Explanation:
Probably a large interest in math and how things works. Many engineering students question the world around them and have a lot of curiosity. It also depends on the type of engineering
Answer:
T=194.3C
X=1
P=542.5kPa
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties
First we find the specific volume knowing the temperature and quality of state 1
v(p=400kPa, x=0.75)=0.3471m^3/kg
As a constant volume process is presented, it keeps constant in state 2, so we find the temperature and quality with p = 600kPa and v = 0.3471m ^ 3 / kg
T(P=600kPa, v=0.3471m ^ 3 / kg)=194.3C
x(P=600kPa, v=0.3471m ^ 3 / kg)=1
for the second part we find the pressure, with a quality of 1 and a volume of
v= 0.3471m ^ 3 / kg
p=542.5kPa
Answer:
hello your question is incomplete attached below is the missing diagram to the question and the detailed solution
Answer : principal stresses : 0.82 MPa, -33.492 MPa
shear stress = 17.157 MPa
∅ = 9.09 ≈ 10°
Explanation:
The principal stress ( б1 ) = 0.82 MPa
( б2 ) = -33.492 MPa
The shear stress = 17.157 MPa
∅ = 9.09 ≈ 10°
attached below is the detailed solution and the Mohr's circle
Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:
![A=\pi*D^2/4](https://tex.z-dn.net/?f=%20A%3D%5Cpi%2AD%5E2%2F4%20)
Replacing the diameter the area results:
![A= 17.76 in^2](https://tex.z-dn.net/?f=%20A%3D%2017.76%20in%5E2%20)
Therefore the the stress results:
![σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi](https://tex.z-dn.net/?f=%20%CF%83%20%3D%2070%2F17.76%20lb%2Fin%5E2%20%3D%2039.11%20lb%2Fin%5E2%3D%2039.11%20psi%20)