designed and developed with parts or components that operates on hydrocarbon such as fuel and motor engine oil. Also, conventional gasoline vehicles do not require the use of high voltage electrical systems and as such poses less threat to technicians, which is in contrast with hybrid electric vehicles (HEVs) or EVs.
This ultimately implies that, the safety risks for technicians who work on hybrid electric vehicles (HEVs) or EVs are different from those who work on conventional gasoline vehicles due to high voltage electrical systems that are being used in the former.
In conclusion, technicians who work on hybrid electric vehicles (HEVs) or EVs are susceptible (vulnerable) to being electrocuted to death when safety risks are not properly adhered to unlike technicians working on conventional gasoline vehicles.
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Which phase of DevSecOps emphasizes reliability, performance, and scaling
Answer:
"Test Phase " is the correct choice.
Explanation:
DevSecOps seems to be a community as well as experience of corporate data science which encompasses software design, regulation, including operational activities. This same main feature of DevSecOps has always been to strengthen customer achievement as well as expedition importance by computerizing, supervising as well as implementing data protection at all stages of the development including its development tools.The testing method throughout the test phase would then help make sure that the controller is designed mostly under the responsibilities forecasted. The test focuses on either the reaction times, dependability, use of resources but instead interoperability of applications.The DevSecOps can be described as a software development life cycle which has seen security introduced into the continous development and operations pipeline. Hence, the phase of DevSecOps which emphasizes reliability, performance and scaling is the Security phase
DevSecOps can be broken down into a continous pipeline of processes which include Development, Security and Operation. By integrating security into the continous software development process, it ensures that software security which is aimed at reducing the vulnerability of data and information is undertaken throughout the entire development lifecycle.Therefore, ensuring that applications are reliable and performs well without having to sacrifice security in the process.
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Why some types of aggregate are susceptible to damage from repeated freezing and thawing? Explain.
Answer:
Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.
When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.
Switches are placed only in the _ of a circuit?
the switch should always be placed immediately adjacent to the non-grounded terminal of the power supply.
Compute the repeat unit molecular weight of PTFE. Also compute the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000
Answer:
a) the repeat unit molecular weight of PTFE MW = 100.015 g/mole
b) the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000 щₙ = 1000150 g/mole
Explanation:
Given that;
PTFE which is also called Polytetrafluoroethylene
structure of repeat unit of Polytetrafluoroethylene
(C2F4)n
a)
To compute the repeat unit molecular weight
we say
MW = 2( atomic weight of C ) + 4( atomic weight of F)
MW = 2 (12.0107) + 4 ( 18.9984)
MW = 100.015 g/mole
therefore the repeat unit molecular weight of PTFE MW = 100.015 g/mole
b)
To compute the number-average molecular weight for PTFE of which the degree of polymerization is 10,000
we say
DP = щₙ / MW
where щₙ is the number of average molecular weight,
MW is the repeat unit molecular weight give as 100.015 g/mole
DP is degree of polymerization which is 10,000
Now we substitute
10,000 = щₙ / 100.015
щₙ = 10,000 × 100.015
щₙ = 1000150 g/mole
Therefore the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000 щₙ = 1000150 g/mole
Give the principle, construction and working of Bourdan tube pressure gauge.
Answer:
Working Principle Of Bourdon GaugeIf a tube having oval cross section is subjected to pressure its cross section tends to change from oval to circular.
Construction of a Bordon GaugeBourdon tube gauges consist of a circular tube.
One end of the tube is fixed while the other end is free to undergo elastic deformation under the effect of pressure.
Fixed end is open and pressure which is to be measured is applied at the fixed end.
Free end is closed and undergoes deformation under the effect of pressure.
Due to applied pressure the circular tube tends to uncoil and become straight along the dotted line.
Working of Bourdon Gauge
As the pressure is applied at the fixed end free end undergoes deformation.
The free end is attached with sector which further meshes with the pinion on which pointer is mounted.
