Physics - Fluid - Surface Tension Force

Main Menu

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 0^o
       so the cos θ factor in figure 7, is equal to unity.  Then,
(b) To what depth will mercury ( at 15^oC) be
       depressed in the same tube?  In this case the contact
       angle is 135^o ; 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)
Acetone	20	0.0237
Alcohol,methyl	20	0.0226
Benzene	20	0.0288
Water	0	0.0756
Water	20	0.0728
Water	30	0.0712
Water	100	0.0589
Mercury	15	0.487

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.

Brake Booster

Physics volume & mass flow rate

Main Menu

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) A water line necks down from a pipe with a 12.5 mm.
       radius to a pipe with a 9 mm radius. If the speed of the
       water in the 12.5 mm pipe is 1.8 m.s^-1, what is the speed
       in the smaller pipe?
(b) What is the volume flow rate?
(c) What is the mass flow rate?

Strategy - the rule "Must have" of Mr.Zhang ®

ΔM	=	ρΔV	=	ρAV&#916t
ΔM &#916t	=	ρAV	→	(1) The rate of flow of mass
ρV1A1 = ρV2A2			→	The equation of continuty.
V1A1 = V2A2			→	(2) Density is constant.

Solution

(a) Compute the speed in the smaller pipe by solving the equation (2)

V2

=

V1A1 A2

V2

=

V1 π(r1)2 π(r2)2

V2	=	1.8 m.s-1(12.5 x 10-3)2 (9 x 10-3)2
Ans. The speed in the smaller pipe is 3.47 m.s-1,therefore, it make sense.
How doest it make sense?

(b) Compute the volume flow rate is;
V1A1	=	V2A2
	=	π(12.5x10-3m)2(1.8 m.s-1)
	=	8.8 x 10-4 m3.s-1

Ans. The volume flow rate is 8.8 x 10^-4m³s^-1.

(c) The mass flow rate is;
Compute it from eqaution(2);
ρV1A1	=	ρV2A2
	=	(1.0x103 kg.m-3)(8.8x10-4 m3.s-1)

Ans. The mass flow rate is 0.88 kg.s^-1.

Brake Booster

Pressure between oil and water

Main Menu

Problem# h2phy1 - Pressure converted inside the Aorta

Problem# h2phy2 - Pressure inside vein

Problem# h2phy3 - Pressure on the floor

Problem# h2phy4 - Pressure under the sea

Problem# h2phy5 - Pressure between oil and water

The U-tube contains water of density&nbspρ_w in the right arm
and oil of unknown density ρ_x in the left.   Measurement gives
L = 135 mm. and d = 12.3 mm. What is the density of the oil ?

Strategy - the rule "Must have" of Mr.Zhang ®

Interface area is joint of the two substances. P = (Po) + (Pg) Absolute pressure = Atmospheric pressure + Gauge pressure       Material or Object         Density (kg/m3)     Water: 20 οC and 1 atm 0.998x103   1 atm = 760 mmHg = 760 torr = 29.9 in.Hg =101.325 kPa = 14.7 lb/in.2

Solution Pint = Po +&nbspρwgl          (right side)  →(1) Pint = Po +&nbspρoilg(d+l)    (left side)  →(2) (1)=(2);   Po +&nbspρwgl = Po +&nbspρoilg(d+l)     ρwgl = ρoilg(d+l)     ρwl = ρoil(d+l)     (ρw)/(d+l) = ρoil ∴The density of the oil is 916.5 kg/m3

Brake Booster

Physics-High-school-Fluid

Main Menu

Problem# h2phy1 - Pressure inside the Aorta

Problem# h2phy2 - Pressure on the floor

A living room has floor dimension of 3.5 m and 4.2 m and height of 2.4 m.
a. What does the air in the room weight?
b. What force does the atmosphere exert on the floor of the room?

Strategy - the rule "Must have" of Mr.Zhang ® material for solving this problem ; (1) w=mg    (2) m=ρv    (3) ρ=F⊥/A

Material or Object         Density (kg/m3)     Air: 20 οC and 1 atm 1.21

Solution a) The air in the room weighs;  w=mg    →(1)  m=ρv    →(2) We take (2) substitue in (1) ∴  W=ρairvg ∴  W = (1.21 kg/m3)(3.5 m x 4.2 m x 2.4 m)(9.8 m/s2 ∴  W = 418 kg m/s2 ∴  W = 418 N

Ans.   ∴The air in the room weighs 418 N

b) What force does the atmosphere exert on the floor of the room?  ρ=F⊥/A  ρatm=F⊥/A  F⊥=ρatmA  F⊥=1.01x105 N/m2)(3.5 m x4.2 m)  F⊥=1.5x106 N

Ans.   ∴The atmosphere exerts 1.5x106 N on the floor of the floor Brake Booster