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Nadusha1986 [10]
3 years ago
5

Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

ensity is 855 kg/m3 and the viscosity is 2.1 × 10−2 Pa · s. What is the velocity in the pipe?It is desired to maintain the same Reynolds number of 2100 and the same velocity as in part (a) using a second fluid with a density of 925 kg/m3 and a viscosity of 1.5 × 10−2 Pa · s. What pipe diameter should be used?
Engineering
1 answer:
alexdok [17]3 years ago
8 0

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number (Re) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

Re=\frac{\rho v D}{\mu}

where \rhois the density of the fluid, \mu is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm

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The correct answer is 'velocity'of liquid flowing out of an orifice is proportional to the square root of the 'height'  of liquid above the center of the orifice.

Explanation:

Torricelli's theorem states that

v_{exit}=\sqrt{2gh}

where

v_{exit} is the velocity with which the fluid leaves orifice

h is the head under which the flow occurs.

Thus we can compare the given options to arrive at the correct answer

Velocity is proportional to square root of head under which the flow occurs.

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3 years ago
A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?
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Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

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v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

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Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

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Q5
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The C++ code that would draw all the iterations in the selection sort process on the array is given below:

<h3>C++ Code</h3>

#include <stdio.h>

#include <stdlib.h>

int main() {

   int i, temp1, temp2;

   int string2[16] = { 0, 4, 2, 5, 1, 5, 6, 2, 6, 89, 21, 32, 31, 5, 32, 12 };

   _Bool check = 1;

   while (check) {

       temp1 = string2[i];

       temp2 = string2[i + 1];

       if (temp1 < temp2) {

           string2[i + 1] = temp1;

           string2[i] = temp2;

           i = 0;

       } else {

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           if (i = 15) {

               check = !check;

           }

       }

   }

   

   return 0;

}

Read more about C++ programming here:

brainly.com/question/20339175

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