Gas Specific Heat: When to use Cv versus Cp

[Ijoinedbecausecovid]

How do you know when to use Cv vs. Cp?

I read the below and it makes complete sense to me, but if I'm given an air flow heater problem how do I know if the system is at fixed volume or fixed pressure?

"Cv for a gas is the change in internal energy (U) of a  system with respect to change in temperature at a fixed volume of the  system i.e. Cv =(∂ U/∂ T)v  whereas Cp  for a gas is the change in the enthalpy (H) of the system with respect  to change in temperature at a fixed pressure of the system i.e Cp = (∂ H/∂ T)p.
We  know that, ΔH = ΔU + PΔV (+ VΔP, ΔP=0 for constant pressure) . So the  enthalpy term is  greater than the internal energy term because of  the PΔV term i.e in case of a constant pressure process more energy is  needed, to be provided to the system as compared to that of a constant  volume process to achieve the same temperature rise, as some energy is  utilized in the expansion work of the system. And the relation that  correlates these two is Cp = Cv + R
But since liquids and solids can practically assumed to be incompressible, Cp and Cv for them have almost same values and hence only a single value of specific heat is used for them."

 

[Slay the P.E.]

How do you know when to use Cv vs. Cp?

I read the below and it makes complete sense to me, but if I'm given an air flow heater problem how do I know if the system is at fixed volume or fixed pressure?

"Cv for a gas is the change in internal energy (U) of a  system with respect to change in temperature at a fixed volume of the  system i.e. Cv =(∂ U/∂ T)v  whereas Cp  for a gas is the change in the enthalpy (H) of the system with respect  to change in temperature at a fixed pressure of the system i.e Cp = (∂ H/∂ T)p.
We  know that, ΔH = ΔU + PΔV (+ VΔP, ΔP=0 for constant pressure) . So the  enthalpy term is  greater than the internal energy term because of  the PΔV term i.e in case of a constant pressure process more energy is  needed, to be provided to the system as compared to that of a constant  volume process to achieve the same temperature rise, as some energy is  utilized in the expansion work of the system. And the relation that  correlates these two is Cp = Cv + R
But since liquids and solids can practically assumed to be incompressible, Cp and Cv for them have almost same values and hence only a single value of specific heat is used for them."

Great question. For ideal gases it can be shown that both internal energy, u, and enthalpy, h depend on temperature only (you can find this proof in any Thermo textbook).  Since u and h depend only on temperature for an ideal gas, the specific heats Cv and Cp also depend on temperature only. Thus, for ideal gases, the partial derivatives in the definitions provided in your post can be replaced by ordinary derivatives:

In other words: For ideal gases Cv = du/dT and Cp = dh/dT. Furthermore, if you assume Cv and Cp are constant then these relationships can be easily integrated to yield:

Δu = (Cv)ΔT  and   Δh = (Cp)ΔT  

So, where does this leave us? Well... if you need to calculate a change in internal energy, then you do it by using (Cv)ΔT -- no matter what the process is. If you need Δu for a constant pressure process then Δu is still (Cv)ΔT.

If you need to calculate a change in enthalpy, then you do it by using (Cp)ΔT -- no matter what the process is. If you need Δh for a constant volume process then Δh is still (Cp)ΔT.  

If you are given an air flow heater, then you are dealing with an OPEN system. That is a system which has mass flow through it (unlike, say a tank with closed inlet and outlet valves which would be a CLOSED system). Now, take a look at the energy balance for a generic open system: 

​

You will note that what comes into play for such systems is the enthalpy at the inlet, hi and the enthalpy at the exit, he. So, in OPEN systems (such as your steady-flow heater), you are usually interested in how much the enthalpy changes as the gas flows from the inlet to the exit (he - hi), regardless of the nature of the process. Since you are working with an ideal gas then (he - hi) =  (Cp)(Te - Ti) regardless of how the heater is accomplishing its task.

Let me know if this isn't clear. I can come up with some easy examples to walk through if needed.

 

[Dean Agnostic]

This brings back so many memories from Fluid Mechanics, Introduction to Environmental Engineering, Thermodynamics and Physics classes. 

 

[Ijoinedbecausecovid]

Wow!! Thank you so much! This is a way better answer than any I got during undergrad and grad school. I think I follow everything you've laid out. It may take a while to wrap my mind around it conceptually, but it seems like the big take away that I will need for the PE (I'm taking the MDM test so a problem like this is unlikely to appear and should be simple enough if it does) is that an open system will require Cp and a closed system will require Cv. Is that a good general assumption?

 

[Audi Driver P.E.]

Great question. For ideal gases it can be shown that both internal energy, u, and enthalpy, h depend on temperature only (you can find this proof in any Thermo textbook).  Since u and h depend only on temperature for an ideal gas, the specific heats Cv and Cp also depend on temperature only. Thus, for ideal gases, the partial derivatives in the definitions provided in your post can be replaced by ordinary derivatives:

In other words: For ideal gases Cv = du/dT and Cp = dh/dT. Furthermore, if you assume Cv and Cp are constant then these relationships can be easily integrated to yield:

Δu = (Cv)ΔT  and   Δh = (Cp)ΔT  

So, where does this leave us? Well... if you need to calculate a change in internal energy, then you do it by using (Cv)ΔT -- no matter what the process is. If you need Δu for a constant pressure process then Δu is still (Cv)ΔT.

If you need to calculate a change in enthalpy, then you do it by using (Cp)ΔT -- no matter what the process is. If you need Δh for a constant volume process then Δh is still (Cp)ΔT.  

If you are given an air flow heater, then you are dealing with an OPEN system. That is a system which has mass flow through it (unlike, say a tank with closed inlet and outlet valves which would be a CLOSED system). Now, take a look at the energy balance for a generic open system: 

View attachment 17381​

You will note that what comes into play for such systems is the enthalpy at the inlet, hi and the enthalpy at the exit, he. So, in OPEN systems (such as your steady-flow heater), you are usually interested in how much the enthalpy changes as the gas flows from the inlet to the exit (he - hi), regardless of the nature of the process. Since you are working with an ideal gas then (he - hi) =  (Cp)(Te - Ti) regardless of how the heater is accomplishing its task.

Let me know if this isn't clear. I can come up with some easy examples to walk through if needed.

I sure wish I had you as an instructor during undergrad. This is excellent!