A Copper Cube Under Pressure

Q. Taking the bulk modulus of copper as 140 × 10⁹ N/m², what is the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa, in cm³? expects 0.050 cm³

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Q. Taking the bulk modulus of copper as 140 × 10⁹ N/m² $B = 140 \times 10^{9}$ N/m², what is the volume contraction of a solid copper cube, 10 cm $a = 10$ cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa $P_{g} = 7.0 \times 10^{6}$ Pa, in cm³?

The playground

Inputs

edge

a

10 cm

pressure

P_{g}

7e+06 Pa

modulus

B

1.4e+11 N/m²

Outputs

contraction

Δ V

The solution

Solved with BulkModulus

Given

$a = 10$ cm
$P_{g} = 7.0 \times 10^{6}$ Pa
$B = 140 \times 10^{9}$ N/m²

Find $Δ V$ .

Working

$a = 10$ cm $= 0.1$ m
$B = 140 \times 10^{9}$ N/m² $= 1.4 \times 10^{11}$ Pa
bulk modulus is pressure over volume strain(cube) $B = \frac{P_{g}}{Δ V_V}$
$1.4 \times 10^{11} = \frac{7 \times 10^{6}}{Δ V_V}$
$Δ V_V = 5 \times 10^{−5}$
a cube s volume(cube) $V = a^{3}$
$V = {0.1}^{3}$
$V = 0.001$ m³
fractional compression is the fall over the volume(cube) $Δ V_V = \frac{Δ V}{V}$
$5 \times 10^{−5} = \frac{Δ V}{0.001}$
$Δ V = 0.05$ cm³
$Δ V = 5 \times 10^{−8}$ m³ $= 0.05$ cm³
$Δ V \approx 0.050$ cm³to 2 significant figures

Conclusion

The cube loses the same fraction of its volume as the pressure is of copper's bulk modulus; copper is so stiff that the loss is a tiny sliver of the whole.

The Ruby

1require_relative "../physics"23question :a_copper_cube_under_pressure do4  kind :numerical5  describes "Taking the bulk modulus of copper as %{modulus}, %{contraction}, %{edge} on an edge, when subjected " \6            "to a hydraulic pressure of %{pressure}, in cm³?"7  given edge: "10 cm"8  given pressure: "7.0e6 Pa"9  given modulus: "140e9 N/m2"10  asks contraction: "what is the volume contraction of a solid copper cube"11end

1require_relative "../questions/a_copper_cube_under_pressure"2require_relative "../science/mechanical_properties_of_solids/bulk_modulus"34solve :a_copper_cube_under_pressure, using: BulkModulus do5  objects :cube67  take edge: cube.edge8  take pressure: cube.gauge_pressure9  take modulus: cube.bulk_modulus1011  contraction = cube.fall_in_volume12  answer contraction:, in: "cm3", figures: 21314  conclusion "The cube loses the same fraction of its volume as the pressure is of copper's bulk modulus; " \15             "copper is so stiff that the loss is a tiny sliver of the whole."16end

1require_relative "../solutions/a_copper_cube_under_pressure"23play_with :a_copper_cube_under_pressure do4  title "A Copper Cube Under Pressure"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 cube	$P$	Pa	0 … ∞
thrust on cube	$F$	N	—
area of cube	$A$	m²	—
gauge pressure on ground	$P_{g,g}$	Pa	—
depth of ground	$h_{g}$	m	—
gauge pressure on cube	$P_{g}$	Pa	—
depth of cube	$h$	m	—
bulk modulus of ground	$B_{g}$	Pa	—
fractional compression of ground	$Δ {V_V}_{g}$	—	—
bulk modulus of cube	$B$	Pa	—
fractional compression of cube	$Δ V_V$	—	—
volume of ground	$V_{g}$	m³	0 … ∞
edge of ground	$a_{g}$	m	—
volume of cube	$V$	m³	0 … ∞
edge of cube	$a$	m	—
fall in volume of ground	$Δ V_{g}$	m³	—
fall in volume of cube	$Δ V$	m³	—
density when squeezed of ground	$ρ_{B,g}$	kg/m³	—
density when squeezed of cube	$ρ_{B}$	kg/m³	—
mass of cube	$m$	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(cube)

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(cube)

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

pressure adds the surface s(ground)

P = P_{a} + P_{g}

pressure adds the surface s(cube)

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(cube)

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

a cube s volume(ground)

V = a^{3}

a cube s volume(cube)

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(cube)

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(cube)