Solutions Manual for Transport Phenomena In Biological Systems 2nd Edition by Truskey

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Solution to Problems in Chapter 17, Section 17.10 17.1. In words, the conservation relation is: &Rate of Energy # & Net Rate of Energy# &Rate of Work # &Rate of Energy # $Accumulation ! = $Transfer Across ! + $Done on the ! + $Production ! $ ! $ ! $ ! $ ! %$ within the system"! $%System Surfaces "! %$System "! %$ within the system"!

Using a rectangular control volume and the definition of the system energy per unit mass (Equation (17.2.3)) and energy flux (Equation (17.2.4))

!x!y!z"

#Ê = ex x $ ex #t

(

x + !x

) !y!z + ( e

y y

$ ey

y+ !y

) !x!z + (e

z z

$ ez

z + !z

) !x!y + (W! + Q! ) !x!y!z t

* p

Dividing by the volume element ΔxΔyΔz, taking the limit as the volume goes to zero and using the definition of the derivative yields: !

"Ê "e "ey "ez =# x # # + W! t + Q! *p "t "x "y "x

(S17.1.1)

Using the definition of the divergence of a vector (Equation (A.3.10), Equation (S17.1.1) becomes !

"Ê = #$ie + W! t + Q! *p "t

(S17.1.2)

Using the definition of e, Equation (17.2.4), the divergence of e is:

(

)

(

)

!ie = !i " Êv + q = " Ê!iv + vi!" Ê + !iq

(S17.1.3)

For an incompressible fluid, !iv = 0 and ρ is a constant. As a result, Equation (S17.1.3) reduces to: (S17.1.4) !ie = !i " Êv + q = "vi!Ê + !iq Inserting Equation (S17.1.4) into Equation (S17.1.2) "Ê = # !vi$Ê # $iq + W! t + Q! *p "t

(S17.1.5)

$ "Ê ' !& + vi#Ê ) = *#iq + W! t + Q! *p % "t (

(S17.1.5)

!

Moving both terms with the system energy to the left hand side of Equation (S17.1.5) yields:

Lastly, the total rate of work represents work done by fluid stresses ( !i(" iv ) = #!i( pv ) + !i($ iv ) ), body forces ( Fiv ) and other types of mechanical work by the body ( W! ). Inserting these terms into Equation (S17.1.5) yields Equation (17.2.6) $ "Ê ' !& + vi#Ê ) = *#iq * #i( pv ) + #i(+ iv ) + Fiv + W! + Q! *p % "t (

(S17.1.6)

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17.2. The work is:

W = ! Findx = ! Fdx since the force and unit outward normal are

both positive. Normally, a protein is present in a specific conformation which is much less than the maximum length, know as the contour length, L. The contour length is the length of the polymer if each chain element were aligned along a line.

Substituting for the wormlike chain model: '2 '1 ! k BT $ x ! ! k BT $ ! x$ x$ x$ x2 $ ! !

W =# 0.25 1 ' ' 0.25 + dx = 0.25L 1 ' ' 0.25x + #" & #" & & (# # L &# L% L &% L% 2L &% " Lp % 0 " " p %"

'1 ! kBTL $ ! x$ x x2 $ !

W =# & # 0.25 #" 1 ' L &% ' 0.25 L + 2L2 & % " Lp % " This result is shown in the graph below. At low extensions, the relation is linear.

However, as x approaches L, the work increases dramatically. The flexibility of the polymer arises from the arrangement of the chains. As the polymer elongates, more work must be done to extend the polymer to overcome the tendency for the chains to move freely and to extend each element.

228

17.3. Note: The equation listed in the problem statement should be: 2

! v = " : #v = µ $& dvz ') ! % dr ( The shear stress tensor for a Newtonian fluid is:

(

! = µ "v + ( "v )

T

(

)

(S17.3.1)

)

! v = " : #v = µ #v + ( #v )T : #v !

(S17.3.2)

Using the summation convention for vectors and tensors # 3 3 # "v "v j & & # 3 3 "v & T k µ !v + ( !v ) : !v = µ % ) ) % i + e e ek el ( i j ( :% )) ( ' $ i =1 j =1 $ "x j "xi ' ' $ k =1 l =1 "xl

(

)

Since

(

ei e j : ek el = e j iek

) ( e ie ) = !

! il

, Equation (S17.3.3) becomes: 2 3 3 ## "vi & # "v j & # "v j & & T µ !v + ( !v ) : !v = µ ) ) % % (% ( + %$ "x (' ( i =1 j =1 $ $ "x j ' $ "xi ' i ' For fully developed steady, laminar flow in a cylindrical tube of radius R, # "v "v & ! = µ % z + r ( er ez $ "r "z ' Using the symmetry property of the shear stress, τij = τji: ! v = µ #% "vz er ez + "vr ez er &( : "vz er ez ! $ "r ' "r "z

(

i

l

jk

(S17.3.3)

)

"v # "v & "v # "v "v & # "v & = µ % z er ez + z ez er ( : z er ez = µ % z z ( = µ % z ( $ "r ' "r $ "r "r ' $ "r ' "r

2

(S17.3.4)

(S17.3.5)

For laminar flow in a tube " r2 % vz = 2 vz $ 1 ! 2 ' R & #

(S17.3.6a)

!vz 2 vz r = (S17.3.6b) !r R2 The velocity gradient is maximum at r = R. Thus, the maximum value of viscous dissipation is: 2

" 4 vz % 16 µ vz !v = µ ! $ R ' = R2 # &

2

(S17.3.7)

In terms of flow rate

!v !

For the data given:

max

16 µQ 2 = 2 6 " R

(S17.3.8)

229

!v !

max

!v !

max

=

(

)(

16 0.01 g cm -1 s-1 83.3 cm 3 s-1

" (1.5 cm ) 2

)

2

6

= 22.22 g cm -1 s-3 = 2.22 kg m -1 s-3

= 2.22 Pa s-1 = 2.22 J m -3 s-1 = 2.22 W m -3

To compute the maximum heating in blood arising from viscous dissipation, assume steady radial conduction with viscous dissipation. From Equations (17.2.8), (17.2.9), (17.2.12) and (S17.3.6b), the following result is obtained.

k d ! dT $ 16 µQ 2 r 2 ! r = ' ( = ' (S17.3.9) # & v r dr " dr % ) 2 R8 The boundary conditions are that for r = 0, the flux is zero and at r = R, T = T0. Integrating Equation (S17.3.9) once yields:

dT 4 µQ 2 r 3 C =! + dr k" 2 R 8 r

(S17.3.10)

From the boundary condition at r = 0, C = 0. Integrating Equation (S17.3.10) yields: !v # r4 µQ 2 r 4 max T = ! 2 8 + C2 = ! + C2 (S17.3.11) k" R kR 2 From the boundary condition at r = R, C2 is

C2 = T0 +

!v !

max

R2

k

(S17.3.12)

The temperature profile is:

!v !

R2 # r4 & T = T0 + 1 " %$ k R 4 (' The maximum temperature difference occurs between r = 0 and r = R: !v # R2 max !Tmax = T ( r = 0 ) " T0 = k max

(S17.3.13)

(S17.3.14)

For the value of the viscous dissipation obtained above and the thermal conductivity of blood (Table 17.2): 2 2.22 W m -3 ( 0.015 m ) !Tmax = = 0.00078 K (S17.3.15) 0.642 W m -1K -1

(

)

Thus, viscous dissipation has a very minor effect on the temperature of blood and can be neglected.

230

17.4. For steady conduction for a spherical surface of radius R, Equation (17.2.14c) simplifies to:

k d ! 2 dT $ (S17.4.1) #r & =0 r 2 dr " dr % The boundary conditions are at r = R, T = T0 and as r —> ∞, T = T∞. Integrating equation (S17.4.1) twice yields: C1 + C2 r From the boundary condition as r —> ∞, C2 = T∞. At r = R C1 = (T0 ! T" ) R T=

(S17.4.2) (S17.4.3)

The temperature profile is: T = (T0 ! T" )

R + T" r

(S17.4.4)

To obtain the Nusselt number, compute the flux at r = R and apply the definition of the heat transfer coefficient:

(T ! T" ) = h T ! T dT =k 0 m( 0 ") dr r = R R The heat transfer coefficient for conduction is: k k hm = = 2 R D Using this result in the definition of the Nusselt number: !k

Num =

hm D kD =2 =2 k kD

17.5. The definition of β is given by Equation (17.4.7) 1 # "V & != % ( V $ "T ' P

(S17.4.5)

(S17.4.6)

(S17.4.7)

(S17.5.1)

From the ideal gas relationship, PV = nRT. For a fixed number of moles, V=nRT/P and the derivative in Equation (S17.5.1) is:

nR " !V % $# '& = !T P P

(S17.5.2)

Thus, nR 1 = VP T since T = PV/nR for an ideal gas.

