1 Heating Heating of liquids is required during the mashing and boiling steps in the brewing process Wort boiling has the highest energy requirement of any of the steps in the brewing process. Depending upon the types of packaging operations a brewery might have, wort boiling can account for as much as 60% of the total heating (and/or steam) demand in the brewery. Because of this people who design, own and operate breweries have expended a lot of effort to reduce energy demand, improve energy efficiency, and improve energy recovery from wort boiling operations Fluid Dynamics Lecture XXX - P 1
2 Heating There are several ways that modern breweries can heat and boil wort in a brew kettle Traditional direct-fired kettles Kettles with internal heating systems (e.g. steam coils) Kettles with external jackets (e.g. steam jackets) Internal or external wort boilinig system (e.g. calandria) Each type of equipment has advantages and disadvantages Fluid Dynamics Lecture XXX - P 2
3 Heating – Direct Fired KettlesDirect-fired kettles are exactly what the name says – a kettle that has fire directly beneath it. In a modern direct-fired brewery, the fire is almost always derived from burning natural gas, but historically, the fuel for the fire may have been wood, coal or peat Firebox Enclosure Brick Support Currents Boil Vessel Grade Fluid Dynamics Lecture XXX - P 3
4 Heating – Direct Fired KettlesBecause the heat is localized at the bottom of the kettle, the volume of wort that can be boiled in a single batch is usually a maximum of about 200 bbls. A disadvantage of using direct-fired kettles is that they are relatively inefficient in transferring heat from the fire to the wort. Additionally, the heating surfaces near the flame become very hot. This promotes carmalization and darkening and burning of the wort and the burnt wort requires frequent (every 2-5 batches?) cleaning. Also, high evaporation rates are required in order to produce an effective, turbulent, vigorous boil. Evaporation rates of 10%/hr are common. Fluid Dynamics Lecture XXX - P 4
5 Heating – Internally Heated KettlesInternally heated kettles typically use steam coils that are actually inside the boiling vessel Saturated steam, produced by an external boiler, is allowed to pass through the coils, condense, and then give up heat to the wort within the vessel. Boil Vessel Boil Vessel Steam in Condensate Return out Currents Coil Fluid Dynamics Lecture XXX - P 5
6 Heating – Internally Heated KettlesInternally heated kettles typically use steam coils that are actually inside the boiling vessel Use of steam coils allows use of larger boiling vessels because there is a larger surface area available for heat transfer. Because the coil is completely surrounded by wort, heat transfer is more efficient Additionally, coil heat-transfer surface temperatures are lower than with direct-fire systems, so there is less carmalization and wort burning Fluid Dynamics Lecture XXX - P 6
7 Heating – Internally Heated KettlesDisadvantages of using internal steam coils are largely related to cleaning difficulty using conventional CIP systems. Build-up of “wort gunk” on coils reduces heat transfer efficiency and may require manual cleaning if excessive. Also, because wort circulation relies on thermal currents and boiling, turbulence over the coils may not always be adequate to prevent some carmalization of wort from happening Another disadvantage is the potential for corrosion of the coils and the possibility of introducing steam/condensate into the wort, or having wort back-flow into the steam delivery or condensate return piping. Fluid Dynamics Lecture XXX - P 7
8 Heating –Kettles With Steam JacketsAn alternative to immersed-coil steam heating is to use a steam-jacketed vessel This overcomes the difficulties associated with the need to clean the internal coils Boil Vessel Steam in Condensate Return out Currents Steam Jacket Surrounding the Vessel Body Fluid Dynamics Lecture XXX - P 8
9 Heating –Kettles With Steam JacketsJacketed vessels, like direct-fired vessels, have the similar problem of achieving adequate, efficient heat transfer to obtain a satisfactory boil. Although jacketed vessels have a lower tendency to foul than either direct-fired vessels or immersion-coil heated vessels, they will still require cleaning every 6-12 batches to ensure that effective heat transfer is maintained. Because there is no cleaning issue with an immersion coil on a jacket-heated vessel, CIP systems are much more effective in cleaning than with an immersed-coil system Fluid Dynamics Lecture XXX - P 9
