Mechanical Engineering - Steam Nozzles and Turbines - Discussion
Discussion Forum : Steam Nozzles and Turbines - Section 1 (Q.No. 11)
11.
The isentropic enthalpy drop in moving blade is two-third of the isentropic enthalpy drop in fixed blades of a turbine. The degree of reaction will be
Discussion:
15 comments Page 1 of 2.
Basit said:
7 years ago
DOR = enthalpy drop in rotor/(total energy drop in rotor).
Total energy = enthalpy drop in rotor + kinetic energy drop in rotor
Enthalpy drop in rotor = 2/3 enthalpy drop in stator ( nozzle or fixed blades) ===>>equ i
Kinetic energy drop in rotor is equal to kinetic energy gain in stator which is equal to enthalpy drop in stator.
So,
The kinetic energy drop = enthalpy drop in stator ====>> equ ii.
So we can solve and get an answer = .4.
Total energy = enthalpy drop in rotor + kinetic energy drop in rotor
Enthalpy drop in rotor = 2/3 enthalpy drop in stator ( nozzle or fixed blades) ===>>equ i
Kinetic energy drop in rotor is equal to kinetic energy gain in stator which is equal to enthalpy drop in stator.
So,
The kinetic energy drop = enthalpy drop in stator ====>> equ ii.
So we can solve and get an answer = .4.
Lakshmi Kant said:
1 decade ago
Degree of reaction or reaction ratio (R) is defined as the ratio of static pressure drop in the rotor to the static pressure drop in the stage or as the ratio of static enthalpy drop in the rotor to the static enthalpy drop in the stage.
i.e. R = 0.4 for this problem.
i.e. R = 0.4 for this problem.
Sharad said:
9 years ago
0.4 is the right answer.
D.R = Change of enthalpy in moving the blade to the change of enthalpy of moving blade plus fix blade.
Give that change of enthalpy of moving blade 2/3 of fixed blade,
So, D.R = 0.4.
D.R = Change of enthalpy in moving the blade to the change of enthalpy of moving blade plus fix blade.
Give that change of enthalpy of moving blade 2/3 of fixed blade,
So, D.R = 0.4.
Kanji dodiya said:
8 years ago
Degree of reaction = isentropic enthalpy drop in moving blade/(isentropic enthalpy drop in fixed blade+isentropic enthalpy drop in moving blade).
= (2/3)/[1+(2/3)].
= (2/3)/(5/3),
= 2/5,
= 0.4 ans.
= (2/3)/[1+(2/3)].
= (2/3)/(5/3),
= 2/5,
= 0.4 ans.
R.Bhavanisankar said:
6 years ago
Degree of reaction = isentropic enthalpy drop in moving blade/(isentropic enthalpy drop in fixed blade+isentropic enthalpy drop in moving blade).
= (2/3)/[1+(2/3)].
= (2/3)/(5/3),
= 2/5,
= 0.4 ans.
= (2/3)/[1+(2/3)].
= (2/3)/(5/3),
= 2/5,
= 0.4 ans.
Manikanta said:
1 year ago
The degree reaction is,
Enthalpy dropped in moving blades = 2,
Total enthalpy dropped in turbine = 3 (total enthalpy dropped 2+3=5//
Then 2/5. =0.4.
Enthalpy dropped in moving blades = 2,
Total enthalpy dropped in turbine = 3 (total enthalpy dropped 2+3=5//
Then 2/5. =0.4.
Umakant sahu said:
1 decade ago
Degree of reaction = isentropic enthalpy drop in moving blade of turbine/isentropic enthalpy drop in fixed blad of turbine.
LAV said:
9 years ago
Rd = Hm/(Hf+Hm).
So,
Rd = (2/3)Hf/{(2/3)Hf+Hf}.
= 0.4.
Because, Given that.
Hm = 2/3 Hf.
So,
Rd = (2/3)Hf/{(2/3)Hf+Hf}.
= 0.4.
Because, Given that.
Hm = 2/3 Hf.
Guddi kumari said:
1 decade ago
Yes, degree of reaction=isentropic entropy drop in moving blade of turbine/enthalpy drop in fixed blade turbine.
Kavishwar gaikwad said:
1 decade ago
DoR = Enthalpy drop in moving/enthalpy drop in moving+enthalpy drop in fixed blade.
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