!=

(S17.5.3)

231

17.6. For this problem, assume unsteady conduction in a tissue of thickness 2L. Based upon analogy with unsteady diffusion in a region of half thickness of L, the time to reach steady state is 2L2/α. While specific thermal diffusivities for tissue are not provided in Table 17.2, a reasonable value, between water and fat, is 1.1 x 10-7 m2 s-1. For the half-thickness of 125 µm = 1.25 x 10-4 m, the time to reach steady state is 0.284 s. So, one would expect uniform temperatures in well perfused tissues.

17.7. Note: The phase change during freezing is discussed in Section 17.3.4, not Section 17.3.3. dX . X is given by Equation (17.3.26b). Thus, dt dX ! =C S (S17.7.1) dt t C is dimensionless and is provided by solving Equation (17.3.31) or Equation (17.3.33). Values of C are tabulated in Table 17.3 for several different values of Tm-T0 and αS is given in Table 17.2 as 1.06 x 10-6 m2 s-1. For a value of Tm-T0 =10 C, C = 0.183 and the derivative in Equation (S17.7.1) is (1.8448 x 10-4)t-1/2 m s-1.

The rate of growth of the ice front is

17.8. This problem is a modification of the problem presented in Example 6.6. Thus, Equation (6.7.25) applies for the distribution of vapor concentration in a column of height δ. d " 1 dx % (S17.8.1) =0 dy $# 1 ! x dy '& The boundary conditions are that, at y = h, x = xa which is the vapor pressure at the given temperature and pressure. At y = h + δ, x = xs, the relative humidity in the air. Integrating Equation (S17.8.1) once yields: dx = C1 (1 ! x ) (S17.8.2) dy Integrating again, ln (1 ! x ) = !C1 y + C2 (S17.8.3) Applying the boundary conditions: ln (1 ! xs ) = !C1h + C2

ln (1 ! xa ) = !C1 ( h + " ) + C2 Subtracting (S17.8.4b) from Equation (S17.8.4a) 1 # 1 " xa & C1 = ln % ! $ 1 " xa (' Inserting Equation (S17.8.4c) in Equation (S17.8.4b) and solving for C2 yields; " 1 ! xs % " h + ( % ln (1 ! xa ) = ! ln $ $ ' + C2 # 1 ! xs '& # ( &

232

(S17.8.4a) (S17.8.4b) (S17.8.4c)

(S17.8.4d)

" 1 ! xs % " h + ( % C2 = ln (1 ! xa ) + ln $ $ ' # 1 ! xa '& # ( &

(S17.8.4d)

The solution is: " 1! x % y " 1 ! xs % " h + ( % " 1 ! xs % ln $ = ! ln +$ ' ln ( $# 1 ! xa '& # ( & $# 1 ! xa '& # 1 ! xa '& Add the term ln((1-xa)/(1-xs)) to each side: " 1! x % " 1 ! xa % " 1 ! xa % y " 1 ! xs % " h % " 1 ! xs % ln $ + ln = ! ln + + 1 ln + ln $ ' $# 1 ! x '& $# 1 ! x '& ( $# 1 ! xa '& # ( & $# 1 ! xa '& # 1 ! xa '& s s Collect terms " 1 ! x % y " 1 ! xa % h " 1 ! xa % y ! h " 1 ! xa % ln $ = ln ! ln = ln $ ( # 1 ! xs '& ( $# 1 ! xs '& ( $# 1 ! xs '& # 1 ! xs '& Raising each side to the power e:

" 1 ! x % " 1 ! xa % $# 1 ! x '& = $# 1 ! x '& s s

y! h (

(S17.8.4d)

(S17.8.5)

(S17.8.6)

(S17.8.7)

17.9. The vapor flux is given by Equation (17.5.11) # 1 " xa & cD N y= h = w,air ln % ! $ 1 " xs (' where xs is the partial pressure of water in air at saturation (vapor pressure/total air pressure) and xa is the partial pressure of water/total air pressure. The quantity xa can be expressed as xHxs, where xH is the relative humidity. Using the data for Problem 17.10 and a total air pressure of 101,325 Pa. The quantity c = ptot/RT = 101,325 Pa/(8.314 N m K-1 mol-1)(298 K) = 40.90 mole m-3. The diffusivity of water in air is provided in the text, p. 797, as 2.6 x 10-5 m2 s-1. Thus, xs = 0.0310 at 25 C and 0.0728 at 40 C. For 20% relative humidity at 25 C.

( 40.90 mol m )( 2.6 x 10 = -3

N y= h

!5

m 2s-1

) ln " 1 ! 0.20 ( 0.031)) % = 0.0020 mol m

$# 0.0136 m 1 ! 0.031 For 80% relative humidity, the flux is 0.00050 mol m-2 s-1.

'&

-2 -1

s

17.10. The error can be computed from the ratio of Equations (17.5.12) to Equation (17.5.13): " 1 ! xs % " 1 ! xs % error = xs ! xa ln $ = x 1 ! x ln ( ) s H $# 1 ! x x '& # 1 ! xa '& s H At 25 C the error is -0.000226 and at 40 C the error rises to -0.00128. Thus, Equation (17.5.13) is a good approximation.

17.11. Since the enthalpy of vaporization is a function of temperature, application of Equation (17.5.25) or Equation (17.5.26) is done iteratively. That is, the enthalpy of vaporization is updated, once the temperature at the air-sweat interface is calculated. The flux for the 233

evaporating liquid is temperature independent and was found to be 0.001 mol m-2 s-1 for 60% relative humidity. For the calculation reported in the text, Equation (17.5.5a) was used and !H vap was determined for a temperature of 25 C. Using T equal to 37 C, !H vap = 54047.6 J mol-1. The temperature drop is 0.444 C and the energy flux is 54.05 J m-2 s-1. Updating the values at T = 309.7 K, the temperature drop is 0.444 C and the energy flux is 54.15 J m-2 s-1.

These values are within 1% of the values obtained for T = 310.00 K. 17.12. Use Equation (17.4.3) to calculate the Nusselt number. The Prandtl number does not vary significantly with temperature and a value of 0.72 is commonly used for air. The kinemtic viscosity of air 0.1327 cm2 s-1 = 1.327 x 10-5 m2 s-1. As noted on page 797, a characteristic diameter for a typical female is 0.304 m. The following table lists values of Re, Nu, h and q for various wind speeds. The energy flux can be quite substantial and is reduced significantly by clothing. v, miles/h 1 2 5 10 25

v, m/s 0.447 0.894 2.235 4.470 11.176

Re 10241 20482 51206 102412 256029

Nu 54.64 80.86 140.33 220.05 420.46

h, W m-2 K-1 4.49 6.65 11.54 18.10 34.58

q, W m-2 143.80 212.79 369.30 579.08 1106.48

17.13. Start with the definition of the Grashof number, Equation (17.4.22) ! 2 g"#TL3 Gr = µ2 The definition of β in terms of the density is given by Equation (17.4.6) ! " !0 # !0 $%T Let !" = "0 # " . Thus, !" # "0 $!T . Assuming that density in the definition of the Grashof number is the value at the reference temperature, ρ0, the Grashof number becomes:

!0 2 g"#TL3 !0 2 g#! L3 !0 g#! L3 == = µ2 µ 2 !0 µ2 17.14. For free convection, Equation (17.4.5) is used for flow over a sphere. The viscosity ratio is 0.900 and Pr = 0.72. v, miles/h v, m/s Diameter, m Re Nu adult 10 4.47 0.178 60050 164.29 child 10 4.47 0.124 41820 133.60 Gr =

For free convection, the Grashof number is calculated using Equation (17.4.22) with L equal to the diameter and β = 1/T where T is the air temperature (273.15 K). Equation (17.4.24) is used to determine the Nusselt number for a flat plate. The correlation for spheres is found in reference [18], page 301.

Nu = 2.0 + 0.43(Pr Gr)1/ 4

234

Diameter, m Gr Nu, flat plate Nu, sphere 0.178 42697290 38.57 34.02 0.124 14422151 29.40 26.41

adult child

For radiation, the energy flux is given by Equation (17.2.19c). Treating the absorptivity and emissivity as the same, the flux equals q = σe(Tb4-Tair4). A heat transfer coefficient can be defined as h=q/ΔT and a Nussel number determined. Results are: qrad

h 193.44

Nu adult 5.23

Nu child 37.22

25.93

Comparing results, the free convection and radiation terms are comparable and are about 20% of the value for forced convection.

