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3 years ago

9

A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the

water speed doubles and the cross-sectional area of the pipe triples, what happens to the volume flow rate of the water passing through it?

1 answer:

Arlecino [84]3 years ago

8 0

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

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ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in

Mama L [17]

Answer:

The bending stress is 502.22 MPa

Explanation:

The diameter of the pinion is equal to:

$d_{p} =mN_{p}$

Where

m = module = 5

Np = number of teeth of pinion = 26

$d_{p} =5*26=130mm$ = 0.13 m

The pitch line velocity is equal to:

$V_{t} =\frac{d_{p}*2*\pi *w_{p} }{120}$

Where

wp = speed of the pinion = 1800 rpm

$V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s$

The factor B is equal to:

$B=\frac{(12-Q_{v})^{2/3} }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396$

The factor A is equal to:

A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82

The dynamic factor is:

$K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832$

The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables

JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

The diametrical pitch is equal to:

$P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}$

The bending stress is equal to:

$\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}} \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa$

4 0

3 years ago

The thrust F of a screw propeller is known to depend upon the diameter d,speed of advance \nu ,fluid density p, revolution per s

musickatia [10]

Answer:

<em>screw thrust = ML</em> $T^{-2}$ <em> </em>

Explanation:

thrust of a screw propeller is given by the equation = p $V^{2}$ $D^{2}$ x $\frac{ND}{V}$ Re

where,

D is diameter

V is the fluid velocity

p is the fluid density

N is the angular speed of the screw in revolution per second

Re is the Reynolds number which is equal to puD/μ

where p is the fluid density

u is the fluid velocity, and

μ is the fluid viscosity = kg/m.s = M $L^{-1}$ $T^{-1}$

<em>Reynolds number is dimensionless so it cancels out</em>

The dimensions of the variables are shown below in MLT

diameter is m = L

speed is in m/s = L $T^{-1}$

fluid density is in kg/ $m^{3}$ = M $L^{-3}$

N is in rad/s = L $L^{-1}$ $T^{-1}$ =

If we substitute these dimensions in their respective places in the equation, we get

thrust = M $L^{-3}$ $(LT^{-1}) ^{2}$ $L^{2}$ $\frac{T^{-1} L}{LT^{-1} }$

= M $L^{-3}$ $L^{2}$ $T^{-2}$

<em>screw thrust = ML</em> $T^{-2}$ <em> </em>

This is the dimension for a force which indicates that thrust is a type of force

6 0

3 years ago

In the first-order process,a blue dye reacts to form a purple dye. The amount of blue dye at the end of 1 hr is 480 g and the en

Tasya [4]

Answer:

The answer is 960 kg

Explanation:

Solution

Given that:

Assume the initial dye concentration as A₀

We write the expression for the dye concentration for one hour as follows:

ln (C₁) = ln (A₀) -kt

Here

C₁ = is the concentration at 1 hour

t =time

Now

Substitute 480 g for C₁ and 1 hour for t

ln (480) = ln (A₀) -k(1) ------- (1)

6.173786 = ln (A₀) -k

Now

We write the expression for the dye concentration for three hours as follows:

ln (C₃) = ln (A₀) -k

Here

C₃ = is the concentration at 3 hour

t =time

Thus

Substitute 480 g for C₃ and 3 hour for t

ln (120) = ln (A₀) -k(3) ------- (2)

4.787492 = ln (A₀) -3k

Solve for the equation 1 and 2

k =0.693

Now

Calculate the amount of blue present initially using the expression:

Substitute 0.693 for k in equation (2)

4.787492 = ln (A₀) -3 (0.693)

ln (A₀) =6.866492

A₀ =e^6.866492

= 960 kg

Therefore, the amount of the blue dye present from the beginning is 960 kg

6 0

4 years ago

Consider a junction that connects three pipes A, B and C. What can we say about the mass flow rates in each pipe for steady flow

Elis [28]

Answer:

The statement regarding the mass rate of flow is mathematically represented as follows $\Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}$

Explanation:

A junction of 3 pipes with indicated mass rates of flow is indicated in the attached figure

As a basic sense of intuition we know that the mass of the water that is in the pipe junction at any instant of time is conserved as the junction does not accumulate any mass.

The above statement can be mathematically written as

$Mass_{Junction}=Constant\\\\\Rightarrow Mass_{in}=Mass_{out}$

this is known as equation of conservation of mass / Equation of continuity.

Now we know that in a time 't' the volume that enter's the Junction 'O' is

1) From pipe 1 = $V_{1}=Q_{1}\times t$

1) From pipe 2 = $V_{2}=Q_{2}\times t$

Mass leaving the junction 'O' in the same time equals

From pipe 3 = $V_{3}=Q_{3}\times t$

From the basic relation of density, volume and mass we have

$\rho =\frac{mass}{Volume}$

Using the above relations in our basic equation of continuity we obtain

$\rho \times V_{3}=\rho \times V_{1}+\rho \times V_{2}\\\\Q_{3}\times t=Q_{1}\times t+Q_{2}\times t\\\\\Rightarrow Q_{3}=Q_{1}+Q_{2}$

Thus the mass flow rate equation becomes $\Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}$

6 0

4 years ago

Citrus2011 [14]

Answer:

Explanation:

Code:

import java.io.File;

import java.io.FileWriter;

import java.io.IOException;

import java.util.Scanner;

public class Knapsack {

public static void knapsack(int wk[], int pr[], int W, String ofile) throws IOException

{

int i, w;

int[][] Ksack = new int[wk.length + 1][W + 1];

for (i = 0; i <= wk.length; i++) {

for (w = 0; w <= W; w++) {

if (i == 0 || w == 0)

Ksack[i][w] = 0;

else if (wk[i - 1] <= w)

Ksack[i][w] = Math.max(pr[i - 1] + Ksack[i - 1][w - wk[i - 1]], Ksack[i - 1][w]);

else

Ksack[i][w] = Ksack[i - 1][w];

}

}

int maxProfit = Ksack[wk.length][W];

int tempProfit = maxProfit;

int count = 0;

w = W;

int[] projectIncluded = new int[1000];

for (i = wk.length; i > 0 && tempProfit > 0; i--) {

if (tempProfit == Ksack[i - 1][w])

continue;

else {

projectIncluded[count++] = i-1;

tempProfit = tempProfit - pr[i - 1];

w = w - wk[i - 1];

}

FileWriter f =new FileWriter("C:\\Users\\gshubhita\\Desktop\\"+ ofile);

f.write("Number of projects available: "+ wk.length+ "\r\n");

f.write("Available employee work weeks: "+ W + "\r\n");

f.write("Number of projects chosen: "+ count + "\r\n");

f.write("Total profit: "+ maxProfit + "\r\n");

for (int j = 0; j < count; j++)

f.write("\nProject"+ projectIncluded[j] +" " +wk[projectIncluded[j]]+ " "+ pr[projectIncluded[j]] + "\r\n");

f.close();

}

}

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

{

Scanner sc = new Scanner(System.in);

System.out.print("Enter the number of available employee work weeks: ");

int avbWeeks = sc.nextInt();

System.out.print("Enter the name of input file: ");

String inputFile = sc.next();

System.out.print("Enter the name of output file: ");

String outputFile = sc.next();

System.out.print("Number of projects = ");

int projects = sc.nextInt();

int[] workWeeks = new int[projects];

int[] profit = new int[projects];

File file = new File("C:\\Users\\gshubhita\\Desktop\\" + inputFile);

Scanner fl = new Scanner(file);

int count = 0;

while (fl.hasNextLine()){

String line = fl.nextLine();

String[] x = line.split(" ");

workWeeks[count] = Integer.parseInt(x[1]);

profit[count] = Integer.parseInt(x[2]);

count++;

}

knapsack(workWeeks, profit, avbWeeks, outputFile);

}

}

Console Output:

Enter the number of available employee work weeks: 10

Enter the name of input file: input.txt

Enter the name of output file: output.txt

Number of projects = 4

Output.txt:

Number of projects available: 4

Available employee work weeks: 10

Number of projects chosen: 2

Total profit: 46

Project2 4 16

Project0 6 30

8 0

3 years ago