Answer:
Explanation:
From the question we are told that:
Thickness
Internal Pressure
Shear stress
Elastic modulus
Generally the equation for shear stress is mathematically given by
Where
r_i=internal Radius
Therefore
Generally
Generally the equation for outer diameter is mathematically given by
Therefore
Assuming that the thin cylinder is subjected to integral Pressure
Outer Diameter is
Answer:
The mass flow rate of steam m=5.4 Kg/s
Explanation:
Given:
At the inlet of turbine P=10 MPa ,T=500 C
AT the exit of turbine P=10 KPa ,x=0.9
Required power=5 MW
From steam table
<u> At 10 MPa and 500 C:</u>
h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)
<u>At 10 KPa:</u>
=2675.1 KJ/Kg,
=417.51 KJ/Kg
= 7.3 KJ/Kg-K ,
=1.3 KJ/Kg-K
So enthalpy of steam at the exit of turbine
h=
Now by putting the values
h= 417.51+0.9(2675.1- 417.51) KJ/Kg
h=2449.34 KJ/Kg
Lets take m is the mass flow rate of steam
So
m=5.4 Kg/s
So the mass flow rate of steam m=5.4 Kg/s