Is enthelpy in a non-flow process?

  • Thread starter Thread starter Ali Durrani
  • Start date Start date
  • Tags Tags
    Process
AI Thread Summary
Enthalpy can be associated with non-flow processes, as it is a state function defined as the sum of internal energy and the product of pressure and volume, independent of flow. In closed systems, changes in enthalpy reflect heat added or work done, regardless of whether the process is at constant pressure or not. Heat capacities at constant pressure (Cp) and constant volume (Cv) are experimentally defined but are applicable in various processes, including those not at constant conditions. For ideal gases, changes in enthalpy can be expressed as a function of temperature alone, making the relationship valid across different processes. Understanding enthalpy mathematically is crucial for solving thermodynamic problems, rather than seeking a physical interpretation.
Ali Durrani
Messages
55
Reaction score
5
Right i have two questions,
Is enthalpy term associated with a non flow process? because what i have understood so far is that Enthalpy actually is the internal energy plus the flow work, and in a non flow process there literally is no flow so how can enthalpy be there? because in U+PV, PV is not defined.

secondly, i would really like to understand heat capacities at constant pressure and constant volume (Cp, Cv) because they are sometimes not used in their respective processes, for instance pump work is a function of enthalpy change of the water and the mass flow rate, and enthalpy is a function of heat capacity at constant pressure but pumping up water is not at all a constant pressure process
 
Engineering news on Phys.org
Regarding definition: enthalpy is a measurement of internal energy, it doesn't have any flow parameters in its formulation. You can determine enthalpy for a fixed volume of fluid in an enclosed tank with no flow in or out for example.

More reading here: https://en.wikipedia.org/wiki/Enthalpy
Wikipedia.org said:
Enthalpy is a measurement of energy in a thermodynamic system. It is the thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume.[1]

More technically, it includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure.[2]
 
When you ask whether enthalpy can be associated with a "non-flow process," what do you mean exactly? Because as far as I can tell enthalpy's definition doesn't have a flow rate anywhere in it...
 
Ali Durrani said:
Right i have two questions,
Is enthalpy term associated with a non flow process? because what i have understood so far is that Enthalpy actually is the internal energy plus the flow work, and in a non flow process there literally is no flow so how can enthalpy be there? because in U+PV, PV is not defined.
You are talking about the distinction between the open system and closed system versions of the first law of thermodynamics. In a closed system operating at constant pressure, the change in enthalpy is equal to the amount of heat added (I'm sure you've seen this before). But, even if the system does not operate at constant pressure, the change in enthalpy is equal to ##\Delta U+\Delta (PV)## between the initial and final thermodynamic equilibrium states. Since enthalpy is a function of state (i.e., a physical property of the material), this is, of course, independent of any process.

In the case of an open system (control volume) operating at steady state, the first law is sometimes written as ##Q-W_s=\Delta H##, where ##W_s## is the "shaft work" (i.e., all work except that required to push mass into and out of the system), and ##\Delta H=m\Delta h##, where m is the amount of mass which has entered and exited the system during a time interval and h is the enthalpy per unit mass of the material entering and leaving.

So change in enthalpy is a feature of both closed systems and open systems.
secondly, i would really like to understand heat capacities at constant pressure and constant volume (Cp, Cv) because they are sometimes not used in their respective processes, for instance pump work is a function of enthalpy change of the water and the mass flow rate, and enthalpy is a function of heat capacity at constant pressure but pumping up water is not at all a constant pressure process
The subscripts p and v and the terms "at constant pressure" and "at constant volume" refer to how these quantities are measured experimentally, and not how they are applied in practice to solve problems. In terms of internal energy and enthalpy, these heat capacities are precisely defined as follows:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_v\tag{1}$$
$$C_p=\left(\frac{\partial H}{\partial T}\right)_p\tag{2}$$
If you want to measure ##C_v##, you do a test at constant volume where ##\Delta U=C_v\Delta T=Q##. So, by measuring the amount of heat added for a given temperature change, you can get ##C_v##. Similarly, if you want to measure ##C_p##, you do a test at constant pressure where ##\Delta H=C_p\Delta T=Q##. So, again, by measuring the amount of heat added for a given temperature change, you can get ##C_p##. So, in both these cases, you measure the amount of heat added in the test. That's where the name heat capacity comes from.

However, aside from these measurements, ##C_v## and ##C_p## are much more generally applicable than just at constant v and p, respectively, and apply to arbitrary process paths via Eqns. 1 and 2.
 
Can you please provide the exact quotes from which these deductions are reached?
 
I just wanted to ask that,

if dh = du + pdV is valid, then we can change volume through different processes and change dh by different amount between two same states.

