61 replies to this topic

[3rd party JKor]

What's CoSSIMS, you ask? I'll tell you. It's a Continuously Stirred Steam Injection Mashing System.More to follow...

[HVB]

Sweet!I have a 65RPM motor that I have thought about using to stir my mash. Have to re-vist that idea this winter when things slow down. I am really interested in the steam mashing. Have seen it on the other sites but steam scares me a bit.

[3rd party JKor]

Yeah, steam is a little scary, but as long as you're careful and put the proper safeguards in place, it's fine. Plus, working with a pressure cooker as your steam generator, which maxes you out at 15 psi, provides a level of safety in and of itself.

[stellarbrew]

Any project involving steam has got to be fun.

[3rd party JKor]

Any project involving steam has got to be fun.

mmmmmmm...hot, wet gas...

[stellarbrew]

mmmmmmm...hot, wet gas...

Yes, and you can pack a respectable amount of enthalpy into 15 psig saturated steam.

[3rd party JKor]

Yes, and you can pack a respectable amount of enthalpy into 15 psig saturated steam.

Yup, the only issue is energy flow rate. Ultimately, you're limited to whatever the energy output of your source is. In this case it's the electric hot plate that will heat my presure cooker, probably ~1500W. From a continuous energy flow perspective, it's no different than using an electric element in the MLT. The benefit you have is that the pressure cooker itself is storing a lot of energy as hot water and steam @ 250F. So for quick, "small" injections of energy to the mash it's very effective.I'm just going to use the pressure cooker I have (for now), but if I was designing a steam generator the ideal size* would be one that can hold enough energy in water and steam to do the biggest mash step you would normally do.*Techincally, the ideal design would be to heat the generator with a power source capable of continuously generating the amount of steam required to raise the mash temp at 2°C/min (or whatever rate you want).

[stellarbrew]

*Techincally, the ideal design would be to heat the generator with a power source capable of continuously generating the amount of steam required to raise the mash temp at 2°C/min (or whatever rate you want).

Or, take advantage of the energy storage capacity of the steam by getting a larger steam vessel. Borrowing an electrical analogy, the steam vessel is acting like a capacitor, as you have indicated.

Edited by stellarbrew, 03 August 2009 - 12:52 PM.

[3rd party JKor]

Or, take advantage of the energy storage capacity of the steam by getting a larger steam vessel. Borrowing an electrical analogy, the steam vessel is acting like a capacitor, as you have indicated.

Steam isn't dense enough to store significant energy in the vapor form, i.e. the volumetric energy density is low. My 20L pressure cooker will hold about 32 kJ of steam at 100°C, if the same volume is filled with water at 100°C, the energy content of the pressure cooker is about 8,000 kJ.

[stellarbrew]

Steam isn't dense enough to store significant energy in the vapor form, i.e. the volumetric energy density is low. My 20L pressure cooker will hold about 32 kJ of steam at 100°C, if the same volume is filled with water at 100°C, the energy content of the pressure cooker is about 8,000 kJ.

My numbers don't agree exactly with yours, although they are on the same order of maginitude. Your earlier point was that you can realize a benefit of using a steam generator, over using direct electrical resistance heating, because of the energy stored in the hot water and steam. I was merely pointing out that you can maximize that benefit by having a large enough volume of hot water and steam in reserve. In order to effectively utilize the heat stored in your hot water (using the steam injection model you have proposed), the energy input to the steam generator must match the latent heat of vaporization, which is the heat required to boil the water once it is at saturation temperature. At 100 C, hfg is 2256 KJ/Kg, and hg is 2676 KJ/Kg. Thus, you can realize a 16% decrease in instantaneous energy requirement, by having an adequate reserve of heated water.Edit: I see your point now, a 20 L vessel will hold more than enough hot water to supply the steam for any mash step you would undertake.

Edited by stellarbrew, 03 August 2009 - 02:30 PM.

[3rd party JKor]

My numbers don't agree exactly with yours, although they are on the same order of maginitude. Your earlier point was that you can realize a benefit of using a steam generator, over using direct electrical resistance heating, because of the energy stored in the hot water and steam. I was merely pointing out that you can maximize that benefit by having a large enough volume of hot water and steam in reserve. In order to effectively utilize the heat stored in your hot water (using the steam injection model you have proposed), the energy input to the steam generator must match the latent heat of vaporization, which is the heat required to boil the water once it is at saturation temperature. At 100 C, hfg is 2256 KJ/Kg, and hg is 2676 KJ/Kg. Thus, you can realize a 16% decrease in instantaneous energy requirement, by having an adequate reserve of heated water.Edit: I see your point now, a 20 L vessel will hold more than enough hot water to supply the steam for any mash step you would undertake.

