EXAMPLE 9.5 Oil and Water GOAL Calculate pressures created by layers of differen

Question

EXAMPLE 9.5 Oil and Water GOAL Calculate pressures created by layers of different fluids Air Po Oil PROBLEM In a huge oil tanker, salt water has flooded an oil tank to a depth of h2 5.00 m. On top of the water is a layer of oil h1 8.00 m deep, as in the cross-sectional view of the tank in the figure. The oil has a density of 0.700 g/cm3. Find the pressure at the bottom of the tank. (Take 1,025 kg/m3 as the density of salt water.) Water Phot STRATEGY P Po + gh must be used twice. First, use it to calculate the pressure P1 at the bottom of the oil layer. Then use this pressure in place of Po in the equation and calculate the pressure Pbot at the bottom of the water layer SOLUTION Use the equation to calculate the pressure at the bottom of the oil layer = 1.01 x 105 Pa +(7.00 x 102 kg/m3 (9.80 m/s2 8.00 m P1 = 1.56 x 105 Pa (2) Pbot=P1 + pgh2 Now adapt the equation to the new starting pressure, and use it to calculate the pressure at the bottom of the water layer = 1.56 x 105 Pa 1.025 x 103 kg/m3 (9.80 m/s2)Y 5.00 m Pbot2.06 x 105 Pa

Explanation / Answer

Practice it:

P = Po + rho_oil*g*h1 + rho_water*g*h2

= 1.013*10^5 + 700*9.8*7.85 + 1025*9.8*4.85

= 2.04*10^5 pa

Excercise :

P = Po + rho_water*g*h1 + rho_mud*g*h2

= 1.013*10^5 + 1025*9.8*11.5 + 1.75*10^3*9.8*3.75

= 2.81*10^5 pa

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