With the aid of a labbled diagram describe the operation of a core type single phase transformer​

Answers

Answer 1
Answer:

Answer:

A simple single-phase transformer has each winding being wound cylindrically on a soft iron limb separately to provide a necessary magnetic circuit, which is commonly referred to as “transformer core”. It offers a path for the flow of the magnetic field to induce voltage between two windings.


Related Questions

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A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool 5 large watermelons, 10 kg each, to 8 C. If the watermelons are initially at 28 C, determine how long it will take for the refrigerator to cool them.
Benzene vapor at 480°C is cooled and converted to a liquid at 25°C in a continuous condenser. The condensate is drained into 1.75-m3 drums, each of which takes 2.0 minutes to fill. Calculate the rate (kW) at which heat is transferred from the benzene in the condenser.
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 2.1. If, after 146 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?
For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration to 0.38 wt% at a point 2.3 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 5.3 mm position for an identical steel and at the same carburizing temperature.

An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value

Answers

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

The annual worth of contract = $ 190,000

The land application has the least cost , hence it is preferred .

or a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What is the maximum load (in N) that may be applied to a specimen having a cross-sectional area of 164 mm2 without plastic deformation

Answers

Answer:

The maximum load (in N) that may be applied to a specimen with this cross sectional area is F=43788 N

Explanation:

We know that the stress at which plastic deformation begins is 267 MPa.

We are going  to assume that the stress is homogeneously distributed. As a definition, we know that stress (P) is force (F) over area(A), as follows:

P=(F)/(A)    (equation 1)

We need to find the maximum load (or force) that may be applied before plastic deformation. And the problem says that the maximum stress before  plastic deformation is 267 MPa. So, we need to find the value of load when P= 267 MPa.

Before apply the equation, we need to convert the units of area in m^(2). So,

A[m^(2)]=A[mm^(2) ]((1 m)/(1000 mm)) ^(2)\nA[m^(2)]=164 mm^(2)((1 m)/(1000 mm))^(2) \nA[m^(2)]=0.000164 m^(2)

And then, from the equation 1,

F=PA\nF=(267 MPa)(0.000164 m^(2))\nF=(267x10^(6) Pa)(0.000164 m^(2))\nF=(267x10^(6) N/m^(2))(0.000164 m^(2))\nF=43788 N

Describe how a feeler gauge can be used to assist in the adjustment of a spark plug electrode gap​

Answers

Answer:

Explanation:

Adjusting the distance between the two electrodes is called gapping your spark plugs. You need a feeler gauge to gap your spark plugs properly If you're re-gapping a used plug, make sure that it's clean (gently scrub it with a wire brush)

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Write a program to accept a one-line string (maximum of 100 characters) from the keyboard. Edit the string entered in Part 1a (with code that you write) to remove all the white space,digits, punctuation, and other special characters, leaving only the letters. Print out the resulting compressed string to the screen.

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Answer:

// This program is written in C++ programming language

// Comments are used for explanatory purpose

/* The aim of this program is to to remove all the white space,digits, punctuation, and other special characters, leaving only the letters. */

// Program starts here

#include <stdio.h>

#include<iostream>

using namespace std;

int main()

{

// Declare Variable of 100 characters

char word[100];

// Prompt user for input

cout<<"Your input goes here (max, 100 characters)";

cin>>word;

// Iterate through string to check for non alphabetic characters

for (int i = 0; word[i] != '\0'; ++i) {

// Check for uppercase and lowercase letters

while (!((word[i] >= 'a' && word[i] <= 'z') || (word[i] >= 'A' && word[i] <= 'Z') || word[i] == '\0')) {

for (int j = i; word[j] != '\0'; ++j) {

word[j] = word[j + 1];

}

word[j] = '\0';

}

}

cout<<"The resulting compressed string: "<<word;

return 0;

}

Answer:

w = str(input("input your values: "))

values = ' '.join(filter(str.isalpha, w))

while len(w) < 100:

       print(values)

       break

Explanation:

The code is written in python

w = str(input("input your values: "))

This code ask the user to input any string values with characters, numbers, line spaces , letters etc.

values = ' '.join(filter(str.isalpha, w))

This code filters the inputted value to bring only letters. All the letter are then joined together

while len(w) < 100:

The code check if the inputted value is less than 100 characters. While it is less than 100 characters. If it is less than 100 character the next code will function.

print(values)

This code prints the joined letters after checking with  a while loop to confirm the length of character is less than 100

break

The break function breaks the code whether it print the values or not.

Generally, the letters will only be printed if the character inputted is less than 100 and later break the while loop or will not print any letter if the character is greater than 100 and later break.

Describe the difference between design guidelines or criteria and design performance. Explain the relationship between the use of guideline/criteria tools and performance tools during the design process

Answers

Answer:

PART A

Design guidelines are sets of procedures to be followed in order to enhance the designing of an object or other things.

Design Performance is the actual process of carrying out the design process of an object using the design guidelines or criteria.

PART B

(1) Design guidelines tools helps to enhance design Performance.

(2) Design guidelines tools helps the designing performance tools to be effective.

Explanation:Design guidelines are the various steps which has special tools used to guide the designer in order to enhance the designing performance tools and ensure that the design process is done devoid of errors.

Design Performance tools are tools which helps to enhance the actual design Activities.

Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)

Answers

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

E(t)=E_o[1-a(T)/(T_m)]

where,

E(t) is the modulus of elasticity at any temperature 'T'

E_o is the modulus of elasticity at absolute zero.

T_(m) is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.