Answer:
endurance length is 236.64 MPa
Explanation:
data given:
d = 37.5 mm
Sut = 760MPa
endurance limit is
Se = 0.5 Sut
= 0.5*760 = 380 MPa
surface factor is
Ka = a*Sut^b
where
Sut is ultimate strength
for AISI 1040 STEEL
a = 4.51, b = -0.265
Ka = 4.51*380^{-0.265}
Ka = 0.93
size factor is given as
Kb =1.29 d^{-0.17}
Kb = 0.669
Se = Sut *Ka*Kb
= 380*0.669*0.93
Se = 236.64 MPa
therefore endurance length is 236.64 MPa
Answer:
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Answer:
Q=486.49 KJ/kg
Explanation:
Given that
V= 0.2 m³
At initial condition
P= 2 MPa
T=320 °C
Final condition
P= 2 MPa
T=540°C
From steam table
At P= 2 MPa and T=320 °C
h₁=3070.15 KJ/kg
At P= 2 MPa and T=540°C
h₂=3556.64 KJ/kg
So the heat transfer ,Q=h₂ - h₁
Q= 3556.64 - 3070.15 KJ/kg
Q=486.49 KJ/kg
Answer:
The stand-by equipment is technically required in case where the we need some urgent equipment for the purpose of maintenance in emergency and if the other equipment system get fails. the term stand by means backup equipment and component.
In the reliability engineering, we always need to provide an extra equipment or component in case of emergency. It is basically used so that it does not affect any type of productivity in the organization. It is also increase the redundancy of the equipment.