piston cylinder has 5kg water at 500kPa and 300c. Heated by 100V and 4A that operates for 500 secs. It's cooled to a saturated liquid. what is the boundary work and heat lost?

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();

}

Answer:

The question is mentioned in the attachment.

Explanation:

Answer:

There is very little possibility of ionic bonding.

Explanation:

First we need to understand the meaning of ionic bonding. ionic bonding means the complete transfer of valance electrons between atoms.

The electronegativity of Al is 1.5 whereas electronegativity of nickel "Ni" is 1.8-1.9. As we can see from these values of electronegativity of Al & Ni that they are closer to each other & the difference of electonegativity of these metals is very little which decreases the possibility of ionic bonding between them so it is not expected that there will be much ionic bonding. Moreover it is noticeable here that these both are metals and they prefer to give up their electrons rather to gain them.

The two elements can not form ionic bonds because ionic bonds are formed when the electronegativity difference is greater than 1.5.

The term intermetallic compound refers to a compound that is composed of two metallic elements. Let us recall that metals often have a low value of electronegativity.

Owing to the fact that the electronegativity difference between nickel and aluminium is small (about 0.1). An ionic bond can not be formed between the two elements because ionic bonds are formed when the electronegativity difference is greater than 1.5.

Learn more about ionic bonds: https://brainly.com/question/867804

Answer:

The answer to the question is 514.17 lbf

Explanation:

Volume of cylindrical tank = πr²h = 3.92699 ft³

Weight of tank = 125 lb

Specific weight of content = 66.4 lb/ft³

Mass of content =  66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

Answer:

Explanation:

The principle and application of differential equation is applied and the steps with detailed and appropriate substitution is as carefully shown in the attached files.

m = mass attached

k = spring constant

c = damping coefficient

Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

Answer:

The triangle() function is an inbuilt function in p5.js which is used to draw a triangle in a plane. This function accepts three vertices of triangle.

Syntax:

triangle(x1, y1, x2, y2, x3, y3)

Below programs illustrates the triangle() function in p5.js:

function setup() {  

   createCanvas(400, 400);  

}  

function draw() {  

   background(220);  

   fill('lightgreen');  

   // A triangle at (100, 250), (250, 170) and (330, 300)    

   triangle(100, 250, 250, 170, 330, 300);  

}  

Answer:

Design rate of superelevation= 3.4%

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Answer:

The answers to the question are

(a) T = 2·π·r/v

(b) 3.3 % change in period of the under-inflated tire compared to the properly inflated tire

(c) Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters.

Explanation:

(a) The period T = 2π/ω

The velocity v = ωr or ω = v/r

Therefore T = 2π/(v/r) = 2πr/v

T = 2·π·r/v

(b) Period of properly inflated tire with radius = 303 mm is 2π303/v

Period of under-inflated tire with radius = 293 mm is 2π293/v

Therefore we have percentage change in period of  of the under-inflated tire compared to the properly inflated tire is given by

(2π303/v -2π293/v)/(2π303/v) = 2π10/v/(2π303/v) = 10/303 × 100 = 3.3 %

(c) The period of the under-inflated tire is 10/303 less than that of the inflated tire. Therefore for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303)  = 303/10 or 30.3 revolutions of either tire which is 30.3×2×π×303 = 57685.296 mm = 57.685 meters

Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters

At 30.3 revolutions the distance covered by the under-inflated

= 55781.49 mm

Subtracting the two distances gives

1903.805 mm

The circumference of the inflated tire = 2×π×303 = 1903.805 mm

A) The mathematical equation that relates the period T of the wheel rotation of the tire radius r and the speed v of the car is; T = 2πr/v

B) The percent change of the period of the underinflated tire compared to the properly inflated tire is; 3.3 %

C) The distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is; 57685.296 mm

(A) The formula for period is;

T = 2π/ω

angular velocity is;

ω = v/r

Thus; T = 2πr/v

where T is period, v is velocity and r is radius

(b) We are told that radius is 303 mm. Thus Period of inflated tire with radius is;

T =  2π ×  303/v

Period of under-inflated tire with radius of 293 mm is;

T = 2π× 293/v

Percentage Change is;

((2π ×  303/v) - (2π× 293/v))/(2π ×  303/v) * 100% = 10/303 * 100% = 3.3 %

(c) From B above we see that the period of the under-inflated tire is 10/303 less than that of the inflated tire.

