The Effect of Back Pressure on a Vortex Tube Part 2, Calculating Btu/Hr.

My previous blog post was about how Vortex Tubes react when there is back pressure due to a restriction on either the hot or cold discharge of the Vortex Tube.  In it I mentioned that there is a formula to calculate what the cooling capacity (Btu/Hr) will be if there is no way to avoid operating the Vortex Tube without back pressure on the discharge. That is the calculation focus of this blog – calculating Btu/hr of a Vortex Tube with back pressure.

To continue with the same example, the calculations from the previous blog are shown below.  Last time the example Vortex Tube was operating at 100 psig inlet pressure, 50% cold fraction, and 10 psi of back pressure. We will need some additional information to determine the Btu/Hr capacity. The additional information needed is the temperature of the supplied compressed air as well as the ambient air temperature desired to maintain.  For the example the inlet compressed air will be 70°F and desired ambient air temperature to maintain will be 90°F.

(100 psig + 14.7 psia) / (10 psig + 14.7 psia) = X / 14.7 psia
4.6437 = X / 14.7
X= 14.7 * 4.6437
X = 68.2628
(Values have been rounded for display purposes)

The calculation above gives the compensated operating pressure (X = 68.2628) which will be needed for the BTU/hr calculation. The rated air consumption value of the Vortex Tube will also need to be known.  A 30 SCFM rated generator will be used for this example, the normal BTU capacity of a Vortex Tube with a 30 SCFM generator is 2,000 BTU/hr.

First, determine the new consumption rate by establishing a ratio of the compensated pressure (68.2628 psi) against the rated pressure (100 psi) at absolute conditions (14.7 psia).

(68.2628 PSIG + 14.7 (atmospheric pressure)) / (100 PSIG (rated pressure) + 14.7) = .7233
.7233 x 30 SCFM  = 21.7 SCFM Input 

Second, the volumetric flow of cold air at the previously mentioned cold fraction (50%) will be calculated.  To do this multiply the cold fraction setting (50%) of the Vortex Tube by the compensated input consumption (21.7 SCFM) of the Vortex Tube.

50% cold fraction x 21.7 SCFM input = 10.85 SCFM of cold air flow

Third, the temperature of air that will be produced by the Vortex Tube will need to be calculated.  For this consult the Vortex Tube performance chart which is shown below. To simplify the example the compensated operating pressure (68.2628 psi) will be rounded to 70 psig and to obtain the 70 psig value the mean between 80 psig and 60 psig performance from the chart will be used.

Cold Fraction

EXAIR Vortex Tube Performance Chart

For the example: A 70 psig inlet pressure at 50% cold fraction will produce approximately an 88°F drop.
Fourth, subtract the temperature drop (88°F) from the temperature of the supplied compressed air temperature (70°F).

70°F Supply air – 88°F drop = -18°F Output Air Temperature

Fifth,  determine the difference between the temperature of the air being produced by the Vortex Tube (-18°F) and the ambient air temperature that is desired (90°F).

90°F ambient – -18°F air generated = 108°F difference.

The sixth and final step in the calculation is to apply the answers obtained above into a refrigeration formula to calculate BTU/hr.

1.0746 (BTU/hr. constant for air) x 10.85 SCFM of cold air flow x 108°F ΔT = 1,259 BTU/hr.

In summary, if a 2,000 BTU/hr. Vortex tube is operated at 100 psig inlet pressure, 50% cold fraction, 70°F inlet air to maintain a 90°F ambient condition with 10 psi of back pressure on the outlets of the Vortex Tube the cooling capacity will be de-rated to 1,259 BTU/hr.  That is a 37% reduction in performance.  If a back pressure cannot be avoided and the cooling capacity needed is known then it is possible to compensate and ensure the cooling capacity can still be achieved.  The ideal scenario for a Vortex Tube to remain at optimal performance is to operate with no back pressure on the cold or hot outlet.

Brian Farno
Application Engineer Manager

The Effect of Back Pressure on a Vortex Tube

Vortex tubes have been considered a phenomena of Physics and boggled minds for many years.  To give a brief run down of how the Vortex Tube works please refer to Figure 1 below.


Figure 1

As seen above, the control valve is determining the amount of air allowed to escape the hot end and sets the cold fraction.  A cold fraction is the percentage of air that exits the cold side versus the hot side. The cold fraction and operating pressure sets the temperature drop on the cold end and temperature rise on the hot end, as well as volumetric flow out of both ends. The control valve is not the only variable that can alter the cold fraction of the Vortex Tube though.

