EQUATIONS (HELIUM-FILLED BALLOON)
Density of helium
4 0.076 0.010 lb/cu ft ( 1)
Density difference, helium inside vs.air
outside of balloon
0.076 0.010 0.066lb/cu ft ( 2)
Volume of balloon(sphere)
4 r V ( 3)
Lifting force of balloon
0.066 3 0.20lb ( 4)
0.020 16 3.2oz ( 5)
Weight of ballon
3. 2 2. 5 0.7oz ( 6)
EQUATIONS (HOT-AIR BALLOON)
Volume of giant balloon
4 r V ( 7)
Ambient air temperature (in ºR.) ( 8)
Giant balloon temperature, ºR.
300º F. 460 760º R. ( 9)
Density of hot air inside giant balloon
0.076 760 530 0.053lb/cu ft ( 10)
Denisty difference, hot air inside vs air
outside giant balloon
0.076 0.053 0.023lb/cu ft ( 11)
Lift of giant balloon
0.023lb/cu ft 32,000 cu ft 740lb ( 12)
Oil & Gas Journal | Apr. 3, 2017 63
ly the same as those he carried out as a child to understand
the workings of the world around him.
While the voice calling for him to stop his work to eat
dinner may now be that of his wife instead of his mother,
the drive to apply basic principles to complex questions is
the same as it was during his childhood in New York City.
The long ride to the refinery afforded the author an opportunity to share with the young engineer a relevant experiment from his childhood to help simplify the problem
that awaited them.
The author’s childhood experiment involved a test to determine how many small lead fishing weights a large red
helium-filled balloon could lift so that it would remain suspended in his bedroom. To conduct the experiment, the author only had to apply principles and calculations that he’d
learned in school.
The first step involved determining the density of helium
inside the balloon. To do this required knowing the molecular weight of helium ( 4 lb), the molecular weight of air ( 29
lb), and the density of air (0.076 lb/cu ft) at sea level and 70°
F., the same conditions as the author’s room.
After calculating the density of helium inside the balloon
as 0.010 lb/cu ft (Equation 1), the author was able to calculate the difference of density between the helium inside and
air outside the balloon as 0.066 lb/cu ft (Equation 2), as well
as the volume of the round balloon as 3 cu ft (Equation 3).
With these determinations, the lifting force of the balloon
was calculated as 0.20 lb (Equation 4). Knowing 16 oz. are
in 1 lb., the author concluded that lift on the balloon would
be 3. 2 oz (Equation 5).
To verify the theoretical lift force of the balloon, the au-
thor used his father’s postal scale to measure 3. 2 oz of paper,
of which the balloon disappointingly could only lift 2. 5 oz.
In reviewing the experiment, the author realized he’d
neglected to consider the weight of the balloon itself. With
both theoretical and actual lift force known, the author was
able to calculate the balloon’s weight at 0.7 oz (Equation 6).
Temperature effect on lift
Helium is expensive, and hydrogen is dangerous due to its
combustibility. So if you want to go on a balloon ride, the
balloon is going to be lifted by hot air.
A balloon 40 ft in diameter will have a volume of 32,000
cu ft (Equation 7).
Ambient temperature is 70° F. The hot air inside the giant
balloon has been heated to 300° F. The elevated temperature
forces the molecules of air away from each other, making the
air less dense in proportion to its temperature. The lower
density is not in proportion to temperature measured in °F.,
however, but in proportion to temperature in Rankine (°R).
To convert from °F. to °R., add 460 to the °F. temperature
(°F. + 460 = °R.).
In our example of the giant balloon, the ambient air temperature would be 530° R. (Equation 8) and the hot air inside the balloon, 760° R. (Equation 9).
With the density of air at 70° F. equivalent to 0.076 lb/
cu ft, the density of the 300° F. (760° R.) hot air inside the
giant balloon equals 0.053 lb/cu ft (Equation 10). The density difference between the hot (300° F.) air inside and cool
( 70° F.) air outside of the giant balloon would be 0.023 lb/cu