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

Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean : 
Standard deviation : 
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula,
, the z-value corresponds to 2.39 will be :-

z-value corresponds to 2.60 will be :-

Using the standard normal table for z, we have
P-value = 

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%
Answer:
A working with machinery be a common type of caught-in and caught-between hazard is described below in complete detail.
Explanation:
“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ =
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm