A Submarine Window

solved PressureAtDepth

Q. At a depth of 1000 m $h_{s} = 1000$ m in an ocean, a submarine has a window of area 0.04 m² $A_{w} = 0.04$ m², and its interior is maintained at sea-level atmospheric pressure $P_{w} = P_{g,s}$ . The density of sea water is 1.03 × 10³ kg/m³ $ρ = 1.03 \times 10^{3}$ kg/m³; take g as 10 m/s² $g = 10$ m/s² and atmospheric pressure as 1.01 × 10⁵ Pa $P_{a} = 1.01 \times 10^{5}$ Pa. What is the force acting on the window, in newtons?

Inputs

depth

h_{s}

1000.00 m

area

A_{w}

0.04 m²

density

ρ

1030 kg/m³

gravity

g

10 m/s²

atmosphere

P_{a}

1.01e+05 Pa

Outputs

force

F_{w}

How it's solved

Solved with PressureAtDepth

Given

$h_{s} = 1000$ m
$A_{w} = 0.04$ m²
$ρ = 1.03 \times 10^{3}$ kg/m³
$g = 10$ m/s²
$P_{a} = 1.01 \times 10^{5}$ Pa
$P_{w} = P_{g,s}$ from “maintained at sea-level atmospheric pressure”

Find $F_{w}$ .

Working

pressure of the fluid above(submarine) $P_{g,s} = ρ \cdot g \cdot h_{s}$
$P_{g,s} = 1030 \cdot 10 \cdot 1000$
$P_{g,s} = 1.03 \times 10^{7}$ Pa
maintained at sea-level atmospheric pressure $P_{w} = P_{g,s}$
$P_{w} = 1.03 \times 10^{7}$
$P_{w} = 1.03 \times 10^{7}$ Pa
pressure is force over area(window) $P_{w} = \frac{F_{w}}{A_{w}}$
$1.03 \times 10^{7} = \frac{F_{w}}{0.04}$
$F_{w} = 412000$ N
$F_{w} \approx 4.12 \times 10^{5}$ Nto 3 significant figures

Conclusion

The air inside pushes back with the atmosphere's share, so what is left pressing on the window is the gauge pressure of the water over its area.

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 :a_submarine_window do4  kind :numerical5  describes "At a depth of %{depth} in an ocean, a submarine has a window of area %{area}, and its interior is " \6            "maintained at sea-level atmospheric pressure. The density of sea water is %{density}; take g as " \7            "%{gravity} and atmospheric pressure as %{atmosphere}. %{force}, in newtons?"8  given depth: "1000 m"9  given area: "0.04 m2"10  given density: "1.03e3 kg/m3"11  given gravity: "10 m/s2"12  given atmosphere: "1.01e5 Pa"13  asks force: "What is the force acting on the window"14end

solution.rb

1require_relative "../questions/a_submarine_window"2require_relative "../science/mechanical_properties_of_fluids/pressure_at_depth"34solve :a_submarine_window, using: PressureAtDepth do5  objects :submarine, :window67  take depth: submarine.depth8  take area: window.area9  take density: rho10  take gravity: g11  take atmosphere: P_a1213  deduce_from "maintained at sea-level atmospheric pressure", window.pressure => submarine.gauge_pressure1415  force = window.thrust16  answer force:, in: "N", figures: 31718  conclusion "The air inside pushes back with the atmosphere's share, so what is left pressing on the " \19             "window is the gauge pressure of the water over its area."20end

playground.rb

1require_relative "../solutions/a_submarine_window"23play_with :a_submarine_window do4  title "A Submarine Window"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 submarine	$P_{s}$	Pa	0 … ∞
thrust on submarine	$F_{s}$	N	—
area of submarine	$A_{s}$	m²	—
pressure of window	$P_{w}$	Pa	0 … ∞
thrust on window	$F_{w}$	N	—
area of window	$A_{w}$	m²	—
gauge pressure on ground	$P_{g,g}$	Pa	—
depth of ground	$h_{g}$	m	—
gauge pressure on submarine	$P_{g,s}$	Pa	—
depth of submarine	$h_{s}$	m	—
gauge pressure on window	$P_{g,w}$	Pa	—
depth of window	$h_{w}$	m	—
mass of submarine	$m_{s}$	kg	0 … ∞
mass of window	$m_{w}$	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_{s} = \frac{F_{s}}{A_{s}}

pressure is force over area(submarine)

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

pressure is force over area(window)

P_{g,g} = ρ \cdot g \cdot h_{g}

pressure of the fluid above(ground)

P_{g,s} = ρ \cdot g \cdot h_{s}

pressure of the fluid above(submarine)

P_{g,w} = ρ \cdot g \cdot h_{w}

pressure of the fluid above(window)

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

pressure adds the surface s(ground)

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

pressure adds the surface s(submarine)

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

pressure adds the surface s(window)

P_{w} = P_{g,s}

maintained at sea-level atmospheric pressure