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egoroff_w [7]
3 years ago
7

Nguyên lý hoạt động của kim phun

Engineering
1 answer:
arsen [322]3 years ago
7 0

Answer:

Fuel from the high-pressure pump comes out with a certain pressure when it passes through the oil pipe and will be led into the high-pressure injector.

Here it will be inserted into the spring and injector cavity. The injector opens only when the oil pressure must be greater than the spring compression. Then the cone will be opened to help fuel through.

Immediately after the injection process ends, the pressure will decrease and the cone will close again. That process repeats itself and depends entirely on the pressure generated by the high-pressure pump.

The compression pressure from the spring can be adjusted through the adjustment knob, which can be tightened or loosened at will. The compression of the spring is therefore also increased and decreased. The nozzle pressure was then also changed.

Explanation:

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Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distanc
Andrei [34K]

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m

b) f = 1 GHz = 10⁹ Hz.

\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10}  e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10}  e=-1.5 \\\\z=\frac{-1.5}{log_{10}  e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm

4 0
3 years ago
PLZ HELP ME PLZ I NEED HELP AS FAST AS POSSIBLE PLZ I WILL MARK THE BEST ANSWER AS BRAINLYIEST The most dangerous time to be on
MatroZZZ [7]

I think the answer is B) 1 a.m. and 5 a.m. I very sorry if it is wrong plz let me know if it is wrong bro thankx

8 0
3 years ago
Read 2 more answers
What are the optical properties of steel
dezoksy [38]

Answer:

A selective surface with large absorption for solar radiation and high reflectance for thermal infrared radiation was produced by use of surface oxidation of stainless steel. The surfaces were studied for use with concentrated light in a solar power plant at temperatures of 400°C and higher.

In order to investigate the relation between surface treatment and optical properties, stainless steels (AISI 304 and 430) which were submitted to different chemical and mechanical surface treatments, were used. To increase the spectral selectivity, these surfaces were treated in air and in vacuum at different temperatures and times. The optical properties of these films were investigated. Visual and infrared spectral absorptances were measured at room temperature. The thermal hemispherical emittance and absorptance were obtained by a calorimetric method at 200°C. It was noticed that these chemically and mechanically treated stainless steel surfaces have good spectral properties without further oxidations. This is very important for high temperature uses. The best values are found for samples 7 and 8 under vacuum and air. These two samples with mechanically ground surfaces retained their selectivity and specularity after several hours oxidation. One can conclude that the surface ground treatment confers good selectivity on the steel surfaces for use in concentrating solar collectors with a working temperature of 500°C.

Sample surfaces were subjected to long temperature ageing tests in order to gain some idea of the thermal stability of the surfaces. The results promise better-performing surface and the production of durable selective finishes at, possibly, lower cost than competing processes.

Explanation:

3 0
3 years ago
The scale of the blueprint tells us the<br> of drawing to real space?
klasskru [66]

Answer:

yes

Explanation:

blueprint of the construction is a prediction of project its is slightly auto cad

8 0
3 years ago
What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has a total equivalent leng
skad [1K]

Answer:

0.067wc

Explanation:

The formula is actual static pressure loss = (total equivalent divided by 100) multiplied by rate of friction

We substitute values

actual static pressure = 0.1

Total equivalent length = 150 ft

0.1 = (150ft/100) multiplied by Rate of friction

Friction rate at 100ft = 0.067

So we have that the required friction needed is 0.067wc

6 0
3 years ago
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