8-101
8-111E A liquid mixture flowing in a tube with a square-edged inlet is subjected to uniform wall heat flux. The friction
coefficient is to be determined.
Assumptions Steady operating conditions exist.
Properties The properties of the ethylene glycol-distilled water mixture are given to be Pr = 13.8, ν = 18.410-6 ft2/s and
b/
s
= 1.12.
Analysis: For the calculation of the non-isothermal fully developed friction coefficient, it is necessary to determine the flow
regime before making any decision regarding which friction coefficient relation to use. The Reynolds number at the specified
location is
( )
6425
gal/min8.448
/sft1
/sft104.18
ft12/622.0)]ft1011.2/()gal/min16.2[(
)/(
Re
3
6
23
2
=
== −
−
DAc
V
since
2322 ft10110.24/ft)12/622.0(4/ −
===
DAc
From Table 8-6, the transition Reynolds number range for this case (square-edged inlet and a heat flux of 8 kW/m2) is 3860
< Re < 5200, which means that the flow in this case is turbulent and Eq. 8–82 is the appropriate equation to use. It gives
( )
0.00859=
=
=−25.0
25.025.0
turb 12.1
6425
0791.0
Re
0791.0 m
s
b
f,
C
Repeating the calculations when the volume flow rate is increased by 50%, we obtain
( )
9639
gal/min8.448
/sft1
/sft104.18
ft12/622.0)]ft1011.2/()gal/min16.2(5.1[
)/(
Re
3
6
23
2
=
== −
−
DAc
V
( )
0.00776=
=
=−25.0
25.025.0
turb 12.1
9639
0791.0
Re
0791.0 m
s
b
f,
C
8-103
8-113 A liquid mixture flowing in a tube is subjected to uniform wall heat flux. The Nusselt number at a specified location is
b/
s = 2.0.
Analysis For a tube with a known diameter and volume flow rate, the type of flow regime is determined before making any
decision regarding which Nusselt number correlation to use. The Reynolds number at the specified location is
( )
4790
/sm1045.3
m0158.0)]m10961.1/()/sm1005.2[(
)/(
Re
2
6
2434
=
== −
−−
DAc
V
since
2422 m10961.14/m)0158.0(4/ −
===
DAc
.
Therefore, the flow regime is in the transition region for all three inlet configurations (thus use the information given in Table
8-9 with x/D = 10) and therefore Eq. 8-87 should be used with the constants a, b, c found in Table 8-8. However, Nulam and
Nuturb are the inputs to Eq. 8-87 and they need to be evaluated first from Eqs. 8-88 and 8-89, respectively. It should be
mentioned that the correlations for Nulam and Nuturb have no inlet dependency.
From Eq. 8-88:
( )
( )
4.35
)0.2(]46.33)(000,60[025.0
10
)46.33)(4790(
24.1
GrPr025.0
PrRe
24.1Nu
14.0
3/1
75.0
14.0
3/1
75.0
lam
=
+
=
+
=
s
b
x
D
From Eq. 8-89:
PrRe023.0Nu
14.00054.0
385.08.0
14.0
0054.0
385.08.0
turb
=
−
−
s
b
D
x
8-105
Review Problems
8-115 The velocity profile in fully developed laminar flow in a circular pipe is given. The radius of the pipe, the mean
velocity, and the maximum velocity are to be determined.
Assumptions The flow is steady, laminar, and fully developed.
Analysis The velocity profile in fully developed laminar flow in a circular pipe is
−= 2
2
max 1)(
R
r
Vru
The velocity profile in this case is given by
)1001(4)( 2
rru −=
Comparing the two relations above gives the pipe radius, the
maximum velocity, and the mean velocity to be
m 0.10=→= RR
100
1
2
Vmax = 4 m/s
m/s 2=== 2
m/s 4
2
max
avg
V
V
R
r
0
Vmax
u(r)=Vmax(1–r2/R2)
8-108
8-118 Repeat Prob. 8-117 for turbulent flow. Turbulent flow of a fluid through an isothermal square channel is considered.
The change in the pressure drop and the rate of heat transfer are to be determined when the free-stream velocity is doubled.
Assumptions 1 The flow is fully developed. 2 The effect of the change in Tlm on the rate of heat transfer is not considered.
Analysis The pressure drop of the fluid for turbulent flow is expressed as
L
D
V
L
DVV
L
V
L
2.0
1.8
2
avg
2.00.2
avg
2
avg
2.0
2
avg
−
−−
−
8-111
8-121 A fluid flows through a tube subjected to uniform heat flux, (a) the convection heat transfer coefficient, (b) the value of
Ts – Tm, and (c) the value of Te – Ti are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant.
Properties The properties of the fluid are given:
= 1000 kg/m3,
= 1.4 × 10−3 kg/m∙s, cp = 4.2 kJ/kg∙K, and k = 0.58
W/m∙K.
