# Isn't the flow work in the first law of thermodynamics calculated twice?

So, from what I know, (assuming there are no changes in kinetic and potential energies, etc.) the first law of thermodynamics says that

$\dot Q= \dot W + \dot H_2-\dot H_1$

(Dots above letters indicate a flow.) Enthalpy is defined as $h_1=u_1+p_1v_1$, where $pv$ is flow work - work needed to push the fluid into the system (and later out of the system, $h_2=u_2+p_2v_2$).

And, the work in an open system is the negative integral of $vdp$: ($-\int vdp$). But, how did we get to that $vdp$? Well, we have work done by expansion, $pdv$ (like in closed systems), plus the work needed to push the fluid in $p_1v_1$, minus the work needed to push it out $p_2v_2$. Graphically this is the area to the left of the curve, as I already mentioned, the integral of $vdp$.

What confuses me is, aren't we calculating that push-in push-out work twice? Once in the enthalpy and then again in the work?

• $\dot Q$ is written by typing dollarsign \dot Q dollarsign where dollarsign = $. See this reference on MathJax for entering equations: docs.mathjax.org/en/latest/start.html Mar 8, 2017 at 2:06 • What is your source for the statement you have there that "work in an open system is negative integral of vdp". At least that seems a special case. Try this learnengineering.org/2013/03/… Mar 8, 2017 at 3:07 • The source is my textbook, it has a picture like the one I posted in my question and this formula: w [J/kg]=p1v1+ integ(pdv) - p2v2 and it says, translated, that in this formula we can see the work done to push the fluid in and out, as well as the work done due the the change in volume. Mar 8, 2017 at 13:56 ## 2 Answers In this steady state, open system (control volume) version of the 1st law of thermodynamics,$\dot{W}\$ is the "shaft work," and does not include the work to push fluid into and out of the system. That is, the work in this version of the 1st law is separated into two distinct parts, the shaft work and the work to push fluid into and out of the system. And the latter is lumped together with the enthalpy of the entering and exit streams.

• How would you then calculate the shaft work? Mar 8, 2017 at 13:52
• It depends on the particular situation. Some examples would be turbines or compressors. Incidentally, the shaft work for an open system is vdp only if the system is operating adiabatically and reversibly. Mar 8, 2017 at 14:35

Assuming an adiabatic process, the 1st Law reads $$\dot H_2- \dot H_1 = \dot W_S$$. I'm using the sign convention that work added to a system is positive, and the subscript '$$S$$' stands for shaft work. In terms of per-mass-flow-rate quantities, you have $$h_2- h_1 = w_S$$.

The difference in enthalpies $$h$$ between states 1 and 2 is re-written as $$\int_1^2 d(u+pv) = u_2-u_1 + \int_1^2 p \,dv + \int_1^2 v \,dp$$ For a reversible, adiabatic (isentropic) state change from 1 to 2 (this does not include flow work!), the balance for internal energy reads $$u_2-u_1 = -\int_1^2 p \,dv$$ i.e. there is a change in internal energy of the medium due to volume change work. The energy balance for the expansion process thus reads $$\int_1^2 v \,dp = w_S$$ i.e. the shaft work generated by a turbine, say, is equal to pressure change work.

Returning to your question: the push-in/push-out work, a.k.a. flow work, is not counted twice. You should interpret the equation $$h_2- h_1 = w_S$$ as follows:

• internal energy and flow work balances with shaft work

Or – considering that the change in internal energy of the medium results from volume change work:

• volume change work and flow work balances with shaft work

It seems to me that you ended up with a wrong conclusion, because you started with the statement "work in an open system is the negative integral of $$v \, dp$$" (never mind the sign convention), which is the result of the analysis, but not its starting point.