Answer:
15,000 psi
Explanation:
The solution / solving is attach below.
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress
= 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress 
Making thickness t the subject; we have


t = 0.0056 m
t = 5.6 mm
For longitudinal stress.



t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
Answer:
Explanation:
Given conditions
1)The stress on the blade is 100 MPa
2)The yield strength of the blade is 175 MPa
3)The Young’s modulus for the blade is 50 GPa
4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.
5)The temperature of the blade is 800°C.
6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)
where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K
Young Modulus, E = Stress,
/Strain, ∈
initial Strain, 


creep rate in the steady state


but Tinitial = 0


solving the above equation,
we get
Tfinal = 2459.82 hr
Answer:
Tso = 28.15°C
Explanation:
given data
t2 = 21 mm
ki = 0.026 W/m K
t1 = 9 mm
kp = 180 W/m K
length of the roof is L = 13 m
net solar radiation into the roof = 107 W/m²
temperature of the inner surface Ts,i = -4°C
air temperature is T[infinity] = 29°C
convective heat transfer coefficient h = 47 W/m² K
solution
As when energy on the outer surface at roof of a refrigerated truck that is balance as
Q =
.....................1
Q =
.....................2
now we compare both equation 1 and 2 and put here value
solve it and we get
Tso = 28.153113
so Tso = 28.15°C