15.32 Cannabis Extraction Thermodynamics 101

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.

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.

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. 

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. 

When you look at a stainless membrane’s 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. 

They are also easier to clean in place, which is important when processing food and pharmaceuticals. 

Plate heat exchanger designs that allow easy access for cleaning on the outside and doesn’t circulate product inside the exchangers, provide the maximum surface area in the smallest space.

Throughout we will refer to “Q”, which is the total heat energy involved.

The LPG and Ethanol that we are heating, and chilling require a known number of btu’s per pound to raise them one degree Fahrenheit, which is known as the Specific Heat.

The formula is Q = lbs of liquid X Specific HT btu’s X Delta T.

Example: Q btu’s = (1 lb 70:30 n-Butane:n-Propane) X 0.39 btu/lb X 164F Delta T = 64 btu

Another important number that comes into play is the Heat of Vaporization during recovery, which is the number of additional btu’s per pound that are required over and above its Specific Heat, to change it from a liquid to a vapor state.

The formula is Q = lbs of liquid X Heat of Vaporization btu’s.

Example: Q btu’s = 1 lb X 171.1 btu/lb = 171.1 btu’s

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 times that of water to compute their weight per cubic inch.

The formula is pounds per cubic inch = 0.0361 lbs/in3 X Specific Gravity.

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

The Boiling Point 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. 

If we reduce the atmospheric pressure on a liquid, its boiling point drops and if we increase atmospheric pressure it rises. 

See boiling points at atmospheric pressure of the LPG and EtOH solvents we use below:

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

To compute Delta T from an ambient of +70 F to a subzero number, simply add the 70 degrees F to the negative number.

Delta T = (+F) + (-F)

Example:  Delta T with dry ice slurry = 70F + 109F sublimation temperature of dry ice  = 179 degrees F

Example: Delta T with LN2 = 70F + 320F = 390F

If using a chiller, just substitute that Delta T.

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.

As the Delta T drops, the transfer rates drop in direct proportion, so actual removal rate is the average of all the rates.

 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.

The “K Value” of 304SS is 14.4 btu/ft2/hr/F @ 68F, which is the number of btu’s it will flow through a square foot of surface area, per hour, per degree of Delta T.

Q = K X Ft2 X Delta T X hours

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

The last number in the formula is Time, which is included in some of the numbers, but not others, so it is important to keep our numbers straight.  

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.

BTU/minute = btu/hr rejection rate X 60 minutes

Example:  2578 btu/hr X 60 minutes = 154,680 btu/hr rejection rate.

Let’s first look at those value and then some examples:

Useful numbers:

1.0          Specific heat, which is the number of btu’s that one (1) pound of specific substance requires to raise its temperature (1) one degree Fahrenheit.

                1.1          n-Butane, Isobutane, Propane = 0.39 btu/lb/F

                1.2          Ethanol = 0.614 btu/lb/F              

2.0          Heat of Vaporization is the additional btu’s required per pound to change from a liquid state to a vapor state.

                2.1          n-Butane  and Isobutane = 165.6 btu/lb

                2.2          Propane =184 btu/lb

                2.3          70/30 mix = 171.1 btu/lb

                2.4          Ethanol = 364 /btu/lb    

3.0          Specific gravity and lbs/in3

                3.1          Water = 1.0 SG = .0361 lbs/in3

                                3.1.1      Weight per cubic inch = 0.0361 lb/in3 X Specific Gravity of Liquid X.

                3.2          n-Butane = 0.601 SG = .0361 X .601 SG = 0.021696 lb/in3

                3.3          Isobutane = .563 SG = .0361 X .563 SG = 0.0203243 lb/in3

                3.4          Propane = .495 SG or .0361 X .495 SG = 0.0178695 lbs/in3

                3.5          70 n-Butane/30 n-Propane =.0361 X .548 SG = 0.019765342 lb/in3

                               3.5.1 (.7 X 0.021696) + (.3 X 0.0178695) = 0.019765342

               3.6          Ethanol = .787 SG = .0361 X .787 = 0 .0284107 lb/in3      

4.0          Boiling points are the point where the internal vapor pressure of a solvent exceeds one atmosphere, or 14.7 psi Absolute.

               4.1          n-Butane = -1C/30.2F

               4.2          Isobutane = -11.7C/10.94F

               4.3          Propane = -41.2C/-42.2F

               4.4          70/30 mix = -41.2C/-42.2F fractional distillation of the Propane

               4.5          Ethanol = 78.4C/173.1F

5.0          Ambient and cooling bath properties

               5.1          Ambient temperature 21C/70F

               5.2          Sublimation temperature of dry ice is -78.5C/-109.3F      

                              5.2.1      Assuming -70C/-94F bath temperature.

               5.3          Dry ice contains approximately 270 btu/lb

               5.4          Boiling point of LN2 is -195.8C/-320F       

6.0          ½” 304SS tubing Heat Exchanger properties

               6.1          K value for 304 SS = 14.4 btu/ft2/hr/F @ 68F, Per Engineering Toolbox:

                              https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

               6.2          Transfer area and transfer rate of ½” X .049 wall tubing per foot of length =  15.16 in2 per foot area = 0 .10528 ft2 per linear foot

                              6.2.1      0.50 OD – (.049” X 2) =.402” ID

                               6.2.2      (.402” ID X 3.14159) x 12” = 15.16 in2 per food area

                               6.2.3      15.16 in2 / 144 in2 = .10528 ft2 per linear foot for heat transfer

                6.3          Transfer rate = 0.10528 ft2 per foot X 14.4 btu/ft2/hr/ F = 1.52 btu/hr/F per linear foot.

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.

                7.1          Specific HT btu = SH 0.39 btu X Delt T.

Ambient temp

Target Temp

Delta T-F

Specific Ht

BTU’s/LB

0C/32F

-50C/-58F

90F

0.39 btu/lb

35.1 bti/lb

21C/70F

-50C/-58F

128F

0.39 btu/lb

50 btu/lb

 

8.0          BTU removal to recover 70:30 mix at a rate of 1lb per minute

               8.1          Specific Heat btu = 1 lb X SH 0.039 btu X Delta T

               8.2          Heat of Vaporization btu = 1 lb X HV 171.1 btu

               8.3          Target temperature -41.2C/-42.2F

Temperature

BTU’s @ Delta Temperature

Celsius

Fahrenheit

Delta T-F

BTU Spec HT

BTU Ht Vap

Total BTU /LB

43.3

110

152.2

59.4

171.1

231

48.9

120

162.2

63.3

171.1

234

54.4

130

172

67.1

171.1

238

60

140

182

71

171.1

242

65.6

150

192

75

171.1

246

71.1

160

202

79

171.1

250

76.7

170

212

82.7

171.1

254

82.2

180

222

86.6

171.1

258

 

                Soooo, how do we use the useful numbers?

9.0          Project Example #1.  Chill LPG from ambient of 70F to -58F for injection. 

               Assumptions:

                9.1          We have a single column 4” D X 36” tall.

                9.2          We wish to flow 3 column volumes injected per cycle

                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

                9.4          We wish to inject below -50C/-58F.

10.0        THEREFORE:

               10.1        Three column volumes of LPG = (4” X 4” X 0.7854) X 36” X 3 volumes =  1357 in3 total injected

               10.2        Weight of 70:30 LPG injected = 1357 in3 X 0.019765342 lb/in3 = 26.8 lbs

               10.3        Number of btu’s 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 .

               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  

11.0        Project Example #2.  Recover 1 lb per hour of 70:30 mix using a pump.

               Assumptions:

               11.1        70% n-Butane and 30% n-Propane

                11.2        Pump exhaust 82C/180F

                11.3        Cooling bath at -70C/-95F.

                11.4       Recovery rate 1 lb per minute or per 60 lbs per hour

12.0        Therefore:

                12.1        BTU removal required to cool 60 lbs 70:30 mix from pump discharge vapors at 82.2C/180F to -41.2C/-42.2F  =

                               (180F + 42.2F) X (60lb X 0.39 SH) = 222.2 Delta T X 23.4 btu =5199 btu/hr from Specific Heat

                 12.2        BTU removal required remove Heat of Vaporization for 60 lbs 70:30 mix LPG =

                                60lb  X HV 171.1 btu/lb = 10,266 btu from Heat of Vaporization.

                 12.3        Total removal required to cool pump 26.8 lb pump discharge = 5199 btu + 10,266 btu = 15,983 hr total

13.0        Project Example # 3.  Recover 1 lb per minute passively using 110F hot water and dry ice slurry.

               Assumptions:

               13.1        70:30 Mix with fractional distillation boiling point of -41.2C/-42.2F

                13.2        1 lb per minute or 60 lb hr recovery rate.

                13.3        Hot bath 43.3C/110F

                13.4        Cold bath -70C/-95F

14.0        Therefore:

                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.

                14.2        60 lbs X HV-171.1 btu =  10,266 btu Heat of Vaporization removal

                14.3        Total removal is 3,561 btu + 10,266 btu = 13,827 btu total

15.0        Project Example # 4.  How many feet of ½” 304SS tubing needed using dry ice slurry?

                ASSUMPTIONS:

                15.1        Start temperature 110F

                15.2        End temperature -43F

                15.3        Dry ice slurry bath at -70C/-95F

                15.4        Remove 13,827 btu in 1 hour

                15.5       Use ½” 304SS tubing

                             15.2.1    Transfer rate of 304CC 1/2" tubing 1.52 btu, per linear foot, per degree F, per hour

16.0        Therefore:

               16.1        Delta T at entry of heat exchanger coil = 110F + 95F = 205F using dry ice slurry

               16.2        Delta T at end of heat exchanger coil = 95F – 43F = 52F

               16.3        Average Delta T during passage = 205F + 52F /2 = 128.5F

               16.4        13,827 btu / (1.52 btu X 128.5F average Delta T) = 13,827 btu/195.32 = 71 feet ½” 304SS tubing required.

