Deep Submergence Rescue Vehicle Intended for rescue missions in the event of a submarine accident, the Deep Submergence Rescue Vehicle can dive to a maximum depth of 5000 ft in the ocean. In dimensions of psi, by how much greater is the water’s pressure at that depth than at the ocean’s surface?

Question

Deep Submergence Rescue Vehicle

Intended for rescue missions in the event of a submarine accident, the Deep Submergence Rescue Vehicle can dive to a maximum depth of 5000 ftin the ocean. In dimensions of psi, by how much greater is the water’s pressure at that depth than at the ocean’s surface?

Approach
To fi nd the difference in pressure, we will apply Equation (6.4),

[latex]\rho_1 =p_0 + pg h [/latex] (6.4)

where the water’s pressure increases in proportion to depth. We read the density of seawater from Table 6.1

Table 6.1 Density and Viscosity Values for Several Gases and Liquids at Room Temperature and Pressure

	Density,ρ		Viscosity, μ
Fluid	[latex]kg/m^3[/latex]	[latex]slug/ft^3[/latex]	kg/(m . s)	slug/(ft . s)
Air	1.2	[latex]2.33×10^{-3}[/latex]	[latex]1.8×10^{-5}[/latex]	[latex]3.8×10^{-7}[/latex]
Helium	0.182	[latex]5.53×10^{-4}[/latex]	[latex]1.9×10^{-5}[/latex]	[latex]4.1×10^{-7}[/latex]
Freshwater	1000	1.94	[latex]1.0×10^{-3}[/latex]	[latex]2.1×10^{-5}[/latex]
Seawater	1026	1.99	[latex]1.2×10^{-3}[/latex]	[latex]2.5×10^{-5}[/latex]
Gasoline	680	1.32	[latex]2.9×10^{-4}[/latex]	[latex]6.1×10^{-6}[/latex]
SAE 30 oil	917	1.78	0.26	[latex]5.4×10^{-3}[/latex]

as 1.99[latex] slugs/ft^3 [/latex] and assume that the density of the seawater is constant.

Accepted Answer

We denote the difference in pressure between the ocean’s surface [latex]( ρ_0) [/latex] and the submarine ([latex]p_1[/latex]) by [latex] \Delta p =p _1-p _0[/latex] . The pressure increase is given by

[latex] \Delta p=\left(1.99\frac{slugs}{ft^3} ight)(32.2\frac{ft}{s^2} )(5000 ft) \longleftarrow[p_1 =p_0 + pg h ] [/latex]

[latex]=3.204 imes10^5 (\frac{ slug}{\cancel{ft^2}.ft} )(\frac{\cancel{ft}}{s^2} )(\cancel{ft})[/latex]

[latex]=3.204 imes10^5\frac{slug}{s^2.ft} [/latex]

We don’t immediately recognize the dimensions of this quantity, and so we manipulate it slightly by multiplying both the numerator and denominator by the dimension of foot:

[latex]\Delta p=3.204 imes10^5\left(\frac{slug.ft}{s^2} ight) (\frac{1}{ft^2} ) [/latex]

[latex]=3.204 imes10^5 \frac{lb}{ft^2} [/latex]

[latex]=3.204 imes10^5 psf [/latex]

We next convert this quantity to the desired unit of psi by using the conversion factor listed in the third row of Table 6.2:

Table 6.2 Conversion Factors Between USCS and SI Units for Pressure

[latex]Pa (N/m^2)[/latex]	[latex]psi (lb/in^2)[/latex]	[latex]psf (lb/ft^2)[/latex]	atm
1	[latex]1.450×10^{-4}[/latex]	[latex]2.089×10^{-2}[/latex]	[latex]9.869×10^{-6}[/latex]
6895	1	144	[latex]6.805×10^{-2}[/latex]
47.88	[latex]6.944×10^{-3}[/latex]	1	[latex]4.725×10^{-4}[/latex]
[latex]1.013×10^{-5}[/latex]	14.70	2116	1

[latex]\Delta p=(3.204 imes10^5psf)\left\lgroup6.944 imes10^{-3}\frac{psi}{psf} ight group [/latex]

[latex] =2225 \cancel {psf} \left(\frac{psi}{\cancel {psf}} ight) [/latex]

[latex]=2225 psi [/latex]

Discussion
At this depth, the water’s pressure is over 150 times greater than the standard atmospheric pressure of 14.7 psi. Over 300,000 lb of force act on each square foot of the Deep Submergence Rescue Vehicle’s hull, and the force over each square inch is equivalent to the weight of a small automobile. The density of the seawater will probably vary across this depth, but this gives a good estimation for the pressure difference.

[latex]Δp = 2225 psi[/latex]

	Density,ρ		Viscosity, μ
Fluid	$kg/m^3$	$slug/ft^3$	kg/(m . s)	slug/(ft . s)
Air	1.20	$2.33×10^{-3}$	$1.8×10^{-5}$	$3.8×10^{-7}$
Helium	0.182	$5.53×10^{-4}$	$1.9×10^{-5}$	$4.1×10^{-7}$
Freshwater	1000	1.94	$1.0×10^{-3}$	$2.1×10^{-5}$
Seawater	1026	1.99	$1.2×10^{-3}$	$2.5×10^{-5}$
Gasoline	680	1.32	$2.9×10^{-4}$	$6.1×10^{-6}$
SAE 30 oil	917	1.78	0.26	$5.4×10^{-3}$

Question 6.3: Deep Submergence Rescue Vehicle Intended for rescue missions...

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