Enthelpy change at constant volume?

[shuuchuu]

I can't seem to figure this out although I suspect I'm making a silly mistake.
Assuming a closed volume of ideal gas that's also thermally insulated apart from the addition of heat of Q joules.
Since it's constant volume, dT = Q / Cvm
also for internal energy U, dU = CvmdT, i.e. dU = Q , also true because dV = 0
now considering enthalpy, h = U + PV => dh = dU + PdV + VdP => dh = dU + VdP
but dP = d(rho.RT) = rho.RdT
so, dh = dU + mRdT = dU + (R/Cv)Q = dU + (gamma - 1)Q
basically dh = gamma.Q

the part I don't understand is how can enthalpy increase by gamma.Q when only Q transfers into the box. The change in enthalpy is greater than the energy going in?

 

[Bill_K]

Energy conservation is expressed by the fact that the change in U is equal to the energy going in. H is just a useful quantity, it doesn't represent the energy.

 

[shuuchuu]

Thanks for the reply.

So if you were pushing a packet of fluid into an existing volume, the actual energy increase in the volume would be CvmT, as opposed to CpmT, with the corresponding work?

Specifically it's for CFD, so if you had a control volume containing fluid at T Kelvin, and the mass flux across a face was 1kg, then the real energy flux would be CvT?

Cheers

 

[Curl]

"pushing" a fluid into a control volume requires flow work which is PV for an ideal gas or for an incompressible substance. This must be taken into account when you are doing CFD, unless the program does it for you.

Also note that CvT is not the "total" internal energy, just what is called "sensible" energy since its the one which gets moved around. THe object has also other energies which are neglected because they never change in most applications.

 

[sardar]

It is important to note that in this scenario, the system is thermally insulated and there is no work being done (dV = 0). This means that the only way for the internal energy to change is through the addition of heat (dU = Q). However, at constant volume, the internal energy and enthalpy are not the same, as shown by the equation dh = dU + (gamma - 1)Q. This means that the change in enthalpy (dh) will be greater than the change in internal energy (dU) when heat is added to the system. This is because enthalpy takes into account not only the internal energy of the system, but also the work done by the system (in this case, PdV = 0) and the energy needed to push back against external pressure (VdP = 0). Therefore, in this scenario, the change in enthalpy will be greater than the energy going in (Q).