Answer:
a) Q = 251.758 kJ/mol
b) creep rate is
Explanation:
we know Arrhenius expression is given as
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 =
solving for Q we get
Q = 251.758 kJ/mol
b) creep rate at 500 degree C
we know
Answer:
a. 9947 m
b. 99476 times
c. 2*10^11 molecules
Explanation:
a) To find the mean free path of the air molecules you use the following formula:
R: ideal gas constant = 8.3144 Pam^3/mol K
P: pressure = 1.5*10^{-6} Pa
T: temperature = 300K
N_A: Avogadros' constant = 2.022*10^{23}molecules/mol
d: diameter of the particle = 0.25nm=0.25*10^-9m
By replacing all these values you obtain:
b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:
c) By using the equation of the ideal gases you obtain:
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.