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ANEK [815]
3 years ago
13

Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribu

ción de temperatura en función de "x" el cual es la distancia desde la cara izquierda de una pared, está dado por la siguiente función: T(x) = 150x2 -30x, donde x está en metros. Calcúlese el flujo de calor por unidad de área cuando x=0 y x=30, para cada caso menciones si se está enfriando o calentando el sólido.
Engineering
1 answer:
aksik [14]3 years ago
4 0

Answer:

Para x=0:

\phi=1.2 W/m^{2}  

Para x=30 cm:

\phi=-2.4 W/m^{2}  

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

\phi=-k\frac{dT}{dx}(1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

\frac{dT(x)}{dx}=300x-30

Remplezando esta derivada en (1):

\phi=-0.04(300x-30)

Para x=0:

\phi=0.04(30)

\phi=1.2 W/m^{2}  

Para x=30 cm:

\phi=-0.04(300*0.3-30)

\phi=-2.4 W/m^{2}    

Espero que te haya ayudado!

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Answer:

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3 years ago
Use the case structure in order to do the following tasks. a) Use a five layer case structure in order to do following for a two
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Answer:

The complete answer along with step by step explanation and output results is provided below.

Explanation:

Task a)

#include<iostream>

using namespace std;

int main()

{

  int op;

  double num1, num2;

   cout<<"Enter num1 and num2"<<endl;

   cin>>num1>>num2;  

  // To provide the option of required 5 cases  

  cout << "Select the operation:"

          "\n1 = Addition"

          "\n2 = Subtraction"

          "\n3 = Multiplication"

          "\n4 = Division"

          "\n5 = Negation\n";

  cin >> op;  // user input the desired operation

  switch(op)  // switch to the corresponding case according to user input

   {

       case 1:

           cout <<"The Addition of num1="<<num1<<" and num2="<<num2<<" is: "<<num1+num2;

           break;

       case 2:

           cout <<"The Subtraction of num1="<<num1<<" and num2="<<num2<<" is: "<<num1-num2;

           break;

       case 3:

           cout <<"The Multiplication of num1="<<num1<<" and num2="<<num2<<" is: "<<num1*num2;

           break;

       case 4:        

        while(num2 == 0) // to check if divisor is zero  

        {

           cout << "\nWrong divisor! Please select the correct divisor again: ";

           cin >> num2; // if divisor is zero then ask user to input num2 again

        }

           cout <<"The division of num1="<<num1<<" and num2="<<num2<<" is: "<<num1/num2;

           break;

       case 5:

           cout <<"The Negation of num1="<<num1<<" and num2="<<num2<<" is: "<<-1*num1<<" "<<-1*num2;

           break;

       default:

           // If the operation is other than listed above then error will be shown

           cout << "Error! The selected operatorion is not correct";

           break;

   }

  return 0;

}

Output:

Test 1:

Enter num1 and num2

2

9

Select the operation:

1 = Addition

2 = Subtraction

3 = Multiplication

4 = Division

5 = Negation

1

The Addition of num1=2 and num2=9 is: 11

Hence the output is correct and working as it was required

Test 2:

Enter num1 and num2

8

0

Select the operation:

1 = Addition

2 = Subtraction

3 = Multiplication

4 = Division

5 = Negation

4

Wrong divisor! Please select the correct divisor again: 2

The Division of num1=8 and num2=2 is: 4

Hence the output is correct and working as it was required

Test 3:

Enter num1 and num2

-2

4

Select the operation:

1 = Addition

2 = Subtraction

3 = Multiplication

4 = Division

5 = Negation

5

The Negation of num1=-2 and num2=4 is: 2 -4

Hence the output is correct and working as it was required

Task b)

#include<iostream>

#include<cmath>   // required to calculate square root

using namespace std;

int main()

{

  int op;

  double num;

   cout<<"Enter a real number > 0"<<endl;

   cin>>num;

  cout << "Press 1 for square root:";

  cin >> op;

  switch(op)  // switch to the corresponding case according to user input

   {

       case 1:

          if (num <= 0) // to check if number is less or equal to zero

        {

           cout << "\nError! number is not valid";

           break;   // if number is not valid then terminate program

        }

           cout <<"The square root of num="<<num<<" is: "<<sqrt(num);

// if number is valid then square root will be calculated

           break;

       default:

           // If the operation is other than listed above then error will be shown

           cout << "Error! The selected operation is not correct";

           break;

   }

  return 0;

}

Output:

Test 1:

Enter a real number > 0

6

Press 1 for square root: 1

The square root of num=6 is: 2.44949

Hence the output is correct and working as it was required

Test 2:

Enter a real number > 0

-4

Press 1 for square root: 1

Error! number is not valid

Hence the output is correct and working as it was required

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What is the purpose of the graphic language?
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Answer:

D = 0.060732 in

Explanation:

given data

sp. wt. = 500 lb/ft³

diameter = 0.036 in

solution

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here \sigma surface tension of water at 60⁰f  = 5.03 × 10^{-3}  lb/ft and y = 500 lb/ft³

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