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Nimfa-mama [501]
3 years ago
11

The kinetic energy correction factor depends on the (shape — volume - mass) of the cross section Of the pipe and the (velocity —

pressure — temperature distribution.
Engineering
1 answer:
butalik [34]3 years ago
3 0

Answer:

The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.

Explanation:

The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform.  In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.

You might be interested in
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
Along with refining craft skills another way to increase the odds for career advancement is to
Xelga [282]

The acquisition of additional certifications with a personal refined craft skills can increase the odds for career advancemen.

<h3>What is a career advancement?</h3>

An advancement is achieved in a career if a professional use their skill sets, determination or perserverance to achieve new career height.

An example of a career advancement is when an employee progresses from entry-level position to management and transits from an occupation to another.

Therefore, the Option A is correct.

Read more about career advancement

<em>brainly.com/question/7053706</em>

7 0
2 years ago
When you see a street with white markings only, what kind of street is it?
Georgia [21]

Answer:

it's a one way street

3 0
4 years ago
4.
Stella [2.4K]
A gas cylinders be cleaning chemical see wedding rods de leathers must be stored in and approve steel cabinet
3 0
3 years ago
A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.
Alexus [3.1K]

Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e  = \frac{V_v}{V_s} =  \frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)

Specific gravity G_{s} = \frac{P_s}{P_w} =  \frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)

% Saturation S = \frac{V_w}{Vv} × \frac{100}{1} =  \frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right} × \frac{100}{1}--------(3)

water content w =  \frac{M_w}{M_s} = \frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

∴ \frac{1}{G_s} = \frac{P_w}{P_s} ------(6)

putting eqn. (6) into (5)

w = \frac{V_w}{V_sG_s} -----(7)

Again, from eqn (1)

V_s = \frac{V_v}{e}

substituting into eqn. (7)

w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v}  = S

∴ w = \frac{Se}{G_s} -----(8)

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, G_{s} = 2.7, e= 0.9

∴w= \frac{0.9*1}{2.7} = 0.333

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

w= \frac{0.01*0.9 }{2.7} = 0.0033

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil (P_w=\frac{1_g}{cm^{3} } )

Given, V_{s} = 100 cm^{3 }, S = 50% = 0.5

%S = \frac{V_w}{V_v} × \frac{100}{1} = \frac{V_w}{eV_s} × \frac{100}{1}

0.50 = \frac{V_w}{0.9* 100}  = 45cm^{3}

mass of water = P_wV_w= 1 * 45 = 45_{g}

7 0
4 years ago
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