Quadratic Quantities
Here we aim at confirming if the theoretical relations of the quadratic quantities are satisfied even after discretized.
Specifically, as discrete counterparts, we consider the discrete momentum balance multiplied by the Jacobian determinant (incorporating the volume integrals) and the velocity:
N3∑k=1N2∑j=1N1+12∑i=12Ju1∂u1∂t=−N3∑k=1N2∑j=1N1+12∑i=12Ju11J¯¯Jhξ1u1ξ1δξ1u1ξ1−N3∑k=1N2∑j=1N1+12∑i=12Ju11J¯¯Jhξ1u1ξ2δξ1u2ξ1−N3∑k=1N2∑j=1N1+12∑i=12Ju11J¯¯Jhξ1u1ξ3δξ1u3ξ1−N3∑k=1N2∑j=1N1+12∑i=12Ju11hξ1δξ1p+N3∑k=1N2∑j=1N1+12∑i=12Ju1√Pr√Ra1Jδξ1(Jhξ11hξ1δξ1u1)+N3∑k=1N2∑j=1N1+12∑i=12Ju1√Pr√Ra1Jδξ2(Jhξ21hξ2δξ2u1)+N3∑k=1N2∑j=1N1+12∑i=12Ju1√Pr√Ra1Jδξ3(Jhξ31hξ3δξ3u1)+N3∑k=1N2∑j=1N1+12∑i=12Ju1¯Tξ1,
N3∑k=1N2−12∑j=12N1∑i=1Ju2∂u2∂t=−N3∑k=1N2−12∑j=12N1∑i=1Ju21J¯¯Jhξ2u2ξ1δξ2u1ξ2−N3∑k=1N2−12∑j=12N1∑i=1Ju21J¯¯Jhξ2u2ξ2δξ2u2ξ2−N3∑k=1N2−12∑j=12N1∑i=1Ju21J¯¯Jhξ2u2ξ3δξ2u3ξ2−N3∑k=1N2−12∑j=12N1∑i=1Ju21hξ2δξ2p+N3∑k=1N2−12∑j=12N1∑i=1Ju2√Pr√Ra1Jδξ1(Jhξ11hξ1δξ1u2)+N3∑k=1N2−12∑j=12N1∑i=1Ju2√Pr√Ra1Jδξ2(Jhξ21hξ2δξ2u2)+N3∑k=1N2−12∑j=12N1∑i=1Ju2√Pr√Ra1Jδξ3(Jhξ31hξ3δξ3u2),
N3−12∑k=12N2∑j=1N1∑i=1Ju3∂u3∂t=−N3−12∑k=12N2∑j=1N1∑i=1Ju31J¯¯Jhξ3u3ξ1δξ3u1ξ3−N3−12∑k=12N2∑j=1N1∑i=1Ju31J¯¯Jhξ3u3ξ2δξ3u2ξ3−N3−12∑k=12N2∑j=1N1∑i=1Ju31J¯¯Jhξ3u3ξ3δξ3u3ξ3−N3−12∑k=12N2∑j=1N1∑i=1Ju31hξ3δξ3p+N3−12∑k=12N2∑j=1N1∑i=1Ju3√Pr√Ra1Jδξ1(Jhξ11hξ1δξ1u3)+N3−12∑k=12N2∑j=1N1∑i=1Ju3√Pr√Ra1Jδξ2(Jhξ21hξ2δξ2u3)+N3−12∑k=12N2∑j=1N1∑i=1Ju3√Pr√Ra1Jδξ3(Jhξ31hξ3δξ3u3).
Similarly, multiplying the discrete internal energy balance by temperature and volume-integrating it yield
N3∑k=1N2∑j=1N1∑i=1JT∂T∂t=−N3∑k=1N2∑j=1N1∑i=1JT1J¯Jhξ1u1δξ1Tξ1−N3∑k=1N2∑j=1N1∑i=1JT1J¯Jhξ2u2δξ2Tξ2−N3∑k=1N2∑j=1N1∑i=1JT1J¯Jhξ3u3δξ3Tξ3+N3∑k=1N2∑j=1N1∑i=1JT1√Pr√Ra1Jδξ1(Jhξ11hξ1δξ1T)+N3∑k=1N2∑j=1N1∑i=1JT1√Pr√Ra1Jδξ2(Jhξ21hξ2δξ2T)+N3∑k=1N2∑j=1N1∑i=1JT1√Pr√Ra1Jδξ3(Jhξ31hξ3δξ3T).
It should be noted that, since the velocities are defined at different positions due to the staggered grid arrangement, quadratic quantities are also evaluated at different locations, which is the reason why we consider three components separately.
With some algebra, we obtain
N3∑k=1N2∑j=1N1+12∑i=12J∂k1∂t+N3∑k=1N2−12∑j=12N1∑i=1J∂k2∂t+N3−12∑k=12N2∑j=1N1∑i=1J∂k3∂t=−N3∑k=1N2∑j=1N1∑i=1√Pr√RaJ(1hξ1δξ1u1)2−N3∑k=1N2−12∑j=12N1+12∑i=12√Pr√RaJ(1hξ2δξ2u1)2−N3−12∑k=12N2∑j=1N1+12∑i=12√Pr√RaJ(1hξ3δξ3u1)2−N3∑k=1N2−12∑j=12N1+12∑i=12√Pr√RaJ(1hξ1δξ1u2)2−N3∑k=1N2∑j=1N1∑i=1√Pr√RaJ(1hξ2δξ2u2)2−N3−12∑k=12N2−12∑j=12N1∑i=1√Pr√RaJ(1hξ3δξ3u2)2−N3−12∑k=12N2∑j=1N1+12∑i=12√Pr√RaJ(1hξ1δξ1u3)2−N3−12∑k=12N2−12∑j=12N1∑i=1√Pr√RaJ(1hξ2δξ2u3)2−N3∑k=1N2∑j=1N1∑i=1√Pr√RaJ(1hξ3δξ3u3)2+N3∑k=1N2∑j=1N1+12∑i=12Ju1¯Tξ1,
and
N3∑k=1N2∑j=1N1∑i=1J∂h∂t=−N3∑k=1N2∑j=1N1+12∑i=121√Pr√RaJ(1hξ1δξ1T)2−N3∑k=1N2−12∑j=12N1∑i=11√Pr√RaJ(1hξ2δξ2T)2−N3−12∑k=12N2∑j=1N1∑i=11√Pr√RaJ(1hξ3δξ3T)2−N3∑k=1N2∑j=11√Pr√Ra(Jhξ1T1hξ1δξ1T)|12+N3∑k=1N2∑j=11√Pr√Ra(Jhξ1T1hξ1δξ1T)|N1+12,
with respect to the squared velocity and the squared temperature, respectively.
To derive them, the individual components are focused below.
The derived two relations have source terms and sink terms, which are elaborated below as well as their implementations.