Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
=
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
=
...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432
Answer:
A)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
Explanation:
Given data :
45.00 % mole of methane
55.00 % of ethane
attached below is a detailed solution
A) calculate - Q(kw)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B ) calculate - Q ( KJ )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
since combustion takes place in a constant-volume batch reactor
Answer:
Stratification is the process of arranging the different elements classification in a specific manner.
In the storage tank, hot water and cold water create a specific layer and coexist between them for the process of stratification. Thermal storage and heat storage exchanger are the types of storage tank, in which the stratification process take place.
The density of the cold water is more as compared to hot water. Then, due to this the cold water sink to the bottom and hot ware rises up in the storage vessel. In this way, the water gets heated and the stratification process take place in the tank.