In Figure 1 and the performance chart below, there is no restriction on the hot end or the cold end outlets. No restriction means no back pressure and the cold air has the easiest path to the area needing cooling. Back pressure can directly affect the performance of a Vortex Tube.  As little as 3 psig of back pressure can begin to alter the temperature drop or rise on the Vortex Tube.  This is due to the fact that Vortex Tubes operate off an absolute pressure differential.  If the outlets have a restriction on them then they are not discharging at atmospheric pressure, 14.7 psi. What kind of items can cause back pressure and can the performance with a back pressure on the outlet be determined?

Back pressure is created by implementing any form of restriction on the hot or cold outlet. This may be undersized tubing to deliver the cold air or a valve that has been installed to try and control the volume of air being blown onto the process as well as many other possibilities.  The best rule of thumb to eliminate back pressure is to keep the tubing on an outlet the same cross sectional dimension as the outlet on the Vortex Tube and try to keep the tubing as short as possible.

If back pressure cannot be prevented, the performance variance of the Vortex Tube can be calculated and possibly compensated for. The variables that are needed to do so are the inlet air pressure of the vortex tube and the amount of back pressure that is being seen on the outlets. If this is different from the hot end to the cold end both will need to be known.  If these are not known they can be measure by installing a pipe Tee and a pressure gauge. This may need to be a sensitive pressure gauge that measures even relatively low psig. (1-15 psig)

Once these variables are known, we want to look at an absolute pressure differential versus the back pressure differential. For example, the Vortex Tube is a operating at 100 psig inlet pressure, 50% cold fraction and 10 psi of back pressure.  We look at the pressure differentials and can use Algebraic method to determine the inlet pressure supply that the tube will actually perform at.

(100 psig + 14.7 psia) / (10 psig + 14.7 psia) = X / 14.7 psia
4.6437 = X / 14.7
X= 14.7 * 4.6437
X = 68.2628
(Values have been rounded for display purposes)

So if there is a 10 psig back pressure on the outlet of a Vortex tube operating a 100 psig inlet pressure the tube will actually carry performance as if the inlet pressure was ~68 psig.   To showcase the alteration in performance we will look at just the temperature drop out of the cold side of the Vortex Tube. (Keep in mind this is a drop from the incoming compressed air temperature.)

Vortex Tube Performance Data

Vortex Tube Performance Chart

As shown in the performance chart above, if the Vortex Tube was operating at 100 psig inlet pressure and 50% cold fraction the temperature drop would be 100°F.  By applying a 10 psi back pressure on the outlet of the Vortex Tube the temperature will be decreased to ~87°F temperature drop.   This will also decrease the volumetric flow of air exiting the Vortex Tube which can also be calculated in order to determine the cooling capacity of the Vortex Tube at the altered state.  Keep an eye out for a follow up blog coming soon to see that calculation.

Brian Farno
Application Engineer Manager

Calculating Humidification of a Room

I had an application where a customer needed to have a room at 80% relative humidity (RH). They produced a nylon backing for carpet, and they needed the high RH to reduce the “stickiness” in the process. Currently he was at 40% RH in a room that was sized at 40ft long by 20ft wide by 20ft high (12.2m long X 6.1m wide X 6.1m high). He wondered if our Atomizing Nozzles could help him. I decided to put on my engineering hat to calculate the amount of water that he would need to increase the moisture content. Other markets that would require higher RH in their ambient air are wood working, dust control, laboratories, and High Voltage applications.

Relative humidity (RH) is the percentage of water vapor as compared to saturation at the same temperature. So, at 100% RH, the ambient air cannot hold any more water. With our atomizing nozzles, we can atomize the water droplets to a very small droplet to help increase the absorption rate into ambient air. This will increase the RH of a room, but I will have to determine what size and how many.

The equation that I use is as follows, Equation 1:

Imperial Units                                                                    S.I. Units

H = V * RAC * (Wf – Wi) / (v * 7000) Imperial         H = V * RAC * (Wf – Wi) / (v * 997.9) Metric


H – mass flow rate of water, Lbs/hr                        H – mass flow rate of water, Kg/hr

V – Volume of Section, ft^3                                     V – Volume of Section, m^3

RAC – Room Air Changes, No. per hour                RAC – Room Air Changes, No. per hour

Wf – Final Water Content, Grains/lb of dry air        Wf – Final Water Content, Grams/Kg of dry air

Wi – Initial Water Content, Grains/lb of dry air        Wi – Initial Water Content, Grams/Kg of dry air

v – Specific Volume of Air, ft^3/lb                            v – Specific Volume of Air, m^3/Kg

Conversion Constant – 7000 Grains/lb                   Conversion Constant – 997.9 Grams/Kg

The customer stated that the room is at 68 deg. F (20 deg C). The humidity sensor is +/- 5%; so, when the RH in the room gets to 75%, it will kick on their system. They also use a standard HVAC unit to heat and cool the room. From these factors, we can determine some of the variables above. With the water content, you can find a chart online to determine the amount of water vapor that is contained in air at a specific temperature and RH. At 68 deg. F (20 deg. C), I was able to find the following information:

Imperial Units                                                       S.I. Units

Wi = 43 Grains/lb of dry air at 40% RH               Wi = 6.1 Grams/Kg of dry air at 40% RH

Wi = 80.5 Grains/lb of dry air at 75% RH            Wi = 11.5 Grams/Kg of dry air at 75% RH

Wf = 85.5 Grains/lb of dry air at 80% RH            Wf = 12.2 Grams/Kg of dry air at 80% RH

v = 13.35 ft^3/lb @ 68 deg. F, 1 atm                   v = 0.8334 M^3/Kg at 20 deg. C, 1 bar (absolute)

V = 40ft X 20ft X 20ft = 16,000 ft^3                     V = 12.2m X 6.1m X 6.1m = 454 m^3

Another factor is the number of air changes in that room. With the HVAC system, it will turn on and off to heat and cool the air.  Some fresh air is brought in during this cycle.  With a typical system, the room air will change between 2 – 4 times an hour.  So, RAC = 4/hour (worse case).  (Other locations may have scrubber systems, continuous air flow systems, etc. and the RAC will be greater).

If we plug in the numbers that we have, we can determine how much water that we will need to spray into the air to increase the RH from 40% to 80%.

Imperial Units

H = V * RAC * (Wf – Wi) / (v * 7000)

H = 16,000 ft^3 * 4/hr * (85.5 – 43 Grains/lb)/(13.35 ft^3/lb * 7000 Grains/lb)

H = 29.1 lb./hr

S.I. Units

H = V * RAC * (Wf – Wi) / (v * 997.9)

H = 454m^3 * 4/hr * (12.2 – 6.1 Grams/Kg)/ (0.8334 m^3/Kg * 997.9 Grams/Kg)

H = 13.3 Kg/hr.

Now that we know the rate of water to put into the ambient air, we have to look at the set up. With the settling time of the water droplets and the location of the humidity sensor, we will have a lead/lag problem.  To help in this situation, I would recommend to turn on the Atomizing Nozzles for 10 – 15 seconds, and wait 2 minutes to re-measure the RH.  This will help to not over saturate the room.  As for the location of the Atomizing Nozzles, you have to make sure that the spray does not contact any structure or other atomizing spray patterns.  This will cause the water to condense and either coat a structure or create rain.  To help with the entire system, I suggested our No Drip External Mix Wide Angle Flat Fan Pattern Atomizing Nozzle. This will eliminate a water valve at each Atomizing Nozzle. When the air pressure is turned off to stop spraying, the No Drip Atomizing Nozzle will seal and not allow any water to drip. To also help with consistent RH in the room, the EB2030SS was my choice. The spray range helps to cover the area especially with multiple units operating.

No Drip Atomizing Nozzle

No Drip Atomizing Nozzle

To determine the number of Atomizing Nozzles, we want to look at the time determination with the controller and the intermittence of operation. With the RAC = 4/hour, the air in the room will change over every 15 minutes.  We want to have a balance between the new air and the existing air.  So, with the time measurement of 2 minutes off and 15 seconds on, we will have 6 humidity checks over 15 minutes.  We can divide the amount of water to be injected into the room by 6 to cover that time span.  Also, we have to factor in that we will not be running the Atomizing Nozzle for the continuous hour.  We will have to adjust the amount for only running for 15 seconds.  So, the intermittent factor will be 0.0042 (the 15 seconds portion of the hour).

In taking into consideration the flow rate required during operation time, we can calculate the amount of flow required for the Atomizing Nozzle as in Equation 2.

Imperial Units                                                               SI Units

Flow rate: Q = H / (D * T * f)                                     Flow rate: Q = H / (D * T * f)

Mass Flow Rate: H = 29.1 lbs/hr                              Mass Flow Rate: H = 13.3 Kg/hr

Density of Water: D = 8.34 lbs/gal                            Density of Water: D = 1 Kg/L

Span division of time: T = 6                                      Span division of time: T=6

Intermittent Factor: f = 0.0042                                  Intermittent Factor: f = 0.0042

Q = 29.1 lbs/hr / (8.34 lbs/gal * 6 * 0.0042)              Q = 13.3 Kg/hr / (1 Kg/L * 6 * 0.0042)

Q = 138.5 gal/hr (GPH)                                            Q = 527.8 L/hr (LPH)

In the catalog, the model EB2030SS will flow 14.0 GPH (53.0 LPH) at 40 PSIG (2.8 Bar) water pressure. This would be in the compressed air pressure range of 50 PSIG (3.4 Bar) to 95 PSIG (6.5 Bar).  If we divide these out, it will tell us how many atomizing nozzles that is needed to humidify the room.

