Pascal's Wine Barometer

solved PressureAtDepth

Q. Toricelli's barometer $P_{a} = 0$ Pa used mercury. Pascal duplicated it using French wine of density 984 kg/m³ $ρ = 984$ kg/m³. Taking g as 9.8 m/s² $g = 9.8$ m/s², what is the height of the wine column for normal atmospheric pressure of 1 atm $P = 1$ atm, in metres?

Inputs

density

ρ

984 kg/m³

gravity

g

9.8 m/s²

atmosphere

P

1 atm

Outputs

height

h

How it's solved

Solved with PressureAtDepth

Given

$ρ = 984$ kg/m³
$g = 9.8$ m/s²
$P = 1$ atm
$P_{a} = 0$ Pa from “Toricelli's barometer”

Find $h$ .

Working

$P = 1$ atm $= 101325$ Pa
pressure adds the surface s(column foot) $P = P_{a} + P_{g}$
$101325 = 0 + P_{g}$
$P_{g} = 101325$ Pa
pressure of the fluid above(column foot) $P_{g} = ρ \cdot g \cdot h$
$101325 = 984 \cdot 9.8 \cdot h$
$h = 10.51 m$
$h \approx 10.5$ mto 3 significant figures

Conclusion

Above the column is empty space, so the wine alone must press as hard as the atmosphere; wine is far lighter than mercury, so its column stands far taller.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :pascal_s_wine_barometer do4  kind :numerical5  describes "Toricelli's barometer used mercury. Pascal duplicated it using French wine of density %{density}. " \6            "Taking g as %{gravity}, %{height} for normal atmospheric pressure of %{atmosphere}, in metres?"7  given density: "984 kg/m3"8  given gravity: "9.8 m/s2"9  given atmosphere: "1 atm"10  asks height: "what is the height of the wine column"11end

solution.rb

1require_relative "../questions/pascal_s_wine_barometer"2require_relative "../science/mechanical_properties_of_fluids/pressure_at_depth"34solve :pascal_s_wine_barometer, using: PressureAtDepth do5  objects :column_foot67  take density: rho8  take gravity: g9  take atmosphere: column_foot.pressure1011  deduce_from "Toricelli's barometer", P_a => "0 Pa"1213  height = column_foot.depth14  answer height:, in: "m", figures: 31516  conclusion "Above the column is empty space, so the wine alone must press as hard as the atmosphere; " \17             "wine is far lighter than mercury, so its column stands far taller."18end

playground.rb

1require_relative "../solutions/pascal_s_wine_barometer"23play_with :pascal_s_wine_barometer do4  title "Pascal's Wine Barometer"5end

The laws it drew on

quantity	written	units	within	from
mass	$m$	kg	0 … ∞	Pressure
gravity	$g$	m/s²	—	Pressure
thrust	$F$	N	—	Pressure
area	$A$	m²	—	Pressure
pressure	$P$	Pa	0 … ∞	Pressure
pressure at the surface	$P_{a}$	Pa	—	PressureAtDepth
density	$ρ$	kg/m³	0 … ∞	PressureAtDepth
depth	$h$	m	—	PressureAtDepth
gauge pressure	$P_{g}$	Pa	—	PressureAtDepth
pressure of ground	$P_{g}$	Pa	0 … ∞
thrust on ground	$F_{g}$	N	—
area of ground	$A_{g}$	m²	—
pressure of column foot	$P$	Pa	0 … ∞
thrust on column foot	$F$	N	—
area of column foot	$A$	m²	—
gauge pressure on ground	$P_{g,g}$	Pa	—
depth of ground	$h_{g}$	m	—
gauge pressure on column foot	$P_{g}$	Pa	—
depth of column foot	$h$	m	—
mass of column foot	$m$	kg	0 … ∞

P_{s} = \frac{F_{s}}{A_{s}}

pressure is force over area

P_{g,p} = ρ \cdot g \cdot h_{p}

pressure of the fluid above

P_{p} = P_{a} + P_{g,p}

pressure adds the surface s

P_{g} = \frac{F_{g}}{A_{g}}

pressure is force over area(ground)

P = \frac{F}{A}

pressure is force over area(column foot)

P_{g,g} = ρ \cdot g \cdot h_{g}

pressure of the fluid above(ground)

P_{g} = ρ \cdot g \cdot h

pressure of the fluid above(column foot)

P_{g} = P_{a} + P_{g,g}

pressure adds the surface s(ground)

P = P_{a} + P_{g}

pressure adds the surface s(column foot)