Analysis (a) The Reynolds number, hydrodynamic and thermal entry lengths are
)m 01.0)(m/s 3.0)(kg/m 1000(
3
avg =
DV
8-113
8-123 Air (1 atm) enters into a 5-mm diameter circular tube, the convection heat transfer coefficient for (a) a 10-cm long tube
and (b) a 50-cm long are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Constant tube surface temperature.
Properties The properties of air at 50°C: cp = 1007 J/kg∙K, k = 0.02735 W/m∙K,
= 1.092 kg/m3,
= 1.963 10−5 kg/m∙s,
= 1.798 10−5 m2/s, and Pr = 0.7228; at Ts = 160°C:
s = 2.420 10−5 kg/m∙s (Table A–15).
Analysis The Reynolds number, hydrodynamic and thermal entry lengths are
)m 005.0)(m/s 5(
avg =
DV
8-115
8-125 Liquid mercury flowing through a tube, the tube length is to be determined using (a) the appropriate Nusselt number
relation for liquid metals and (b) the Dittus-Boelter equation. The results of (a) and (b) are to be compared.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Constant tube surface temperature. 4 Fully
developed flow.
Properties The properties of liquid mercury at Tb = (Ti + Te)/2 = 150°C: cp = 136.1 J/kg∙K, k = 10.0778 W/m∙K,
=
1.12610−3 kg/m∙s, and Pr = 0.0152; at Ts = 250°C: Prs = 0.0119 (Table A-14).
Analysis The Reynolds number is
)kg/s 6.0(44
m
8-117
8-127 Water is heated by passing it through five identical tubes that are maintained at a specified temperature. The rate of
heat transfer and the length of the tubes necessary are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The surface temperature is constant and uniform. 3 The inner surfaces of the
tubes are smooth. 4 Heat transfer to the surroundings is negligible.
Properties The properties of water at the bulk mean fluid temperature of (15+35)/2=25ºC are (Table A–9)
14.6Pr
CJ/kg. 4180
/sm 10891.0
C W/m.607.0
kg/m 997
23–
3
=
=
=
=
=
p
c
k
Analysis (a) The rate of heat transfer is
(b) The water velocity is
kg/s )5/10(
m
5
D
Water
15ºC
10 kg/s
5 tubes
D = 5 cm
35ºC
60ºC
8-118
8-128 Water is heated by passing it through five identical tubes that are maintained at a specified temperature. The rate of
heat transfer and the length of the tubes necessary are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The surface temperature is constant and uniform. 3 The inner surfaces of the
tubes are smooth. 4 Heat transfer to the surroundings is negligible.
Properties The properties of water at the bulk mean fluid temperature of (15+35)/2=25ºC are (Table A–9)
14.6Pr
CJ/kg. 4180
/sm 10891.0
C W/m.607.0
kg/m 997
23–
3
=
=
=
=
=
p
c
k
Analysis (a) The rate of heat transfer is
5
D
Water
15ºC
20 kg/s
5 tubes
D = 5 cm
35ºC
60ºC
8-119
8-129 Water is heated as it flows in a smooth tube that is maintained at a specified temperature. The necessary tube length
and the water outlet temperature if the tube length is doubled are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The surface temperature is constant and uniform. 3 The inner surfaces of the
tube are smooth. 4 Heat transfer to the surroundings is negligible.
Properties The properties of water at the bulk mean fluid temperature of (10+40)/2=25ºC are (Table A–9)
14.6Pr
CJ/kg. 4180
/sm 10891.0
C W/m.607.0
kg/m 997
23–
3
=
=
=
=
=
p
c
k
Analysis (a) The rate of heat transfer is
Water
1500 kg/h
L
D = 1 cm
8-120
8-130E The exhaust gases of an automotive engine enter a steel exhaust pipe. The velocity of exhaust gases at the inlet and
the temperature of exhaust gases at the exit are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The inner surfaces of the pipe are smooth. 3 The thermal resistance of the pipe
is negligible. 4 Exhaust gases have the properties of air, which is an ideal gas with constant properties.
Properties We take the bulk mean temperature for exhaust gases to be 700F since the mean temperature of gases at the inlet
will drop somewhat as a result of heat loss through the exhaust pipe whose surface is at a lower temperature. The properties
of air at this temperature and 1 atm pressure are (Table A-15E)
/sft 10225.6
FBtu/h.ft. 0280.0
lbm/ft 03421.0
24–
3
=
=
=
k
694.0Pr
FBtu/lbm. 2535.0
=
=
p
c
Noting that 1 atm = 14.7 psia, the pressure in atm is
P = (15.5 psia)/(14.7 psia) = 1.054 atm. Then,
/sft 105.906=)/s)/(1.054ft 10225.6(
lbm/ft 03606.0)054.1)(lbm/ft 03421.0(
24–24–
33
=
==
v
Analysis (a) The velocity of exhaust gases at the inlet of the exhaust pipe is
lbm/s 0.05
m
Exhaust
800F
0.05 lbm/s
80F
L = 8 ft
D = 3.5 in