# Thermo Q RE:isentropic efficiency through nozzles

1. Aug 5, 2007

### nickbone59

1. The problem statement, all variables and given/known data
Exhaust gas expands through a nozzle whose isentropic efficiency is 88%. The inlet pressure and temperature are 1.6 bar and 400c respectively. At outlet the pressure has fallen to an ambient level of 1.013 bar. Given that the inlet velocity is negligible, and Cp and $$\gamma$$are 1.15kJ/kgK and 1.3 respectively, determine;

i)the outlet pressure and velocity
ii)the rate of generation of entropy, if the mass flowrate is 2.8 kg/s

2. Relevant equations
s2-s1=cp ln (T2/T1)-R ln(P2/P1)

3. The attempt at a solution
Outlet Temp

T2s/T1=(p2/p1)^($$\gamma$$-1/$$\gamma$$)
Therefore,
T2s=673x(1.013/1.6)^(0.23076923)
=605.6266K

T2-673= - 67.373/0.88= -76.56

Therefore,
T2=673+-76.56
= 596.44K

This doesn't seem right? Is it just the first part of the formula i use, to get T2s?

After this i am stuck. Any help with this will be greatly appreciated!!

2. Oct 10, 2008

### deepthishan

to find outlet velocity (i dont understand how to 'find' outlet pressure since uve already stated it!):

T2s/T1=(p2/p1)^(-1/)
Therefore,
T2s=673x(1.013/1.6)^(0.23076923)
=605.6266K

for isentropic conditions (i.e. no external heat transfer),

Cp*T1 + .5*u1^2 = Cp*T2i + .5*u2i^2
since u1=0,
u2i=420.0m/s

.88=u2^2/u2i^2
therefore, u2=394.0m/s

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3. Oct 11, 2008

### deepthishan

p/rho=R*T

by bernoulli's principle,

u1^2/2 + R*T1 = u2^2/2 + R*T2
u2=394.0m/s (from previous solution)

T1= 1213.4K