Principles Of Heat Transfer 8th edition, Problem 1.67chapter 1.Please comment about the solution PROPERTIES AND CONSTANTSFrom Appendisx 2, Table 10, Thermal conductivity of stainles steel (k)= 142 W(m K)SOLUTIONSince solid angle for complete sphere is 720° or 4z the sarface area reduces by the factor ofand thus720720resistances increase by the factor of -.The tharmal eircuit diagram for the problem is given as followsRIR:720 1720 ra-n720 P-ra7201e 4:th ne 4th rTotal heat loss with the lans ia given byT. T.Ru-R+Re+ R+ R.1R.20 1573 KW4*10 *0.01Total heat los with the lens is given byT.T.RoeR-R+R+ R+ R.720 1R1-7.2- 573 KW4"10 *0.01720 ra0.0125 0.01-7.232.14 KW4"0.35*0.0125 0.01720R0.0145 0.01257.2--79 KW4"0.8"0.0145 "0.01257201-7.21545 KW4*5*0.0145Ru-R+R+R+ R-1158.64 KWTotal heat loss with the lens isT. T. 36.9 22Re 1158.64-W-0.0129 W PROBLEM 2.36The development of contact lenses has transformed the solutions that are available today for visionimpairments. However, wearing them also poses several problems that includes the condition ofdry eyes due to lack of cooling, oxygenation, and moisturizing or lubrication of cornea, amongothers. In a development phase of a new protective liquid for contact lenses, the optometrist wouldlike to know rate of heat loss from the anterior chamber of the eye through the cornea and thecontact lens. The system is modelled as a partial hollow sphere, as shown in schematic diagram.The inner and outer radius of the cornea are respectively r;=10 mm and r=12.5 mm, and the outerradius of the fitted contact lens is ry=14.5 mm. The anterior chamber, which contains the aqueoushumor that provides nutrients to the cornea (as the later tissue has no blood vessel:), exposes theinner surface of the cornea to a temperature of T=36.9°C with a convective heat transfer coefficientof h=10 W/(m² K). The thermal conductivity of transparent tissue of cornea is 0.35 W/(m K), andthat of the contact lens material is 0.8 W/(m K). The outer surface of the contact lens is exposed toroom air at T,-22°C and has a convection coefficient of h=5 W/(m² K). Draw a thermal circuitfor the system showing all of the temperature potentials and thermal resistances. Then estimatethe rate of heat loss from the anterior eye assuming that steady state exists and eye aperturespherical angle is of 100°.GIVENInner radius of comea r=10 mm=0.01 mOuter radius of cornea n=12.5 mm=0.0125 mRadius of contact lens r;=14.5 mm=0.0145 mInner comea temperature T-36.9°CInner convective heat transfer coefficient h=10 W/(m K)Conductivity of cornea tissue (k.)=0.35 W/(m K)Conductivity of lens material (k.)=0.8 W/(m K)Room air temperature Te=22°COuter convection coefficient of h,=5 W/(m K)

The ratio of the temperature difference between two faces of a material to the rate of heat flow…

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Elements Of Electromagnetics

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ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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Principles Of Heat Transfer 8th edition, Problem 1.67chapter 1.

Please comment about the solution

PROPERTIES AND CONSTANTS
From Appendisx 2, Table 10, Thermal conductivity of stainles steel (k)= 142 W(m K)
SOLUTION
Since solid angle for complete sphere is 720° or 4z the sarface area reduces by the factor of
and thus
720
720
resistances increase by the factor of -.
The tharmal eircuit diagram for the problem is given as follows
RI
R:
720 1
720 ra-n
720 P-ra
720
1
e 4:th n
e 4th r
Total heat loss with the lans ia given by
T. T.
Ru-R+Re+ R+ R.
1
R.20 1
573 KW
4*10 *0.01
Total heat los with the lens is given by
T.T.
Roe
R-R+R+ R+ R.
720 1
R
1
-7.2
- 573 KW
4"10 *0.01
720 ra
0.0125 0.01
-7.2
32.14 KW
4"0.35*0.0125 0.01
720
R
0.0145 0.0125
7.2
--79 KW
4"0.8"0.0145 "0.0125
720
1
-7.2
1
545 KW
4*5*0.0145
Ru-R+R+R+ R-1158.64 KW
Total heat loss with the lens is
T. T. 36.9 22
Re 1158.64
-W-0.0129 W

PROBLEM 2.36
The development of contact lenses has transformed the solutions that are available today for vision
impairments. However, wearing them also poses several problems that includes the condition of
dry eyes due to lack of cooling, oxygenation, and moisturizing or lubrication of cornea, among
others. In a development phase of a new protective liquid for contact lenses, the optometrist would
like to know rate of heat loss from the anterior chamber of the eye through the cornea and the
contact lens. The system is modelled as a partial hollow sphere, as shown in schematic diagram.
The inner and outer radius of the cornea are respectively r;=10 mm and r=12.5 mm, and the outer
radius of the fitted contact lens is ry=14.5 mm. The anterior chamber, which contains the aqueous
humor that provides nutrients to the cornea (as the later tissue has no blood vessel:), exposes the
inner surface of the cornea to a temperature of T=36.9°C with a convective heat transfer coefficient
of h=10 W/(m² K). The thermal conductivity of transparent tissue of cornea is 0.35 W/(m K), and
that of the contact lens material is 0.8 W/(m K). The outer surface of the contact lens is exposed to
room air at T,-22°C and has a convection coefficient of h=5 W/(m² K). Draw a thermal circuit
for the system showing all of the temperature potentials and thermal resistances. Then estimate
the rate of heat loss from the anterior eye assuming that steady state exists and eye aperture
spherical angle is of 100°.
GIVEN
Inner radius of comea r=10 mm=0.01 m
Outer radius of cornea n=12.5 mm=0.0125 m
Radius of contact lens r;=14.5 mm=0.0145 m
Inner comea temperature T-36.9°C
Inner convective heat transfer coefficient h=10 W/(m K)
Conductivity of cornea tissue (k.)=0.35 W/(m K)
Conductivity of lens material (k.)=0.8 W/(m K)
Room air temperature Te=22°C
Outer convection coefficient of h,=5 W/(m K)

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