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matrenka [14]
3 years ago
10

A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The

compressor has a cooling jacket where water flows at 1500 kg/hr and undergoes a 20K temperature rise. Assuming air is an ideal gas, calculate the work furnished by the compressor, and also determine the minimum work required for the same state change of air.

Engineering
1 answer:
AleksandrR [38]3 years ago
8 0

Answer:

The work furnished by the compressor is 69.77kJ/s

The minimum work required for the state to change is 55.26kW

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

You might be interested in
The shape of the wood beam is a rectangle. In a paragraph explain the tradeoffs an engineer would make in selecting a wood with
oksian1 [2.3K]

Answer:

Wood is heavy

Explanation:

7 0
3 years ago
Biologists use a sequence of letters A, C, T, and G to model a genome. A gene isa substring of a genome that starts after a trip
kogti [31]

Answer:

You did not mention the programming language for implementation so i am writing a JAVA code.

import java.util.Scanner; // to get input from user

public class Genome{

public static void main(String[] args) { //start of main() function body

Scanner input = new Scanner(System.in); //creates Scanner object

System.out.print("Enter a genome string: ");

//prompts user to enter a genome string

String genome = input.nextLine();

//reads the input genome string and stores it into genome variable

boolean gene_found = false;

//variable gene_found of boolean type that has two value true or false

int startGene = 0; // stores starting of the gene string

for (int i = 0; i < genome.length() - 2; i++) {

//loop moves through genome string until the third last gene character

String triplet = genome.substring(i, i + 3);

//stores the triplet of genome substring

if (triplet.equals("ATG")) {

//if value in triplet is equal to ATG

startGene = i + 3;

//3 is added to i-th position of the genome string

}

else if (((triplet.equals("TAG")) || (triplet.equals("TAA")) || (triplet.equals("TGA"))) &&(startGene != 0))

//checks if the genome ends with one the given triplets TAG TAA and TGA

{ String gene = genome.substring(startGene, i);

gene stores substring of genome string from startGene to the position i

if (gene.length() % 3 == 0)

//if the the mod of gene length is 0 then the gene is found

{gene_found = true;

System.out.println(gene); //returns the found gene

startGene = 0;} } }

if (!gene_found) //if gene is not found returns the message below

System.out.println("no gene is found"); }  }

Explanation:

This program first asks user to enter a genome string.

The loop starts from the first character of the entered string and this loop continues to execute until the value of i is 2 less than the genome input string length.

triplet variable stores first 3 characters of the genome string in first iteration and then moves through the string taking 3 characters each. This is done by dividing genome string to substring of 3 characters.

If condition checks if the 3 characters of genome string matches ATG using equals() function. If it is true this means start of genome is reached and these triplets are stored in startGene.

Else condition checks the end of the genome as the genome ends before one of TAG, TAA or TGA triplets. So this is checked here.

gene variable holds the triplet value stored in startGene and the value stored in index position i which means it holds the start of the genome till the end of the genome sequence. The end which is pointed by i variable is 2 less than the genome length and it is stored in gene variable.

After the loop ends the substring stored in gene variable is checked for a valid genome sequence by mod operator. If the length of the value stored in gene variable mod 0 is equal to 0 this means genome sequence is found and this string sequence stored in gene is displayed else the no gene is found message is displayed on output screen.

7 0
3 years ago
A helicopter moves horizontally in the x direction at a speed of 120 mi/h. Knowing that the main blades rotate clockwise when vi
devlian [24]

Answer:

The instantaneous axis of rotation=

x = 0 ; z = 8.4 ft

Explanation:

Given:

Speed of helicopter, Vo= 120 mi/h, converting to ft/sec, we have:

\frac{5280 * 120}{60*60}

= 176 ft/s

Angular velociyy, w = 220 rpm, converting to rad/sec, we have: \frac{200*2*pi}{60} =20.95 rad/s

The helicopter moves horizontally in the x direction at a speed of 120 mi/h, this means that the helicopter moves in the positive x direction at 120mi/h

To find the instantaneous axis of rotation of the main blades, we have:

Where Vc = 20.95 rad/s

Vo = 176 ft/s

z = \frac{V_0}{V_c} = \frac{176ft/s}{20.95rad/s}

= 8.4 ft

Therefore the axis of rotation=

x = 0 ; z = 8.4 ft

4 0
3 years ago
Given an integer k, a set C of n cities c1, . . . , cn, and the distances between these cities dij = d(ci , cj ), for 1 ⤠i &lt
Snezhnost [94]

Answer:

See explaination

Explanation:

2-Approximation Algorithm

Step 1: Choose any one city from the given set of cities C arbitrarily and put it in to a set H which is initially empty.

Step 2: For every city c in set C that is currently not present in set H compute min_distc = Minimum[ d(c, c1), d(c, c2), d(c, c3), ..... . . . . d(c, ci) ]

where c1, c2, ... ci are the cities in set H

and d(x, y) is the euclidean distance between city x and city y

Step 3: H = H ∪ {cx} where cx is the city have maximum value of min_dist over all possible cities c, computed in Step-2.

Step 4: Step-2 and Step-3 are iterated for k-1 times so that k cities are included int set H.

The set H is the required set of cities.

Example

Assume:-

C = {0, 1, 2, 3}

d(0,1) = 10, d(0,2) = 7, d(0,3) = 6, d(1,2) = 8, d(1,3) = 5, d(2,3) = 12

k = 3

Solution:-

Initially H = { }

Step-1: H = {0}

Step-2: Cities c \not\in H are {1, 2, 3}

min_dist1 = min{dist(0,1)} = min{10} = 10

min_dist2 = min{dist(0,2)} = min{7} = 7

min_dist3 = min{dist(0,3)} = min{6} = 6

Step-3: Max{10, 7, 6} = 10

Step-4: cx = 1

Step-5: H = H ∪ cx = {0} \cup {1} = {0, 1}

Step-6: Cities c \not\in H are {2, 3}

min_dist2 = min{dist(0,2), dist(1,2)} = min{7, 8} = 7

min_dist3 = min{dist(0,3), dist(1,3)} = min{6, 5} = 5

Step-7: Max{7, 5} = 7

Step-8: cx = 2

Step-9: H = H \cup cx = {0, 1} \cup {2} = {0, 1, 2}

Result: The set H is {0, 1, 2}.

6 0
3 years ago
According to the rules of dimension how should written notes appear
Arturiano [62]

Explanation:

Leaders for notes should be straight, not curved, and point to the center of circular  views of holes wherever possible. Leaders should slope at 45°, 30° or 60° with  horizontal but may be made at any convenient angle except vertical or horizontal.

7 0
3 years ago
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