Answer:
The elastic modulus of the steel is 139062.5 N/in^2
Explanation:
Elastic modulus = stress ÷ strain
Load = 89,000 N
Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2
Stress = load/area = 89,000/0.64 = 139.0625 N/in^2
Length of steel bar = 4 in
Extension = 4×10^-3 in
Strain = extension/length = 4×10^-3/4 = 1×10^-3
Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2
Answer:
The correct option is;
c. the exergy of the tank can be anything between zero to P₀·V
Explanation:
The given parameters are;
The volume of the tank = V
The pressure in the tank = 0 Pascal
The pressure of the surrounding = P₀
The temperature of the surrounding = T₀
Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change
The exergy balance equation is given as follows;
![X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}](https://tex.z-dn.net/?f=X_2%20-%20X_1%20%3D%20%5Cint%5Climits%5E2_1%20%7B%7D%20%5C%2C%20%5Cdelta%20Q%20%5Cleft%20%281%20-%20%5Cdfrac%7BT_0%7D%7BT%7D%20%5Cright%20%29%20-%20%5BW%20-%20P_0%20%5Ccdot%20%28V_2%20-%20V_1%29%5D-%20X_%7Bdestroyed%7D)
Where;
X₂ - X₁ is the difference between the two exergies
Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system
Therefore, the exergy of the tank can be anything between zero to P₀·V.
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.