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Mariulka [41]
3 years ago
6

Create a function (prob3_5) that will take inputs of vectors x and y in feet, scalar N, scalars L and W in feet and scalars T1 a

nd T2 in degrees Fahrenheit. It will output a matrix T which is the temperature of each x and y locations. T will have the number of columns equal to the number of elements in x and rows equal to the number of elements in y. Though this can be done without loops (perhaps more efficiently), your program must use a nested loop.
Engineering
1 answer:
Shalnov [3]3 years ago
3 0

Answer:

clear, clc

prob3_5([1,2,3],[6,5,7],12,11,22,55,76)

function T=prob3_5(x,y,N,L,W,T1,T2)

w=zeros(1,length(x));

for n=1:2:N

for i=1:length(x)

w(i)=w(i)+(2/pi)*(2/n)*sin(n*pi*x(i)/L).*sinh(n*pi*y(i)/L)/sinh(n*pi*W/L);

end

end

T=(T2-T1)*w+T1;

end

Explanation:

Please input the commands into MATLAB

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Which of the following describes design components that deal with the outward appearance or beauty of an object?
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Aesthetic elements are the components that are added to the design to be considered pleasing to the eye.

<h3>What are aesthetic elements?</h3>

They are those characteristics of an object that deal with the outward appearance or beauty of an object, that is, they are those elements that make it valuable, appreciable, relevant or transcendent.

To do this, the qualities must be in the design of the object but must also be perceived by the consumer, the aesthetic being what we like to perceive in objects.

Therefore, we can conclude that aesthetic elements are the components that are added to the design to be considered pleasing to the eye.

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2 years ago
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Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

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6 0
2 years ago
The specific volume of a system consisting of refrigerant-134a at 1.0 Mpa is 0.01 m /kg. The quality of the R-134a is: (a) 12.6%
Flura [38]

Answer:

option c is correct

47.2%

Explanation:

given data

consisting of refrigerant = 134 a

volume V = 0.01 m³/kg

pressure P = 1MPa = 1000 kPa

to find out

quality of the R 134a

solution

we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a

that is

Vf = 0.0008701 m³/kg

Vv = 0.0203 m³/kg

so we will apply here formula that is

quality = (V - Vf) / (Vv - Vf)    ............1

put here value

quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )

quality = 0.4698

so quality is 47 %

SO OPTION C IS CORRECT

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3 years ago
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