Answer:
The correct option is;
c. the exergy of the tank can be anything between zero to P₀·V
Explanation:
The given parameters are;
The volume of the tank = V
The pressure in the tank = 0 Pascal
The pressure of the surrounding = P₀
The temperature of the surrounding = T₀
Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change
The exergy balance equation is given as follows;
![X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}](https://tex.z-dn.net/?f=X_2%20-%20X_1%20%3D%20%5Cint%5Climits%5E2_1%20%7B%7D%20%5C%2C%20%5Cdelta%20Q%20%5Cleft%20%281%20-%20%5Cdfrac%7BT_0%7D%7BT%7D%20%5Cright%20%29%20-%20%5BW%20-%20P_0%20%5Ccdot%20%28V_2%20-%20V_1%29%5D-%20X_%7Bdestroyed%7D)
Where;
X₂ - X₁ is the difference between the two exergies
Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system
Therefore, the exergy of the tank can be anything between zero to P₀·V.
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:
Technician B
Explanation:
Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.
Answer:
endurance length is 236.64 MPa
Explanation:
data given:
d = 37.5 mm
Sut = 760MPa
endurance limit is
Se = 0.5 Sut
= 0.5*760 = 380 MPa
surface factor is
Ka = a*Sut^b
where
Sut is ultimate strength
for AISI 1040 STEEL
a = 4.51, b = -0.265
Ka = 4.51*380^{-0.265}
Ka = 0.93
size factor is given as
Kb =1.29 d^{-0.17}
Kb = 0.669
Se = Sut *Ka*Kb
= 380*0.669*0.93
Se = 236.64 MPa
therefore endurance length is 236.64 MPa
Answer:
The percentage ductility is 35.5%.
Explanation:
Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.
Step1
Given:
Diameter of shaft is 10.2 mm.
Final area of the shaft is 52.7 mm².
Calculation:
Step2
Initial area is calculated as follows:
A = 81.713 mm².
Step3
Percentage ductility is calculated as follows:
D = 35.5%.
Thus, the percentage ductility is 35.5%.
