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3 years ago

Describe the operation of a pneumatic robot controlled by a drum

Engineering

1 answer:

Naddika [18.5K]3 years ago

7 0

Answer:hopefully this helps

Explanation:A MIDI drum machine was connected to a box of electronics containing a series of relays, which in turn operate solenoid valves, causing the pneumatic rams to function. The rams were connected to various mechanisms that strike the drums in accordance to the rhythm programmed on the Groovebox. The smoke machine was also controlled via MIDI.

There was also some circuitry added to control the length of the pulse sent to the solenoid valve, so that the time before the air flow in the ram was reversed could be adjusted to create an optimum percussive sound. The rod seals were machined to make the rams act faster

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Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e

marshall27 [118]

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

$\Delta x=\frac{PL^3}{3EI}$

Hence we can write

$P=\frac{3EI\cdot \Delta x}{L^3}$

Comparing with the standard spring equation $F=kx$ we find the cantilever analogous to spring with $k=\frac{3EI}{L^3}$

Now the angular frequency of a spring is given by

$\omega =\sqrt{\frac{k}{m}}$

where

'm' is the mass of the load

Thus applying values we get

$\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}$

$\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec$

8 0

3 years ago

Read 2 more answers

An insulated mixing chamber receives 0.5 kg/s of steam at 3 MPa and 300°C through one inlet, and saturated liquid water at 3 MPa

maxonik [38]

Answer:

$\dot m_{2} = 0.199\,\frac{kg}{s}$

Explanation:

The mixing chamber can be modelled by applying the First Law of Thermodynamics:

$\dot W_{in}+\dot m_{1}\cdot h_{1} +\dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:

$\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - z\cdot h_{3} = 0$

Besides, the following expression derived from the Principle of Mass Conservation is presented below:

$1 + y = z$

Then, the expression is simplified afterwards:

$\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - (1+y)\cdot h_{3} = 0$

$\frac{\dot W_{in}}{\dot m_{1}} +h_{1} - h_{3} + y\cdot (h_{2}-h_{3}) = 0$

Specific enthalpies are obtained from steam tables and described as follows:

State 1 (Superheated vapor)

$h = 2994.3\,\frac{kJ}{kg}$

State 2 (Saturated liquid)

$h = 1008.3\,\frac{kJ}{kg}$

State 3 (Liquid-Vapor mixture)

$h = 2444.22\,\frac{kJ}{kg}$

The ratio of the stream at state 2 to the stream at state 1 is:

$y = \frac{\frac{\dot W_{in}}{\dot m_{1}}+h_{1}-h_{3}}{h_{3}-h_{2}}$

$y = \frac{\frac{10\,kW}{0.5\,\frac{kg}{s} }+2994.3\,\frac{kJ}{kg}-2444.22\,\frac{kJ}{kg} }{2444.22\,\frac{kJ}{kg}-1008.3\,\frac{kJ}{kg} }$

$y = 0.397$

The mass flow rate of the saturated liquid is:

$\dot m_{2} = y\cdot \dot m_{1}$

$\dot m_{2} = 0.397\cdot (0.5\,\frac{kg}{s} )$

$\dot m_{2} = 0.199\,\frac{kg}{s}$

5 0

3 years ago

Why is transverse unequal length wishbone preferred to that of equal length?

Damm [24]

Answer:

It allows the wheels to move of the neutral position without the tires scrubbing

Explanation:

The double wishbone suspension is independent and it can be used in both the front wheels and it affords very good control of the outward or inward tilting of the front wheel and it keeps the wheels perpendicular to the road surface

However, when equal length parallel wishbone are installed, it gives rise to scrubbing of the tires as the wheels turn in the tracks

The development of unequal length non-parallel transverse or converging wishbones with A-arms suspension resolved the tire scrubbing effect on the wheels when moving out of the neutral position.

8 0

4 years ago

(a) Define a slip system. (b) Cite one slip system for the simple cubic crystal structure. (c) One slip system for the BCC cryst

Lemur [1.5K]

Answer:

I'm gonna say D.

4 0

3 years ago

A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a

prohojiy [21]

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

= 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

= 2622.6 / 25 = 104.904 ft^3/sec

8 0

3 years ago