What is wrong in this thinking?

[I removed the previous question, because I thought my Statement was not completely understood]
 
Last edited:
Diptangshu said:
I just wanted to ask that,

if dh = du + pdV is valid, then we can change volume through different processes and change dh by different amount between two same states.

What is wrong in this thinking?

[I removed the previous question, because I thought my Statement was not completely understood]
This equation is incorrect. it should read $$dh=du+d(pv)$$
 
Chestermiller said:
This equation is incorrect. it should read $$dh=du+d(pv)$$

Sorry Sir, I mistakenly wrote dV instead dv.

So in this d(pv) any work can be included?

Starting from expansion compression work in closed system to flow work in Open system?

What kind of Works does this d(pv) include? What is exact significance of (pv)?
Can you explain in brief?
 
Diptangshu said:
Sorry Sir, I mistakenly wrote dV instead dv.

So in this d(pv) any work can be included?

Starting from expansion compression work in closed system to flow work in Open system?

What kind of Works does this d(pv) include? What is exact significance of (pv)?
Can you explain in brief?
It doesn't imply any kind of work or any kind of process. It is simply the definition of the enthalpy of a material. This is a physical property relationship for the substance, and applies to any kind of process that the substance experiences.
 
  • #10
Chestermiller said:
It doesn't imply any kind of work or any kind of process. It is simply the definition of the enthalpy of a material. This is a physical property relationship for the substance, and applies to any kind of process that the substance experiences.
So (pv) itself is a property of the system?

From where I read, may be older text, (pv) is referred to as flow energy or displacement energy, the energy due to the bulk of a moving fluid or the energy content of a static fluid due to the volume it occupies.

Is that a correct thing?
 
  • #11
Diptangshu said:
So (pv) itself is a property of the system?

From where I read, may be older text, (pv) is referred to as flow energy or displacement energy, the energy due to the bulk of a moving fluid or the energy content of a static fluid due to the volume it occupies.

Is that a correct thing?
This is correct in the mind of the author. It was provided in your text because students often ask the teacher for a physical interpretation of enthalpy. This is their best attempt at providing an interpretation. In my judgment, it is only confusing to students, and no physical interpretation is really needed. Enthalpy is simply a convenient function (i.e., parameter) to work with in solving many different kinds of thermodynamics problems. My advice is not to waste your valuable time trying to assign a physical interpretation to enthalpy. Just understand its meaning mathematically, and then use it as needed to solve problems.
 
Last edited:
  • #12
Chestermiller said:
This is correct in the mind of the author. It was provided in your text because students often ask the teacher for a physical interpretation of enthalpy. This is their best attempt at providing an interpretation. In my judgment, it is only confusing to students, and no physical interpretation is really needed. Enthalpy is simply a convenient function (i.e., parameter) to work with in solving my different kinds of thermodynamics problems. My advice is not to waste your valuable time trying to assign a physical interpretation to enthalpy. Just understand its meaning mathematically, and then use it as needed to solve problems.
Thank You Sir.
 
  • #13
I still have a slight doubt. I couldn't understand the significance of enthalpy in non flow process. What does enthalpy mean in a non flow process? Because as there is no flow in closed system then how can we account for it?

And also in gas turbines the compressor or turbine work which is actually an isoentropic process, why do we take change in enthalpy as Cp (T2-T1) . Because it's not a constant pressure process.
Hence my question, do we always write enthalpy As
Cp (T2-T1)?
 
  • #14
Soji_jacob1997 said:
I still have a slight doubt. I couldn't understand the significance of enthalpy in non flow process. What does enthalpy mean in a non flow process? Because as there is no flow in closed system then how can we account for it?
Enthalpy is a physical property of a material that is independent of any kind of process that the material experiences. It applies to the equilibrium states of the material. It is defined by the equation: $$H\equiv U+PV$$where U is the internal energy of the material, V is the volume of material, and P is the pressure of the material.
And also in gas turbines the compressor or turbine work which is actually an isoentropic process, why do we take change in enthalpy as Cp (T2-T1) . Because it's not a constant pressure process.
Hence my question, do we always write enthalpy As
Cp (T2-T1)?
The specific heat capacity at constant pressure of a material is defined by the equation:$$C_p\equiv \left(\frac{\partial H}{\partial T}\right)_P$$where H is the specific enthalpy. So ordinarily, it would be generally incorrect to write $$dH=C_pdT$$However, in the case of an ideal gas, the enthalpy is a function only of temperature T, and not pressure P (or specific volume V). So, irrespective of any process, it is always valid to express enthalpy changes in this way for an ideal gas.
 
Back
Top