Answer:
Q=67.95 W
T=119.83°C
Explanation:
Given that
For air
Cp = 1.005 kJ/kg·°C
T= 20°C
V=0.6 m³/s
P= 95 KPa
We know that for air
P V = m' R T
95 x 0.6 = m x 0.287 x 293
m=0.677 kg/s
For gas
Cp = 1.10 kJ/kg·°C
m'=0.95 kg/s
Ti=160°C ,To= 95°C
Heat loose by gas = Heat gain by air
[m Cp ΔT] for air =[m Cp ΔT] for gas
by putting the values
0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )
T=119.83°C
T is the exit temperature of the air.
Heat transfer
Q=[m Cp ΔT] for gas
Q=0.95 x 1.1 x ( 160 -95 )
Q=67.95 W
Answer:
public static int average(int j, int k) {
return (int)(( (long)(i) + (long)(j) ) /2 );
}
Explanation:
The above code returns the average of two integer variables
Line 1 of the code declares a method along with 2 variables
Method declared: average of integer data type
Variables: j and k of type integer, respectively
Line 2 calculates the average of the two variables and returns the value of the average.
The first of two integers to average is j
The second of two integers to average is k
The last parameter ensures average using (j+k)/2
Answer:

Explanation:
The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:




The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

The rate of heat transfer in the boiler is:


Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as


Integrating both sides we get

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get

Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of

Thus the remaining 125 meters will be covered with a constant speed of

in time equalling 
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds