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I am a little confused, from the first law of thermodynamics (energy conservation)

$$\Delta E = \delta Q - \delta W $$

If the amount of work done is a volume expansion of a gas in, say a piston cylinder instrument at constant pressure,

$$\Delta E = \delta Q - pdv$$

Here $p$ is the constant pressure and $dv$ is the change in (specific) volume.

So, when do I take into account

$$\delta W = d(pv) = pdv + vdp$$

I am assuming that for cases of boundary work, at constant pressure, the $vdp$ term is zero.

So under what conditions should I consider the $vdp$ term?

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Where are you getting $\delta W = d(pv)$ from? I'm skeptical that it's true since it would seem to imply that if you were to increase the temperature of an ideal gas sample in a closed, rigid box (and thus increase its pressure while keeping volume constant), then the gas would do work, but a gas does no work in such a situation. –  joshphysics Jan 23 '13 at 20:04
    
@joshphysics why don't I see what you see? So when do we use vdp? I was always under the impression that $$\delta W = d(pv) = p dv + vdp$$. Generaly for constant pressure processes we would neglect $$vdp$$... –  drN Jan 23 '13 at 20:08
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I am also skeptical about $\delta W=d(pv)$, since work is not an exact differential form (that's why you write $\delta$ instead of $d$), while $d(pv)$ is exact: it's the differential of $pv$, which is a state function. –  Bzazz Jan 23 '13 at 20:14
    
@Bzazz you are right. so, $$\delta W = p dv$$ with an integral from state 1 to 2 is correct? –  drN Jan 23 '13 at 20:17
    
Usually it is in systems like the one you mentioned. I don't have exceptions in mind at the moment, but I'm not an expert. –  Bzazz Jan 23 '13 at 20:53

4 Answers 4

up vote 2 down vote accepted

$PdV$ is boundary work. $VdP$ is isentropic shaft work in pumps (as you have identified above), gas turbines, etc. Now you must realize that even in a pump or turbine the mechanism of work is still $Pdv$, i.e., the gas pushing on the blade out of its way. But, then there is work required to maintain the flow in and out of the device/control volume, which requires flow work $PV$ so the net reversible work from a steady-flow device turns out to be shaft $vdP$.

Why flow work $PV$? To push a packet of fluid with volume $V$ forward into a device you have to do work against the pressure of the fluid already in the device, i.e., overcome the back force of that fluid. This implies the work you do in pushing your new packet of length $L$ and cross-section area $A$ into the device is: \begin{align*} \int Fdx = \int_{0}^{L}PAdx = PV \end{align*} It must be noted that in a steady-flow device (unlike in a piston) the back pressure $P$ is constant.

Now consider the device (e.g., turbine to be a control volume). The energy of the fluid going in is its internal energy and the work invested into the fluid to enter the device: $U_{entry}+P_{entry}V_{entry}=H_{entry}$. Similarly for exit from the device. The net change across the device is $\Delta H$. For a differential device (or across a small change) this is $dH$. The work output from the shaft of then device is the $\delta W= dH$.

Now if the device is isentropic, i.e., adiabatic-reversible. The Gibbs equation provides: \begin{align*} &dH=TdS+VdP=VdP\\ &\delta W =dH=VdP \qquad (\text{isentropic}\; dS=0) \end{align*}

Therefore $VdP$ is isentropic shaft work from a flowing device.

Important points: 1) Both internal energy and enthalpy are state variables, therefore can be measured for a system static or flowing. This is why sometimes there is a tendency to use $U$ and $H$ incorrectly. The true purpose of $H$ is to capture the work required to push/maintain a flow against a back pressure, i.e., it incorporates the $PV$ part. Therefore when you write an energy balance with flows coming in and out, the energy crossing boundary is not just $U$ but $H$ and this distinction must be kept in mind.

2) $VdP$ is isentropic steady-flow shaft work. The isentropic is key here.

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Nice, thanks. I think the point of the answer by Bryson S. is that $H_{exit}=H_{entry}-W$ for an adiabatic process, where $W$ is the (positive) work output of the shaft, so $H$ decreases as shaft work is performed. –  Art Brown May 10 at 1:07

$\delta W=\mathrm{d}(PV)$ is wrong. We always have $\delta W=P\mathrm{d}V$ (unless there are other interactions like magnetic field). In fact, the reason that many books opt to denote infinitesimal work as $\delta W$ instead of $\mathrm{d}W$ is to emphasize that work is not an exact differential.

$V\mathrm{d}P$ manifests in these formulas $$\mathrm{d}H=T\mathrm{d}S+V\mathrm{d}P$$ $$\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}P$$ where $H$ and $G$ are not internal energy, but enthalpy and Gibbs free energy. These two concepts are more useful than internal energy in isobaric processes.

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Sankaran is correct in that the magnitude of the net reversible shaft work $\delta w$ is $vdP$, but he is incorrect in ascribing to this quantity a positive sign. In reality, the shaft work done by the system must be $-vdP $, here's why...

If, by definition, a differential change in enthalpy $dh = TdS + vdP$, then for an adiabatic process where $\delta q = TdS = 0$, the differential change in enthalpy must be

$dh = (0) + vdP = vdP$.

Now, if the work done by the system is positive ($\delta w > 0)$ its change in enthalpy must be negative $(dh < 0)$. These two facts in conjunction must mean that the net reversible shaft work done by the system is:

$\delta w = -vdP = -dh$

I ran into this problem myself in trying to derive the specific work produced by an isentropic turbine, and you will definitely get the wrong answer if you start with the premise that the differential shaft work is simply $vdP$.

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I hate answering my own question but I'd like to share this with you as it is definitive:

$$\mathrm{d}h = \mathrm{d}u + \mathrm{d} (p v)$$ $$h_2 - h_1 = u_2 - u_1 + p(v_2 - v_1) + v (p_2 - p_1)$$ Now, for a pump working in the compressed liquid (subcooled liquid zone), it is noted that the change in specific volume $v$ is minimal when the pump adds pressure energy to the liquid. It is minimal because liquid is incompressible.

This leaves us with the following terms:

$$h_2 - h_1 = u_2 - u_1 + v (p_2 - p_1)$$ $$\Delta h = C (T_2 - T_1) + v(p_2 - p_1)$$

An other assumption that is made that the fluid flow through the pump does not raise the temperature much which would allow us to drop the $C (T_2 - T_1)$ term where $C$ is the specific heat of the liquid.

  • This leaves us with the equation that was troubling me:

$$h_2 - h_1 = v (p_2 -p_1)$$

So this is valid for a pump working in the compressed liquid region of the $P-v$ or $T-v$ diagram.

If you'd like to comment further, please do so. This has been quite enlightening for me already! :)

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Is $h$ here enthalpy? –  Siyuan Ren Jan 24 '13 at 15:26
    
@C.R. yes, $h$ is the enthalpy? Would you like to edit the answer as you noticed that it wasn't defined? –  drN Jan 24 '13 at 15:38

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