Fysik - Ballonproces
Starten
- Mini ballon
- Størrelse
- Tid
- Mislykkedes
Erfaring
- Oplysninger
- Mini blå booster
- Flamme
- Skorsten
- Idégenering
Værksted
- Plastik
- Skorsten
- Forsøg og aftestninger
- Fin justeringer
Konkurrence
- Fejlede
- Holdt i lang tid!
- Flammen var svag
- For lidt ilt
Nødballon - Opdrift
Data
- 55 gram
- Temp. i lokalet: 18 grader
- Temp. inden i: 120 grader
- 100 cm høj
- 40 cm i diameter
Databehandling
- Volume
- Cylinder
- SI-enheder
- 1 m høj
- 0,4 m i diameter
V=\pi\cdot r^{2}\cdot h
V
=
π
⋅
r
2
⋅
h
V=\pi\cdot 0,2^{2}\cdot 1
V
=
π
⋅
0
,
2
2
⋅
1
V=0,126 m^{3}
V
=
0
,
1
2
6
m
3
Databehandling
- Opdrift
- Luft uden om
- Ukendt variable
F_{op}=V_{ballon}\cdot\rho_{kold}\cdot g
F
o
p
=
V
b
a
l
l
o
n
⋅
ρ
k
o
l
d
⋅
g
\rho_{kold}=\frac{M}{R}\cdot\frac{P}{T}
ρ
k
o
l
d
=
R
M
⋅
T
P
\rho_{kold}=\frac{0,028\frac{kg}{mol}}{8,31\frac{Pa\cdot m^{3}}{mol\cdot K}}\cdot\frac{101324Pa}{(273+18)K}
ρ
k
o
l
d
=
8
,
3
1
m
o
l
⋅
K
P
a
⋅
m
3
0
,
0
2
8
m
o
l
k
g
⋅
(
2
7
3
+
1
8
)
K
1
0
1
3
2
4
P
a
\rho_{kold}=1,17\frac{kg}{^{m^{3}}}
ρ
k
o
l
d
=
1
,
1
7
m
3
k
g
Databehandling
- Opdrift
- Tre kendte variabler
F_{op}=V_{ballon}\cdot\rho_{kold}\cdot g
F
o
p
=
V
b
a
l
l
o
n
⋅
ρ
k
o
l
d
⋅
g
F_{op}=0,126m^{3}\cdot 1,17\frac{kg}{m^{3}}\cdot 9,82\frac{N}{kg}
F
o
p
=
0
,
1
2
6
m
3
⋅
1
,
1
7
m
3
k
g
⋅
9
,
8
2
k
g
N
F_{op}=1,45N
F
o
p
=
1
,
4
5
N
Databehandling
- Tyngdekraft af ballon
- En ukendt variable
F_{t}=m\cdot g
F
t
=
m
⋅
g
F_{t}=(m_{plast}+m_{varm})\cdot g
F
t
=
(
m
p
l
a
s
t
+
m
v
a
r
m
)
⋅
g
F_{t}=(m_{plast}+V_{ballon}\cdot\rho_{varm})\cdot g
F
t
=
(
m
p
l
a
s
t
+
V
b
a
l
l
o
n
⋅
ρ
v
a
r
m
)
⋅
g
Databehandling
F_{t}=(m_{plast}+V_{ballon}\cdot\rho_{varm})\cdot g
F
t
=
(
m
p
l
a
s
t
+
V
b
a
l
l
o
n
⋅
ρ
v
a
r
m
)
⋅
g
\rho_{varm}=\frac{0,028\frac{kg}{mol}}{8,31\frac{Pa\cdot m^{3}}{mol\cdot K}}\cdot\frac{101324Pa}{(273+120)K}
ρ
v
a
r
m
=
8
,
3
1
m
o
l
⋅
K
P
a
⋅
m
3
0
,
0
2
8
m
o
l
k
g
⋅
(
2
7
3
+
1
2
0
)
K
1
0
1
3
2
4
P
a
\rho_{varm}=0,87\frac{kg}{m^{3}}
ρ
v
a
r
m
=
0
,
8
7
m
3
k
g
F_{t}=(0,055kg+0,126m^{3}\cdot 0,87\frac{kg}{m3})\cdot 9,82\frac{N}{kg}
F
t
=
(
0
,
0
5
5
k
g
+
0
,
1
2
6
m
3
⋅
0
,
8
7
m
3
k
g
)
⋅
9
,
8
2
k
g
N
F_{t}=1,55N
F
t
=
1
,
5
5
N
Databehandling
- Temp. inden i er måske højere?
- Vi ændrer temp. fra 120 til 200 grader
\rho_{varm}=\frac{0,028\frac{kg}{mol}}{8,31\frac{Pa\cdot m^{3}}{mol\cdot K}}\cdot\frac{101324Pa}{(273+200)K}
ρ
v
a
r
m
=
8
,
3
1
m
o
l
⋅
K
P
a
⋅
m
3
0
,
0
2
8
m
o
l
k
g
⋅
(
2
7
3
+
2
0
0
)
K
1
0
1
3
2
4
P
a
\rho_{varm}=0,72\frac{kg}{m^{3}}
ρ
v
a
r
m
=
0
,
7
2
m
3
k
g
F_{t}=(0,055kg+0,126m^{3}\cdot 0,72\frac{kg}{m3})\cdot 9,82\frac{N}{kg}
F
t
=
(
0
,
0
5
5
k
g
+
0
,
1
2
6
m
3
⋅
0
,
7
2
m
3
k
g
)
⋅
9
,
8
2
k
g
N
F_{t}=1,43N
F
t
=
1
,
4
3
N
Diskussion
- Fejlkilder
- Røg - CO2
- Volumen
- Temperaturen
- Vægt
Konklusion
- Fejlkilder
- Røg - CO2
- Volumen
- Temperaturen
- Vægt
FARVEL OG TAK!
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