Aciduated water is added to lower the degrees Brix (\(^\circ\textrm{Bx}\)) of grape juice for winemaking. Note that per 5 gallons of must there are about 3 (Bordeaux grapes) to 3.5 (Rhône grapes & Zinfandel) gallons of liquid, i.e. about a fraction 0.6 to 0.7 of the must volume. These choices are available as dropdown options if you click once or twice in some browsers. Since this input value is inexact, it's best to start by mixing in a portion of the water and retesting before adding the entire addition.
The dilution of juice or must calculation takes as input the initial volume of the must or juice, \(V_{initial}\), the fraction of the must or juice that is juice, \(f\), the initial degrees Brix (\(^\circ\textrm{Bx}\)) of the juice, \(B_{initial}\), and the target \(^\circ\textrm{Bx}\) of the juice after diluting, \(B_{final}\). These are used to calculate the volume of aciduated water, \(V_{water}\), to add to the must or juice to reach \(B_{final}\), and the mass of tartaric acid, $M_{tartaric-acid}$, to add to the water before mixing it in the must. The volume of aciduated water to add is calculated as $$V_{water} = \left(\frac{B_{initial}}{B_{final}} - 1\right) \cdot f \cdot V_{initial}$$ To derive this equation first note that the final volume of juice, \(V_{juice-final}\), will be the sum of the initial volume of juice, \(V_{juice-initial}\), and the volume of aciduated water added, \(V_{water}\), $$V_{juice-final} = V_{juice-initial} + V_{water}$$ When aciduated water is added to juice, the mass of sugar dissolved in the solution will remain constant, so the initial mass of sugar in the solution, \(M_{sugar-initial}\), will equal the final mass of sugar in the solution, \(M_{sugar-final}\), $$M_{sugar-initial} = M_{sugar-final}$$ We seek to rewrite this equation in terms of the measured values of \(^\circ\textrm{Bx}\), and volume of solution, and then use it with the volume addition equation to eliminate \(V_{juice-final}\) and solve for \(V_{water}\). $^\circ\textrm{Bx}$ is a unit characterizing all dissolved solutes in a solution. For the current winemaking purposes, solutes other than sugar can be ignored, and we can assume that degrees Brix (\(^\circ\textrm{Bx}\)) represents the mass percent of sucrose in the solution, $$1^\circ\textrm{Bx} = \frac{1\textrm{g sucrose}}{100\textrm{g of solution}}\cdot 100$$ The mass of sugar dissolved is then the measured Brix multiplied by the mass of solution. The mass of a solution, \(M_{solution}\), is equal to the volume density (mass per unit volume) of the solution, \(\rho_{solution}\), multiplied by the volume of solution \(V_{solution}\), $M_{solution} = \rho_{solution} \cdot V_{solution}$ Then with $B$ as the $^\circ\textrm{Bx}$ of the solution, the mass of sugar in the solution can be written as $$M_{sugar} = B \cdot M_{solution} / 100 = B \cdot \rho_{solution} \cdot V_{solution} /100$$ Using this in the sucrose mass conservation equation, with the initial and final measured \(^\circ\textrm{Bx}\) represented by \(B_{initial}\) and \(B_{final}\), gives $$B_{initial} \cdot \rho_{juice-initial} \cdot V_{juice-initial} = B_{final} \cdot \rho_{juice-final} \cdot V_{juice-final}$$ Rearranging gives $$V_{juice-final} = \frac{\rho_{juice-initial}}{\rho_{juice-final}} \cdot \frac{B_{initial}}{B_{final}} \cdot V_{juice-initial}$$ Solving for \(V_{water}\) in the volume addition equation and plugging in this relation for \(V_{juice-final}\) then gives $$V_{water} = \left(\frac{\rho_{juice-initial}}{\rho_{juice-final}} \cdot \frac{B_{initial}}{B_{final}}- 1\right) \cdot V_{juice-initial}$$ For the case of must, $$V_{juice-initial} = f \cdot V_{must-initial}$$ where \(f\) represents the fraction of the must that is juice and \(V_{must-initial}\) is the initial volume of must $$V_{water} = \left(\frac{\rho_{juice-initial}}{\rho_{juice-final}} \cdot \frac{B_{initial}}{B_{final}}- 1\right) \cdot f \cdot V_{must-initial}$$ For pure juice \(f = 1\) and $V_{juice-initial} = V_{must-initial}$. For the typical range of values and the accuracy needed, the difference in densities of the initial and final solutions can be ignored, \(\rho_{juice-initial} \approx \rho_{juice-final}\). Canceling the densities then gives the result $$V_{water} \approx \left(\frac{B_{initial}}{B_{final}} - 1\right) \cdot f \cdot V_{must-initial}$$ The mass of tartaric acid to add is calculated by multiplying the grams of acid per liter of water by the volume of water in liters $$M^{in-grams}_{tartaric-acid} = \frac{\textrm{grams tartaric acid}}{1\textrm{L water}} \cdot V^{in-liters}_{water}$$