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Nimfa-mama [501]
3 years ago
11

The kinetic energy correction factor depends on the (shape — volume - mass) of the cross section Of the pipe and the (velocity —

pressure — temperature distribution.
Engineering
1 answer:
butalik [34]3 years ago
3 0

Answer:

The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.

Explanation:

The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform.  In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.

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A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th
Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ    

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588  = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

 cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 =  2552.64 kJ    

b)  \Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)

Where:

x_A and x_B are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, x_A = 248 /(248 + 42.8) = 0.83

x_B = 42.8/(248 + 42.8) = 0.1472

∴ \Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472) =  1066.0279 J/K

5 0
3 years ago
500 flights land each day at San Jose’s airport. Assume that each flight has a 5% chance of being late, independently of whether
BARSIC [14]

Answer:

a.0.0199

b.0.1765

c.0.0785

d.0.1268

e.Yes

Explanation:

It is given that   X follows a  Binomial distribution with (n= 500, p = 0.05)

The  probabilities  are computed using the EXCEL .

a) The required probability here is:

P(X less of equal to  15)

= binom.dist(15,500,0.05,TRUE)

=0.0199

Therefore the probability is 0.0199 .

b) The required probability here is:

P(X greater or equal to 30) = 1 - P(X less or equal to  29)  

=1 - binom.dist(29,500,0.05,TRUE)

=0.1765

Therefore the probability is 0.1765

c) P(X = 26 )

= binom.dist(26,500,0.05,FALSE)  

=0.0785

Therefore the probability is 0.0785

d) The required probability here is computed as:

P(10 less or equal to X less or equal to 20 ) = P(X less or equal to 19) - P(X less or equal to 10)

= binom.dist(19,500,0.05,TRUE) - binom.dist(10,500,0.05,TRUE)

=0.1268

Therefore the probability 0.1268

e) Yes . Therefore the probability because that is the assumption used to apply binomial distribution .

6 0
3 years ago
In your Reader/Writer Notebook, write a short first-person narrative from the perspective of the bank clerk, describing her phys
guajiro [1.7K]
BAHHHHAHHH HOFHOFYOCIGC
5 0
3 years ago
Technician A says that the carpet padding is designed to help reduce noise and vibrations.
Firdavs [7]

Answer:

Technicians A is right for the answer

4 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
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