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mojhsa [17]
2 years ago
7

A sample of cast iron with .35 wt%C is slow cooled from 1500C to room temperature. What is the fraction of proeutectoid Cementit

e phase in this material?
Engineering
1 answer:
Marizza181 [45]2 years ago
6 0

Answer:

So the fraction of proeutectoid cementite is 44.3%

Explanation:

Given that

cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.

We know that

Fraction of  proeutectoid cementite phase gievn as

{w_p}'=\dfrac{{C_o}'-0.022}{0.74}

Now by putting the values

{w_p}'=\dfrac{0.35-0.022}{0.74}

{w_p}'=0.443

So the fraction of proeutectoid cementite is 44.3%

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1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by
zvonat [6]

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299.36 feet

Explanation:

To \ find  \   the  \ distance \  of  \ the  \ ball \  from  \ the \ home  \ plate.  \\ \\ From  \ the  \ given  \ information:

Height \ h = 4 \ ft

Initial \ speed \ V_o = 98 \ ft/s ec

The  \ angle \  \theta = 45^0

Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s

U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}

U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}

So;

S_y = u_y t - \dfrac{1}{2}gt^2

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By solving:

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Thus;

horizontal \ distance = U_x t

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5 0
2 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
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Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

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a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

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w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

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