The number of pulses per second from IGBTs is referred to as
1 answer:
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Answer:
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
int main() {
string name[5];
int age[5];
int i,j;
for ( i = 0; i<=4; i++ ) {
cout << "Please enter student's name:";
cin >> name[i];
cout << "Please enter student's age:";
cin >> age[i];
}
for (i=0;i<=4;i++){
cout<<"Age of "<< name[i]<<" is "<<age[i]<<endl;
}
}
Output of above program is displayed in figure attached.
Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 × = 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × / 3
so d = 0.0133 m
so diameter is 14 mm