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3 years ago

Select four examples of fluid or pneumatic power systems.

Engineering

2 answers:

Dennis_Churaev [7]3 years ago

8 0

Answer:

Students learn about the fundamental concepts important to fluid power, which includes both pneumatic (gas) and hydraulic (liquid) systems. Both systems contain four basic components: reservoir/receiver, pump/compressor, valve, cylinder.

Explanation:

Firlakuza [10]3 years ago

3 0

Answer:Air brakes on buses and trucks are formally known as compressed air brake systems. Exercise machines can be built on pneumatic systems. Compressed-air engines, also called pneumatic motors, do mechanical work by expanding compressed air.

Explanation:

Fluid power is the use of fluids under pressure to generate, control, and transmit power. Fluid power is subdivided into hydraulics using a liquid such as mineral oil or water, and pneumatic using a gas such as air or other gases

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Explanation:

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2 years ago

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3 years ago

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

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3 years ago

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2 years ago

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yarga [219]

Answer:

<h2>True Most Especially in the field of Automotive Engineering</h2>

Explanation:

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