{"id":584,"date":"2023-01-11T00:13:34","date_gmt":"2023-01-11T00:13:34","guid":{"rendered":"https:\/\/graywolfslair.com\/?p=584"},"modified":"2023-03-21T19:07:52","modified_gmt":"2023-03-21T19:07:52","slug":"15-32-cannabis-extraction-thermodynamics-101","status":"publish","type":"post","link":"https:\/\/graywolfslair.com\/index.php\/15-diy-equipment\/15-32-cannabis-extraction-thermodynamics-101\/","title":{"rendered":"15.32 Cannabis Extraction Thermodynamics 101"},"content":{"rendered":"\n

Now that more brothers and sisters are using closed loop LPG and EtOH extraction systems at subzero temperatures, questions regularly arise as to how much heating or cooling capacity is required for an application, as well as how to calculate heat exchanger transfer rates in our applications.<\/p>\n\n\n\n

Thermodynamics as the name suggests, is a way to predict how much thermal energy flows from one point to another, based on the temperature difference between the two, as well as the medium through which the thermal energy is flowing.<\/p>\n\n\n\n

In cannabis extraction, we typically chill both our LPG and EtOH processes to below zero temperatures, as well as to chill recovery tanks and their ancillaries. <\/p>\n\n\n\n

We achieve that a number of different ways, but what most have in common is that a lower temperature coolant is on one side of a stainless membrane (wall)  and the solution to be chilled is on the other.  That membrane can be a tank jacket, a tubing wall, or a plate exchanger wall, and the tubing can be either submerged in a chilled bath or in a counter flow heat exchanger. <\/p>\n\n\n\n

When you look at a stainless membrane\u2019s resistance to the flow of thermal energy, you have both the resistance of the membrane itself, as well as the resistance of the boundary layers on both sides, so tube style heat exchangers maintaining rapid flow are more efficient from a transfer standpoint, though they have less time in which to make the transfer, so it takes a longer coil. <\/p>\n\n\n\n

They are also easier to clean in place, which is important when processing food and pharmaceuticals. <\/p>\n\n\n\n

Plate heat exchanger designs that allow easy access for cleaning on the outside and doesn\u2019t circulate product inside the exchangers, provide the maximum surface area in the smallest space.<\/p>\n\n\n\n

Throughout we will refer to \u201cQ\u201d,<\/strong> which is the total heat energy involved.<\/p>\n\n\n\n

The LPG and Ethanol that we are heating, and chilling require a known number of btu\u2019s per pound to raise them one degree Fahrenheit, which is known as the Specific Heat.<\/strong><\/p>\n\n\n\n

The formula is Q = lbs of liquid X Specific HT btu\u2019s X Delta T.<\/strong><\/p>\n\n\n\n

Example: Q btu\u2019s = (1 lb 70:30 n-Butane:n-Propane) X 0.39 btu\/lb X 164F Delta T = 64 btu<\/strong><\/p>\n\n\n\n

Another important number that comes into play is the Heat of Vaporization<\/strong> during recovery, which is the number of additional btu\u2019s per pound that are required over and above its Specific Heat, to change it from a liquid to a vapor state.<\/p>\n\n\n\n

The formula is Q = lbs of liquid X Heat of Vaporization btu\u2019s.<\/strong><\/p>\n\n\n\n

Example: Q btu\u2019s = 1 lb X 171.1 btu\/lb = 171.1 btu\u2019s<\/strong><\/p>\n\n\n\n

We of course need to know how many pounds of LPG or alcohol are involved and as both are liquids, we can use their Specific Gravities<\/strong> times that of water to compute their weight per cubic inch.<\/p>\n\n\n\n

The formula is pounds per cubic inch = 0.0361 lbs\/in3 X Specific Gravity.<\/strong><\/p>\n\n\n\n

Example: Weight of n-butane per cubic inch = 0.0361 lb\/in3 X Specific Gravity of 0.601 = 0.0217 lb\/in3<\/strong><\/p>\n\n\n\n