Using the latent heat of vaporization and the energy output of your heat source, you can calculate the amount of steam that's generated at a certain pressure. The problem is that if you inject steam at a higher rate than you can produce it, the pressure drops. The water in the steam generator is then superheated and will convert into a gas as fast as it can to make up for the pressure drop. This will continue happening until the pressure in the generator equalizes with the pressure in the MLT (which is essentially 1 atmosphere). If your electrical energy source is powerful enough (or your steam injection is slow enough) the pressure in the generator will be maintained.The point I was trying to make about the water is that it's a better energy storage medium than steam. If you have a system where you can't generate enough heat at your source to maintain your your steam output, the more water you have the more energy you'll be able to put out, because it's acting as an energy sink.I haven't sat down and done the calculations of how much steam can be generated by a 1500W electrical element, so I'm not really sure of all the details...but I'll get to it.

Edited by JKoravos, 03 August 2009 - 03:33 PM.

[Jimmy James]

Yeah, steam is a little scary, but as long as you're careful and put the proper safeguards in place, it's fine.

You could probably say the same about nuclear fission. Looking forward to seeing what you cook up.

[3rd party JKor]

You could probably say the same about nuclear fission. Looking forward to seeing what you cook up.

[stellarbrew]

Using the latent heat of vaporization and the energy output of your heat source, you can calculate the amount of steam that's generated at a certain pressure. The problem is that if you inject steam at a higher rate than you can produce it, the pressure drops. The water in the steam generator is then superheated and will convert into a gas as fast as it can to make up for the pressure drop. This will continue happening until the pressure in the generator equalizes with the pressure in the MLT (which is essentially 1 atmosphere). If your electrical energy source is powerful enough (or your steam injection is slow enough) the pressure in the generator will be maintained.The point I was trying to make about the water is that it's a better energy storage medium than steam. If you have a system where you can't generate enough heat at your source to maintain your your steam output, the more water you have the more energy you'll be able to put out, because it's acting as an energy sink.I haven't sat down and done the calculations of how much steam can be generated by a 1500W electrical element, so I'm not really sure of all the details...but I'll get to it.

Yeah, the steady state gage pressure in your steam drum will be equal to the pressure drop through the distrubution system (frictional and sudden expansion losses). Any superheating will be a transient phenomenon, since at steady state it must exist at equilibrium with the saturated water in your steam drum. You are correct that it will settle out pretty near atmospheric pressure if you are using steam faster than you are making it. It is possible to calculate what the transient heat contribution to the mash would be.What you need is a throttling valve in your distribution system, so that you can balance the steam flow with the input heat. The normal mechanism of control for a small steam generator is to use the throttling valve to maintain drum pressure, and modulate heat input to control steam flow. I recommend setting your drum pressure to the highest safe pressure for best efficiency.Yes I agree with that water will hold more energy per volume than low pressure steam. However, the rate you may use that stored energy is limited by the your ability to supply heat to provide the latent heat of vaporization. If you are limited by a 1500 W element, and you choose an operating drum pressure of 1.0 bar (14.5 psig), then the amount of steam (mass flow rate) which can be supplied on a continuous basis (up until you run out of saturated water), is equal to the input heat rate divided by the heat of vaporization (h_fg) of saturated water at 1.0 bar.m_dot=(1.5 KJ/s)/(2258 KJ/Kg)=6.64E-4 Kg/sYou can check my numbers and see if you agree. I am, after all, presently enjoying my second consecutive Oskar Blues Gordon.What other hobby kind you find someone else who is also a math/science freak and enjoys this stuff?

[3rd party JKor]

Yeah, the steady state gage pressure in your steam drum will be equal to the pressure drop through the distrubution system (frictional and sudden expansion losses). Any superheating will be a transient phenomenon, since at steady state it must exist at equilibrium with the saturated water in your steam drum. You are correct that it will settle out pretty near atmospheric pressure if you are using steam faster than you are making it. It is possible to calculate what the transient heat contribution to the mash would be.

As I go through the build I'll make the heat transfer calculations. I'm not going to go too crazy with the calculations, though, the system "is what it is". I'm not really doing system design, since my heat sources and vessels are already defined by cost and availability. I can make calculations to model the behavior of the system, but at that point it's really just a simulation.

