1
$\begingroup$

Can first law of thermodynamics defined for a closed system be applied to the steady flow energy equation? Why? I came across the derivation of $Vdp$ work and Every book applied the first law defined for closed system to steady flow energy equation.

enter image description here

enter image description here

Please clarify

$\endgroup$
  • $\begingroup$ First law of thermodynamics is valid everywhere.What exactly do you want to ask? $\endgroup$ – Rishabh Jain Jun 15 '17 at 8:36
  • $\begingroup$ I don't understand how first law defined for a closed system dQ = dU + dw...here considering only pdV work..is applied to steady flow energy equation which is an open system...you can see in the second image...it says using property relation eq 7.41.....but 7.41 was defined for a closed system... $\endgroup$ – user158324 Jun 15 '17 at 10:51
  • $\begingroup$ Have you learned about the open system (control volume) version of the first law of thermodynamics. This version of the first law is derived, based on the closed system version. You are aware that, in the open system version, the work is split into two separate parts: (a) work to push fluid into or out of the control volume and (b) shaft work, correct? $\endgroup$ – Chet Miller Jun 15 '17 at 11:42
  • $\begingroup$ @chester I totally understand the first law for open and closed system... Its a fundamental thing...but what I ask is that this particular relation TDs=dh + Vdp is defined for a non flow process then why is it being applied to a flow process...is it because its steady process and since property at any point in space does not change with time... then we consider control volume as control mass ? $\endgroup$ – user158324 Jun 15 '17 at 12:36
  • $\begingroup$ This equation is a general physical property relationship that is satisfied by the changes in s, h, and p between neighboring thermodynamic equilibrium states, and is independent of any process. $\endgroup$ – Chet Miller Jun 15 '17 at 19:32
1
$\begingroup$

OH , i get it now. You are right.We can't apply the first law if you don't consider the entire system and if the energy can leak in or out by other means ,say a more energetic fluid comes in. But in steady state flow ,no particular trait of a fluid changes with time.This means that the property of a fluid at a particular spatial point is constant in time(to be more mathematical,it's partial derivative with respect to time is zero).This means that the only way the energy of fluid can change in a constraint volume is by change in kinetic energy,change in potential energy or by work done.

For detailed explanantion,go to https://wiki.ucl.ac.uk/display/MechEngThermodyn/First+law+applied+to+flow+processes

$\endgroup$
  • $\begingroup$ So are you saying that despite the "identity" of the fluid particle changing (because fluid comes in and goes out) we treat it as a closed system because although the identity changes but all the properties are same....i.e they don't change over time for a particular position. $\endgroup$ – user158324 Jun 15 '17 at 12:17
  • $\begingroup$ The fluid particles with same properties are coming in at a particular point at all times .This is what is meant by steady state.What do you mean by identity? I mentioned a general derivation for flow processes. For only the steady state derivation,go to ecourses.ou.edu/cgi-bin/… $\endgroup$ – Rishabh Jain Jun 15 '17 at 12:23
  • $\begingroup$ By identity I meant that suppose fluid particle A is present at a particular point in space and after some time it is replaced by B...this is control volume...but in control mass A remains A...and here because A and B have same properties and it is a steady process therefore....we are treating control volume as control mass...this is what I understand... $\endgroup$ – user158324 Jun 15 '17 at 12:29
  • $\begingroup$ In the derivation I don't have any issues until work or heat is written in terms of enthalpy... But when dh is replaced by TDs+Vdp then I don't understand because Tds = dh + Vdp is defined for a closed NON FLOW process(see image 1) $\endgroup$ – user158324 Jun 15 '17 at 12:32
  • $\begingroup$ this is not control mass but control volume and dh=Tds+Vdp in the first as well as second image.Plus, this derivation showed you how can generaize the first law for steady flows $\endgroup$ – Rishabh Jain Jun 15 '17 at 14:50
0
$\begingroup$

It helps if you consider the first and second law expressed in terms of energy and entropy densities that depend explicitly on space and time, rather than differentials.

The first laws is: $$\frac{\partial e}{\partial t} + \nabla\cdot(e\boldsymbol{v}) = \boldsymbol{T}:\nabla\boldsymbol{v}+\nabla\cdot\boldsymbol{q} - Q,$$ where $e$ is the energy density; $\boldsymbol{v}$ the velocity of the fluid or material; $\boldsymbol{T}$ is the stress, e.g. the pressure or viscosity in a fluid; $\boldsymbol{q}$ is the surface flux of heat, and $Q$ is the volume sink of heat. The term $\boldsymbol{T}:\nabla\boldsymbol{v}$ represent the work being done on the fluid or material, and reduces to $p\partial V/\partial t$ in the case of a fluid with no viscosity. If you integrate this equation over a control volume you get a formula that says, in words, $$\frac{\mathrm{d}(\text{total internal energy within volume})}{\mathrm{d}t} + (\text{flux of energy through volume boundary}) = \text{power} + \text{heating}.$$ The "power" is the work done per unit time, and the "heating" is the heat transferred per unit time. The flux of energy through the boundary appears if the system is open, so matter that moves through the boundary is carrying its internal energy in or out.

The second law has a similar expression: $$\frac{\partial s}{\partial t} + \nabla\cdot(s\boldsymbol{v}) \geqslant \nabla\cdot\left(\frac{\boldsymbol{q}}{T}\right) - \frac{Q}{T},$$ where $s$ is the entropy density and $T$ the temperature. Again, if you integrate over a control volume you obtain $$\frac{\mathrm{d}(\text{total entropy within volume})}{\mathrm{d}t} + (\text{flux of entropy through volume boundary}) \geqslant \frac{\text{heating}}{\text{temperature}}.$$ The flux of entropy through the boundary appears if the system is open, so matter that moves through the boundary is carrying its entropy in or out.

I'm sorry if this looks complicated, but if you devote some time to it, even skipping the technical details, it can give you a better grasp of what the first and second laws say and how to use them in unfamiliar situations.

Check out C. A. Truesdell: Rational Thermodynamics (2nd ed., Springer 1984), or I. Samohýl & M. Pekař: The Thermodynamics of Linear Fluids and Fluid Mixtures (Springer 2014).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.