Heat Transfer to Blood Vessels – A Tutorial By: Mahsa Farshi Taghavi

January 2019

The material in this tutorial is based on standard curriculum of K. N. Toosi University of

Technology, Faculty of Electrical Engineering. For more information, please write to

mahsatghv@gmail.com.

mailto:mahsatghv@gmail.com

1

Introduction

LITTLE is known definitely about the effect of the temperature of human blood. From a physiological

standpoint the ideal temperature for blood is 37°C. However, blood can function with reduced ability to

absorb oxygen and desorb carbon dioxide in a range of temperatures from 30 to 45°C. Blood can survive

for a long period of time if frozen (- 5 to - 10°C) and thawed in an appropriate manner. Above 45°C the

proteins in the blood will start to precipitate out of the plasma and will coagulate. Enzymes in the blood

do not function well above 45°C. Change of blood temperature of this degree is unlikely to occur within a

body as the normal body temperature regulation system will activate in an attempt to reverse this effect.

Many situations now exist in which the blood is removed from a body to be processed (for example,

oxygenation, hemodialysis, etc.) and returned to the body. While in the body, temperature control of the

blood is almost never a problem; however, whenever the blood is out of the body, temperature control

becomes important in order to prevent damage. The information about the temperature field for blood

flowing in a tube is, therefore, of practical significance.

Fluid entering a tube with uniform velocity and temperature profiles undergoes development during its

course of flow. The development of velocity and temperature profiles is due to the growth of

hydrodynamic and thermal boundary layers on the wall of a tube. As fluid progresses, the boundary layers

grow until they intersect with the layers from the opposite side. After the point of intersection. the flow

is considered fully developed. The portion of the tube wherein the boundary layers are developing is called

the entrance region. The velocity and temperature profiles may develop at different rates. Whether the

velocity profile developed faster or slower than the temperature profile depends on the Prandtl number

of the fluid [1].

One of the most difficult problems of estimating heat transfer in living systems is the assessment of the

effect of blood circulation. If we consider only the capillaries, we can assume, as will be shown later, that

the blood is essentially at tissue temperature. Consequently, the overall effect on the energy transport,

can be estimated on the basis of an enhanced thermal conductivity with the increase related in some way

to the blood perfusion rate in the tissue. For the bigger blood vessels, the assumption of local thermal

equilibrium gets progressively worse as the size and the velocities increase [2].

2

Heat Transfer to Blood Flow in a Blood Vessel

Two significant questions can be asked about heat transfer to blood flowing in a vessel:

1) In what distance from the entrance does the blood reach tissue temperature?

2) What is the blood temperature leaving the blood vessel of a given length in relation to that of the

surrounding tissue?

To answer these questions, consider the heat transfer in the tissue around the blood vessel as shown in

Fig.1. In order to reach a meaningful solution, we assume that at some arbitrary radius, R2, the tissue is

held at a constant temperature To, by some unspecified, physiological processes occurring in the rest of

the body.

We assume that the heat transfer inside the vessel is carried out only through convection and in the tissue

only through conduction. In this case, the heat transferred from the blood to the tissue can be calculated

according to the following relations:

𝑄 =𝑇𝑏 − 𝑇𝑂

𝑅𝑐𝑜𝑛𝑣. + 𝑅𝑐𝑜𝑛𝑑. (1)

Figure 1 Idealized geometry of heat transfer around a blood vessel [2]

The conductive resistance for the cylinder is obtained according to the following relations:

𝑄𝑐𝑜𝑛𝑑.° = −𝑘𝐴

𝑑𝑇

𝑑𝑟= −𝑘2𝜋𝑟𝑙

𝑑𝑇

𝑑𝑟 (2)

𝑄𝑐𝑜𝑛𝑑.° ∫

𝑑𝑟

𝑟

𝑅2

𝑅1

= −2𝑘𝜋𝑙 ∫ 𝑑𝑇𝑇𝑜

𝑇1

(3)

𝑄𝑐𝑜𝑛𝑑.° ln (

𝑅2𝑅1) = −2𝑘𝜋𝑙(𝑇𝑜 − 𝑇1) (4)

3

𝑄𝑐𝑜𝑛𝑑.° =

𝑇1 − 𝑇𝑜

ln (𝑅2𝑅1)

2𝑘𝜋𝑙

(5) → 𝑅𝑐𝑜𝑛𝑑. =ln (𝑅2𝑅1)

2𝑘𝜋𝑙 (6)

And also, the convective resistance for the cylinder is obtained according to the following relations:

𝑄𝑐𝑜𝑛𝑣.° = ℎ𝑏𝐴(𝑇𝑏 − 𝑇1) = ℎ𝑏2𝜋𝑅1𝑙(𝑇𝑏 − 𝑇1) (7)

𝑄𝑐𝑜𝑛𝑣.° =

(𝑇𝑏 − 𝑇1)

12𝜋ℎ𝑏𝑅1𝑙

(8) → 𝑅𝑐𝑜𝑛𝑣. =1

2𝜋ℎ𝑏𝑅1𝑙 (9)

Therefore:

𝑄 =𝑇𝑏 − 𝑇𝑂

𝑅𝑐𝑜𝑛𝑑. + 𝑅𝑐𝑜𝑛𝑣.=

𝑇𝑏 − 𝑇𝑂

12𝜋ℎ𝑏𝑅1𝑙

+ln (𝑅2𝑅1)

2𝑘𝑡𝜋𝑙

(10)

On the other hand, it can be written:

𝑄 =𝑇 − 𝑇𝑂

ln (𝑅2𝑟)

2𝑘𝑡𝜋𝑙

(11)

Consequently:

𝑄 =𝑇𝑏 − 𝑇𝑂

12𝜋ℎ𝑏𝑅1𝑙

+ln (𝑅2𝑅1)

2𝑘𝑡𝜋𝑙

=𝑇 − 𝑇𝑂

ln (𝑅2𝑟)

2𝑘𝑡𝜋𝑙

(12)

As a result, the radial temperature distribution becomes:

→ 𝑇 − 𝑇𝑂𝑇𝑏 − 𝑇𝑂

=

ln (𝑅2𝑟)

2𝑘𝑡𝜋𝑙

12𝜋ℎ𝑏𝑅1𝑙

+ln (𝑅2𝑅1)

2𝑘𝑡𝜋𝑙

(13)

→ 𝑇 − 𝑇𝑂𝑇𝑏 − 𝑇𝑂

=ln (𝑅2𝑟)

𝑘𝑡ℎ𝑏𝑅1

+ ln (𝑅2𝑅1) 𝑅1 ≤ 𝑟 ≤ 𝑅2 (14)

4

and the heat transfer is:

𝑄

𝐿= 𝑞𝑙 =

2𝜋𝑘𝑡𝑘𝑡ℎ𝑏𝑅1

+ ln (𝑅2𝑅1)(𝑇𝑏 − 𝑇𝑂) (15)

In order to obtain the heat transferred and temperature distribution, the aforementioned statement

should be obtained. For this, usually all the unknown variables are obtained as an empirical expression

and are not individually calculated. In the heat transfer, the Nusselt dimensionless number is adopted to

do this.

As a reason mentioned, we represent the amount of heat transferred with the heat transfer coefficient

equivalent:

𝑞𝑙 = ℎ𝑒2𝜋𝑅1(𝑇𝑏 − 𝑇𝑂) (16)

Consequently:

𝑞𝑙 = ℎ𝑒2𝜋𝑅1(𝑇𝑏 − 𝑇𝑂) =2𝜋𝑘𝑡

𝑘𝑡ℎ𝑏𝑅1

+ ln (𝑅2𝑅1)(𝑇𝑏 − 𝑇𝑂) (17)

1

ℎ𝑒=1

ℎ𝑏+𝑅1 ln (

𝑅2𝑅1)

𝑘𝑡 (18)

Now, we convert this expression to the Nusselt number.

1

ℎ𝑒=1

ℎ𝑏+𝑅1 ln (

𝑅2𝑅1)

𝑘𝑡 ×𝑘𝑏𝐷→

1

𝑁𝑢𝑒1=

1

𝑁𝑢𝐷2+𝑘𝑏𝑅1 ln (

𝑅2𝑅1)

𝐷𝑘𝑡 (19)

→ 1

𝑁𝑢𝑒=

1

𝑁𝑢𝐷(1 +

𝑁𝑢𝐷2

𝑘𝑏𝑘𝑡ln (𝑅2𝑅1)) (20)

→ 1

𝑁𝑢𝑒=

1

𝑁𝑢𝐷(1 + 𝑁𝑢𝐷

𝑘𝑏𝑘𝑡ln√

𝑅2𝑅1) (21)

1 𝑁𝑢𝑒 =

ℎ𝑒𝐷

𝑘𝑏

2 𝑁𝑢𝐷 =ℎ𝑏𝐷

𝑘𝑏

5

→ 𝑁𝑢𝑒 =1

1𝑁𝑢𝐷

+𝑘𝑏𝑘𝑡ln√

𝑅2𝑅1

(22)

We now look at the term obtained for the Nusselt dimensionless number and examine the two extreme

conditions R2/Rx = 1 and R2/R\ > > 1. In the first case:

𝑅2𝑅1= 1 → 𝑁𝑢𝑒 = 𝑁𝑢𝐷 (23)

the local Nusselt number with constant wall temperature can be approximated within less than 3.5

percent by:

𝑁𝑢𝐷 = 4 + 0.48624 ln2 (𝐺𝑧𝑙18) 𝑓𝑜𝑟 𝐺𝑧𝑙 > 18 (24)

𝑁𝑢𝐷 = 4 𝑓𝑜𝑟 𝐺𝑧𝑙 < 18 (25)

Where 𝐺𝑧𝑙 = 𝑅𝑒𝑃𝑟𝐷/𝑥

In the second case when R2/R\ >> 1:

𝑅2𝑅1≫ 1 → 𝑁𝑢𝑒 =

1

0.25 +𝑘𝑏𝑘𝑡ln√

𝑅2𝑅1

(26)

Here 𝑁𝑢𝐷 = 4 was used as a constant, small correction factor. It should be noted that 1 <𝑘𝑏

𝑘𝑡< 2.5 and

from a practical standpoint 𝑅2

𝑅1< 10. Therefore, the smallest constant value for Nue is:

𝑁𝑢𝑒,𝑚𝑖𝑛 ≅ 0.32 (27)

And Also, in steady flow for the variation of the mean Nusselt number (based on diameter) Nu𝐷̅̅ ̅̅ ̅̅ , with

Graetz number, Gz = RePrD/L = 8.333 VD2/L (mm units), at constant wall temperature can be

approximated by

Nu𝐷̅̅ ̅̅ ̅̅ = 4 + 0.155𝑒1.58 𝑙𝑜𝑔𝐺𝑧 𝑓𝑜𝑟 𝐺𝑧 < 103 (28)

where the log is based on 10. For very long tubes with fully developed thermal entrance length, Gz < 1.5,

Nu𝐷̅̅ ̅̅ ̅̅ reaches its minimum, constant value of 4, which is just slightly higher than the theoretical value for

a Newtonian fluid. Since the Nusselt number for constant heat flux is much higher, we can use equation

(28) as a conservative estimate for heat transfer.

6

Figure 2 Schematic temperature profile in a single blood vessel corresponding to Fig.1 [2]

We can obtain the governing differential equation by applying the energy balance to the differential

length, dx, of the blood vessel as shown in Fig. 2. The result is:

𝑑𝑄 = ℎ𝑒(𝑇𝑜 − 𝑇𝑏)𝑑𝐴𝑠 = 𝑚𝑏° 𝐶𝑏𝑑𝑇𝑏 (29)

ℎ𝑒(𝑇𝑜 − 𝑇𝑏)2𝜋𝑅1𝑑𝑥 = 𝜌𝑏𝑉𝑏𝜋𝑅12𝐶𝑏𝑑𝑇𝑏 (30)

𝑑𝑇𝑏𝑑𝑥

+2ℎ𝑒

𝜌𝑏𝑉𝑏𝐶𝑏𝑅1(𝑇𝑏 − 𝑇𝑜) = 0 (31)

or in dimensionless form:

𝑑𝜃

𝑑𝑋+ 𝜆𝜃 = 0 (32)

Where:

𝜃 =𝑇𝑏 − 𝑇𝑜𝑇𝑏1 − 𝑇𝑜

, 𝑋 =𝑥

2𝑅1=𝑥

𝐷 (33)

𝜆 =2ℎ𝑒

𝜌𝑏𝑉𝑏𝐶𝑏𝑅1=4𝑁𝑢𝑒𝑅𝑒𝑃𝑟

, 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑋 (34)

The solution of the differential equation (32) for the first case, R2/R1 = 1 and Nue = NuD as given by

equations (24-25), is:

𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

= exp {−1.945

𝐺𝑧[10.226 + 𝑙𝑛2 (

𝐺𝑧

18) + 2 𝑙𝑛 (

𝐺𝑧

18)]} 𝑓𝑜𝑟 𝐺𝑧 > 18 (35)

7

𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

= 0.8057 exp [−16

𝐺𝑧] 𝑓𝑜𝑟 𝐺𝑧 ≤ 18 (36)

For the second case, R2/R1 >> 1 and Nue=0.32:

𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

= exp [−1.28

𝐺𝑧] (37)

An alternate way to calculate the temperature is to consider the average heat transfer coefficient as a

constant to be evaluated from equation (28). Then, assuming Nu𝑒 = Nu𝐷̅̅ ̅̅ ̅̅ :

𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

= exp [−16 + 0.62 exp(1.58 𝑙𝑜𝑔𝐺𝑧)

𝐺𝑧] (38)

The heat transfer effectiveness is:

𝜖 = 1 −𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

=𝑇𝑏1 − 𝑇𝑏2𝑇𝑏1 − 𝑇𝑂

(39)

Now we can answer the two questions stated before. To find the distance required for the blood to reach

tissue temperature, let us assume that a 95 percent heat transfer effectiveness can be considered as

practically equilibrium. Then:

𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

= 0.05 (40)

The corresponding dimensionless distance is 1/Gze3:

From equation (36) Gze=5.76 1/Gze=0.1737 (41)

From equation (37) Gze =0.427 1/Gze = 2.34 (42)

From equation (38) Gze=6.05 1/Gze=0.1653 (43)

So:

To the answer the first question that, what distance from the entrance does the blood reach tissue

temperature:

1) In the first step, according to vessel diameter and the predicted range of the Gz, an equation among

the 35-38 equations is selected and the Gz number is calculated (36-38 relations can be used also).

2) In the second step, using the Gz number, the required length is calculated.

𝐺𝑧 =𝑅𝑒𝑃𝑟𝐷

𝐿 (44) → 𝐿 =

𝑅𝑒𝑃𝑟𝐷

𝐺𝑧 (45)

3 Gze = Graetz number corresponding to length where blood temperature reaches equilibrium with tissue

8

And to answer the second question that, what is the blood temperature leaving the blood vessel of a given

length in relation to that of the surrounding tissue:

By knowing the diameter and length of the vessel, Gz number is calculated at first. and corresponding to

the amount of Gz, one of the 35-38 equations is chosen and then the outlet temperature of the blood can

be calculated.

Fig. 3 shows the effectiveness as a function of the Graetz number from equations (35-38). Since the results

of equations (35-36) and (38) are virtually identical, the latter correlation is preferred because of its

relative simplicity.

Figure 3 Heat transfer effectiveness of a single blood vessel. εmin from equation (37), εmax from equations (35-36), and 𝜖 ̅from equation (38)

Table 1 shows data on the systemic circulatory system of a dog. Although part of the original data was

taken over 90 yr ago by Mall [3], it has been quoted by virtually everyone using this type of information.

The orders of magnitudes are correct enough for our purposes. When estimated from Fig. 3, the

effectivenesses fall into three categories.

1) The largest vessels, the aorta and vena cava, with Gz > 1000 have very little heat exchange with the

tissue. It should be noted that in man there are much larger main blood vessels than in the dog described

in Table 1; and, consequently, there are also more blood vessels that fall into this category.

2) The smallest vessels, the arterioles, capillaries, and venules, with Gz < 0.4 are essentially ideal heat

exchangers with the blood leaving at tissue temperature. The maximum L/D for thermal equilibrium is

only 1.2 in the arterioles from equation (42).

9

3) The intermediate blood vessels fall in a relatively narrow band of 6 < Gz < 54 with the maximum

effectivenesses between 0.39 and 0.92. If we consider the minimum effectivenesses, εmin < 0.1, this group

could be considered as part of the first. Since there is metabolic heat generation in the tissue, εmax is

probably a better estimation of the heat transfer than 6min; and this group remains a separate one. Thus

the blood supply from all intermediate and small arteries to the tissue can be at distinctly different

temperature levels than the tissue itself, requiring some form of accounting in the energy balances.

Problem

The blood with the temperature of 40˚C enter the large arteries vessels with 3mm and 200mm

diameter and length respectively, in the 37˚C tissue. Does the exhaust blood turn into the same

temperature with the tissue (Is 95% of the heat transfer happen?)?

