Calculate the number of vacancies per cubic meter at 1000∘C for a metal that has an energy for vacancy formation of 1.22 eV/atom
1 answer:
Answer:
C
Explanation:
N = Na.P/A------(1)
Na = avogadro's number = 6.02210²³
P = density
A = atomic weight of metal
When we substitute into equation 1 above we get
1.0x10²⁹atoms/m³
From here we calculate the number of vacancies
T = 1000⁰c = 1273K
The formula to use is
Nv= Nexo(Qt/K.T) -----(2)
Qt = 1.22eV
K = Boltzmann's constant = 8.6210x10^-5
When we substitute values into equation 2
We get Nv = 1.49 x 10²⁴m-3
Therefore option c is correct
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Complete question:
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Answer:
Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection
so that critical flow is subject to detection
Explanation:
We are given:
Plane strain fracture toughness K
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:
We already know
K= design
a = length of surface creak
Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.
Therefore
Substituting figures in the expression above, we have:
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection
Answer:
a) , b)
Explanation:
a) The coefficient of performance of a reversible refrigeration cycle is:
Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)
b) The respective coefficient of performance is determined:
But:
The temperature at hot reservoir is found with some algebraic help:
Answer:
Final length of the rod = 13.90 in
Explanation:
Cross Sectional Area of the polythene rod, A = 0.04 in²
Original length of the polythene rod, l = 10 inches
Tensile modulus for the polymer, E = 25,000 psi
Viscosity,
Weight = 358 lbs - f
time, t = 1 hr = 3600 sec
Stress is given by:
Based on Maxwell's equation, the strain is given by:
Strain = Extension/(original Length)
0.39022 = Extension/10
Extension = 0.39022 * 10
Extension = 3.9022 in
Extension = Final length - Original length
3.9022 = Final length - 10
Final length = 10 + 3.9022
Final length = 13.9022 in
Final length = 13.90 in