Study of the EEqn.H in rhoPimpleDyMFoam.

Horacio Aguerre · May 27, 2013, 08:03

Hi,

I'm working with the solver "rhoPimpleDyMFoam" from the OpenFOAM 2-2-x version. In specific i was trying to replicate the polytropic process in an adiabatic compression.
The case i studied is a pistone-cylinder sistem where air is compressed by the pistone patch which is a moving wall. The pistone velocity is adjusted to 0,1(m/s) to aproximate a reversible process.

When i choose energy "he" as internal energy i could reproduce the polytropic process in an adiabatic compression.

However when i choose energy "he" as entalphy I couldn't achieve the theoretical expression $pV^{1,4}=constant$ for air. Instead the system looks as it would be non adiabatic or like the conversion between work and internal energy was not made properly.

Therefore i started to study the EEqn.H.

I compare the theoretical balance of total energy in terms of entalphy with what is written in the EEqn.H.

Balance of a tensorial property in an arbritary moving domain.

$\dfrac{d}{dt} \int_{V}\rho \phi dV + \int_{S}\rho \phi\left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}=-\int_{S}\rho \mathbf{Q_{\phi}} \cdot \mathbf{ds}+\int_{V}S_{\phi}dV$

Balance of mechanical energy.

$\phi \equiv \frac{\mathbf{v} \cdot \mathbf{v}}{2} = K$

$\dfrac{d}{dt} \int_{V}\rho K dV + \int_{S}\rho K \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S}\left(p\mathbf{v}+ \boldsymbol{\tau} \cdot \mathbf{v}\right) \cdot \mathbf{ds} + \int_{V} \left[p \left(\boldsymbol{\nabla} \cdot \mathbf{v}\right)+\boldsymbol{\tau}:\boldsymbol{\nabla v}\right] dV+ \int_{V}\rho\left(\mathbf{v} \cdot \mathbf{g}\right) dV$

Balance of internal energy.

$\phi \equiv \widehat{u}$

$\dfrac{d}{dt} \int_{V}\rho \widehat{u} dV + \int_{S}\rho \widehat{u} \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S} \mathbf{q} \cdot \mathbf{ds}-\int_{V} \left[p \left(\boldsymbol{\nabla} \cdot \mathbf{v}\right)+\boldsymbol{\tau}:\boldsymbol{\nabla v}\right] dV$

The balance of internal energy can be expressed in terms of entalphy.

$\widehat{h}=\widehat{u}+\frac{p}{\rho}\implies \widehat{u}=\widehat{h}-\frac{p}{\rho}$

Replacing in the balance of internal energy.

$\dfrac{d}{dt} \int_{V}\rho \left(\widehat{h}-\frac{p}{\rho}\right) dV + \int_{S}\rho \left(\widehat{h}-\frac{p}{\rho}\right) \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S} \mathbf{q} \cdot \mathbf{ds}-\int_{V} \left[p \left(\boldsymbol{\nabla} \cdot \mathbf{v}\right)+\boldsymbol{\tau}:\boldsymbol{\nabla v}\right] dV$

$\dfrac{d}{dt} \int_{V}\rho \widehat{h} dV+ \int_{S}\rho \widehat{h} \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}-\dfrac{d}{dt} \int_{V}p dV - \int_{S} p \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S} \mathbf{q} \cdot \mathbf{ds}-\int_{V} \left[p \left(\boldsymbol{\nabla} \cdot \mathbf{v}\right)+\boldsymbol{\tau}:\boldsymbol{\nabla v}\right] dV$

Balance of total energy.

The sum of the balances of mechanical and internal energy results in the balance of total energy.

$\dfrac{d}{dt} \int_{V}\rho \left(\widehat{h}+ K \right) dV + \int_{S}\rho \left(\widehat{h}+ K\right) \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds} -\dfrac{d}{dt} \int_{V}p dV - \int_{S} p \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S} \mathbf{q} \cdot \mathbf{ds}-\int_{S}\left(p\mathbf{v}+ \boldsymbol{\tau} \cdot \mathbf{v}\right) \cdot \mathbf{ds} + \int_{V}\rho\left(\mathbf{v} \cdot \mathbf{g}\right) dV$

If the terms of work of viscous forces and the gravity are neglected, the balance is:

$\dfrac{d}{dt} \int_{V}\rho \left(\widehat{h}+ K \right) dV + \int_{S}\rho \left(\widehat{h}+ K\right) \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds} -\dfrac{d}{dt} \int_{V}p dV + \int_{S} p \mathbf{v_{b}} \cdot \mathbf{ds}$ $= -\int_{S} \mathbf{q} \cdot \mathbf{ds}$

rhoPimpleDyMFoam: EEqn.H

The balance of total energy in the EEqn.H file with

considered as entalphy $\widehat{h}$ is:

Code:

fvm::ddt(rho, he) + fvm::div(phi, he) + fvc::ddt(rho, K) + fvc::div(phi, K)
- dpdt - fvm::laplacian(turbulence ->alphaEff(), he) == fvOptions(rho, he)

Where, $\phi=\rho_{f} \left(\mathbf{v}-\mathbf{v_{b}}\right)_{f} \cdot \mathbf{S_{f}}$ , is relative to the mesh.

If fvOptions(rho, he) are considered as zero, the term $\int_{S} p \mathbf{v_{b}} \cdot \mathbf{ds}$ from the final balance is no considered in EEqn.H.

Then i try adding in the EEqn.H the therm $\int_{S} p \mathbf{v_{b}} \cdot \mathbf{ds}$ .

Code:

fvm::ddt(rho, he) + fvm::div(phi, he) + fvc::ddt(rho, K) + fvc::div(phi, K)
- dpdt + fvc::div(fvc::meshPhi(U),p) - fvm::laplacian(turbulence ->alphaEff(), he) == fvOptions(rho, he)

The result was correct. As in the first case (he=internal energy) the adiabatic compression was reproduced with the relation $pV^{1,4}=constant$ .

The question is,

should the term $\int_{S} p \mathbf{v_{b}} \cdot \mathbf{ds}$ be introduced in any way in fvOptions(rho, he) or it must be explicit in the EEqn.H?

Thanks,

Horacio

May 27, 2013, 08:03	Study of the EEqn.H in rhoPimpleDyMFoam.	#1
Horacio Aguerre New Member Join Date: May 2013 Location: Santa Fé, Argentina Posts: 5 Rep Power: 13	Hi, I'm working with the solver "rhoPimpleDyMFoam" from the OpenFOAM 2-2-x version. In specific i was trying to replicate the polytropic process in an adiabatic compression. The case i studied is a pistone-cylinder sistem where air is compressed by the pistone patch which is a moving wall. The pistone velocity is adjusted to 0,1(m/s) to aproximate a reversible process. When i choose energy "he" as internal energy i could reproduce the polytropic process in an adiabatic compression. However when i choose energy "he" as entalphy I couldn't achieve the theoretical expression $pV^{1,4}=constant$ for air. Instead the system looks as it would be non adiabatic or like the conversion between work and internal energy was not made properly. Therefore i started to study the EEqn.H. I compare the theoretical balance of total energy in terms of entalphy with what is written in the EEqn.H. Balance of a tensorial property in an arbritary moving domain. $\dfrac{d}{dt} \int_{V}\rho \phi dV + \int_{S}\rho \phi\left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}=-\int_{S}\rho \mathbf{Q_{\phi}} \cdot \mathbf{ds}+\int_{V}S_{\phi}dV$ Balance of mechanical energy. $\phi \equiv \frac{\mathbf{v} \cdot \mathbf{v}}{2} = K$ $\dfrac{d}{dt} \int_{V}\rho K dV + \int_{S}\rho K \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S}\left(p\mathbf{v}+ \boldsymbol{\tau} \cdot \mathbf{v}\right) \cdot \mathbf{ds} + \int_{V} \left[p \left(\boldsymbol{\nabla} \cdot \mathbf{v}\right)+\boldsymbol{\tau}:\boldsymbol{\nabla v}\right] dV+ \int_{V}\rho\left(\mathbf{v} \cdot \mathbf{g}\right) dV$ Balance of internal energy. $\phi \equiv \widehat{u}$ $\dfrac{d}{dt} \int_{V}\rho \widehat{u} dV + \int_{S}\rho \widehat{u} \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S} \mathbf{q} \cdot \mathbf{ds}-\int_{V} \left[p \left(\boldsymbol{\nabla} \cdot \mathbf{v}\right)+\boldsymbol{\tau}:\boldsymbol{\nabla v}\right] dV$ The balance of internal energy can be expressed in terms of entalphy. $\widehat{h}=\widehat{u}+\frac{p}{\rho}\implies \widehat{u}=\widehat{h}-\frac{p}{\rho}$ Replacing in the balance of internal energy. $\dfrac{d}{dt} \int_{V}\rho \left(\widehat{h}-\frac{p}{\rho}\right) dV + \int_{S}\rho \left(\widehat{h}-\frac{p}{\rho}\right) \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S} \mathbf{q} \cdot \mathbf{ds}-\int_{V} \left[p \left(\boldsymbol{\nabla} \cdot \mathbf{v}\right)+\boldsymbol{\tau}:\boldsymbol{\nabla v}\right] dV$ $\dfrac{d}{dt} \int_{V}\rho \widehat{h} dV+ \int_{S}\rho \widehat{h} \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}-\dfrac{d}{dt} \int_{V}p dV - \int_{S} p \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S} \mathbf{q} \cdot \mathbf{ds}-\int_{V} \left[p \left(\boldsymbol{\nabla} \cdot \mathbf{v}\right)+\boldsymbol{\tau}:\boldsymbol{\nabla v}\right] dV$ Balance of total energy. The sum of the balances of mechanical and internal energy results in the balance of total energy. $\dfrac{d}{dt} \int_{V}\rho \left(\widehat{h}+ K \right) dV + \int_{S}\rho \left(\widehat{h}+ K\right) \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds} -\dfrac{d}{dt} \int_{V}p dV - \int_{S} p \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds}$ $=-\int_{S} \mathbf{q} \cdot \mathbf{ds}-\int_{S}\left(p\mathbf{v}+ \boldsymbol{\tau} \cdot \mathbf{v}\right) \cdot \mathbf{ds} + \int_{V}\rho\left(\mathbf{v} \cdot \mathbf{g}\right) dV$ If the terms of work of viscous forces and the gravity are neglected, the balance is: $\dfrac{d}{dt} \int_{V}\rho \left(\widehat{h}+ K \right) dV + \int_{S}\rho \left(\widehat{h}+ K\right) \left(\mathbf{v}-\mathbf{v_{b}}\right) \cdot \mathbf{ds} -\dfrac{d}{dt} \int_{V}p dV + \int_{S} p \mathbf{v_{b}} \cdot \mathbf{ds}$ $= -\int_{S} \mathbf{q} \cdot \mathbf{ds}$ rhoPimpleDyMFoam: EEqn.H The balance of total energy in the EEqn.H file with considered as entalphy $\widehat{h}$ is: Code: fvm::ddt(rho, he) + fvm::div(phi, he) + fvc::ddt(rho, K) + fvc::div(phi, K) - dpdt - fvm::laplacian(turbulence ->alphaEff(), he) == fvOptions(rho, he) Where, $\phi=\rho_{f} \left(\mathbf{v}-\mathbf{v_{b}}\right)_{f} \cdot \mathbf{S_{f}}$ , is relative to the mesh. If fvOptions(rho, he) are considered as zero, the term $\int_{S} p \mathbf{v_{b}} \cdot \mathbf{ds}$ from the final balance is no considered in EEqn.H. Then i try adding in the EEqn.H the therm $\int_{S} p \mathbf{v_{b}} \cdot \mathbf{ds}$ . Code: fvm::ddt(rho, he) + fvm::div(phi, he) + fvc::ddt(rho, K) + fvc::div(phi, K) - dpdt + fvc::div(fvc::meshPhi(U),p) - fvm::laplacian(turbulence ->alphaEff(), he) == fvOptions(rho, he) The result was correct. As in the first case (he=internal energy) the adiabatic compression was reproduced with the relation $pV^{1,4}=constant$ . The question is, should the term $\int_{S} p \mathbf{v_{b}} \cdot \mathbf{ds}$ be introduced in any way in fvOptions(rho, he) or it must be explicit in the EEqn.H? Thanks, Horacio mturcios777, fredo490, mgg and 2 others like this.

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