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October 27, 2000, 05:12 
Isentropic expansion

#1 
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In producing supersonic flow in a nozzle by assuming no friction or heat loss how does the flow become isentropic? No hat loss should mean adiabatic but then why do they call it isentroic also and say the entropy along a path is constant?


October 27, 2000, 07:03 
Re: Isentropic expansion

#2 
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Did you study the second principle of thermodynamics ?
dS=dQ/T 

October 27, 2000, 08:45 
Re: Isentropic expansion

#3 
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An isentropic flow is a reversible adiabatic flow. When we say isentreopic flow we also mean that the flow is smooth (reversible). If the flow is merely adiabatic, there is certainly no heat exchange with the surroundings but the entropy of the particles can certainly change if there are shocks for example. This is also true of a thermally insulated nozzle flow with friction: flow is adiabatic but not isentropic because it is not reversible.
Ravi 

October 27, 2000, 08:51 
Re: Isentropic expansion

#4 
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In your equation, I am missing one term which corresponds to thermalenergy production due to viscous dissipation (w_viscous).
ds=dq/T + w_viscous/T By this equation, isentropic flow means 1.) adiabatic walls and no vicous effects (friction) or 2.) rate of thermalenergy production by viscous dissipation equals the the rate of heat through the walls. 

October 27, 2000, 09:36 
Re: Isentropic expansion

#5 
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Hi,
no shocks, no friction = no dissipation The corollary of second principle of thermodynamics (known as principle of increase of entropy S) says: (dS)_adiabatic >= 0 The equality sign holds when no dissipation is involved (i.e. reversible process). If dissipation plays a role, then the inequality applies. Ravi said the same using different words. Cheers Maurizio 

October 28, 2000, 10:44 
Re: Isentropic expansion

#6 
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(1). Flow through a supersonic nozzle in real world will have viscous loss through the boundary layer, along with the wall heat transfer, not to mention the shock waves in the supersonic flow section and the exit section. (2). So, the real world problem is not easy to solve. (3). In the hypothetical (idealized, or theoretical) world, these effect can be ignored mathematically, one by one. And in the thermodynamic hs diagram, you can plot a point on it based on the assumed state of the inlet flow (uniform, constant pressure, enthalpy, etc.) then draw a vertical line (constant entropy line)from the point. Each point along this line will represent the end state of the flow in an isentropic process between the inlet and the end state. (4). The process along the constants line is called isentropic process. It is "reversible" because you can move from stateA to stateB, then back to stateA. It is not real and it does not exist. It is there only on paper in hs diagram. So, you can not have wall boundary layer, viscous loss, wall heat transfer, internal mixing, internal shock waves,.... So, it has nothing to do with the real world. It is almost useless. (5). But, it will give you something simple to work with, as a guide or reference.


October 30, 2000, 07:48 
Re: Isentropic expansion

#7 
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I miss nothing in my equation since 98.4 F says that the flow is frictionless. Moreover he's not talking about the presence of shocks in the nozzle field so I didn't find the necessity to talk about it. Bye


November 4, 2000, 16:57 
Re: Isentropic expansion

#8 
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I think the question to ask yourself is not what makes it "become isentropic," but what makes it "become nonisentropic", i.e., what would be the possible modes of entropy production? If there are none, its an isentropic process. Some modes of entropy production are heat transfer, turbulence, and shock waves.


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