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Alex73 [517]
2 years ago
14

Help me!!

Engineering
1 answer:
kati45 [8]2 years ago
6 0

Answer:  System Consists Of 1 Kg Of CO2 (Cp = 46.4 J Moll K:') Gas Initially At 1 Bar And 300K. The System Undergo

Explanation:

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A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
4 years ago
Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards
user100 [1]

Answer:

The first step is to;

a. import Scanner class present in java.util package for take input in arrays.

b. When final variable is initialized it can not be changed so fix the size to 4.

c. create array to hold 4 integer elements in array.

d. Now ensure users enter integer values separated by space in single line & assign elements at each index of array starting from index 0.

e. Next is to print the array elements in forward directions first.

reset i to 0 to print all elements of array from index 0.

f. Then ensure to traverse array and display each element one by one separated by space in same line using for loop.

g. Therefore to move to next line .

reset i to array.length -1. loop will go from last elements index till index 0.

h. Finally print elements in array backwards from last elements of array.

The second step is given here,

import java.util.Scanner;

public class CourseGradePrinter {

public static void main(String args) {

//using scanner to take input from user..

Scanner scnr = new Scanner(System.in);

//fix the array size to 4.

final int NUM_VALS = 4;

//array created of size 4 to hold int data

int[] courseGrades = new int[NUM_VALS];

//declare i for user input and traversing array

int i;

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

//user enters integer values

courseGrades[i] = scnr.nextInt();

}

/* Your solution goes here */

//display the array elements forwards now

i=0;

Ensure to print elements in array.

4 0
3 years ago
Suppose a steam locomotive is rated at 7500 horsepower. If its efficiency is 6%, how much wood must be burned in a 3-hour trip?
Sati [7]

Answer:

50,625kg of wood

Explanation:

Power input = power output ÷ efficiency

Power output = 7500 horsepower = 7500×750W = 5625000W, efficiency = 6% = 0.06

Power input = 5625000/0.06 = 93750000W

Energy input = power input × time

time = 3hour = 3×3600 = 10800sec

Energy input = 93750000 × 10800 = 1.0125×10^12J

Mass = Energy ÷ Energy density = (1.0125×10^12J) ÷ (20×10^6J/kg) = 50,625kg of wood

3 0
3 years ago
Problem #1: A centrifugal compressor has a mass flow rate of 1.6 kg/s while rotating at 11,000 rpm. The impellor is a radial imp
saveliy_v [14]

Answer:

a)  The slip coefficient is 0.9

b) Blade tip speed is 345.57 m/s

c) Stagnation temperature exiting the impellor is 416.84 k

d) Exit flow velocity is 84.88 m/s

e) Exit flow angle is 76.2°

f) Exit static temperature is 353.84 k

g) Impellor exit Mach number is 0.943

Explanation:

Flow at entry is axial α₁ = 0

Tagential velocity at entry V_{w1}= 0

Blade at exit is radial  β₂ = 0

μ₂ = V_{w2}

3 0
3 years ago
The electricity generated by wind turbines annually in kilowatt-hours per year is given in a file. The amount of electricity is
Anika [276]

Answer:

Steps:

1. Create a text file that contains blade diameter (in feet), wind velocity (in mph) and the approximate electricity generated for the year

2.  load the data file for example, in matlab, use ('fileame.txt') to load the file

3. create variables from each column of your data

  for example, in matlab,  

     x=t{1}

     y=t{2}

4. plot the wind velocity and electricity generated.

   plot(x, y)

5. Label the individual axis and name the graph title.

    title('Graph of wind velocity vs approximate electricity generated for the year')

     xlabel('wind velocity')

     ylabel('approximate electricity generated for the year')

5 0
3 years ago
Read 2 more answers
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