Does Molecular Weight of a Gas affect its lifting properties at the same velocity over the same wing?Why is max endurance different for jet and props?What does the 'hp' rating actually signify, with respect to an aircraft engine?For a plane with a pusher propeller, how does a root-stalled wing affect its performance?How does aircraft weight affect landing distance?How (and why) does engine thrust changes with airspeed?How much does weight really affect the operation cost?How does maximum speed vary with altitude?Would this hybrid impeller design create enough thrust for a VTOL aircraft?
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Does Molecular Weight of a Gas affect its lifting properties at the same velocity over the same wing?
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Does Molecular Weight of a Gas affect its lifting properties at the same velocity over the same wing?
Why is max endurance different for jet and props?What does the 'hp' rating actually signify, with respect to an aircraft engine?For a plane with a pusher propeller, how does a root-stalled wing affect its performance?How does aircraft weight affect landing distance?How (and why) does engine thrust changes with airspeed?How much does weight really affect the operation cost?How does maximum speed vary with altitude?Would this hybrid impeller design create enough thrust for a VTOL aircraft?
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$begingroup$
Let's say we test the coefficient of lift of the same wing at the same AOA and the same flow velocity at the same pressure.
The difference is wind tunnel A will contain helium (molecular weight 4) gas and wind tunnel B will contain air (average molecular weight around 29).
Which test will produce a higher lift coefficient? Why?
aircraft-performance
asked 8 hours ago
Robert DiGiovanniRobert DiGiovanni
4,6581 gold badge4 silver badges22 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's say we test the coefficient of lift of the same wing at the same AOA and the same flow velocity at the same pressure.
The difference is wind tunnel A will contain helium (molecular weight 4) gas and wind tunnel B will contain air (average molecular weight around 29).
Which test will produce a higher lift coefficient? Why?
aircraft-performance
asked 8 hours ago
Robert DiGiovanniRobert DiGiovanni
4,6581 gold badge4 silver badges22 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's say we test the coefficient of lift of the same wing at the same AOA and the same flow velocity at the same pressure.
The difference is wind tunnel A will contain helium (molecular weight 4) gas and wind tunnel B will contain air (average molecular weight around 29).
Which test will produce a higher lift coefficient? Why?
aircraft-performance
asked 8 hours ago
Robert DiGiovanniRobert DiGiovanni
4,6581 gold badge4 silver badges22 bronze badges
$endgroup$
Let's say we test the coefficient of lift of the same wing at the same AOA and the same flow velocity at the same pressure.
The difference is wind tunnel A will contain helium (molecular weight 4) gas and wind tunnel B will contain air (average molecular weight around 29).
Which test will produce a higher lift coefficient? Why?
aircraft-performance
aircraft-performance
asked 8 hours ago
Robert DiGiovanniRobert DiGiovanni
4,6581 gold badge4 silver badges22 bronze badges
asked 8 hours ago
Robert DiGiovanniRobert DiGiovanni
4,6581 gold badge4 silver badges22 bronze badges
asked 8 hours ago
Robert DiGiovanniRobert DiGiovanni
4,6581 gold badge4 silver badges22 bronze badges
asked 8 hours ago
Robert DiGiovanniRobert DiGiovanni
4,6581 gold badge4 silver badges22 bronze badges
asked 8 hours ago
Robert DiGiovanniRobert DiGiovanni
4,6581 gold badge4 silver badges22 bronze badges
4,6581 gold badge4 silver badges22 bronze badges
add a comment |
add a comment |
1 Answer
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$begingroup$
The wing will produce much less lift in helium than in air; all else being equal, lift is proportional to the molar mass of the gas in question.
The lift equation states that lift $L$ is proportional to the air density $rho$ (rho):
$$L = frac12 rho v^2 S C_L,$$
where $v$ is the true airspeed, $S$ is the wing area, and $C_L$ is the coefficient of lift.
Meanwhile, one form of the ideal gas law tells us how the air density $rho$ is related to the molar mass $M$:
$$rho = fracMPRT,$$
where $P$ is the (static) pressure, $R$ is the gas constant, and $T$ is the absolute temperature.
Combining both of these equations, we find that
$$L = fracM P v^2 S C_L2 R T.$$
In words, lift is
- the molar mass of the gas,
- times the static pressure,
- times the square of the airspeed,
- times the wing area,
- times the coefficient of lift,
- divided by the absolute temperature,
- divided by a constant (the gas constant $R$),
- divided by $2$.
answered 8 hours ago
Tanner SwettTanner Swett
3,1961 gold badge12 silver badges42 bronze badges
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add a comment |
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1 Answer
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$begingroup$
The wing will produce much less lift in helium than in air; all else being equal, lift is proportional to the molar mass of the gas in question.