The Boiling Point<\/strong> of a liquid is the temperature that the internal vapor pressure of a liquid exceeds one atmosphere pressure (14.7 psi Absolute) and it changes to a vapor state. <\/p>\n\n\n\n

If we reduce the atmospheric pressure on a liquid, its boiling point drops and if we increase atmospheric pressure it rises. <\/p>\n\n\n\n

See boiling points at atmospheric pressure of the LPG and EtOH solvents we use below:<\/p>\n\n\n\n

The temperature difference across the heat exchanger membrane is referred to as its Delta Temperature<\/strong>, or Delta T<\/strong>.  For my examples, I use both dry ice slurry and LN2, with the LN2 in a counterflow or cold finger configuration.<\/p>\n\n\n\n

To compute Delta T from an ambient of +70 F to a subzero number, simply add the 70 degrees F to the negative number.<\/p>\n\n\n\n

Delta T = (+F) + (-F)<\/strong><\/p>\n\n\n\n

Example:  Delta T with dry ice slurry = 70F + 109F sublimation temperature of dry ice  = 179 degrees F<\/strong><\/p>\n\n\n\n

Example: Delta T with LN2 = 70F + 320F = 390F<\/strong><\/p>\n\n\n\n

If using a chiller, just substitute that Delta T.<\/p>\n\n\n\n

Consider that when the fluid enters the entry port of the heat exchanger, that the Delta T is greater there than at the end when both sides of the heat exchanger are closer to the same temperature.<\/p>\n\n\n\n

As the Delta T drops, the transfer rates drop in direct proportion, so actual removal rate is the average of all the rates.<\/p>\n\n\n\n

 Example:  When the 110F vapor enters the coil, the Delta T is 205F and when it leaves the Delta T is only 52F and the average Delta T would be the average rate of 205 + 52\/2 = 128.5F Delta T.<\/p>\n\n\n\n

The \u201cK Value\u201d of 304SS<\/strong> is 14.4 btu\/ft2\/hr\/F @ 68F, which is the number of btu\u2019s it will flow through a square foot of surface area, per hour, per degree of Delta T.<\/p>\n\n\n\n

Q = K X Ft2 X Delta T X hours<\/strong><\/p>\n\n\n\n

Example, one square foot for one hour at a 179F DT:  Q = K-14.4 btu\/hr X 1 ft2 X 179F DT X 1 hour = 2578 btu\/hr<\/strong><\/p>\n\n\n\n

The last number in the formula is Time<\/strong>, which is included in some of the numbers, but not others, so it is important to keep our numbers straight.  <\/p>\n\n\n\n

For instance, if a heat exchanger will cool one pound of 70:30 mix 179 degrees F in one hour, it would take an exchanger with 60 times more capacity to do the same job in one minute.<\/p>\n\n\n\n

BTU\/minute = btu\/hr rejection rate X 60 minutes<\/strong><\/p>\n\n\n\n

Example:  2578 btu\/hr X 60 minutes = 154,680 btu\/hr rejection rate.<\/strong><\/p>\n\n\n\n

Let\u2019s first look at those value and then some examples:<\/p>\n\n\n\n

Useful numbers:<\/strong><\/p>\n\n\n\n

1.0          Specific heat<\/strong>, which is the number of btu\u2019s that one (1) pound of specific substance requires to raise its temperature (1) one degree Fahrenheit.<\/p>\n\n\n\n

                1.1          n-Butane, Isobutane, Propane = 0.39 btu\/lb\/F<\/p>\n\n\n\n

                1.2          Ethanol = 0.614 btu\/lb\/F              <\/p>\n\n\n\n

2.0          Heat of Vaporization<\/strong> is the additional btu\u2019s required per pound to change from a liquid state to a vapor state.<\/p>\n\n\n\n

