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zavuch27 [327]
4 years ago
15

Nitrogen can be liquefied using a Joule-Thomson expansioni process. This is done by rapidlyl and adiabatically expandign cold ni

trogen gas from high pressure to a low pressure. If nitrogen at 135 K and 20 MPa undergoes a Joule-Thomson expansion to 0.4 MPa,
a. Estimate the fraction of vapor and liquid present after the expansion, and the emperature of this mixture using the pressure-enthalpy diagram for nitrogen.

b. Repeat the calculation assuming nitrogen to be an ideal gas with Cp* = 29.3 J/molK
Engineering
1 answer:
solniwko [45]4 years ago
8 0
T is desired to produced liquefied methane as illustrated in problem 1; however, theconditions are now changed so that the gas is initially available at 1 bar and 200 K, and leaving the cooler will be at 100 bar and 200 K. The flash drum is adiabatic, oerates at 1 bar, and each compressor stage canbe assumed to operate reversibly and adiabatically. A three-stage compressor will be used with the first stage compressing the gas fro 1 bar to 5 bar, each stage the gas will be isobarically cooled to 200 K.
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The base class Pet has attributes name and age. The derived class Dog inherits attributes from the base class Pet class and incl
Nonamiya [84]

Answer:

Explanation:

class Pet:

   def __init__(self):

       self.name = ''

       self.age = 0

   def print_info(self):

       print('Pet Information:')

       print('   Name:', self.name)

       print('   Age:', self.age)

class Dog(Pet):

   def __init__(self):

       Pet.__init__(self)

       self.breed = ''

def main():

   my_pet = Pet()

   my_dog = Dog()

   pet_name = input()

   pet_age = int(input())

   dog_name = input()

   dog_age = int(input())

   dog_breed = input()

   my_pet.name = pet_name

   my_pet.age = pet_age

   my_pet.print_info()

   my_dog.name = dog_name

   my_dog.age = dog_age

   my_dog.breed = dog_breed

   my_dog.print_info()

   print('   Breed:', my_dog.breed)

main()

3 0
3 years ago
Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its l
ycow [4]

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

3 0
3 years ago
Read 2 more answers
An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/
Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

<u>Step 1:</u> Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

<u>Step 3:</u> The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755  / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min  = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

3 0
3 years ago
Una empresa realizó en el ejercicio de compras al contado por valor
Tanzania [10]

Answer:

englishhhh pleasee

Explanation:

we dont understand sorry....

8 0
4 years ago
Given a matrix, clockwise-rotate elements in it. Please add code to problem3.cpp and the makefile. Use the code in p3 to test yo
rusak2 [61]

Answer:

/* C Program to rotate matrix by 90 degrees */

#include<stdio.h>

int main()

{

int matrix[100][100];

int m,n,i,j;

printf("Enter row and columns of matrix: ");

scanf("%d%d",&m,&n);

 

/* Enter m*n array elements */

printf("Enter matrix elements: \n");

for(i=0;i<m;i++)

{

 for(j=0;j<n;j++)

 {

  scanf("%d",&matrix[i][j]);

 }

}

 

/* matrix after the 90 degrees rotation */

printf("Matrix after 90 degrees roration \n");

for(i=0;i<n;i++)

{

 for(j=m-1;j>=0;j--)

 {

  printf("%d  ",matrix[j][i]);

 }

 printf("\n");

}

 

return 0;

 

}

5 0
3 years ago
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