Deformation of the pointer is transferred to pointer via this mechanism.
As a result point undergoes deflection and shows the pressure reading on calibrated dial.
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Answer:
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Explanation:
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Class hello Sarah{
Pubic static void(“hello Sarah”)
Name=input(“hello Sarah”)
Print(“hello Sarah”).
Answer:
class helloSarah {
public static void main(String [] args) {
Name = "Hello Sarah";
System.out.print(Name);
}
}
Explanation:
The question seem incomplete; However, since the program is incorrect, I'll assume the question is to correct the given code;
The program segment was written in Java and the correct program is in the answer section
See bold texts for line by line explanation
This line declares the class name
class helloSarah {
This line declares the main method of the program
public static void main(String [] args) {
This line initializes string variable Name
String Name = "Hello Sarah";
This line prints the initialized variable
System.out.print(Name);
}
}
Answer:
class helloSarah {
public static void main(String [] args) {
Name = "Hello Sarah";
System.out.print(Name);
]
Explanation:
Which phase of DevSecOps emphasizes reliability, performance, and scaling
Answer:
Test Phase
Explanation:
DevSecOps is an organizational software engineering culture and practice that unifies software development (Dev), security (Sec) and operations (Ops). The main characteristic of DevSecOps is to improve customer outcomes and mission value by automating, monitoring, and applying security at all phases of the software lifecycle.
There are nine phases of the software lifecycle which are: plan, develop, build, test, release, deliver, deploy, operate, and monitor.
The Performance test in the test phase will ensure that applications will perform well under the expected workload. The test focus is on application response time, reliability, resource usage and scalability.
In development phases, database design, development, and testing activities generate database artifacts, which are data models, database schema files, trigger definitions, view definition, test data, test data generation scripts, test scripts, etc. These database artifacts must be under configuration management control. During test phase, database functional test is like application code unit test and functional test to validate the schema, triggers, and data compliance. The non-functional test includes load testing, stress test, and performance test. The security test focuses on vulnerability scan, user authentication and authorization, unauthorized access to data, data encryption, privilege elevation, SQL injection, and denial of service.
In a bid eliminate the vulnerability experienced during the traditional development process, DevSecOps emphasizes reliability, performance and Scaling with the integration of Security phase.
The integration of Security infrastructure into the Development operation(DevOps) process ensures that security challenges experienced by softwares are tackled immediately hence ensuring reliability and reduced vulnerability.
DevSecOps ensures that performance isn't sacrificed for security, hence, softwares are continously checked for security at every phase of the development process during testing.
Therefore, the security phase of the DevSecOps pipeline ensures that satisfactory security and Performance levels are met.
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1. At the end of the day, a bakery gives everything that is unsold to food banks for the needy. If it has 12 apple pies left at the end of a given day, in how many different ways can it distribute these pies among six food banks for the needy?
2. In how many different ways can the bakery distribute the 12 apple pies if each of the six food banks is to receive at least one pie?
Answer:
The answer ix below
Explanation:
There are 12 apple pies left.
Given that:
n = number of apple pies left = 12
x = number of food banks = 6
1) For the 12 apple pies to be distributed among 6 food banks. The number of ways in which this can be done is:
C(n + x - 1, x - 1) = C(12 + 6 -1, 6 - 1) = C(17, 5) = [tex]\frac{17!}{(17-5)!5!} =\frac{17!}{12!5!}=6188 \ ways[/tex]
12 apple pies can be distributed among 6 food banks in 6188 ways
2) For the 12 apple pies to be distributed among 6 food banks if each food bank must receive one pie, 6 pies would be remaining. The number of ways in which this can be done is:
C((n - x) + x - 1, n - x) = C(12 - 6 + 6 - 1, 12 - 6) = C(11, 6) = [tex]\frac{11!}{(11-6)!6!} =\frac{11!}{6!5!}=462 \ ways[/tex]
12 apple pies can be distributed among 6 food banks if each food bank must receive one pie in 462 ways
A 400-MVA, 240-kV/24-kV, three-phase Y-A transformer has an equivalent series impedance of 1.2 + j6 N per phase referred to the high-voltage side. The transformer is supplying a three-phase load of 400-MVA, 0.8 power factor lagging at a terminal voltage of 24 kV (line to line) on its low-voltage side. The primary is supplied from a feeder with an impedance of 0.6 + ji.2 A per phase. Determine the line-to-line voltage at the high-voltage ter- minals of the transformer and the sending-end of the feeder.
