What else will change, if you change the point of view

Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

$V_{gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{fe}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

$m_{fe}$ , $m_{gr}$ - Masses of the ferrite and graphite phases, measured in grams.

$\rho_{fe}, \rho_{gr}$ - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

$\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%$

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is:

$\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%$

$\%V_{gr} = 10.197\,\%V$

The volume percentage of graphite is 10.197 per cent.

5 0

3 years ago

Define Plastic vs elastic deformation.

Snowcat [4.5K]

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

3 0

3 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz

Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s and ρ= 850 kg/m³

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ = ρv

=850 x 0.00062

= 0.527 kg/m·s

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

$Q = \dfrac{\Delta P \pi D^4}{128 \mu L}$

$Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}$

Q = 5.06 x 10⁻⁸ m³/s

4 0

3 years ago

Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the

Musya8 [376]

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$$P(h$

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore, $x_1=368 \ veh/h$

$=\frac{368}{1000} = 0.368$

Given, $t_1=2+x_1$

= 2 + 0.368

= 2.368 min

At user equilibrium, $t_2=t_1$

∴ $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

= 1368 veh/h

7 0

3 years ago

Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r

9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m

Explanation:

see attached file below

3 0

3 years ago