                2.1          n-Butane  and Isobutane = 165.6 btu\/lb<\/p>\n\n\n\n

                2.2          Propane =184 btu\/lb<\/p>\n\n\n\n

                2.3          70\/30 mix = 171.1 btu\/lb<\/p>\n\n\n\n

                2.4          Ethanol = 364 \/btu\/lb    <\/p>\n\n\n\n

3.0          Specific gravity and lbs\/in3<\/strong><\/p>\n\n\n\n

                3.1          Water = 1.0 SG = .0361 lbs\/in3<\/strong><\/p>\n\n\n\n

                                <\/strong>3.1.1      Weight per cubic inch = 0.0361 lb\/in3 X Specific Gravity of Liquid X.<\/p>\n\n\n\n

                3.2          n-Butane = 0.601 SG<\/strong> = .0361 X .601 SG = 0.021696 lb\/in3<\/strong><\/p>\n\n\n\n

                3.3          Isobutane = .563 SG = .0361 X .563 SG = 0.0203243 lb\/in3<\/strong><\/p>\n\n\n\n

                3.4          Propane = .495 SG or .0361 X .495 SG = 0.0178695 lbs\/in3<\/strong><\/p>\n\n\n\n

                3.5          70 n-Butane\/30 n-Propane =.0361 X .548 SG = 0.019765342 lb\/in3<\/strong><\/p>\n\n\n\n

                               3.5.1 <\/strong>(.7 X 0.021696) + (.3 X 0.0178695) = 0.019765342<\/p>\n\n\n\n

               3.6          Ethanol = .787 SG = .0361 X .787 = 0 .0284107 lb\/in3<\/strong>      <\/p>\n\n\n\n

4.0          Boiling points<\/strong> are the point where the internal vapor pressure of a solvent exceeds one atmosphere, or 14.7 psi Absolute.<\/p>\n\n\n\n

               4.1          n-Butane = -1C\/30.2F<\/p>\n\n\n\n

               4.2          Isobutane = -11.7C\/10.94F<\/p>\n\n\n\n

               4.3          Propane = -41.2C\/-42.2F<\/p>\n\n\n\n

               4.4          70\/30 mix = -41.2C\/-42.2F fractional distillation of the Propane<\/p>\n\n\n\n

               4.5          Ethanol = 78.4C\/173.1F<\/p>\n\n\n\n

5.0          Ambient and cooling bath properties<\/strong><\/p>\n\n\n\n

               5.1          Ambient temperature 21C\/70F<\/p>\n\n\n\n

               5.2          Sublimation temperature of dry ice is -78.5C\/-109.3F      <\/p>\n\n\n\n

                              5.2.1      Assuming -70C\/-94F bath temperature.<\/u><\/strong><\/p>\n\n\n\n

               5.3          Dry ice contains approximately 270 btu\/lb<\/p>\n\n\n\n

               5.4          Boiling point of LN2 is -195.8C\/-320F       <\/p>\n\n\n\n

6.0          \u00bd\u201d 304SS tubing Heat Exchanger properties<\/strong><\/p>\n\n\n\n

               6.1          K value for 304 SS<\/strong> = 14.4 btu\/ft2\/hr\/F @ 68F, Per Engineering Toolbox:<\/p>\n\n\n\n

https:\/\/www.engineeringtoolbox.com\/thermal-conductivity-d_429.html<\/a><\/p>\n\n\n\n

               6.2          Transfer area and transfer rate <\/strong>of \u00bd\u201d X .049 wall tubing per foot of length = <\/strong> 15.16 in2 per foot area = 0<\/strong> .10528 ft2 per linear foot<\/strong><\/p>\n\n\n\n

                              6.2.1      0.50 OD \u2013 (.049\u201d X 2) =.402\u201d ID<\/p>\n\n\n\n

                               6.2.2      (.402\u201d ID X 3.14159) x 12\u201d = 15.16 in2 per food area<\/p>\n\n\n\n

