Problem Set 7 (2.006, Spring 08)
Problem 1
a) The ENTU single stream heat exchanger expression is used to calculate the temperature of the blood
leaving the foot (T
f,out
).
()
63
3
min
*
min
max
min
25 10
1000
/
4000 /
1.667
/
60
1.5
/
/0
,
0
.
9
0
1.667
/
hb
p
foot
m
C
C
Vc
kg m
J kgK
W K
s
UA
WK
CC C
N
T
U
CW
K
ρ
−
⎛⎞
×
==
=
=
⎜⎟
⎝⎠
=
=
=
&
Use
1e
x
p
NTU
ε
=−
−
,
configuration #1 from Table 12.1 of the course note. Then
h
C
=
,,
min
fi
n
fo
u
t
TT
C
−
,
,
,
42
1 exp
0.90
42 ( 7)
12.9
o
f out
o
o
n
i
c
e
o
f out
CT
C
C
TC
−
−
−
−−
−
∴
=
The heat lost to the ice per foot is
1.667
/
42
12.9
48.5
oo
hf
i
n f
o
u
t
QCT T
C
C
W
=
−=
&
b) In this case we have two heat exchangers.
The first heat exchanger is the single stream heat exchanger that
models the heat transfer from the duck’s foot to the ice.
The second heat exchanger is a counter flow heat
exchanger that models heat transfer between the arteries.
To find a value for T
c,out
we first guess a value for the blood temperature leaving the duck’s foot (T
f,out
in the
figure). We then use that temperature and the ENTU expression for a counter flow concentric tube heat
exchanger to calculate the blood temperature entering the duck’s foot, T
f,in
.
Next, the calculation done in part
(a) is repeated to calculate T
f,out
.
This temperature is then compared to the guessed temperature.
The desired
result is obtained when the guessed temperature equals the calculated temperature exiting the duck’s foot.
Once T
f,out
is converged, then calculate T
c,out
.
T
f,in
=T
h,out
T
f,out
=T
c,in
T
c,out
T
h,in=
42
o
C
T
f, in
T
f, out
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View Full DocumentThe two arteries have matching heat capacities therefore the counter flow heat exchanger expression (Using
L’hospital’s rule on the correlation #3 in Table 12.1 of the course note. Or the formula in the right column for
the case
C*
=1) reduces to:
()
min
,,
,
,
0.073
/
,0
.
0
4
3
8
11
.
6
6
7
/
0.0438
0.0420
10
.
0
4
3
8
leg
h in
f in
c out
f out
h in
f out
h in
f out
UA
NTU
W K
NTU
NTU
C
W K
TT
T
T
ε
==
=
=
+
−−
⇒=
=
=
=
+
The iterative process can be done in excel. For example, if you start with
,
12.9
o
f out
TC
=
, you get
,
40.78
o
fi
n
=
and
,
12.43
o
fo
u
t
=
. By repeating you will get
,
12.42
o
f out
=
(and
,
40.76
o
n
=
).
Then
,
,
0.0420
13.66
o
c out
f out
h in
f out
T
T
C
=+
−
=
c) The heat loss from the duck’s foot to the ice under the condition of a finite heat transfer between the two
arteries in the duck’s leg is
1.667
/
40.76
12.42
47.2
oo
hf
i
n f
o
u
t
QCT T
WK
C
C
W
=−
=
−
=
&
d)
A geometric model for the two arteries in the duck’s leg is illustrated in the figure 2b.
In calculating the value for (UA)
leg
we ignore any thermal resistance in the tissue that makes up the wall of the
artery.
This is not a good assumption since we expect that tissue to have a thermal conductivity close to that
of water. However, we make this assumption here since the thickness of the walls is not given to us.
The
calculated (UA)
leg
will be an overestimation the actual value.
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 Spring '08
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