Problem 134P

Consider a U–tube filled with mercury as shown in Fig. P3–142. The diameter of the right arm of the U–tube is D = 2 cm. and the diameter of the left arm is twice that. Oil with a specific gravity of 2.72 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil that can be added into the left arm.

Solution

Introduction

Consider a U tube filled with mercury, the volume of the mercury in the tube is constant and while pouring an oil in to the left arm will reduce the level of mercury in the left arm and will increase the mercury volume in the right arm.

If the volume of mercury drop an amount x in the left arm and let h is the corresponding rise in the right arm.

STEP 1

Diameter of the right arm, D=2cm

Diameter of the left arm = 2D=4cm

Volume of the left arm, Vleft =x =x=h

Vleft =Vright

The pressure at A and B are equal

Patm+ρoilg(hoil +1 ) =Patm+ρHgg(hHg )

It will result

SGoilρwg(hoil+x)=SGHgρwg(5x)

1=

1=

X =

STEP 2

Specific gravity of the liquid SGoil =2.72

Specific gravity of mercury, SGHg =13.6

Level of oil in the left arm, hoil =12 cm

X = =0.5 cm

STEP 3

So the maximum amount of oil that can added to the left arm

Vmax =(2)2(hoil +x)

Vmax =(2)2(hoil +x)

Vmax =(3.14)(2)2(12 +0.5)

Vmax =157.08 cm3

Vmax =0.157 L