What you need is a throttling valve in your distribution system, so that you can balance the steam flow with the input heat. The normal mechanism of control for a small steam generator is to use the throttling valve to maintain drum pressure, and modulate heat input to control steam flow. I recommend setting your drum pressure to the highest safe pressure for best efficiency.

I'll probably have a flow control valve in the line and I'll "set it and forget it". My goal isn't necessarily to maintain the pressure in the cooker, it's to transfer the energy to the mash as fast as possible. If you think about it, mashing is a transient process, so it's OK for the steam generator to behave in a transient manner as well. If I exhaust all the excess energy in the PC during a step, it's OK because it will have time to recover during the rest.

Yes I agree with that water will hold more energy per volume than low pressure steam. However, the rate you may use that stored energy is limited by the your ability to supply heat to provide the latent heat of vaporization. If you are limited by a 1500 W element, and you choose an operating drum pressure of 1.0 bar (14.5 psig), then the amount of steam (mass flow rate) which can be supplied on a continuous basis (up until you run out of saturated water), is equal to the input heat rate divided by the heat of vaporization (h_fg) of saturated water at 1.0 bar.m_dot=(1.5 KJ/s)/(2258 KJ/Kg)=6.64E-4 Kg/sYou can check my numbers and see if you agree. I am, after all, presently enjoying my second consecutive Oskar Blues Gordon.

Your calculation is correct with respect to supplying continuous steam at a predetermined pressure, but that's not really what I'm trying to do. What I'm trying to do is build up as much energy as I can in the PC, and then release it as fast as I can.Let's say that saturated water at 1 bar is defined as an energy of 0 KJ/kg. I'm defining it this way because there's no pressure differential between the head space of the PC and the mash manifold (this means I'm also assuming the pressure at the mash manifold is 1 bar, even though it is slightly higher due to the static head of the mash liquid...but we're assuming here), so I can't transfer any heat to the mash via pressurized steam.By closing off the system and allowing the pressure/temperature to rise you are able to add energy to the water (and the steam, but in a mostly full pot the energy contained in the steam is negligible). Using steam tables you can see that the enthalpy of water at 15psig is 88.4 kJ/kg (sorry for mixing SI with English units ) greater than at 0 psig. Assuming I filled my 23 quart PC up with 20L of water, I'd have about 20kg of water in my system. So, If I was to heat up my PC to 250F/15psig, I would have about 1770 kJ of energy stored in my PC with respect to the zero energy condition. Now, If I was to completely turn off my electric heating element (and my system was perfectly insulated), I would still be able to transfer that 1770 kJ to my mash without adding any additional heat. The simple process of the water in the PC coming to equilibrium with the pressure at the mash manifold would release that energy. Furthermore, (sorry, I'm in full out geek mode now ) if I go to ProMash and open up the Mash Designer, I can input my recipe and dough-in rate. I'll use 20lbs of grain and 1.25 qts per pound. Mash-in with 25 qts of water at 161 to get a mash temp of 150. Now, if I want to ramp up to mashout at 168, ProMash tells me I need 11.86 qts (= 11.22 L = 11.22 kg*) of boiling water. The differential in enthalpy in boiling water and water at 168F is 103 kJ/kg. Therefore, back calculating from ProMash's data for mash enthalpy, we can see that it takes (11.22 kg * 103 kJ/kg =) 1156 kJ to raise the specified mash from 150F to 168F. Well within the 1770kJ capacity of the full PC of 250F water.We could do the same for raising from a protein rest at 122F to a saccharification rest at 152F. For this case, the required energy input is 2,030kJ. In this case there is a deficit of (2030-1770=) 260kJ. We would need all the energy in the PC plus (260kJ/1.5kW=) 173.3 seconds of heat energy from the heating element.Of course, these calculations are all in a perfect world, with perfect insulation etc. The real system will have significant heat losses.

What other hobby kind you find someone else who is also a math/science freak and enjoys this stuff?

I'm sure there are other hobbies out there with just as many science geeks, the difference is we end up with a crap load of beer as a reward for our geekiness. In other words, we freakin' rule. *I don't know if ProMash corrects for density when stating water volumes at different temperatures, or if the reference it to room temp/STP. 11.86 quarts of boiling water actually weighs 10.75 kg.

Edited by JKoravos, 03 August 2009 - 07:06 PM.

[djinkc]

After watching Drew brewing at MBC, steam jacketed kettles are just awesome. I thought about steam injection at my place and decided not to mess with it. Plus, it would have been a major pain being all electric. But, don't let me stop you.....And, we never see any pics of this stuff................

