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kvasek [131]
3 years ago
15

The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.91 and 0.09, respectively. Is this a hypoe

utectoid or hypereutectoid alloy? Why?
Engineering
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

hypoeutectoid

Explanation:

ferrite:  pure form of iron

cementite: It is iron carbide with 93.3% iron and 6.67% carbon

hypoeutectoid: Eutectoid steel with carbon fraction less than 0.8%

hypereutectoid: Eutectoid Steel with carbon content more than 0.8%

For the mentioned iron-carbide alloy,

% of carbon in iron-carbide alloy= percentage of cementite × percentage of carbon in cementite

% of carbon in iron-carbide alloy= 0.09× 0.0667

                                                     = 0.6%

so the alloy is hypoeutectoid

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A two-phase mixture of water and steam with a quality of 0.63 and T = 300F expands isothermally until only saturated vapor rema
VMariaS [17]

Answer:

Explanation:

Hello!

To solve this problem you must follow the following steps, which are fully registered in the attached image.

1. Draw the complete outline of the problem.

2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties.

3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.

3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.

4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values ​​and finds the work done.

5. draw the pV diagram using the 300F isothermal line

5 0
3 years ago
Hello, I have a question, I would be glad if you can help.
Anastasy [175]

Answer:

  infinite

Explanation:

The tangent of the angle the arm makes with the vertical will be the ratio of the centripetal acceleration to the acceleration due to gravity on the center of mass of the arm. The angle can only be 90° (fully-open arms) if that ratio is infinite.

The speed must be infinite for the arms to be fully open.

3 0
2 years ago
X+3=2<br>x=??<br><br><br><br>No spamming​
PtichkaEL [24]

Answer:

x+3=2

x=2-3꧁

꧁꧁꧂꧂꧂꧂

x=-1

4 0
3 years ago
1. Two technicians are discussing tire rotation. Technician A says that you always follow the tire-rotation procedure outlined i
siniylev [52]

Answer:

don't know

Explanation:

huhuh

8 0
3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
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