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3 years ago

Hi. I would like to know why one side of an island can get more rain (more rain forms), while the other gets less.

Engineering

2 answers:

adelina 88 [10]3 years ago

6 0

The elevation might be higher causing more rain

sdas [7]3 years ago

4 0

Answer:

It's because the moisture in the air condenses much quicker at a higher elevation. When the air condenses, rain clouds are formed. The side the air comes from is called the windward side, which is where there is more rainfall.

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A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at -30C by rejecting

chubhunter [2.5K]

Answer:

a) x = 0.4795

b) QL = 5.85 KW

c) COP = 2.33

d) QL_max = 12.72 KW

Explanation:

Solution:-

- Assuming the steady state flow conditions for both fluids R-134a and water.

- The thermodynamic properties remain constant for respective independent intensive properties.

- We will first evaluate the state properties of the R-134a and water.

- Compressor Inlet, ( Saturated Vapor ) - Ideal R-134a vapor cycle

P1 = 60 KPa, Tsat = -36.5°C

T1 = -34°C , h1 = hg = 230.03 KJ/kg

Qin = 450 W - surrounding heat

- Condenser Inlet, ( Super-heated R-134a vapor ):

P2 = 1.2 MPa , Tsat = 46.32°C

T2 = 65°C , h2 = 295.16 KJ/kg

- Condenser Outlet, ( Saturation R-134a point ):

P3 = P2 = 1.2 MPa , Tsat = 46.32°C

T3 = 42°C , h3 = hf = 111.23 KJ/kg

- R-134a is throttled to the pressure of P4 = compressor pressure = P1 = 60 KPa by an "isenthalpic - constant enthalpy pressure reduction" expansion valve.

- Inlet of Evaporator - ( liquid-vapor state )

P4 = P1 = 60 KPa, hf = 3.9 KJ/kg , hfg = 223.9 KJ/kg

h4 = h3 = 111.23 KJ/kg

- The quality ( x ) of the liquid-vapor R-134a at evaporator inlet can be determined:

x4 = ( h4 - hf ) / hfg

x4 = ( 111.23 - 3.9 ) / 223.9

x4 = 0.4795 Answer ( a )

- Water stream at a flow rate flow ( mw ) = 0.25 kg/s is used to take away heat from the R-134a.

- Condenser Inlet, ( Saturated liquid water ):

Ti = 18°C , h = hf = 75.47 KJ/kg

- Condenser Outlet, ( Saturated liquid water ):

To = 26°C , h = hf = 108.94 KJ/kg

- Since the heat of R-134a was exchanged with water in the condenser. The amount of heat added to water (Qh) is equal to amount of heat lost from refrigerant R-134a.

- Apply thermodynamic balance on the R-134a refrigerant in the condenser:

Qh = flow (mr) * [ h2 - h3 ]

Where,

flow ( mr ) : The flow rate of R-134a gas in the refrigeration cycle

flow ( mr ) = Qh / [ h2 - h3 ]

flow ( mr ) = 8.3675 / [ 295.16 - 111.23 ]

flow ( mr ) = 0.0455 kg/s

- The cooling load of the refrigeration cycle ( QL ) is determined from energy balance of the cycle net work input ( Compressor work input ) - "Win" and the amount of heat lost from R-134a in condenser ( Qh ).

- Apply the thermodynamic balance for the compressor:

Win = flow ( mr )*[ h2 - h1 ] - Qin

Win = 0.0455*[ 295.16 - 230.03] KW - 0.45 KW

Win = 2.513 KW

- The cooling load ( QL ) for the refrigeration cycle can now be calculated. Apply thermodynamic balance for the refrigeration cycle:

QL = Qh - Win

QL = 8.3675 - 2.513

QL= 5.85 KW .... Refrigeration Load, Answer ( b )

- The COP of the refrigeration cycle is calculated as the ratio of useful work and total work input required:

COP = QL / Win

COP = 5.85 / 2.513

COP = 2.33 Answer ( c )

- For a compressor to be working at 100% efficiency or ideal then the maximum COP for the refrigeration cycle would be:

COP_max = [ TL ] / [ Th - TL ]

