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3 years ago

Asperities are microscopic peaks and valleys on

Engineering

2 answers:

Rudiy273 years ago

8 0

Answer:

On a surface.

Explanation:

Asperities are microscopic peaks and valleys on a surface. This simply means that, when asperities contacts, surfaces also make contact. Asperities are the roughness or unevenness of a surface coming in contact during wear or friction.

Dovator [93]3 years ago

6 0

Answer:

Surfaces presumed to be smooth

Explanation:

Asperity is an area that exhibit surface unevenness or roughness which consists of surface peaks and valleys that are present in microscopic level on even very smooth. In lubrication, as the major part of the lubricated surfaces are kept apart by the lubricating layer, asperities are able to come in contact with other asperities .

Surfaces, presumed to be smooth actually contains peaks and valleys, observable on a microscopic scale, known as asperities.

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What are the general principles of DFA? What are the steps to minimize the number of parts for an assembly?

BigorU [14]

Answer Explanation : The general principles for design for assembly (DFA) are,

MINIMIZE NUMBER OF COMPONENT
USE STANDARD COMMERCIALLY AVAILABLE COMPONENTS
USE COMMON PARTS ACROSS PRODUCT LINES
DESIGN FOR EASE OF PART FABRICATION
DESIGN PARTS WITH TOLERANCE THAT ARE WITHIN PROCESS CAPABILITY
MINIMIZE USE OF FLEXIBLE COMPONENT
DESIGN FOR EASE OF ASSEMBLY
USE MODULAR DESIGN
REDUCE ADJUSTMENT REQUIRED

STEPS TO MINIMIZE THE NUMBER OF PARTS

USE OF INCORPORATE HINGS
USE OF INTEGRAL SPRINGS
USE OF SNAP FITS
USE OF GUIDES BEARINGS
USE OF COVERS

6 0

3 years ago

Plane wall of material A with internal heat generation is insulated on one side and bounded by a second wall of material B, whic

viktelen [127]

Sorry❤

Have a nice day ✨

8 0

3 years ago

What is an isochoric process? b) Can heat be exchanged in an isochoric process? c) A 100L container holding an ideal gas at an i

oksano4ka [1.4K]

Answer:

a)A constant volume process is called isochoric process.

b)Yes

c)Work =0

Explanation:

Isochoric process:

A constant volume process is called isochoric process.

In constant volume process work done on the system or work done by the system will remain zero .Because we know that work done give as

work = PΔV

Where P is pressure and ΔV is the change in volume.

For constant volume process ΔV = 0⇒ Work =0

Yes heat transfer can be take place in isochoric process.Because we know that temperature difference leads to transfer of heat.

Given that

Initial P=10 MPa

Final pressure =15 MPa

Volume = 100 L

Here volume of gas is constant so the work work done will be zero.

4 0

4 years ago

An air-standard dual cycle has a compression ratio of 9.1 and displacement of Vd = 2.2 L. At the beginning of compression, p1 =

jok3333 [9.3K]

Answer:

a) T₂ is 701.479 K

T₃ is 1226.05 K

T₄ is 2350.34 K

T₅ is 1260.56 K

b) The net work of the cycle in kJ is 2.28 kJ

c) The power developed is 114.2 kW

d) The thermal efficiency, $\eta _{dual}$ is 53.78%

e) The mean effective pressure is 1038.25 kPa

Explanation:

a) Here we have;

$\frac{T_{2}}{T_{1}}=\left (\frac{v_{1}}{v_{2}} \right )^{\gamma -1} = \left (r \right )^{\gamma -1} = \left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}$

Where:

p₁ = Initial pressure = 95 kPa

p₂ = Final pressure =

T₁ = Initial temperature = 290 K

T₂ = Final temperature

v₁ = Initial volume

v₂ = Final volume

$v_d$ = Displacement volume =

γ = Ratio of specific heats at constant pressure and constant volume cp/cv = 1.4 for air

r = Compression ratio = 9.1

Total heat added = 4.25 kJ

1/4 × Total heat added = $c_v \times (T_3 - T_2)$

3/4 × Total heat added = $c_p \times (T_4 - T_3)$

$c_v$ = Specific heat at constant volume = 0.718×2.821× 10⁻³

$c_p$ = Specific heat at constant pressure = 1.005×2.821× 10⁻³

v₁ - v₂ = 2.2 L

$\left \frac{v_{1}}{v_{2}} \right =r \right = 9.1$

v₁ = v₂·9.1

∴ 9.1·v₂ - v₂ = 2.2 L = 2.2 × 10⁻³ m³

8.1·v₂ = 2.2 × 10⁻³ m³

v₂ = 2.2 × 10⁻³ m³ ÷ 8.1 = 2.72 × 10⁻⁴ m³

v₁ = v₂×9.1 = 2.72 × 10⁻⁴ m³ × 9.1 = 2.47 × 10⁻³ m³

Plugging in the values, we have;

${T_{2}}= T_{1} \times \left (r \right )^{\gamma -1} = 290 \times 9.1^{1.4 - 1} = 701.479 \, K$

From;

$\left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}= \left (r \right )^{\gamma -1}$ we have;

$p_{2} = p_{1}} \times \left (r \right )^{\gamma } = 95 \times \left (9.1 \right )^{1.4} = 2091.13 \ kPa$

1/4×4.25 = $0.718 \times 2.821 \times 10^{-3}\times (T_3 - 701.479)$

∴ T₃ = 1226.05 K

Also;

3/4 × Total heat added = $c_p \times (T_4 - T_3)$ gives;

3/4 × 4.25 = $1.005 \times 2.821 \times 10^{-3} \times (T_4 - 1226.05)$ gives;

T₄ = 2350.34 K

$\frac{T_{4}}{T_{5}}=\left (\frac{v_{5}}{v_{4}} \right )^{\gamma -1} = \left (\frac{r}{\rho } \right )^{\gamma -1}$

$\rho = \frac{T_4}{T_3} = \frac{2350.34}{1226.04} = 1.92$

$T_{5} = \frac{T_{4}}{\left (\frac{r}{\rho } \right )^{\gamma -1}}= \frac{2350.34 }{\left (\frac{9.1}{1.92 } \right )^{1.4-1}} =1260.56 \ K$

b) Heat rejected = $c_v \times (T_5 - T_1)$

$Therefore \ heat \ rejected = 0.718 \times 2.821 \times 10^{-3}\times (1260.56 - 290) = 1.966 kJ$

The net work done = Heat added - Heat rejected

∴ The net work done = 4.25 - 1.966 = 2.28 kJ

The net work of the cycle in kJ = 2.28 kJ

c) Power = Work done per each cycle × Number of cycles completed each second

Where we have 3000 cycles per minute, we have 3000/60 = 50 cycles per second

Hence, the power developed = 2.28 kJ/cycle × 50 cycle/second = 114.2 kW

$Thermal \ efficiency, \, \eta _{dual} = \frac{Work \ done}{Heat \ supplied} = \frac{2.28}{4.25} \times 100 = 53.74 \%$

The thermal efficiency, $\eta _{dual}$ = 53.78%

e) The mean effective pressure, $p_m$ , is found as follows;

$p_m = \frac{W}{v_1 - v_2} =\frac{2.28}{2.2 \times 10^{-3}} = 1038.25 \ kPa$

The mean effective pressure = 1038.25 kPa.

3 0

4 years ago

Your Java program will be reading input from a file name strInput.txt. Each record contains String firstname String lastName Str

stiks02 [169]

Answer:

The program requires that you have the specified input files and it reads from each file at a time and processes salary in digits, states the city, state and bonus with respective first and last name as requested in the question. Note that you must have access to the mentioned output files for the program to work properly. Below is the java version of the program.

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

class Driver

{

public static void main(String[] args) throws FileNotFoundException

{

Scanner sc = new Scanner(new File("strInput.txt"));

PrintWriter pd = new PrintWriter(new File("strOutputD"));

PrintWriter prf = new PrintWriter(new File("strOutputRF"));

String firstname = "", lastname = "", strSalary = "", status = "", cityState = "", city = "", state = "";

double salary = 0, bonus = 0;

int incorrectRecords = 0;

int dRecords = 0;

int fRecords = 0;

while(sc.hasNextLine())

{

firstname = sc.next();

lastname = sc.next();

strSalary = sc.next();

status = sc.next();

cityState = sc.next();

if(!status.equals("D") && !status.equals("F"))

{

System.out.println("Records is neither D nor F. Skipping this...");

incorrectRecords++;

continue;

}

else if(status.equals("D") || status.equals("F"))

{

char c = ' ';

int i = 0;

for(i=0; i<strSalary.length() && c != '.'; i++)

{

c = strSalary.charAt(i);

if(!Character.isDigit(c))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

if(c == '.')

{

if(i+1 == strSalary.length()-1)

{

if(!Character.isDigit(strSalary.charAt(i)))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

if(!Character.isDigit(strSalary.charAt(i+1)))

{

System.out.println("Char at position " + (i+1+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

else

{

System.out.println("Period is in the wrong position. Expected at " + (strSalary.length()-3) + " but found at " + (i+1));

continue;

}

city = cityState.split(",")[0];

state = cityState.split(",")[1];

salary = Double.parseDouble(strSalary);

if(status.equals("D"))

{

bonus = salary * 0.125;

dRecords++;

pd.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

else

{

bonus = salary * 0.18;

fRecords++;

prf.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

System.out.println("No of D records : " + dRecords);

System.out.println("No of F records : " + fRecords);

System.out.println("No of incorrect records : " + incorrectRecords);

}

6 0

3 years ago