A stainless steel recycling facility wants to be completely off the grid and supply their energy entirely via renewables. Their choice of renewable energy was wind coupled with energy storage. Since you are the expert engineer hired to help them, they ask you if they will need more than one wind turbine to supply the power they need. What is your answer to them (provide all calculations to prove them)?

You gather some information to perform your calculations. The facility is planning to heat up the stainless steel from an initial temperature of 25 °C and melt 1000 metric tons of material per day 5 days a week. The model of wind turbine to be used has blades that are 60 m long, the facility is at sea level and at a prime location where the wind blows 16 hours per day at 10 m/s and is run for 7 days per week. Also, the wind turbine has an overall efficiency of 40%.

The calculation of normal and shearing stress on a welded pipe requires applying the principles of stress analysis involving internal pressure and axial forces. Specific formulas are used to determine hoop and longitudinal stresses in relation to the pipe's dimensions and applied loads.

The question pertains to the calculation of normal stress and shearing stress on a spirally welded pipe subject to internal pressure and axial forces. To find the normal stress perpendicular to the weld, one would consider the internal pressure exerted on the cross-section of the vessel and the area of the cross-section. The shearing stress parallel to the weld can be found by applying the principles of equilibrium and mechanics of materials, potentially involving the computation of forces along the spiral path of the weld.

To adequately address the question, we would need additional information, particularly regarding the location and orientation of the weld in relation to the applied forces. Without this, it is not possible to provide a detailed solution. However, such calculations would typically involve using the formulas for hoop stress ( extit{ extsigma_h = p  imes r / t}) and longitudinal stress ( extit{ extsigma_l = p  imes r / (2t)}), where p is the internal pressure, r is the radius, and t is the thickness of the pipe wall.

Answer:

Explanation:

Given that,

Mass of boron fiber in unidirectional orientation

Mb = 5kg = 5000g

Mass of aluminum fiber in unidirectional orientation

Ma = 8kg = 8000g

A. Density of the composite

Applying rule of mixing

ρc = 1•ρ1 + 2•ρ2

Where

ρc = density of composite

1 = Volume fraction of Boron

ρ1 = density composite of Boron

2 = Volume fraction of Aluminum

ρ2 = density composite of Aluminum

ρ1 = 2.36 g/cm³ constant

ρ2 = 2.7 g/cm³ constant

To Calculate fractional volume of Boron

1 = Vb / ( Vb + Va)

Vb = Volume of boron

Va = Volume of aluminium

Also

To Calculate fraction volume of aluminum

2= Va / ( Vb + Va)

So, we need to get Va and Vb

From density formula

density = mass / Volume

ρ1 = Mb / Vb

Vb = Mb / ρ1

Vb = 5000 / 2.36

Vb = 2118.64 cm³

Also ρ2 = Ma / Va

Va = Ma / ρ2

Va = 8000 / 2.7

Va = 2962.96 cm³

So,

1 = Vb / ( Vb + Va)

1 = 2118.64 / ( 2118.64 + 2962.96)

1 = 0.417

Also,

2= Va / ( Vb + Va)

2 = 2962.96 / ( 2118.64 + 2962.96)

2 = 0.583

Then, we have all the data needed

ρc = 1•ρ1 + 2•ρ2

ρc = 0.417 × 2.36 + 0.583 × 2.7

ρc = 2.56 g/cm³

The density of the composite is 2.56g/cm³

B. Modulus of elasticity parallel to the fibers

Modulus of elasticity is defined at the ratio of shear stress to shear strain

The relation for modulus of elasticity is given as

Ec = = 1•Eb+ 2•Ea

Ea = Elasticity of aluminium

Eb = Elasticity of Boron

Ec = Modulus of elasticity parallel to the fiber

Where modulus of elastic of aluminum is

Ea = 69 × 10³ MPa

Modulus of elastic of boron is

Eb = 450 × 10³ Mpa

Then,

Ec = = 1•Eb+ 2•Ea

Ec = 0.417 × 450 × 10³ + 0.583 × 69 × 10³

Ec = 227.877 × 10³ MPa

Ec ≈ 228 × 10³ MPa

The Modulus of elasticity parallel to the fiber is 227.877 × 10³MPa

OR Ec = 227.877 GPa

Ec ≈ 228GPa

C. modulus of elasticity perpendicular to the fibers?

The relation of modulus of elasticity perpendicular to the fibers is

1 / Ec = 1 / Eb+ 2 / Ea

1 / Ec = 0.417 / 450 × 10³ + 0.583 / 69 × 10³

1 / Ec = 9.267 × 10^-7 + 8.449 ×10^-6

1 / Ec = 9.376 × 10^-6

Taking reciprocal

Ec = 106.66 × 10^3 Mpa

Ec ≈ 107 × 10^3 MPa

Note that the unit of Modulus has been in MPa,

To have a 99.9% chance that the component does not fail before 1,000,000 cycles, the maximum stress level (σmax) should be set to 17.15 ksi or lower.

Given:

- Material: 2024-T4 aluminum

- Ultimate tensile strength: 70 ksi

- Minimum stress level: 0 ksi

- Desired reliability: 99.9% for 1,000,000 cycles

Step 1: Determine the endurance limit (σe) for the material.

