Select four types of engineers who might be involved in the development of a product such as an iPhone.

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De

Anettt [7]

Answer:

$Q=4.98\times 10^{-3}\ m^3/s$

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

For Copper tube is 3/4 standard type K drawn tube

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

$\mu =0.00164\ Pa.s$

Pressure difference given as

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

Where

L is length of tube

μ is dynamic viscosity

Q is volume flow rate

d is inner diameter of tube

ΔP is pressure drop

Now by putting the values

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

$130\times 1000=\dfrac{128\times 0.00164\times 50\times Q}{\pi\times 0.0189^4}$

$Q=4.98\times 10^{-3}\ m^3/s$

So flow rate is $Q=4.98\times 10^{-3}\ m^3/s$

7 0

3 years ago

Sexual Harassment falls under Title IX. What are the two types of Sexual Harassment, not examples, but there are only two kinds.

Anit [1.1K]

Answer:

According to the Equal Employment Opportunity Commission (EEOC), there are two types of sexual harassment claims: "quid pro quo" and "hostile work environment." The EEOC provides guidance on defining sexual harassment and establishing employer liability. hope that helps

Explanation:

4 0

3 years ago

Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or

n200080 [17]

Answer:

Here is the JAVA program:

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed that is 4 and assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two lists origList[] and offsetAmount[] are assigned values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // product variable to store the product of 2 lists

for(i = 0; i <= origList.length - 1; i++){

/* loop starts with i at 0th position or index and ends when the end of the origList is reached */

/* multiples each element of origList to corresponding element of offsetAmount and stores result in the form of character string in product*/

product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }

System.out.println(product); }} //displays the product of both lists

Output:

80 180 80 400

Explanation:

If you want to print the product of origList with corresponding value in offsetAmount in vertical form you can do this in the following way:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

origList[i] *= offsetAmount[i];

System.out.println(origList[i]); } }}

Output:

80

180

80

400

The program along with the output is attached as screenshot with the input given in the example.

8 0

3 years ago

Suzanne Brett wants to borrow $55,000 from the bank. The interest rate is 6.5% and the term is for 5 years.

Otrada [13]

Answer: $14575

$55000

6.5%

5 years

Total Payment Amount: $72875

Yearly payment :$72875/5= $14575

8 0

2 years ago

Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

Zarrin [17]

The total amount of daily heat transfer is 1382.38 M w.

The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.

<u>Explanation:</u>

Given data,

$T_{\infty}$ = 10° C

$h_{0}$ = 250 w/ $m^{2}$ k

Pipe length = 20 m

Inner diameter $d_{1}$ = 6 cm, $r_{1}$ = 3 cm

Outer diameter $d_{2}$ = 8 cm, $r_{2}$ = 4 cm

The thickness of insulation is 4 cm.

$r_{3}$ = $r_{2}$ + 4

= 4+4

$r_{3}$ = 8 cm

$h_{0}$ is the heat transfer coefficient of convection inside, $h_{i}$ is the heat transfer coefficient of convection outside.

The heat transfer rate between ambient and steam is

$q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}$ watt

= $\begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned}$ watt