17.15. Note, there is a typographical error in the text and Equation (17.5.25) should be: + # )Ĉ p & # )Ĉ p &. exp N y-h " exp N * ( ) y= h y= h % ( % (0 # !H vap N y= h h & -, $ Cvap ka ' $ Cvap ka ' 0/ (17.5.25) Ta = Tair + % " !T ( kl + # )Ĉ p &. $ ' k a # hN y= h )Ĉ p & " -1 " exp % N y= h* ( 0 kl %$ Cvap k a (' $ Cvap ka ' 0/ -, Begin with Equation (17.5.21) for air and Equation (17.5.24) for the liquid. " !Ĉ p % a1Cvap k a Ta = exp $ N y= h y ' + a 2 N y= h !Ĉ p # Cvap ka & Tl = a3 y + a4 The boundary conditions are: y=0 Tl = Tb y=h Tl = Ta dT dT ka a ! kl l dy y= h dy y = h+δ

Ta = Tair

(17.5.21) (17.5.24)

(S17.15.1a) (S17.15.1b)

= "H vap N y= h

(S17.15.1c)

y= h

(S17.15.1a)

From the boundary condition at y = 0 a 4 =Tb and Tl = a3 y + Tb

(S17.15.2a) (S17.15.2b)

From the boundary condition at y = h+δ,

235

a 2 =Tair !

Ta = Tair +

$ "Ĉ p ' exp & N y= h ( h + # )) N y= h "Ĉ p % Cvap ka ( a1Cvap k a

" !Ĉ p % " !Ĉ p %a1Cvap k a * N y= h y ' ( exp $ N y= h ( h + ) )' / , exp $ N y= h !Ĉ p ,+ # Cvap ka & # Cvap ka & /.

(S17.15.3a)

(S17.15.3b)

Equating Equations (S17.15.2b) and (S17.15.3b) at y = h, " !Ĉ p % " !Ĉ p %a1Cvap k a * a3h + Tb = Tair + N y= h h ' ( exp $ N y= h ( h + ) )' / (S17.15.4a) , exp $ N y= h !Ĉ p ,+ # Cvap ka & # Cvap ka & /. a3 = !

$ #Ĉ p '+ $ #Ĉ p '. a1Cvap k a "T + exp & N y= h h ) -1 ! exp & N y= h* ) 0 h hN y= h #Ĉ p % Cvap ka ( -, % Cvap ka ( 0/

(S17.15.4b)

where !T=Tb -Tair .

The liquid temperature is Tl = !"T

$ #Ĉ p '+ $ #Ĉ p '. y ya1Cvap k a + exp & N y= h h ) -1 ! exp & N y= h* ) 0 + Tb (S17.15.5) h hN y= h #Ĉ p % Cvap ka ( -, % Cvap ka ( 0/

Use Equations (S17.15.3b) and (S17.15.5) to compute the derivatives of the temperature. The boundary condition, Equation (S17.15.1c), becomes: " !Ĉ p % k (T a1k a exp $ N y= h h ' + l h # Cvap ka & " !Ĉ p %+ " !Ĉ p %. ) exp $ N y= h h ' -1 ) exp $ N y= h* ' 0 = (H vap N y= h hN y= h !Ĉ p # Cvap ka & -, # Cvap ka & 0/

(S17.15.6)

kl a1Cvap k a

Solving for a1:

# "Ĉ p &# k )T & exp % !

N y= h h ( % )H vap N y= h ! l ( h ' $ Cvap ka '$ (S17.15.7) a1 = # "Ĉ p &. kl Cvap k a + ka !

N y= h* ( 0 -1 ! exp % hN y= h "Ĉ p -, $ Cvap ka ' 0/ Inserting this expression for a1 into Equation (S17.15.3b) yields the final result for the air temperature.

