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2 years ago

Question 3

Engineering

1 answer:

Dmitry [639]2 years ago

6 0

Answer:

Explanation:

The tour length will be the sum of the lengths of its segments. Here, we will figure the length using the greedy algorithm.

Starting at B=2, the nearest unvisited city is C=3, a distance of 1.

From C, the nearest unvisited city is A=0, a distance of 3.

From A, the nearest unvisited city is D=10, a distance of 10.

The return trip from D to B is a distance of 8.

The total length of the tour is ...

1 + 3 + 10 + 8 = 22

You might be interested in

How can I find the quotient of 27 divided 91.8

gayaneshka [121]

Step 1:
Start by setting it up with the divisor 8 on the left side and the dividend 27 on the right side like this:

8 ⟌ 2 7

Step 2:
The divisor (8) goes into the first digit of the dividend (2), 0 time(s). Therefore, put 0 on top:

0
8 ⟌ 2 7

Step 3:
Multiply the divisor by the result in the previous step (8 x 0 = 0) and write that answer below the dividend.

0
8 ⟌ 2 7
0

Step 4:
Subtract the result in the previous step from the first digit of the dividend (2 - 0 = 2) and write the answer below.

0
8 ⟌ 2 7
- 0
2

Step 5:
Move down the 2nd digit of the dividend (7) like this:

0
8 ⟌ 2 7
- 0
2 7

Step 6:
The divisor (8) goes into the bottom number (27), 3 time(s). Therefore, put 3 on top:

0 3
8 ⟌ 2 7
- 0
2 7

Step 7:
Multiply the divisor by the result in the previous step (8 x 3 = 24) and write that answer at the bottom:

0 3
8 ⟌ 2 7
- 0
2 7
2 4

Step 8:
Subtract the result in the previous step from the number written above it. (27 - 24 = 3) and write the answer at the bottom.

0 3
8 ⟌ 2 7
- 0
2 7
- 2 4
3

You are done, because there are no more digits to move down from the dividend.

The answer is the top number and the remainder is the bottom number.

Therefore, the answer to 27 divided by 8 calculated using Long Division is:

3

8 0

3 years ago

Here you go!!!!!!!!!!!!!!!!!1

sweet [91]

Answer:

Im confused, what does this mean

Explanation:

i mean, thx lol

3 0

3 years ago

The magic square is an arrangement of numbers in a square grid in such a way that the sum of the numbers in each row, and in eac

Yuki888 [10]

Answer:

See the explanation for the answer;

Explanation:

Matlab code is as given below

-------------------------------------------------------------------------------------Start of code

% Program to create a magic square and verify it

clc

M= magic(5)

% To find sum of elements of each row

r1= sum(M(1,:)) % Sum of row 1

r2= sum(M(2,:)) % Sum of row 2

r3= sum(M(3,:)) % Sum of row 3

r4= sum(M(4,:)) % Sum of row 4

r5= sum(M(5,:)) % Sum of row 5

% To find sum of each coloumn

c1= sum(M(:,1)) % Sum of coloumn 1

c2= sum(M(:,2)) % Sum of coloumn 2

c2= sum(M(:,3)) % Sum of coloumn 3

c2= sum(M(:,4)) % Sum of coloumn 4

c2= sum(M(:,5)) % Sum of coloumn 5

% To find sum of diagonal

d1= sum(diag(M)) % Sum of principal diagonal elements

d2= sum(diag(fliplr(M)))

-------------------------------------------------------------------------------End of code

Following results are obtained when executed.

M =

17 24 1 8 15

23 5 7 14 16

4 6 13 20 22

10 12 19 21 3

11 18 25 2 9

r1 =

r2 =

r3 =

r4 =

r5 =

c1 =

c2 =

d1 =

d2 =

3 0

3 years ago

A fair die is thrown, What is the probability gained if you are told that 4 will

Semmy [17]

Answer:

1/6

Explanation:

A dice has 6 sides, the probability of 4 appearing is 1/6.