[3rd party JKor]

Taking a break from the discussion of steam thermodynamics ...here's what I'm going for with this system:Your typical RIMS/HERMS/whatever system is based around control of the mash temp. Of course, this makes sense because it's generally the most difficult thing to do in the brew process. Controlling the mash temp is important to me, but I'm going for full system control here. The system will be virtually touch-free. At the beginning of the brew day I will weigh out my grains, hops and any other kettle additions (whirlfloc, etc.), load the grains into the mill hopper and the hops/etc. in the the timed doser. From there, I won't touch the process until the yeast is ready to pitch. Additionally, the system is going to be set up for CIP, so cleaning will be minimal. This is going to be built in my basment and will be powered completely by electricity. The system will be controlled via process control software loaded onto a PC. The software I have is a very powerful suite designed for data acquisition and process control. The expandability of the control hardware is essentially unlimited. The preliminary system design that I put together consists of 37 control points including 6 thermocouples, 3 pressure sensor, 2 flow meters, 10 solenoids, 3 level floats, 1 limit switch and 12 relays that control various pumps, motors and fans. Ther process will be controlled from milling all the way through the end of primary ferment. Transferring to kegs and drinking the beer is still going to be manual. :angry:OK, any questions?

Edited by JKoravos, 03 August 2009 - 07:59 PM.

[3rd party JKor]

After watching Drew brewing at MBC, steam jacketed kettles are just awesome. I thought about steam injection at my place and decided not to mess with it. Plus, it would have been a major pain being all electric. But, don't let me stop you.....And, we never see any pics of this stuff................

No burn cream needed, we're just having a friendly conversation among brew geeks.

[3rd party JKor]

I'll just add on to the thread as thoughts pop into my head...I'd like to do the whole thing in stainless with 1/2" tube and Swagelok fittings, but things like stainless solenoids are pricey and hard to find at surplus. 1/2"NPT to Swagelok adapter runs about $16 a pop and there are probably at least 20 in my system, so we're probably talking $300-$500 in tube fittings alone unless I can find a surplus outlet.

[stellarbrew]

Let's say that saturated water at 1 bar is defined as an energy of 0 KJ/kg. I'm defining it this way because there's no pressure differential between the head space of the PC and the mash manifold (this means I'm also assuming the pressure at the mash manifold is 1 bar, even though it is slightly higher due to the static head of the mash liquid...but we're assuming here), so I can't transfer any heat to the mash via pressurized steam.By closing off the system and allowing the pressure/temperature to rise you are able to add energy to the water (and the steam, but in a mostly full pot the energy contained in the steam is negligible). Using steam tables you can see that the enthalpy of water at 15psig is 88.4 kJ/kg (sorry for mixing SI with English units ) greater than at 0 psig. Assuming I filled my 23 quart PC up with 20L of water, I'd have about 20kg of water in my system. So, If I was to heat up my PC to 250F/15psig, I would have about 1770 kJ of energy stored in my PC with respect to the zero energy condition. Now, If I was to completely turn off my electric heating element (and my system was perfectly insulated), I would still be able to transfer that 1770 kJ to my mash without adding any additional heat. The simple process of the water in the PC coming to equilibrium with the pressure at the mash manifold would release that energy. Furthermore, (sorry, I'm in full out geek mode now ) if I go to ProMash and open up the Mash Designer, I can input my recipe and dough-in rate. I'll use 20lbs of grain and 1.25 qts per pound. Mash-in with 25 qts of water at 161 to get a mash temp of 150. Now, if I want to ramp up to mashout at 168, ProMash tells me I need 11.86 qts (= 11.22 L = 11.22 kg*) of boiling water. The differential in enthalpy in boiling water and water at 168F is 103 kJ/kg. Therefore, back calculating from ProMash's data for mash enthalpy, we can see that it takes (11.22 kg * 103 kJ/kg =) 1156 kJ to raise the specified mash from 150F to 168F. Well within the 1770kJ capacity of the full PC of 250F water.We could do the same for raising from a protein rest at 122F to a saccharification rest at 152F. For this case, the required energy input is 2,030kJ. In this case there is a deficit of (2030-1770=) 260kJ. We would need all the energy in the PC plus (260kJ/1.5kW=) 173.3 seconds of heat energy from the heating element.Of course, these calculations are all in a perfect world, with perfect insulation etc. The real system will have significant heat losses.

Yeah, that makes sense. As the pressure in the cooker is dropped from 15 psig to atmospheric pressure, the water will boil to release it's surplus enthalpy, in order to reach an equilibrium at atmospheric pressure. The pressure cooker becomes a flash tank. Will this be an original, or has this been done before?