PHY–302 K.
Solutions for Problem set # 12.
Textbook problem 10.55:
Pressure drop due to viscosity in a round pipe of radius R is given by the Poiseuille equation:
∆P
8ηv̄
8ηF
=
=
2
L
R
πR4
(1)
where η is viscosity of the fluid, F is the volumetric flow rate, and v̄ = F/πR2 is the average
velocity of the flow.
Human aorta has approximately round cross-section. In healthy adults, its diameter varies
between about 2 and 3.5 cm; for this exercise, we use 2R = 2.4 cm (i.e., R = 1.2 cm) from
the textbook example 10–11 (page 269). The blood flow through aorta depends on what the
person is doing (resting or exercising); for this exercise we take
v̄ = 40 cm/s
=⇒
F = πR2 × v̄ = 180 cm3 /s = 11 L/min.
(2)
from the same textbook example 10–11. Finally, the viscosity of whole blood at body temperature is η = 4 · 10−3 Pa · s (from table 10–3 on page 275 of the textbook). Plugging all
these data into the Poiseuille equation (1), we get
8(4 · 10−3 Pa · s)(0.4 m/s)
∆P
=
≈ 90 Pa/m = 0.9 Pa/cm.
L
(0.012 m)2
(3)
Note: over the whole length of the aorta — about 50 cm — the blood pressure drop due to
viscosity is only 45 Pa or 1/3 of a millimeter of mercury.
Textbook problem 10.58:
Squeezing blood through a thin needle requires a pressure difference: the blood pressure in
the tube connected to the needle should be higher than the blood pressure in the patient’s
vein. Specifically, according to Poiseuille equation (1), we need
∆P ≡ Ptube − Pvein =
8ηFL
πR4
(4)
where R is the inner diameter of the needle, L is its length, η is the viscosity of blood, and
F is the desired flow rate. For the needle in question, L = 4.0 cm and 2R = 0.40 mm, we
1
want F = 4.0 cm3 /min = 0.067 cm3 /s, and the blood viscosity is η = 4.0 · 10−3 Pa · s (from
textbook table 10–3). Therefore,
∆P =
8(4.0 · 10−3 Pa · s)(0.067 cm3 /2)(4.0 cm)
= 17.0 kPa.
π(0.040 cm/2)4
(5)
To create this pressure difference, the bottle of donor’s blood is raised above the needle,
which creates the hydrostatic pressure
tube
Pnear
needle = Pbottle + ρgh
(6)
where ρ = 1050 kg/m3 is the density of blood (from table 10–1 on page 256 of the textbook).
The bottle is made of very soft plastic, so the pressure inside it is equal to the atmospheric
pressure, Pbottle = Patm . Consequently, the gauge pressure in the tube near the needle is
gauge
= Ptube − Patm = ρgh.
Ptube
(7)
To be precise, we should correct this formula by the Bernoulli term − 21 ρv 2 (where v is the
speed of the blood in the tube) and by the effects of viscosity in the tube, but for a typical
tube radius Rtube ∼ 2 mm, both effects are too small to matter.
The gauge pressure of blood in the patient vein is
gauge
Pvein
= 18 Torr ≡ 18 mm.Hg = 2.4 kPa.
(8)
hence, to create the pressure difference (5) along the needle, the blood pressure in the tube
feeding the needle should be
gauge
gauge
Ptube
= Pvein
+ ∆P = 17.0 kPa + 2.4 kPa = 19.4 kPa.
(9)
Comparing this value to the hydrostatic formula (7), we can find the height h of the blood
bottle above the needle as
gauge
Ptube
19.4 · 103 Pa
h =
=
= 1.88 m,
ρg
1050 kg/m3 × 9.8 N/kg
or about 6 feet and 2 inches.
2
(10)
Non-textbook problem #1:
There are three forces acting on a red blood cell: its weight mg, the buoyant force from the
plasma
FB = ρplasma gVcell = ρplasma g
ρplasma
m
= mg ×
,
ρcell
ρcell
(11)
and the velocity-dependent viscous drag force
FD (v) = CLη × v
(12)
where L is the red cell’s linear size and C is the viscous drag coefficient for its shape. For a
spherical cell of radius R, CL = 6πR, i.e.
FD (v) = 6πRη × v.
Also, a spherical cell has volume V =
4π 3
3 R
(13)
and hence weight
4π 3
R ρcell g
3
(14)
4π 3
R ρplasma g.
3
(15)
mg =
and buoyant force
FB =
When the cell reaches its terminal velocity vt , it’s not accelerating any more, so the net
force acting on it should vanish,
Fnet = mg − FB − FD (vt ) = 0.
