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PHYS3511-Biological PhysicsWinter 2015, Assignment #7 Assigned on Monday 16 March 2015. Due on Monday 23 March 2015. Question 1 A) Read section 13.1 (pp 320-322) and focus on Figure 13.3. Estimate the total length of all arteries, veins, and capillaries from the data of Figure 13.3. Compare to the 60000 to 100000 km found online. B) Read example 13.6 on pp. 337-338. Table 13.2 calculates the effective resistance of the arterioles, aorta, and capillaries. Using data from figure 13.3, calculate the resistance of the remaining type of vessels. Calculate the effective resistance of the blood circulation system. Use the Poseuille’s law to estimate the pressure drop, ΔP , over the whole circulation system. Compare your result with the pressure drop of 2000 Pa, from when blood left the heart from the aorta to when it returns to the heart through the veins. A) Aorta: n aorta = 1, volume/cross-section 180cm 3 / 5.3cm 2 = 34 cm , aorta = 34 cm . Large Arteries: one large artery A Lartery = π 0.4 cm ( ) 2 = 0.5cm 2 ; all area A all = 20cm 2 ; number n Larteries = A all / A Lartery = 40 ; 1 large artery Lartery = V / A all = 250cm 3 / 20cm 2 = 12.5cm ; total length of arteries n Lartery Lartery = 500cm . Small Arteries: one small artery A Sartery = π 0.15cm ( ) 2 = 0.071 cm 2 ; all area A all = 20cm 2 ; number n Sarteries = A all / A Sartery = 280 ; 1 small artery Sartery = V / A all = 250cm 3 / 20cm 2 = 12.5cm ; total length n Sartery Sartery = 3500cm . Arterioles: one arterioles A arteriole = π 0.001 cm ( ) 2 = 3.14 × 10 6 cm 2 ; all area A all = 500cm 2 ; number n arteriole = A all / A arteriole = 1.6 × 10 8 ; 1 arteriole arteriole = V / A all = 125cm 3 / 500cm 2 = 0.25cm ; total length of arterioles n arteriole Sarteriole = 4 × 10 7 cm . Capillaries: one capillary A capillary = π 0.00045cm ( ) 2 = 6.4 × 10 7 cm 2 ; all area A all = 3500cm 2 ; number n capillary = A all / A capillary = 5.5 × 10 9 ; 1 capillary capillary = V / A all = 300cm 3 / 3500cm 2 = 0.086cm ; total length of capillaries n capillary capillary = 4.7 × 10 8 cm . Venule: one venule A venule = π 0.00125cm ( ) 2 = 4.9 × 10 6 cm 2 ; all area A all = 2700cm 2 ; n venule = A all / A venule = 5.9 × 10 8 ; 1 venule venule = V / A all = 550cm 3 / 2700cm 2 = 0.2cm ; total length of venules n venule venule = 1.2 × 10 8 cm . Small veins: one small vein A Svein = π 0.075cm ( ) 2 = 0.018cm 2 ; all area A all = 100cm 2 ; number n Svein = A all / A Svein = 5600 ; 1 small vein Svein = V / A all = 1550cm 3 /100cm 2 = 1.55cm ; total length of n Svein Svein = 8700cm . Large veins: one small vein A Lvein = π 0.8cm ( ) 2 = 2cm 2 ; all area A all = 30cm 2 ; number n Lvein = A all / A Lvein = 15 ; 1 large vein Lvein = V / A all = 900cm 3 / 30cm 2 = 30cm ; total length of n Lvein Lvein = 450cm .

Transcript of Assigned on Monday 16 March 2015. Due on Monday 23 March ... · PHYS3511-Biological PhysicsWinter...

Page 1: Assigned on Monday 16 March 2015. Due on Monday 23 March ... · PHYS3511-Biological PhysicsWinter 2015, Assignment #7 Assigned on Monday 16 March 2015. Due on Monday 23 March 2015.

