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Gemiola [76]
4 years ago
15

The following median grain size data were obtained during isothermal liquid phase sintering of an 82W-8Mo-8Ni-2Fe alloy. What is

the time dependence for the grain growth in the system and what is the probable grain-coarsening mechanism?Ignore the initial grain size. Time, min 30 120 480 960 grain size, um 6.0 13.4 21.3 31.6

Engineering
1 answer:
Morgarella [4.7K]4 years ago
8 0

Answer:

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

( d^{2}- d_{0} ^{2} = kt )

Explanation:

The plot attached below shows the time dependence of the growth of grain.

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

the ideal growth follows this principle = d^{2} - d^{2} _{0}  = kt

d = final grain size

d_{0} = initial grain size

k = constant ( temperature dependent )

t = 0

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A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid
dmitriy555 [2]

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

T_{min} = 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, d_c = 0.5 mm

Number of passes necessary for this reduction, n = \frac{\triangle d}{d_c}

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec

minimum time required to reduce the depth of the plate by 20 mm:

T_{min} = number of passes * Time/pass

T_{min} = n * Time/pass

T_{min} = 40 * 40

T_{min} =  1600 = 26 mins 40 secs

3 0
4 years ago
Read 2 more answers
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slava [35]

Answer:

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Explanation:

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4 0
3 years ago
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2. Similar to problem 1, assume your computer system has a 32-bit byte-addressable architecture where addresses and data are eac
andreev551 [17]

Question:

The question is not complete. The question to answer was not added. See below the possible question and the answer.

a. How many blocks are in the cache with this new arrangement?

b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

Answer:

(a) Number of blocks =  512 blocks

(b) Tag is 18

(c)  Total size of the cache = 8388608 bytes

Explanation:

a .

block size = 32 bytes

cache size = 16384 bytes

No.of blocks = 16384 / 32

No.pf blocks = 512 blocks

b.

Total address size = 32 bits

Address bits = Tag + Line index +block offset

Block Size = 32 bytes.

So block size = 25 bytes.

Hence Offset is 5

No . of Cache blocks = 512 blocks = 29 blocks

Hence line offset is 9

We know that Address bits = Tag + Line index +block offset

So , 32 =tag+9+5

tag = 32-(9+5)

So Tag is 18

c.

Data bits = 32 bits

Tag=18 bits

Valid bit is 1 bit

so Total cache size = 25+218+20

                                  = 223

                                  =8388608 bytes

7 0
4 years ago
A viscometer may be created by using a pair of closely fitting concentric cylinders and filling the gap between the cylinder wit
kari74 [83]

Answer:

The viscosity of the fluid is 6.05 Pa-s

Explanation:

The viscosity obtained from a rotating type viscometer is given by the equation

\mu =\frac{3Tt}{2\pi \omega R_o(R_o^3-R_i^3)}

where

T is the torque applied in the viscometer

't' is the thickness of gap

\omega is the angular speed of rotation

R_o,R_i are the outer and the inner raddi respectively

Since it is given that

\omega =100rpm=\frac{100\times 2\pi }{60}=10.47rad/sec

Applying the values in the above relation we get

\mu =\frac{3\times 0.021\times 0.02\times 10^{-3}}{2\pi \times 10.47\times 37.5\times 10^{-3}(37.52^3-37.5^3)\times 10^{-9}}=6.04Pa-s

5 0
3 years ago
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr
seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F =  mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625

Weight transfer is given as follows;

W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

F_R × 1.3 = 1.7 × F_F

F_R + F_F = 18000

F_R + \dfrac{1.3 }{1.7} \times  F_R = 18000

F_R = 18000*17/30 = 10200 N

F_F = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel F_R  = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel F_F =  7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

\mu = \dfrac{F}{R} =  \dfrac{11250}{18000} = 0.625

The least coefficient of friction, μ = 0.625.

3 0
4 years ago
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