Take caution, and slow down, it could run out in the middle of the road. Try going around it if possible.
Answer:
No it is not a problem
Explanation:
It is not a problem because the stress intensity factor K would approach infinity as you get close to a crack tip and the intensity factor would approach Zero as you get too far away from the crack tip and this is simply because a crack is a notch with zero tip radius .
and The application of stress intensity factor k in respect to present fatigue crack tip is termed " linear elastic fracture mechanics "
Answer: the answer for the question is a
Explanation:
Answer:
Given:
P₁ = 500 kPa
T₁ = 860 K
P₂= 100 kPa
T₂ = 460 K
Let's take entropy properties of T1 and T2 from ideal properties of air,
at T = 860K, s(T₁) = 2.79783 kJ/kg.K
at T = 460K, s(T₂) = 2.13407 kJ/kg.K
using entropy balance equation:
![\frac{\sigma _cv}{m} = s(T_2)- s(T_1) - R In [\frac{P_2}{P_1}]](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Csigma%20_cv%7D%7Bm%7D%20%3D%20s%28T_2%29-%20s%28T_1%29%20-%20R%20In%20%5B%5Cfrac%7BP_2%7D%7BP_1%7D%5D%20)
= - 0.2018 kJ/kg. K
In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).