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3 years ago

Sandwich materials typically use a high density core with non-structural cover plates. a)True b)- False

Engineering

1 answer:

Lorico [155]3 years ago

4 0

Answer: False

Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.

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An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere wherein t

weeeeeb [17]

Answer:

the temperature T at which the treatment is carried is out is 1274.24 K

Explanation:

Fick's second Law posits that the rate of change of concentration of diffusing species is directly proportional to the second derivative of the concentration.

Using the expression of the Fick's second Law:

$\mathbf{\frac{C_x-C_o}{C_s-C_o} = 1- erf(\frac{x}{2\sqrt{Dt} })}$

where;

$C_o$ = initial concentration

$C_x$ = the depth of the concentration

$C_s$ = surface concentration

$erf(\frac{x}{2\sqrt{Dt} })}$ = Gaussian error function.

Let variable z be used for the expression of the Gaussian error function. $erf(\frac{x}{2\sqrt{Dt} })}$

Then, from the above equation: replacing $C_x$ with 0.35 ; $C_o$ with 0.2 and $C_s$ with 1.0; we have:

$\mathbf{\frac{0.35-0.2}{1.0-0.2} = 1- erf( z)}$

erf (z) = 0.8125

we obtain the error function value close to 0.8125 from the error function table and we did the interpolation to obtain the exact value of variable corresponding to 0.8125.

The table below shows the tabular form of the error function value close to 0.8125 .

Value for z Value for erf (z)

0.9 0.797

z 0.8125

0.950 0.8209

From above; we can find the value of variable corresponding to the error function 0.8125 .

i.e

$\frac{z-0.9}{0.95-0.9} =\frac{0.8125-0.797}{0.8209-0.797}$

z = 0.932

However, the temperature dependence relation for the diffusion coefficient D can be expressed as:

$z = \frac{x}{\sqrt{Dt} }$

where;

z = 0.932

x = 3.5 mm = 0.0035 m

t = 50 h = 180000 sec

$0.932 = \frac{0.35}{2\sqrt{D*180000} }$

D = $1.958*10^{-11} m^2/s$

Finally, the temperature T at which the treatment is carried is out is calculated as:

$\mathbf{D=D_o \ exp \ (-\frac{Q_d}{RT}) }$

From the table ‘Diffusion data’, we obtain the values of temperature-independent pre exponential and activation energy for diffusion of carbon in FCC Fe.

$D_o = 2.3*10^{-5} \ m^2/s$

$Q_d = 148, 000 \ J/mol$

Replacing all values needed for the above equation; we have:

$1.958*10^{-11}= (2.3*10^{-5})exp(\frac{-148,000}{(8.31)T})$

$8.51*10^{-7}=exp(\frac{-17,810}{T})$

$In(8.51*10^{-7})=(\frac{-17,810}{T})$

-13.977 = -17,810/T

T = -17,810/ - 13.977

T = 1274.24 K

Hence, the temperature T at which the treatment is carried is out is 1274.24 K

5 0

3 years ago

Describe the role of C-S-H in providing strength for cement. Discuss which compounds produce C-S-H and why balancing the amounts

wariber [46]

Answer:

Following are the solution to this question:

Explanation:

Whenever a chemical reaction occurs between water and cement the heat is released, and a $CaOSiO_2H_2O$ (C-S-H gel) gel constructs gel is also recognized as "tobermorite gel."

This one gel acts like a pack of gum and also has a cement quality, that holds its particles intact and therefore contributes to the overall compression mix. An increase in supply explicitly causes the movement in the outcome of power. C3S and C2S are both the compounds of Bouge that produce hydration C-S-H gel.

It mixture must be balanced as $Ca(OH)_2$ with C-S-H gel also is given as a byproduct. It $Ca(OH)_2$ , that cause sudden with sulphate and form $CaSO_4$ , is an unacceptable substance. Sulfate attack or later deterioration of its cement is caused by this $CaSO_4$ .

All C3S and C2S generate various amounts of C-S-H gel so, the required strength can be maintained without compromising on real term durability.

8 0

2 years ago

Describe two characteristics that bridges and skyscrapers have in common.

Elan Coil [88]

Answer:

skyscrapers and bridges both have structural builds accounting for the influence of wind and they both cost big money

Explanation:

7 0

3 years ago

A centrifugal pump is required to pump water to an open water link situated 4 km away from the location of the pump through a pi

11111nata11111 [884]

Answer:

P= 5.5 bar

Explanation:

Given that

L= 4000 m

d= 0.2 m

Friction factor(F) = 0.01

speed V= 2 m/s

Head = 5 m

Head loss due to friction

$h_f=\dfrac{FLV^2}{2gd}$

$h_f=\dfrac{0.01\times 4000\times 2^2}{2\times 9.81\times 0.2}$

$h_f=40.77m$

So the total head(H) = 5 + 40.77 + 10.3 =56.07

Where 10.3 m is the atmospheric head.

We know that

P=ρ g H

So total Pressure

P= 1000 x 9.81 x 56.07 Pa

$P=5.5\times 10^5\ Pa$

P= 5.5 bar

5 0

3 years ago

A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th

Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

Part 2

For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

And we can quantify the decrease using the relative change:

$\% Change = \frac{5s}{55 s} *100 = 9.09\%$ of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

8 0

3 years ago