Balloon Performance Model
by Peter Cuneo
(This article has been reproduced with the permission of the Author and the Balloon Federation of America - Gas Division)
Is there a reliable way to predict how much ballast will be required to
level out a balloon descending at a given rate? What factors play a part
in making this prediction? How many minutes does take to level out and
how many feet of altitude are lost during that time?
A list of factors that might be important in answering these questions include: balloon envelope size, MSL altitude, lifting gas type, atmospheric conditions, ballasting sequence (whole bags or small scoops) and envelope shape (spherical or natural). In this article I have chosen to study two factors from the above list (altitude and lifting gas type) while limiting the discussion to 1,000 cu. m. (35,325 cu. ft.) envelopes. I will present one previously published formula followed by a computer model’s interpretation of two different flight profiles.
Wörner’s Formula
The published formula for ballasting of a 1,000 cu. m. balloon, found in the Wörner gas balloon manual, on page 23 of the June, 1994 edition is:
B = D2 * 2.16
Example: For a 1,000 cu.m. envelope falling at 400 ft/min, square 4 (this is D in hundreds of ft/min) and then multiply by 2.16. The result is 35 pounds of ballast (B) needed to level off the descent.
The only factor from the above list that is considered in this equation is envelope size. No estimates are made of the time or altitude lost in reaching level flight. Wörner gives similar equations for other balloon sizes. (Note: I have converted Wörner’s constants from metric to English units.)
Computer Model Results
To gain further insight into this topic, I next resorted to that great substitute for true vision, a computer model. This model has been developed slowly over several years with help from Julian Nott and others. Performance validation has been done by comparing results to NSBF’s similar but much more elaborate program SINBAD and to examples from the Scientific Ballooning Handbook. Equations were taken from the book, Aerostatics by Warner, 1926. The computer models the effect of adiabatic cooling and atmospheric drag but not inversions, solar heating or other non-standard atmospheric conditions.
The following profiles are both for a 1,000 cu. m. balloon, initially filled to 95% capacity, with a drag coefficient of 1.0 (correct for a spherical balloon).
First Scenario
Figure 1 compares computer generated, simulated barograph traces for helium and hydrogen balloons. The equivalent of 20 pounds of lift is valved off at ten minutes into the flight while the balloons are in equilibrium, flying level at 5,000 ft MSL. Both balloons are allowed to descend, initially dropping rapidly but soon slowing and reaching a constant descent rate (20ft/min for helium and 40ft/min for hydrogen) with no ballasting.
The smaller, bottom chart shows the temperature difference between the ambient air and the lifting gas (in °C) with a positive value indicating the lifting gas is warmer than the outside air. Both balloons start at equilibrium but adiabatic warming causes a rise in gas temperature as the gas is compressed in the lower, denser atmosphere. Due to chemical differences, the helium and hydrogen both warm more rapidly than the surrounding air. The balloon becomes superheated and transient extra lift is generated. This lift slowly dissipates as the gas cools back to the ambient air temperature just as warm leftovers cool in a refrigerator.
Eventually as the balloons continue to fall, a balance is reached between adiabatic warming and natural cooling of the warm gas surrounded by the cooler atmosphere. A final, constant descent rate is achieved that is dependent on how fast this natural cooling happens. I have assumed a cooling rate that dissipates 90% of the temperature difference in a two hour period. This is based my experience of the time it takes for a balloon to stabilize after undergoing cooling at sunset.
Second Scenario
In figure 2, two balloons rise after launching from 7,000ft MSL and reach an altitude of 10,000ft MSL at eight minutes into the flight. Gas is then valved off to achieve a 500ft/min descent two minutes later. At this point, ten minutes into the flight, enough weight is ballasted to cause the balloon to eventually reach equilibrium. This is not the usual scenario for ballasting during a gas flight but might represent a response to an emergency situation. It is a simple first case for study.
Results of the model show that the hydrogen balloon is rising at 107 ft/minute when 87 pounds of lift are valved off and then two minutes later, 65 pounds of ballast are jettisoned. For helium the numbers are 143 ft/min, 98 pounds of lift and 53 pounds of ballast.
In the lower trace, the temperature imbalance reaches an extreme value of -5°C for the hydrogen system and -10°C for helium, but after ballasting, both systems level off as the temperature balance is restored.
Two questions posed at the start of the article can now be answered (for this flight profile only!). Each system falls about 2,500 feet while recovering from the initial 500 foot per minute descent. The helium system returns to level flight in about eleven minutes while for hydrogen leveling takes about sixteen.
To generalize, this scenario was repeated for six descent rates and six peak altitudes for helium and hydrogen. Figure 3 shows the amount of ballast required to regain level flight for each case.
Several comments are appropriate here: 1) Envelope size is constant for all runs. Thus it was necessary to reduce system weight to achieve higher
altitudes; 2) The hydrogen system weighed more than the equivalent helium system because of its greater lift; 3) For comparison, the dots at the far right side of each chart (at the 20,000ft column) show values predicted by the Wörner formula for each descent rate; 4) For slow descent rates, negative ballast values indicate that instead of ballasting, additional valving was required to stabilize the system. This occurs when the descending, warming gas in the envelope provides enough lift to reverse the descent and induce a slight ascent.
Several conclusions can be drawn from figure 3: 1) for this scenario, less ballast is required at higher altitudes to stop equal decent rates; 2) hydrogen requires more ballasting and valving than helium under equal conditions. 3) Though not shown on the chart, leveling times ranged from 13 to 19 minutes for hydrogen and 10 to 14 minutes for helium.
Overall Conclusions
These two results generate the following thoughts:
1) Any temperature imbalance between the ambient air and the lifting gas has an impact on results. This imbalance comes from prior flight maneuvers. Altitude changes cause temperature imbalances that only slowly dissipate over time.
2) The rate at which any temperature imbalance naturally dissipates varies between systems and can significantly change results.
3) The Wörner formula is roughly correct but is subject to variations from outside factors.
4) A relatively small increase in descent rate requires a significantly larger amount of ballast. This confirms the quadratic in Wörner’s formula.
5) A new study starting with the systems at temperature equilibrium and in level flight would be interesting and should yield different results.
