Concept of enthalpy

Replies

• 

  Voltaire

  Enthalpy
• 

  Kaustubh Katdare

  Wiki article will give you all the details you need. If you have specific questions, feel free to ask them here 😀
• 

  ritupam

  Thank You Sir. The note was quite useful.
  
  But I want to clarify one thing whether enthalpy is defined for flow processes only, or is the concept applied for non flow processes also.
• 

  Voltaire

  I am curious about why you asked this question...
  The short answer is yes - the same applies; but there is a difference.
  Let H = enthalpy[J/kg], U = internal energy[J/kg], v = velocity[m/s], z = elevation from reference height[m], g = constant[m/s²], Q = heat flow[J/s], W = work done by the system (mechanical + electrical)[J/s], W[sub]s[/sub] = shaft work[J/s], ṁ = mass flow per unit time[kg/s].
  From this we define W'[sub]s[/sub] = W/ṁ and Q' = Q/ṁ
  
  To appreciate the notion of work done we consider a piston with constant area A and stroke length ℓ which is filled with a gas under pressure P. We now apply a force to the piston to compress the gas from an initial volume to a final volume. The volume of the gas is given by Aℓ while the force of the piston on the gas is given by PA. The distance the piston travels can also be expressed as ℓ = V/A
  
  In Newtonian physics W = F.s = m.g.s or m.a.s
  To apply this to our piston, F = PA and s = ℓ = V/A. It follows that dW = PAd(V/A) = PdV
  →W = ∑[sup]V2[/sup][sub]V1[/sub]PdV
  
  Enthalpy is a derived energy quantity and is the sum of a) internal energy U (or E in some texts) and b) the PV product. Interestingly - and I think this is pertinent to your question - Daubert notes that: The PV product has units of energy but only truly represents energy in flow processes.' This is because enthalpy is a state function i.e. it can be stated as a function of temperature and pressure alone.
  
  In constant pressure flow processes the energy balance is:
  Energy in = U[sub]1[/sub]ṁ + ṁv²[sub]1[/sub]/2 + ṁgz[sub]1[/sub] + ṁP[sub]1[/sub]V[sub]1[/sub] + Q
  Energy out = U[sub]2[/sub]ṁ + ṁv²[sub]2[/sub]/2 + ṁgz[sub]2[/sub] + ṁP[sub]2[/sub]V[sub]2[/sub] + W[sub]s[/sub]
  Energy accumulation = 0
  
  Equating and dividing by ṁ:
  U[sub]1[/sub] + v²[sub]1[/sub]/2 + gz[sub]1[/sub] + P[sub]1[/sub]V[sub]1[/sub] + Q' = U[sub]2[/sub] + v²[sub]2[/sub]/2 + gz[sub]2[/sub] + P[sub]2[/sub]V[sub]2[/sub] + W[sub]s[/sub]'
  By definition H = U + PV and hence, for flow processes...
  H[sub]1[/sub] + v²[sub]1[/sub]/2 + gz[sub]1[/sub] + Q' = H[sub]2[/sub] + v²[sub]2[/sub]/2 + gz[sub]2[/sub] + W[sub]s[/sub]'
  
  Please tell me if this answers your question.
• 

  ritupam

  Thank You Sir.
  
  The reason for asking this question is that Sir, in the general equation for enthalpy, H=U+PV, the term pv is called flow work which arises because in a flow process, at the inlet, the fluid outside the system does work on the system by pushing the fluid into the control volume. Similarly sir, if we consider at the exit, the exhaust flow work will be P2V2.
  
  Thus it seems that the PV term is solely related to flow processes only. It does not seem to arise in non-flow processes where enthalpy seem to me to be equivalent to internal energy. So it seems we can conclude that enthalpy in flow processes is what internal energy in non-flow processes.
  
  But sir, I have come across many numericals on non-flow processes where both enthalpy and internal energy are to be calculated, whereas my understanding tells me that enthalpy and internal energy should be the same in non-flow processes because of the absence of the PV term.
  
  Thus I find that there must be some missing link in my understanding. I would like you to clarify this doubt.
• 

  Voltaire

  No, I think you understand it well. You have obviously read and appreciated the subject matter better than most
  5 stars
• 

  ritupam

  Can anyone please help me in this regard. Thank You.
• 

  Voltaire

  Here it is again
  Non-flow: Q = H[sub]2[/sub] - H[sub]1[/sub]
  Flow: W = H[sub]1[/sub] - H[sub]2[/sub]