Thus,  for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303)  = 30.3 revolutions of either tire.

Thus, the distance is;

d = 30.3 × 2π × 303

d = 57685.296 mm

Using the same concept, at 30.3 revolutions, the distance covered by the under-inflated tire is calculated to be; 55781.49 mm

Difference in distance = 57685.296 mm -  55781.49 mm

Difference in distance = 1903.805 mm

Read more about motion of cars at; https://brainly.com/question/1315051

The maximum transmitted power throughout the solid circular rod is approximately 86.066 horse power.

Let consider that the solid circular rod experiments an uniform torsion and that the material represents a ductile material.

The solid circular rod rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, and all losses throughout the system are negligible. Net power ([tex]\dot W[/tex]), in horse power, is calculated as the product of the torque ([tex]T[/tex]), in kilopound-force-inches, and the angular speed. That is to say:

[tex]\dot W = T \cdot \omega[/tex]   (1)

In addition, we add a formula that relates the torque with the shear stress experimented by the solid circular rod:

[tex]\tau = \frac{T \cdot D}{2 \cdot J}[/tex]   (2)

Where [tex]J[/tex] is the torsion module, in quartic inches, whose formula for a solid circular cross section is described below:

[tex]J = \frac{\pi \cdot D^{4}}{32}[/tex]   (3)

The torsion module is calculated herein: ([tex]D = 2.5\,in[/tex])

[tex]J = 3.835\, in^{4}[/tex]

The maximum allowable torsion is found by isolating the corresponding variable in (2): ([tex]J = 3.835 in^{4}[/tex], [tex]D = 2.5\,in[/tex], )

[tex]T_{max} = \frac{2 \cdot J \cdot \tau_{max}}{D}[/tex]

[tex]T_{max} = 31.907\, kip \cdot in\,(2.659 \,kip \cdot ft)[/tex]

Then, the maximum transmitted power is determined by (1):

[tex]\dot W = (2.659 kip \cdot ft) \cdot (2)\cdot (\pi) \cdot (170 rpm)\\\dot W \approx 2840.188 \,\frac{kip \cdot ft}{min} \\\dot W \approx 86.066 \,h.p.[/tex]

The maximum transmitted power throughout the solid circular rod is approximately 86.066 horse power. [tex]\blacksquare[/tex]

To learn more on power transmission, we kindly invite to check this verified question: https://brainly.com/question/12750305

Answer:

Explanation:20% of food energy is converted into mechanical energy

Answer:

The answer is in the attachment. Thanks

Explanation:

Answer: The theoretical stopping distance is 1.245 miles (6572.193 ft).

Explanation:

First, the physical model is created by using Principle of Energy Conservation and Work-Energy Theorem. It is assumed that surface is horizontal, so there are no changes associated with potential energy. The car has an initial kinetic energy, which is completely dissipated by braking.

[tex]K_{1} = \Delta W_{loss}[/tex]

[tex]\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (\mu_{2} \cdot \Delta s_{2}+\mu_{2'} \cdot \Delta s_{2'})[/tex]

The distance is isolated from previous equation:

[tex]\Delta s_{2'} = \frac{1}{\mu_{2'}}\cdot (\frac{1}{2}\cdot \frac{v^{2}}{g} - \mu_{2} \cdot \Delta s_{2})[/tex]

By replacing variables, the distance is calculated herein:

[tex]\Delta s_{2'} = \frac{1}{\frac{0.02}{0.80}}\cdot (\frac{1}{2}\cdot \frac{(102.667\frac{ft}{s} )^2}{32.174 \frac{ft}{s^{2}}} - (0.02) \cdot (100 ft))[/tex]

[tex]\Delta s_{2'} = 6472.193 ft\\\Delta s_{2'} = 1.226 mi[/tex]

The theoretical stopping distance is:

[tex]\Delta s = \Delta s_{2} + \Delta s_{2'}\\\Delta s = 6572.193 ft\\\Delta s = 1.245 mi[/tex]

Answer:

Attached is a solution hope it helps:

Answer:

What do i have to do

Explanation:

what do i do

attached below is the answer

I would recommend installing both a 100-Base T network and a Wi-Fi network in the new office building for Eureka.

Based on the information provided, I would recommend installing both a 100-Base T network as well as a Wi-Fi network in the new office building for Eureka.

A 100-Base T network, also known as Ethernet, provides reliable and high-speed wired connections, perfect for the staff members who spend a significant amount of time on their computers searching the Internet. It ensures a stable and fast connection for their work.

On the other hand, a Wi-Fi network allows for flexible and convenient wireless connectivity, which is beneficial for staff members who frequently use their phones or move around the office space. It provides them with the freedom to connect wirelessly wherever they are in the building.

https://brainly.com/question/33827192

#SPJ3

Answer:

Sampling rate = 90

23 and 45 Hz

Explanation:

The Nyquist criteria states that the sampling frequency must be at least twice of the highest frequency component.

(a) We have 10, 12, 13, 23, and 45 Hz signals

The highest frequency component in this list is 45 Hz

So minimum sampling frequency needed to avoid aliasing would be

Sampling rate = 2*45 = 90 sps

Hence a sampling rate of 90 samples per second would be required to avoid aliasing.

(b) if a sampling rate of 40 sps is used then

40 = 2*f

f = 40/2 = 20 Hz

Hence 23 and 45 Hz components will be aliased.

Answer:

Please find attached

Explanation:

Answer:

Explanation:

we will begin by carefully following a step by step order;

Total time in a week = 40 hours = 2400 minutes

Total time Administrator is utilized = 50 * 20 = 1000 minutes

Total time Senior accountant is utilized= 20% * 50 * 40 = 400 minutes

Total time Junior accountant is utilized = 40%*50*15 = 600 minutes

For Total cases, Total capacity = Total time / Bottlenck time (Administrator) = 2400 / 20 = 120 cases

C-3) Capacity of new customers at 20:80 ratio = 20% * 120 = 24 cases

C-4) Capacity of repeat customers at 20:80 ratio = 20%*120 = 96 cases

c-1 = Flow rate = min ( demand, capacity) for new customers = 10 new cases / week (20%*50)

c-2 = Flow rate = 40%*50 = 40 cases / week

c3 = 24 cases

c4 = 96 cases

cheers i hope this helps

Answer:

The gm required is 0.1 [tex]\frac{mA}{V}[/tex]  

Explanation:

Knowing

Vpeak = 0.50 V

Rd = 50 kΩ

Av = 5 V/V

To find the transconductance we should to use the equation

Av = gm * Rd

so,

gm = Av/Rd

gm = 5/50 = 0.1 [tex]\frac{mA}{V}[/tex]  

Answer:

Explanation:

Answer:

Explanation:

Answer:  

Explanation:  

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.  

(a). here we are asked to determine the Temperature and Pressure.  

Given that the properties of Air;  

ha = 230.02 KJ/Kg  

Ta = 230 K  

Pra = 0.5477  

From the energy balance equation for a diffuser;  

ha + Va²/2 = h₁ + V₁²/2  

h₁ = ha + Va²/2 (where V₁²/2 = 0)  

h₁ = 230.02 + 220²/2 ˣ 1/10³  

h₁ = 254.22 KJ/Kg  

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg  

from this we have;  

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]  

Pr₁ = 0.77759  

therefore T₁ = 254.15K  

P₁ = (Pr₁/Pra)Pa  

= 0.77759/0.5477 ˣ 26  

P₁ = 36.91 kPa  

now we calculate Pr₂  

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349  

⇒ now we obtain properties of air at  

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg  

calculating the enthalpy of air at state 2  

ηc = h₁ - h₂ / h₁ - h₂  

0.85 = 254.22 - 505.387 / 254.22 - h₂  

h₂ = 549.71 KJ/Kg  

to obtain the properties of air at h₂ = 549.71 KJ/Kg  

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913  

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at  

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁  

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄  

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃       i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄  + 0  = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

Answer:

The missing Table is;

Month  1  2  3  4  5  6  7

Value  24  13  21  14  20  23  15

MSE = 20.56

Forecast = 19.33

Explanation:

Table Formation

Month Value Forecast Error  Absolute Error Error Square

1          24    

2          13    

3          21    

4          14         19.33333 -5.33333 5.333333 28.44444

5          20           16                       4              4             16

6          23          18.33333 4.666667 4.666667 21.77778

7          15           19                   -4                       4             16

    20.55

    19.33

Table Explanation

In the above table, we first calculate the forecast column. A forecast is taken as the sum of the three entries in order and then taking out the average.

Error is calculated by subtracting the Actual value from the forecast value.

Then we take the absolute of the forecast value.

