Technician A states that a scan tool can read

Steam enters an adiabatic condenser (heat exchanger) at a mass flow rate of 5.55 kg/s where it condensed to saturated liquid wat

Evgen [1.6K]

Answer:

The minimum mass flow rate will be "330 kg/s".

Explanation:

Given:

For steam,

$m_{s}=5.55 \ kg/s$

$\Delta h=2491 \ kg/kj$

For water,

$\Delta T=10^{\circ}C$

$(Cp)_{w}=4.184 \ kJ/kg^{\circ}C$

They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,

⇒ $m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T$

On putting the estimated values, we get

⇒ $5.55\times 2491=M_{w}\times 4.184\times 10\\$

⇒ $13825.05=M_{w}\times 41.84$

⇒ $M_{w}=330 \ kg/s$

7 0

3 years ago

A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum

adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress $\sigma$ = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder there exist both the circumferential stress and the longitudinal stress.

Circumferential stress $\sigma = \dfrac{pd}{2t}$

Making thickness t the subject; we have

$t = \dfrac{pd}{2* \sigma}$

$t = \dfrac{560000*3}{2*150000000}$

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

$\sigma = \dfrac{pd}{4t}$

$t= \dfrac{pd}{4*\sigma }$

$t = \dfrac{560000*3}{4*150000000}$

t = 0.0028 mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm

8 0

3 years ago

computer language C++ (Connect 4 game)( this is all the info that was givin no input or solution) I used the most recent version

Mariana [72]

Answer:

C++ code explained below

Explanation:

#include "hw6.h"

//---------------------------------------------------

// Constructor function

//---------------------------------------------------

Connect4::Connect4()

{

ClearBoard();

}

//---------------------------------------------------

// Destructor function

//---------------------------------------------------

Connect4::~Connect4()

{

// Intentionally empty

}

//---------------------------------------------------

// Clear the Connect4 board

//---------------------------------------------------

void Connect4::ClearBoard()

{

// Initialize Connect4 board

for (int c = 0; c < COLS; c++)

for (int r = 0; r < ROWS; r++)

board[r][c] = ' ';

// Initialize column counters

for (int c = 0; c < COLS; c++)

count[c] = 0;

}

//---------------------------------------------------

// Add player's piece to specified column in board

//---------------------------------------------------

bool Connect4::MakeMove(int col, char player)

{

// Error checking

if ((col < 0) || (col >= COLS) || (count[col] >= ROWS))

return false;

// Make move

int row = count[col];

board[row][col] = player;

count[col]++;

return true;

}

//---------------------------------------------------

// Check to see if player has won the game

//---------------------------------------------------

bool Connect4::CheckWin(char player)

{

// Loop over all starting positions

for (int c = 0; c < COLS; c++)

for (int r = 0; r < ROWS; r++)

if (board[r][c] == player)

{

// Check row

int count = 0;

for (int d = 0; d < WIN; d++)

if ((r+d < ROWS) &&

(board[r+d][c] == player)) count++;

if (count == WIN) return true;

// Check column

count = 0;

for (int d = 0; d < WIN; d++)

if ((c+d < COLS) &&

(board[r][c+d] == player)) count++;

if (count == WIN) return true;

// Check first diagonal

count = 0;

for (int d = 0; d < WIN; d++)

if ((r+d < ROWS) && (c+d < COLS) &&

(board[r+d][c+d] == player)) count++;

if (count == WIN) return true;

// Check second diagonal

count = 0;

for (int d = 0; d < WIN; d++)

if ((r-d >= 0) && (c+d < COLS) &&

(board[r-d][c+d] == player)) count++;

if (count == WIN) return true;

}

return false;

}

//---------------------------------------------------

// Print the Connect4 board

//---------------------------------------------------

void Connect4::PrintBoard()

{

// Print the Connect4 board

for (int r = ROWS-1; r >= 0; r--)

{

// Draw dashed line

cout << "+";

for (int c = 0; c < COLS; c++)

cout << "---+";

cout << "\n";

// Draw board contents

cout << "| ";

for (int c = 0; c < COLS; c++)

cout << board[r][c] << " | ";

cout << "\n";

}

// Draw dashed line

cout << "+";

for (int c = 0; c < COLS; c++)

cout << "---+";

cout << "\n";

// Draw column numbers

cout << " ";

for (int c = 0; c < COLS; c++)

cout << c << " ";

cout << "\n\n";

}

//---------------------------------------------------

// Main program to play Connect4 game

//---------------------------------------------------

int main()

{

int choice;

int counter = 0;

srand (time(NULL));

Connect4 board;

cout << "Welcome to Connect 4!" << endl << "Your Pieces will be labeled 'H' for human. While the computer's will be labeled 'C'" << endl;

board.PrintBoard();

cout << "Where would you like to make your first move? (0-6)";

cin >> choice;

while (board.MakeMove(choice,'H') == false){

cin >> choice;

}

counter++;

while (board.CheckWin('C') == false && board.CheckWin('H') == false && counter != 21){

while (board.MakeMove(rand() % 7, 'C') == false){}

board.PrintBoard();

cout << "Where would you like to make your next move?" << endl;

cin >> choice;

board.MakeMove(choice,'H');

while (board.MakeMove(choice,'H') == false){

cin >> choice;

}

counter++;

}

if (board.CheckWin('C')){

cout << "Computer Wins!" << endl;}

else if (counter == 21){cout << "Tie Game!" << endl;}

else {cout << "Human Wins!" << endl;}

board.PrintBoard();

}

4 0

3 years ago

A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f

KIM [24]

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) $\frac{dm}{2}$ ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = $\frac{L}{\pi dm}$ ...............2

here lead L = n × p

so tan(α) = $\frac{2\times 2}{\pi 19}$

α = 3.83°

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) $\frac{19\times 10^{-3}}{2}$

T = 1.59 Nm

so

power = $\frac{2\pi N \ T }{60}$ .................3

put here value

power = $\frac{2\pi \times 300\times 1.59}{60}$

power = 49.95 W

and

as φ > α

so it is self locking screw

8 0

3 years ago

What is Back EMF? How does it limits the speed of a permanent magnet DC?

ss7ja [257]

Answer and Explanation:

The DC motor has coils inside it which produces magnetic field inside the coil and due to thus magnetic field an emf is induced ,this induced emf is known as back emf. The back emf always acts against the applied voltage. It is represented by $E_b$

The back emf of the DC motor is given by $E_b=\frac{NP\Phi }{60A}$

Here N is speed of the motor ,P signifies the number of poles ,Z signifies the the total number of conductor and A is number of parallel paths

As from the relation we can see that back emf and speed ar dependent on each other it means back emf limits the speed of DC motor

8 0

3 years ago