Answer:
The original length of the specimen ![l_{o} = 104.7 mm](https://tex.z-dn.net/?f=l_%7Bo%7D%20%3D%20104.7%20mm)
Explanation:
Original diameter
= 30 mm
Final diameter
= 30.04 mm
Change in diameter Δd = 0.04 mm
Final length
= 105.20 mm
Elastic modulus E = 65.5 G pa = 65.5 ×
M pa
Shear modulus G = 25.4 G pa = 25.4 ×
M pa
We know that the relation between the shear modulus & elastic modulus is given by
![G = \frac{E}{2(1 + \mu)}](https://tex.z-dn.net/?f=G%20%3D%20%5Cfrac%7BE%7D%7B2%281%20%2B%20%5Cmu%29%7D)
![25.5 = \frac{65.5}{2 (1 + \mu)}](https://tex.z-dn.net/?f=25.5%20%3D%20%5Cfrac%7B65.5%7D%7B2%20%281%20%2B%20%5Cmu%29%7D)
![\mu = 0.28](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.28)
This is the value of possion's ratio.
We know that the possion's ratio is given by
![\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B%5Cfrac%7B0.04%7D%7B30%7D%20%7D%7B%5Cfrac%7Bchange%20%5C%20in%20%5C%20length%7D%7Bl_%7Bo%7D%20%7D%20%7D)
![{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}](https://tex.z-dn.net/?f=%7B%5Cfrac%7Bchange%20%5C%20in%20%5C%20length%7D%7Bl_%7Bo%7D%20%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B0.04%7D%7B30%7D%20%7D%7B0.28%7D)
0.00476
![\frac{l_{1} - l_{o} }{l_{o} } = 0.00476](https://tex.z-dn.net/?f=%5Cfrac%7Bl_%7B1%7D%20-%20l_%7Bo%7D%20%20%7D%7Bl_%7Bo%7D%20%20%7D%20%3D%200.00476)
Final length
= 105.2 m
Original length
![l_{o} = \frac{105.2}{1.00476}](https://tex.z-dn.net/?f=l_%7Bo%7D%20%3D%20%5Cfrac%7B105.2%7D%7B1.00476%7D)
![l_{o} = 104.7 mm](https://tex.z-dn.net/?f=l_%7Bo%7D%20%3D%20104.7%20mm)
This is the original length of the specimen.
Answer:
a) Q = 251.758 kJ/mol
b) creep rate is ![= 1.751 \times 10^{-5} \% per hr](https://tex.z-dn.net/?f=%3D%201.751%20%5Ctimes%2010%5E%7B-5%7D%20%5C%25%20per%20hr)
Explanation:
we know Arrhenius expression is given as
![\dot \epsilon =Ce^{\frac{-Q}{RT}](https://tex.z-dn.net/?f=%5Cdot%20%5Cepsilon%20%3DCe%5E%7B%5Cfrac%7B-Q%7D%7BRT%7D)
where
Q is activation energy
C is pre- exponential constant
At 700 degree C creep rate is
% per hr
At 800 degree C creep rate is
% per hr
activation energy for creep is
= ![= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7BC%5Ctimes%20e%5E%7B%5Cfrac%7B-Q%7D%7BR%28800%2B273%29%7D%7D%7D%7BC%5Ctimes%20e%5E%7B%5Cfrac%7B-Q%7D%7BR%28700%2B273%29%7D%7D%7D)
![\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%5C%25%7D%7B5.5%20%5Ctimes%2010%5E%7B-2%7D%5C%25%7D%20%3D%20e%5E%7B%5B%5Cfrac%7B-Q%7D%7BR%28800%2B273%29%7D%5D%20-%5B%5Cfrac%7B-Q%7D%7BR%28800%2B273%29%7D%5D%7D)
![\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B0.01%7D%7B5.5%5Ctimes%2010%5E%7B-4%7D%7D%20%3D%20ln%20%5Be%5E%7B%5Cfrac%7BQ%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B1073%7D%20-%20%5Cfrac%7B1%7D%7B973%7D%5D%7D%5D)
solving for Q we get
Q = 251.758 kJ/mol
b) creep rate at 500 degree C
we know
![C = \epsilon e^{\frac{Q}{RT}}](https://tex.z-dn.net/?f=C%20%3D%20%5Cepsilon%20e%5E%7B%5Cfrac%7BQ%7D%7BRT%7D%7D)
![=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr](https://tex.z-dn.net/?f=%3D-%201%5C%25%20e%7B%5Cfrac%7B251758%7D%7B8.314%28500%2B273%7D%7D%20%3D%201.804%20%5Ctimes%2010%5E%7B12%7D%20%5C%25%20per%20hr)
![\epsilon_{500} = C e^{\frac{Q}{RT}}](https://tex.z-dn.net/?f=%5Cepsilon_%7B500%7D%20%3D%20C%20e%5E%7B%5Cfrac%7BQ%7D%7BRT%7D%7D)
![= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}](https://tex.z-dn.net/?f=%3D%201.804%20%5Ctimes%2010%5E%7B12%7D%20%20e%7B%5Cfrac%7B251758%7D%7B8.314%28500%2B273%7D%7D)
![= 1.751 \times 10^{-5} \% per hr](https://tex.z-dn.net/?f=%3D%201.751%20%5Ctimes%2010%5E%7B-5%7D%20%5C%25%20per%20hr)
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:
design hour volumes will be 4000 to 6000
Explanation:
given data
AADT = 150000 veh/day
solution
we get here design hour volumes that is express as
design hour volumes = AADT × k × D ..............1
here k is factor and its range is 8 to 12 % for urban
and D is directional distribution i.e traffic equal divided by the direction
so here design hour volumes will be 4000 to 6000