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kykrilka [37]
3 years ago
12

A three-phase motor rated 25 hp, 480 V, operates with a power factor of 0.74 lagging and supplies the rated load. The motor effi

ciency is 96%. Calculate the motor input power, reactive power and current.
Engineering
1 answer:
coldgirl [10]3 years ago
3 0

Answer:

the motor input power is 19.42 KW

the Reactive power is 17.65 KVAR

Current is 31.56 A

Explanation:

Given that;

V = 480V

h.p = 25 hp

p.f = 0.74 lagging

n_motor = 96%

so output = 25hp

and we know that;

1hp = 746 watt

watt = hp × 1hp

so output in watt = 25 × 746 = 18650 Watt = 18.65 KW

n_motor = (output / input) × 100

96 = 1865 / Input

96Input = 1865

Input = 1865 / 96

Input = 19.42 KW

Therefore the motor input power is 19.42 KW

P = √( 3 × V × I × cos∅)

19.42 = √( 3 ×480 × I × 0.74)

I = 31.56 A

Therefore Current is 31.56 A

Q = √( 3 × V × I × sin∅)

we know that

cos∅ = 0.74

so ∅ = cos⁻¹(0.74) = 42.26

so we substitute

Q = √( 3 × 480 × 31.56 × sin(42.26))

 = 17.65 KVAR

Therefore the Reactive power is 17.65 KVAR

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A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L
zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

3 0
4 years ago
HELP HELP HELP
Fantom [35]

Summary

Students learn about the variety of materials used by engineers in the design and construction of modern bridges. They also find out about the material properties important to bridge construction and consider the advantages and disadvantages of steel and concrete as common bridge-building materials to handle compressive and tensile forces.

This engineering curriculum aligns to Next Generation Science Standards (NGSS).

Engineering Connection

When designing structures such as bridges, engineers carefully choose the materials by anticipating the forces the materials (the structural components) are expected to experience during their lifetimes. Usually, ductile materials such as steel, aluminum and other metals are used for components that experience tensile loads. Brittle materials such as concrete, ceramics and glass are used for components that experience compressive loads.

Learning Objectives

After this lesson, students should be able to:

List several common materials used the design and construction of structures.

Describe several factors that engineers consider when selecting materials for the design of a bridge.

Explain the advantages and disadvantages of common materials used in engineering structures (steel and concrete).

Educational Standards

NGSS: Next Generation Science Standards - Science

Common Core State Standards - Math

International Technology and Engineering Educators Association - Technology

State Standards

Suggest an alignment not listed above

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Worksheets and Attachments

Strength of Materials Worksheet (doc)

Strength of Materials Worksheet (pdf)

Strength of Materials Worksheet Answers (doc)

Strength of Materials Worksheet Answers (pdf)

Strength of Materials Math Worksheet (doc)

Strength of Materials Math Worksheet (pdf)

Strength of Materials Math Worksheet Answers (doc)

Strength of Materials Math Worksheet Answers (pdf)

More Curriculum Like This

MIDDLE SCHOOL Activity

Breaking the Mold

Explanation:

pabrainlest Poe ty

8 0
3 years ago
IN JAVA,
Citrus2011 [14]

Answer:

Explanation:

Code:

import java.io.File;

import java.io.FileWriter;

import java.io.IOException;

import java.util.Scanner;

public class Knapsack {

 

  public static void knapsack(int wk[], int pr[], int W, String ofile) throws IOException

  {

      int i, w;

      int[][] Ksack = new int[wk.length + 1][W + 1];

     

      for (i = 0; i <= wk.length; i++) {

  for (w = 0; w <= W; w++) {

  if (i == 0 || w == 0)

  Ksack[i][w] = 0;

  else if (wk[i - 1] <= w)

  Ksack[i][w] = Math.max(pr[i - 1] + Ksack[i - 1][w - wk[i - 1]], Ksack[i - 1][w]);

  else

  Ksack[i][w] = Ksack[i - 1][w];

