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KonstantinChe [14]
3 years ago
15

6-What is the difference between the critical point and the triple point?

Engineering
2 answers:
hichkok12 [17]3 years ago
8 0

Answer:

The triple point represents the combination of pressure and temperature that facilitates all phases of matter at equilibrium. The critical point terminates the liquid/gas phase line .

Explanation:

Svetradugi [14.3K]3 years ago
8 0

Explanation:

<h2><u>Critical Point :</u></h2>
  • It's the end point of equilibrium curve .
  • It's typically higher than standard temperature.
  • pressure is also higher than standard pressure.

<h2><u>T</u><u>r</u><u>i</u><u>p</u><u>l</u><u>e</u><u> </u><u>Point</u><u> </u><u>:</u></h2>
  • It's point where the three equilibrium curves meet.
  • Temperature is lower than standard temperature.
  • pressure is lower than standard pressure.

<u>A</u><u>d</u><u>d</u><u>i</u><u>t</u><u>i</u><u>o</u><u>n</u><u>a</u><u>l</u><u> </u><u>points</u><u> </u><u>:</u>

  • The triple point represents combination of pressure .
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Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate
Soloha48 [4]

Answer:

radius = 0.045 m

Explanation:

Given data:

density of oil = 780 kg/m^3

velocity = 20 m/s

height = 25 m

Total energy is = 57.5 kW

we have now

E = kinetic energy+ potential energy +  flow work

E = \dot m ( \frac{v^2}{2] +  zg + p\nu)

E = \dot m( \frac{v^2}{2] +  zg + p_{atm} \frac{1}{\rho})

57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})

solving for flow rate

\dot m = 99.977we know that [tex]\dot m  = \rho AV

\dot m  = 780 \frac{\pi}{4} D^2\times 16

solving for d

99.97 = 780 \times \frac{\pi}{4} D^2\times 16

d = 0.090 m

so radius = 0.045 m

3 0
3 years ago
Explain why advances in technology can actually create more problems for engineers.
andrew11 [14]

Explanation:

Engineering is science in practical terms. It is the application of scientific findings in problem solving and creating a better world.

How does technological advancements create more problems for engineers?

  • Loss of job to automation: the world is driving at automating work processes through the use of  specially designed and crafted machinery. Work is now properly being done using machines with little to no human input in the whole process. This is a huge let off for engineers. Engineers have to compete with machines which are their own inventions for jobs now.
  • Fast paced work environment: machines can handle work more efficiently and faster than the people making them. There is an increasing race between engineers and their own inventions today for better product delivery. Unless a machine is faulty, they are more productive and efficient than man. This can cause engineers to want to catch up with their own inventions leading to a work life of stress.
  • Environmental problems they cannot solve: most inventions use components from the environment. They release effluents that are very difficult to be properly disposed or stored. This is a huge problem for engineers and can lead to ethical calls from the government and the populace. In short, they can create problems they are expected to solve but cannot solve.
  • Social problems: engineers can be portrayed as terrible beings for their own inventions. This leads to psychological problems on a good and creative invention. For example, rare earth metals in DR Congo are instrumental in making solar panels, but mining of these metals have forced several thousands of people into hard and intense labor on mines; there is a call on technological firms to stop exploiting people this way for their own gains.
  • Misuse of technology: any good technology can be put into the wrong use. A nuclear reaction can be packaged into a bomb and also, it can be the center of electricity generation on a commercial scale. How can engineers solve this kind of problem? Technological inventions are subjective in their usage.

Learn more:

New technology brainly.com/question/5768621

#learnwithBrainly

3 0
3 years ago
(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific
lina2011 [118]

Answer:

The answer to the question is 514.17 lbf

Explanation:

Volume of cylindrical tank = πr²h = 3.92699 ft³

Weight of tank = 125 lb

Specific weight of content = 66.4 lb/ft³

Mass of content =  66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

5 0
3 years ago
When they say in the United States that a car’s tire is filled “to 32 lb,”​ they mean that <br>its internal pressure is 32 lbf/i
arsen [322]

Answer:

0.71 lbf

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.

P = 32 lbf/in² + 14.7 lbf/in²

P = 46.7 lbf/in²

Absolute temperature is in Kelvin or Rankine:

T = 75 + 459.67 R

T = 534.67 R

Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:

PV = nRT

(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)

n = 0.02442 lb-mol

The molar mass of air is 29 lbm/lb-mol, so the mass is:

m = (0.02442 lb-mol) (29 lbm/lb-mol)

m = 0.708 lbm

The weight of 1 lbm is lbf.

W = 0.708 lbf

Rounded to two significant figures, the weight of the air is 0.71 lbf.

3 0
3 years ago
Test de evaluare
ivolga24 [154]
C why it’s c bc they I just got it right
7 0
2 years ago
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