7 0

3 years ago

Consider an ideal cogeneration steam plant to generate power and process heat. Steam enters the turbine from the boiler at 7 Mpa

igomit [66]

Answer:

1. The diagram T-s or H-s is attached to this answer.

2. The fraction of the steam extracted is 4.088Kg/s

3. The net Power produced per kg of steam exiting the boiler is 1089.5KJ/Kg.

4. The mass flow rate of steam supplied by the boiler is 16.352Kg/s

5. the net power produced by the plant is 11016.2KJ/s.

6. The utilization factor is 0.218.

Explanation:

To analyze this problem we need to find all the thermodynamic coordinates of the system. In the second image attached to this answer, we can see the entire ideal cogeneration steam plant system.

From a water thermodynamic properties chart, we can obtain the information for each point.

+ Steam enters the turbine from the boiler at 7 Mpa and 500 degrees C:

h₆=3410.56KJ/Kg

s₆=6.7993 KJ/Kg

This is an ideal cogeneration steam system, therefore: s₆=s₇=s₈

+One-fourth of the steam is extracted from the turbine at 600-kPa:

h₇(s₇) = 2773.74 KJ/Kg (overheated steam)

+The remainder of the steam continues to expand and exhausts to the condenser at 10 kPa.

h₈(s₈)=2153.58 KJ/Kg (this is wet steam with title X=0.8198)

h₁(P=10Kpa)= 191.83 KJ/Kg (condensed water) s₁=0.64925KJ/Kg

-This flow is pumped to 600KPa, so:

s₂=s₁

h₂(s₂)=192.585KJ/Kg

+The steam extracted for the process heater is condensed in the heater:

h₃(P=600KPa)=670.42KJ/Kg (condensed water)

+The steam extracted for the process heater is condensed in the heater and mixed with the feed-water at 600 kPa:

The mixing process of the flow of point 2 and 3 is an adiabatic process, therefore:

$\dot{Q}_4=\dot{Q}_2+\dot{Q}_3=\dot{m}_2 h_2+\dot{m}_3h_3\\\dot{m}_4 h_4=\dot{m}_2 h_2+\dot{m}_3h_3=\dot{m}_2 0.75h_4+\dot{m}_30.25h_4\\h_4=0.75h_2+0.25h_3=312.043KJ/Kg$

s₄=1.02252

+the mixture is pumped to the boiler pressure of 7 Mpa:

s₅=s₄

h₅(s₅)=323.685KJ/Kg

1)Now we have all the thermodynamic coordinates and we can draw the diagram of the system.

2) To determine the fraction of steam, the mass flow that is extracted from the turbine at state 7, we use the information that this flow is used to generate 8600KJ/s in a process of heat. Therefore:

$P=8600KJ/s=\dot{m}_{3-7}(h_7-h_3)\\\dot{m}_{3-7}=8600KJ/s/(h_7-h_3)=4.088Kg/s$

3)The net power produced per kg of steam exiting the boiler can be obtained as the rest between all the power obtained in the turbine less the power used in the pumps:

$P_{turb}/Kg=(h_6-h_7)+0.75(h_7-h_8)=1101.94KJ/Kg\\P_{pump1}/Kg=h_2-h_1=0.755KJ/kg\\P_{pump2}/kg=h_5-h_4=11.642KJ/Kg\\P_{net}/kg=P_{turb}-P_{pump1}-P_{pump2}=1089.543KJ/Kg$

4) To determine the mass flow rate of steam that must be supplied by the boiler, we only have to remember that the flow used in point 2) is a fourth of the total flow. therefore:

$0.25\dot{m}_{tot}=\dot{m}_{3-7}\\\dot{m}_{tot}=4\dot{m}_{3-7}=16.352Kg/s$

5)The net power supplied by the plant is the net power calculated in point 3) less the power used in the heat process 7-3:

$P_{net sys}=P_{net}/kg \cdot \dot{m}_t-6800Kj/s=11016.2KJ/s$

6) The utilization factor is obtained as the division between net power supplied by the plant and the power used to heat the steam. In this case:

$UF=P_{net sys}/P_{boil}=P_{net sys}/[\dot{m}_t(h_6-h_5)]=0.21$

3 0

3 years ago