  }

  }

     

      int maxProfit = Ksack[wk.length][W];

      int tempProfit = maxProfit;

      int count = 0;

      w = W;

      int[] projectIncluded = new int[1000];

      for (i = wk.length; i > 0 && tempProfit > 0; i--) {

         

      if (tempProfit == Ksack[i - 1][w])

      continue;    

      else {

          projectIncluded[count++] = i-1;

      tempProfit = tempProfit - pr[i - 1];

      w = w - wk[i - 1];

      }

     

      FileWriter f =new FileWriter("C:\\Users\\gshubhita\\Desktop\\"+ ofile);

      f.write("Number of projects available: "+ wk.length+ "\r\n");

      f.write("Available employee work weeks: "+ W + "\r\n");

      f.write("Number of projects chosen: "+ count + "\r\n");

      f.write("Total profit: "+ maxProfit + "\r\n");

     

  for (int j = 0; j < count; j++)

  f.write("\nProject"+ projectIncluded[j] +" " +wk[projectIncluded[j]]+ " "+ pr[projectIncluded[j]] + "\r\n");

  f.close();

      }    

  }

 

  public static void main(String[] args) throws Exception

  {

      Scanner sc = new Scanner(System.in);

      System.out.print("Enter the number of available employee work weeks: ");

      int avbWeeks = sc.nextInt();

      System.out.print("Enter the name of input file: ");

  String inputFile = sc.next();

      System.out.print("Enter the name of output file: ");

      String outputFile = sc.next();

      System.out.print("Number of projects = ");

      int projects = sc.nextInt();

      int[] workWeeks = new int[projects];

      int[] profit = new int[projects];

     

      File file = new File("C:\\Users\\gshubhita\\Desktop\\" + inputFile);

  Scanner fl = new Scanner(file);

 

  int count = 0;

  while (fl.hasNextLine()){

  String line = fl.nextLine();

  String[] x = line.split(" ");

  workWeeks[count] = Integer.parseInt(x[1]);

  profit[count] = Integer.parseInt(x[2]);

  count++;

  }

 

  knapsack(workWeeks, profit, avbWeeks, outputFile);

  }

}

Console Output:

Enter the number of available employee work weeks: 10

Enter the name of input file: input.txt

Enter the name of output file: output.txt

Number of projects = 4

Output.txt:

Number of projects available: 4

Available employee work weeks: 10

Number of projects chosen: 2

Total profit: 46

Project2 4 16

Project0 6 30

8 0
3 years ago
A floor system has W24 x 55 sections spaced 8’-0"" o.c. supporting a floor dead load of 50 psf and a live load of 80 psf. Determ
galben [10]

Answer:

Load sup[port by beam is 1040 lb/ft

Explanation:

Given data:

dead load of floor is 50 psf

live load of floor is 80 psf

load per meter can be determined as

Load/mt length = load intensity × effective width

total load  = deal load + live load

                  = 50 + 80 = 130 psf

load /mt length =  130 × 8

                         = 1040 p/ft = 1.04 k /ft

hence load sup[port by beam is 1040 lb/ft

4 0
4 years ago
Helium gas is compressed from 90 kPa and 30oC to 450 kPa in a reversible, adiabatic process. Determine the final temperature and
ra1l [238]

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data :

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

Determine final temperature for an Isentropic  process

T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}  ----------- ( 1 )

T2 = 303 ( \frac{450}{90})^{(1.667- 1)/1.667} =  576.9K

Work done in a piston-cylinder device can be calculated using this formula

w_{in} = c_{v} ( T2 - T1 )    ------- ( 2 )

where : cv = 3.1156 kJ/kg.k  for helium gas

             T2 = 576.9K ,    T1 = 303 K

substitute given values Back to equation 2

w_{in}  = 853.4 kJ/kg

work done in a steady flow compressor can be calculated using this

w_{in} = c_{p} ( T2 - T1 )

where : cp ( constant pressure of helium gas )  = 5.1926 kJ/kg.K

             T2 = 576.9 k , T1 = 303 K

substitute values back to equation 3

w_{in} = 1422.3 kJ / kg

4 0
3 years ago
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