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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.

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La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación

gregori [183]

Answer:

I = 1205.69 Lx

Explanation:

The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀

I = I₀ sin θ

in this case with the initial data we can calculate the initial irradiance

I₀ = $\frac{I}{sin \ \theta }$

I₀ = 1600 /sin 53

I₀ = 2003.42 lx

for when the angle is θ = 37º

I = 2003.42 sin 37

I = 1205.69 Lx

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3 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit

8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, $A_1 = 0.0044 m^2$

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

$KE = \frac{V_2^2 - V_1^2}{2} \\$ .........(1)

Where Inlet velocity, V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

$KE = \frac{50^2 - 80^2}{2} \\$

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a steady state flow.

$q - w = h_2 - h_1 + KE + g(z_2 - z_1)$

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, $W = \dot{m}w$ ..........(3)

Mass flow rate, $\dot{m} = 12 kg/s$

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, $h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg$ , x₂ = 0.92

specific enthalpy at the outlet, h₂ = $h_1 + x_2 h_{fg}$

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

$A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2$

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4 years ago

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 $mm^4$

So the shear stress at point A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

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3 years ago

Q.17) A 50-acre catchment containing cropland is converted ot a Qatar mail

Ghella [55]

Answer:

Option D

Explanation:

A post development hydrograph will have lower concentration time and lower infiltration losses and hence sooner peak and higher peak and more runoff or higher area under graph. Therefore, all the answers are correct hence option D

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3 years ago

What is a ton of refrigeration?

AURORKA [14]

Explanation:

The unit refrigeration is generally is given in terms of tons.In refrigeration compressor consume some amount of work to produce the cooling effect with the help of evaporator and condenser.

In the simple words ton is the cooling load of refrigeration system.

1 ton = 3.5 KW

1 ton = 12,000 BTU/hr

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3 years ago