DescriptionStructural analysis is the determination of the effects of loads on physical structures and their components.
Answer:
<h2>120°C</h2>
Explanation:
Step one:
given data
T_{wi} = 20^{\circ}C
T_{Ai}=1000K
T_{Ae}= 400kPa
P_{Wi}=200kPa
P_{Ai}=125kPa
P_{We}=200kPa
P_{Ae}=100kPa
m_A=2kg/s
m_W=0.5kg/s
We know that the energy equation is
![m_Ah_{Ai}+m_Wh_W=m_Ah_{Ae}+m_Wh_{We}](https://tex.z-dn.net/?f=m_Ah_%7BAi%7D%2Bm_Wh_W%3Dm_Ah_%7BAe%7D%2Bm_Wh_%7BWe%7D)
making
the subject of formula we have
![h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})](https://tex.z-dn.net/?f=h_%7BWe%7D%3Dh_%7BWi%7D%2B%5Cfrac%7Bm_A%7D%7BmW%7D%28h_A-h_%7BAe%7D%29)
from the saturated water table B.1.1 , corresponding to ![T_{wi}= 20c](https://tex.z-dn.net/?f=T_%7Bwi%7D%3D%2020c)
![h_{Wi}=83.94kJ/kg](https://tex.z-dn.net/?f=h_%7BWi%7D%3D83.94kJ%2Fkg)
from the ideal gas properties of air table B.7.1 , corresponding to T=1000K
the enthalpy is:
![h_{Ai}=1046.22kJ/kg](https://tex.z-dn.net/?f=h_%7BAi%7D%3D1046.22kJ%2Fkg)
from the ideal gas properties of air table B.7.1 corresponding to T=400K
![h_{Ae}=401.30kJ/kg](https://tex.z-dn.net/?f=h_%7BAe%7D%3D401.30kJ%2Fkg)
Step two:
substituting into the equation we have
![h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})](https://tex.z-dn.net/?f=h_%7BWe%7D%3Dh_%7BWi%7D%2B%5Cfrac%7Bm_A%7D%7BmW%7D%28h_A-h_%7BAe%7D%29)
![h_{We}=83.94+\frac{2}{0.5}(2046.22-401.30)\\\\h_{We}=2663.62kJ/kg](https://tex.z-dn.net/?f=h_%7BWe%7D%3D83.94%2B%5Cfrac%7B2%7D%7B0.5%7D%282046.22-401.30%29%5C%5C%5C%5Ch_%7BWe%7D%3D2663.62kJ%2Fkg)
from saturated water table B.1.2 at
we can obtain the specific enthalpy:
![h_g=2706.63kJ/kg](https://tex.z-dn.net/?f=h_g%3D2706.63kJ%2Fkg)
we can see that
, hence there are two phases
from saturated water table B.1.2 at ![P_{We}=200kPa](https://tex.z-dn.net/?f=P_%7BWe%7D%3D200kPa)
![T_{We}=120 ^{\circ} C](https://tex.z-dn.net/?f=T_%7BWe%7D%3D120%20%5E%7B%5Ccirc%7D%20C)
Answer:
Technician B only
Explanation:
During rotor reconditioning, which is the process also known as machining and sanding, where sanding is the involves the application of between 120 and 150 grit sandpaper while using a non-excessive force that is applied non-directionally for up to 60 seconds on each side such that the surface roughness meets OE standards. The rotors are then cleaned by washing after they are serviced before they can then be installed.
Answer:
Volume=2160 m^3
Depth=2160/A m
Explanation:
Detailed explanation of the answer is given in the attached files.
Answer:
45.3 MN
Explanation:
The forging force at the end of the stroke is given by
F = Y.π.r².[1 + (2μr/3h)]
The final height, h is given as h = 100/2
h = 50 mm
Next, we find the final radius by applying the volume constancy law
volumes before deformation = volumes after deformation
π * 75² * 2 * 100 = π * r² * 2 * 50
75² * 2 = r²
r² = 11250
r = √11250
r = 106 mm
E = In(100/50)
E = 0.69
From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula
F = Y.π.r².[1 + (2μr/3h)]
F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]
F = 35.3 * [1 + 0.2826]
F = 35.3 * 1.2826
F = 45.3 MN