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Answer:
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
Explanation:
Given data;
Let,
critical stress required for initiating crack propagation Cc = 112MPa
plain strain fracture toughness = 27.0MPa
surface length of the crack = a
dimensionless parameter = Y.
Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m
Also for 6.2mm length of surface crack;
Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m
The dimensionless parameter
Cc = Kic/(Y*√pia*a)
Y = Kic/(Cc*√pia*a)
Y = 27/(112*√pia*4.4*10-³)
Y = 2.05
Now,
Cc = Kic/(Y*√pia*a)
Cc = 27/(2.05*√pia*3.1*10-³)
Cc = 135.78MPa
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
For more understanding, I have provided an attachment to the solution.
Answer:
a. 3.17*10¹⁹ b. 3.17*10¹⁴
Explanation:
a. Area A = 1cm²
Thickness h = 0.1cm
Energy of photon E = hf=hc/λ
Where f = frequency; c=speed of light; λ=wavelength = 6300Α=6300*10⁻¹⁰;
Planck's constant h = 6.626*10⁻³⁴ joule-seconds; speed of light c = 3*10⁸
Therefore E = (6.626*10⁻³⁴)*3*10⁸/6300*10⁻¹⁰=3.155*10¹⁹J
1Watt of light releases 3.17*10¹⁸ photons per second.
Volume of sample = Area * Thickness = 1*0.1=0.1cm³
Therefore, number of electron hole pairs that are generated per unit volume per unit time = 3.17*10¹⁸/0.1 = 3.17*10¹⁹photon/cm³-s
b. Steady state excess carrier concentration = 3.17*10¹⁹*10μs
=3.17*10¹⁹*10*10⁻⁶
=3.17*10¹⁴/cm³
Answer:

Explanation:
Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:



The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:



The properties of the steam are obtained from steam tables:
Intial State






Final State






The entropy change of the steam during the process is:

