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Problem

الكلية كلية الهندسة     القسم  الهندسة البيئية     المرحلة 3
أستاذ المادة عدي عدنان جهاد الخيكاني       05/10/2012 16:53:41
Problem 1. Determine the capillary depression of mercury in a 4 mm ID glass
Tube. Assume surface tension as 0.45 N/m and ? =115°?.
The specific weight of mercury = 13550 × 9.81 N/m3, Equating the surface force and the
pressure force, [h × ? × ?D2/4] = [? × D × ? × cos ?], Solving for h,
h = {4 × ? × cos ?}/{? × D} = [4 × 0.45 × cos 115]/[13550 × 9.81 × 0.004]
Hence actual pressure indicated = 762 + 3.92 = 765.92 mm of mercury.
= – 1.431 × 10–3 m or – 1.431 mm, (depression)
Problem 2. A glass tube of 8 mm ID is immersed in a liquid at 20°C. The specific
weight of the liquid is 20601 N/m 3. The contact angle is 60°. Surface tension is 0.15 N/m.
Calculate the capillary rise and also the radius of curvature of the meniscus.
Capillary rise, h = {4 ×?? × cos???}/{? × D} = {4 × 0.15 × cos 60}/{20601 × 0.008}
= 1.82 × 10 -3 m or 1.82 mm.
The meniscus is a doubly curved surface with equal radius as the section is circular
P i - P o) = ? × {(1/R 1) + (1/R 2)} = 2 ?/R
R = 2?/(P - P ), (P - P ) = specific weight × h
i o i o
So, R = [2 × 0.15]/ [1.82 × 10 -3 × 2060] = 8 × 10 -3 m or 8 mm.


Problem 3. A mercury column is used to measure the atmospheric pressure. The height of column above the mercury well surface is 762 mm. The tube is 3 mm in dia. The contact angle is 140°. Determine the true pressure in mm of mercury if surface tension is 0.51 N/m. The space above the column may be considered as vacuum. In this case capillary depression is involved and so the true pressure = mercury column + capillary depression.
The specific weight of mercury = 13550 × 9.81 N/m 3, equating forces,
[h ×?? ×?? D 2/4] = [? × D ×??? × cos???].
So
h = {4 ×?? × cos???}/{? × D}
h = (4 × 0.51) × cos 140]/[13550 × 9.81 × 0.003]
= - 3.92 × 10 -3 m or - 3.92 mm, (depression)




Problem 4. A hollow cylinder of 150 mm OD with its weight equal to the buoyant
forces is to be kept floating vertically in a liquid with a surface tension of 0.45 N/m 2. The
contact angle is 60°. Determine the additional force required due to surface tension.
In this case a capillary rise will occur and this requires an additional force to keep the
cylinder floating.
Capillary rise, h = {4 ×?? × cos???}/{? × D}.
As

(P i - P o) = h × specific weight, (P i - P o) = {4 ×?? × cos???}/D

(P i - P o) = {4 × 0.45 × cos 60}/{0.15} = 6.0 N/m2

Force = Area × (P i - P o) = {? × 0.015 2/4} × 6 = 0.106 N

As the immersion leads to additional buoyant force the force required to kept the cylinder
floating will be double this value.
So the additional force = 2 × 0.106 = 0.212 N.


المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .