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stira [4]
3 years ago
6

High strength steels are being used to reduce weight on cars. Explain why using a high strength steel would allow you to reduce

the weight of a component Pick a high strength steel and heat treatment for that steel and provide the steel's yield strength and elongation (or fracture strength) Describe at least two concerns you would have when replacing regular steel with high strength steel.
Engineering
1 answer:
gulaghasi [49]3 years ago
4 0

Engaging the frequently tough requirements of vehicle safety, weight reduction for combustible economy, and manufacturability has influenced the steel industry to create a unique variety of 'super steels' for the automobiles of the future.

<h3><u>Explanation</u>:</h3>

• That steel, though, is far more advanced than the materials of just a few years ago.

• At the forefront of these is Advanced High Strength Steel, AHSS, developed by World Auto Steel’s member companies, which is demonstrating to be something of a vision in automobile production.  

• The standard engineering trade-off in steel preference involves considering the need for ultimate strength against flexibility and work-ability – stronger steels tend to be stiffer and less ductile, making them more difficult to develop into cars and more laborious to weld.

• AHSS can retain greatest of the ductility and work-ability of lower grades of steel, while offering much greater strength.

• Where a typical mild steel might have a tensile strength of 300MPa, AHSS can exceed 1500MPa while retaining a highest elongation of 25%, compared to about 40% for mild steel. The intrigue is in the micro-structure, containing a martensite, bainite, austenite phase rather than ferrite, pearlite, or cementite.

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- What will happen if high voltage from the HV battery or motor-generator is shorting to frame ground?
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3 years ago
A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p
Stella [2.4K]

Answer:

Cut-off ratio\dfrac{V_3}{V_2}=6.15

Cxpansion ratio\dfrac{V_4}{V_3}=3.25

The exhaust temperatureT_4=1997.5R

Explanation:

Compression ratio CR(r)=20

\dfrac{V_1}{V_2}=20

P_2=P_3=920 psia

T_1=520 R ,T_{max}=T_3,T_3=3200 R

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{520}=20^{1.4 -1}

T_2=1723.28R

\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}

\dfrac{V_3}{V_2}=\dfrac{3200}{520}

So cut-off ratio\dfrac{V_3}{V_2}=6.15

\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}

Now putting the values in above equation

\dfrac20=\dfrac{V_4}{V_3}\times 6.15

\dfrac{V_4}{V_3}=3.25

So expansion ratio\dfrac{V_4}{V_3}=3.25.

\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}

\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}

T_4=1997.5R

So the exhaust temperatureT_4=1997.5R

3 0
4 years ago
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