Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus 
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
![w_{b, out} =m([tex]u_{1} -u_{2](https://tex.z-dn.net/?f=w_%7Bb%2C%20out%7D%20%3Dm%28%5Btex%5Du_%7B1%7D%20-u_%7B2)
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:they are information, materials, tools and machines, captial, time, energy, and people
Explanation:
Complete Question
Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.
Answer:
The stress is 
Explanation:
From the question we are told that
The critical yield resolved shear stress is 
First we obtain the angle 
between the slip direction [121] and [111]
![\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]](https://tex.z-dn.net/?f=%5Clambda%20%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B%28u_1%20u_2%20%2B%20v_1%20v_2%20%2B%20w_1%20w_2%7D%7B%5Csqrt%7Bu_1%5E2%20%2B%20v_1%20%5E2%2B%20w_1%5E2%7D%29%5Csqrt%7B%28%20u_2%5E2%20%2B%20v_2%5E2%20%2B%20w_2%20%5E2%29%7D%20%7D%20%20%5D)
Where 
are the directional indices
![\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%20cos%20%5E-%5B%20%5Cfrac%7B%281%29%20%28-1%29%20%2B%20%282%29%20%281%29%20%2B%20%281%29%20%281%29%7D%7B%5Csqrt%7B%28%281%29%5E2%20%2B%282%29%5E2%20%2B%20%281%29%5E2%29%7D%5Csqrt%7B%28%28-1%29%5E2%20%2B%20%281%29%5E2%20%2B%20%281%29%5E2%20%29%20%7D%20%20%7D%20%5D)
![= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]](https://tex.z-dn.net/?f=%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B2%7D%7B%5Csqrt%7B6%7D%20%5Csqrt%7B3%7D%20%20%7D%20%5D)
Next is to obtain the angle 
between the direction [121] and [101]
![\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]](https://tex.z-dn.net/?f=%5CO%20%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B%28u_1%20u_3%20%2B%20v_1%20v_3%20%2B%20w_1%20w_3%7D%7B%5Csqrt%7Bu_1%5E2%20%2B%20v_1%20%5E2%2B%20w_1%5E2%7D%29%5Csqrt%7B%28%20u_3%5E2%20%2B%20v_3%5E2%20%2B%20w_3%20%5E2%29%7D%20%7D%20%20%5D)
Substituting 1 for 
, 2 for 
, 1 for 
, 1 for 
, 0 for 
, and 1 for 
![\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]](https://tex.z-dn.net/?f=%5CO%20%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B1%2A%201%20%2B%202%2A0%20%2B%201%2A1%20%7D%7B%5Csqrt%7B1%5E2%20%2B%202%5E2%20%2B%201%5E2%20%7D%20%5Csqrt%7B%281%5E2%20%2B%200%5E2%20%2B%201%5E2%20%29%7D%20%20%7D%20%5D)
![\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]](https://tex.z-dn.net/?f=%5CO%20%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B2%7D%7B%5Csqrt%7B6%7D%20%5Csqrt%7B2%7D%20%20%7D%20%5D)
The stress is mathematically represented as
