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3 years ago

Describe the process that you would use to hot forge an automotive connecting rod, indicating why each of the steps is used.

Engineering

1 answer:

Verizon [17]3 years ago

8 0

Answer:

Hot forging is a process which is carried at a temperature that is higher than the recrystalization temperature.

Explanation:

A connecting rod is used in a reciprocating engine which links the piston to the crankshaft. Connecting rods are made of steel which are hot forged.

The various steps that are used to hot forged a connecting rod are :

1. Rods are made to cut in the required size from the billet by billet shearing machine or saw band.

2. Heating of the billets in the furnace upto its recrystalization temperature.

3. Placing the billets in both upper and lower dies and doing the forging operation.

4. Rolling forging : it is important for the quality of the forged component.

5. Finishing and trimming : finishing is done to improve the surface quality and provide a smooth finish.

6. Inspection : Visual inspection is done for any defects.

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Write the formula for calculating mechanical advantage for Levers, pulleys and wheels and axles.

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2 years ago

Determine the initial void ratio, the relative density and the unit weight (in pounds per cubic foot) of the specimens for each

Elina [12.6K]

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Please note that your question is incomplete so I gave you a general overview to help you better understand the concept

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brainly.com/question/15220801

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3 years ago

Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. D

Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ F$

Change in temperature, $\Delta T = T_2-T_1$

$\Delta T = -2.43-87.745$

$\Delta T=-90.175^\circ F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

$=11.401+0.32198(82.898)$

$=38.09297 \ Btu/lbm$

Change in internal energy

$\Delta u= u_2-u_1$

= 38.09297-40.5485

= -2.4556

5 0

3 years ago

I am making composites of silicone rubber and copper particles; by mixing thermally conductive particles into the thermally insu

Gelneren [198K]

Answer:

The composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

Explanation:

Let us assume that the total mass of composite is 100 lbm So as per the given conditions

15 lbm is copper and 85 lbm is rubber.
Density of rubber is 70 lbm/ft3
Specific gravity of Copper is 9

So As per the formula of specific gravity

$S_{cu}=\frac{\rho_{cu}}{\rho_w}$

here density of water is 62.4 lbm/ft3

Solving for Density of Copper gives

$S_{cu}=\frac{\rho_{cu}}{\rho_w}\\9=\frac{\rho_{cu}}{62.4}\\\rho_{cu}=9 \times 62.4\\\rho_{cu}=561.5 lbm/ft3$

For composition on volume basis, volume of individual components and composite are calculated as

$V_{cu}=\frac{m_{cu}}{\rho_{cu}}\\V_{cu}=\frac{15}{561.5}\\V_{cu}=0.0267 ft^3\\\\V_{r}=\frac{m_{r}}{\rho_{r}}\\V_{r}=\frac{85}{70}\\V_{r}=1.214 ft^3\\\\V_{c}=V_{r}+V_{cu}\\V_{c}=1.214+0.0267 \\V_{c}=1.2409 ft^3$

The composition is given as

$c_{cu}=\frac{V_{cu}}{V_{c}}\\c_{cu}=\frac{0.0267}{1.2409} \times 100 \%\\c_{cu}=2.154 \%\\\\c_{r}=\frac{V_{r}}{V_{c}}\\c_{r}=\frac{1.214}{1.2409} \times 100 \%\\c_{r}=97.83 \%$

So the composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

6 0

3 years ago