The lift equation states that lift $L$ is proportional to the air density $rho$ (rho):
$$L = frac12 rho v^2 S C_L,$$
where $v$ is the true airspeed, $S$ is the wing area, and $C_L$ is the coefficient of lift.
Meanwhile, one form of the ideal gas law tells us how the air density $rho$ is related to the molar mass $M$:
$$rho = fracMPRT,$$
where $P$ is the (static) pressure, $R$ is the gas constant, and $T$ is the absolute temperature.
Combining both of these equations, we find that
$$L = fracM P v^2 S C_L2 R T.$$
In words, lift is
- the molar mass of the gas,
- times the static pressure,
- times the square of the airspeed,
- times the wing area,
- times the coefficient of lift,
- divided by the absolute temperature,
- divided by a constant (the gas constant $R$),
- divided by $2$.
answered 8 hours ago
Tanner SwettTanner Swett
3,1961 gold badge12 silver badges42 bronze badges
$endgroup$
add a comment |
$begingroup$
The wing will produce much less lift in helium than in air; all else being equal, lift is proportional to the molar mass of the gas in question.
The lift equation states that lift $L$ is proportional to the air density $rho$ (rho):
$$L = frac12 rho v^2 S C_L,$$
where $v$ is the true airspeed, $S$ is the wing area, and $C_L$ is the coefficient of lift.
Meanwhile, one form of the ideal gas law tells us how the air density $rho$ is related to the molar mass $M$:
$$rho = fracMPRT,$$
where $P$ is the (static) pressure, $R$ is the gas constant, and $T$ is the absolute temperature.
Combining both of these equations, we find that
$$L = fracM P v^2 S C_L2 R T.$$
In words, lift is
- the molar mass of the gas,
- times the static pressure,
- times the square of the airspeed,
- times the wing area,
- times the coefficient of lift,
- divided by the absolute temperature,
- divided by a constant (the gas constant $R$),
- divided by $2$.
answered 8 hours ago
Tanner SwettTanner Swett
3,1961 gold badge12 silver badges42 bronze badges
$endgroup$
add a comment |
$begingroup$
The wing will produce much less lift in helium than in air; all else being equal, lift is proportional to the molar mass of the gas in question.
The lift equation states that lift $L$ is proportional to the air density $rho$ (rho):
$$L = frac12 rho v^2 S C_L,$$
where $v$ is the true airspeed, $S$ is the wing area, and $C_L$ is the coefficient of lift.
Meanwhile, one form of the ideal gas law tells us how the air density $rho$ is related to the molar mass $M$:
$$rho = fracMPRT,$$
where $P$ is the (static) pressure, $R$ is the gas constant, and $T$ is the absolute temperature.
Combining both of these equations, we find that
$$L = fracM P v^2 S C_L2 R T.$$
In words, lift is
- the molar mass of the gas,
- times the static pressure,
- times the square of the airspeed,
- times the wing area,
- times the coefficient of lift,
- divided by the absolute temperature,
- divided by a constant (the gas constant $R$),
- divided by $2$.
answered 8 hours ago
Tanner SwettTanner Swett
3,1961 gold badge12 silver badges42 bronze badges
$endgroup$
The wing will produce much less lift in helium than in air; all else being equal, lift is proportional to the molar mass of the gas in question.
The lift equation states that lift $L$ is proportional to the air density $rho$ (rho):
$$L = frac12 rho v^2 S C_L,$$
where $v$ is the true airspeed, $S$ is the wing area, and $C_L$ is the coefficient of lift.
Meanwhile, one form of the ideal gas law tells us how the air density $rho$ is related to the molar mass $M$:
$$rho = fracMPRT,$$
where $P$ is the (static) pressure, $R$ is the gas constant, and $T$ is the absolute temperature.
Combining both of these equations, we find that
$$L = fracM P v^2 S C_L2 R T.$$
In words, lift is
- the molar mass of the gas,
- times the static pressure,
- times the square of the airspeed,
- times the wing area,
- times the coefficient of lift,
- divided by the absolute temperature,
- divided by a constant (the gas constant $R$),
- divided by $2$.
answered 8 hours ago
Tanner SwettTanner Swett
3,1961 gold badge12 silver badges42 bronze badges
answered 8 hours ago
Tanner SwettTanner Swett
3,1961 gold badge12 silver badges42 bronze badges
answered 8 hours ago
Tanner SwettTanner Swett
3,1961 gold badge12 silver badges42 bronze badges
answered 8 hours ago
Tanner SwettTanner Swett
3,1961 gold badge12 silver badges42 bronze badges
3,1961 gold badge12 silver badges42 bronze badges
add a comment |
add a comment |
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