Answer:
Explanation:
given data :
output voltage ( Van(t) ) = (Vd /2) + (0.85 Vd/2 sin ( w1 t ) )
w1 = 2
60 rad/sec
find the value of da(t) by inputting the value of Van (t) into
da = Van(t) / Vd
hence: da(t) = 0.5 + 0.425 sin ((2
60)t)
attached below is the plot of the da(t) against time
Answer:
-⅓ cos³ x + C
Explanation:
∫ cos² x sin x dx
If u = cos x, then du = -sin dx.
∫ -u² du
Integrate using power rule:
-⅓ u³ + C
Substitute back:
-⅓ cos³ x + C
Answer:
This doesn't represent an equilibrium state of stress
Explanation:
∝ = 1 , β = 1 , y = 1
x = 0 , y = 0 , z = 0 ( body forces given as 0 )
Attached is the detailed solution is and also the conditions for equilibrium
for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution
Answer:
![\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%200.8708%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
Explanation:
The adiabatic throttling process is modelled after the First Law of Thermodynamics:
![m\cdot (h_{in} - h_{out}) = 0](https://tex.z-dn.net/?f=m%5Ccdot%20%28h_%7Bin%7D%20-%20h_%7Bout%7D%29%20%3D%200)
![h_{in} = h_{out}](https://tex.z-dn.net/?f=h_%7Bin%7D%20%3D%20h_%7Bout%7D)
Properties of water at inlet and outlet are obtained from steam tables:
State 1 - Inlet (Liquid-Vapor Mixture)
![P = 1500\,kPa](https://tex.z-dn.net/?f=P%20%3D%201500%5C%2CkPa)
![T = 198.29\,^{\textdegree}C](https://tex.z-dn.net/?f=T%20%3D%20198.29%5C%2C%5E%7B%5Ctextdegree%7DC)
![h = 2726.9\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h%20%3D%202726.9%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
![s = 6.3068\,\frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=s%20%3D%206.3068%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
![x = 0.967](https://tex.z-dn.net/?f=x%20%3D%200.967)
State 2 - Outlet (Superheated Vapor)
![P = 200\,kPa](https://tex.z-dn.net/?f=P%20%3D%20200%5C%2CkPa)
![T = 130\,^{\textdegree}C](https://tex.z-dn.net/?f=T%20%3D%20130%5C%2C%5E%7B%5Ctextdegree%7DC)
![h = 2726.9\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h%20%3D%202726.9%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
![s = 7.1776\,\frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=s%20%3D%207.1776%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
The change of entropy of the steam is derived of the Second Law of Thermodynamics:
![\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%207.1776%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D%20-%206.3068%5C%2C%20%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
![\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%200.8708%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)