What is the downstream temperature after throttling air?

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Discussion Overview

The discussion revolves around the thermodynamic analysis of throttling air from a higher pressure to a lower pressure, specifically focusing on the downstream temperature, entropy change, rate of entropy generation, and lost work. The context includes theoretical and practical implications of throttling processes in thermodynamics.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that for an ideal gas undergoing throttling through an adiabatic valve, the change in enthalpy is zero, implying no change in temperature.
  • Another participant questions the assumption of adiabatic conditions and points out that if the temperature does not change, the process could be considered both adiabatic and isothermal.
  • A later reply clarifies that throttles are typically approximated as adiabatic and explains that the expansion cooling is balanced by viscous heating, resulting in no net temperature change.
  • There is a query regarding the entropy change for the surroundings, with a participant suggesting that the focus should be on the rate of entropy generation within the throttle itself.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions of adiabatic and isothermal conditions during throttling, and there is no consensus on the implications of these assumptions for the analysis of entropy changes.

Contextual Notes

Participants have not fully resolved the assumptions regarding the nature of the throttling process and its impact on temperature and entropy generation. The discussion highlights the complexity of applying thermodynamic principles to real-world scenarios.

Who May Find This Useful

This discussion may be of interest to students and professionals in thermodynamics, particularly those studying fluid mechanics and energy systems involving throttling processes.

gfd43tg
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Homework Statement


Ten kmol per hour of air is throttled from upstream conditions of 25°C
and 10 bar to a downstream pressure of 1.2 bar. Assume air to be an ideal gas with Cp= (7/2)R.

(a)What is the downstream temperature?
(b)What is the entropy change of the air in J mol-1K-1?
(c)What is the rate of entropy generation in W K-1?
(d)If the surroundings are at 20°C, what is the lost work?


Homework Equations





The Attempt at a Solution


For my entropy balance, I am not sure how to get dS_surr/dt
 

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Maylis said:

Homework Statement


Ten kmol per hour of air is throttled from upstream conditions of 25°C
and 10 bar to a downstream pressure of 1.2 bar. Assume air to be an ideal gas with Cp= (7/2)R.

(a)What is the downstream temperature?
(b)What is the entropy change of the air in J mol-1K-1?
(c)What is the rate of entropy generation in W K-1?
(d)If the surroundings are at 20°C, what is the lost work?


Homework Equations





The Attempt at a Solution


For my entropy balance, I am not sure how to get dS_surr/dt

If air is an ideal gas, and it's passing through an adiabatic throttle valve, the change in enthalpy per mole is zero, and thus the change in temperature is zero. You used the correct formula to get the change in entropy per unit mass, but you need to correct the temperature term (which is now zero). The rate of entropy generation is just the entropy change per mole times the flow rate.

Chet
 
How do you know its adiabatic? Also, if the temperature doesn't change, this is both an adiabatic and isothermal process??
 
Last edited:
Maylis said:
How do you know its adiabatic? Also, if the temperature doesn't change, this is both an adiabatic and isothermal process??
Throttles are usually approximated as adiabatic. Regarding adiabatic and isothermal, yes this change for the air is both. What happens mechanistically is that the expansion cooling to the lower pressure is precisely balanced by the viscous heating in the throttle valve. The net effect is no temperature change. Look up throttling in your thermo book under the version of the first law applicable to open systems operating at steady state.

Chet
 
why is dS/dt for the surroundings zero?
 
Maylis said:
why is dS/dt for the surroundings zero?

Because the question really implies "what is the rate of entropy generation within the throttle."

Chet
 

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