All categories

3 years ago

Supercharged engine what it does to the car

Engineering

2 answers:

Makovka662 [10]3 years ago

8 0

They are correct because it the fuel part^

geniusboy [140]3 years ago

3 0

Answer:

A supercharger is an air compressor that increases the pressure or density of air supplied to an internal combustion engine. This gives each intake cycle of the engine more oxygen, letting it burn more fuel and do more work, thus increasing power.

Explanation:

You might be interested in

A gas expands in a piston-cylinder assembly from p1 = 8 bar, V1 = 0.02 m3 to p2 = 2 bar. The relation between pressure and volum

Charra [1.4K]

Answer:

The heat transfer is 29.75 kJ

Explanation:

The process is a polytropic expansion process

General polytropic expansion process is given by PV^n = constant

Comparing PV^n = constant with PV^1.2 = constant

n = 1.2

(V2/V1)^n = P1/P2

(V2/0.02)^1.2 = 8/2

V2/0.02 = 4^(1/1.2)

V2 = 0.02 × 3.2 = 0.064 m^3

W = (P2V2 - P1V1)/1-n

P1 = 8 bar = 8×100 = 800 kPa

P2 = 2 bar = 2×100 = 200 kPa

V1 = 0.02 m^3

V2 = 0.064 m^3

1 - n = 1 - 1.2 = -0.2

W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ

∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ

Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ

7 0

3 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5

IgorC [24]

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

$\epsilon = \deltaL /L$ (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

$\sigma = E/ \epsilon$ (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

$L= \delta L/ \epsilon$

$\epsilon =\sigma /E$

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

$\sigma=F/A$

$\sigma=\frac{F} {\pi*d^2/4}$

$\sigma=\frac{2170 N}{\pi*4.5 mm^2/4}$

$\sigma= 136000000 Pa= 136 Mpa$

Then de specific defotmation:

$\epsilon =136 MPa / 108 GPa = 1.26*10^{-3}$

Finally the maximum specimen lenght for a elongation 0f 0.45 mm is:

$L= 0.45 mm/ 1.26*10^{-3} = 358 mm = 0.358 m$

4 0

3 years ago

One of the arc welding process is most often used today for welding metal thicker than blank gauge

Otrada [13]

Answer:

I wish I know what this was sorry

3 0

3 years ago

Read 2 more answers

5. Create a short poem or haiku (a poem with 5 syllables in the first line, 7 syllables in the

Zanzabum

Answer:

I should write a poem,

a poem about problem,

that i'm now trying to solve,

I dont know what to involve

Explanation:

Its not haiku, cause I am not natural Englishman and I didnt want to risk any mistakes or so. It shows the problem of writing a poem about a problem. btw Im not kidding if u thought so

6 0

3 years ago

Chaplets are used to support a sand core inside a sand mold cavity. The design of the

ziro4ka [17]

Answer:

a) 2 chaplets

b) 9 chaplets

Explanation:

Before we can determine the minimum number of chaplets that should be placed beneath and above the core, we must know the mass of the sand used to make the surface of the mold cavity as well as the mass of the steel metal poured inside the mold.

Density is defined as the ratio of mass of a substance to its density.

Density = Mass/Volume

For the STEEL:

Density of steel = 7.82g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of steel = 7.82×5450

Mass of steel = 42619g

Mass of steel in kg = 42.619kg

For the SAND:

Density of sand = 1.6g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of sand = 1.6×5450

Mass of sand = 8720g

Mass of sand in kg = 8.72kg

a) Since the the chaplet support the sand from beneath the core, and each chaplet weighs 45N, we need to know the amount of force possessed by the sand.

Since the mass of the sand is 87.2kg

Weight = mass × acceleration due to gravity

Weight = 8.72×9.81

Weight of sand used to mold the core = 85.54N

Since 1 chaplet weighs 45N

This means that (85.54N/45N) i.e 1.9 which is approximately 2 chaplets must be placed beneath the core to sustain it before the steel metal is poured.

b) Since the metal poured in the core is steel, this means that the chaplet placed above the core must be able to withstand the strength of the steel.

Weight of steel = mass of steel × acceleration due to gravity

Weight of steel = 42.619×9.81

Weight of steel = 418.09N

Since one chaplet weigh 45N, the amount of chaplets that must be placed above the core is;

418.09/45

= 9.3

Therefore the minimum number of chaplets that should be placed above the core after pouring steel metal is 9chaplets.

4 0

3 years ago