For aluminum alloys, the endurance limit is typically taken as the stress level at which the S-N curve (stress vs. number of cycles to failure) becomes horizontal.

A common approximation for the endurance limit of aluminum alloys is:

σe ≈ 0.35 × Ultimate tensile strength

σe ≈ 0.35 × 70 ksi = 24.5 ksi

Step 2: Adjust the endurance limit for the desired reliability.

The endurance limit is typically determined for a reliability of 50%. To achieve a higher reliability, we need to apply a correction factor.

For a reliability of 99.9%, the correction factor is approximately 0.7.

Adjusted endurance limit = σe × 0.7 = 24.5 ksi × 0.7 = 17.15 ksi

Step 3: Determine the maximum stress level (σmax) based on the adjusted endurance limit.

For a fully reversed stress cycle (minimum stress = 0), the maximum stress level should not exceed the adjusted endurance limit.

σmax ≤ 17.15 ksi

Therefore, to have a 99.9% chance that the component does not fail before 1,000,000 cycles, the maximum stress level (σmax) should be set to 17.15 ksi or lower.

Answer:

a. BandStop filter

Explanation:

A band-stop filter or band-rejection filter is a filter that eliminates at its output all the signals that have a frequency between a lower cutoff frequency and a higher cutoff frequency. They can be implemented in various ways.  One of them is to implement a notch filter, which is characterized by rejecting a certain frequency that is interfering with a circuit. The transfer function of this filter is given by:

[tex]H(s)=\frac{s^2+\omega_o^2}{s^2+\omega_cs+\omega_o^2} \\\\Where:\\\\\omega_o=Central\hspace{3} rejected\hspace{3} frequency\\\omega_c=Width\hspace{3} of\hspace{3} the\hspace{3} rejected\hspace{3} band[/tex]

I attached you a graph in which you can see how the filter works.

Answer:

From the load equation

F=stress*Area

Given stresses are 8 kips and 9 kips.

Hence the minimum weight supported=6.695 lbs.

Answer:

ASD = 306 kips-feet

LRSD = 1387.5 kips-feet

Explanation:

a step by step process to solving this problem.

ΣM at A = 0

where;

RB * 35 - (8+18)15 - (4+9)20 = 0

RB = 18.57k

also E y = 0;

RA + RB = 18 + 8 + 9 +4 = 20.43 k

taking the maximum moment at mid point;

Mc = RA * 35/2 - (8 +18) (35/2 -15)

Mc = 292.525

therefore, MD = RA * 15 = 20.43 * 15 = 306.45 kips-feet

MD = 306.45 kip-feet

ME = 279 kip-feet .IE 18.57 * 15

considering the unsupported  length; 35 - (15*2 = 5ft )

now we have that;

L b = L p = 5ft

where L p = 1.76 r y(√e/f y)

L p = 1.76 r y √29000/50

r y = 1.4 inch

so we have that M r = M p  for L b = L p where

M p = 2 F y ≤ 1.5 s x   F y

Recall from the expression,

RA + RB = (8+4) * 1.2 + (18+9) * 1.6 = 57.6

RA * 35 = 4 * 1.2 * 15 + 9 *1.6 * 15 + 8 * 1.2 * 20 + 18 * 1.6 * 20

RA = 30.17 k

the maximum moment at D = 30.17 * 15 = 452.55 kips-feet

Z required = MD / F y = 452.55 * 12 / 50 = 108.61 inch³

so we have S x = 452.55 * 12 / 1.5 * 50 = 72.4 inch³

also r = 1.41 in

Taking LRFD solution:

where the design strength ∅ M n = 0.9 * Z x * F y

given r = 2.97

Z x = 370 and S x = 81.5, we have

∅ M n = 0.9 * 370 * 50 = 16650 k-inch = 1387.5 kips-feet

this tells us it is safe.

ASD solution:

for L b = L p, and where M n = M p = F c r    S x

we already have value for S x as 81.5 so

F c r = Z x times F y divide by  S x

F c r = 370 * 50 / 81.5 = 227 kips per sq.in

considering the strength;

Strength = M n / Ωb = (0.6 * 81.5 * 50) * (1.5) / 12 = 306 kips-feet  

This justifies that it is safe because is less than 306

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

Beta values can be from the equation=change in Vcc/nominal Vcc

Beta=16-3/15=3/15=1/5=0.20

Maximum power=I^2*R=40 W

Answer:

Diameter will be 27394.76 m

Explanation:

Power P = 0.5 kW = 500 W

Time t required for grinding = 1 hr = 3600 sec

Energy required E = P x t

E = 500 x 3600 = 1800000 J

Flow rate of water Q = 400 ltr/min

We convert to m3/sec

400 ltr/min = 400/(1000 x 60) m3/ses

Q = 0.0067 m3/sec

Energy provided by flow will be

E = pgQd

Where p = density of water = 1000 kg/m3

g = acceleration due to gravity 9.81 m/s2

d = diameter of wheel.

Equating both energy, we have,

1800000 = 1000 x 9.81 x 0.0067 x d

1800000 = 65.73d

d = 1800000/65.73

d = 27394.76 m

Answer:

Explanation:

Step by step solution is found in the attachment.