                               6.2.3      15.16 in2 \/ 144 in2 = .10528 ft2 per linear foot for heat transfer<\/p>\n\n\n\n

                6.3          Transfer rate<\/strong> = 0.10528 ft2 per foot X 14.4 btu\/ft2\/hr\/ F = 1.52 btu\/hr\/F per linear foot.<\/strong><\/p>\n\n\n\n

7.0          BTU removal to chill one pound of 70:30 mix from ambient of either 0C\/32F, or from 21C\/70F to -50C\/58F.<\/strong><\/p>\n\n\n\n

                <\/strong>7.1          Specific HT btu = SH 0.39 btu X Delt T.<\/p>\n\n\n\n

Ambient temp<\/strong><\/td>Target Temp<\/strong><\/td>Delta T-F<\/strong><\/td>Specific Ht<\/strong><\/td>BTU\u2019s\/LB<\/strong><\/td><\/tr>
0C\/32F<\/td>-50C\/-58F<\/td>90F<\/td>0.39 btu\/lb<\/td>35.1 bti\/lb<\/td><\/tr>
21C\/70F<\/td>-50C\/-58F<\/td>128F<\/td>0.39 btu\/lb<\/td>50 btu\/lb<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

8.0          BTU removal to recover 70:30 mix at a rate of 1lb per minute<\/strong><\/p>\n\n\n\n

               8.1          Specific Heat btu = 1 lb<\/strong> X SH 0.039 btu X Delta T<\/p>\n\n\n\n

               8.2          Heat of Vaporization btu = 1 l<\/strong>b X HV 171.1 btu<\/p>\n\n\n\n

               8.3          Target temperature -41.2C\/-42.2F<\/p>\n\n\n\n

Temperature<\/strong><\/td>BTU\u2019s @ Delta Temperature<\/strong><\/td><\/tr>
Celsius<\/strong><\/td>Fahrenheit<\/strong><\/td>Delta T-F<\/strong><\/td>BTU Spec HT<\/strong><\/td>BTU Ht Vap<\/strong><\/td>Total BTU \/LB<\/strong><\/td><\/tr>
43.3<\/td>110<\/td>152.2<\/td>59.4<\/td>171.1<\/td>231<\/td><\/tr>
48.9<\/td>120<\/td>162.2<\/td>63.3<\/td>171.1<\/td>234<\/td><\/tr>
54.4<\/td>130<\/td>172<\/td>67.1<\/td>171.1<\/td>238<\/td><\/tr>
60<\/td>140<\/td>182<\/td>71<\/td>171.1<\/td>242<\/td><\/tr>
65.6<\/td>150<\/td>192<\/td>75<\/td>171.1<\/td>246<\/td><\/tr>
71.1<\/td>160<\/td>202<\/td>79<\/td>171.1<\/td>250<\/td><\/tr>
76.7<\/td>170<\/td>212<\/td>82.7<\/td>171.1<\/td>254<\/td><\/tr>
82.2<\/td>180<\/td>222<\/td>86.6<\/td>171.1<\/td>258<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

                Soooo, how do we use the useful numbers?<\/p>\n\n\n\n

9.0          Project Example #1.  Chill LPG from ambient of 70F to -58F for injection. <\/strong><\/p>\n\n\n\n

               Assumptions:<\/strong><\/p>\n\n\n\n

                9.1          We have a single column 4\u201d D X 36\u201d tall.<\/p>\n\n\n\n

                9.2          We wish to flow 3 column volumes injected per cycle<\/p>\n\n\n\n

                9.3          We plan to inject one column volume each 1.5 minutes, 3 volumes X 1.5 = 4.5 minutes = 0.075 hours total per injection cycle<\/p>\n\n\n\n

                9.4          We wish to inject below -50C\/-58F.<\/p>\n\n\n\n

10.0        THEREFORE:<\/strong><\/p>\n\n\n\n

               10.1        Three column volumes of LPG = (4\u201d X 4\u201d X 0.7854) X 36\u201d X 3 volumes =  1357 in3 total injected<\/strong><\/p>\n\n\n\n