Answer: the line-to-line voltage at the high-voltage terminals of the transformer and the sending-end of the feeder is 249.71∠1.8° kV
Explanation:
First we find the phase voltage per phase at the primary side connected in Y, so we say
V₂ = 240K/√3 = 138.56 kV
Now we find the primary current
I₁ = ((400 × 10⁶) / 3(138.56 × 10³)) ∠ -cos⁻¹ (0.8)
I₁ = 962.28∠ -36.87° A
To find the voltage V₁, we say
V₁ = ( 1.2 + j6) I₁ + V₂
we substitute
V₁ = ( 1.2 + j6) 962.28∠ -36.87° + 138.56 × 10³
V₁ = 143∠1.57° kV
Now we find the phase voltage at the sending end
Vₓ = ( 0.6 + J1.2 )I₁ + V₁
Vₓ = ( 0.6 + J1.2 ) 962.28∠ -36.87° + 143∠1.57° K
Vₓ = 144.17∠1.8° kV
So to Determine the line to line voltage at the sending end, we say:
Vₓ (line to line) = √3 × 144.17∠1.8° kV
Vₓ (line to line) = 249.71∠1.8° kV
Consider a 2.4-kW hooded electric open burner in an area where the unit costs of electricity and natural gas are $0.10/kWh and $1.20/therm (1 therm = 105,500 kJ), respectively. The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.
Answer:
electric: 2.4 kW; $0.1370 per kWh utilizedgas: 4.61 kW; $0.1078 per kWh utilizedExplanation:
Electric burner
For a 2.4 kW electric burner, the rate of energy consumption is 2.4 kW.
If efficiency is 73%, the cost of utilized energy is ...
($0.10 /kWh) / 0.73 ≈ $0.1370 per kWh utilized
__
Gas burner
If the utilized energy provided by the gas burner is the same as the utilized energy of the electric burner, then the rate of energy consumption will be ...
2.4 kW(0.73)/(0.38) = 4.61 kW
In terms of kWh, the cost of gas is ...
$1.20/(105,500 kJ)·(1 kJ/(kW·s))·(3600 s/h) = $0.04095 /kWh
If efficiency is 38%, the cost of utilized gas energy is ...
($0.04095 /kWh) / 0.38 ≈ $0.1078 per kWh utilized
_____
The cost of gas is about 21% less per utilized joule.
_____
Comment on rates
The "unit rate" will depend on the unit chosen. In order to avoid unnecessary units conversions, we have elected to stick with kWh as the unit of energy for both cases. You may be asked for different units. We trust you or Google can make the necessary conversions.
Define water hammer. Give four effects of water hammer.
Answer:
Water Hammer is a knocking sound in an water pipe which occurs when the tap is turned off briskly
Explanation:
Consider a 5 m long, air-filled section of a coaxial transmission line, given that the radius of the inner conductor is 10 cm and the inner radius of the outer conductor is 20 cm? If the current is 100 mA, how much magnetic energy is strored in the air medium between the conductors. This problem is similar to Example 2 in the Time Varying Fields folder in the Example Bank.
a) 200.89 nj
b) 3.47 nj
c) 10.45nj
d) 80.9nj
Answer: b) 3.47 nj
Explanation:
Given that;
length l = 5m
radius of inner conductor r = 10cm = 0.1m
radius of outer conductor D = 20cm = 0.2m
current I = 100A = 100×10⁻³ = 0.1
medium between conductor in air u₀ = 4π × 10⁻⁷
Energy in a coaxial cable transmission line is
w = u₀ /2π I² en(b/a)
we substitute
L = 4π × 10⁻⁷ /4π ×10⁻²× 5 en (20/10)
L =3.4657 × 10⁻⁹ J
L = 3.4657 nJ ≈ 3.47 nJ
Initially, a pump pressure of __________ pounds per square inch should be used to maintain a sprinkler or standpipe system.
Answer:
of 150 pounds per square inch
Explanation:
Note that the unit for measuring water pressure is called pounds per square inch (psi)
In the case of sprinklers and standpipe systems, a pressure of 150 pounds per square inch was used initially.
Which of the following Components Can't. be used as an amplifier Anless its Control electrode Conducts a Current? a) Ann-Channel Jfetb) An npn transistorc) A p. Channel ufet d) A mosfet
Answer:
d) A mosfet
Explanation:
MOSFET is the most common type of insulated gate Field Effect Transistor (FET), used in electronic circuits and it stands for Metal Oxide Semiconductor Field Effect Transistor.
To configure MOSFET to act as an amplifier, a small AC signal is applied, which is superimposed on to DC bias at the gate input, then the MOSFET will act as a linear amplifier.
Therefore, the correct option is (d) A mosfet
Consider a 2.4-kW hooded electric open burner in an area where the unit costs of electricity and natural gas are $0.10/kWh and $1.20/therm (1 therm = 105,500 kJ), respectively. The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.
Answer:
1.75 kW
$0.137 kWh
4.61 kW
$3.16 therm
Explanation:
Utilized power input of the burner is
P(ui) = total power input * efficiency
P(ui) = 2400 W * 0.73
P(ui) = 1752 W or 1.75 kW
Unit cost of utilized energy is
C(ui) = Unit cost of electricity/efficiency
C(ui) = $0.1 / 0.73 kWh
C(ui) = $0.137 kWh
Power input to the gas burner is
P(gi) = Utilized power input of the burner / efficiency of the burner
P(gi) = 1.75 / 0.38
P(gi) = 4.61 kW
Unit cost of utilized energy is
C(gi) = Unit cost of gas /efficiency
C(gi) = $1.2 / 0.38 kWh
C(gi) = $3.16 therm
A three-phase, 60-Hz, completely transposed 345-kV, 200-km line has two 795,000-cmil (403-mm2) 26/2 ACSR conductors per bundle and the following positive-sequence line constants: z 0.032 + 10.35 /km y j4.2 x 10-6 S/km Full load at the receiving end of the line is 700 MW at 0.99 p.f. leading and at 95% of rated voltage. Assuming a medium-length line, determine the following:
a. ABCD parameters of the nominal π circuit
b. Sending-end voltage Vs, current Is, and real power Ps
c. Percent voltage regulation
d. Transmission-line efficiency at full load
Answer:
B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv
sending end current : Is = 1.241 ∠ 15.5⁰ KA
real power = 730.5 Mw
C) percent voltage regulation = 8.7%
D) Transmission line efficiency = 95.8%
Explanation:
attached is the detailed solution to the problem
Given data:
l = 200 km
z = 0.032 + j0.35 Ω/km
y = j4.2 * 10^-6 S/km
A) find the total series impedance and shunt admittance
B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv
sending end current : Is = 1.241 ∠ 15.5⁰ KA
real power = 730.5 Mw
C) percent voltage regulation = 8.7%
D) Transmission line efficiency = 95.8%
One of the testing equipments used for inspection is? Test tubes Measuring jar Strain gauges None of the mentioned
Answer:
The answer is the strain gauges.
Explanation:
Inspection systems work or are performed to measure the characteristics of a product, to verify if it meets specified requirements, all using benchmarks and test equipment.
The strain gauges are part of the test equipment used for inspection. These are sensors that measure deformation, pressure and load in resistance tests of materials.