PHYS3511-Biological PhysicsWinter 2015, Assignment #7 Assigned on Monday 16 March 2015. Due on Monday 23 March 2015. Question 1 A) Read section 13.1 (pp 320-322) and focus on Figure 13.3. Estimate the total length of all arteries, veins, and capillaries from the data of Figure 13.3. Compare to the 60000 to 100000 km found online. B) Read example 13.6 on pp. 337-338. Table 13.2 calculates the effective resistance of the arterioles, aorta, and capillaries. Using data from figure 13.3, calculate the resistance of the remaining type of vessels. Calculate the effective resistance of the blood circulation system. Use the Poseuille’s law to estimate the pressure drop, ΔP , over the whole circulation system. Compare your result with the pressure drop of 2000 Pa, from when blood left the heart from the aorta to when it returns to the heart through the veins. A) Aorta: naorta = 1, volume/cross-section ≡ 180cm3 / 5.3cm2 = 34cm , ℓaorta = 34cm .

Large Arteries: one large artery ALartery = π 0.4cm( )2 = 0.5cm2 ; all area Aall = 20cm2 ;

number nLarteries = Aall / ALartery = 40 ; 1 large artery

ℓ Lartery =V / Aall = 250cm3 / 20cm2 = 12.5cm ; total length of arteries nLarteryℓ Lartery = 500cm .

Small Arteries: one small artery ASartery = π 0.15cm( )2 = 0.071cm2 ; all area Aall = 20cm2 ;

number nSarteries = Aall / ASartery = 280 ; 1 small artery

ℓSartery =V / Aall = 250cm3 / 20cm2 = 12.5cm ; total length nSarteryℓSartery = 3500cm .

Arterioles: one arterioles Aarteriole = π 0.001cm( )2 = 3.14 ×10−6cm2 ; all area Aall = 500cm

2 ; number narteriole = Aall / Aarteriole = 1.6 ×108 ; 1 arteriole

ℓarteriole =V / Aall = 125cm3 / 500cm2 = 0.25cm ; total length of arterioles

narterioleℓSarteriole = 4 ×107cm .

Capillaries: one capillary Acapillary = π 0.00045cm( )2 = 6.4 ×10−7cm2 ; all area

Aall = 3500cm2 ; number ncapillary = Aall / Acapillary = 5.5 ×10

9 ; 1 capillary

ℓ capillary =V / Aall = 300cm3 / 3500cm2 = 0.086cm ; total length of capillaries

ncapillaryℓ capillary = 4.7 ×108cm .

Venule: one venule Avenule = π 0.00125cm( )2 = 4.9 ×10−6cm2 ; all area Aall = 2700cm2 ;

nvenule = Aall / Avenule = 5.9 ×108 ; 1 venule ℓ venule =V / Aall = 550cm

3 / 2700cm2 = 0.2cm ; total length of venules nvenuleℓ venule = 1.2 ×10

8cm .

Small veins: one small vein ASvein = π 0.075cm( )2 = 0.018cm2 ; all area Aall = 100cm2 ;

number nSvein = Aall / ASvein = 5600 ; 1 small vein

ℓSvein =V / Aall = 1550cm3 /100cm2 = 1.55cm ; total length of nSveinℓSvein = 8700cm .

Large veins: one small vein ALvein = π 0.8cm( )2 = 2cm2 ; all area Aall = 30cm2 ; number

nLvein = Aall / ALvein = 15 ; 1 large vein ℓ Lvein =V / Aall = 900cm3 / 30cm2 = 30cm ; total

length of nLveinℓ Lvein = 450cm .

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Venae cavae: one vanae cavae AVcavae = π 1.6cm( )2 = 8cm2 ; all area Aall = 18cm2 ;

number nVcavae = Aall / AVcavae = 2 ; 1 Venae cavae

ℓVcavae =V / Aall = 250cm3 /18cm2 = 13.9cm ; total length of nVcavaeℓVcavae = 27.8cm .

Adding the total length 34cm + 500cm + 3500cm + 4 ×107cm + 4.7 ×108cm+1.2 ×107cm + 8700cm + 450cm + 27.8cm

This gives ∼ 6 ×108cm or 6000 km, which is smaller than the 100000 km online value. B) Using equation 13.11

R = 8ℓ /πr4( )η , where η = 2.5 ×10−3Pa i s

Aorta:

Raorta =8ℓaortaπraorta

4

⎛⎝⎜

⎞⎠⎟η = 8 × 0.34m

π × 1.3×10−2m( )4⎛

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 7.6 ×104 Pa i s

m3

Note that in the textbook it is assume that the aorta has a thickness of 0.2 cm, so the inner radius of r = 0.011m , which gives Raorta = 1.5 ×10

5Pa i s im−3 (see table 13.2)

Large Arteries:

RLartery =8 × 0.125m

π × 4 ×10−3m( )4⎛

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 3.1×106 Pa i s

m3 . There

are nLartery = 40 , which we will assume are in parallel, giving

RLarterytotal =

nLarteryRLartery

⎝⎜⎞

⎠⎟

−1

= 7.8 ×105Pa i s im−3 .

Small Arteries:

RSartery =8 × 0.125m

π × 1.5 ×10−3m( )4⎛

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 1.6 ×108 Pa i s

m3 . There

are nSarteries = 280 , which we will assume are in parallel, giving

RSarterytotal =

nSarteryRSartery

⎝⎜⎞

⎠⎟

−1

= 5.6 ×105Pa i s im−3

Arterioles:

Rarteriole =8 × 2.5 ×10−3mπ × 1×10−5m( )4

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 1.6 ×1015 Pa i s

m3 . There are

narterioles = 1.6 ×108 , which we will assume are in parallel, giving

Rarterioletotal = narteriole

Rarteriole

⎛⎝⎜

⎞⎠⎟

−1

= 1×107Pa i s im−3 . NOTE: In the textbook the inner radius of

r = 8 ×10−6m is used, which gives Rarteriole = 3.9 ×1015Pa i s im−3 and

Rarterioletotal = 2.4 ×107Pa i s im−3

Capillaries:

Rcapillaries =8 × 8.6 ×10−4m

π × 4.5 ×10−6m( )4⎛

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 1.3×1016 Pa i s

m3 . There

are ncapillary = 5.5 ×109 , which we will assume are in parallel, giving

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Rcapillarytotal =

ncapillaryRcapillary

⎝⎜⎞

⎠⎟

−1

= 2.4 ×106Pa i s im−3 . NOTE: In the textbook the inner radius of

r = 3.5 ×10−6m is used, which gives Rcapillary = 3.6 ×1016Pa i s im−3 and

Rcapillarytotal = 7.2 ×106Pa i s im−3

Venules:

Rvenule =8 × 2 ×10−3m

π × 1.25 ×10−5m( )4⎛

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 5.2 ×1014 Pa i s

m3 . There are

nvenule = 5.9 ×108 , which we will assume are in parallel, giving

Rvenuletotal = nvenule

Rvenule

⎛⎝⎜

⎞⎠⎟

−1

= 8.9 ×105Pa i s im−3 .

Small Veins:

RSvein =8 ×1.55 ×10−2mπ × 7.5 ×10−4m( )4

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 3×108 Pa i s

m3 . There are

nSvein = 5600 , which we will assume are in parallel, giving

RSveintotal = nSvein

RSvein

⎛⎝⎜

⎞⎠⎟

−1

= 5.4 ×104Pa i s im−3 .

Large Veins:

RLvein =8 × 0.3m

π × 8 ×10−3m( )4⎛

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 4.7 ×105 Pa i s

m3 . There are

nLvein = 15 , which we will assume are in parallel, giving

RLveintotal = nLvein

RLvein

⎛⎝⎜

⎞⎠⎟

−1

= 3.1×104Pa i s im−3 .

Vanae Cavae:

RVcavae =8 × 0.139m

π × 1.6 ×10−2m( )4⎛

⎝⎜⎜

⎠⎟⎟× 2.5 ×10−3Pa i s = 1.3×104 Pa i s

m3 . There

are nVcavae = 2 , which we will assume are in parallel, giving

RVcavaetotal = nVcavae

RVCavae

⎛⎝⎜

⎞⎠⎟

−1

= 6.5 ×103Pa i s im−3 .

