Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 2, Problem 101EP
To determine

The capillary drop of mercury in the tube.

Expert Solution & Answer
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Answer to Problem 101EP

The capillary drop in mercury is 0.874in.

Explanation of Solution

Given information:

The diameter of glass tube is 0.018in, the contact angle with the glass is 140°, and the temperature of mercury is 68°F.

Write the expression for the capillary drop.

  h=4σscosϕρgD....... (I)

Here, capillary drop is h, the contact angle is ϕ, the density of air is ρ, acceleration due to gravity is g, and diameter of the tube is D.

Refer to Table 2.4 in the text book, to obtain the value of surface tension of mercury glass at 68°F(20°C) as σs=0.440N/m.

Write the formula for interpolation of two variables.

  ρ=(x2x1)(y3y1)(x3x1)+y1....... (II)

Here, the temperature is denoted by variables x, density is denoted by variables y, and the density is ρ.

Calculation:

Refer to Table A-8E, "Properties of liquid metals" to obtain the values of x2 as 68°F, x1 as 50°F, x3 as 100°F.

Refer to Table A-8E, "Properties of liquid metals" to obtain the values of y1 as 847.2lbm/ft3, y3 as 842.9lbm/ft3.

Prepare the table for temperature and density of mercury.

    Temperature, °FDensity, lbm/ft3
    50(x1)  847.2(y1)
    68(x2)?(ρ)
    100(x3)842.9(y3)

Substitute 50°F for x1, 68°F for x2, 100°F for x3, 847.2lbm/ft3 for y1 and 842.9lbm/ft3 in Equation (II).

  ρ=(842.9lbm/ ft 3847.2lbm/ ft 3)(68°F50°F)(100°F50°F)+847.2lbm/ft3=(4.3lbm/ft3)0.36+847.2lbm/ft3=1.548lbm/ft3+847.2lbm/ft3=845.65lbm/ft3

Substitute 0.440N/m for σs, 140° for ϕ, 845.65lbm/ft3 for ρ, 32.2ft/s2 for g and 0.18in in Equation (I).

  h=4(0.440N/m)(cos140°)(845.65lbm/ ft 3)(32.2ft/ s 2)(0.018in)=4(( 0.440N/m )×( 0.2248lbf 1N )×( 1m 3.28084ft ))(cos140°)(845.65lbm/ ft 3)(32.2ft/ s 2)(0.018in× 1ft 12in)(32.2lbmft/ s 21lbf)=4(0.030148lbf/ ft 3)(cos140°)(845.65lbm/ ft 3)(32.2ft/ s 2)(0.0015ft)(32.2lbmft/ s 21lbf)=0.07282ft

  h=0.07282ft(12ft1in)=0.874in

Here, the negative sign indicates the capillary drop instead of rise.

Conclusion:

The capillary drop in mercury is 0.874in.

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Chapter 2 Solutions

Fluid Mechanics: Fundamentals and Applications

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