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coefficient is
larger as we would expect, but amounts to only a 10% increase.
PROBLEM 8.39
KNOWN: Flow rate and temperature of atmospheric air entering a duct of prescribed diameter, length
and surface temperature.
FIND: (a) Air outlet temperature and duct heat loss for the prescribed conditions and (b) Calculate and
plot q and \u2206p for the range of diameters, 0.1 \u2264 D \u2264 0.2 m, maintaining the total surface area, As = \u3c0DL, at
the same value as part (a). Explain the trade off between the heat transfer rate and pressure drop.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy and
potential energy changes, (4) Uniform surface temperature, (5) Fully developed flow conditions.
PROPERTIES: Table A.4, Air ( mT \u2248 310 K, 1 atm): \u3c1 = 1.128 kg/m3, cp = 1007 J/kg\u22c5K, µ = 189 × 10-
7
 N\u22c5s/m2, k = 0.027 W/m\u22c5K, Pr = 0.706.
ANALYSIS: (a) With
( )D 7 2
4m 4 0.04kg sRe 17,965
D 0.15m 189 10 N s m\u3c0 µ \u3c0 \u2212
×
= = =
× \u22c5
\ufffd
the flow is turbulent. Assuming fully developed conditions throughout the tube, it follows from the
Dittus-Boelter correlation, Eq. 8.60, that
( ) ( )4 / 5 0.44 / 5 0.4 2Dk 0.027 W m Kh 0.023Re Pr 0.023 17,965 0.706 9.44 W m KD 0.15m
\u22c5
= = = \u22c5 .
Hence, from the energy balance relation, Eq. 8.42b,
( )m,o s s m,i
p
DLT T T T exp h
mc
\u3c0\uf8eb \uf8f6
= \u2212 \u2212 \u2212\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8\ufffd
( ) ( )
( )
2
m,o
0.15m 10 m 9.44 W m K
T 15 C 45 Cexp 29.9 C
0.04 kg s 1007 J kg K
\u3c0\uf8eb \uf8f6\u22c5\uf8ec \uf8f7
= + \u2212 =\uf8ec \uf8f7
\u22c5\uf8ec \uf8f7\uf8ed \uf8f8
$ $ $ <
From the overall energy balance, Eq. 8.37, it follows that
( ) ( )p m,o m,iq mc T T 0.04 kg s 1007 J kg K 29.9 60 C 1212 W= \u2212 = × \u22c5 \u2212 = \u2212$\ufffd . <
From Eq. 8.22a, the pressure drop is
2
mup f L
2D
\u3c1\u2206 =
Continued...
PROBLEM 8.39 (Cont.)
and for the smooth surface conditions, Eq. 8.21 can be used to evaluate the friction factor,
( )( ) ( )( )2 2Df 0.790ln Re 1.64 0.790ln 17,965 1.64 0.0269\u2212 \u2212= \u2212 = \u2212 =
Hence, the pressure drop is
( )23 21.128kg m 2.0 m sp 0.0269 10 m 4.03N m
2 0.15m
\u2206 = × =
×
<
where um = cm A\u3c1\ufffd = ( )3 2 20.04 kg s 1.128kg m 0.15 m 4 2.0 m s\u3c0× = .
(b) For the prescribed conditions of part (a), As = \u3c0DL = \u3c0(0.15 m) × 10 m = 4.712 m2, using the IHT
Correlations Tool, Internal Flow for fully developed Turbulent Flow along with the energy balance
equation, rate equation and pressure drop equations used above, the heat rate q and \u2206p are calculated and
plotted below.
100 125 150 175 200
Tube diameter, D (mm)
5
10
15
Tu
be
 le
ng
th
, L
 (m
)
 
100 125 150 175 200
Tube diameter, D (mm)
0
10
20
30
40
50
de
lta
P 
(P
a)
From above, as D increases, L decreases so that
As remains unchanged. The decrease in heat
rate with increasing diameter is nearly linear,
while the pressure drop decreases markedly.
This is the trade off: increased heat rate requires
a more significant increase in pressure drop, and
hence fan blower power requirements.
