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Rufina [12.5K]
1 year ago
7

It is desired to develop a strong Aluminum-based alloy with a shear strength on the order of G/100 by precipitation hardening, w

here G is the shear modulus. Calculate the necessary precipitate spacing, and estimate the required percentage of the precipitate phase. Assume that the precipitate phase consists of spherical particles of radius of R=40 nm with centers uniformly distributed in simple cubic lattice. Does it matter whether the precipitates strain the matrix locally around the interface or not?
Engineering
1 answer:
borishaifa [10]1 year ago
5 0

Answer:

a) L = 50.01 nm

b) 99.98%

Explanation:

<u>Determine the necessary precipitate spacing  and estimate the required percentage of precipitate phase </u>

Given that the shear strength = G/100 by precipitation hardening

G = shear modulus

Radius of particles = 40 nm

attached below is the detailed solution

a) precipitate spacing ( L ) = √( 2π / f )  * R₁

                                       = √ ( 2π / 0.00719956 ) * 1.693 nm

hence L = 50.01 nm

b) Determine The percentage of precipitate phase that is required

( precipitation strength  /  peak strength ) * 100

= ( 18.4583 / 18.4604 ) * 100

= 99.98%

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In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly from zero at y = 0 to 32 m/s at y =
monitta

Answer:

Ψ = 10(y^2) + c

<em><u>y = 1.067m</u></em>

Explanation:

since the flow is one dimensional in positive X direction, the only velocity component is in X, which is denoted by u

while u is a function of y

we find the u in terms of y; u varies linearly wih y

we use similiraty to find the relation

32/1.6 =<em>u/y</em>

<em><u>u = 20y</u></em>

<em><u>Ψ = ∫20ydy</u></em>

<em><u>Ψ = 10(y^2) + c</u></em>

<em><u>(b)</u></em>

<em><u>the flow is half below y = 1.6*(2/3)=1.067 m</u></em>

<em><u>this is because at two third of the height of a triangle lies the centroid of triangle. since the velocity profile forms a right angled triangle , its height is 1.6 m . the flow is halved at y = 1.067m</u></em>

3 0
2 years ago
Use the predicate specifications(x, y): x beats yF (x): x is an (American) football teamQ(x, y): x is quarterback of yL(x,y): x
attashe74 [19]

Answer:

a) ∀y∃x(Q(x, y))

b) (B(Jayhawks, W ildcats)→¬∀y(L(Jayhawks, y)))

c) ∃x(B(Wildcats, x) ∧ B(x, Jayhawks))

Explanation:

a) The statement can be rewritten as "For all football teams, there exists a quarterback" which is written in logical symbols.

b) The statement is an implication and thus have a premise and a conclusion. The premise states "Jayhawks beat the Wildcats" which is translated using B(x, y). The conclusion can be rewritten as "It is not the case that Jayhawks lose to all football teams".

c) The statement is a simple conjunction which can be written as "There exists a team x such that the Wildcats beats x and x beats Jayhawks"

7 0
2 years ago
Six years ago, an 80-kW diesel electric set cost $160,000. The cost index for this class of equipment six years ago was 187 and
exis [7]

Answer:

new boiler total cost = $229706.825

new boiler total cost = $127512

Explanation:

given data

power p1 = 80 kW

cost C = $160000

cost index CI 1 = 187

cost index CI 2= 194

cost capacity factor f = 0.6

power p2 = 120 kW

current cost = $18000

to find out

total cost and cost of 40 kW

solution

we consider here CN cost for new boiler and CO cost for old boiler

and x is capacity of new boiler

first we find old boiler current cost that is

current cost CO = C × \frac{CI 1 }{CI 2 }   .............1

put here value

current cost = 160000 × \frac{194 }{187 }

new current cost = $165989.304

and

use here power sizing technique for 124 kW

CN/CO = (\frac{p2}{p1} )^{f}    ...............2

put here value and find CN

CN/CO = (\frac{p2}{p1} )^{f}  

CN / 165989.304 = (\frac{120}{80} )^{0.6}  

CN = 211706.825

so new cost = $211706.825

so

total cost for new boiler is

total cost = new cost + current cost

total cost = 211706.825 + 18000

new boiler total cost = $229706.825

and

for 40 kW new cost will be

use equation 2

CN/CO = (\frac{p2}{p1} )^{f}

CN / 165989.304 = (\frac{40}{80} )^{0.6}  

CN = 109512

so new cost is $109512

so

total cost for new boiler is

total cost = new cost + current cost

total cost = 109512 + 18000

new boiler total cost = $127512

7 0
2 years ago
A railcar with an overall mass of 78,000 kg traveling with a speed vi is approaching a barrier equipped with a bumper consisting
sergij07 [2.7K]

Answer:

v₀ = 2,562 m / s  = 9.2 km/h

Explanation:

To solve this problem let's use Newton's second law

              F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v

              F dx = m v dv

We replace and integrate

            -β ∫ x³ dx = m ∫ v dv

            β x⁴/ 4 = m v² / 2

We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max

        -β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)

         x_max⁴ = 2 m /β   v₀²

         

Let's look for the speed that the train can have for maximum compression

         x_max = 20 cm = 0.20 m

         

         v₀ =√(β/2m)   x_max²

Let's calculate

          v₀ = √(640 106/2 7.8 104)    0.20²

          v₀ = 64.05  0.04

          v₀ = 2,562 m / s

          v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)

          v₀ = 9.2 km / h

5 0
2 years ago
Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is
GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

   public static void main(String[] args) {

       final int MAXCHEESE = 10;

       String[] names = new String[MAXCHEESE];

       double[] prices = new double[MAXCHEESE];

       double[] amounts = new double[MAXCHEESE];

       // Three Special Cheeses

       names[0] = "Humboldt Fog";

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       names[1] = "Red Hawk";

       prices[1] = 40.50;

       names[2] = "Teleme";

       prices[2] = 17.25;

       System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

       System.out.println(names[0] + ": $" + prices[0] + " per pound");

       System.out.println(names[1] + ": $" + prices[1] + " per pound");

       System.out.println(names[2] + ": $" + prices[2] + " per pound");

       Random ranGen = new Random(100);

       // error at initialising i

       // i should be from 0 to MAXCHEESE value

       for (int i = 0; i < MAXCHEESE; i++) {

           names[i] = "Cheese Type " + (char) ('A' + i);

           prices[i] = ranGen.nextInt(1000) / 100.0;

           amounts[i] = 0;

           System.out.println(names[i] + ": $" + prices[i] + " per pound");

       }        

   }

}

7 0
2 years ago
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