Imperial: 138.5 GPH/14.0 GPH = 9.9 or 10 Atomizing Nozzles.

SI units: 527.8 LPH/53.0 LPH = 9.9 or 10 Atomizing Nozzles.

The last thing to determine is the amount of time that would be required to maintain the 80% RH when the controller calls for more humidification. At 75% RH, we can use Equation 1 to determine the amount required to reach 80%.  As we plug in the initial Water Content, Wi, at 75% RH as 80.5 Grains/lb of dry air (11.5 Grams/Kg of dry air), we will get an H value of 3.42 lb/hr (1.55 Kg/hr).  With each Atomizing Nozzle putting out 14.0 GPH (53.0 LPH) of water, we can determine the time to atomize the 3.42 lbs (1.53 Kg) of water during the operational time.  The control will be much better as the air is changing with the new incoming air and the existing air.  Thus, we have in Equation 3:

Imperial Units                                                                SI Units

Time (sec): T = 3600 * m/ (N * Qa * D)                        Time (sec): T = 3600 * m/ (N * Qa * D)

Mass of water: m = 3.42 lb                                          Mass of water: m= 1.53 Kg

No. of Nozzles: N = 10                                                 No. of Nozzles: N = 10

Atomizing Flow Rate: Qa = 14.0 GPH                          Atomizing Flow Rate: Qa = 53.0 LPH

Density of Water: D = 8.34 lb/gal                                  Density of Water: D = 1 Kg/L

T = 3600 * 3.42 lb / (10 * 14 GPH * 8.34 lb/gal)            T = 3600 * 1.55 Kg / (10 * 53 LPH * 1 Kg/L)

T = 10.5 seconds                                                          T = 10.5 seconds

With some other humidification devices like steam generators, companies have to capitalize the system. With the Atomizing Nozzles, my customer was able to keep the cost down and control the RH at a high level for his manufacturing process.  In turn, he was able to increase productivity and reduce downtime.  If you need to increase the level of moisture in an area, you can always contact one of the Application Engineers at EXAIR for help.

John Ball
Application Engineer
Twitter: @EXAIR_jb

Calculating Compressed Air Cost & Savings Made Easy

If you have ever looked through our catalog, website, blog, twitter feeds, or even our Facebook page, you will see that we can almost always put a dollar amount behind the amount of compressed air you saved by installing EXAIR’s Intelligent Compressed Air Products.   No matter which platform we use to deliver the message, we use the same value for the cost of compressed air which is $.25 per 1,000 Standard Cubic Feet of compressed air. This value is derived from average commercial and industrial energy costs nationwide, if you are on either coast this value may increase slightly. On the positive side, if your cost for compressed air is a bit more, installing an EXAIR product will increase your savings.

So where does this number come from?   I can tell you this much, we didn’t let the marketing department or anyone in Accounting make it up.   This is a number that the Engineering department has deemed feasible and is accurate.

To calculate the amount we first look to what the cost per kilowatt hour is you pay for energy.  Then we will need to know what the compressor shaft horsepower  of the compressor is, plus the run time percentage, the percentage at full-load, and the motor efficiency.

If you don’t have all of these values, no worries.   We can get fairly close by using the industry accepted standard mentioned above, or use some other general standards if all you know is the cost of your electricity.

The way to calculate the cost of compressed air is not an intense mathematical equation like you might think.  The best part is, you don’t even have to worry about doing any of the math shown below because you can contact us and we can work through it for you.

If you prefer to have us compare your current compressed air blow off or application method to one of our engineered products, we can do that AND provide you a report which includes side by side performance comparisons (volume of flow, noise, force) and dollar savings. This refers to our free Efficiency Lab service.

EXAIR's Efficiency Lab is a free service to all US customers.

EXAIR’s Efficiency Lab is a free service to all US customers.

If you already know how much air you are using, you can use the Air Savings Calculators (USD or Euro) within our website’s knowledge base. Just plug in the numbers (EXAIR product data is found on our website or just contact us) and receive air savings per minute, hour, day and year. We also present a simple ROI payback time in days.

Now, back to the math behind our calculation.
Cost ($) =
(bhp) x (0.746) x (#of operating hours) x ($/kWh) x (% time) x ( % full load bhp)
Motor Efficiency

— Compressor shaft horsepower (generally higher than motor nameplate Hp)
0.746 – conversion between hp and KW
Percent Time — percentage of time running at this operating level
Percent full-load bhp — bhp as percentage of full load bhp at this operating level
Motor Efficiency — motor efficiency at this operating level

For an average facility here in the Midwest $0.25/1,000 SCF of compressed air is accurate.   If you would like to attempt the calculation and or share with us your findings, please reach out to us.   If you need help, we are happy to assist.

Brian Farno
Application Engineer Manager


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