This question is not complete, the complete question is;
The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005
Answer:
the length of the pipe is 11583 in or 965.25 ft
Explanation:
Given the data in the question;
Static pressure ratio; p1/p2 = 10
friction coefficient f = 0.005
diameter of pipe, D =4 inch
first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so
4fL
/ D = 57.915
we substitute
(4×0.005×L) / 4 = 57.915
0.005L = 57.915
L = 57.915 / 0.005
L = 11583 in
Therefore, the length of the pipe is 11583 in or 965.25 ft
Answer:
The solution code is written in Java.
- public class Main {
-
- public static void main(String[] args) {
-
- Scanner inNum = new Scanner(System.in);
- System.out.print("Enter number of toss: ");
- int num = inNum.nextInt();
-
- for(int i=0; i < num; i++){
- System.out.println(toss());
- }
- }
-
- public static String toss(){
- String option[] = {"heads", "tails"};
- Random rand = new Random();
- return option[rand.nextInt(2)];
- }
- }
Explanation:
Firstly, we create a function <em>toss()</em> with no parameter but will return a string (Line 14). Within the function body, create an option array with two elements, "heads" and "tails" (Line 15). Next create a Random object (Line 16) and use <em>nextInt()</em> method to get random value either 0 or 1. Please note we need to pass the value of 2 into <em>nextInx() </em>method to ensure the random value generated is either 0 or 1. We use this generate random value as an index of <em>option </em>array and return either "heads" or "tails" as output (Line 17).
In the main program, we create Scanner object and use it to prompt user to input an number for how many times to toss the coin (Line 6 - 7). Next, we use the input num to control how many times a for loop should run (Line 9). In each round of the loop, call the function <em>toss() </em>and print the output to terminal (Line 10).
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ ) 
/ 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
Answer:
The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.
Explanation:
