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4 years ago

Turpentine flows through a 12-nominal schedule 40 pipe. What is the flow rate that corresponds to a Reynolds number of 2000?

Engineering

1 answer:

r-ruslan [8.4K]4 years ago

4 0

Answer:

flow rate is 8.0385 × $x^{-4}$ m³/s or 12.741 gpm

Explanation:

given data

12-nominal schedule 40 pipe

Reynolds number = 2000

to find out

What is the flow rate

solution

we know the diameter of 12-nominal schedule 40 pipe is

Diameter = 12.75 inch

D = 0.32385 m

and

dynamic viscosity of Turpentine is = 0.001375 Pa-s

and Density of Turpentine is 870 kg/m³

Reynolds number is express as

Re = $\frac{\rho*V*D}{\mu}$

here ρ is density and D is diameter and V is velocity and µ is viscosity

so put here all value

2000 = $\frac{870*V*0.3238}{0.001375}$

V = 9.7619 × $x^{-3}$ m/s

and

flow rate is

Q = V × A

here A is area and Q is flow rate

Q = 9.7619 × $x^{-3}$ × $\frac{\pi }{4} * 0.3238^2$

Q = 8.0385 × $x^{-4}$ m³/s

so flow rate is 8.0385 × $x^{-4}$ m³/s or 12.741 gpm

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In some states, repair shops can accept used oil from do-it-yourselfers or ____________.

MrMuchimi

Answer:

C. Homeowners

Explanation:

I just took my automotive pollution prevention certification and passed by answering homeowners for this question.

7 0

3 years ago

Write a Lottery class that simulates a lottery. The class should have an array of five integers named lotteryNumbers. The constr

Bezzdna [24]

Answer:

Output:-

Enter the five digit lottery number

Enter the digit 1 : 23

Enter the digit 2 : 44

Enter the digit 3 : 43

Enter the digit 4 : 66

Enter the digit 5 : 33

YOU LOSS!!

Computer Generated Lottery Number :

|12|38|47|48|49|

Lottery Number Of user:

|23|33|43|44|66|

Number Of digit matched: 0

Code:-

import java.util.Arrays;

import java.util.Random;

import java.util.Scanner;

public class Lottery {

int[] lotteryNumbers = new int[5];

public int[] getLotteryNumbers() {

return lotteryNumbers;

}

Lottery() {

Random randomVal = new Random();

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

lotteryNumbers[i] = randomVal.nextInt((50 - 1) + 1);

}

int compare(int[] personLottery) {

int count = 0;

Arrays.sort(lotteryNumbers);

Arrays.sort(personLottery);

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

if (lotteryNumbers[i] == personLottery[i]) {

count++;

}

return count;

}

public static void main(String[] args) {

int[] personLotteryNum = new int[5];

int matchNum;

Lottery lnum = new Lottery();

Scanner input = new Scanner(System.in);

System.out.println("Enten the five digit lottery number");

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

System.out.println("Enter the digit " + (i + 1) + " :");

personLotteryNum[i] = input.nextInt();

}

matchNum = lnum.compare(personLotteryNum);

if (matchNum == 5)

System.out.println("YOU WIN!!");

else

System.out.println("YOU LOSS!!");

System.out.println("Computer Generated Lottery Number :");

System.out.print("|");

for (int i = 0; i < lnum.getLotteryNumbers().length; i++) {

System.out.print(lnum.getLotteryNumbers()[i] + "|");

}

System.out.println("\n\nLottery Number Of user:");

System.out.print("|");

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

System.out.print(personLotteryNum[i] + "|");

}

System.out.println();

System.out.println("Number Of digit matched: " + matchNum);

}

Explanation:

5 0

3 years ago

A dead-man system should shut off fuel flow within ____ of the system’s maximum flow rate.

AysviL [449]

Answer:

I think it is 10 percent because

3 0

4 years ago

Read 2 more answers

Cos square there + cos square theta multiply cot square theta = cot square theta

Ivahew [28]

Explanation:

We need to prove that : $\cos^2\theta+\cos^2\theta {\cdot} \cot^2\theta=\cot^2\theta$

Taking LHS : $\cos^2\theta+\cos^2\theta {\cdot} \cot^2\theta$

Taking $\cos^2\theta$ common as follows :

$\cos^2\theta(1+ \cot^2\theta)$ ...(1)

We know that :

$cosec^2\theta-\cot^2\theta=1\\\\cosec^2\theta=1+\cos^2\theta$ ....(2)

Use equation (2) in equation (1) as follows :

$\cos^2\theta{\cdot} cosec^2\theta$

We know that : $cosec\theta=\dfrac{1}{\sin\theta}$

So,

$\cos^2\theta{\cdot} \dfrac{1}{\sin^2\theta}\\\\=\cot^2\theta$

=RHS

Hence, LHS = RHS

8 0

3 years ago

An insulated box containing helium gas falls from a balloon 4.5 km above the earth's surface. calculate the temperature rise in

Pepsi [2]

Answer:

The rise in temperature of helium gas is 14.25 K.

Explanation:

Given that

H= 4.5 km

The box is insulated ,at 4.5 km height helium gas have potential energy and when it come down then this potential energy will convert into the internal energy of helium gas.

So by heat balance

$mgh=mC_v\Delta T$

$gh=C_v\Delta T$

$C_v=3.15\frac{KJ}{kg.K}$ for helium gas

Now by putting the values

$gh=C_v\Delta T$

10 x 4.5 =3.15 x ΔT

ΔT=14.25 K

So the rise in temperature of helium gas is 14.25 K.

6 0

3 years ago