Answer:
Explanation:
Inductance = 250 mH = 250 / 1000 = 0.25 H
capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)
ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A
a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms
b) capacitive reactance = 
= 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms
c) impedance = 
= 240 / 0.11 = 2181.82 ohms
Answer:
//Define the header file
#ifndef PLAYER_H
#define PLAYER_H
//header file.
#include <string>
//Use the standard namespace.
using namespace std;
//Define the class Player.
class Player
{
//Declare the required data members.
string name;
int score;
public:
//Declare the required
//member functions.
void setName(string par_name);
void setScore(int par_score);
string getName();
int getScore();
}
//End the definition
//of the header file.
#endif
Player.cpp:
//Include the "Player.h" header file,
#include "Player.h"
//Define the setName() function.
void Player::setName(string par_name)
{
name = par_name;
}
//Define the setScore() function.
void Player::setScore(int par_score)
{
score = par_score;
}
//Define the getName() function.
string Player::getName()
{
return name;
}
//Define the getScore() function.
int Player::getScore()
{
return score;
}
Answer:
Explanation:
Steam at outlet is an superheated steam, since 
. From steam tables, the specific enthalpy is:
The throttle valve is modelled after the First Law of Thermodynamics:
Hence, specific enthalpy at inlet is:
The quality in the steam line is:
Answer:
A). Dry unit weight = 1657.08Kg/m3
B). Porosity = 0.37
C). Void ratio = 0.593
D). 0.712
Explanation:
Total unit weight, Y = 120pcf =1922.2 Kg/m3
Specific gravity of solids, Gs = 2.64
Water content, w = 16%
A). Dry unit weight
Yd = Y/(1+w)
= 1922.2/(1+0.16) = 1657.08Kg/m3
B). Porosity
However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3
Void ratio = 2.64×1000/1657.08 = 0.593
And porosity = e/(1+e) =0.593/(1+0.593) = 0.37
C). void ratio, e = 0.593
D). Degree of saturation, S = m×Gs/e where m =water content
S = 0.16×2.64/0.593 = 0.712
Answer:
hello your question is incomplete attached below is the missing part and also attached is the solution
answer: a) 0.4801
b) 5.398 kw
c) 2.14
d) 12.72
Explanation:
The quality of the refrigerant at the evaporator inlet
h4 = hf4 + x4(hfx4)
Refrigeration load
Ql = m(h1-h4)
COP of the refrigerator
Ql / m(h2-h1) - Qm
Theoretical maximum refrigeration load
( Ql )max = COPr.rev * [m(h2-h1) - Qin]
