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:

Answer:
The condition does not hold for a compression test
Explanation:
For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.
<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test
Answer:
Sell his crop, use his crop as food, and sell his crop
Explanation:
Answer:
maximum stress is 2872.28 MPa
Explanation:
given data
radius of curvature = 3 ×
mm
crack length = 5.5 ×
mm
tensile stress = 150 MPa
to find out
maximum stress
solution
we know that maximum stress formula that is express as
......................1
here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack
so put here all value in equation 1 we get
σm = 2872.28 MPa
so maximum stress is 2872.28 MPa
Answer:
The primary piston activates one of the two subsystems. The hydraulic pressure created, and the force of the primary piston spring, moves the secondary piston forward.