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3 years ago

What are the nine Historical periods?

Engineering

1 answer:

Rufina [12.5K]3 years ago

7 0

Answer:

https://quizlet.com/148993376/the-nine-distinct-periods-of-time-flash-cards/

Explanation:

you can find them all : )

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System grounding on a power system means electrically connecting the __?__ of every wye-connected transformer or generator to ea

irinina [24]

Answer: Neutrals

Explanation: System grounding on a power system is a term used to describe the entire processes involved when a neutral is used as the conductor to connect to the solid earth. This ensures that power is generated. This is usually done using either an inductor, an impendance or a resistor. It is very important and necessary to carry out a proper grounding of a power system in order to ensure the safety of the equipment and the personnel etc

6 0

3 years ago

A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The

Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

s1 = s-P1 = 0.6492 KJ/kg.K

v1 = v-P1 = 0.001010 m^3 / kg

P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

$w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}$

- From the following relation we can determine ( h2 ) as follows:

h2 = h1 + wp

h2 = 191.81 + 8.0699

h2 = 199.88 KJ/kg

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

P3 = 8 MPa

T3 = ? ( assume fluid exist in the saturated vapor phase )

h3 = hg-P3 = 2758.7 KJ/kg

s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

$q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}$

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

P4 = 10 KPa

s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

- Compute the quality of the mixture at condenser inlet by the following relation:

$x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947$

- Determine the isentropic ( h4s ) at this state as follows:

$h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}$

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

$h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\$

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

$w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}$

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

$W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}$

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

$n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31$

Answer: The thermal efficiency of the cycle is 0.31

7 0

3 years ago

The angle of twist can be computed using the material’s shear modulus if and only if: (a)- The shear stress is still in the elas

ollegr [7]

Answer:

The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region

Explanation:

The shear modulus (G) is the ratio of shear stress to shear strain. Like the modulus of elasticity, the shear modulus is governed by Hooke’s Law: the relationship between shear stress and shear strain is proportional up to the proportional limit of the material. The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region.

3 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 four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

3 years ago