October 31

Fluids Problem (Buoyancy)

Study Problem

A piece of equipment weights 300 pounds on dry land. When the equipment is fully submerged in water the equipment weighs only 65 pounds.

What is the volume of the equipment?
A) 3.77 ft3
B) 1.96 ft3
C) 0.78 ft3
D) 1.28 ft3

What is the specific gravity of the equipment?
A) 1.03
B) 0.78
C) 1.28
D) 0.82

If the equipment was submerged in API 42 Oil, what would the equipment weight?
A) 246 lbs
B) 79 lbs
C) 193 lbs
D) 107 lbs

Solution

This fluids study problems explains how to calculate volume, specific gravity and weight of an object when placed in water and crude oil.

Calculating buoyant force of water

F_b\ =\ W_{air} - W_{submerged}
F_b = 300\ lbs - 65\ lbs = 235\ lbs

Calculating the volume of water displaced

V\ =\ \frac {weight}{\rho}
V\ = \frac {235\ lbs}{62.4\ \frac{lbs}{ft^3}} =\ 3.77\ ft^3

The volume is 3.77 cubic feet
Choice A is correct

Calculating specific gravity of the equipment

SG\ =\ \frac{wt_{object}}{wt_{water}}
SG\ =\ \frac{300}{235} = 1.28

The specific gravity of the equipment is 1.28
Choice C is correct

Convert API to SG

SG\ =\ \frac{141.5}{131.5 + API^{\circ}}
SG\ =\ \frac {141.5}{131.5 + 42} = 0.82

Calculate density

\rho_{oil} = (\rho_{water})(SG_{oil})
\rho_{oil} = (62.4\ \frac{lbs}{ft^3})(0.82) = 51.2\ \frac{lbs}{ft^3}

Calculate the weight of oil displaced

wt\ =\ (\rho)(V)
wt\ =\ (51.2\ \frac{lbs}{ft^3})(3.77\ ft^3)\ =\ 193\ lbs

Calculate the weight of the equipment in oil

Wt_{equipment\ submerged}\ =\ wt_{equipment} - wt_{oil\ displaced}
Wt_{equipment\ submerged}\ =\ 300\ lbs\ -\ 193\ lbs\ =\ 107\ lbs

The weight of the equipment in oil is 107 lbs
Choice D is correct


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Posted October 31, 2014 by Patrick Matherne in category "Hydraulics and Fluids