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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:

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Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

After some algebraic handling, the temperature of the cold reservoir is determined:

T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}

T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}}  \right)

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\dot Q_{H} = (12.5)\cdot (2.15\,kW)

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A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
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Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

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k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

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printf("address of intvar = %p\n", (void *)(&intvar));

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