5.6 Questions

5.6 Questions

07 October 2011

11:53

Collins, p.107

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Note: ρfresh water = 1,000kg/m3; g = 10N/kg

ps. 1,000mbar = 1 bar = 100,000Pa!

ANSWERS Collins, p107

Q5

pressure difference = height × density × g

∆p = h × ρ × g

∆p = pressure of the fluid (N/m2 or Pa) = 250kPa - 100kPa = 150kPa (x1000) = 150 000

h = height FROM SURFACE LEVEL (m) = ?

ρ = density of the fluid (kg/m3) = 1000

g = gravitational field strength (N/kg) = 10

height => h = 150 000/ (1000 x 10) = 15m OR NOTE IN FRESH WATER: EVERY 10m descend = +100kPa
If he was diving in water that was slightly denser than fresh water then we see from the equation that the pressure difference would be divided by a greater number, therefore the answer would be less meaning that the diver would dive less below the surface.

Q6

pressure difference = height × density × g

∆p = h × ρ × g

∆p = pressure of the fluid (N/m2 or Pa) = ?

h = height FROM SURFACE LEVEL (m) = 50

ρ = density of the fluid (kg/m3) = 0.42g/cm3 = 0.00042kg/cm3 => (1cm3 = 1/100^3) = 0.00042 x 100^3 = 420 OR NOTE: 1 g/cm3 = 1000kg/m3

g = gravitational field strength (N/kg) = 1.4

NOTE: 1 g/cm3 = 1000kg/m3

∆p =50 x 420 x 1.4 = 29400Pa OR 29.4kPa

Total pressure (atmospheric pressure on Titan = 1600mbar (1 mbar = 0.1kPa = 100Pa) => 1600 x 0.1 = 160kPa

TP = 160 +29.4 = 189.4kPa