Answer:
mechanical power used to overcome frictional effects in piping is 2.37 hp
Explanation:
given data
efficient pump = 80%
power input = 20 hp
rate = 1.5 ft³/s
free surface = 80 ft
solution
we use mechanical pumping power delivered to water is
.............1
put here value
= (0.80)(20)
= 16 hp
and
now we get change in the total mechanical energy of water is equal to the change in its potential energy
..............2
and that can be express as
..................3
so
![\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]](https://tex.z-dn.net/?f=%5CDelta%20%7BE_%7Bmech%7D%7D%20%3D%20%2862.4lbm%2Fft%5E3%29%281.5ft%5E3%2Fs%29%2832.2ft%2Fs%5E2%29%2880ft%29%5B%5Cfrac%7B1lbf%7D%7B32.2lbm%5Ccdot%20ft%2Fs%5E2%7D%5D%5B%5Cfrac%7B1hp%7D%7B550lbf%20%5Ccdot%20ft%2Fs%7D%5D)
......4
solve it we get
hp
so here
due to frictional effects, mechanical power lost in piping
we get here
put here value
= 16 -13.614
= 2.37 hp
so mechanical power used to overcome frictional effects in piping is 2.37 hp
Answer:
The British Standards Institution, is the national standards body of the United Kingdom. BSI produces technical standards on a wide range of products and services and also supplies certification and standards-related services to businesses
Answer:
import java.util.Scanner;
public class FindMatchValue {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_VALS = 4;
int[] userValues = new int[NUM_VALS];
int i;
int matchValue;
int numMatches = -99; // Assign numMatches with 0 before your for loop
matchValue = scnr.nextInt();
for (i = 0; i < userValues.length; ++i) {
userValues[i] = scnr.nextInt();
}
/* Your solution goes here */
numMatches = 0;
for (i = 0; i < userValues.length; ++i) {
if(userValues[i] == matchValue) {
numMatches++;
}
}
System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);
}
}
Answer:
Light = A xor B
Explanation:
If switches A and B produce True or False, then Light will be True for ...
Light = A xor B
