Mixing Waters of Different Temperature It is important to understand how much hot water to add to a batch of colder water in order to obtain a particular temperature; you might desire to mash using something other than a single-temperature infusion mash (i.e. step-mash schedule) : Fluid Dynamics Lecture XXX - P 1

Mixing Waters of Different Temperature To achieve these “temperature steps”, hot water is added to the mash, and then the temperature is held relatively constant for a period of time in order allow enzymes to work at an optimal temperature for a particular kind of enzyme Hot water additions Fluid Dynamics Lecture 1- P 2

Mixing Waters of Different Temperature Here’s a sketch of the mash-tun that illustrates what’s happening when a hot-water addition occurs: Hot Water Addition: Temp = 200 °F Mass (or Volume) = ??? Mash Water & Grain Bed: Tinitial = 122 °F Massinitial (or Volume) = 2500 lbs (300 gal) Tfinal = 140 °F Massfinal (or Volume) = ??? We want to figure out how much hot water is needed to achieve desired mash temperature Fluid Dynamics Lecture 1- P 3

Mixing Waters of Different Temperature Perhaps the easiest and most practical way to do this is to monitor the temperature of the mash as you are adding hot water, and stop the addition when desired temperature is achieved. But you will still need to: Be sure that your mash system is well-mixed to ensure as uniform a temperature as possible within the mash tun Ensure temperature indicator is accurate Ensure the hot addition water is hot enough to do the job without requiring excessive water volume It’s always a good idea to build experience with your particular system, but it is also important to calculate the water volume and temperature requirements before mashing in order to be sure that your system can do what you need it to do. Fluid Dynamics Lecture 1- P 4

Mixing Waters of Different Temperature To determine the amount of water needed to achieve a particular temperature when mixed with another volume of water, it is useful to understand that heat will be transferred between the different-temperature waters The temperature of the new mixture can be calculated by assuming an adiabatic system (heat not lost or gained from the overall system) and understanding that the heat lost by the hotter water will be equal to the heat gained by the cooler water (and grain bed). To simplify the discussion and keep the calculations simple, let’s assume that we are working only with water Fluid Dynamics Lecture 1- P 5

Mixing Waters of Different Temperature The amount of heat required to increase the temperature of a particular amount of water by a specified amount is given by: Where: Q = heat required, BTU m = mass of water, lbs Cp = heat capacity of water, BTU/lb-°F DT = temperature change of the water, °F 𝑸=𝒎 𝑪 𝒑 𝜟𝑻 Fluid Dynamics Lecture 1- P 6

Mixing Waters of Different Temperature Since the amount of heat gained by an amount of cooler water is equal to the amount that must be supplier by some amount of hotter water, the following relationship is valid Where: Q1 = heat gained by cooler water, BTU m1 =mass of cooler water, lbs Cp1 = heat capacity of cooler water, BTU/lb- °F DT1 = temperature change of the cooler water, °F Q2 = heat provided by hotter water, BTU m2 =mass of hotter water, lbs Cp2 = heat capacity of hotter water, BTU/lb- °F DT2 = temperature change of the hotter water, °F 𝑄 1 = 𝑄 2 𝑄 1 = 𝑚 1 𝐶 𝑝1 Δ 𝑇 1 = 𝑄 2 = 𝑚 2 𝐶 𝑝2 Δ 𝑇 2 Fluid Dynamics Lecture 1- P 7

Mixing Waters of Different Temperature It is important to understand that the heat capacity terms, Cp1 & Cp2, are, for practical purposes, equal and have a value that is very close to 1 BTU/lb-°F It is also important to understand that the DT terms, DT1 & DT2, are equivalent to the absolute value of (Tfinal – Tinitial) for the water in question Example: if Tinitial = 122°F and Tfinal = 140°F, then DT = (140°F - 122°F) = 18°F Fluid Dynamics Lecture 1- P 8

Mixing Waters of Different Temperature We can now do a bit of algebra and rearrange the equations to calculate the amount of hot water that is needed to raise the temperature to a particular level: 𝑄 1 = 𝑚 1 𝐶 𝑝1 Δ 𝑇 1 = 𝑄 2 = 𝑚 2 𝐶 𝑝2 Δ 𝑇 2 𝑚 1 𝐶 𝑝1 Δ 𝑇 1 = 𝑚 2 𝐶 𝑝2 Δ 𝑇 2 𝑚 1 𝐶 𝑝1 Δ 𝑇 1 𝐶 𝑝2 Δ 𝑇 2 =𝑚2 Fluid Dynamics Lecture 1- P 9

Mixing Waters of Different Temperature Since Cp2 ≈ Cp1 , we can cancel them out in the numerator and denominator: And then expand the DT terms to get a useful equation: Where: m1 =mass of cooler water in lauter tun, lbs m2 =mass of hotter water being added, lbs Tcoldfinal = final temp of originally-cooler water Tcoldinitial = initial temp of originally-cooler water Thotfinal = final temp of originally-hotter water Thotinitial = initial temp of originally-hotter water 𝑚 1 Δ 𝑇 1 Δ 𝑇 2 =𝑚2 𝑚 2 = 𝑚 1 𝑇 𝑐𝑜𝑙𝑑𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑐𝑜𝑙𝑑𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑇 ℎ𝑜𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑇 ℎ𝑜𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 Fluid Dynamics Lecture 1- P 10