If it doesn’t occur, calculate the blood temperature in the exit of the vessel.

Answer:

For solve this problem, we should calculate the required length of the vessel that blood needs to the

become isotherm with the tissue and then compare that with the vessel length.

Using equation (38), Graetz number cloud be calculated:

𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

= exp [−16 + 0.62 exp(1.58 𝑙𝑜𝑔𝐺𝑧)

𝐺𝑧]

0.05= exp [−16 + 0.62 exp(1.58 𝑙𝑜𝑔𝐺𝑧)

𝐺𝑧]

𝐺𝑧 = 6.05

10

In the next step, using the following relations, the required length is calculated:

𝐺𝑧 =𝑅𝑒𝑃𝑟𝐷

𝐿 → 𝐿 =

𝑅𝑒𝑃𝑟𝐷

𝐺𝑧

From the table 1:

𝑅𝑒 = 130, 𝑃𝑟 = 25

→ 𝐿 =130 × 25 × 3

6.05= 1611.57 𝑚𝑚

So the blood could not reach to the tissue temperature. therefore, the blood temperature at the outlet is

determined corresponding to the following calculations:

𝐺𝑧 =𝑅𝑒𝑃𝑟𝐷

𝐿=30 × 25 × 3

20048.7

𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

= exp [−16 + 0.62 exp(1.58 𝑙𝑜𝑔𝐺𝑧)

𝐺𝑧] = exp [−

16 + 0.62 exp(1.58 𝑙𝑜𝑔48.7)

48.7]

𝑇𝑏2 − 𝑇𝑂𝑇𝑏1 − 𝑇𝑂

= 0.59948

𝑇𝑏2 − 37

40 − 37= 0.59948 → 𝑇𝑏2 = 38.8℃

11

Heat Transfer Between Parallel Blood Vessels

In this section, we just represent relations that help to determine the amount of heat transferred and the

outlet temperature of the streams and for more details you can refer to reference 2. It is assumed that

other blood vessels and the skin surface are far enough away not to have any thermal effect on the two

blood vessels considered.

Figure 4 heat flow between blood vessels when they are close together

The relations are developed for establishing the heat transfer between given lengths of two parallel blood

vessels. Since parallel arteries or veins with blood running in the same direction would have almost

identical temperatures, the important configuration consists of an artery and a vein with the blood

running in counter flow [2].

Figure 5 Arrangement of Countercurrent blood vessels

12

Steps to obtain the f heat transfer and the temperature of outlet blood streams

1- In the first step, by knowing the blood streams data, the minimum and maximum streams are

determined. And then according to following relation, heat capacity rate ratio, C is calculated.

𝐶 =𝑚𝑚𝑖𝑛𝑚𝑚𝑎𝑥

(46)

2- In the second step, using bellow equations, b1, b2 and finally B are calculated.

𝑏1 =(𝑙/𝑅1)

2 − (𝑅2/𝑅1)2 + 1

2(𝑙/𝑅1) (47)

𝑏2 =(𝑙/𝑅2)

2 − (𝑅1/𝑅2)2 + 1

2(𝑙/𝑅2) (48)

𝐵 = [(𝑏1 + √𝑏12 − 1 ) (𝑏2 +√𝑏2

2 − 1)] (49)

Where “l” is a centerline distance.

If the two blood vessels are of the same size:

𝑏1 = 𝑏2 = 𝑏 =1

2𝑅=1

𝐷 (50)

𝐵 = [(𝑏 + √𝑏2 − 1)]2 (51)

3- In the next step, the number of transfer units, N, is determined according to equation 52.

𝑁 ≅2𝜋𝑘𝑏𝐿

𝑚𝑚𝑖𝑛𝑐𝑏 (1 +𝑘𝑏𝑘𝑡𝑙𝑛𝐵)

(52)

4- Then, the heat transfer effectiveness, ε, is calculated. For a counter flow heat exchanger:

𝜖 =1 − exp [−𝑁(1 − 𝐶)]

1 − 𝐶 exp [−𝑁(1 − 𝐶)] (53)

when C = 1, this reduces to:

𝜖 =𝑁

1 + 𝑁 (54)

13

5- After calculating mentioned parameters, the amount of heat transferred can be calculated:

𝑞 = 𝑚𝑚𝑖𝑛𝑐𝑏𝜖(𝑇𝑎1 − 𝑇𝑣1) (55)

6- Finally, the overall temperature changes are calculated and the outlet temperature of streams are

determined.

