Adiabatic steady state flow in a nozzle

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Discussion Overview

The discussion revolves around the calculation of exit temperature in an adiabatic nozzle flow, focusing on the application of energy balance equations and pressure-temperature relations. Participants explore the conditions under which these equations are valid, particularly in the context of isentropic versus non-isentropic flows.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents an energy balance equation to find the exit temperature of air in an adiabatic nozzle, yielding a temperature of 184 degrees Celsius.
  • The same participant questions the validity of a pressure-temperature relation, noting a significant discrepancy in temperature results when applying it, and asks if it is only valid for reversible processes.
  • Another participant clarifies that the energy balance equation is applicable for inviscid adiabatic flows and does not require reversibility, while the pressure-temperature relation is only valid for isentropic flows.
  • A further response emphasizes that the energy balance equation always works under certain assumptions, while the pressure-temperature relation requires specific conditions such as isentropic flow and constant specific heats.
  • It is noted that the participant had inverted the terms in the pressure-temperature relation.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the pressure-temperature relation and the conditions required for its use. There is no consensus on the validity of the equations in the context presented, indicating ongoing debate.

Contextual Notes

The discussion highlights limitations regarding the assumptions necessary for the equations used, particularly the need for isentropic conditions in the pressure-temperature relation and the implications of non-ideal flow conditions.

jason.bourne
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air enters an adaibatic nozzle steadily at 300 kPa, 200 degree celsius and 30m/s and leaves at 100 kPa and 180 m/s.
find the exit temperature?

when m using energy balance equation: h1 + [ (v1)^2 / 2 ] = h2 + [(v2)^2 / 2 ]
i get the answer 184 degree celsius.

but when i use the other relation i.e, pressure temperature relation

[ T1 / T2 ] = [ P1 / p2 ]^ ((k-1)/k) (i used k = 1.4)

i get very different values of temperature. is this equation not valid?
why m i getting different answers?

is this pressure temperature relation only valid for reversible processes?
 
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The first equation (with h1 and h2 in it) works for inviscid adiabatic flows (with no work done on the fluid by propellers, etc). It does not require flow to be reversible. As a result, it will work even across the shock wave.
The second equation works only for isentropic flows - i.e. reversible, adiabaric flows.

Thus, these two equations will yield the same answers (assuming p1, p2, h1, h2 , T1 are somehow known or given) only for isentropic flows.
 
jason.bourne said:
air enters an adaibatic nozzle steadily at 300 kPa, 200 degree celsius and 30m/s and leaves at 100 kPa and 180 m/s.
find the exit temperature?

when m using energy balance equation: h1 + [ (v1)^2 / 2 ] = h2 + [(v2)^2 / 2 ]
i get the answer 184 degree celsius.

but when i use the other relation i.e, pressure temperature relation

[ T1 / T2 ] = [ P1 / p2 ]^ ((k-1)/k) (i used k = 1.4)

i get very different values of temperature. is this equation not valid?
why m i getting different answers?

is this pressure temperature relation only valid for reversible processes?

The first equation is the based on the full energy balance (with some assumptions that drop out some terms). It always works.

The second is for an Ideal gas, isentropic, and constant specific heat assumption.

Based on your problem description, the system does not meet the criteria for the second relation. If your system was given as isentropic, ideal gas, with constant specific heats, then you could use the second relation you listed.

BTW you have the T1/T2 and P1/P2 inverted (T2 and P2 are the numerators).

CS
 
thank you so much
 

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