WILL BRAINLIEST IF CORRECT!!!!! Some one help ASAP.

Select the properties and typical applications for the high carbon steels.

yanalaym [24]

Answer:

Option-(A)

Explanation:

Typical applications for the high carbon steels includes the following;

It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.

3 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio

diamong [38]

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

$x^2$ / Dt = constant

where D is constant

then

$x^2$ / t = constant

$x^2_1$ / t₁ = $x^2_2$ / t₂

$x^2_1$ t₂ = t₁ $x^2_2$

t₂ = t₁ $x^2_2$ / $x^2_1$

t₂ = ( $x^2_2$ / $x^2_1$ )t₁

t₂ = $($ $x_2$ / $x_1$ $)^2$ × t₁

so we substitute

t₂ = $($ 0.0049 / 0.0018 $)^2$ × 11.3 hrs

t₂ = 7.41 × 11.3 hrs

t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

8 0

2 years ago

Determine the work input and entropy generation during the compression of steam from 100 kPa to 1 MPa in (a) an adiabatic pump a

Goshia [24]

Answer:

See attachment below

Explanation:

6 0

3 years ago

(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary e

aksik [14]

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

5 0

3 years ago