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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
[email protected].
mailto:[email protected]
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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].
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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)
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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)
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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
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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.
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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)
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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
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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).
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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
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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℃
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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
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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)
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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
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