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Alekssandra [29.7K]
3 years ago
8

Explain the importance of water quality in aquaculture business.

Engineering
2 answers:
mars1129 [50]3 years ago
5 0

Answer:

Mantain the image of the business

Explanation:

Good water quality gives good images to others and furthers business

Scorpion4ik [409]3 years ago
4 0

Answer:

this is very important the agricultural because there can be a risk of hight salt concentrations that can limit the amount of water plants can take

You might be interested in
List all possible fracture mechanisms under which the unidirectional composites fail. Briefly explain and describe the related m
professor190 [17]

Answer:

Ususushehehehhuuiiïbbb

Explanation:

Yyshehshehshshsheyysysueueue

7 0
2 years ago
What are the different types of documents used to communicate engineering designs?
Ipatiy [6.2K]

Answer:

COMMON ENGINEERING DOCUMENTS

Inspection or trip reports.

Research, laboratory, and field reports.

Specifications.

Proposals.

Progress reports.

ect...

Explanation:

7 0
3 years ago
A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f
dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14,  pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, head\ loss  = \frac{f L V^2}{( 2 g D)}

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

V = \frac{Q}{A}

    = \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s

obtained Darcy friction factor  

calculate Reynold number (Re) ,

Re = \frac{\rho V D}{\mu}

where,\rho = density of water

\mu = Dynamic viscosity of water at 15 degree  C = 0.001 Ns/m2

so reynold number is

Re = \frac{1000\times 2.015\times 0.318}{0.001}

            = 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is  0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}

HL = 1.64 m

7 0
3 years ago
The second programming project involves writing a program that accepts an arithmetic expression of unsigned integers in postfix
Tpy6a [65]

Answer:

Explanation:

Note: In case of any queries, just comment in box I would be very happy to assist all your queries

SourceCode:

// MyGUI.java:

// Import packages

import java.awt.FlowLayout;

import java.awt.GridLayout;

import java.awt.event.ActionEvent;

import java.awt.event.ActionListener;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.EmptyStackException;

import java.util.Stack;

import javax.swing.JButton;

import javax.swing.JFrame;

import javax.swing.JLabel;

import javax.swing.JOptionPane;

import javax.swing.JPanel;

import javax.swing.JTextField;

import javax.swing.SwingConstants;

// Declaare and define the class MyGUI

abstract class MyGUI extends JFrame implements ActionListener {

JTextField userInput;

JLabel inputDescLbl, resultLbl;

JPanel inputPanel, resultPanel;

JButton evlBtn;

Stack<Object> stk;

// Define the constructor MyGUI

MyGUI() {

super("Tree Address Generator");

inputPanel = new JPanel(new FlowLayout());

resultPanel = new JPanel(new FlowLayout());

setLayout(new GridLayout(2, 1));

userInput = new JTextField(20);

inputDescLbl = new JLabel("Enter Postfix Expression:");

evlBtn = new JButton("Construct Tree");

evlBtn.addActionListener(this);

resultLbl = new JLabel("Infix Expression:", SwingConstants.LEFT);

add(inputPanel);

add(resultPanel);

inputPanel.add(inputDescLbl);

inputPanel.add(userInput);

inputPanel.add(evlBtn);

resultPanel.add(resultLbl);

stk = new Stack<Object>();

}

}

//Stack.java:

// Declare and define the class Stack

class Stack {

private int[] a;

private int top, m;

public Stack(int max) {

m = max;

a = new int[m];

top = -1; }

public void push(int key) {

a[++top] = key; }

public int pop() {

return (a[top--]); }

}

// Declare and define the class Evaluation()

class Evaluation {

public int calculate(String s) {

int n, r = 0;

n = s.length();

Stack a = new Stack(n);

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

char ch = s.charAt(i);

if (ch >= '0' && ch <= '9')

a.push((int) (ch - '0'));

else if (ch == ' ')

continue;

else {

int x = a.pop();

int y = a.pop();

switch (ch) {

case '+':

r = x + y;

break;

case '-':

r = y - x;

break;

case '*':

r = x * y;

break;

case '/':

r = y / x;

break;

default:

r = 0;

}

a.push(r);

}

}

r = a.pop();

return (r);

}

}

// PostfixToInfix.java:

// Import packages

import java.util.Scanner;

import java.util.Stack;

// Declare and define the class PostfixToInfix

class PostfixToInfix {

// Determine whether the character entered is an operator or not

private boolean isOperator(char c) {

if (c == '+' || c == '-' || c == '*' || c == '/' || c == '^')

return true;

return false;

}

// Declare and define the convert()

public String convert(String postfix) {

Stack<String> s = new Stack<>();

for (int i = 0; i < postfix.length(); i++) {

char c = postfix.charAt(i);

if (isOperator(c)) {

String b = s.pop();

String a = s.pop();

s.push("(" + a + c + b + ")");

} else

s.push("" + c);

}

return s.pop();

}

// Program starts from main()

public static void main(String[] args) {

PostfixToInfix obj = new PostfixToInfix();

Scanner sc = new Scanner(System.in);

// Prompt the user to enter the postfix expression

System.out.print("Postfix : ");

String postfix = sc.next();

// Display the expression in infix expression

System.out.println("Infix : " + obj.convert(postfix));

}

}

Output:

e Console X terminated PostfixTolnfix [Java Application] C:\Program Files\Java\jrel.8.0_121\bin\javaw.exe Postfix : ABD++C-D/ .

3 0
3 years ago
Consider an adiabatic throttling valve with water entering at pressure of 1.5 MPa, a temperature of 150°C and a velocity of 4.5
Marta_Voda [28]

Answer:

So the exit velocity of water is 4.5 m/s

Explanation:

Given that

Water entering pressure = 1.5 MPa

Temperature = 150°C

Velocity = 4.5 m/s

From first law of thermodynamics for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+W

Here given that valve is adiabatic so Q= 0

In valve W= 0

Wen also also know that throttling process is an constant enthaply process so

h_1=h_2

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+W

h_1+\dfrac{V_1^2}{2}+0=h_1+\dfrac{V_2^2}{2}+0

So from above equation we can say that

V_2=V_1

So the exit velocity of water is 4.5 m/s

5 0
3 years ago
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