100 120 140 160 180 200
Tube diameter, D (mm)
0
0.5
1
1.5
2
H
ea
t r
at
e,
 q
 (k
W
)
COMMENTS: (1) To check the calculations, compute q from Eq. 8.44, where mT\u2206 &quot; is given by Eq.
8.45. It follows that mT\u2206 &quot; = -27.1°C and q = -1206 W. The small difference in results may be attributed
to round-off error.
(2) For part (a), a slight improvement in accuracy may be obtained by evaluating the properties at mT =
318 K: h = 9.42 W/m2\u22c5K, Tm,o = 303 K = 30°C, q = -1211 W, f = 0.0271 and \u2206p = 4.20 N/m2.
PROBLEM 8.40
KNOWN: Inlet temperature, pressure and flow rate of air. Tube diameter and length. Pressure of
saturated steam.
FIND: Outlet temperature and pressure of air. Mass rate of steam condensation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Outer surface of annulus is adiabatic, (3) Negligible potential
energy, kinetic energy and flow work changes for air, (4) Fully-developed flow throughout the tube,
(5) Smooth tube surface, (6) Constant properties.
PROPERTIES: Table A-4, air ( )mT 325 K, p 5 atm :\u2248 = ( ) 35 1atm 5.391kg / m ,\u3c1 \u3c1= × =
pc 1008 J / kg K,= \u22c5 
7 2196.4 10 N s / m ,µ \u2212= × \u22c5 k 0.0281W / m K,= \u22c5 Pr 0.703.= Table A-6, sat. steam (p
= 2.455 bars): Ts = 400 K, hfg = 2183 kJ/kg.
ANALYSIS: With a uniform surface temperature, the air outlet temperature is
( ) im,o s s m,i
p
D LT T T T exp h
m c
\u3c0\uf8eb \uf8f6
= \u2212 \u2212 \u2212\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8\ufffd
With ( ) 7D iRe 4m / D 0.12 kg / s / 0.05m 196.4 10 kg / s m 38,980,\u3c0 µ \u3c0 \u2212= = × \u22c5 =\ufffd the flow is
turbulent, and the Dittus-Boelter correlation yields
( ) ( )4 / 5 0.4 24 / 5 0.4fd D
i
k 0.0281W / m K
h h 0.023 Re Pr 0.023 38,980 0.703 52.8 W / m K
D 0.05m
\u22c5
\u2248 = = = \u22c5
\uf8eb \uf8f6 \uf8eb \uf8f6\uf8ec \uf8f7\uf8ec \uf8f7 \uf8ed \uf8f8\uf8ed \uf8f8
( )
2
m,o
0.05m 5m 52.8 W / m KT 127 C 110 C exp 99 C
0.03kg / s 1008J / kg K
\u3c0\uf8eb \uf8f6× × × \u22c5\uf8ec \uf8f7= ° \u2212 ° \u2212 = °\uf8ec \uf8f7× \u22c5\uf8ed \uf8f8
<
The pressure drop is ( )2m ip f u / 2 D L,\u3c1\u2206 = where, with 2 3 2c iA D / 4 1.963 10 m ,\u3c0 \u2212= = ×
m cu m / A 2.83 m / s,\u3c1= =\ufffd and with DRe 38, 980,= Fig. 8.3 yields f \u2248 0.022. Hence,
( )23 2 42.83m / s 5mp 0.022 5.391kg / m 95 N / m 9.4 10 atm
2 0.05m
\u2212\u2206 \u2248 × = = ×
×
<
The rate of heat transfer to the air is
( ) ( )p m,o m,iq m c T T 0.03kg / s 1008J / kg K 82 C 2480 W= \u2212 = × \u22c5 ° =\ufffd
and the rate of condensation is then
3
c 6fg
q 2480 W
m 1.14 10 kg / s
h 2.183 10 J / kg
\u2212
= = = ×
×
\ufffd <
COMMENTS: (1) With ( )m m,i m,oT T T / 2 331K,= + = the initial estimate of 325 K is reasonable
and iteration is not necessary. (2) For a steam flow rate of 0.01 kg/s, approximately 10% of the
outflow would be in the form of saturated liquid, (3) With L/Di = 100, it is reasonable to assume fully
developed flow throughout the tube.
