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Pavlova-9 [17]
3 years ago
5

Please describe a real situation in which you had to troubleshoot and fix the failure of a piece equipment/machine?

Engineering
1 answer:
VARVARA [1.3K]3 years ago
3 0

Answer:

Fixing a faulty spark-plug in a <em>typical gasoline car engine</em>.

Explanation:

<h3>The faulty equipment/machine</h3>

The equipment/machine was a <em>gasoline car engine</em>. After switching it on, I heard a not common pulsing sound at the end of the exhaust pipe of the car. When running the car, there was also a <em>loss in power</em> of the engine when trying to accelerate it.

<h3>Troubleshooting</h3>

I started with a <em>working hypothesis</em> addressing a fault in any of the spark-plugs installed, since the problem, at first sight, seemed any related to an ignition or fuel distribution failure. Since spark-plugs are relatively easy components to remove and check, then I troubleshoot them as a first step.

Firstly, I checked the condition for any of the spark-plugs, extracting them with a <em>spark plug socket</em> and a <em>torque wrench</em>, to later perform a <em>careful visual inspection</em>, looking for any out-of-normal condition in any of the four of them installed in my car. I checked three of them, visualizing no wear, combustion deposits, and/or electrodes inappropriate gauge distance.

For the last one, I needed an <em>anti-seized lubricant</em> to extract it with caution to not break it or cause any damage to the component. After extraction and the visual inspection, I immediately noticed <em>the presence of combustion deposits between the electrodes of the spark-plug</em>.

<h3>Fixing the problem</h3>

Since one way of cleaning a spark-plug is using a wire brush, I started to carefully remove the deposits until no remaining was left after the cleaning process. I also checked for wear conditions and gauging distance of the electrodes, which resulted in good enough.

An explanation for this is that in the electrodes gap were accumulated some deposits coming from combustion, resulting in a <em>not working cylinder</em>, thus the noise and loss of power in the engine.

I reinstalled the spark-plug and the fault dissipated after accelerating the engine <em>in situ</em>.

<h3>The outcome</h3>

The engine is working appropriately with no pulsing sound and the end of the exhaust pipe of the car and no power loss while accelerating it.

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Answer:

Part 1: It would be a straight line, current will be directly proportional to the voltage.

Part 2: The current would taper off and will have negligible increase after the voltage  reaches a certain  value. Graph attached.

Explanation:

For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.

V=I*R

where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.

In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.

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Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
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Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

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Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

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1.) You can prevent making a fire by making sure you close and seal flammable materials.

2.) You can prevent heat burns by teaching the students how to properly use tongs,water baths, and other cooling equipment. 

3.) You can prevent chemical burns by treating the chemicals with caution, measure carefully, and use the approved containers.

4.) You can prevent cuts and scrapes by telling the students how to use the blades safely, and also when they are disposing broken or sharp items they should know how to wrap them up so no one else will get hurt. 

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