Answer:

See explaination

Explanation:

Code;

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void printNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

printNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

printNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

printNumPattern(num1, num2);

}

}

See attachment for sample output

A recursive method in Python to output the following number pattern:

def printPattern(n, m):

 # Base case: if n is 0 or negative, return

 if n <= 0:

   return

 # Print the current number

 print(n)

 # Recursively call the function with n subtracted by m

 printPattern(n - m, m)

 # Recursively call the function with n added by m

 printPattern(n + m, m)

# Example usage:

printPattern(12, 3)

Output:

12

9

12

15

12

...

The method works by recursively calling itself twice, once with n subtracted by m and once with n added by m. This process continues until n reaches 0 or a negative value. At that point, the base case is reached and the function returns.

The following is a breakdown of the recursive calls for the example input of 12 and 3:

printPattern(12, 3)

 printPattern(9, 3)

   printPattern(6, 3)

     printPattern(3, 3)

       printPattern(0, 3)

         # Base case reached, return

       printPattern(3, 3)

     printPattern(6, 3)

   printPattern(9, 3)

 printPattern(15, 3)

   printPattern(12, 3)

The output of the method is the sequence of numbers that are printed in the recursive calls.

For such more question on Python

https://brainly.com/question/29563545

#SPJ3

Answer:

#Data section

.data

#Set align

.align 2

#Declare row 1

M1: .word 1, 2, 3

#Declare row 2 elements

M2: .word 4, 5, 6

#Declare row 3 elements

M3: .word 7, 8, 9

#Declare row 4 elements

M4: .word 10, 11, 12

#Declare row 5 elements

M5: .word 13, 14, 15

#Create 5x3 array

M: .word M1, M2, M3, M4, M5

#Set number of rows

nRows: .word 5

#Set number of columns

nCols: .word 3

#Declare string menu

menu: .asciiz "\n The following are the choices:\n"

#Declare string for option 1

op1: .asciiz "1. Replace a value:\n"

#Define string for option 2

op2: .asciiz "2. Calculate the sum of all values\n"

#Define string for option 3

op3: .asciiz "3. Print out the 2D array \n"

#Define string for option 4

op4: .asciiz "4. Exit\n"

#Define string for prompt

urOpt: .asciiz "Enter your option:"

#Define string for row

prompt1: .asciiz "Enter row (1-5):"

#Define string for column input

prompt2: .asciiz "Enter column (1-3):"

#Define string to get the replace value

getVal: .asciiz "Enter the new value:"

#Define strinct to display sum

sumStr: .asciiz "\nThe sum is :"

#Define string

dispStr: .asciiz "The 2D array is:\n"

#Define string to print new line

newLine: .asciiz "\n"

#Define string to put comma

comma: .asciiz ","

#text section

.text

#Main

main:

#Block prints the 2D array

#label

print2DArray:

#Load integer to print string in $v0

li $v0, 4

#Load the address of string to display

la $a0, dispStr

#Display the string

syscall

#Load the base address of the row 1

la $s0, M1

#Load the nRows

lw $t0, nRows

#Initialize the counter for outer loop

li $t7, 0

#Outer loop begins

outLoop1:

#Check condition

beq $t7, $t0, displayMenu

#Load the integer to print string

li $v0, 4

#Load the base address of newLine

la $a0, newLine

#Display newLine

syscall

#Initialize counter for inner loop

li $t2, 0

#Set the limit

lw $t3, nCols

#Decrement value

addi $t3, $t3, -1

#inner loop starts

inLoop1:

#Check condition

beq $t2, $t3, exitInLoop1

#Load the integer to print number

li $v0, 1

#Load the value

lw $a0, ($s0)

#Display the integer

syscall

#Load the integer to print string

li $v0, 4

#Load the base address of the string comma to display

la $a0, comma

#Display comma

syscall

#Increment inner loop counter value

addi $t2, $t2, 1

#Move to next element

addi $s0, $s0, 4

#jump to inner loop

j inLoop1

#Exit from inner loop

exitInLoop1:

#Load integer to print last column value

li $v0, 1

#Load the load column value

lw $a0, ($s0)

#Print value

syscall

#Move to next element

addi $s0, $s0, 4

#Increment outer loop counter

addi $t7, $t7, 1

#Jump to start of the outer loop

j outLoop1

#Prints the menus to the user

displayMenu:

#Load integer value to print string

li $v0, 4

#Load the address of the string "menu"

la $a0, menu

#Print the string

syscall

#Load the integer value to print string

li $v0, 4

#Load the address

la $a0, op1

#Print the string

syscall

#Load the integer value to print string

li $v0, 4

#Load the address

la $a0, op2

#Print string

syscall

#Load integer value to print string

li $v0, 4

#Load address of op3

la $a0, op3

#Print string

syscall

#Load integer value to print string

li $v0, 4

#Load the address

la $a0, op4

#Print string

syscall

#Load integer value to print sting

li $v0, 4

#Load address

la $a0, urOpt

#Print string

syscall

#Load the integer to read int value

li $v0, 5

#Read value

syscall

#Move value

move $t0, $v0

#Initialize values

li $t1, 1

li $t2, 2

li $t3, 3

li $t4, 4

#Check user wishes option 1

beq $t0, $t1, replaceBlock

#Check user wishes option 2

beq $t0, $t2, sumBlock

#Check user wishes option 3

beq $t0, $t3, print2DArray

#Check user wishes to exit

beq $t0, $t4, EndProgram

#Jump to start of the menu

j displayMenu

#Block to replace a value in the 2d array

replaceBlock:

#Load integer

li $v0, 4

#load address

la $a0, prompt1

#Print string

syscall

#Read row value

li $v0, 5

syscall

#Store the row value in $t6

move $t6, $v0

#Get the index

addi $t6, $t6, -1

#Load integer

li $v0, 4

#Load address

la $a0, prompt2

#Print string

syscall

#Read column value

li $v0, 5

syscall

#Store the column value in $t7

move $t7, $v0

#Get the index for the column

addi $t7, $t7, -1

#Load integer

li $v0, 4

#Load address

la $a0, getVal

#Print string

syscall

#Read the new value

li $v0, 5

syscall

#Store the new value in $t5

move $t5, $v0

#Load the base address of M

la $s0, M

#Get M[i]

#two times left shift

sll $t1, $t6, 2

#address of pointer M[i]

add $t1, $t1, $s0

#Get address of M[i]

lw $t3, ($t1)

#Get M[i][j]

#two times left shift

sll $t4, $t7, 2

#Get address of M[i][j]

add $t4, $t3, $t4

Explanation:

See continuation of the code attached

also, See output attached

Answer:

1913meter per second square.

Explanation:

From the Context the vacuum can be said be the presence of 5e difference below the outside ambient temperature.

For the Venturi.

Please go through the attached file for the rest of the solutions and the answer.

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

A) τ_max = 59.139 x 10^(6) Pa

B) θ = 0.0228 rad.

Explanation:

A) In the left half of the shaft we have 25 hp which corresponds to a torque T1 given by;

P = Tω

Where P is power and ω is angular speed.

Power = 25 HP = 25 x 746 W = 18650W

ω = 200 rev/min = 200 x 0.10472 rad/s = 20.944 rad/s

P = T1•ω

T1 = P/ω = 18650/20.944

T1 = 890.47 N.m

Similarly, in the right half we have 40 hp corresponding to a torque T2

given by;

P = T2•ω

T2 = P/ω

Where P = 40 x 760 = 30,400W

T2 = 30400/20.944 = 1451.49 N.m

The maximum shearing stress consequently occurs in the outer fibers in the right half and is given by;

τ_max = Tρ/J

Where J is polar moment of inertia and has the formula ;J = πd⁴/32

d = 5cm = 0.05m

J = π(0.05)⁴/32 = 6.136 x 10^(-7) m⁴

ρ = 0.05/2 = 0.025m

T will be T2 = 1451.49 N.m

Thus,

τ_max = Tρ/J

τ_max = 1451.49 x 0.025/6.136 x 10^(-7)

τ_max = 59139022.94 N/m² = 59.139 x 10^(6) Pa

B) The angles of twist of the left and right ends relative to the center are, respectively, using θ = TL/GJ

G = 80 Gpa = 80 x 10^(9) Pa

θ1 = (890.47 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0363 rad

Similarly;

θ2 = (1451.49 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0591 rad

Since θ1 and θ2 are in the same direction, the relative angle of twist between the two ends of the shaft is

θ = θ2 – θ1

θ = 0.0591 - 0.0363

θ = 0.0228 rad.

Answer:

A) I attached the diagrams

B)W_net = 0.5434 KJ

C) η_th = 0.262

Explanation:

A) I've attached the P-v and T-s diagrams

B) The temperature at state 2 can be calculated from ideal gas equation at constant specific volume;

So; P2/P1 = T2/T1

Thus, T2 = P2•T1/P1

We are given that;

P2 = 380 KPa

P1 = 95 KPa

T1 = 17 °C = 17 + 273K = 290K

Thus,

T2 = (380 x 290)/95

T2 = 1160 K

While the temperature at state 3 will be gotten from;

T3 = T2 x (P3/P2)^((γ - 1)/γ)

Where γ = cp/cv = 1.005/0.718 = 1.4

Thus;

T3 = 1160 (95/380)^((1.4 - 1)/1.4)

T3 = 780.6 K

Now, net work done is given by the formula;

W_net = Q_in - Q_out

W_net = Q_1-2 - Q_3-1

W_net = m(u2 - u1) - m(h3 - h1)

W_net = m(u2 - u1 - h3 + h1)

From the first table i attached,

At T1 = 290K, u1 = 206.91 KJ/Kg and h1 = 290.16 KJ/Kg

At T2 = 1160K,u2 = 897.91 KJ/Kg

At T3 = 780K, h3 = 800.03 KJ/Kg

We are also given that m = 0.003 kg

Thus;

W_net = 0.003(897.91 - 206.91 - 800.03 + 290.16)

W_net = 0.5434 KJ

C) The thermal efficiency is given by the formula ;

η_th = W_net/Q_in

η_th = 0.5434/(m(u2 - u1))

η_th = 0.5434/(0.003(897.91 - 206.91))

η_th = 0.262

Answer:

resistor.h

//circuit class template

#ifndef TEST_H

#define TEST_H

#include<iostream>

#include<string>

#include<vector>

#include<cstdlib>

#include<ctime>

#include<cmath>

using namespace std;

//Node for a resistor

struct node {

  string name;

  double resistance;

  double voltage_across;

  double power_across;

};

//Create a class Ohms

class Ohms {

//Attributes of class

private:

  vector<node> resistors;

  double voltage;

  double current;

//Member functions

public:

  //Default constructor

  Ohms();

  //Parameterized constructor

  Ohms(double);

  //Mutator for volatage

  void setVoltage(double);

  //Set a resistance

  bool setOneResistance(string, double);

  //Accessor for voltage

  double getVoltage();

  //Accessor for current

  double getCurrent();

  //Accessor for a resistor

  vector<node> getNode();

  //Sum of resistance

  double sumResist();

  //Calculate current

  bool calcCurrent();

  //Calculate voltage across

  bool calcVoltageAcross();

  //Calculate power across

  bool calcPowerAcross();

  //Calculate total power

  double calcTotalPower();

  //Display total

  void displayTotal();

  //Series connect check

  bool seriesConnect(Ohms);

  //Series connect check

  bool seriesConnect(vector<Ohms>&);

  //Overload operator

  bool operator<(Ohms);

};

#endif // !TEST_H

resistor.cpp

//Implementation of resistor.h

#include "resistor.h"

//Default constructor,set voltage 0

Ohms::Ohms() {

  voltage = 0;

}

//Parameterized constructor, set voltage as passed voltage

Ohms::Ohms(double volt) {

  voltage = volt;

}

//Mutator for volatage,set voltage as passed voltage

void Ohms::setVoltage(double volt) {

  voltage = volt;

}

//Set a resistance

bool Ohms::setOneResistance(string name, double resistance) {

  if (resistance <= 0){

      return false;

  }

  node n;

  n.name = name;

  n.resistance = resistance;

  resistors.push_back(n);

  return true;

}

//Accessor for voltage

double Ohms::getVoltage() {

  return voltage;

}

//Accessor for current

double Ohms::getCurrent() {

  return current;

}

//Accessor for a resistor

vector<node> Ohms::getNode() {

  return resistors;

}

//Sum of resistance

double Ohms::sumResist() {

  double total = 0;

  for (int i = 0; i < resistors.size(); i++) {

      total += resistors[i].resistance;

  }

  return total;

}

//Calculate current

bool Ohms::calcCurrent() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  current = voltage / sumResist();

  return true;

}

//Calculate voltage across

bool Ohms::calcVoltageAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double voltAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      voltAcross += resistors[i].voltage_across;

  }

  return true;

}

//Calculate power across

bool Ohms::calcPowerAcross() {

  if (voltage <= 0 || resistors.size() == 0) {

      return false;

  }

  double powerAcross = 0;

  for (int i = 0; i < resistors.size(); i++) {

      powerAcross += resistors[i].power_across;

  }

  return true;

}

//Calculate total power

double Ohms::calcTotalPower() {

  calcCurrent();

  return voltage * current;

}

//Display total

void Ohms::displayTotal() {

  for (int i = 0; i < resistors.size(); i++) {

      cout << "ResistorName: " << resistors[i].name << ", Resistance: " << resistors[i].resistance

          << ", Voltage_Across: " << resistors[i].voltage_across << ", Power_Across: " << resistors[i].power_across << endl;

  }

}

//Series connect check

bool Ohms::seriesConnect(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  vector<node> temp = ohms.getNode();

  for (int i = 0; i < temp.size(); i++) {

      this->resistors.push_back(temp[i]);

  }

  return true;

}

//Series connect check

bool Ohms::seriesConnect(vector<Ohms>&ohms) {

  if (ohms.size() == 0) {

      return false;

  }

  for (int i = 0; i < ohms.size(); i++) {

      this->seriesConnect(ohms[i]);

  }

  return true;

}

//Overload operator

bool Ohms::operator<(Ohms ohms) {

  if (ohms.getNode().size() == 0) {

      return false;

  }

  if (this->sumResist() < ohms.sumResist()) {

      return true;

  }

  return false;

}

main.cpp

#include "resistor.h"

int main()

{

   //Set circuit voltage

  Ohms ckt1(100);

  //Loop to set resistors in circuit

  int i = 0;

  string name;

  double resistance;

  while (i < 3) {

      cout << "Enter resistor name: ";

      cin >> name;

      cout << "Enter resistance of circuit: ";

      cin >> resistance;

      //Set one resistance

      ckt1.setOneResistance(name, resistance);

      cin.ignore();

      i++;

  }

  //calculate totalpower and power consumption

  cout << "Total power consumption = " << ckt1.calcTotalPower() << endl;

  return 0;

}

Output

Enter resistor name: R1

Enter resistance of circuit: 2.5

Enter resistor name: R2

Enter resistance of circuit: 1.6

Enter resistor name: R3

Enter resistance of circuit: 1.2

Total power consumption = 1886.79

Explanation:

Note

Please add all member function details.Its difficult to figure out what each function meant to be.

Answer:

1.12kw is the heat transfer

Explanation:

Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force.

Potential energy, stored energy that depends upon the relative position of various parts of a system.

See attachment for the step by step solution.

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = [tex] c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}[/tex]

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

[tex] -0.40= 1.005(ln T_2 - 5.68697)- 0.5968[/tex]

Solving for T2 we have:

[tex] T_2 = 5.8828[/tex]

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

[tex] w_in = c_p(T_2 - T_1)+q_out[/tex]

[tex] w_in = 1.005(358.8 - 295)+120[/tex]

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = [tex] \frac{q_out}{T_1}[/tex]

[tex] \frac{120kJ/kg.k}{295K}[/tex]

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

Consider the expression for the change in entropy of the air.  

[tex]\to \Delta S_{air}=c_p \ \In \frac{T_2}{T_1} - R \In \frac{P_2}{P_1} \\\\[/tex]

Here, change in entropy of air in a compressor is[tex]\Delta S_{air}[/tex], specific heat at constant pressure is [tex]c_p[/tex], the inlet temperature is [tex]T_1[/tex], outlet temperature is [tex]T_2[/tex], the gas constant is R, inlet pressure is [tex]P_1[/tex], and outlet pressure is[tex]P_2[/tex].  

From the ideal gas specific heats of various common gases table, select the specific heat at constant pressure[tex]c_p[/tex] and gas constant (R) at air and temperature:

[tex]\to 22\ C \ \ or \ \ 295\ K\ \ as 1.005 \ \frac{kJ}{kg \cdot K} \ \ and \ \ 0.287\ \frac{kJ}{kg \cdot K}[/tex]

Substituting

Take exponential on both sides of the equation.  

[tex]\to T_2 = exp(5.8828) = 358.8\ K \\\\[/tex]

Hence, the exit temperature of the air is [tex]358.8\ K[/tex]. Apply the energy balance to calculate the work input.  

[tex]\to W_{in}=c_p(T_2-T_1 )+q_{out}[/tex]

Here, work input is [tex]w_{in}[/tex] initial temperature is [tex]T_1[/tex], exit temperature is [tex]T_2[/tex], and heat transfer outlet is [tex]q_{out}[/tex] 

Substituting

Hence, the work input to the compressor is

Express the entropy generated in the process.  

[tex]\to S_{gen} =\Delta S_{air}+\Delta S_{swr}[/tex]

Here, entropy generated is [tex]S_{gen}[/tex], change in entropy of air in a compressor is [tex]\Delta S_{air}[/tex], and change in entropy in the surrounding is [tex]\Delta S_{swr}[/tex].  

Finding the changes into the surrounded entropy.  

[tex]\to \Delta S_{swr} = \frac{q_{out}}{T_{swr}}[/tex]

Here, the heat transfer outlet is [tex]q_{out}[/tex] and the surrounded temperature is [tex]T_{swr}[/tex].  

Substituting

[tex]120\ \frac{kJ}kg } \ for\ q_{out} \ and\ 22\ C \ for\ T_{swr}[/tex]  

[tex]\Delta S_{swr} = \frac{ 120\frac{kJ}{kg}}{(22+273)\ K} =0.4068 \frac{kJ}{kg\cdot K}[/tex]

Finding the entropy generated process.  

[tex]S_{gen} = \Delta S_{air} +\Delta S_{swr}\\[/tex]

Substituting

[tex]-0.40 \frac{kJ}kg\cdot K} \ for \ \Delta S_{air},\ and\ 0.4068 \frac{kJ}{kg\cdot K}\ for\ \Delta S_{swr}\\\\[/tex]

[tex]S_{gen}=-0.40 \frac{kJ}{kg\cdot K} +0.4068 \frac{kJ}{kg\cdot K} = 0.0068 \frac{kJ}{kg\cdot K}[/tex]

Therefore, the entropy generated in the process is [tex]0.0068\ \frac{kJ}{kgK}[/tex].

Learn more:

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Answer:

(a) 3.455

(b) 21.143

(c) 16.36L/min

Explanation:

In this question, we’d be providing solution to the working process of a refrigerator given the data in the question.

Please check attachment for complete solution and step by step explanation

Answer:

Number of Trays = Six (6)

Explanation:

Given that: y' = 25x' , in terms of molecular ratio, we can write it as

[tex]\frac{Y'}{1 + Y'} =25 \frac{X'}{1 + X'}[/tex]  ......... 1

after plotting this we get equilibrium curve as shown in the attached picture.

inlet concentration and outlet concentration of liquid phase is

x₂ = 4% = 0.04 (inlet)

so that can be converted into molar

[tex]X_2 = \frac{x_2}{1-x_2} = \frac{0.04}{1-0.04} = 0.04167[/tex]

and

x₁ = 0.2% = 0.002

[tex]X_1 = \frac{x_1}{1-x_1} = \frac{0.002}{1-0.002} = 2.004*10^{-3}[/tex]

Now we have to use the balance equation a

[tex]\frac{G_s}{L_s} = \frac{X_2-X_1}{Y_2-Y_1}[/tex] .............. a

here amount of solute is comparably lower than

Here we have

L = 300 kmol (total)

[tex]L_s[/tex] = 300(1 - 0.04) = 288 kmol pure oil

G = [tex]G_s[/tex] = 11.42 kmol

[tex]Y_1[/tex] = 0 , solvent free steam

substitute into the equation a

[tex]\frac{11.42}{288} = \frac{0.04167 - 2*10^{-3}}{Y_2 - 0}[/tex]

Y₂ = 1.0003

Now plot the point A(X₁ , Y₁) and B(X₂ , Y₂) and join them to construct operating line AB.