17.0        Project Example #5.  How many feet of counterflow heat exchanger required using LN2?:

                ASSUMPTIONS:

                17.1        Start temperature 110F

                17.2        End temperature -43F

                17.3        LN2 boiling point is -195.8C/-320F

                17.4        Remove 13,827 btu in 1 hour

                17.5        Use ½” 304SS center tube of counterflow

                               17.5.1    Transfer rate of 1.52 btu, per linear foot, per degree F, per hour

18.0        Therefore:

                18.1        Delta T at entry of heat exchanger coil = 110F + 320F = 430F using LN2

                18.2        Delta T at end of heat exchanger coil = 320 -  43 = 277F

                18.3        Average Delta T during passage = 430F + 277F /2 = 353.5F Average Delta T

                18.4        13,827 btu / (1.52 btu X 353.5F) =  13,827 btu/ 537 = 26 feet ½” 304SS tubing required.

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

                ASSUME:

                19.1        5 gallons EtOH

                19.2        Specific HT EtOH 0.614 btu/lb/F

                19.3        EtOH Specific Gravity .587 and it weighs 0 .0284107 lb/in3

                19.4        21C/70F Ambient temperature

                19.5        -70C/-94F bath temperature

                19.6        Dry ice has 270 btu/lb cooling potential

                19.7        231 in3/gallon

20.0        Therefore:

                20.1        (5 gal X 231 in3) X .0361 X .787 = 32.8 lbs Ethanol

                20.2        Delta T = 70F + 94F = 164F

                20.3        32.8 lbs EtOH X SH-0.614 X Delta T 164F = 3303 btu

                20.4        3303 btu/270 btu/lb = 12.24 lbs dry ice required

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

                ASSUME:

                21.1        100# Tank @ 80% fill.

                21.2        Specific gravity of n-Butane .601

                21.3        Heat of vaporization n-Butane is 165.6 BTU.

                21.4        Specific heat of n-Butane is 0.39 BTU

                21.5        Boiling point n-Butane is -1C/30.2F

22.0        Therefore:

                22.1        100# Tank is water so 100# X .601 SG X .80 fill = 48# n-Butane

                22.2        70F – 30.2F = 39.8 Delta Temperature

                22.3        48# X 0.39 SpHt X 39.8 Delta T = 745 BTU

                22.4        745 BTU / 165.6 BTU HtVp = 4.5 lbs evaporated

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

                ASSUME:

                23.1        100# Tank @ 80% fill.

                23.2        Specific gravity of n-Propane .495

                23.3        Heat of vaporization n-Butane is 184 BTU.

                23.4        Specific heat of n-Butane is 0.39 BTU

                23.5        Boiling point n-Butane is -41.2C/-42.2F

24.0        Therefore:

                24.1        100# Tank is water so 100# X .495 SG X .80 fill = 39.6# n-Propane

                24.2        70F + 42.2F  = 112.2F Delta Temperature

                24.3        39.6# X 0.39 SpHt X 112.2 Delta T = 1733 BTU needed

                24.4        1733BTU / 184 BTU HtVp = 9.4 lbs evaporated.

25.0        Project EXAMPLE # 9 Calculate the BTU losses with 1” foam or vacuum panel insulation: 

                ASSUME:

                25.1        R = 1/K

                25.2        K = 1/R

                25.3        K Value of 300 series stainless steel 14.4 btu/hr/ft2/F

                25.4        R Value for 1” foam insulation = 3.6 to 4.2 or assume about ~4.0

                25.5        R Value for 1” vacuum panel insulation = 25

                25.6        Delta Temperature of 44.6C/112.2F

                25.7        4” X 36” column with 6” X 34 X 0.109T wall jacket.

26.0        Therefore:

                26.1        6 X 6 X .7854 X 34 “ / 144 in2 = 6.7 ft2 outside surface of jacket.

                26.2        R Value for stainless = 1/14.4 btu = R-0.0694 btu/hr/ft2/F/inT

                26.3        R Value of SS/foam composite = (R-0.0694 X 0.109T) +  R-4.0 = R-4.008

                26.4        K Value of SS/foam composite = 1 / 4.008 =K-0.250 btu/hr/ft2/F

                26.5        BTU hour lost at Delta T 44.6C/112.2F = 28.1 btu/hr/ft2

                26.6        28.1 btu/hr/ft2 X 6.7ft2 = 188.3 bty/hr loss from column

                26.7        R Value of SS/vacuum composite = (R-0.0694 X 0.109T) +  R-25.0 = R-25.008

                26.8        K Value of SS/vacuum composite = 1 /25.008 =K-.040 btu/hr/ft2/F

                26.9        BTU hour lost at Delta T 44.6C/112.2F = 112.2DT X .040 btu/hr/ft2 = 4.49 btu/hr/ft2

                26.10     4.49 btu/hr/ft2 X 6.7 ft2 = 30.1 btu/hr loss from column

               

 


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