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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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A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

$= 71.82 \ kN/m^2$

$\sim 72 \ kN/m^2$

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 =$ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

= 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

= 13.14

= 13

8 0

3 years ago

At what times should you use your headlights?

SVETLANKA909090 [29]

Answer:

Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.

Explanation:

hope this helps

8 0

3 years ago

Read 2 more answers

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

max2010maxim [7]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

7 0

3 years ago

A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f

jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

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3 years ago

Most deformation occurs along plate boundaries because ________.

Svetlanka [38]

B) the plates are in constant motion and as a result the boundaries are where they interact

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3 years ago