               10.2        Weight of 70:30 LPG injected = 1357 in3 X 0.<\/strong>019765342 lb\/in3 = 26.8 lb<\/strong>s<\/p>\n\n\n\n

               10.3        Number of btu\u2019s required removed to chill 26.8 lbs from 21C\/70F to -50C\/-58F =  26.8 lbs X 0.39 Specific Heat X (70F + 58F) = 133.8 btu .<\/strong><\/p>\n\n\n\n

               10.4        To remove it in 0.075 hours (4.5 minutes) requires the equivalent heat exchanger capacity of 133.8 btu\/0.075 hours = 1784 btu\/hr  <\/strong><\/p>\n\n\n\n

11.0        Project Example #2.  Recover 1 lb per hour of 70:30 mix using a pump.<\/strong><\/p>\n\n\n\n

               Assumptions:<\/strong><\/p>\n\n\n\n

               11.1        70% n-Butane and 30% n-Propane<\/p>\n\n\n\n

                11.2        Pump exhaust 82C\/180F<\/p>\n\n\n\n

                11.3        Cooling bath at -70C\/-95F.<\/p>\n\n\n\n

                11.4       Recovery rate 1 lb per minute or per 60 lbs per hour<\/p>\n\n\n\n

12.0        Therefore:<\/strong><\/p>\n\n\n\n

                12.1        BTU removal required to cool 60 lbs 70:30 mix from pump discharge vapor<\/strong>s at 82.2C\/180F to -41.2C\/-42.2F  =<\/p>\n\n\n\n

                               (180F + 42.2F) X (60lb X 0.39 SH) = 222.2 Delta T X 23.4 btu =5199 btu\/hr from Specific Heat<\/strong><\/p>\n\n\n\n

                 12.2        BTU removal required remove Heat of Vaporization for 60 lbs 70:30 mix LPG =<\/p>\n\n\n\n

                                60lb  X HV 171.1 btu\/lb = 10,266 btu from Heat of Vaporization.<\/strong><\/p>\n\n\n\n

                 12.3        Total removal required to cool pump 26.8 lb pump discharge = 5199 btu + 10,266 btu = 15,983 hr total<\/strong><\/p>\n\n\n\n

13.0        Project Example # 3.  Recover 1 lb per minute passively using 110F hot water and dry ice slurry.<\/strong><\/p>\n\n\n\n

               Assumptions:<\/strong><\/p>\n\n\n\n

               13.1        70:30 Mix with fractional distillation boiling point of -41.2C\/-42.2F<\/p>\n\n\n\n

                13.2        1 lb per minute or 60 lb hr recovery rate.<\/p>\n\n\n\n

                13.3        Hot bath 43.3C\/110F<\/p>\n\n\n\n

                13.4        Cold bath -70C\/-95F<\/p>\n\n\n\n

14.0        Therefore:<\/strong><\/p>\n\n\n\n

                14.1        (110F + 42.2 F) X (60 lb X .039 btu SH) =  152.2F Delta T X 23.4 btu = 3561 btu Specific Heat removal.<\/strong><\/p>\n\n\n\n

                14.2        60 lbs X HV-171.1 btu =  10,266 btu Heat of Vaporization removal<\/strong><\/p>\n\n\n\n

                14.3        Total removal is 3,561 btu + 10,266 btu = 13,827 btu total<\/strong><\/p>\n\n\n\n

15.0        Project Example # 4.  How many feet of \u00bd\u201d 304SS tubing needed using dry ice slurry?<\/strong><\/p>\n\n\n\n

                ASSUMPTIONS:<\/strong><\/p>\n\n\n\n

                <\/strong>15.1        Start temperature 110F<\/p>\n\n\n\n

                15.2        End temperature -43F<\/p>\n\n\n\n

                15.3        Dry ice slurry bath at -70C\/-95F<\/p>\n\n\n\n

                15.4        Remove 13,827 btu in 1 hour<\/p>\n\n\n\n

                15.5       Use \u00bd\u201d 304SS tubing<\/p>\n\n\n\n

                             15.2.1    Transfer rate of 304CC 1\/2″ tubing 1.52 btu, per linear foot, per degree F, per hour<\/p>\n\n\n\n

16.0        Therefore:<\/strong><\/p>\n\n\n\n

               16.1        Delta T at entry of heat exchanger coil = 110F + 95F = 205F using dry ice slurry<\/p>\n\n\n\n

               16.2        Delta T at end of heat exchanger coil = 95F \u2013 43F = 52F<\/p>\n\n\n\n

               16.3        Average Delta T during passage = 205F + 52F \/2 = 128.5F<\/p>\n\n\n\n

               16.4        13,827 btu \/ (1.52 btu X 128.5F average Delta T) = 13,827 btu\/195.32 = 71<\/strong> feet \u00bd\u201d 304SS tubing required.<\/strong><\/p>\n\n\n\n

17.0        Project Example #5.  How many feet of counterflow heat exchanger required using LN2?:<\/strong><\/p>\n\n\n\n

                ASSUMPTIONS:<\/strong><\/p>\n\n\n\n

                <\/strong>17.1        Start temperature 110F<\/p>\n\n\n\n

                17.2        End temperature -43F<\/p>\n\n\n\n

                17.3        LN2 boiling point is -195.8C\/-320F<\/p>\n\n\n\n

                17.4        Remove 13,827 btu in 1 hour<\/p>\n\n\n\n

                17.5        Use \u00bd\u201d 304SS center tube of counterflow<\/p>\n\n\n\n

                               17.5.1    Transfer rate of 1.52 btu, per linear foot, per degree F, per hour<\/p>\n\n\n\n

18.0        Therefore:<\/strong><\/p>\n\n\n\n

                18.1        Delta T at entry of heat exchanger coil = 110F + 320F = 430F using LN2<\/p>\n\n\n\n

                18.2        Delta T at end of heat exchanger coil = 320 –  43 = 277F<\/p>\n\n\n\n

                18.3        Average Delta T during passage = 430F + 277F \/2 = 353.5F Average Delta T<\/strong><\/p>\n\n\n\n

                18.4        13,827 btu \/ (1.52 btu X 353.5F) =  13,827 btu\/ 537 = 26 feet \u00bd\u201d 304SS tubing required.<\/strong><\/p>\n\n\n\n

19.0        Project EXAMPLE #6:  Cool 5 gallons of EtOH from 21C\/70F to -70C\/-94F by dropping in dry ice.<\/strong><\/p>\n\n\n\n

                ASSUME:<\/strong><\/p>\n\n\n\n

                <\/strong>19.1        5 gallons EtOH<\/p>\n\n\n\n

                19.2        Specific HT EtOH 0.614 btu\/lb\/F<\/p>\n\n\n\n

                19.3        EtOH Specific Gravity .587 and it weighs 0 .0284107 lb\/in3<\/p>\n\n\n\n

                19.4        21C\/70F Ambient temperature<\/p>\n\n\n\n

                19.5        -70C\/-94F bath temperature<\/p>\n\n\n\n

                <\/strong>19.6        Dry ice has 270 btu\/lb cooling potential<\/p>\n\n\n\n

                19.7        231 in3\/gallon<\/p>\n\n\n\n

20.0        Therefore:<\/strong><\/p>\n\n\n\n

                20.1        (5 gal X 231 in3) X .0361 X .787 = 32.8 lbs Ethanol<\/strong><\/p>\n\n\n\n

                20.2        Delta T = 70F + 94F = 164F<\/p>\n\n\n\n

                20.3        32.8 lbs EtOH X SH-0.614 X Delta T 164F = 3303 btu<\/p>\n\n\n\n

                20.4        3303 btu\/270 btu\/lb = 12.24 lbs dry ice required<\/strong><\/p>\n\n\n\n

