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seraphim [82]
3 years ago
11

Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon al

loy that initially contains 0.10 wt% C. The surface concentration is to be maintained at 0.90 wt% C, and the treatment is to be conducted at 1100°C. Use the diffusion data for γ-Fe.
Engineering
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

idk

Explanation:

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he is the chief officer of crime invistegating department.

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ASAP PLEASEEEE HELP : Eccentricity, Inclination, True anomaly, Argument of perigee, Right ascension or the ascending node, Semi-
natita [175]

Answer:

d

Explanation:

Eternity because it could be culprited by the True Anomaly and the Argument of perigee

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3 years ago
A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identi
vladimir2022 [97]

Answer: the thermal conductivity of the sample is 22.4 W/m . °C

Explanation:

We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.

ASSUMPTIONS

1. Steady operating conditions exist since the temperature readings do not change with time.

2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.

3. The apparatus possess thermal symmetry

ANALYSIS

The electrical power consumed by resistance heater and converted to heat is:

Wₐ = V<em>I</em> = ( 110 V ) ( 0.4 A ) = 44 W

Q = 1/2Wₐ = 1/2 ( 44 A )

Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so

A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²

Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be

Q = kA ( ΔT/L ) → k = QL / AΔT

k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )

k = 22.4 W/m . °C

3 0
3 years ago
Lawn maintenance is an alternative energy source<br> -true<br> -false
Reika [66]

Answer:

false

Explanation:

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How would you expect an increase in the austenite grain size to affect the hardenability of a steel alloy? Why?
seraphim [82]

Answer:

The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.

3 0
3 years ago
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