Mechanical Engineering - Steam Nozzles and Turbines - Discussion

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

[A]. 0.4
[B]. 0.56
[C]. 0.67
[D]. 1.67

Answer: Option A

Explanation:

No answer description available for this question.

Umakant Sahu said: (Jun 12, 2013)  
Degree of reaction = isentropic enthalpy drop in moving blade of turbine/isentropic enthalpy drop in fixed blad of turbine.

Guddi Kumari said: (Dec 18, 2013)  
Yes, degree of reaction=isentropic entropy drop in moving blade of turbine/enthalpy drop in fixed blade turbine.

Lakshmi Kant said: (Jan 17, 2014)  
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.

Nikunj said: (Feb 20, 2015)  
How its 0.4?

Kavishwar Gaikwad said: (Mar 10, 2015)  
DoR = Enthalpy drop in moving/enthalpy drop in moving+enthalpy drop in fixed blade.

M.Bhavani Ganesh said: (May 31, 2015)  
= (2/3)/((2/3)+1) = 2/5 = 0.4.

Lav said: (Mar 22, 2016)  
Rd = Hm/(Hf+Hm).

So,
Rd = (2/3)Hf/{(2/3)Hf+Hf}.
= 0.4.

Because, Given that.
Hm = 2/3 Hf.

Sharad said: (Jul 19, 2016)  
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.

Vijay Prajapati said: (Dec 22, 2016)  
Your explanation is right @Lav.

Kanji Dodiya said: (Mar 18, 2017)  
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.

Ramu said: (Feb 13, 2018)  
How fixed blade is 1?

Basit said: (Apr 14, 2018)  
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.

R.Bhavanisankar said: (Mar 28, 2019)  
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.

Gaurav said: (Nov 4, 2019)  
@Kanji and @Bhavanishankar are correct.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.