236

# )Ĉ p & # )Ĉ p &.

Cvap k a + N y= h ( y-h )( " exp % N y= h* ( 0 - exp % $ Cvap ka ' $ Cvap ka ' 0/ k !T & N y= h )Ĉ p -, # Ta = Tair + % !H vap N y= h " l ( $ h ' # )Ĉ p &. kC k + k a " l vap a -1 " exp % N y= h* ( 0 hN y= h )Ĉ p -, $ Cvap ka ' 0/ (S17.15.8a) Rearrange to yield the correct form of Equation (17.5.25)

# !H vap N y= h h & Ta = Tair + % " !T ( kl $ '

+ # )Ĉ p & # )Ĉ p &.

N y= h ( y-h )( " exp % N y= h* ( 0 - exp % $ Cvap ka ' $ Cvap ka ' 0/ -, (17.5.25) + # )Ĉ p &. k a # hN y= h )Ĉ p & " -1 " exp % N y= h* ( 0 kl %$ Cvap k a (' $ Cvap ka ' 0/ -,

At y = h, Equation (17.5.25) is:

# !H vap N y= h h & Ta = Tair + % " !T ( kl $ '

The group

hN y= h !Ĉ p Cvap k a

+ # )Ĉ p &.

N y= h* ( 0 -1 " exp % $ Cvap ka ' 0/ -, + # )Ĉ p &. k a # hN y= h )Ĉ p & " 1 " exp N * % C k y= h ( 0 kl %$ Cvap k a (' $ vap a ' 0/ -,

(S17.15.8b)

hN y= h !Ĉ p can be rewritten as the following by using Equation (17.5.17): Cvap k a =

hCvap vy !Ĉ p Cvap k a

=

hvy !Ĉ p ka

= Peair

*+1 " exp ( Peair) / h ) ,# !H vap N y= h h & Ta = Tair + % " !T ( kl $ ' k a ( Pe ) " *1 " exp ( Pe ) / h ) , air air + kl

(S17.15.8c)

The thermal Peclet number for air is 0.20, which is larger than the value for sweat, but still much less than 1. For the case of conduction only, energy transport through the liquid is unchanged.

Equation (17.5.17) for the air simplifies to:

d 2Ta =0 dy 2

(S17.15.9)

After integration we obtain:

Ta = a1 y + a2 At y = h+δ, a2 = Tair ! a1 ( h + " )

237

Ta = a1 ( y ! ( h + " )) + Tair

Tl = a3 y + Tb Equating the air and sweat temperatures at y = h: a3h + Tb = !a1" + Tair

" Tair ! Tb " #T + = !a1 ! h h h h " #T ' $ Tl = & !a1 ! y + Tb h h )( % Use these results for Ta and Tl to compute the derivatives in Equation (S17.15.1c) # ! "T & vap ka a1 + kl % a1 + (S17.15.10a) ( = "H N y= h $ h h ' Solving for a1: a3 = !a1

!% (T " a1 $ ka + kl ' = (H vap N y= h ) kl # h& h !T !H vap N y= h " kl h a1 = #' $ &% ka + kl )( h The resulting expression for the air temperature is: $ $ !T ' vap & !T & kl h " !H N y= h ) Ta = Tair + & ) (( h + # ) " y ) = Tair + & # $ ' & & ) ka + kl ) & & % % ( h( %

(S17.15.10b)

(S17.15.10c)

h!H vap N y= h ' " ) kl ) (( h + # ) " y ) $ ka ' ) ) &% k )( h + # ( l

for y = h # % !T Ta ( y = h ) = Tair + % % % $

h!H vap N y= h & " ( kl () # ka & ( ( %$ k (' h + ) ' l

(S17.15.11)

For values of h (0.005 m) and δ (0.0136 m) provided in Section 17.5 and thermal conductivities ! of air and water in Table 17.2, = 0.985 . Thus, the approximation presented in " ka % $# k '& h + ! l Equation (17.5.26) is reasonable. Further, Equation (17.5.26) arises as a limiting value of Equation (17.5.25) when ka/klPe << 1.

238

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If vaporization does not occur, then !H vap = 0 and Equation (17.5.27) results.

17.16. From Table 2.4, the blood vessel diameters range from 6 x 10-6 m to 5 x 10-5 m.

Corresponding mean velocities range from 2 x 10-4 to 0.001 m s-1. The Pe ranges from 0.0068 to 0.284. Blood vessel densities range from 2.0 x 108 vessels m-2 to 2.22 x 109 vessels m-2. The ratio of thermal conductivities between blood and tissue range from 1.5 to 3. From Equation (17.7.4), keff/ktissue ranges from 1.00 to 5.21.

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