Since the balloon is open to the air, the pressure P in the balloon
is the same as the ambient pressure.
According to
the force diagram shown on the left, the sum of all forces should
be zero when the balloon is still in the air.
FB - (Gb + Gp + Gair) =
0
FB = ρcool
airgVballoon
= ρcool airgπ (4/3)(D/2)3
ρcool air = P/(RT)
=
90(103)/(287(15+273.15))
=
1.089 kg/m3
Rearranging the equation to give the expression of mass of the hot air.
mair = FB/g
- mb - 3(mp)
With all the data known
mair = (1.089)(9.8)(4/3)π(10)3 /9.8
- 80 -
3(65)
= 4287 kg
(2) The average temperature in the balloon
After the mass of hot air is determined, the average temperature of
the hot air can be determined using the ideal-gas equation of state.
PV = mairRT
T = PV/
(mairR)
= 90(103)(4/3)(3.14)(10)3/
((4287)(287))
= 306.5 K = 33.34 oC
(3) The movement of the balloon if the local air temperature is 30 oC
The density of the air decreases with the temperature increases. Hence
the buoyancy force will decrease and the balloon will move downward.
Repeating the solution above, the temperature of the hot air
equals
50.38 oC if the balloon keeps still again. |