10 Heating –External Wort Boiling SystemsA more modern design uses an external heater (external wort boiler) which removes wort from the kettle and passes it through a heat exchanger for heating These wort boilers achieve high rates of heat transfer by exploiting the two-phase flow and nucleate boiling situation within the system Currents Heat Exchanger Pump Steam in Condensate Return out Fluid Dynamics Lecture XXX - P 10
11 Heating –External Wort Boiling SystemsThese systems operate at relatively low steam pressures ( bar) to heat the wort in the heat exchanger Turbulence in the boil is enhanced by the pump-induced movement of the wort through the heating loop Because of these efficiencies, the classic 90 minute boil with 10%/hr evaporation loss can often be reduced to a 60 minute boil with only 5-6%/hr evaporative loss, without loss of wort/beer quality This represents a significant improvement in efficiency and energy utilization Fluid Dynamics Lecture XXX - P 11
12 Heating –External Wort Boiling SystemsAdditionally, because of the design, pre-heating of he wort can start when about 15% of the total kettle contents have been introduced into the vessel. This allows the kettle to heat in parallel with filling and be at a boil almost immediately when filled, thus improving vessel utilization Also, since low-pressure steam in used, the rate of fouling of the heat-transfer surface is decreased Because of this, the batches between required “deep cleanings” may be achieved. This decreases brewhouse downtime and increases throughput. Fluid Dynamics Lecture XXX - P 12
13 Heating –External Wort Boiling SystemsOne disadvantage of using external wort boiling is that the action of pumping increases shear forces on the wort. This can damage floc formation (trub or hot-break particles) and increase filtration/sedimentation times Another disadvantage is the relative complexity of an external-boil system vs. other types, and the increased capital cost and maintenance costs that go along with increased system complexity. Fluid Dynamics Lecture XXX - P 13
14 Agenda 1st topic Learning Objectives Page 42nd topic Cleaning & Sanitation Page 6 3rd topic Clean-In-Place Systems Page 38 4th topic Heating Equipment Page 57 5th topic Chilling Equipment Page 71 6th topic Temperature From Mixing Page 83
15 Mixing Waters of Different TemperatureIt 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 15
16 Mixing Waters of Different TemperatureTo 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 16
17 Mixing Waters of Different TemperatureHere’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 17
18 Mixing Waters of Different TemperaturePerhaps 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 18
19 Mixing Waters of Different TemperatureTo 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 19
20 Mixing Waters of Different TemperatureThe 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 20
21 Mixing Waters of Different TemperatureSince 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 21
22 Mixing Waters of Different TemperatureIt 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 22
23 Mixing Waters of Different TemperatureWe 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 23
24 Mixing Waters of Different TemperatureSince 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 24
25 Mixing Waters of Different TemperatureLet’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 25
26 Mixing Waters of Different TemperatureFor 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 26
27 Mixing Waters of Different TemperatureIf 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 27
28 Mixing Waters of Different TemperatureWe get: 𝑚 2 = 𝑚 1 𝑇 𝑐𝑜𝑙𝑑𝑓𝑖𝑛𝑎𝑙 − 𝑇 𝑐𝑜𝑙𝑑𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑇 ℎ𝑜𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑇 ℎ𝑜𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑚 2 = 𝑙𝑏𝑠 140°𝐹−122°𝐹 °𝐹−200°𝐹 𝑚 2 = 𝑙𝑏𝑠 18°𝐹 60°𝐹 𝑚 2 =750 𝑙𝑏𝑠 We need 750 lbs of hot water Fluid Dynamics Lecture 1- P 28
29 Mixing Waters of Different TemperatureSince 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 = lbs 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 lbs (89.9 gal) new water to 140°F Fluid Dynamics Lecture 1- P 29
30 Mixing Waters of Different TemperatureWe 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 30
31 Mixing Waters of Different TemperatureAs 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 𝑙𝑏𝑠 2500 𝑙𝑏𝑠+750 𝑙𝑏𝑠 200°𝐹 =140°𝐹 Fluid Dynamics Lecture 1- P 31
32 Mixing Waters of Different TemperatureSince 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 32