Air undergoes a polytropic process in an adiabatic nozzle with n = 1.2. Inlet state of air is 800 kPa, 123^{0}C with a velocity of 32 m/s and exit state is 300 kPa. Find out the velocity of air at nozzle exit.

- a)654 m/s
- b)348 m/s
- c)244 m/s
- d)436 m/s

Correct answer is 'B'. Can you explain this answer?

Related Test: Basic Thermodynamics

By
Ganesh Sundarapalli
·
Sep 19, 2018 ·Mechanical Engineering

6 Answers

Sabir Hussain
answered
Jul 20, 2018

From P, T, n relationship, we can find T2. So dh can be found out using CpdT. Using dh value in SFEE exit velocity will be found out as 347.94 m/s

Vivek Sharma
answered
Jun 12, 2018

Just equate : V2^2 + V1^2 = 2Ã—CpÃ—(T1-T2) and T2 you will get from adiabatic equation as 340 K

Rapuru Bhavanisankar
answered
Aug 20, 2020

Exit velocity of air=348m/s

proof: For polytropic process

T2/T1 ={p2/p1}^(n-1/n)

T2=

here n=1.2 (polytropic index) given,

p1 and p2 are also given which are initial and final pressure respectively,

initial temperature t1=123 C

T1=123+273=396K

from above formula we get final temperature T2=336.28K

h1-h2=((v2^2)/2000)-((v1^2)/2000)

here v is velocity

h1-h2=mCpâˆ†T=60.02kJ

Cp=1.005 for air

By above formulae

velocity V2=347.94m/s

proof: For polytropic process

T2/T1 ={p2/p1}^(n-1/n)

T2=

here n=1.2 (polytropic index) given,

p1 and p2 are also given which are initial and final pressure respectively,

initial temperature t1=123 C

T1=123+273=396K

from above formula we get final temperature T2=336.28K

h1-h2=((v2^2)/2000)-((v1^2)/2000)

here v is velocity

h1-h2=mCpâˆ†T=60.02kJ

Cp=1.005 for air

By above formulae

velocity V2=347.94m/s

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