These parts are commonly called carburetor emulsion tubes. These tubes maintain the air-fuel ratio at different speeds.
The carburetor is a device of the combustion engine power supply system that mixes fuel and air in order to facilitate internal combustion.
The carburetor emulsion tubes are tubes that maintain the air-fuel ratio at different velocities.
These tubes (carburetor emulsion tubes) are small brass cylinders where the metering needle slides into them.
Learn more about carburetors here:
brainly.com/question/4237015
Answer:
The heat of the arc melts the surface of the base metal and the end of the electrode. The electric arc has a temperature that ranges from 3,000 to 20,000 °C
Explanation:
Welding fumes are complex mixtures of particles and ionized gases.
Answer:
#include <iostream>
using namespace std;
void PrintPopcornTime(int bagOunces) {
if(bagOunces < 3){
cout << "Too small";
cout << endl;
}
else if(bagOunces > 10){
cout << "Too large";
cout << endl;
}
else{
cout << (6 * bagOunces) << " seconds" << endl;
}
}
int main() {
PrintPopcornTime(7);
return 0;
}
Explanation:
Using C++ to write the program. In line 1 we define the header "#include <iostream>" that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.
Answer:
a) The Net power developed in this air-standard Brayton cycle is 43.8MW
b) The rate of heat addition in the combustor is 84.2MW
c) The thermal efficiency of the cycle is 52%
Explanation:
To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.
Considering this:
Now we can calculate the enthalpy of each work point:
h₁=281.4KJ/Kg
h₂=695.41KJ/Kg
h₃=2105KJ/Kg
h₄=957.14KJ/Kg
The net power developed:
The rate of heat:
The thermal efficiency:
