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a) We need to find the steady-state temperature distribution in the cylinder, T(r).
We will use the heat transport equation (cylindrical coordinates) on the region shown in
the figure to the right:
��� ����� + ��� +����� + � ����� = � ��
�� ��
���� +
��
���� � +
������� + ��
• Steady state: ���� = 0
• No convection: = = � = 0 • T = f(r) only:
��� = 0, ��
�� = 0 • Uniform heat generation: �� = � (units: ����������)
The heat transport equation reduces to: !! ��
!�!� = −�
#$�) = − %& �' + (� ln � + ('
We need two boundary conditions to solve for the constants of integration (� and ('. We
will define the temperature at the cylinder surface as Ts, so that #$+) = #,. At the cylinder centerline (� = 0), #$0) should be finite.
Boundary conditions: at � = 0, #$0) is finite at � = +, #$+) = #, (� = 0 (' = #- + %.�
&
/$0) = − %&1 �' + #- + %.�
& = 23456 �7 −
0434� + /8
b) Given:
R = 1 cm
L = 5 cm
Ta = 10°C = 283 K
S = 10-3
W/cm3
h = 0.2 W/m2/K
We need to find the surface (skin) temperature Ts for the mouse.
We know that at the surface:
� = ℎ$#- − #:) = −� !�!;<.
ℎ$#- − #:) = %.'
/8 = 234= + /> = ?@AC = ?D°F
Given:
Twater = 2°C
V = 0.5 m/sec
kw = 0.57 W/m/K
Pr = 12.2
ν = 0.016 cm2/sec
R = 0.1525 m (radius of the cylinder approximating the body)
L = 1.73 m
A = 1.8 m2
Rcore = 0.1 m (radius of the core held at 37 °C)
Tcore = 37°C
kbody = 0.41 W/m/K
lsuit = 0.007 m (thickness of the wetsuit)
ksuit = 0.024 W/m/K (thermal conductivity of the wetsuit)
a) We need to find the steady-state rate of heat loss G� > from the person and their skin
temperature /86HI.
We can find the total rate of heat loss J�: as: J�: = K�
ℜ , where M# = #N�� − #�:�� is the temperature drop and ℜ is the thermal resistance. In
order to solve for J�:, we must first find ℜ. The tissue outside the central core and the water can be modeled as two resistances in series, one modeling the thermal resistance of
the tissue around the central core, the other modeling the transfer of heat from the skin to
the flowing water:
ℜ = ℜ�O--�� +ℜ�:��
First, we will find the resistance (ℜ�O--��) of the tissue surrounding the central core.
We will use the steady-state 1-D (radial) heat transport equation, with no convection and
no heat generation:
0 = �P�!Q�
RR� S�
R#R�T
#$�) = (� ln � + ('
Boundary conditions: at � = +N��, # = #N�� at � = +, # = #- OU
(� = − �VWXYZ�[\]^_`$./.[\]^) (' = #N�� +
�VWXYZ�[\]^_`$./.[\]^) ln +N��
#$�) = #N�� − �VWXYZ�[\]^_`$./.[\]^) $ln+ − ln+N��)
�$)Z�[\]^�VWXYZ�[\]^ =
_`$/.[\]^)_`$./.[\]^)
The rate of hear transfer from the skin J�: is: J�: = −�P�!Qb !�
!;<. = −$2d+e)�P�!Q ��VWXYZ�[\]^_`$./.[\]^)��. =
ℜ�O--�� ℜ�:��
= 'fg h\ij_`$./.[\]^) $#N�� − #- OU)
Thus, the tissue resistance is found to be:
ℜ�O--�� = �[\]^Z�VWXYk�l = _`$./.[\]^)
'fg h\ij = 9.469 × 10Z's/t
The resistance to the transport of heat from the ski to the water is found as:
ℜ�:�� = �VWXYZ�ulv^]k�w = K�
xyzK� =�xyz
We need to find ℎ{. We model this heat transfer process (water flowing over the body) as
crossflow over a cylinder. The following relationship is used (Biotransport: Principles
and Applications by Roselli and Diller, Springer Science + Business Media, Chapter 12):
|}{{{{ = xy$'.) u = 0.57+��.����.�� = 401.7
where +� = �$'.)� = 9.531 × 10&
Rearranging, we obtain:
ℎ{ = u'. $0.57+��.����.��) = 751 �
���
ℜ�:�� = �xyz = 7.40 × 10Z&°s/t
The total thermal resistance is:
ℜ = ℜ�O--�� +ℜ�:�� = _`$./.[\]^)'fg h\ij + �
xyz = 9.543 × 10Z's/t
The steady-state rate of heat loss J�: from the person is:
G� > = �/� = /��0�Z/�>��0
� = ?��. A�
The person’s skin temperature #- OU is: /86HI = /��0� − G� >��H88�� = 4. 4�°F
b) We need to find the steady-state rate of heat loss G� > from the person and their skin
temperature /86HI, when they are wearing an insulating wet suit.
We will use a resistance-in-series model again, with an additional resistance due to the
presence of the suit:
ℜ = ℜ�O--�� +ℜ-�O� +ℜ�:��
We need to find the resistance of the wet suit. We will treat the suit as a hollow cylinder
of thickness �-�O�. Using analysis analogous to that used for the tissue, we find that the resistance of the suit is:
ℜ-�O� = _`�$.��V�Xv)/.�'fg V�Xv = 0.1720°s/t
ℜ�O--�� ℜ�:�� ℜ-�O�
The total thermal resistance is:
ℜ = ℜ�O--�� +ℜ-�O� +ℜ�:�� = = _`$./.[\]^)
'fg h\ij + _`�$.��V�Xv)/.�'fg V�Xv + �
xyz = 0.2675°s/t
(Here we have neglected the small change in the value of ℜ�:�� due to the increase of the “cylinder” radius when the person wears the suit. Since the water has very little heat
transfer resistance, and because the suit is relatively thin, this approximation is a very
good one.) The steady-state rate of heat loss J�: from the person is then:
G� > = �/� = /��0�Z/�>��0
� = 7?@. ��
The person’s skin temperature #- OU is: /86HI = /��0� − G� >��H88�� = 45. �°F
