Explanation:
Superheater has two types of parts which are:
- The primary super-heater
- The secondary super-heater
Primary super-heater is first heater which is passed by the steam after steam comes out of steam drum.
After steam is heated on super primary heater, then the steam is passed on secondary super-heater so to be heated again. Thus, on secondary super-heater, the steam formed is hottest steam among others.
Steam from secondary super-heater which becomes the superheated steam, flow to rotate the High-Pressure Turbine.
Answer:
<u>note:</u>
<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>
Answer:
Change in length = 0.1257 mm
Change in diameter= -0.03771mm
Explanation:
Given
Diameter, d = 15 mm
Length of rod, L = 200mm
F = Force= 300N
d = 0.015m
Ep=2.70 GPa, np=0.4.
First, we have to calculate the normal stress using
σ = F/A where F = Force acting on the Cross-sectional area
A = Area
Area is calculated as πd²/4 where d = 0.015m
A = 22/7 * 0.015²/4
A = 0.000176785714285m²
A = 1.768E-4m²
So, stress. σ = 300N/1.768E-4m²
σ = 1696832.579185520Pa
σ = 1.697MPa
Calculating E(long)
E(long) = σ /Ep
E(long) = 1.697E-3/2.70
E(long) = 0.0006285
At this point, we fan now calculate the change in length of the element;
∆L = E(long) * L
∆L = 0.0006285 * 200mm
∆L = 0.1257mm
Calculating E(lat)
E(lat) = -np * E(long)
E(lat) = -4 * 0.0006285
E(lat) = -0.002514
At this point, we can now calculate the change in diameter of the element;
∆D = E(lat) * D
∆L = -0.002514 * 15mm
∆L = -0.03771mm