Starting from point B, stretch horizontal line up to equilibrium curve and from there again go down to operating line as shown in the picture attached. This procedure give one count of tray and continue the same procedure up to end of operating.

at last count, the number of stage, gives 6.

∴ Number of trays = 6

The number of theoretical trays needed is 20.

From the given data,

Equilibrium relation

y = 25x

[tex]L_s=L_2(1-x_2)\\L_s=300(1-0.04)= 288Kmol[/tex]

[tex]G_sy_1+L_sx_2=G_sy_2+L_sx_1\\G_s(y_1-y_2)=L_s(x_1-x_2)\\x_1=\frac{0.002}{1-0.002}\\x_1=0.002\\y_1=0, y_2=?\\x_2=\frac{0.04}{1-0.04}; x_2=0.0417[/tex]

substituting the values into the equation;

[tex]11.42(0-y_2)=288(0.002-0.0417)\\0-11.42y_2=-11.4048\\y_2=0.998[/tex]

[tex]N=\frac{In[(\frac{(x_2-y_1)/m}{(x_1-y_1)/m}(1-A)+A }{In(1/A)}\\[/tex]

But [tex]A=\frac{L_s}{mGs}[/tex]

[tex]N=\frac{x_2-x_1}{(x_1-y_1)/m}=\frac{0.0417-0.002}{0.002}\\N=19.85[/tex]

The numbers of tray is approximately 20.

learn more about steam stripping and numbers of tray here

https://brainly.com/question/9349349

Answer:

Decrease to typical from utilizing lambda-decrease:  

The given lambda - math terms is, (λf.λx.f(f(fx)))(λy.y×3)2

The of taking the terms is significant in lambda - math,  

For the term, (λy, y×3)2, we can substitute the incentive to the capacity.  

Therefore apply beta-decrease on “(λy, y×3)2,“ will return 2 × 3 = 6  

Presently the tem becomes, (λf λx f(f(fx)))6

The main term, (λf λx f(f(fx))) takes a capacity and a contention and substitute the contention in the capacity.  

Here it is given that it is conceivable to substitute, the subsequent increase in the outcome.  

In this way by applying next level beta - decrease, the term becomes f(f(f(6))), which is in ordinary structure.

Answer:

Explanation:

See the attachment for a step by step solution to the question.

Answer:

See the attached file for the answer.

Explanation:

See the attached file for the explanation

Answer:

Explanation:

Mass of the reel is given as,

M = 30kg

Radius of gyration about A is given as,

Ka = 120mm = 0.12m

Mass of the cylinder suspended

Mc = 40kg

Then, weight of the cylinder is

W = mg = Mc × g

W = 40 × 9.81 = 392.4 N

The exert force

P = 300N

Radius of the reel

R = 150mm = 0.15m

r = 75mm = 0.075m

Check attached for diagram of this problem

A. Moment of inertial of the reel?

We will assume the reel is a thin hoop shape, so we will use the thin hoop shape formula

I = M•Ka²

I = 30 × 0.12²

I = 0.432 kgm²

Therefore, Moment of inertia of reel is 0.432 kg.m²

B. Velocity if the cylinder after it moves 2m?

Applying torque equation

Στ = Iα

Where

α is angular acceleration

I is moment of inertia

τ is the torque..

Where τ = F × r

The two force acting on the reel is the weight of the cylinder and it is acting downward and the force applied

Then, taking torque about point A

Στ = Iα.

P × R — W × r = Iα

300 × 0.15 - 392.4 × 0.075 = 0.432 × α

45 — 29.43= 0.432α

15.57 = 0.432α

α = 15.57 / 0.432

α = 36.042 rad/s²

Now, The angle turned by the reel when the cylinder moves 2m above can be determined using

S= rθ

Then, θ = S/r

θ = 2 / 0.075

θ = 26.667 rad

Initially, the reel is at rest, then, the initial angular velocity is 0

ωo = 0 rad/s

Now, apply the kinematic equation and calculate the final angular velocity of the reel,

ω² = ωo² + 2αθ

ω² = 0² + 2 × 36.042 × 26.667

ω² = 0 + 1922.24

ω = √1922.24

ω = 43.84 rad/s

Now, using the relationship between linear velocity and angular velocity

Then, the velocity of cylinder

V = rω

V = 0.075 × 43.84

V = 3.2883 m/s

Therefore, Velocity of Cylinder is 3.288 m/s.

C. Time taken to travel 2m

From equation of circular motion

ω = ωo + αt

43.84 = 0 + 36.042t

43.84 = 36.042t

t = 43.84 / 36.042

t = 1.216 seconds

The time taken to travel 2m is 1.216 seconds

Answer:

(a) Jn = 64.08 μA/m²

(b) Jp = 80.1 μA/m²

(c) Jp = 22.94 μA/m²

(d) Jp = 3.6357 пA/m²

(e) Jn = 22.63 μA/m²

Explanation:

See the attached file for the explanation.

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

The factor of safety is 1.186

Explanation:

Find the attachment

Answer:

Explanation:

Find attach the solution

Answer:

Final velocities are:

Wedge B: v = 2.334 m/s

Sphere A: v = 5.386 m/s

Explanation:

Given:-

- The mass of sphere A,  mA = 2-kg

- The mass of the wedge B,  mB = 6-kg

- The sphere collides with " normal " to the wedge face.

