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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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About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa
Nana76 [90]

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

K.E = \frac{2KZe^2}{r}

where;

  • Z is the atomic number of aluminium  = 13
  • e is charge
  • r is distance of closest approach = thickness of aluminium
  • k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

7 0
2 years ago
Elements of parallel computing
Leya [2.2K]

\huge{\orange}\fcolorbox{purple}{cyan}{\bf{\underline{\green{\color{pink}Answer}}}}

<h3><u>E</u><u>lements of parallel </u><u>computing:</u></h3>

  • Computer systems organization.
  • Computing methodologies.
  • General and reference.
  • Networks.
  • Software and its engineering.
  • Theory of computation.

6 0
2 years ago
Which statement about lean manufacturing is true when you compare it to mass production?
Len [333]
Where are the statements then bbs lol
6 0
3 years ago
Say you have a random, unordered list containing 4096 four-digit numbers. Describe the most efficient way to: sort the list and
Debora [2.8K]

Answer:

Answer explained below

Explanation:

It is given that numbers are four-digit so maximum value of a number in this list could be 9999.

So we need to sort a list of integers, where each integer lies between [0,9999].

For these given constraints we can use counting sort which will run in linear time i.e. O(n).

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

Psuedo Code:

countSort(int numList[]) {

int count[10000];

count[i] = 0; for all i;

for(int num in numList){

count[num]+= 1;

}

return count;

}

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

Searching in this count array will be just O(1).

E.g. Lets say we want to search if 3 was present in the original list.

Case 1: it was present in the original list:

Then the count[3] would have been incremented by our sorting algorithm. so in case element exists then count value of that element will be greater than 0.

Case 2: it was not present:

In this case count[3] will remain at 0. so in case element does not exist then count of that element will be 0.

So to search for an element, say x, we just need to check if count[x]>0.

So search is O(1).

Run times:

Sorting: O(n)

Search: O(1)

6 0
3 years ago
Does anyone have cumulative exams today or tomorrow?(There so boring!)
IRINA_888 [86]

Answer:

I do!!

Explanation:

I have to sit for 3 hours lol‍♀️

3 0
3 years ago
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