For the unity negative feedback system G(s) = K(s+6)/ (s + 1)(s + 2)(s + 5) It's known that the system is operating with a domin
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Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 /
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
=
given that = 480 mm
we substitute
=
×
= 501.998 mm
Now we find the elongation;
Elongation =
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
Now, we will find (enthalpy at the outlet for the isentropic process):
Now, the isentropic efficiency of the turbine can be given as follows: