What Degree Do You Need To Become a Solar Engineer? (2 or more sentences please)
1 answer:
You might be interested in
Answer:
6.37 inch
Explanation:
Thinking process:
We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.
To determine the pressure drop in the pipe:
Using the Bernoulli equation for mass conservation:
thus
The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.
Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F
from the tables
Re= 2.01 × 10⁵
Hence, f = 0.018
Therefore, pressure drop, (P1-P2)/p = 2.70 ft
This occurs at ae presure change of 1.17 psi
Correlating with the chart, we find that the diameter will be D= 0.513
= <u>6.37 in Ans</u>
Answer:
Check the explanation
Explanation:
Points to consider:
We need to take the input from the user
We need to find the manhatan distance and euclidian using the formula
(x1, y1) and (x2, y2) are the two points
Manhattan:
Euclidian Distance:
Code
#include<stdio.h>
#include<math.h>
struct Point{
int x, y;
};
int manhattan(Point A, Point B){
return abs(A.x - B.x) + abs(A.y- B.y);
}
float euclidean(Point A, Point B){
return sqrt(pow(A.x - B.x, 2) + pow(A.y - B.y, 2));
}
int main(){
struct Point A, B;
printf("Enter x and Y for first point: ");
int x, y;
scanf("%d%d", &x, &y);
A.x = x;
A.y = y;
printf("Enter x and Y for second point: ");
scanf("%d%d", &x, &y);
B.x = x;
B.y = y;
printf("Manhattan Distance: %d\n", manhattan(A, B));
printf("Euclidian Distance: %f\n", euclidean(A, B));
}
Sample output
Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e
r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Answer:
Gc(s) =
Explanation:
comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.
attached is the detailed solution and the plot in Matlab
