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3 years ago

1. When the pneumatic device was demonstrated, explain something that was different than what you expected.

Engineering

2 answers:

Vera_Pavlovna [14]3 years ago

5 0

Answer:

1. Effect of air pressure

2. air- powered wheel chair

3. Pneumatic valves

Explanation:

1. In any pneumatic device, the mipact of air pressure to produce the moving effect on an heavy object is unexpected.

2. pneumatice demultiplexer when air in comprressed tank is allowed released to cause movement of the chair.

3. In industries, a pneumatic valve operates by force of air when actuated. A signal causes actuation of coil. When coil is energized, compressed high pressure air is allowe to enter in a small cylinder and cause operation of valve

Alex Ar [27]3 years ago

4 0

Answer (1)

<em>The difference in the pneumatic device to what i was expecting </em>was<em> </em>that i was expecting a device with a very complicated working system but was amazed at how simple but yet a little complex the pneumatic device was.

Answer (2)

<em>A typical device used help a disabled person </em>the pneumatic wheel chair. It uses compressed air to lift and lower the seat of the wheel chair. The compressed air is stored in a compressor below the seat and connected to the seat through piston cylinders. In order to lift, actuators activate valves that open and close at certain place to force air upward lifting the chair. Coming down is usually done by slowly leaking out the air and letting the chair descend under the weight of the user.

Answer:

<em>A pneumatic device used an automated piece of machinery </em>is the pneumatic press. The pneumatic press is a compressed air operated device in most manufacturing setting, frequently used in the automotive parts manufacturing industries. It uses compressed air to force down a ram which is usually used for baling scraps or for shaping metals.

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A HSS152.4 × 101.6 × 6.4 structural steel [E = 200 GPa] section (see Appendix B for crosssectional properties) is used as a colu

Temka [501]

Answer:

(a) 126.66 kN (b) 31.665 kN (c) 258.49 kN (d) 506.64 kN

Explanation:

Solution

Given

A HSS152.4 × 101.6 × 6.4 structural steel is used as a column

Actual length of the column , L= 6 m

The elasticity modules, E = 200 GPa

The factor of safety with respect to failing buckling . F.S =2

Geometric properties of structural steel shapes for, A HSS152.4 × 101.6 × 6.4

the moment of inertial about y axis Iy =4 .62 * 10^ 6 mm ^4

For

(a) If the end condition is pinned - pinned

The effective length factor, K =1

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 1* 6* 10 ^3)

= 253319.85N

= 253.32N

The maximum safe load , Pallow = 253.32 /2 = 126.66kN

hence, the maximum safe for the column is 126.66kN

(b)If the end condition is fixed free-free

the effective length factor, K= 2

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 2 * 6 * 10 ^3)²

= 63329.96N

=63.33kN

The maximum safe load, P allow = Pcr/F.S

P allow = 63.33/2

31.665 kN

Therefore the maximum safe for the column is 31.665 kN

The effective length factor K =0.7

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.7 * 6 * 10 ^3)²

=516979.2 8N

=516.98 kN

The maximum safe load, P allow = Pcr/F.S

P allow = 516.98 kN/2

=258.49 kN

Therefore the maximum safe for the column is 258.49 kN

(d) If the end condition is fixed -fixed

The effective factor, K =0.5

The critical buckling load , Pcr = π²EI/(KL)²

Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.5 * 6 * 10 ^3)²

=1013279.4 N

=1013.28 kN

The maximum safe load, P allow = Pcr/F.S

P allow = 1013.28 / 2

= 506.64 kN

P allow = 506.64 kN

Therefore the maximum safe for the column is 506.64 kN

8 0

4 years ago

The Hubble space telescope is a bus-sized satellite powered by huge ___ panels

Artemon [7]

Answer:

Solar

Explanation:

The Hubble space telescope is a bus-sized satellite powered by huge solar panels.

4 0

3 years ago

Read 2 more answers

A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at

Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, $T_{s}=280^{o}C$

Temperature of air, $T_{\infinity}=20^{o}C$

Velocity of air flow $V=5.5m/s$

The film temperature is determined as:

$T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C$

from the table, properties of air at 1 atm pressure

At $T_{f}=150^{o}C$

Thermal conductivity, $K = 0.03443 W/m^oC$ ; kinematic viscosity $v=2.860 \times 10^{-5} m^2/s$ ; Prandtl number $Pr=0.70275$

The reynolds number for the flow is determined as:

$Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92$

sice the obtained reynolds number is less than $2\times10^5$ , the flow is said to be laminar.

The nusselt number is determined from the relation given by:

$Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}$

$Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11$

The covective heat transfer coefficient is given by:

$Nu_{cyl}=\frac{hD}{k}$

Rewrite and solve for $h$

$h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K$

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

$Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m$

The rate of heat transfer from the wire to the air per meter length is $Q=340.42W/m$

6 0

3 years ago

A system consists of a disk rotating on a frictionless axle

kakasveta [241]

The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.

<h3>What is the angular momentum?</h3>

The angular momentum of the device earlier than and after the clay sticks can be the same.

Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.

Li = Lf where;
li is the preliminary second of inertia
If is the very last second of inertia
wi is the preliminary angular velocity
wf is the very last angular velocity
Li is the preliminary angular momentum
Lf is the very last angular momentum

Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.