Pressure drop w.r.to time through a orifice dia(D) in a container of volume (V)

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SUMMARY

The discussion focuses on calculating the pressure drop over time through an orifice in a container with a volume of 100,000 cubic cm and an initial pressure of 10 bar, aiming to determine when the pressure will reach 5 bar. The analysis involves concepts of adiabatic expansion and conservation of energy, using the equation P(t+dt)(V+Af.dt)γ = P(t)Vγ, where γ is the specific heat ratio (1.4 for gases). The user also considers the effects of ambient pressure (1 bar) on the calculations, highlighting the importance of accurately accounting for all variables in pressure drop scenarios.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law and specific heat ratios
  • Knowledge of fluid dynamics, particularly flow through orifices
  • Basic calculus for differential equations and rate of change analysis
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  • Study the principles of adiabatic expansion in gases
  • Learn about the ideal gas law and its applications in pressure calculations
  • Research fluid dynamics related to orifice flow and pressure drop
  • Explore conservation of energy principles in thermodynamic systems
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Engineers, physicists, and students in fluid mechanics or thermodynamics who are analyzing pressure changes in gas systems or designing containers with orifices.

Jothiram80
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Let initial preasure (P) = 10 bar,volume of container (V)=100000 cubic cm.
At what time period pressure will become 5 bar.
 
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Jothiram80 said:
Let initial preasure (P) = 10 bar,volume of container (V)=100000 cubic cm.
At what time period pressure will become 5 bar.
What kind of a substance is in the container? A gas or a liquid? If it is a gas, is it an ideal gas? Is it escaping against atmospheric pressure of 1 bar?

AM
 
I make no guarantees for the correctness of this, but here's what I got.
Let vol of tank be V, pressure P(t), mass remaining in tank m(t), outlet area A, exit velocity f(t).
In time dt, a volume A*f.dt, mass (Afm/V).dt, exits.
Assuming adiabatic expansion:
P(t+dt)(V+Af.dt)γ = P(t)Vγ, where γ = 1.4
whence
P' = -PγAf/V
m' = -fAm/V
By conservation of energy
(P-Pa)Af.dt = Af3m.dt/2V
where Pa is ambient pressure.
I forgot about Pa at first, and obtained
(γ-1)At = ((P0/P)(1-γ)/2γ - 1)√(2Vm0/P0)
where P0 = P(0), m0 = m(0).
I might try to correct that tomorrow.
 

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