(16)
In light of eqs. (13), (14), and (15), this gives us
4π 3
4π 3
R ρcell g −
R ρplasma g − 6πRη × vt = 0
3
3
(17)
and hence
vt =
4π 3
3 R g(ρcell
− ρplasma )
2R2 g(ρcell − ρplasma )
=
.
6πRη
9ηplasma
(18)
Numerically, for R = 6µm, ρcell = 1125 kg/m3 , ρplasma = 1025 kg/m3 , and ηplasma = 1.5 ·
3
10−3 Pa · s, we get
vt = 5.3 · 10−6 m/s = 19 mm/hr.
(19)
PS: When typing up this problem, I have confused the radius of a red blood cell with its
diameter. In real life, the human red blood cells have diameters between 6 and 8 µm and
hence radii between 3 and 4 µm. Plugging R = 3.5 µm into eq. (18) would give the terminal
velocity vt = 6.5 mm/hr.
Of course, in real life the red blood cells are disk-shaped rather than spherical, and sometimes a few cells stick together, which lowers the viscous drag. Consequently, the erythrocyte
sedimentation rate of a healthy adult varies between 10 and 20 mm/hr, depending on age and
sex of the person.
Non-textbook problem #2:
The forces acting on the Huygens probe about to land on Titan are its weight mg in Titan’s
gravity and the aerodynamic drag on its parachute,
FD (v) = C × 21 ρair A × v 2
(20)
where A = πR2 is the parachute’s area and C = 1.50 is its drag coefficient. Note that unlike
the previous problem, the drag on the parachute is due to turbulence rather than viscosity.
That’s why it depends on the density of the Titan’s ‘air’ rather than its viscosity; also it grows
with the parachute’s speed as v 2 rather than v.
The acceleration the probe depends on its velocity v as
ma(v) = Fnet (v) = mg − FD (v) = mg −
CρA
× v2.
2
(21)
As the probe’s velocity increases, the acceleration slows down. After a while, the probe reaches
the terminal velocity vt for which
a(vt ) = 0
=⇒
mg −
4
CρA
× vt2 = 0.
2
(22)
Solving this equation for the vt , we get
vt2 =
2(318 kg)(1.35 m/s2 )
2mg
=
= 14.4 m2 /s2
Cρ(A = πR2 )
1.50(5.5 kg/m3 )π(3.03 m/2)2
=⇒
vt = 3.8 m/s.
(23)
Since the Huygens probe had plenty of time to reach the terminal velocity, it landed on Titan’s
surface with speed v = vt = 3.8 m/s = 8.5 MPH.
Textbook problem 13.7:
◦
Concrete has thermal expansion coefficient α = 12 · 10−6 ( C)−1 . Thus, a concrete slab of
length L0 = 12 m made at temperature T0 = 20◦ C will expand in hot weather T1 = 50◦ C by
∆L1 = L0 × α(T1 − T0 ) = 4.3 mm
(24)
and contract in cold weather T2 = −30◦ C by
−∆L2 = −L0 × α(T2 − T0 ) = 7.2 mm.
(25)
Consequently, the gaps between adjacent slabs will shrink in hot weather by 4.3 mm and
increase in cold weather by 7.2 mm,
`gap (50◦ C) = `gap (20◦ C) − 4.3 mm,
`gap (−30◦ C) = `gap (20◦ C) + 7.2 mm.
(26)
To avoid horizontal stresses in concrete, the slabs should not interfere with their neighbors’
thermal expansion. Thus, when the slabs expand in hot weather, they should not bump into
each other — that’s what the gaps are for. The gaps shrink in hot weather, but they should not
close until the temperature reaches the maximum expected by this highway, namely T1 = 50◦ C.
Thus
`gap (50◦ C) = 0
(27)
`gap (20◦ C) = 4.3 mm
(28)
and consequently
5
and
`gap (−30◦ C) = 4.3 mm + 7.2 mm = 11.5 mm.
(29)
In particular, when the concrete slabs are laid down at 20◦ C, the highway builders should
leave gaps of 4.3 mm between each slab.
Textbook problem 13.90:
As the temperature changes, the length of the platinum bar between two marks changes by
∆L = L0 × α∆T.
(30)
∆L = L0 α∆T ≤ ∆Lmax = 1 µm,
(31)
To keep this change small enough —
— we need to limit temperature changes to
|∆T | =
Lmax
1 · 10−6 m
= 0.11◦ C.
=
αL0
1 m × 9 · 10−6 (C◦ )−1
(32)
That is, the temperature of the bar should not change by more than ±0.11◦ C (or ±0.2◦ F)
from the temperature at which the marks on the bar were made.
