Answer
No;
The two flows are not dynamically similar
Explanation: Given
T∞,1 = 800k
V∞,1 = 200m/s
p∞,1 = 1.739kg/m³
T∞,2 = 200k
V∞,2 = 100m/s
p∞,2 = 1.23kg/m³
Size1 = 2 * Size2 (L1 = 2L2) Assumptions Made
α ∝√T
μ∝√T Two (2) conditions must be met if the two flows are to be considered similar.
Condition 1: Similar Parameters must be the same for both flows
Condition 2: The bodies and boundaries must be genetically true. Condition 2 is true
Checking for the first condition...
Well need to calculate Reynold's Number for both flows
And Check if they have the same Reynold's Number Using the following formula
Re = pVl/μ
Re1 = p1V1l1/μ1 Re2 = p2V2l2/μ2 Re1/Re2 = p1V1l1/μ1 ÷ p2V2l2/μ2
Re1/Re2 = p1V1l1/μ1 * μ2/p2V2l2
Re1/Re2 = p1V1l1μ2/p2V2l2μ1
Re1/Re2 = p1V1l1√T2 / p1V1l1√T1
Re1/Re2 = (1.739 * 200 * 2L2 * √200) / (1.23 * 100 * L2 * √800)
Re1/Re2 = 9837.2/3479
Re1/Re2 = 2,828/1
Re1:Re2 = 2.828:1
Re1 ≠ Re2,
So condition 1 is not satisfied Since one of tbe conditions is not true, the two flows are not dynamically similar
Answer:
A) б1 = 28 ksi and б2 = -6.02 ksi
B) 1.25
Explanation:
Given data :
Torsional stress = 13 ksi
Alternating bending stress = 22ksi
A) determine yielding factor of safety according to the distortion energy theory
б1,2 = 
± √(22/2)² + 13²
= 11 ± 17
therefore б1 = 28 ksi hence б2 = -6.02 ksi
B) determine the fatigue factor of safety
with properties ; Se = 35ksi, Sy = 60 ksi, Sut = 85 ksi
( б1 - б2 )² + ( б2 - б3 )² + ( б3 - б1 )² ≤ 2 ( Sy / FOS ) ²
( 28 + 6.02 ) ² + ( 6.02 - 0 )² + ( 0 - 28 )² ≤ 2 ( 60 / FOS ) ²
solving for FOS = 1.9
Next we can determine FOS with the use of Goodman criterion
бm / Sut + бa / Se = 1 / FOS
= 0 / 85 + 28/35 = 1 / FOS
making FOS the subject of the equation ; hence FOS = 1.25
