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3 years ago

7

Nguyên lý hoạt động của kim phun

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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3 years ago

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.11 m2 and whose thickn

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Answer:

Explanation:

Given that,

The area of glass $A_g$ = $0.11m^2$

The thickness of the glass $t_g=4mm=4\times10^-^3m$

The area of the styrofoam $A_s=11m^2$

The thickness of the styrofoam $t_s=0.20m$

The thermal conductivity of the glass $k_g=0.80J(s.m.C^o)$

The thermal conductivity of the styrofoam $k_s=0.010J(s.m.C^o)$

Inside and outside temperature difference is ΔT

The heat loss due to conduction in the window is

$Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j$

The heat loss due to conduction in the wall is

$Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j$

The net heat loss of the wall and the window is

$Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j$

The percentage of heat lost by the window is

$=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%$

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4 years ago

The ventilating fan of the bathroom has a volumetric flow rate of 1,901 L/s. If the air density is 1.23 kg/m3, calculate the mas

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Answer:

$\dot m_{air} = 8417.628\,\frac{kg}{h}$

Explanation:

Mass flow rate of air is:

$\dot m _{air} = \rho_{air}\cdot \dot V_{fan}$

$\dot m_{air} = (1.23\,\frac{kg}{m^{3}})\cdot (1901\,\frac{L}{s} )\cdot (\frac{1\,m^{3}}{1000\,L} )\cdot (\frac{3600\,s}{1\,h} )$

$\dot m_{air} = 8417.628\,\frac{kg}{h}$

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