Given:
outer radius, R' = 10 m
inner diameter, d = 2 m
inner radius, R = = 1 m
surface temperature, T' =
Thermal conductivity of soil, K = 0.52 W/mK
Solution:
To calculate the thermal temperature of conductor, T, we know amount of heat, Q is given by :
Q =
500 =
T = 68.86 +20 =
Therefore, outside surface temperature is
Answer:
i) S–N plot is attached
ii) fatigue strength = 100 MPa
iii) fatigue life = 5.62 x 10^(5) cycles
Explanation:
i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)
ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.
So let's find the log of this and trace it on the graph attached.
Log(4 × 10^(6)) = 6.6
From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa
iii) The question says we should find the fatigue life for 120 MPa.
Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.
Thus,the fatigue life will be the inverse log of 5.75.
Thus, fatigue life = 10^(5.75)
Fatigue life = 5.62 x 10^(5)
Answer:
a. 4.279 MPa
b. 3.198 MPa to 4.279 MPa
c. 0.939 MPa
d. Below 3.198 MPa
Explanation:
From the given parameters
M = 1.75 MPa
M at 1.6 MPa gives A/A* = 1.2502
M at 1.8 MPa gives A/A* = 1.4390
Therefore, by interpolation, we have M = 1.75 MPa gives A
However, we shall use M = 1.75 MPa and A
Similarly,
P/P₀ = 0.1878
a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have
A/A* = 1.387. by interpolation M
Therefore P = P₀ × P
Which shows that the nozzle is choked for back pressures lower than 4.279 MPa
b) Where there is a normal shock at the exit of the nozzle, we have;
M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa
Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406
Where the normal shock occurs at the nozzle exit, we have
Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as = 4.279 MPa
Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa
c) At M = 1.75 MPa and P
d) Where the back pressure is less than 3.198 MPa according to isentropic flow relations supersonic flow will exist at the exit plane