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vodomira [7]
1 year ago
5

technician a says that dirt bypassing the filter on many common rail injectors can cause an injector to stick open and continuou

sly fuel a cylinder. technician b says that any dirt particles may potentially damage the injector, but would not allow an injector to full fuel or the engine to run away. who is correct?
Engineering
1 answer:
NNADVOKAT [17]1 year ago
3 0

Technician a is correct because he says that Many common rail injectors filters can be bypassed by dirt, which can lead to an injector sticking open and continuously fueling a cylinder.

Coalescence is used to separate the water and fuel. To the fuel injector cleaning kit, fasten your air compressor. Diesel engines run at compression ratios that are greater than those of gasoline engines. greater ratio compared to gasoline engines. increased thermal expansion as a result. more fuel energy that is transformed into usable power. The great benefit of using a dry cylinder sleeve is that by quickly installing new sleeves, the cylinder block can be quickly restored to its original specifications. Vacuum drying can be used to get rid of small amounts of water. A nozzle is used to spray the fuel into the vacuum chamber of engines. Air and unsolved free water are taken out of the oil. The fuel is evenly dispersed, which facilitates efficient drying.

Learn more about injectors here:

brainly.com/question/27969202

#SPJ4

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I have a Dutch oven that looks like this what do I do?
dangina [55]

Answer:

baking soda and vinegar dish soap

Explanation:

it will create a bubbles and let it sit for 3 hours and it will go away

3 0
3 years ago
Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at
andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:  

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that;  distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using  this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0
3 years ago
Analyse what effect the building of an airport may have on the decision of how to use an area of land nearby. (6)​
Sonja [21]
An effect might be a customer not wanting to buy it specifically because it’s by an airport, or maybe the customer wants to buy it because it’s right next to the airport, and a lot of people go to the airport so therefore they might go to the building next to the airport.
5 0
3 years ago
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 5 mm diameter, 40 m long horiz
ra1l [238]

Answer:

Flow rate is 1.82\times 10^{-8} m^{3}/s

Explanation:

Given information

Density of oil, \rho_{oil}= 850 Kg/m^{3}

kinematic viscosity, v= 0.00062 m^{2} /s

Diameter of pipe, D= 5 mm= 0.005 m

Length of pipe, L=40 m

Height of liquid, h= 3 m

Volume flow rate for horizontal pipe will be given by

\bar v=\frac {\triangle P\pi D^{4}}{128\mu L} where \mu is dynamic viscosity and \triangle P is pressure drop

At the bottom of the tank, pressure is given by

P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}

Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still 25015.5 N/m^{2}

Dynamic viscosity, \mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s

Now the volume flow rate will be

\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s

Proof of flow being laminar

The velocity of flow is given by

V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104  m/s

Reynolds number, Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648

Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.

5 0
4 years ago
6.39 A particular machine part is subjected in service to a maximum load of 10 kN. With the thought of providing a safety factor
Serggg [28]

Answer:

2.275 %

Explanation:

see the attached picture for detailed answer.

8 0
3 years ago
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