Total resistance of the circulatory system is Rtot = Raorta + RLarterytotal + ...≈1.5 ×107Pa i s im−3

Average flow in human is Q = 5L imin−1 = 8.3×10−5m3 i s−1 , which gives a pressure drop of Δp =QRtot = 8.3×10

−5m3 i s−1 ×1.5 ×107Pa i s im−3 = 1250Pa , which is somewhat smaller than the 2000 Pa stated at the beginning of the question. Question 2 Problem P13.18 of Chapter 13 textbook (page 347). A patient is to be injected with 0.5 L of an electrolyte solution in 1⁄2 hour. Assuming that the solution is elevated by 1.0 m above the arm and the needle is 2.5 cm long, what inner radius should the needle have? Use water parameters for the solution, and assume that the pressure in the patient’s vein is atmospheric pressure.

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ΔVΔt

=5 ×10−4m3

30min× 60s/ min= 2.78 ×10−7m3 / s

ΔVΔt

=π8η

r4 ΔpL

, with L = 2.5 ×10−2cm , η = 1.01×10−3N is / m2

Δp = ρgh = 1000kg / m3( ) 9.8m / s2( ) 1m( ) = 9800Pa

r = ΔV

Δt8ηLπΔp

⎛⎝⎜

⎞⎠⎟

1/4

= 2.78 ×10−7 m3

s8 ×1.01×10−3N isim−2 × 2.5 ×10−2m

π × 9800Pa⎛⎝⎜

⎞⎠⎟

Question 3 Do Multiple-Choice MC 15.6, 15.7 and 15.11 of chapter 15 of page 400. MC 15.6 ANSWER E MC 15.7 ANSWER E MC 15.11 ANSWER C Question 4 Problem 15.12 of Chapter 15 in textbook (page 401) Use equation 15.14 for the intensity I = 1/ 2( )cρA2ω 2 , where A is the amplitude of the

wave. Solving A = 1/ω( ) 2I / cρ( ) . Take c = 340m / s (Table 15.1), and

ρ = 1.293kg im−3 (example 15.3). The frequency is not given, but let’s assume that it is the frequency that is most sensitive to the human ear f ∼ 3000Hz→ω = 1.88 ×104 s−1 ,

A = 1.88 ×104 s−1( )−1 2 ×1×10−12 J im−2 i s−1 / 340m i s−1 ×1.293kg im−3( ) = 3.5 ×10−12m

Question 5 Problem 15.13 of Chapter 15 in textbook (page 401)

A) Sound intensity is simply I ≡ Energy

area × time= 2.11×10−11J5 ×10−4m2 × 4s

= 1.055 ×10−8 Jm2 i s

.

B) From equation 15.11 Δpmax = cρω( )A→ A = Δpmax / cρω( ) . Substituting into equation

15.14 (see question 4) I = 1/ 2( )cρA2ω 2 = 1/ 2( ) Δpmax( )2 / cρ , and solving

Δpmax = 2Icρ = 2 ×1.055 ×10−8 J im−2 i s−1 × 340m i s−1 ×1.2kg im−3 = 2.9 ×10−3Pa Question 6 Problem 15.19 of Chapter 15 in textbook (page 401) Using section 15.4.2 (figure 15.18, table 15.2) the first harmonic is λ1 = 2L , with corresponding frequency f1 = c / λ1 = c / 2L( )→ L = c / 2 f1( ) = 340m i s−1 / 2 ×18500Hz( ) , which gives L = 9.2 ×10−3m . The radius of the eardrum is about 4.6 mm. Question 7 Read case study15.6 (pp. 391-392). Estimate the length, L, of the whale’s vocal tube for the case when A) it is filled with water, and B) it is filled with air. Based on your answers, do you think the vocal tune is filled with air or water. Briefly justify. Analysis in the case study determine a closed tube resonance fn = nc / 2L( ) A) speed of sound in seawater is c = 1400m i s−1 , and the first harmonic is f1 = 20Hz , which gives L = c / 2 f1 = 35m . B) speed of sound in air is c = 340m i s−1 , and the first harmonic is f1 = 20Hz , which gives L = c / 2 f1 = 8.5m . Although a blue whale (length of about 30 m) is big, a 35 m vocal tube is much too large, but it can probably accommodate a canal of about 8m. And it is well known that whale’s and dolphin’s vocal tubes are air filled.

