3. A steel pipe of outside diameter 20 mm and thickness 3 mm is

A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co

podryga [215]

Answer:

a) $T_{H} = 1.967\,^{\circ}F$ , b) $COP_{R} = 9.105$ , c) $T_{H} = 115.934\,^{\circ}F$ , d) $COP_{R} = 6.995$ , e) $T_{H} = 25.129\,^{\circ}C$

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

$COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}$

$10 = \frac{419.67\,R}{T_{H}-419.67\,R}$

The temperature of the hot reservoir is:

$10\cdot T_{H} - 4196.7 = 419.67$

$T_{H} = 461.637\,R$

$T_{H} = 1.967\,^{\circ}F$

b) The coefficient of performance is:

$COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}$

$COP_{R} = 9.105$

c) The temperature of the hot reservoir can be determined with the help of the following relation:

$\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}$

$\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5 = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5\cdot T_{H} - 2398.35 = 479.67$

$T_{H} = 575.604\,R$

$T_{H} = 115.934\,^{\circ}F$

d) The coefficient of performance is:

$COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}$

$COP_{R} = 6.995$

e) The temperature of the cold reservoir is:

$8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}$

$8.9\cdot T_{H} - 2386.535 = 268.15$

$T_{H} = 298.279\,K$

$T_{H} = 25.129\,^{\circ}C$

8 0

3 years ago

What the phat is this

Alex17521 [72]

Answer:

It's Brainly ;)

8 0

3 years ago

Read 2 more answers

Question 2: (a) In your own words, clearly distinguish and differentiate between Ethics in Engineering and Ethics in Computing (

zlopas [31]

Engineering ethics is not without abstraction, but in contrast with computing, it is animated by a robust and active movement concerned with the seamless identification of ethics with practice.

<h3 /><h3>What is engineering?</h3>

This is a branch of science and technology concerned with the design, building, and use of engines, machines, and structures that uses scientific principles.

Comparing ethics in engineering and ethics in computing:

Engineering ethics are a set of rules and guidelines. While computing ethics deals with procedures, values and practices.
In engineering ethics, engineers must adhere to these rules as a moral obligation to their profession While in computing ethics, the ethics govern the process of consuming computer technology.
Following these ethics for the two professions will NOT cause damage, but disobeying them causes damage.

Some practical examples in the computing field:

Avoid using the computer to harm other people such as creating a bomb or destroying other people's work.
Users also should not use a computer for stealing activities like breaking into a bank or company.
Make sure a copy of the software had been paid for by the users before it is used.

Some practical examples in the engineering field:

Integrity for oneself.
Respect for one another.
Pursuit of excellence and accountability.

Hence, Engineering ethics is the field of system of moral principles that apply to the practice of engineering and following them is important to the profession.

Read more about <em>engineering</em> here:

brainly.com/question/17169621

#SPJ1

7 0

2 years ago

Technician a s ays both an ohmmeter and a self-powered test light may be used to test for continuity. technician b says both may

amm1812

Both A and B technicians are correct because both might be used to test fuses, according to technician B.

<h3>What is continuity?</h3>

The behavior of a function at a certain point or section is described by continuity. The limit can be used to determine continuity.

From the question:

We can conclude:

The technician claims that you may check for continuity using both an ohmmeter and a self-powered test light. Both might be used to test fuses, according to technician B.

Thus, both A and B technicians are correct because both might be used to test fuses, according to technician B.

Technician A says both an ohmmeter and a self-powered test light may be used to test for continuity. Technician B says both may be used to test fuses. Who is correct?

Learn more about the continuity here:

brainly.com/question/15025692

#SPJ1

5 0

1 year ago

A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance

Talja [164]

Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

4 0

3 years ago