21.0        Project EXAMPLE #7: Cool tank of n-Butane from 21C\/70F to -1C\/30.2F by evaporating away X-lbs.<\/strong><\/p>\n\n\n\n

                ASSUME:<\/strong><\/p>\n\n\n\n

                <\/strong>21.1        100# Tank @ 80% fill.<\/p>\n\n\n\n

                21.2        Specific gravity of n-Butane .601<\/p>\n\n\n\n

                21.3        Heat of vaporization n-Butane is 165.6 BTU.<\/p>\n\n\n\n

                21.4        Specific heat of n-Butane is 0.39 BTU<\/p>\n\n\n\n

                21.5        Boiling point n-Butane is -1C\/30.2F<\/p>\n\n\n\n

22.0        Therefore:<\/strong><\/p>\n\n\n\n

                22.1        100# Tank is water so 100# X .601 SG X .80 fill = 48# n-Butane<\/p>\n\n\n\n

                22.2        70F \u2013 30.2F = 39.8 Delta Temperature<\/p>\n\n\n\n

                22.3        48# X 0.39 SpHt X 39.8 Delta T = 745 BTU<\/p>\n\n\n\n

                22.4        745 BTU \/ 165.6 BTU HtVp = 4.5 lbs evaporated<\/strong><\/p>\n\n\n\n

23.0        Project EXAMPLE #8: Cool tank of n-Propane from 21C\/70F to -41.2C\/-42.2F by evaporating away X-lbs.<\/strong><\/p>\n\n\n\n

                ASSUME:<\/strong><\/p>\n\n\n\n

                <\/strong>23.1        100# Tank @ 80% fill.<\/p>\n\n\n\n

                23.2        Specific gravity of n-Propane .495<\/p>\n\n\n\n

                23.3        Heat of vaporization n-Butane is 184 BTU.<\/p>\n\n\n\n

                23.4        Specific heat of n-Butane is 0.39 BTU<\/p>\n\n\n\n

                23.5        Boiling point n-Butane is -41.2C\/-42.2F<\/p>\n\n\n\n

24.0        Therefore:<\/strong><\/p>\n\n\n\n

                24.1        100# Tank is water so 100# X .495 SG X .80 fill = 39.6# n-Propane<\/p>\n\n\n\n

                24.2        70F + 42.2F  = 112.2F Delta Temperature<\/p>\n\n\n\n

                24.3        39.6# X 0.39 SpHt X 112.2 Delta T = 1733 BTU needed<\/p>\n\n\n\n

                24.4        1733BTU \/ 184 BTU HtVp = 9.4 lbs<\/strong> evaporated<\/strong>.<\/p>\n\n\n\n

25.0        Project EXAMPLE # 9 Calculate the BTU losses with 1\u201d foam or vacuum panel insulation: <\/strong><\/p>\n\n\n\n

                ASSUME:<\/strong><\/p>\n\n\n\n

                25.1        R = 1\/K<\/p>\n\n\n\n

                25.2        K = 1\/R<\/p>\n\n\n\n

                25.3        K Value of 300 series stainless steel 14.4 btu\/hr\/ft2\/F<\/p>\n\n\n\n

                25.4        R Value for 1\u201d foam insulation = 3.6 to 4.2 or assume about ~4.0<\/p>\n\n\n\n

                25.5        R Value for 1\u201d vacuum panel insulation = 25<\/p>\n\n\n\n

                25.6        Delta Temperature of 44.6C\/112.2F<\/p>\n\n\n\n

                25.7        4\u201d X 36\u201d column with 6\u201d X 34 X 0.109T wall jacket.<\/p>\n\n\n\n

26.0        Therefore:<\/strong><\/p>\n\n\n\n

                <\/strong>26.1        6 X 6 X .7854 X 34 \u201c \/ 144 in2 = 6.7 ft2 outside surface of jacket.<\/p>\n\n\n\n

