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bogdanovich [222]
3 years ago
6

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

to 0.35 wt% at a point 2.0 mm from the surface. Estimate the time necessary to achieve the same concentration at a 6.0-mm position for an identical steel and at the same carburizing temperature.
Engineering
1 answer:
Amiraneli [1.4K]3 years ago
5 0

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

so t_2=135\ hour

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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
An orchestra is having a recording done of 2 performances in the same concert hall. The first show is sold out. They struggled t
konstantin123 [22]

Answer:

yes, the recordings sound is same

Explanation:

given data

recording done = 2 performances

1st  show = sold out

2nd show =  lightly attended

to find out

recordings sound the same and why

solution

as per given in

  • 1st show is sold out it mean in this case concert hall is full so that recording sound should be high here
  • 2nd case only few people are attended and struggle for ticket  and orchestra

it mean it sound performance so in both case recording sound will be same

because we do not other all are sitting at front row or they sit as they want

4 0
3 years ago
You are traveling upstream on a river at dusk. You see a buoy with the number 5 and a flashing green light . What should you do?
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Answer:

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this is my brothers account he wants to get 5 brainliest

8 0
3 years ago
What is the difference between science and engineering?
jek_recluse [69]

Answer:

The following table is among some of the discrepancies between some of the different topics.

Explanation:

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  • Science would be a set of information and established frameworks whereas the use of these models and information to construct structures as well as mechanisms becomes engineering.
4 0
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kolezko [41]

Answer & Explanation:

"The force <em>on a cutting tool</em> are 2600N vertically downward" sounds a little unusual, since most of the time, the tool is above the object to be cut in such a way that the force acting "ON the tool" is upwards.  We will accept the statement as it is (downwards).

Since the two forces are acting at right angles to each other, the resultant can be found using Pythagoras theorem, namely

resultant = sqrt(2600^2+2100^2) = 3342 N (approx.)

The angle can be found using the arcTangent function, or

angle = arcTangent(2600/2100) = 51.07 degrees below the horizontal, since the 2600 N force is acting downwards.

3 0
3 years ago
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