Textbook problem 13.86:
In equilibrium, the weight of the fluid displaced by the floating body should equal to the
body’s own weight. Consequently, the submerged part of the body’s volume is related to the
total volume of the body as the body’s density to the fluid’s density,
ρbody
Vsubm
=
.
Vtot
ρfluid
(33)
When the temperature raises, both mercury and iron expand in volume and their densities
6
decrease,
ρ(T ) =
ρ(T0 )
≈ ρ(T0 ) × (1 − β(T − T0 )).
1 + β(T − T0 )
(34)
But the mercury expands faster than iron — β(Hg) = 180·10−6 /◦C while β(Fe) = 35·10−6 /◦C
— so the ratio ρ(Fe)/ρ(Hg) becomes larger:
1 + βHg (T − T0 )
ρFe (T )
ρFe (T0 )
ρFe (T0 ) =
×
≈
× 1 + (βHg − βFe ) × (T − T0 ) . (35)
ρHg (T )
ρHg (T0 )
1 + βFe (T − T0 )
ρHg (T0 )
According to eq. (33), this means that for an iron cube floating in mercury
Vsubm
Vsubm
(T ) =
(T0 ) × 1 + (βHg − βFe ) × (T − T0 )
Vtot
Vtot
(36)
— as the temperature increase, the submerged fraction of the cube’s volume increases, and
the cube floats lower in mercury than at the lower temperature. Specifically, for T increasing
from 0◦ C to 25◦ C,
◦
(βHg − βFe ) × (T − T0 ) = (180 − 35) · 10−6 ( C)−1 × 25◦ C = 3.6 · 10−3 = 0.36%,
(37)
so the submerged fraction of the iron cube’s volume increases by 0.36%.
Note: this percentage increase is relative to the original fraction
Vsubm ◦
ρFe (0◦ C)
7881 kg/m3
(0 C) =
=
= 0.5797 = 57.97% :
Vtot
ρHg (0◦ C)
13595 kg/m3
(38)
As the temperature rises, the submerged fraction increases to
57.97% × (1 + 0.0036) = 57.97% + 0.21% = 58.18%.
(39)
Thus, relative to the total cube volume, the submerged fraction increases by only 0.21%.
7
Non-textbook problem #3:
The volume of alcohol in the thermometer is
V = Vb + πr2 × h
(40)
where Vb is the (inside) volume of the bulb, r is the (inside) radius of the cylindrical tube, and
h is the height of alcohol column inside the tube. We assume the glass does not expand, so
Vb and r are constant and only the height h changes with the temperature-dependent volume
of alcohol according to
∆V = πr2 × ∆h
=⇒
∆h =
∆V
.
πr2
(41)
As the temperature rises, the volume of alcohol changes by
∆V = V0 × β × ∆T
(42)
which makes the alcohol height rise by
∆h =
V0
× β × ∆T.
πr2
(43)
Originally, h0 = 3.0 cm and
πr2 × h0 = 9.4 · 10−4 cm3 Vb = 1.00 cm3 .
(44)
V0 = Vb + πr2 × h0 ≈ Vb
(45)
Consequently,
which allows us to simplify eq. (43) to
∆h ≈
Vb
× β × ∆T.
πr2
(46)
For the thermometer in question,
Vb
1.00 cm3
◦
×
β
=
× 1.09 · 10−3 ( C)−1 = 3.47 cm/◦ C,
2
2
πr
π(0.0100 cm)
(47)
so when the temperature increases from 10◦ C to 30◦ C, the alcohol in the thermometer raises
8
by
∆h = (3.47 cm/◦ C) × (20◦ C) = 69.4 cm,
(48)
form h0 = 3.0 cm to h = h0 + ∆h = 72.3 cm.
Non-textbook problem #4:
According to the universal gas law, for any fixed amount of gas
PV
= nR = const.
T
(49)
As the helium-filled balloon raises from the ground to the 10,000 ft altitude, helium’s pressure
and temperature change according to the pressure and temperature of the surrounding air.
Consequently, the volume of the balloon changes according to
P1 V1
T2
P1
P2 V2
=
=⇒ V2 = V1 ×
×
.
T2
T1
T1
P2
(50)
Note that the temperatures T1 and T2 in this formula are absolute temperatures (counting
from absolute zero), so we need to translate them from degrees Fahrenheit to degrees Celsius
and hence to Kelvins. Thus,
◦
T1 = 77 F =
77 − 32
= 25
1.8
◦
C =
25 + 273.15 = 298.15 K
(51)
23 − 32
◦
= −5 C =
1.8
−5 + 273.15 = 268.15 K
(52)
while
◦
T2 = 23 F =
and hence
T2
269.15 K
=
= 0.8994.
