Answer:
Explanation:
For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.
Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
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)