Answer:
0.245 m^3/s
Explanation:
Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s
Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Explanation:
F, W and B are the fresh feed, brine and total water obtained
w = 2 x 10^4 L/h
we know that
F = W + B
we substitute
F = 2 x 10^4 + B
F = 20000 + B .................EQUA 1
solute
0.035F = 0.05B
B = 0.035F/0.05
B = 0.7F
now we substitute value of B in equation 1
F = 20000 + 0.7F
0.3F = 20000
F = 20000/0.3
F = 66666.67 kg/hr
B = 0.7F
B = 0.7 * F
B = 0.7 * 66666.67
B = 46,666.669 kg/hr
the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Answer:
Explanation:
Reward him if he does his homework/work
If he doesnt do his homework, take things that he loves off of him. and tell him if he does his homework/work he will get them things back
Answer:
// Program is written in Java Programming Language
// Comments are used for explanatory purpose
import java.util.*;
public class FlipCoin
{
public static void main(String[] args)
{
// Declare Scanner
Scanner input = new Scanner (System.in);
int flips;
// Prompt to enter number of toss or flips
System.out.print("Number of Flips: ");
flips = input.nextInt();
if (flips > 0)
{
HeadsOrTails();
}
}
}
public static String HeadsOrTails(Random rand)
{
// Simulate the coin tosses.
for (int count = 0; count < flips; count++)
{
rand = new Random();
if (rand.nextInt(2) == 0) {
System.out.println("Tails"); }
else {
System.out.println("Heads"); }
rand = 0;
}
}
