The answer is D hope this helps have a good day !!:)
Answer:
A) 
B) 
Explanation:
Given data:
P-1 = 100 lbf/in^2
degree f
effeciency = 80%
from steady flow enerfy equation
where h1 and h2 are inlet and exit enthalpy
for P1 = 100 lbf/in^2 and T1 = 500 degree F
for P1 = 40 lbf/in^2
exit enthalapy h_2
from above equation
[1 Btu/lbm = 25037 ft^2/s^2]
b) amount of entropy
at ![h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2](https://tex.z-dn.net/?f=h_2%20%3D%201197.77%20Btu%2Flbm%20%5B%5Ctex%5D%20%20and%20%5Btex%5DP_2%20%3D%2040%20lbf%2Fin%5E2)
Answer:
the correct answer is option B. W
Answer: 133.88 MPa approximately 134 MPa
Explanation:
Given
Plane strains fracture toughness, k = 26 MPa
Stress at which fracture occurs, σ = 112 MPa
Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m
Critical internal crack length, l' = 6 mm = 6*10^-3 m
We know that
σ = K/(Y.√πa), where
112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]
112 MPa = 26 MPa / Y.√(3.142 * 0.043)
112 = 26 / Y.√1.35*10^-2
112 = 26 / Y * 0.116
Y = 26 / 112 * 0.116
Y = 26 / 13
Y = 2
σ = K/(Y.√πa), using l'instead of l and, using Y as 2
σ = 26 / 2 * [√3.142 * (6*10^-3/2)]
σ = 26 / 2 * √(3.142 *3*10^-3)
σ = 26 / 2 * √0.009426
σ = 26 / 2 * 0.0971
σ = 26 / 0.1942
σ = 133.88 MPa