Mixing Waters of Different Temperature Let’s work an example to illustrate how this equation is used to calculate the amount of hot water that must be added to increase the temperature of the mash liquid. Here’s the system (assume we are working only with water): Hot Water Addition: Temp = 200 °F Mass = ??? Mash Water & Grain Bed: Tinitial = 122 °F Massinitial = 2500 lbs Tfinal = 140 °F Massfinal = Massinitial + Mass of hot water added = ??? Fluid Dynamics Lecture 1- P 11

Mixing Waters of Different Temperature For this system we have: m1 = 2500 lbs Tcoldfinal = 140°F Tcoldinitial = 122°F Thotfinal = 140°F Thotinitial = 200°F Hot Water Addition: Thotinitial = 200 °F Mass = ??? Mash Water & Grain Bed: Tcoldinitial = 122 °F Massinitial = 2500 lbs Tfinal = 140 °F Massfinal = Massinitial + Mass of hot water added = ??? Fluid Dynamics Lecture 1- P 12

Mixing Waters of Different Temperature If we plug the values m1 = 2500 lbs Tcoldfinal = 140°F Tcoldinitial = 122°F Thotfinal = 140°F Thotinitial = 200°F Into the equation: 𝑚 2 = 𝑚 1 𝑇 𝑐𝑜𝑙𝑑𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑐𝑜𝑙𝑑𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑇 ℎ𝑜𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑇 ℎ𝑜𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 Fluid Dynamics Lecture 1- P 13

Mixing Waters of Different Temperature We get: 𝑚 2 = 𝑚 1 𝑇 𝑐𝑜𝑙𝑑𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑐𝑜𝑙𝑑𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑇 ℎ𝑜𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑇 ℎ𝑜𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑚 2 = 2500 𝑙𝑏𝑠 140°𝐹−122°𝐹 140°𝐹−200°𝐹 𝑚 2 = 2500 𝑙𝑏𝑠 18°𝐹 60°𝐹 𝑚 2 =750 𝑙𝑏𝑠 We need 750 lbs of hot water Fluid Dynamics Lecture 1- P 14

Mixing Waters of Different Temperature Since the final weight of the water in the mash tun = m1 + m2, the final weight of the water in the mash tun will be m1 + m2 = 2500 lbs + 750 lbs = 3250 lbs Recall that the conversion factor for water between lbs and gallons is: So we can also say that we had to add: of 200°F water In order to accomplish our goal. 1 𝑔𝑎𝑙 𝑤𝑎𝑡𝑒𝑟=8.34 𝑙𝑏𝑠 𝑤𝑎𝑡𝑒𝑟 750 𝑙𝑏𝑠 𝑤𝑎𝑡𝑒𝑟 1 𝑔𝑎𝑙 𝑤𝑎𝑡𝑒𝑟 8.34 𝑙𝑏𝑠 𝑤𝑎𝑡𝑒𝑟 =89.9 𝑔𝑎𝑙 𝑤𝑎𝑡𝑒𝑟 We need 750 lbs (89.9gal) of 200°F water to raise the temperature of 2500 lbs (300 gal) of 122°F water + 750 lbs (89.9 gal) new water to 140°F Fluid Dynamics Lecture 1- P 15

Mixing Waters of Different Temperature We can also calculate the final temperature of a mixture of known amounts of hotter and cooler water (m1, m2 Tcoldinitial & Thotinitial known) by using a weighted average method: 𝑚 1 𝑚 1+ 𝑚 2 𝑇 𝑐𝑜𝑙𝑑𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑚 2 𝑚 1 + 𝑚 2 𝑇 ℎ𝑜𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =𝐹𝑖𝑛𝑎𝑙 𝑇𝑒𝑚𝑝 𝑜𝑓 𝑀𝑖𝑥𝑡𝑢𝑟𝑒 Fluid Dynamics Lecture 1- P 16

Mixing Waters of Different Temperature As an example, use: m1 = 2500 lbs Tcoldinitial = 122°F m2 = 750 lbs Thotinitial = 200°F And substitute into: To get: 𝑚 1 𝑚 1+ 𝑚 2 𝑇 𝑐𝑜𝑙𝑑𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑚 2 𝑚 1 + 𝑚 2 𝑇 ℎ𝑜𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =𝐹𝑖𝑛𝑎𝑙 𝑇𝑒𝑚𝑝 𝑜𝑓 𝑀𝑖𝑥𝑡𝑢𝑟𝑒 2500 𝑙𝑏𝑠 2500 𝑙𝑏𝑠+750 𝑙𝑏𝑠 122°𝐹 + 750 𝑙𝑏𝑠 2500 𝑙𝑏𝑠+750 𝑙𝑏𝑠 200°𝐹 =140°𝐹 Fluid Dynamics Lecture 1- P 17

Mixing Waters of Different Temperature Since mass and volume are converted between one another by using a constant value for the density, this same “weighted average” approach is applicable if volume is know instead of mass v1 = volume of cooler water (gallons) v2 = volume of hotter water (gallons) 𝑣 1 𝑣 1+ 𝑣 2 𝑇 𝑐𝑜𝑙𝑑𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑣 2 𝑣 1 + 𝑣 2 𝑇 ℎ𝑜𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =𝐹𝑖𝑛𝑎𝑙 𝑇𝑒𝑚𝑝 𝑜𝑓 𝑀𝑖𝑥𝑡𝑢𝑟𝑒 Fluid Dynamics Lecture 1- P 18