A Litre of Water Compressed

Q. Taking the bulk modulus of water as 2.2 × 10⁹ N/m², how much should the pressure on a litre of water be changed to compress it by 0.10 %, in pascals? expects 2.2 × 10⁶ Pa

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

Q. Taking the bulk modulus of water as 2.2 × 10⁹ N/m² $B = 2.2 \times 10^{9}$ N/m², how much should the pressure on a litre of water be changed to compress it by 0.10 % $Δ V_V = 0.10$ %, in pascals?

The playground

Inputs

modulus

B

2.2e+09 N/m²

compression

Δ V_V

0.1%

Outputs

pressure

P_{g}

The solution

Solved with BulkModulus

Given

$B = 2.2 \times 10^{9}$ N/m²
$Δ V_V = 0.10$ %

Find $P_{g}$ .

Working

$B = 2.2 \times 10^{9}$ N/m² $= 2.2 \times 10^{9}$ Pa
bulk modulus is pressure over volume strain(water) $B = \frac{P_{g}}{Δ V_V}$
$2.2 \times 10^{9} = \frac{P_{g}}{0.001}$
$P_{g} = 2.2 \times 10^{6}$ Pa
$P_{g} \approx 2.2 \times 10^{6}$ Pato 2 significant figures

Conclusion

The pressure needed is the fraction of compression times the bulk modulus, whatever the amount of water; water is stiff, so even a small squeeze takes many atmospheres.

The Ruby

1require_relative "../physics"23question :a_litre_of_water_compressed do4  kind :numerical5  describes "Taking the bulk modulus of water as %{modulus}, %{pressure} to compress it by %{compression}, in " \6            "pascals?"7  given modulus: "2.2e9 N/m2"8  given compression: "0.10 %"9  asks pressure: "how much should the pressure on a litre of water be changed"10end

1require_relative "../questions/a_litre_of_water_compressed"2require_relative "../science/mechanical_properties_of_solids/bulk_modulus"34solve :a_litre_of_water_compressed, using: BulkModulus do5  objects :water67  take modulus: water.bulk_modulus8  take compression: water.fractional_compression910  pressure = water.gauge_pressure11  answer pressure:, in: "Pa", figures: 21213  conclusion "The pressure needed is the fraction of compression times the bulk modulus, whatever the " \14             "amount of water; water is stiff, so even a small squeeze takes many atmospheres."15end

1require_relative "../solutions/a_litre_of_water_compressed"23play_with :a_litre_of_water_compressed do4  title "A Litre of Water Compressed"5end

The laws

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
bulk modulus	$B$	Pa	—	BulkModulus
fractional compression	$Δ V_V$	—	—	BulkModulus
edge	$a$	m	—	BulkModulus
volume	$V$	m³	0 … ∞	BulkModulus
fall in volume	$Δ V$	m³	—	BulkModulus
density when squeezed	$ρ_{B}$	kg/m³	—	BulkModulus
pressure of ground	$P_{g}$	Pa	0 … ∞
thrust on ground	$F_{g}$	N	—
area of ground	$A_{g}$	m²	—
pressure of water	$P$	Pa	0 … ∞
thrust on water	$F$	N	—
area of water	$A$	m²	—
gauge pressure on ground	$P_{g,g}$	Pa	—
depth of ground	$h_{g}$	m	—
gauge pressure on water	$P_{g}$	Pa	—
depth of water	$h$	m	—
bulk modulus of ground	$B_{g}$	Pa	—
fractional compression of ground	$Δ {V_V}_{g}$	—	—
bulk modulus of water	$B$	Pa	—
fractional compression of water	$Δ V_V$	—	—
volume of ground	$V_{g}$	m³	0 … ∞
edge of ground	$a_{g}$	m	—
volume of water	$V$	m³	0 … ∞
edge of water	$a$	m	—
fall in volume of ground	$Δ V_{g}$	m³	—
fall in volume of water	$Δ V$	m³	—
density when squeezed of ground	$ρ_{B,g}$	kg/m³	—
density when squeezed of water	$ρ_{B}$	kg/m³	—
mass of water	$m$	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

B_{b} = \frac{P_{g,b}}{Δ {V_V}_{b}}

bulk modulus is pressure over volume strain

V_{b} = {a_{b}}^{3}

a cube s volume

Δ {V_V}_{b} = \frac{Δ V_{b}}{V_{b}}

fractional compression is the fall over the volume

ρ_{B,b} = \frac{ρ}{1 - Δ {V_V}_{b}}

the same mass in less room

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

pressure is force over area(ground)

P = \frac{F}{A}

pressure is force over area(water)

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(water)

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

pressure adds the surface s(ground)

P = P_{a} + P_{g}

pressure adds the surface s(water)

B_{g} = \frac{P_{g,g}}{Δ {V_V}_{g}}

bulk modulus is pressure over volume strain(ground)

B = \frac{P_{g}}{Δ V_V}

bulk modulus is pressure over volume strain(water)

V_{g} = {a_{g}}^{3}

a cube s volume(ground)

V = a^{3}

a cube s volume(water)

Δ {V_V}_{g} = \frac{Δ V_{g}}{V_{g}}

fractional compression is the fall over the volume(ground)

Δ V_V = \frac{Δ V}{V}

fractional compression is the fall over the volume(water)

ρ_{B,g} = \frac{ρ}{1 - Δ {V_V}_{g}}

the same mass in less room(ground)

ρ_{B} = \frac{ρ}{1 - Δ V_V}

the same mass in less room(water)