𝑓𝑜𝑟 𝑚𝑚𝑖𝑛: 𝜖(𝑇𝑎1 − 𝑇𝑣1) (56)

𝑓𝑜𝑟 𝑚𝑚𝑎𝑥: 𝐶𝜖(𝑇𝑎1 − 𝑇𝑣1) (57)

Example

consider a 100-mm long main arterial branch (see Table 1) exchanging heat with a main venous branch

of the same length located at a centerline distance of 5 mm. Assume entering temperatures of Ta1 = 37

C and Tv1 = 33 C, and a tissue conductivity of 𝑘𝑡 = 0.4 × 10−3 W/mm℃.

𝑚𝑎 = 6.657 × 10−2 𝑔/𝑠, 𝑚𝑣 = 7.187 × 10

−2 𝑔/𝑠

Answer: to solve this problem, we follow mentioned steps:

1-

𝐶 =𝑚𝑚𝑖𝑛𝑚𝑚𝑎𝑥

=6.657 × 10−2

7.187 × 10−2= 0.926

2- R1 and R2 are obtain using Table 1.

𝑏𝑎 =(𝑙/𝑅1)

2 − (𝑅2/𝑅1)2 + 1

2(𝑙/𝑅1)=(5/0.5)2 − (1.2/0.5)2 + 1

2(5/0.5)= 4.762

𝑏𝑣 =(𝑙/𝑅2)

2 − (𝑅1/𝑅2)2 + 1

2(𝑙/𝑅2)=(5/1.2)2 − (0.5/1.2)2 + 1

2(5/1.2)= 2.1825

𝐵 = [(𝑏𝑎 + √𝑏𝑎2 − 1 ) (𝑏𝑣 + √𝑏𝑣

2 − 1)] = [(4.762 + √4.7622 − 1 ) (2.1825 + √2.18252 − 1)]

= 38.8242

3

𝑁 ≅2𝜋𝑘𝑏𝐿

𝑚𝑚𝑖𝑛𝑐𝑏4(1 +𝑘𝑏5

𝑘𝑡𝑙𝑛𝐵)

=2𝜋 × 0.492 × 10−3 × 100

6.657 × 10−2 × 3.8(1 +0.492 × 10−3

0.4 × 10−3𝑙𝑛38.8242)

= 0.222

4 Ref [4] 5 Ref [5]

14

4-

𝜖 =1 − exp [−𝑁(1 − 𝐶)]

1 − 𝐶 exp [−𝑁(1 − 𝐶)]=

1 − exp [−0.222(1 − 0.926)]

1 − 0.926 exp [−0.222(1 − 0.926)]= 0.183

5-

𝑞 = 𝑚𝑚𝑖𝑛𝑐𝑏𝜖(𝑇𝑎1 − 𝑇𝑣1) = 6.657 × 10−2 × 3.8 × 0.183 × (37 − 33) = 0.185

6-

𝑇𝑎1 − 𝑇𝑎2 = 𝜖(𝑇𝑎1 − 𝑇𝑣1) = 0.183 × (37 − 33) = 0.732 → 𝑇𝑎2 = 36.27℃

𝑇𝑣2 − 𝑇𝑣1 = 𝐶𝜖(𝑇𝑎1 − 𝑇𝑣1) = 0.926 × 0.183 × (37 − 33) = 0.678 → 𝑇𝑣2 = 33.678℃

15

[1] Victor, S. A., and V. L. Shah. "Steady state heat transfer to blood flowing in the entrance region of a

tube." International Journal of Heat and Mass Transfer 19, no. 7 (1976): 777-783.

[2] Chato, J. C. "Heat transfer to blood vessels." Journal of biomechanical engineering 102, no. 2 (1980):

110-118.

[3] Mall, Franklin Paine. Die blut-und lymphwege im dünndarm des hundes. Vol. 14, no. 3. S. Hirzel, 1887.

[4] Gamgee, A. “The Specific Heat of Blood” Journal of anatomy and physiology vol. 5,Pt 1 (1870): 139-41.

[5] Ghassemi, M., A. Shahidian, G. Ahmadi, and S. Hamian. "A new effective thermal conductivity model

for a bio-nanofluid (blood with nanoparticle Al2O3)." International Communications in Heat and Mass

Transfer 37, no. 7 (2010): 929-934.