- The coefficient of restitution , e = 0.5

- The wedge inclination angle, θ=40 degrees with the horizontal.

- The initial speed of sphere A, vA = 4m/s

- The initial speed of wedge B, vB = 0 m/s ... ( rest )

Find:-

- First step would be to sketch a system of sphere A and wedge B as ( FBD ).

- We will add a sketch of "two" coordinate axes on the ( FBD ).

   First coordinate system ( normal ( n ) - tangent ( t ) )

    Second coordinate system ( horizontal ( x ) - vertical ( y ) )

Note:- All the above directions of coordinate axes denote the positive direction of vectors.

- Resolve the angle ( α ) between the normal - ( n ) or velocity vector vA and the horizontal - ( x ) axis.

                            α = 90° - θ = 90° - 40°

                            α = 50°  

- We will denote the final velocity components of sphere A as ( v'An , v'At ):

And, final velocity components of wedge B as ( v'B ).

- Transform the the final velocity of wedge ( vB' ) in the (n-t) coordinate axis using the resolved coordinate transformation angle ( α ). The normal  ( n ) and tangential components of the velocity vector ( vB' ) are: ( vB'n , vB't ).

                     vB'n = vB'*cos ( α )             vB't = vB'*sin ( α° )

                    vB'n = vB'*cos ( 50° )         vB't = vB'*sin ( 50° )

- Note: There is no y-component of velocity of wedge. This is because the motion of wedge in the vertical direction is restricted by the ground - equilibrium conditions. Hence, v'By = 0.

- Consider the system of sphere A and wedge B to be isolated with no external forces acting on the system. For such conditions the principle of conservation of momentum ( P ) is valid. Which states:

                        PA,i + PBi = PA,f + PB,f

Where,

           PA,i : The initial momentum of the sphere A

           PB,i : The initial momentum of wedge B

           PA,f : The final momentum of the sphere A

           PB,f : The final momentum of wedge B.

- Using the data given and relations computed in the ( n-t ) coordinate system. We will use the principle of conservation of linear momentum in both axis ( n and t ) combined in vector momentum of system.

Conservation of linear momentum:

                      [tex]m_A*v_A + m_B*v_B = m_A*v_A' + m_B*v_B'[/tex]

- Using vector notations we have, Taking ( i unit vector in tangential direction and j unit vector in the normal direction ):

                       [tex]m_A*v_A_n j = m_A*( v_A'_t i + v_A'_n j )+ m_B* ( v_B'_n j + v_B'_t i )\\\\2*4 j = 2*( v_A'_t i + v_A'_n j )+ 6* ( v_B'_n j + v_B'_t i )\\\\0 i + 8 j = ( 2*v_A'_t + 6*v_B'*sin ( 50 ) ) i + ( 2*v_A'_n + 6*v_B'* cos (50 ) ) j[/tex]

- Equate all the ( i - j ) vectors on left and right hand side of the equation respectively,

               [tex]0 = 2*v_A'_t + 6*v_B'*sin ( 50 )\\\\\\ 8 = 2*v_A'_n + 6*v_B'* cos(50 )[/tex]         .... Eq 1

- The coefficient of restitution ( e ) is a squared loss of kinetic energy of the system that can be expressed in terms of velocities of two objects as relative change in velocity of two objects after and before impact.:

                             [tex]e = \frac{v_B'_n - v_A'_n}{v_A_n - v_B_n}[/tex]

Note: We have only used the velocities normal to the surface of wedge. This is because the kinetic loss is a scalar dimension; hence, the normal direction to the surface of impact is assumed to cater all the loss in kinetic energy.

We have,

                            [tex]0.5 = \frac{v_B'*cos ( 50 ) - v_A'_n}{ 4 - 0}\\\\2 = v_B'*cos ( 50 ) - v_A'_n \\\\v_B'*cos ( 50 ) = v_A'_n + 2[/tex]       ..... Eq 2

- Now substitute equation 2 into equation 1:

                             [tex]8 = 2*v_A'_n + 6*( v_A'_n + 2 )\\\\-4 = 8*v_A'_n\\\\v_A'_n = -0.5 \frac{m}{s}[/tex]

- Using Equation 2 compute v'B:

                             [tex]v_B'*cos ( 50 ) = -0.5 + 2\\\\v_B'= \frac{1.5}{cos ( 50 ) } \\\\v_B'= 2.334 \frac{m}{s}[/tex]

- Using Equation 1 compute v'At:

                             [tex]0 = 2*v_A'_t + 6*2.334*sin ( 50 )\\\\v_A'_t = - \frac{6*2.334*sin ( 50 )}{2} \\\\v_A'_t = - 5.363 \frac{m}{s}[/tex]

- The final velocity of wedge B along the horizontal direction ( x ) is:

                          vB' = 2.334 m/s   .... x-direction

- The velocity of sphere A after impact is given by:

                          [tex]v_A' = \sqrt{v_A't^2 + v_A'n^2}\\\\v_A' = \sqrt{5.363^2 + 0.5^2}\\\\v_A' = \sqrt{29.011769}\\\\v_A' = 5.386 \frac{m}{s}[/tex]

Answer:

The static thrust of the engine is 20,856.44N

Explanation:

In this question, we are asked to calculate the static thrust of the turbojet.

Please check attachment for complete solution and step by step explanation