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.
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.
GAURAV said:
6 years ago
@Kanji and @Bhavanishankar are correct.
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.
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.
Ramu said:
8 years ago
How fixed blade is 1?
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.
Vijay prajapati said:
9 years ago
Your explanation is right @Lav.
(1)
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.
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.
M.Bhavani Ganesh said:
1 decade ago
= (2/3)/((2/3)+1) = 2/5 = 0.4.
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