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Paul [167]
3 years ago
11

Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia

meter of the fin is 3 cm. The thermal conductivity of the fin is 150 W/m·K. The heat transfer coefficient is 123 W/m2·K. Estimate the fin temperature in °C at a distance of 10 cm from the base
Engineering
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

98°C

Explanation:

Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7

22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²

Temperature change, t = (50 - 25)°C = 25°C = 298K

Hence, Temperature =  150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =

∴ Temperature change = 2.00K

But temperature, T= (373 - 2)K = 371 K

In °C = (371 - 273)K = 98°C         

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IceJOKER [234]

Answer:

Applying Kirchhoff's current law at inverting terminal

Explanation:

Detailed explanation is given in the attached document.

5 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
The undisturbed soil at given borrow pit is found to have the following property:
tankabanditka [31]

Answer:

Check the explanation

Explanation:

Determine the weight o ids in the each truck °slim the relation,  

W,=\frac{W}{1+w}

Here, W is net weight of soil and water on the truck and w is water content  

substitute 72 7 kN for W and 15% for w.

you will need to also determine the number of truck loads required using the relation:

Number of truck loads required = \frac{W_{sc} }{W_{s}}

Kindly check the attached image below for the full explanation to the question above.

7 0
3 years ago
In wet mill ethanol plants, a total energy of 74,488 Btu (British thermal units, a common energy unit) is used to produce 1 gall
V125BC [204]

Answer:

The energy yield for one gallon of ethanol is 2.473 %.

Explanation:

The net energy yield (\% e), expressed in percentage for one gallon of ethanol is the percentage of the ratio of the difference of the provided energy (E_{g}), measured in Btu, and the energy needed to produce the ethanol (E_{p}), measured in Btu, divided by the energy needed to produce the ethanol. That is:

\% e =\frac{E_{g}-E_{p}}{E_{p}} \times 100\,\% (1)

If we know that E_{g} = 76330\,Btu and E_{p} = 74488\,Btu, then the net energy yield of 1 gallon of ethanol:

\%e = \frac{76330\,Btu-74488\,Btu}{74488\,Btu}\times 100\,\%

\%e = 2.473\,\%

The energy yield for one gallon of ethanol is 2.473 %.

4 0
3 years ago
A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration a = {6ti + 12t^2k} ft/s^
Sindrei [870]

Answer:

The particle's position at t = 2 is

r = (11ft, 2ft, 21ft) = (11, 2, 21) ft

Explanation:

r₀ = (3, 2, 5) ft

a = (6t, 0, 12t²) ft/s²

a = dv/dt

dv/dt = (6tî + 0j + 12t²ķ)

dv = (6tî + 0j + 12t²ķ) st

Integrating the left hand side from 0 to v (the particle was originally at rest) and the right hand side from 0 to t,

We obtain,

v = (3t²î + 0j + 4t³ķ) ft/s

v = dr/dt

dr/dt = (3t²î + 0j + 4t³ķ)

dr = (3t²î + 0j + 4t³ķ) st

Integrating the left hand side from r₀ (the original position) to r and the right hand side from 0 to t,

r - r₀ = (t³î + 0j + t⁴ķ) ft

r = (t³î + 0j + t⁴ķ) + r₀

r = (t³î + 0j + t⁴ķ) + (3î + 2j + 5ķ)

At t=2s, t³ = 8 and t⁴ = 16

r = (8î + 0j + 16ķ) + (3î + 2j + 5ķ)

r = (11î + 2j + 21ķ) ft

r = (11ft, 2ft, 21ft)

Hope this helps!

4 0
3 years ago
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