Then we take the square of the absolute of the forecasted values.

Now we calculate MSE as;

MSE = ( 28.44 + 16 + 21.78 + 16 ) / 4

MSE = 20.56

Forecast for 8th month  = ( 20 + 23 + 15 ) / 3

Forecast for 8th month  = 19.33

Answer:

maximum length to which it may be stretched without causing plastic deformation is 200 + 0.4782 = 2.4782 mm

Explanation:

given data

stress = 275 MPa

modulus of elasticity = 115 GPa

solution

we will apply here modulus of elasticity  formula that is

modulus of elasticity = stress ÷ strain   ....................1

here we consider original length is 200 mm

so strain is = [tex]\frac{\triangle L}{original\ length }[/tex]  ..........2

put here value in equation 1 we get

115 ×  [tex]10^{9}[/tex] = [tex]\frac{275 \times 10^{6}}{\frac{x}{200} }[/tex]

solve it we get

x = [tex]\frac{11}{23}[/tex]  

x = 0.4782

so maximum length to which it may be stretched without causing plastic deformation is 200 + 0.4782 = 2.4782 mm

a) Heat transfer to the working fluid in the steam generator: 3303.9 kJ/kg

b) Thermal efficiency: 100%

c) Heat transfer from the working fluid in the condenser to the cooling water: 2083.6 kJ/kg

Given parameters:

- Superheated vapor enters turbine: P₁ = 10 MPa, T₁ = 480°C

- Condenser pressure: P₃ = 6 kPa

- Extraction pressure for closed feedwater heater: P₂ = 0.7 MPa

- Isentropic efficiencies of pump and turbine stages: ηᵥ = ηₜ = 80%

a) Heat Transfer in the Steam Generator:

1. From the steam tables, at P₁ = 10 MPa:

  - h₁ = 3478.1 kJ/kg (enthalpy of superheated vapor)

2. At P₂ = 0.7 MPa:

  - h₂ = 209.4 kJ/kg (enthalpy of extracted steam)

3. The enthalpy at the inlet of the closed feedwater heater is the same as h₂:

  - h₃ = 209.4 kJ/kg

4. At P₃ = 6 kPa, as condensate:

  - h₄ = 191.8 kJ/kg (enthalpy of saturated liquid)

5. The enthalpy at the outlet of the closed feedwater heater is the same as h₄:

  - h₅ = 191.8 kJ/kg

6. Pump work per unit mass of water: [tex]W_{\text{pump}}[/tex]= h₅ - h₃

  - [tex]\( W_{\text{pump}} = 191.8 - 209.4 = -17.6 \, \text{kJ/kg} \)[/tex]

7. Actual work output from turbine per unit mass of steam: [tex]W_{\text{act}}[/tex]= h₁ - h₂

  [tex]\( W_{\text{act}} = 3478.1 - 209.4 = 3268.7 \, \text{kJ/kg} \)[/tex]

8. Isentropic work output from turbine per unit mass of steam: [tex]\( W_{\text{isentropic}} = \frac{W_{\text{act}}}{\eta_t} \)[/tex]

   [tex]\( W_{\text{isentropic}} = \frac{3268.7}{0.8} = 4085.9 \, \text{kJ/kg} \)[/tex]

9. Heat input per unit mass of steam:  [tex]Q_{\text{in}}[/tex] = h₁ - h₅

  [tex]\( Q_{\text{in}} = 3478.1 - 191.8 = 3286.3 \, \text{kJ/kg} \)[/tex]

10. Heat transfer in the steam generator: [tex]\( Q_{\text{gen}} = Q_{\text{in}} - |W_{\text{pump}}| \)[/tex]

  - [tex]\( Q_{\text{gen}} = 3286.3 - |-17.6| = 3303.9 \, \text{kJ/kg} \)[/tex]

  - Rounded to one decimal place: 3303.9 kJ/kg

b) Thermal Efficiency:

Thermal efficiency (η) is given by: [tex]\( \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} \)[/tex]

where[tex]\( W_{\text{net}} = W_{\text{act}} - |W_{\text{pump}}| \)[/tex]

[tex]\( \eta = \frac{3268.7 - |-17.6|}{3286.3} \times 100 \)\\\( \eta = \frac{3286.3}{3286.3} \times 100 \)\\\( \eta = 1 \times 100 \)\\\( \eta = 100\% \)[/tex]

c) Heat Transfer in the Condenser:

