Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
Answer:
Explanation:
Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

Obtain the following properties at 10kPa from the table "saturated water"

Calculate the enthalpy at exit of the turbine using the energy balance equation.

Since, the process is isentropic process 

Use the isentropic relations:

Calculate the enthalpy at isentropic state 2s.

a.)
Calculate the isentropic turbine efficiency.

b.)
Find the quality of the water at state 2
since
at 10KPa <
<
at 10KPa
Therefore, state 2 is in two-phase region.

Calculate the entropy at state 2.

Calculate the rate of entropy production.

since, Q = 0

It’s D. This is because having oil changes often, makes the care for your car better. I hope this helps.