All categories

3 years ago

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.

You might be interested in

I need help with this I dont know the word

DiKsa [7]

Would it be Unit?

...............

5 0

3 years ago

Read 2 more answers

2. A counter flow tube-shell heat exchanger is used to heat a cold water stream from 18 to 78oC at a flow rate of 1 kg/s. Heatin

Anastaziya [24]

Answer:

a) $L = 220\,m$ , b) $U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

Explanation:

a) The counterflow heat exchanger is presented in the attachment. Given that cold water is an uncompressible fluid, specific heat does not vary significantly with changes on temperature. Let assume that cold water has the following specific heat:

$c_{p,c} = 4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C}$

The effectiveness of the counterflow heat exchanger as a function of the capacity ratio and NTU is:

$\epsilon = \frac{1-e^{-NTU\cdot(1-c)}}{1-c\cdot e^{-NTU\cdot (1-c)}}$

The capacity ratio is:

$c = \frac{C_{min}}{C_{max}}$

$c = \frac{(1\,\frac{kg}{s} )\cdot(4.186\,\frac{kW}{kg^{\textdegree}C} )}{(1.8\,\frac{kg}{s} )\cdot(4.30\,\frac{kW}{kg^{\textdegree}C} )}$

$c = 0.541$

Heat exchangers with NTU greater than 3 have enormous heat transfer surfaces and are not justified economically. Let consider that $NTU = 2.5$ . The efectiveness of the heat exchanger is:

$\epsilon = \frac{1-e^{-(2.5)\cdot(1-0.541)}}{1-(2.5)\cdot e^{-(2.5)\cdot (1-0.541)}}$

$\epsilon \approx 0.824$

The real heat transfer rate is:

$\dot Q = \epsilon \cdot \dot Q_{max}$

$\dot Q = \epsilon \cdot C_{min}\cdot (T_{h,in}-T_{c,in})$

$\dot Q = (0.824)\cdot (4.186\,\frac{kW}{^{\textdegree}C} )\cdot (160^{\textdegree}C-18^{\textdegree}C)$

$\dot Q = 489.795\,kW$

The exit temperature of the hot fluid is:

$\dot Q = \dot m_{h}\cdot c_{p,h}\cdot (T_{h,in}-T_{h,out})$

$T_{h,out} = T_{h,in} - \frac{\dot Q}{\dot m_{h}\cdot c_{p,h}}$

$T_{h,out} = 160^{\textdegree}C + \frac{489.795\,kW}{(7.74\,\frac{kW}{^{\textdegree}C} )}$

$T_{h,out} = 96.719^{\textdegree}C$

The log mean temperature difference is determined herein:

$\Delta T_{lm} = \frac{(T_{h,in}-T_{c, out})-(T_{h,out}-T_{c,in})}{\ln\frac{T_{h,in}-T_{c, out}}{T_{h,out}-T_{c,in}} }$

$\Delta T_{lm} = \frac{(160^{\textdegree}C-78^{\textdegree}C)-(96.719^{\textdegree}C-18^{\textdegree}C)}{\ln\frac{160^{\textdegree}C-78^{\textdegree}C}{96.719^{\textdegree}C-18^{\textdegree}C} }$

$\Delta T_{lm} \approx 80.348^{\textdegree}C$

The heat transfer surface area is:

$A_{i} = \frac{\dot Q}{U_{i}\cdot \Delta T_{lm}}$

$A_{i} = \frac{489.795\,kW}{(0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C} )\cdot(80.348^{\textdegree}C) }$

$A_{i} = 9.676\,m^{2}$

Length of a single pass counter flow heat exchanger is:

$L =\frac{A_{i}}{\pi\cdot D_{i}}$

$L = \frac{9.676\,m^{2}}{\pi\cdot (0.014\,m)}$

$L = 220\,m$

b) Given that tube wall is very thin, inner and outer heat transfer areas are similar and, consequently, the cold side heat transfer coefficient is approximately equal to the hot side heat transfer coefficient.

$U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

5 0

3 years ago

The depletion in the Ozone layer is caused by:

andrew11 [14]

D. Chlorofluorocarbon

This is a man-made carbon that causes the gradual thinning(deception) in the earth’s Ozone layer.

Hope this helped and pls mark as brainliest!

~ Luna

3 0

3 years ago

Read 2 more answers

D. projectile

Colt1911 [192]

Answer: parabola

Explanation:

•Parabolic Trajectory:

In conclusion, projectiles travel with a parabolic trajectory due to the fact that the downward force of gravity accelerates them downward from their otherwise straight-line, gravity-free trajectory.

5 0

3 years ago

What is the average distance (in terms of R) between the mobile on the fringe of the serving cell and the second and third tier

kramer

Answer:

do you need all work shown for this?

4 0

3 years ago