Where,

TL : The absolute temperature of heat sink, refrigerated space

TH : The absolute temperature of heat source, water inlet

COP_max = [ -30+273 ] / [ (18+273) - (-30+273) ]

COP_max = 5.063

- The theoretical ideal refrigeration load ( QL max ) would be:

COP_max = QL_max / Win

QL_max = Win*COP_max

QL_max = 2.513*5.063

QL_max = 12.72 KW Answer ( d )

5 0

4 years ago

Implement

kolbaska11 [484]

Answer:

#include <iostream>

using namespace std;

// Pixel structure

struct Pixel

{

unsigned int red;

unsigned int green;

unsigned int blue;

Pixel() {

red = 0;

green = 0;

blue = 0;

}

};

// function prototype

int energy(Pixel** image, int x, int y, int width, int height);

// main function

int main() {

// create array of pixel 3 by 4

Pixel** image = new Pixel*[3];

for (int i = 0; i < 3; i++) {

image[i] = new Pixel[4];

}

// initialize array

image[0][0].red = 255;

image[0][0].green = 101;

image[0][0].blue = 51;

image[1][0].red = 255;

image[1][0].green = 101;

image[1][0].blue = 153;

image[2][0].red = 255;

image[2][0].green = 101;

image[2][0].blue = 255;

image[0][1].red = 255;

image[0][1].green = 153;

image[0][1].blue = 51;

image[1][1].red = 255;

image[1][1].green = 153;

image[1][1].blue = 153;

image[2][1].red = 255;

image[2][1].green = 153;

image[2][1].blue = 255;

image[0][2].red = 255;

image[0][2].green = 203;

image[0][2].blue = 51;

image[1][2].red = 255;

image[1][2].green = 204;

image[1][2].blue = 153;

image[2][2].red = 255;

image[2][2].green = 205;

image[2][2].blue = 255;

image[0][3].red = 255;

image[0][3].green = 255;

image[0][3].blue = 51;

image[1][3].red = 255;

image[1][3].green = 255;

image[1][3].blue = 153;

image[2][3].red = 255;

image[2][3].green = 255;

image[2][3].blue = 255;

// create 3by4 array to store energy of each pixel

int energies[3][4];

// calculate energy for each pixel

for (int i = 0; i < 3; i++) {

for (int j = 0; j < 4; j++) {

energies[i][j] = energy(image, i, j, 3, 4);

}

// print energies of each pixel

for (int i = 0; i < 4; i++) {

for (int j = 0; j < 3; j++) {

// print by column

cout << energies[j][i] << " ";

}

cout << endl;

}

// function prototype

int energy(Pixel** image, int x, int y, int width, int height) {

// get adjacent pixels

Pixel left, right, up, down;

if (x > 0) {

left = image[x - 1][y];

if (x < width - 1) {

right = image[x + 1][y];

}

else {

right = image[0][y];

}

else {

left = image[width - 1][y];

if (x < width - 1) {

right = image[x + 1][y];

}

else {

right = image[0][y];

}

if (y > 0) {

up = image[x][y - 1];

if (y < height - 1) {

down = image[x][y + 1];

}

else {

down = image[x][0];

}

else {

up = image[x][height - 1];

if (y < height - 1) {

down = image[x][y + 1];

}

else {

down = image[x][0];

}

// calculate x-gradient and y-gradient

Pixel x_gradient;

Pixel y_gradient;

x_gradient.blue = right.blue - left.blue;

x_gradient.green = right.green - left.green;

x_gradient.red = right.red - left.red;

y_gradient.blue = down.blue - up.blue;

y_gradient.green = down.green - up.green;

y_gradient.red = down.red - up.red;

int x_value = x_gradient.blue * x_gradient.blue + x_gradient.green * x_gradient.green + x_gradient.red * x_gradient.red;

int y_value = y_gradient.blue * y_gradient.blue + y_gradient.green * y_gradient.green + y_gradient.red * y_gradient.red;

// return energy of pixel

return x_value + y_value;

}

Explanation:

Please see attachment for ouput

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zheka24 [161]

Answer:

Explanation:

The detailed analysis and step by step calculation is as shown in the attachment.

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