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Ludmilka [50]
3 years ago
14

Water flows at a rate of 0.040 m3/s in a 0.12-m-diameter pipe that contains a sudden contraction to a 0.06-m-diameter pipe. Dete

rmine the pressure drop across the contraction section. How much of this pressure difference is due to losses and how much is due to kinetic energy changes?
Engineering
1 answer:
LUCKY_DIMON [66]3 years ago
4 0

Answer:cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

Explanation:

ccccccccccccccccccccccccccccccccccccccccccccccc

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A 0.06-m3 rigid tank initially contains refrigerant- 134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve t
lesantik [10]

Answer:

a) 0.50613

b) 22.639 kJ

Explanation:

From table A-11 , we will make use of the properties of Refrigerant R-13a at 24°C

first step : calculate the  volume of R-13a  ( values gotten from table A-11 )

V = m1 * v1 = 5 * 0.0008261 = 0.00413 m^3

next : calculate final specific volume ( v2 )

v2 = V / m2 = 0.00413 / 0.25 ≈ 0.01652 m^3/kg

<u>a) Calculate the mass of refrigerant that entered the tank </u>

v2 = Vf + x2 * Vfg

v2 = Vf +  [ x2 * ( Vg - Vf ) ] ----- ( 1 )

where:  Vf = 0.0008261 m^3/kg, V2 = 0.01652 m^3/kg , Vg = 0.031834 m^3/kg  ( insert values into equation 1 above )

x2 = ( 0.01652 - 0.0008261 ) / 0.031834

     = 0.50613 ( mass of refrigerant that entered tank )

<u>b) Calculate the amount of heat transfer </u>

Final specific internal energy = u2 = Uf + ( x2 + Ufg ) ----- ( 2 )

uf = 84.44 kj/kg , x2 = 0.50613 , Ufg = 158.65 Kj/kg

therefore U2 = 164.737 Kj/kg

The mass balance  ( me ) = m1 - m2 --- ( 3 )

energy balance( Qin ) = ( m2 * u2 ) - ( m1 * u1 ) + ( m1 - m2 ) * he

therefore Qin = 41.184 - 422.2 + 403.655  = 22.639 kJ

3 0
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
Igoryamba

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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thier only motto and goal is to work for society and not make any profits A.small business entreprenuership B.scalable start up
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Answer:

social entrepreneurship

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Fault location techniques are used in power systems for accurate pinpointing of the fault position.

This paper presents a comparative study between two fault location methods in distribution network with Distributed Generation (DG). Both methods are based on computing the impedance using fundamental voltage and current signals. The first method uses one-end information and the second uses both ends

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