Isentropic work in closed versus open systems 2

[AthlonXPme]

I am preparing the Chemical PE and found these is a minor difference in calculating the isentropic work for closed versus open system in the reference book. The formula for isentropic work in closed system misses the specific heat ratio k on the numerator as compared with the formula for open system. What's the reason behind ? I thought the formula should be the same for both systems.

 

[1503-44]

Probably something to do with

Closed system = no mass transfer
Open system allows mass transfer

Specific heat capacity
To find the molar specific heat capacity of the gas involved, the following equations apply for any general gas that is calorically perfect. The property γ is either called the adiabatic index or the heat capacity ratio. Some published sources might use k instead of γ.

https://en.m.wikipedia.org/wiki/Heat_capacity_ratio

is the ratio of the heat capacity at constant pressure (CP) to heat capacity at constant volume (CV). It is sometimes also known as the isentropic expansion factor and is denoted by γ (gamma) for an ideal gas[note 1] or κ (kappa), the isentropic exponent for a real gas. The symbol γ is used by aerospace and chemical engineers.

𝛾=𝐶p/𝐶v =-(dP/P)/(dV/V)
At constant vol, dV=0, 𝛾=○○


Molar isochoric specific heat:
(Vol remains constant)
Cv =𝑅/(𝛾−1)

Molar isobaric specific heat:
(Press remains constant)
C𝑝=𝛾 𝑅/(𝛾−1)

PVⁿ= constant

https://en.m.wikipedia.org/wiki/Isobaric_process

Isothermal n=1
PV¹= constant

Isobaric n=1
PV⁰= constant

Polytropic 1<n<○○
PV^1o= constant

Isochoric
Vol = constant
PV^oo= constant

Isentropic
Entropy =constant
n=𝛾
PV^𝛾= constant

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Heaviside1925]

I may be off base here, but when I learned thermo, I look at these equations as 2nd law statements. Based on your equations shown, the statement is, that it is possible for more work to be done in an open system as opposed to a closed in an isentropic system. So that leads me to the question of, where would this addition work come from in an isentropic system. If you look at the T-S diagram below and follow the lines of constant entropy and move from one isobar to the next, you have changed the enthalpy. To me it seems intuitive that one would expect a higher delta work in an open system than a closed, which is what the addition heat capacity ratio would seem to imply. In other words, there is an additional exchange of enthalpy that is occurring.

 

[Snickster]

In a closed system the work on compressing a non-flowing fluid is the integral of (P dV) = Force x Distance = P*A*dX

Where PV^k = Constant or P = C/V^k

So the integral is (C/V^k)dV - The solution to this integral is the equation for closed flow in your table.

In an open system the work is the integral of (V dP). This can be seen by the steady flow energy equation as follows:

TdS - W = dU + d(PV) + dKE + dPE = dU + PdV +VdP + dKE + dPE

But in a non-flow closed system we know that any heat added or removed Q = TdS = dU + PdV which is also true for a flowing system - so in a process with no heat added process such as an isentropic compression as the pressure increases and the dU increases due to compression, the integral of PdV is negative, because V2 is smaller than V1, but of equal value as positive dU so they all cancel - so PdV is still the pure compression work (that goes into dU) but VdP is the total work of compression and pushing the gas out of the cylinder once the compression has been completed, and pulling new gas back into the cylinder for compression - note that PdV is called the area under the curve and VdP is called the area behind the curve in thermo books.

so TdS and dU + PdV cancels out of the above equation for steady flow and the resulting steady flow equation is:

VdP = dKE + dPE + W

Typically change in kinetic energy and potential energy are negligible so the equation reduces to:

Integral of VdP = Work Steady Flow

The soliution of this integral is the equation shown for an open system in your table.

 

[1503-44]

Very nice explanation. Makes perfect sense. Thanks.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."