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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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Answer:

Explanation:

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7 0
3 years ago
The term variation describes the degree to which an object or idea differs from others of the same type or from a standard.
AfilCa [17]
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5 0
3 years ago
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Nancy wants to buy a cooking stove that’s electrically insulated and resistant to heat. What material should she choose for the
astra-53 [7]

Answer:

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3 0
3 years ago
It is easy to say that an organization should hire, reward, and dismiss employees based on their character as well as their know
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5 0
3 years ago
Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water
olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate  = 2 L/s

water temperature 30 degree celcius

Q = A\times V

2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity

V = \frac{20\times 4}{9\times \pi} = 2.83 m/s

Re = \frac{\rho\times V\times D}{\mu}

at 30 degree celcius = \mu = 0.798\times 10^{-3}Pa-s , K  = 0.6154

Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}

Re = 106390

So ,this is turbulent flow

Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}

Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419

\frac{h\times 0.03}{0.615}  = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0
4 years ago
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