วันศุกร์ที่ 28 มีนาคม พ.ศ. 2557

Physics - Fluid - Surface Tension Force


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Fluid Problems
# h2phy1 - Pressure converted inside Aorta 
# h2phy2 - Pressure inside vein
# h2phy3 - Pressure on the floor
# h2phy4 - Pressure under the sea # h2phym1- Hydraulic Lift - 1
# h2phy5 - Pressure between oil and water # h2phym2- Hydraulic Lift - 2
# h2phy6 - Buoyant Force # h2phym3- Hydraulic Lift - 3
# h2phy7 - Flow rate & Velocity
# h2phy8 - Surface Tension Force(1)
# h2phy9 - Volume & Mass Flow rate
# h2phy10 - Surface Tension Force(2)


(a) To what height will water (at 20 oC) rise in a glass tube
       with a bore radius of 0.1 mm?
       The contact angle for water in a glass tube is 0o
       so the cos θ factor in figure 7, is equal to unity.  Then,
(b) To what depth will mercury ( at 15oC) be
       depressed in the same tube?  In this case the contact
       angle is 135o ; therefore,
(density of Mercury=13.6x103 kg.m-3)


Strategy - the rule "Must have" of Mr.Zhang ®
Coefficient of surface tension(γ) =
Tension Force(FT)
Circumference(L)
 →  SI unit N.m-1
Ftension → (Forced required to stop movable side from sliding)
L(circumference ) =2πr ; because of the round shape of the surface.
γ 2πr=FT
We break down FT to FT (cosθ);
γ 2πr(cos θ)=FT (cosθ)  → (1) Upward force  ↑
mg=ρvg
                                                                 =(πr2h)(ρg)→ (2)Downward force  ↓

Table 8 Values of the Surface Tension for various
liquids in contact with air
Liquid Temperature(oC) -γ-(J/m2 or N/m)
Acetone200.0237
Alcohol,methyl200.0226
Benzene200.0288
Water00.0756
Water200.0728
Water300.0712
Water1000.0589
Mercury150.487

Solution
 (a) Compute to find the height by bringing equation (1)=(2) according to the equlibrium
γ 2πr(cos θ)= (πr2h)(ρg)
γ 2πr(cos θ)= (πr2h)(ρg)
h=
2γcos θ
ρgr
h=
2(0.0728 N.m-1)
(103kg/m3)(9.8m/s2)(10-4m)
=0.15 m
Ans. Water (at 20 oC) rise 15 cm height in a glass tube.

(b) To what depth will mercury ( at 15oC) be
       depressed in the same tube?  In this case          the contact angle is 135o ; therefore, 
      (density of Mercury=13.6x103 kg.m-3)
h=
2γcos135o
ρgr
h=
2(0.487 N.m-1)(-0.707)
(13.6x103 kg.m-3)(9.8 m.s-2(10-4 m)
h= -5.16 cm.
Ans. Mercury (at 15 oC) will be depressed 5.16 cm.




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