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maxonik [38]
3 years ago
11

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

. After being compressed, the air is at 450 °C. Determine
(a) the final pressure in [MPa],
(b) the increase in total internal energy in [kJ], and
(c) the total work required in [kJ].
Note that for air R-287 J/kg.K and c.-716.5 J/kg.K, and ?-
Engineering
1 answer:
Helen [10]3 years ago
4 0

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

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Choose the best data type for each of the following so that any reasonable value is accommodated but no memory storage is wasted
stiks02 [169]

Answer:

Explanation:

Part (a):

Statement : The number of siblings you have

Suitable Data type : Byte

Typical Value : From -128 and up to 127

Explanation: Byte data type is the most suitable since it can covers minimum and maximum number of siblings one can have.

Part (b):

Statement : Your final grade in this class

Suitable Data type : Char

Typical Value : 1 byte

Explanation: Grades is in the form of alphabetical letter which is either A, B, C, D, F or E which can be stored in character data type.

Part (c):

Statement : Population of Earth

Suitable Data type : Long

Maximum Value : 9223372036854775807

Explanation: Long Data takes up to 8 bytes and can store up to 9223372036854775807 which can cater for more than 36 billion. The population of earth is only around 7 billion currently making Long data type the most suitable data type to store earth population.

Part (d):

Statement : Population of US Country

Suitable Data type : Integer

Typical Value :2147483647

Explanation: Integer data type takes up to 4 bytes and can store up to  2147483647 making it suitable to store U.S population.

Part (e):

Statement : The number of passengers on bus

Suitable Data type : Byte

Typical Value :From -128 up to 127

Explanation: The typical maximum number of passengers of a bus are only around 72. Byte data type is the most suitable since it can cater the number up to 127.

Part (f):

Statement : Player's score in a Scrabble game

Suitable Data type : Short

Typical Value : 32767

Explanation: The maximum point can be scored in the Scrabble game is only 830 therefore the most suitable data type for this case is the short data type.

Part (g):

Statement : One team's score in a Major League Baseball game

Suitable Data type : Byte

Typical Value : From -128 up to 127

Explanation: The maximum point can be scored in the Base ball game is only 49 therefore the most suitable data type for this case is the Byte data type since it can cater up to 127.

Part (h):

Statement : The year an historical event occurred

Suitable Data type : Short

Maximum Value: 32767

Explanation: The historic event year can be any number from 1 to 2020 therefore the most suitable data type is the short data type.

Part (i):

Statement : The number of legs on an animal

Suitable Data type : Short

Maximum Value: 32767

Explanation: The most number of legs found are 750 legs therefore the most suitable data type is the short data type which can cater up to 32767.

Part (j):

Statement : The Price of an automobile

Suitable Data type : Float

Maximum Value: 340282350

Explanation: The most expensive car is around 15 million therefore the most suitable data type is the float data type which can cater up to 340 million.  

3 0
3 years ago
3. If nothing can ever be at absolute zero, why does the concept exist?
Tanzania [10]

The absolute zero in temperature refers to the minimal possible temperature. It is the temperature at which the molecules of a system stop moving, so it is a really useful reference point.

<h3>Why absolute zero can't be reached?</h3>

It would mean that we need to remove all the energy from a system, but to do this we need to interact with the system in some way, and by interacting with it we give it "some" energy.

Actually, from a quantum mechanical point of view, the absolute zero has a residual energy (so it is not actually zero) and it is called the "zero point". This happens because it must meet <u>Heisenberg's uncertainty principle</u>.

So yes, the absolute zero can't be reached, but there are really good approximations (At the moment there is a difference of about 150 nanokelvins between the absolute zero and the smallest temperature reached). Also, there are a lot of investigations near the absolute zero, like people that try to reach it or people that just need to work with really low temperatures, like in type I superconductors.

So, concluding, why does the concept exist?

  • Because it is a reference point.
  • It is the theoretical temperature at which the molecules stop moving, defining this as the <u>minimum possible temperature.</u>

If you want to learn more about the absolute zero, you can read:

brainly.com/question/3795971

3 0
2 years ago
3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the
gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

                          V_fat = x_fat*V

- The volume of the muscle is given by:

                          V_muscle = V - V_fat

                                            = V - x_fat*V

                                            =V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

                         mass_fat + mass_muscle = Total average mass

                         p_fat*V_fat + p_muscle*V_muscle = p_avg*V

                         p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

                         Weight reading (Wsa) = m = p_1*V

                         p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

                         Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

                         Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

                         Where, p_2 = p_1 - p_water

                         p_2 = Wsw / V

- The average density p_avg:

                         p_avg = 0.5*(p_1 + p_2)  

                         p_avg = 0.5*(Wsa / V + Wsw / V)  

                         p_avg = 0.5*(Wsa + Wsw) / V                      

- Plug in the mass equation:

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

                         x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

                   x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

                         

6 0
3 years ago
8.19 - Airline Reservations System (Project Name: Airline) - A small airline has just purchased a computer for its new automated
e-lub [12.9K]

Answer:

The App is written in C++ language using dev C++.

Explanation:

/******************************************************************************

You can run this program in any C++ compiler like dev C++ or any online C++ compiler

*******************************************************************************/

#include <iostream>

using namespace std;

class bookingSeat// class for airline reservation system

{

  private:

   

   

  bool reserveSeat[10];// 10 seats (1-5) for first class and 6-10 for economy class

  int firstClassCounter=0;//count first class seat

  int economyClassCounter=5;//count economy class seat

  char seatPlacement;/* switch between economy and first clas seat----- a variable for making decision based on user input*/

  public:  

  void setFirstClassSeat()//

  {

      if(firstClassCounter<5)// first class seat should be in range of 1 to 5

      {

          reserveSeat[firstClassCounter]=1; /*set first class seat..... change index value to 1 meaning that it now it is reserved*/

          cout<<"Your First Class seat is booked and your seat no is "<<firstClassCounter+1; //display seat number reserved

          firstClassCounter++; //increament counter

      }

      else//in case seats are ful

      {

          cout<<"\nSeats are full";

          if(economyClassCounter==10 && firstClassCounter==5)

          {

              cout<<"\n Next flight leaves in 3 hours.";

          }

          else

          {

              cout<<"\nIt’s acceptable to be placed to you in the first-class section  y/n ";//take input from user

              cin>>seatPlacement;//user input

              if(seatPlacement=='y')//if customer want to reserve seat in first class

              {

                  setEconomyClassSeat();// then reserve first class seat

              }

              else

              {

                  cout<<"\n Next flight leaves in 3 hours.";

               }

               

          }

      }

       

  }

  void setEconomyClassSeat()//set economy class seat

  {

    if(economyClassCounter<10)//seat ranges between 6 and 10

      {

          reserveSeat[economyClassCounter]=1;// reserve economy class seat

          cout<<"Your Economy class seat is booked and your seat no is "<<economyClassCounter+1;//display reservation message about seat

          economyClassCounter++;//increament counter

      }

      else// if economy class seats are fulled

      {

          cout<<"\nSeats are full";

          if(economyClassCounter==10 && firstClassCounter==5)//check if all seats are booked in both classes

          {

              cout<<"\n Next flight leaves in 3 hours.";

          }

          else

          {

              cout<<"\nIt’s acceptable to be placed to you in the first-class section  y/n ";//take input from user

              cin>>seatPlacement;//user input

              if(seatPlacement=='y')//if customer want to reserve seat in first class

              {

                  setFirstClassSeat();// then reserve first class seat

              }

              else

              {

                  cout<<"\n Next flight leaves in 3 hours.";

               }

               

          }

      }

  }

   

   

};

int main()

{   int checkseat=10;// check seat

   int classType;//class type economy or first class

   bookingSeat bookseat;//object declaration of class bookingSeat

   while(checkseat<=10)//run the application until seats are fulled in both classes

   {

       cout<<"\nEnter 1 for First Class and 2 for Economy Class ";

       cin>>classType;//what user entered

       switch (classType)//decide which seat class to be reserved  

       {

           case 1://if user enter 1 then reserve first class seat

           bookseat.setFirstClassSeat();

           break;

           case 2://if user enter 2 then reserve the economy class seat

           bookseat.setEconomyClassSeat();

           

       }

       

   }

   

   return 0;

}

8 0
2 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
ankoles [38]

Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ_{fs} = F_{f} L / \pi  R^{3}

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

  = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

  = 187 / 3.141593 (0.0056)³

  = 338943767.745358

  = 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

  = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

  = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)

  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

  =  97344  / 11

F =  8849 N

4 0
3 years ago
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