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du(p,T) = $\left(\dfrac{\partial u}{\partial p}\right)_{T}dp + \left(\dfrac{\partial u}{\partial T}\right)_{p}dT $ is the specific internal energy differential.

The idea is to numerically integrate this differential along the two paths, to obtain the state function with known partials that come from the fluid properties. I first integrate along the p-direction, holding T constant.

$du(p,T_{0}) = \left(\dfrac{\partial u}{\partial p}\right)_{T}(p, T_{0})dp$

Then I numerically integrate along the T-direction at each $p_{0}$ holding p constant.

$du(p_{0},T) = \left(\dfrac{\partial u}{\partial T}\right)_{p}(p_{0}, T)dT$

This will yield $\bar u(p,T)$ where $\left(\dfrac{\partial \bar u}{\partial p}\right)_{T} \ne\left(\dfrac{\partial u}{\partial p}\right)_{T}$ along with a very good approximation of the temperature partial

$\left(\dfrac{\partial \bar u}{\partial T}\right)_{p} \sim \left(\dfrac{\partial u}{\partial T}\right)_{p}$

I tried midpoint, trapazoidal, and simpsons rule to see if the approximate pressure partial would converge to the exact partial and nothing mattered. I also refined the grid in both dimensions, and it didn't affect the solution much. It appears the numerical error is coming in through the temperature partial term. It should be noted the partial error in absolute terms is small on the order of 1e-5 [(kJ/kg/kPa)] where the exact absolute value of this partial has a similiar order of magnitude. Is there any way to reduce this numerical error? I'm not sure if percent error is just high because of the exact number being close to zero, or if the theory is wrong.

Thanks!

1 Answers 1

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To obtain a better approximate function $\bar u$, you can solve the below equation on every point on the 2D lattice.

$\left(\dfrac{\partial \bar u}{\partial p}\right)_{T}dp + \left(\dfrac{\partial \bar u}{\partial T}\right)_{p}dT $ = $\left(\dfrac{\partial u}{\partial p}\right)_{T}dp + \left(\dfrac{\partial u}{\partial T}\right)_{p}dT $

Expand around $(p_{i}, T_{j})$

$\left(\dfrac{\partial \bar u}{\partial p}\right)_{T}(p_{i}, T_{j})\Delta p_{i} + \left(\dfrac{\partial \bar u}{\partial T}\right)_{p}(p_{i}, T_{j})\Delta T_{j} $ = $\left(\dfrac{\partial u}{\partial p}\right)_{T}(p_{i}, T_{j}) \Delta p_{i} + \left(\dfrac{\partial u}{\partial T}\right)_{p}(p_{i}, T_{j}) \Delta T_{j}$

Where $\left(\dfrac{\partial \bar u}{\partial p}\right)_{T}(p_{i}, T_{j})$ and $\left(\dfrac{\partial u}{\partial T}\right)_{p}(p_{i}, T_{j})$ need to be approximated using a finite difference. This should lead to a large linear system that can be solved.