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2 years ago

15

Discuss the microstrucral transformation of eutectic reaction, hypo-eutecticand hyper eutectic. Including temperature, reactions

and microstrucral development/phases(draw diagrams)

1 answer:

RSB [31]2 years ago

7 0

Answer:

where do i draw

Explanation:

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3 years ago

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Explanation:

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4 years ago

A thick casting with a thermal diffusivity of 5 x 10-6 m2/s is initially at a uniform temperature of 150oC. One surface of the c

fredd [130]

Answer:

$T_o = 141.81 ^0C$

Explanation:

Given that;

Thermal diffusivity $\alpha = 5 \times 10 ^{-6} m^2/s$

Thermal conductivity $k = 20 \ W/m.K$

Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K

Initial temperature = 150 ° C = (150+273) K = 423 K

Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K

Time = 40 seconds

Length = 20mm = 0.02 m

The objective is to determine the temperature at the surface at a depth of 20 mm after 40 seconds.

$Bi = \dfrac{hL}{k}$

$Bi = \dfrac{20*0.02}{20}$

Bi == 0.02

$\tau = \dfrac{\alpha t}{L^2}$

$\tau= \dfrac{5*10^{-6 }* 40}{0.020^2}$

$\tau = 0.5$

For a wall at 0.2 Bi

$A_1 = 1.0311$

$\lambda _1 = 0.4328$

Therefore;

$\dfrac{T_o - T_{\infty}}{T_i - T_{\infty}}= A_1 e ^{-( \lambda_1^2 \ \tau)$

$\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.438^2 \times 0.5 )$

$\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )$

$\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085$

$\dfrac{T_o - 293 }{130}= 0.937$

$T_o - 293= 0.937 \times 130$

$T_o - 293= 121.81$

$T_o = 121.81+ 293$

$T_o = 414.81 \ K$

$T_o = (414.81 - 273)^0C$

$T_o = 141.81 ^0C$

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4 years ago

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xeze [42]

Explanation:

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3 years ago

Tiny hydrogen bubbles are being used as tracers to visualize a flow. All the bubbles are generated at the origin (x = 0, y = 0).

notka56 [123]

Answer:

for t = 1 : ( x1 , y1 ) = (1,2)

for t = 2 : ( x2, y2 ) = (2,4)

for t = 3 : ( x3, y3 ) = ( 0,-3)

for t = 4 : ( x4, y4 ) = ( 0,-4)

Explanation:

Attached below is the detailed solution and the sketch

we have to make assumptions for the various values of t and integrate accordingly

for t = 1 : ( x1 , y1 ) = (1,2)

for t = 2 : ( x2, y2 ) = (2,4)

for t = 3 : ( x3, y3 ) = ( 0,-3)

for t = 4 : ( x4, y4 ) = ( 0,-4)

Given data :

origin ( x = 0, y = 0 )

u = 1 m/s, v = 2 m/s,

$0 \leq t < 2s$ where u = 0 m/s v = -1 m/s

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3 years ago