PROBLEM 8.41
KNOWN: Duct diameter and length. Thermal conductivity of insulation. Gas inlet temperature and
velocity and minimum allowable outlet temperature. Temperature and velocity of air in cross flow.
FIND: Minimum allowable insulation thickness.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible potential and kinetic energy and flow work changes for gas flow
through duct, (2) Fully developed flow throughout duct, (3) Negligible duct wall conduction
resistance, (4) Negligible effect of insulation thickness on outer convection coefficient and thermal
resistance, (5) Properties of gas may be approximated as those of air.
PROPERTIES: Table A-4, air (p = 1 atm). Tm,i = 1600K: (\u3c1i = 0.218 kg/m3). mT = (Tm,i +Tm,o)/2
= 1500K: (\u3c1 = 0.232 kg/m3, cp = 1230 J/kg\u22c5K, µ = 557 × 10-7 N\u22c5s/m2, k = 0.100 W/m\u22c5K, Pr = 0.685).
Tf \u2248 300K (assumed): \u3bd = 15.89 × 10-6 m2/s, k = 0.0263 W/m\u22c5K, Pr = 0.707.
ANALYSIS: From Eqs. (8.46a) and (3.19),
m,o s
m,i p tot p
T T UA1150 K 10.852 exp exp
T T 1350 K m c R mc
\u221e
\u221e
\uf8eb \uf8f6 \uf8eb \uf8f6
\u2212
\u2212
= = = \u2212 = \u2212\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ec \uf8f7 \uf8ec \uf8f7
\u2212 \u2212 \uf8ed \uf8f8 \uf8ed \uf8f8\ufffd \ufffd
Hence, with ( ) ( )3 2m c im u A 0.218 kg / m 10 m / s 1m / 4 1.712 kg / s,\u3c1 \u3c0= = × × =\ufffd
( ) ( )[ ]1 1 3tot pR m c ln 0.852 1.712 kg / s 1230 J / kg K 0.160 2.96 10 K / W\u2212 \u2212 \u2212= \u2212 = \u2212 × \u22c5 × \u2212 = ×\uf8ee \uf8f9\uf8f0 \uf8fb\ufffd
The total thermal resistance is
( ) ( ) ( )1 1o itot conv,i cond,ins conv,o i i o o
ins
ln D / D
R R R R h D L h D L
2 k L
\u3c0 \u3c0
\u3c0
\u2212 \u2212
= + + = + + (1)
With ( ) ( )7 2D,i iRe 4m / D 4 1.712 kg / s / 1m 557 10 N s / m 39,130,\u3c0 µ \u3c0 \u2212= = × × × × \u22c5 =\ufffd the Dittus-Boelter
correlation yields
( ) ( )4 / 5 1/ 3 24 / 5 1/ 3i Dk 0.100 W / m Kh 0.023 Re Pr 0.023 39,130 0.685 9.57 W / m KD 1m
\u22c5
= = = \u22c5
\uf8eb \uf8f6 \uf8eb \uf8f6\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
The internal resistance is then
( ) ( ) 11 2 4conv,i i iR h D L 9.57 W / m K 1m 100m 3.33 10 K / W\u3c0 \u3c0 \u2212\u2212 \u2212= = \u22c5 × × × = ×
With ReD \u2248 VDi/\u3bd = 15 m/s × 1m/15.89 × 10
-6
 m
2/s = 9.44 × 105, the Churchill-Bernstein correlation
yields
Continued \u2026..
PROBLEM 8.41 (Cont.)
( )
4 /51/ 2 1/3 5/8
2DD
o 1/ 42 /3
0.62 Re Prk Reh 0.3 1 30.9 W / m K
D 282,000
1 0.4 / Pr
\uf8f1 \uf8fc\uf8f4 \uf8ee \uf8f9 \uf8f4\uf8eb \uf8f6\uf8f4 \uf8f4\uf8eb \uf8f6 \uf8ef \uf8fa\u2248 + + = \u22c5\uf8f2 \uf8fd\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ef \uf8fa\uf8ed \uf8f8 \uf8ed \uf8f8\uf8f4 \uf8f4\uf8ee \uf8f9 \uf8f0 \uf8fb+\uf8f4