A circular tank with a 1.50 m radius is filled with two fluids, a 4.00 m layer o

Question

A circular tank with a 1.50 m radius is filled with two fluids, a 4.00 m layer of water and a 3.00 m layer of oil. Use Doil= 8.24 × 103 kg/m3 and Dwater= 1.00 × 103 kg/m3 , and Datm = 1.01 × 105 N/m2 . A. What are the gauge and absolute pressures 1.00 m above the bottom of the tank? B. A block of material in the shape of a cube (m = 100 kg and side length = 42.0 cm) is released at the top of the oil layer. Where does the block come to rest? Justify your answer. If it comes to rest between two layers, specify which layers and what portion of the block sits in each layer. [Note: V = cube a ] 3 C. A small 1.00 cm radius opening is made in the side of the tank 0.500 m up from its base (block was removed). What volume of water drains from the tank in 10.0 s?

Explanation / Answer

Solution (A):

At a depth of 1m above the bottom of tank:

Height of water column above = 3m

Height of oil column above = 3m

Now, pressure p of a liquid column of height h is given by:

p = d g h

where

d = density of liquid in kg/m3

g = acceleration due to gravity (9.8 m/s2)

Given:

Density of water column above = 1000 kg/m3

Density of oil column above = 8240 kg/m3

Thus, at the required depth,

pressure of water column above

= 1000 × 9.8 × 1 = 9800 N/m2

pressure of oil column above

= 8240 × 9.8 × 3

= 242256 N/m2

So, total pressure at the desired depth

= pressure of water column + pressure of oil column

= 9800 N/m2 + 242256 N/m2

= 252056 N/m2

= 2.52 × 105 N/m2 (two significant places)

This is the absolute pressure at the required depth.

Gauge pressure = absolute pressure - atmospheric pressure

= 2.52 × 105 N/m2 - 1.01 × 105 N/m2

=1.51 × 105N/m2

(Please note: All values must be in SI units)

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