                26.2        R Value for stainless = 1\/14.4 btu = R-0.0694 btu\/hr\/ft2\/F\/inT<\/p>\n\n\n\n

                26.3        R Value of SS\/foam composite<\/strong> = (R-0.0694 X 0.109T) +  R-4.0 = R-4.008<\/p>\n\n\n\n

                26.4        K Value of SS\/foam composite = 1 \/ 4.008 =K-0.250 btu\/hr\/ft2\/F<\/p>\n\n\n\n

                26.5        BTU hour lost at Delta T 44.6C\/112.2F = 28.1 btu\/hr\/ft2<\/strong><\/p>\n\n\n\n

                26.6        28.1 btu\/hr\/ft2 X 6.7ft2 = 188.3 bty\/hr loss from column<\/strong><\/p>\n\n\n\n

                26.7        R Value of SS\/vacuum composite<\/strong> = (R-0.0694 X 0.109T) +  R-25.0 = R-25.008<\/p>\n\n\n\n

                26.8        K Value of SS\/vacuum composite = 1 \/25.008 =K-.040 btu\/hr\/ft2\/F<\/p>\n\n\n\n

                26.9        BTU hour lost at Delta T 44.6C\/112.2F = 112.2DT X .040 btu\/hr\/ft2 = 4.49 btu\/hr\/ft2<\/strong><\/p>\n\n\n\n

                26.10     4.49 btu\/hr\/ft2 X 6.7 ft2 = 30.1 btu\/hr loss from column<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Now that more brothers and sisters are using closed loop LPG and EtOH extraction systems at subzero temperatures, questions regularly arise as to how much […]<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[50],"tags":[],"class_list":["post-584","post","type-post","status-publish","format-standard","hentry","category-15-diy-equipment"],"yoast_head":"\n15.32 Cannabis Extraction Thermodynamics 101 - GrayWolf's Lair<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/graywolfslair.com\/index.php\/15-diy-equipment\/15-32-cannabis-extraction-thermodynamics-101\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"15.32 Cannabis Extraction Thermodynamics 101 - GrayWolf's Lair\" \/>\n<meta property=\"og:description\" content=\"Now that more brothers and sisters are using closed loop LPG and EtOH extraction systems at subzero temperatures, questions regularly arise as to how much […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/graywolfslair.com\/index.php\/15-diy-equipment\/15-32-cannabis-extraction-thermodynamics-101\/\" \/>\n<meta property=\"og:site_name\" content=\"GrayWolf's Lair\" \/>\n<meta property=\"article:author\" content=\"https:\/\/www.facebook.com\/james.ellis.9081\" \/>\n<meta property=\"article:published_time\" content=\"2023-01-11T00:13:34+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-03-21T19:07:52+00:00\" \/>\n<meta name=\"author\" content=\"GrayWolf\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"GrayWolf\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"12 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/graywolfslair.com\\\/index.php\\\/15-diy-equipment\\\/15-32-cannabis-extraction-thermodynamics-101\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/graywolfslair.com\\\/index.php\\\/15-diy-equipment\\\/15-32-cannabis-extraction-thermodynamics-101\\\/\"},\"author\":{\"name\":\"GrayWolf\",\"@id\":\"https:\\\/\\\/graywolfslair.com\\\/#\\\/schema\\\/person\\\/40d6da885f58d0e767e114277e1c7d1c\"},\"headline\":\"15.32 Cannabis Extraction Thermodynamics 101\",\"datePublished\":\"2023-01-11T00:13:34+00:00\",\"dateModified\":\"2023-03-21T19:07:52+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/graywolfslair.com\\\/index.php\\\/15-diy-equipment\\\/15-32-cannabis-extraction-thermodynamics-101\\\/\"},\"wordCount\":5582,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/graywolfslair.com\\\/#organization\"},\"articleSection\":[\"15. 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