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Question 8 Place theory of pitch perception. The biophysicist G. von Bekesy has am important theory on how the basilar membrane (BM) of the inner ear perceives different pitch (frequency) of sound. The BM is a rolled strip of membrane of about 32 mm long, with radius ranging from 0.04 mm near the base to 0.5 mm near the apex, which can detect sound frequency ranging from 20000Hz at the apex to 200 Hz at the base. The 32 mm length can be divided into 1500 resonance boxes, design to induce resonance at a given frequency. A) Assume that the apex end can resolve 200 Hz, calculate the velocity,

c, and tension, T, of the wave. B) Using the tension of part A) determine the upper limit of the frequency that can be determined at the base. Use the relation c = T / ρl , where ρl = ρA , where A is the cross-section area of the BM, and ρ = 1000kg im−3 mass density of BM, which is assumed to be the same as that of water. C) What modification would you make to the model so that the upper limit is 20000 Hz?

A) Assume a closed tube resonance of section 15.4.2 (figure 15.18, table 15.2), with the first harmonic is λ1 = 2L . With a total length of 3.2 ×10−3m i s−1 , and 1500 resonance boxes, L = 3.2 ×10−2m /1500 = 2.13×10−5m , which corresponds roughly to the distance of a row 4 hair cells, where the resonance occurred, which corresponds to a wavelength of λ1 = 2 2.13×10

−5m( ) = 4.26 ×10−5m . At the apex the frequency is f1 = 200Hz , hence the speed of the traveling wave must be

c = f1λ1 = 200Hz × 4.26 ×10−5m = 8.52 ×10−3m i s−1or about 8.5 mm per second. To

calculate the tension use, A = π 2.5 ×10−4m( )2 = 1.96 ×10−7m2 , which gives a linear

density ρl = ρA = 1000kg im−3 ×1.96 ×10−7m2 = 1.96 ×10−4 kg im−1 . c = T / ρl , or

T = c2ρl = 8.52 ×10−3m i s−1( )2 1.96 ×10−4 kg im−1( ) = 1.4 ×10−8N .

B) At base A = π 2 ×10−5m( )2 = 1.26 ×10−9m2 ,

ρl = ρA = 1000kg im−3 ×1.26 ×10−9m2 = 1.26 ×10−6 kg im−1 , which gives the speed of

c = 1.4 ×10−8N /1.26 ×10−6 kg im−1 = 0.105m i s−1 . This will corresponds to a frequency of f = c / λ1 = 0.105m i s−1 / 4.26 ×10−5m = 2500Hz , which is far below (by a factor of 10) the known upper limit of ∼ 20000Hz . C) The traveling wave resonance occur at the same wavelength, λ1 = 4.26 ×10

−5m , and the only adjustable variable is the wave speed at the resonance region, c = T / ρl . Since the upper limit frequency is too low, one way to increase the upper limit is to allow the

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tension to increase the tension T, as you approach the base. To fit the upper limit the tension must increase by a factor of 10 at the base compare with the apex. Another model (a more common one) is to assume that the speed vary as c = σ / A , where σ is the surface tension, an A is the local area of the basilar membrane. Since σ increases and A decreases from apex to base, the speed will increase towards the base. Final Note on this assignment Question 1, 7, and 8 uses many common-sense approximations, so you should not lose mark if you take a slightly different approach from the solution, as long as you use correct physics concepts. Note the following reading assignment that accompanied assignment 7. I will assume that you read them for the final exam. Assigned on Monday 16 March 2015. Due on Monday 23 March 2015. BioPhysics Reading: Section 13.2, Flow of Ideal Fluid (especially Figure 13.3 and 13.4). Case study 13.1, page 326 to 327, on blood flow. Case study 13.2 on page 329 on arteriosclerosis. Section 13.3.6 on Kirchhof’s law of blood vessels on page 335 to 339. Read all of Chapter 15.