T1
208.15 K
(53)
1000 mbar
P1
=
= 1.450.
P2
690 mbar
(54)
At the same time,
Therefore, the volume of the balloon changes from V1 = 100 m3 to
T2
P1
V2 = V1 ×
×
= 100 m3 × 0.8994 × 1.450 ≈ 130 m3 .
T1
P2
9
(55)
Non-textbook problem #5:
According to the universal gas law
P V = nRT
(56)
the amount of gas (in mols) in volume V under pressure P and absolute temperature T is
PV
.
RT
n =
(57)
The mass of this gas (in grams!) is
m = µ×n =
µP V
RT
(58)
where µ is the molecular weight of the gas. Consequently, the density of the gas is
ρ =
m
µP
=
.
V
RT
(59)
Now let’s apply this formula to the atmosphere of Venus. It consists mostly of CO2 —
whose molecular weight is µ = 44 g/mol — and near the surface has pressure P = 92 bar =
9.2 · 106 Pa, and absolute temperature T = 740 K. Consequently, the density is
ρ =
(44 g/mol) × (9.2 · 106 Pa)
(8.314 J/K/mol) × (740 K)
= 65, 800 g/m3
= 65.8 kg/m3 ,
about 60 times denser than the air on Earth.
10
hh note units! ii
(60)
Non-textbook problem #6:
In thermal equilibrium, each component of the gas mixture has the same average kinetic
energy of a molecule,
1
2 mv
2
=
avg
3
2 kT
(61)
where k = 1.38 · 10−23 J/K is the Boltzmann’s constant and T is the absolute temperature of
the gas. Consequently, for each component (i) of the gas, the mean velocity2 of a molecule is
D
2
v(i)
E
=
avg
3kT
m(i)
(62)
and hence RMS (root-mean-square) molecular velocity is
rms
v(i)
=
rD
2
v(i)
s
E
avg
=
3kT
.
m(i)
(63)
The molecular weight µ of a gas component is the mass of one molecule in atomic mass
units,
m(i) = µi × 1 amu,
(64)
where
1 amu =
1g
= 1.6605 · 10−27 kg.
NA
(65)
Consequently, the RMS molecular velocity is
s
rms
v(i)
=
3kT
.
µi × 1 amu
(66)
Inside human lungs T = 311 K, hence an (i)th component of the gas mixture has
rms
v(i)
p
3kT /1 amu
2785 m/s
=
=
.
√
√
µi
µi
Specifically:
11
(67)
• Helium is a mono-atomic gas of atomic weight µ = 4, hence the RMS speed of a helium
atom in the mix is
rms
=
vHe
2785 m/s
√
= 1393 m/s.
4
(68)
• Oxygen O2 has molecular weight µ = 2 × 16 = 32, hence the RMS speed of an oxygen
molecule in the mix is
rms
=
vO
2
2785 m/s
√
= 492 m/s.
32
(69)
• Water H2 O has molecular weight µ = 2 × 1 + 16 = 18, hence the RMS speed of a water
molecule in the mix is
rms
vH
=
2O
2785 m/s
√
= 656 m/s.
18
(70)
• Carbon dioxide CO2 has molecular weight µ = 12 + 2 × 16 = 44, hence the RMS speed
of a CO2 molecule in the mix is
rms
vCO
=
2
2785 m/s
√
= 420 m/s.
44
(71)
Textbook problem 13.53:
According to the universal gas law
P V = nRT = N kT
(72)
where n is the amount of gas in mols while N = n × NA is the net number of molecules, and
k = R/NA is the Boltzmann’s constant. The same constant appears in the formula for the
average kinetic energy of a molecule in a gas,
1
2 mv
2
=
3
2 kT.
(73)
In terms of the root(mean square) speed of a molecule,
1
2 mv
2
=
2
1
2 m(vrms )
12
=
3
2 kT
(74)
hence
(vrms )2 =
3kT
.
m
(75)
Combining this formula with eq. (72), we get
(vrms )
2
3
PV
3P V
=
× kT =
=
m
N
Nm
(76)
Note that N m in the denominator of this formula is the total mass M of the gas and
Nm
M
=
= ρ
V
V
(77)
is its density. Plugging this formula into eq. (76), we arrive at
2
(vrms )
= 3P ×
V
1
=
Nm
ρ
=
3P
ρ
(78)
and hence
s
vrms =
3P
.
ρ
Quod erat demonstrandum (which is precisely what had to be proved).
13
(79)