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algol [13]
4 years ago
5

Consider the series solution, Equation 5.42, for the plane wall with convection. Calculate midplane (x* = 0) and surface (x* = 1

) temperatures θ* for Fo = 0.1 and 1, using Bi = 0.1, 1, and 10. Consider only the first four eigenvalues. Based on these results, discuss the validity of the approximate solutions, Equations 5.43 and 5.44.

Engineering
1 answer:
VashaNatasha [74]4 years ago
8 0

Answer:

We conclude that the approximate series solution (with only one eigein value) provides systematically high results but by less than 1.5%, for the biot number range from 0.11 to 10. See attached image.

Explanation:

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Python:
stira [4]

Answer:

# Python Program to Print  

# all subsets of given size of a set  

 

import itertools  

 

def findsubsets(s, n):  

   return list(itertools.combinations(s, n))  

 

# Driver Code  

s = {1, 2, 3}  

n = 2

 

print(findsubsets(s, n))

-----------------------------------------------

# Python Program to Print  

# all subsets of given size of a set  

 

import itertools  

# def findsubsets(s, n):  

def findsubsets(s, n):  

   return [set(i) for i in itertools.combinations(s, n)]  

     

# Driver Code  

s = {1, 2, 3, 4}  

n = 3

 

print(findsubsets(s, n))

-------------------------------------------------------------

# Python Program to Print  

# all subsets of given size of a set  

 

import itertools  

from itertools import combinations, chain  

 

def findsubsets(s, n):  

   return list(map(set, itertools.combinations(s, n)))  

     

# Driver Code  

s = {1, 2, 3}  

n = 2

 

print(findsubsets(s, n))

4 0
4 years ago
If copper (which has a melting point of 1085°C) homogeneously nucleates at 849°C, calculate the critical radius given values of
____ [38]

Answer:

The critical radius is -1.30 nm

Explanation:

Temperature for homogenous nucleation of copper, T_{H} = 849^{0} C = 849 + 273 = 1122 K

Melting point of copper, T_{cu} = 1085^{0} C = 1085 + 273 = 1358 K

Latent heat of fusion, H_{f} = -1.77 * 10^{9} J/m^{3}

Surface free energy, \gamma = 0.200 J/m^{2}

Critical radius, r = ?

The formula for the critical radius is given by:

r = \frac{2 \gamma T_{cu} }{H_{f}(T_{cu} - T_{H})  }

r = \frac{2 * 0.2*1358 }{(-1.77 * 10^{9}) (1358 - 1122)  }

r = \frac{543.2 }{(-1.77 * 10^{9}) 236}\\r = -1.30 * 10^{-9} m\\r = -1.30 nm

The critical radius is -1.30 nm

8 0
4 years ago
2. Consider Dekker’s algorithm written for an arbitrary number of processes by changing the statement executed when leaving the
neonofarm [45]

Answer:

Algorith does not work.

Explanation:

One of the ways to obtain the Dekker Algorithm is through a change in the declaration, that is, a declaration that can be executed at the exact moment it leaves the critical section. This way it is possible that the statement,

turn = 1-i / * P0 sets turn to 1 and P1 sets turn 0 * /

It can be changed to,

turn = (turn +1) \% n / * n = number or processes * /

The result will allow to define if it works or not, that is, if it is greater than 2 the algorithm will not be able to work.

Given this consideration we can say that,

<em>- The dead lock does not occur, because the mutual is imposed (if a resource unit has been assigned to a process, then no other process can access that resource).</em>

<em>- There is the possibility of starving if the shift is established in a non-contentious process.</em>

Directly it can be concluded that there is a possibility of starvation so the algorithm could not work, despite the fact that mutual exclusion guarantees that a dead block does not occur.

4 0
3 years ago
A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. IZK is specified to be
8090 [49]

Answer:

The maximum power dissipation of the zener diode 112mV.

Explanation:

The minimum zener current should be:

5 * Iza= 5 * 1=  5 mA.

Since the load current can be at maximum 15 mA, we should select R so that, IL= 15 mA.

A zener current of 5 mA is available, Thus the current should be 20 mA, which leads to,

R = \frac{15 - 5.6}{20 mA} = 470 Ω.

Maximum power dissipated in the diode occours when, IL=0 is

Pmax = 20 * 10^{3} * 5.6 = 112mV.

5 0
4 years ago
code a class named sllqueue that uses a double-ended singly linked list to implement a queue as described in this chapter. provi
julia-pushkina [17]

Answer:

See attached file please.

Explanation:

See attached file for detailed explanation and code.

import java.util.*;

class LinklistImplementQueue {  

public static void main(String[] args)

{

Scanner scan = new Scanner(System.in);

/* Creating object of class SLLQueue */    

SLLQueue lq = new SLLQueue();

char ch;

do

{

System.out.println("\nQueue Operations");

System.out.println("1. ENQUEUE");

System.out.println("2. DEQUEUE");

int choice = scan.nextInt();

.

.

.

.

See attached file for complete code.

Download txt
6 0
3 years ago
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