The minimum safe working distance from exposed electrical conductors

In a river reach, the rate of inflow at any time is 350 cfs and the rate of outflow is 285 cfs. After 90 min, the inflow and out

Semenov [28]

Answer:

change in storage = -310,500 ft^3

intital storage= 3.67 acre ft

Explanation:

Given data:

Rate of inflow = 350 cfs

Rate of outflow = 285 cfs

After 90 min, rate of inflow = 250 cfs

Rate of outflow = 200 cfs

final storage = 10.8 acre-ft

calculating the average inflow and outflow

average inflow $= \frac{(350+250)}{2} = 300 cfs$

average outlow $= \frac{(285+200)}{2} = 242.5 cfs$

total amount of water drain during the period of one hour

= (average outflow - average inflow) *60*90

= (242.5 - 300)*60*90 = -310,500

change in storage is calculate as

= -310,500 ft^3

in cubic meter

= -310500/35.315 = 8792.30 cm^3

in acre-ft

= -310,500/43560 = 7.13 acre ft

initial storage = 10.8 - 7.13 = 3.67 acre ft

3 0

3 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

6 0

3 years ago

A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan

hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

$R = \dfrac{u^2}{15(e+f_s)}$

where;

R= radius

$f_s$ = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

$f_s$ = 0.20

So;

$R = \dfrac{30^2}{15(0.08+0.20)}$

$R = \dfrac{900}{15(0.28)}$

$R = \dfrac{900}{4.2}$

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0

3 years ago

Stream Piracy – Kaaterskill, NY. Check and double-click the Problem 15 folder. The dark blue and orange streams highlight the pr

baherus [9]

Answer:

b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.

Explanation:

From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.

The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.

However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.

The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.

It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.

As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.

5 0

3 years ago

An ideal gas turbine operates using air coming at 355C and 350 kPa at a flow rate of 2.0 kg/s. Find the rate work output

GuDViN [60]

Answer:

The rate of work output = -396.17 kJ/s

Explanation:

Here we have the given parameters

Initial temperature, T₁ = 355°C = 628.15 K

Initial pressure, P₁ = 350 kPa

h₁ = 763.088 kJ/kg

s₁ = 4.287 kJ/(kg·K)

Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately

h₂ = 79.572 kJ/kg

The saturation temperature at the given

T₂ = 79°C

The rate of work output $\dot W$ = $\dot m$ × $c_p$ ×(T₂ - T₁)

Where;

$c_p$ = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)

$\dot m$ = The mass flow rate = 2.0 kg/s

Substituting the values, we have;

$\dot W$ = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s

$\dot W$ = -396.17 kJ/s

7 0

4 years ago