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Shtirlitz [24]
4 years ago
5

The Reynolds number, rhoVD/μ is a very important parameter in fluid mechanics. Determine its value for ethyl alcohol flowing at

a velocity of 4 m/s through a 2-in.-diameter pipe.
Engineering
1 answer:
ElenaW [278]4 years ago
8 0

Answer:

The Reynold Number is 146415.34

Explanation:

The density of ethyl alcohol = ρ = 789 kg/m^3

The dynamic viscosity of ethyl alcohol = μ = 0.001095 kg/m.s

Velocity = V = 4 m/s

Diameter of pipe = D = 2 in = 0.0508 m

The Reynold's number (Re), will be:

Re = ρVD/μ

Re = (789 kg/m^3)(4 m/s)(0.0508 m)/(0.001095 kg/ms)

<u>Re = 146415.34</u>

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Nc3
Masja [62]

please give a better explanation of what you want to be answered.

3 0
4 years ago
A square power screw has a mean diameter of 30 mm and a pitch of 4 mm with single thread. The collar diameter can be assumed to
poizon [28]

Answer:

i) The torque required to raise the load is 15.85 N*m

ii) The torque required to lower the load is 6.91 N*m

iii) The minimum coefficient of friction is -0.016

Explanation:

Given:

dm = mean diameter = 0.03 m

p = pitch = 0.004 m

n = number of starts = 1

The lead is:

L = n * p = 1 * 0.004 = 0.004 m

F = load = 7000 N

dc = collar diameter = 0.035 m

u = 0.05

i) The helix angle is:

tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43

The torque is:

T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm

ii) The torque to lowering the load is:

T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm

iii)

T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\  0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}

Clearing u:

u = -0.016

5 0
4 years ago
____________ is the organization that oversees environmental compliance.
ELEN [110]

Answer:

Environmental Protection Agency (EPA)

Explanation:

Pollution can be defined as the physical degradation or contamination of the environment through an emission of harmful, poisonous and toxic chemical substances.

Environmental Protection Agency (EPA) is the organization that oversees environmental compliance.

In the United States of America, the agency which was established by US Congress and saddled with the responsibility of overseeing all aspects of pollution, environmental clean up, pesticide use, contamination, and hazardous waste spills is the Environmental Protection Agency (EPA). Also, EPA research solutions, policy development, and enforcement of regulations through the resource Conservation and Recovery Act.

Hence, the Environmental Protection Agency (EPA) is the governmental agency set up to ensure that various industries, factories and people comply with laws and regulations concerning the environment.

8 0
3 years ago
If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respect
Natasha2012 [34]

Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance x!

= cos 75⁰ = \frac{x^!}{10ft}

    x! = 10 * cos 75^{o} = 2.59 ft

calculate distance x

= cos 75⁰ = \frac{x}{30ft}

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment  produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * x! + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

      = 4014.5 + 1950.35 + 2030.535

      = 7995.385 ft. Ib ≈ 7.9954 kip.ft

6 0
4 years ago
Write a program that asks the user for the name of a file. The program should display the number of words that the file contains
Flura [38]

Answer:

import java.io.*;

import java.util.Scanner;

public class CountWordsInFile {

   public static void main(String[] args) throws IOException {

       Scanner keyboard = new Scanner(System.in);

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

       String fileName = keyboard.next();

       File file = new File(fileName);

       try {

       

           Scanner scan = new Scanner(file);

           

           int count = 0;

           

           while(scan.hasNext()) {

               scan.next();

               count += 1;

           }

           scan.close();

           System.out.println("Number of words: "+count);        

       } catch (FileNotFoundException e) {

           System.out.println("File " + file.getName() + " not present ");

           System.exit(0);

       }

   }

}

3 0
4 years ago
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