Question 1 solution

PROBLEM 3.110
KNOWN: Temperature sensing probe of thermal conductivity k, length L and diameter D is mounted
on a duct wall; portion of probe Li is exposed to water stream at T∞,i while other end is exposed to
ambient air at T∞,o ; convection coefficients hi and ho are prescribed.
FIND: (a) Expression for the measurement error, ∆ e
T rr = t
T ip − T∞,i , (b) For prescribed T∞,i and
T∞,o , calculate
e
T
∆ rr for immersion to total length ratios of 0.225, 0.425, and 0.625, (c) Compute
and plot the effects of probe thermal conductivity and water velocity (hi) on
e
T
∆ rr .
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in probe, (3) Probe is
thermally isolated from the duct, (4) Convection coefficients are uniform over their respective
regions.
PROPERTIES: Probe material (given): k = 177 W/m⋅K.
ANALYSIS: (a) To derive an expression for
e
T
∆ rr = Ttip - T∞,i , we need to determine the
temperature distribution in the immersed
length of the probe Ti(x). Consider the probe
to consist of two regions: 0 ≤ x ≤
i
Li, the
immersed portion, and 0 ≤ x ≤
o
(L - Li), the
ambient-air portion where the origin
corresponds to the location of the duct wall.
Use the results for the temperature distribution
and fin heat rate of Case A, Table 3.4:
Temperature distribution in region i:
θ
T x
− T
cosh m L − x
+ h m k sinh L − x
i
i ( i )
∞,i
( i ( i i)) ( i i )
( i i )
=
=
(1)
θb,i
o
T − T∞,i
cosh (miLi ) + (hi mik)sinh (mi i
L )
and the tip temperature, Ttip = Ti(Li) at xi = Li, is
t
T ip − T∞,i
cosh (0)+ (hi mik)sinh (0)
= A =
(2)
o
T − T∞,i
cosh (miLi )+ (hi mik)sinh (mi i
L )
and hence
∆ e
T rr = t
T ip − T ,i = A ( o
T − T
∞
∞,i )
(3) <
where To is the temperature at xi = xo = 0 which at present is unknown, but can be found by setting the
fin heat rates equal, that is,
q
 q
(4)
f ,o
f ,i
Continued...

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PROBLEM 3.110 (Cont.)
(h PkA )1/2θ ⋅B = −(h PkA )1/2
o
c
b,o
i
c
θb,i ⋅C
Solving for To, find
θb,o
o
T − T∞,o
=
= −(h PkA )1/ 2
i
c
θb,i ⋅C
θb,i
o
T − T∞,i

1/ 2


1/ 2

 h 


i
C
hi
C
= 
+




+

o
T
T∞,o
T∞,i
1 

(5)

ho
B


ho
B 








where the constants B and C are,
sinh (moLo )+ (ho mok)cosh (moLo )
B =
(6)
cosh (moLo )+ (ho mok)sinh (moLo )
sinh (mi i
L ) + (hi mik)cosh (miLi )
C =
(7)
cosh (miLi ) + (hi mik)sinh (mi i
L )
(b) To calculate the immersion error for prescribed immersion lengths, Li/L = 0.225, 0.425 and 0.625,
we use Eq. (3) as well as Eqs. (2, 6, 7 and 5) for A, B, C, and To, respectively. Results of these
calculations are summarized below.
Li/L
Lo (mm)
Li (mm)
A
B
C
To (°C)
∆Terr (°C)
0.225
155
45
0.2328
0.5865
0.9731
76.7
-0.76
<
0.425
115
85
0.0396
0.4639
0.992
77.5
-0.10
<
0.625
75
125
0.0067
0.3205
0.9999
78.2
-0.01
<
(c) The probe behaves as a fin having
2.5
)
ends exposed to the cool ambient air

(
C
t
i
p
and the hot ambient water whose
2

-

T
f
o
temperature is to be measured. If the
i
n
1.5
thermal conductivity is decreased, heat
r
,

T
r
r
o

e
transfer along the probe length is
r
e
1
t
u
likewise decreased, the tip temperature
r
a
e
p
will be closer to the water temperature.
0.5
m
e
T
If the velocity of the water decreases,
0
the convection coefficient will
0.2
0.3
0.4
0.5
0.6
0.7
decrease, and the difference between the
Immersion length ratio, Li/L
tip and water temperatures will increase.
Base case: k = 177 W/m.K; ho = 1100 W/m^2.K
Low velocity flow: k = 177 W/m.K; ho = 500 W/m^2.K
Low conductivity probe: k = 50 W/m.K; ho = 1100 W/m^2.K

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