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Serhud [2]
4 years ago
13

Consider a dip-coating process where a very long (assume infinitely long) wire(solid) with radius, ri, is being pulled verticall

y upward with axial velocity, V0, in a co-centric configuration through a very long (assume infinitely long) cylinder with radius, ro . Assume this annular flow is steady state, incompressible, fully-developedand laminar. The liquid density, ????, viscosity, µ, and gravity, g, are also known constants.
a. Simplify the Navier-Stokes equation for this flow (include the body force).
b. Write the boundary conditions for this flow.
c. Determine the velocity profile, uz(r) for the dip-coating process considered here by solving the simplified Navier-Stokes equation in section (a)
d. Determine the shear stress on the surface of the wire (Note: there is no pressure-gradient in this flow, only body force).

Engineering
1 answer:
Gekata [30.6K]4 years ago
6 0

Answer:

See explaination and attachment.

Explanation:

Navier-Stokes equation is to momentum what the continuity equation is to conservation of mass. It simply enforces F=ma in an Eulerian frame.

The starting point of the Navier-Stokes equations is the equilibrium equation.

The first key step is to partition the stress in the equations into hydrostatic (pressure) and deviatoric constituents.

The second step is to relate the deviatoric stress to viscosity in the fluid.

The final step is to impose any special cases of interest, usually incompressibility.

Please kindly check attachment for step by step solution.

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What is the the force available at the roadway surface to perform work?
scoray [572]

Answer:

Tractive force or traction

Explanation:

The main purpose of the tractive forces is to improve the ability to <u>transform the engine's energy into the vehicle's movement.</u> There are several systems that have different qualities and uses. Here we explain how they work and what they are for.

In a traction vehicle with one of its axles, its ability to transmit engine power to the ground is limited for two reasons:

-  At least one of the wheels must have adhesion, and this as long as it has a self-locking differential. Otherwise, simply with one lacking grip, we can no longer move forward.

- If there are two wheels that must distribute the power, it will always be easier to saturate the traction capacity of the tire than if we divide the force by four. The example is very simple: if we try to drag an object on the ground by pulling a rubber, it will stretch more than if we pull 4 identical tires, although the force we make is the same.

Contrary to what one might think, all-wheel drive vehicles are nothing recent. What happens is that it did take time to reduce the weight and size, as well as to increase the resistance of the homokinetic joints (they are articulations on the axles to allow the wheel to go up and down with the suspension or turn with the steering) to to adapt these systems to cars.

7 0
3 years ago
1. The system must be able to manage multiple students (max of 15) and their grades for assignments from three different assignm
mr_godi [17]

Answer:

#include<iostream>

#include<string.h>

using namespace std;

class gradebook;

class student

{

friend class gradebook;

char name[20]; // std name

int homework_grades[6], quiz_grades[6], exam_grades[4]; //grades array

float homework_avg, quiz_avg, exam_avg, final_avg;

};

class gradebook

{

student list[16];

public:

char gb_name[20]; // gradebook name

int max, total;

// Constructor

gradebook()

{

strcpy(gb_name,"noname");

max = 5;

total = 0;

for(int i=1;i<=15;i++)

{

strcpy(list[i].name,"noname");

for(int j=1;j<=5;j++)

{

list[i].homework_grades[j] = -1;

list[i].quiz_grades[j] = -1;

}

list[i].exam_grades[1] = list[i].exam_grades[2] = list[i].exam_grades[3] = -1;

}

}

// Member functions

void display_std_name();

void display_grades(int, int);

int add_std(char *new_name);

int remove(int);

//int ch_grade(int, int);

};

// Define all functions

// Remove a student from list and adjust remaining

int gradebook::remove(int std_id)

{

if(total==0 || std_id>total || std_id<=0)

return 0;

if(std_id==total)

{

cout<<"\n“"<<list[std_id].name<<"” has been successfully removed! (New class size: 2)";

strcpy(list[total--].name,"noname");

return 1;

}

cout<<"\n“"<<list[std_id].name<<"” has been successfully removed! (New class size: 2)";

for(int i=std_id; i<total; i++)

{

list[i]=list[i+1];

}

total--;

return 1;

}

// Add student to gradebook

int gradebook::add_std(char *new_name)

{

if(total<15)

{

strcpy(list[++total].name, new_name);

return 1;

}

return 0;

}

// Display student name list

void gradebook::display_std_name()

{

for(int i=1;i<=total;i++)

{

cout<<"\n"<<i<<" "<<list[i].name;

}

}

/* Display grades of students

// (grade_id, student_number_in_list)

homework(1), quiz(2), exam(3)

*/

void gradebook::display_grades(int g_id, int std_id)

{

int i;

if(std_id<=total)

{

if(g_id==1)

{

for(i=1;i<6;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].homework_grades[i]==-1) cout<<"<ungraded>";

}

}

else if(g_id==2)

{

for(i=1;i<6;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].quiz_grades[i]==-1) cout<<"<ungraded>";

}

}

else if(g_id==3)

{

for(i=1;i<4;i++)

{

cout<<"\n"<<i<<". ";

if(list[std_id].exam_grades[i]==-1) cout<<"<ungraded>";

}

}

}

else

cout<<"\nNo such std exists";

}

/****************************** MAIN ****************************/

int main()

{

int main_choice, grade_choice, std_id, new_grade;

char g_name[20], s_name[20];

float a,b,c;

gradebook g;

do

{

cout<<"\n\n-=| MAIN MENU |=-";

cout<<"\n1. Add a student\n2. Remove a student\n3. Change a student’s grade\n4. Change group weights\n5. Change the gradebook name\n6. Display class averages\n7. Display full report\n0. QUIT\n";

cout<<"\nEnter an action: ";

cin>>main_choice;

switch(main_choice)

{

case 0:

break;

case 1:

cout<<"\n-=| ADDING STUDENT |=-";

cout<<"\nPlease enter the student’s name: ";

cin>>s_name;

if(g.add_std(s_name))

cout<<"\n"<<s_name<<" was successfully added to the gradebook!";

else

cout<<"\nStudents cannot be added because the gradebook is full!";

break;

case 2:

cout<<"\n-=| REMOVING STUDENT |=-";

if(g.total!=0)

{

g.display_std_name();

cout<<"\n Enter student to remove: ";

cin>>std_id;

if(!g.remove(std_id))

cout<<"\nCannot remove students";

}

else

cout<<"\nStudents cannot be added because the gradebook is full!";

break;

case 3:

cout<<"\n-=| CHANGING GRADE |=-\n1. Change a homework grade\n2. Change a quiz grade\n3. Change an exam grade";

cout<<"\nWhat type of grade would you like to change: ";

cin>>grade_choice;

g.display_std_name();

cout<<"\nWhich student’s grade would you like to change? ";

cin>>std_id;

//ch_grade(grade_choice, std_id);

g.display_grades(grade_choice, std_id);

break;

/*

case 4:

cout<<"\n-=| CHANGING WEIGHTS |=-";

while(1)

{

cout<<"\nEnter the weights, separated by spaces, in the order of homework, quizzes, and exams (total must add up to 1.0):";

cin>>a>>b>>c;

if(a+b+c==1.00)

break;

else

cout<<"\n Weights do not add up to 1.0, try again...";

}

break;

case 5:

cout<<"\n-=| CHANGING NAME |=-\nPlease enter the new name for the gradebook: ";

cin>>g_name;

cout<<"\n Gradebook name changed from “ ” to "<<g_name;

break;

case 6:

cout<<"\n-=| CLASS AVERAGES |=-";

cout<<"\nHomework average of class: ";

cout<<"\nQuiz average of class: ";

cout<<"\nExam average of class: ";

cout<<"\nFinal average of class: ";

break;

case 7:

break;

*/

default:

cout<<"\n\nInvalid input\n\n";

}

}while(main_choice!=0);

return 0;

}

Explanation:

6 0
3 years ago
Prompt the user to input an integer, a double, a character, and a string, storing each into separate variables. Then, output tho
Likurg_2 [28]

Answer:

See explanation

Explanation:

//Include the

//required header files.

#include <stdio.h>

//Define the

//main() function.

int main(void) {

//Declare the

//required variables.

char input_char;

int input_int;

double input_double;

char input_string[100];

//Prompt the user

//to enter an integer.

printf("Enter integer: ");

//Read and store

//the integer.

scanf("%d", &input_int);

//Prompt the user

//to enter a double value.

printf("Enter double: ");

//Read and store

//the double value.

scanf("%lf", &input_double);

//Prompt the user

//to enter a character.

printf("Enter character: ");

//Read and store

//the character.

scanf(" %c", &input_char);

//Prompt user to

//enter the string

printf("Enter string: ");

//Read and

//store the string.

scanf("%s", input_string);

//(1)

//Display the values.

printf("%d %lf %c %s\n",

input_int, input_double,

input_char, input_string);

//(2)

//Display the values

//in reverse order.

printf("%s %c %lf %d\n",

input_string, input_char,

input_double, input_int);

//(3)

//Cast the double to

//an integer and display it.

printf("%lf cast to an integer is %d",

input_double, (int)(input_double));

//Return from the

//main() function.

return 0;

}

4 0
3 years ago
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as ______.​
DiKsa [7]
I believe it is cold forging?

sorry this answer was very confusing.
4 0
3 years ago
A. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant eq
Sever21 [200]

Answer:

a) Solid solution strengthening and alloying,  Precipitation hardening, work hardening

b) Absence of enough  crystallographic misalignment in the grain boundary region for a small-angle

Explanation:

<u>A) strengthening mechanism</u>

i) Solid solution strengthening and alloying:

In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal

also the shear stress required can be represented as:  Δz = Gb√Ce^3/2

where : C = solute concentration , e = strain on material

ii) Precipitation hardening:

During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments

particle bowing cab be written as :  Δz = Gb / L-2x

iii) work hardening :

Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism

dislocation can be represented as ; Gb √ p

where : G = shear modulus , b = Burgess vector, p = dislocation density

B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle

3 0
3 years ago
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