Heat transfer from working fluid in the condenser to cooling water:

[tex]Q_{\text{out}}[/tex]= h₄ - h₃

[tex]Q_{\text{out}} = 191.8 - 209.4 = -17.6 \, \text{kJ/kg}[/tex]

- Rounded to one decimal place: -17.6 kJ/kg ≈ -2083.6 kJ/kg

a) To find the heat transfer in the steam generator, we considered the enthalpies at different stages and calculated the pump work, actual work output from the turbine, isentropic work, heat input, and finally the heat transfer in the steam generator.

b) The thermal efficiency was calculated using the formula for efficiency, considering the net work output and heat input.

c) Heat transfer in the condenser was determined by comparing the enthalpies at the inlet and outlet of the closed feedwater heater, considering the negative sign due to the direction of heat flow from the working fluid to the cooling water.

Explanation:

Find the difference between the simple interest and the compound interest on Rs 5000 for 2 years at 9% per annum

Answer:

Technicians A

Explanation:

The technician A would also tap the the teeth with a blunt end to break it loose. He also remove the underlying O ring in the front case groove. Technicians A also remove the driven gear bolt that secures the oil pump driven gear to the left balance shaft.

Answer:

Answer is 0.74

Explanation:

Probability that the distance is at most 100 m:

P ( X ≤ 100 )  =  F ( 100 )

              =   (1 − e-^ λ *x)

              =   1 − e −^( ( 0.0134 )* ( 100 ))

              = 1 − e −^ 1.34

              = 1 − 0.2618

             = 0.7381

             ≈ 0.74

Therefore,

P ( X ≤ 100 )≈ 0.74

Answer.

Answer:

0.7400

Explanation:

Let's first look at what we have:

λ = 0.0134                                                                                                                         mean = 1/λ = 74.24                                                                                                      standard deviation = 1/λ = 74.24

We'll use the following formula;                                                                     CDF, C(x) = 1 - e^(-λx)

Probability of a distance not exceeding 100 m

P(X <= 100)

= C(100)

= 1 - e^(-0.01347 * 100)

= 0.7400

Answer:

The answer and explanation to this question is attached.

Answer:

Explanation:

Let first simplified the risk given our specific loss function. if f(x) = i i is not double , then the risk is

R(f(x) = i|x) = ∑ L(f(x) = i , y = j ) P(y = j|x)                                      2

                 =0.P (Y= i|x) +λc ∑ P (Y= j|x)                                      3

                 =λc (1 - P(Y= i|x))                                                        4

When f(x)  = c  + 1, meaning you have choosing doubt , the risk is

R(f(x) = c +1|x) = ∑ L (f(x)= c+1, y=j) P(Y=j|x)                                  5

        =λd∑ P(Y=j|x)                                                                       6

        =λd                                                                                      7

because   ∑ P(Y=j|x)  should sum to 1 since its a proper probability distribution.

Now let fopt : Rd→ {1, . . . , c + 1} be the decision rule which implements (R1)–(R3).We want to show that in expectation the rule foptis at least as good as an arbitrary rulef. Let x ∈ Rdbe a data point, which we want to classify. Let’s examine all the possiblescenarios where fopt(x) and another arbitrary rule f(x) might differ:Case 1: Let fopt(x) = i where i 6= c + 1.– Case 1a: f(x) = k where k 6= i. Then we get with (R1) thatR(fopt(x) = i|x) = λc1 − P(Y = i| x)≤ λc1 − P(Y = k|x)= R(f(x) = k|x).– Case 1b: f(x) = c + 1. Then we get with (R1) thatR(fopt(x) = i|x) = λc1 − P(Y = i| x)≤ λc(1 − (1 −λdλc)) = λd= R(f(x) = c + 1|x).Case 2: Let fopt(x) = c + 1 and f(x) = k where k 6= c + 1. Then:R(f(x) = k|x) = λc(1 − P (Y = k|x)R(fopt(x) = c + 1|x) = λ            

Answer:A

Explanation:

Damp roof is generally applied at basement level which restrict the movement of moisture through walls and floors. Therefore it could be inside or the outside basement walls.

Answer:

The correct option is A

Explanation:

Top down budgeting is a type of budget in which the senior management prepares a high level budget for the company. The senior management then allocates the amounts for the individual departments, which then use this number in preparing their own budget.