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3 years ago

9

If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head

, and power requirements

1 answer:

FinnZ [79.3K]3 years ago

8 0

Answer:

- the capacity of the pump reduces by 35%.

- the head gets reduced by 57%.

the power consumption by the pump is reduced by 72%

Explanation:

the pump capacity is related to the speed as speed is reduces by 35%

so new speed is (100 - 35) = 65% of orginal speed

speed Q ∝ N ⇒ Q1/Q2 = N1/N2

Q2 = (N2/N1)Q1

Q2 = (65/100)Q1

which means that the capacity of the pump is also reduces by 35%.

the head in a pump is related by

H ∝ N² ⇒ H1/H2 = N1²/N2²

H2 = (N2N1)²H1

H2 = (65/100)²H1 = 0.4225H1

so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.

Now The power requirement of a pump is related as

P ∝ N³ ⇒ P1/P2 = N1³/N2³

P2 = (N2/N1)³P1

H2 = (65/100)²P1 = 0.274P1

So the reduction in power is 1 - 0.274 = 0.725 which is 72%

Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.

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3 years ago

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

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Answer:

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Explanation:

given details

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W =256.76 kJ

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$\Delta U = m *c_v*{V_2-V_1}$

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3 years ago

What is the purpose of O-ring and valve seals in a cylinder head?

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2 years ago

Write a program that prompts for a line of text and then transforms the text based on chosen actions. Actions include reversing

nlexa [21]

Answer:

public class TextConverterDemo

{

//Method definition of action1337

public static String action1337(String current)

{

//Replace each L or l with a 1 (numeral one)

current = current.replace('L', '1');

current = current.replace('l', '1');

//Replace each E or e with a 3 (numeral three)

current = current.replace('E', '3');

current = current.replace('e', '3');

//Replace each T or t with a 7 (numeral seven)

current = current.replace('T', '7');

current = current.replace('t', '7');

//Replace each O or o with a 0 (numeral zero)

current = current.replace('O', '0');

current = current.replace('o', '0');

//Replace each S or s with a $ (dollar sign)

current = current.replace('S', '$');

current = current.replace('s', '$');

return current;

}

//Method definition of actionReverse

//This method is used to reverses the order of

//characters in the current string

public static String actionReverse(String current)

{

//Create a StringBuilder's object

StringBuilder originalStr = new StringBuilder();

//Append the original string to the StribgBuilder's object

originalStr.append(current);

//Use reverse method to reverse the original string

originalStr = originalStr.reverse();

//return the string in reversed order

return originalStr.toString();

}

//Method definition of main

public static void main(String[] args)

{

//Declare variables

String input, action;

//Prompt the input message

System.out.println("Welcome to the Text Converter.");

System.out.println("Available Actions:");

System.out.println("\t1337) convert to 1337-speak");

System.out.println("\trev) reverse the string");

System.out.print("Please enter a string: ");

//Create a Scanner class's object

Scanner scn = new Scanner(System.in);

//Read input from the user

input = scn.nextLine();

do

{

/*Based on the action the user chooses, call the appropriate

* action method. If an unrecognized action is entered then

* the message "Unrecognized action." should be shown on a

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* */

System.out.print("Action (1337, rev, quit): ");

action = scn.nextLine();

if (action.equals("1337"))

{

input = action1337(input);

System.out.println(input);

} else if (action.equals("rev"))

{

input = actionReverse(input);

System.out.println(input);

} else if (!action.equals("quit"))

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System.out.println("Unrecognized action.");

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} while (!action.equals("quit"));

System.out.println("See you next time!");

scn.close();

}

}

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3 years ago

Miller Indices:

svetlana [45]

Answer:

A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].

B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.

C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.

Compleated question:

1. Miller Indices:

a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.

b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?

c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.

Explanation:

A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]

2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

<u>In this case, for [1 2 1]:</u>

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=1=1/c_0 \rightarrow c_0=1$

<u>for </u> $[1 2 \overline{4}]$ <u>:</u>

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25$

B) The closest distance between planes with the same Miller indices can be calculated as:

With $\pi:[l m n]$ ,the distance is $d_{lmn}= \displaystyle \frac{a}{\sqrt{l^2+m^2+n^2}}$ with lattice constant a.

<u>In this case, for [1 2 1]:</u>

<u /> $d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a$ <u />

<u>for </u> $[1 2 \overline{4}]$ <u>:</u>

$d_{12\overline{4}}= \displaystyle \frac{a}{\sqrt{1^2+2^2+(-4)^2}}=\frac{a}{\sqrt{21}}=0.22a$

C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:

dir₁=|0 0 1|

dir₂=|0 0.5 1.5|≡|0 1 3|

dir₃=|0 1 1|

dir₄=|0 1.5 0.5|≡|0 3 1|

dir₅=|0 0 1|

5 0

3 years ago