fundamentals of heat and mass transfer solutions manual phần 6 docx

PROBLEM 8.28
KNOWN:
Inlet temperature, flow rate and properties of hot fluid. Initial temperature, volume and
properties of pharmaceutical. Heat transfer coefficient at outer surface and dimensions of coil.
FIND:
(a) Expressions for T
c
(t) and T
h,o
(t), (b) Plots of T
c
(t) and T
h,o
(t) for prescribed conditions.
Effect of flow rate on time for pharmaceutical to reach a prescribed temperature.
SCHEMATIC:
ASSUMPTIONS:
(1) Constant properties, (2) Negligible heat loss from vessel to surroundings, (3)
Pharmaceutical is isothermal, (4) Negligible work due to stirring, (5) Negligible thermal energy
generation (or absorption) due to chemical reactions associated with the batch process, (6) Negligible
kinetic energy, potential energy and flow work changes for the hot fluid, (7) Negligible tube wall
conduction resistance.
ANALYSIS:
(a) Performing an energy balance for a control surface about the stirred liquid, it
follows that
()
()
cc
c c v,c c c c v,c
dU dT

d
cT Vc qt
dt dt dt
ρρ
=∀ = =
(1)
where,
()
()
hp,h h,i h,o
qt m c T T
=−

(2)
or,
()
sm
qt UA T
=∆
(3a)
where
()( )()
h,i c h,o c h,i h,o
m
h,ic h,ic
h,o c h,o c
TT T T TT
T
TT TT
nn

TT TT
−− − −
∆= =

−−


−−


(3b)
Substituting (3b) into (3a) and equating to (2),
()
()
h,i h,o
hp,h h,i h,o s
h,i c
h,o c
TT
mc T T UA
TT
n
TT
−
−=

−


−




Hence,
h,i c
s
h,o c h p,h
TT
UA
n
TTmc

−
=


−



or,
()
()
()
h,o c h,i c s h p,h
T t T T T exp UA /m c
=+ − −

(4)
<

Substituting Eqs. (2) and (4) into Eq. (1),
Continued …
PROBLEM 8.28 (Cont.)
()
()
c
c c v,c h p,h h,i c h,i c s h p,h
dT
c m c T T T T exp UA / m c
dt
ρ

∀= −−−−


()
()
hp,h h,i c
c
s h p,h
ccv,c
mc T T
dT
1 exp UA / m c
dt c
ρ
−

=−−


∀


()
()
()
c
c,i
Tt t
hp,h
c
s h p,h
To
ccv,c
ch,i
mc
dT
1 exp UA / m c dt
c
TT
ρ

−=−−

∀
−
∫∫


()

hp,h
ch,i
s h p,h
c,i h,i c c v,c
mc
TT
n1expUA/mct
TT Vc
ρ

−

−=−−



−




()
()
()
hp,h hp,h
ch,ih,ic,i
ccv,c
mc 1exp UA/mc t
Tt T T T exp
c

ρ


−−


=− − −

∀



(5)
<
Eq. (5) may be used to determine T
c
(t) and the result used with (4) to determine T
h,o
(t).
(b) To evaluate the temperature histories, the overall heat transfer coefficient,
()
1
11
oi
Uh h ,
−
−−
=+
must
first be determined. With

()
2
D
Re 4m / D 4 2.4 kg /s / 0.05m 0.002 N s / m 30,600,
π
µ
π
==× ⋅=
the flow
is turbulent and
()()
4/5 0.3
2
iD
k 0.260W / m K
h Nu 0.023 30,600 20 1140W /m K
D 0.05m
⋅

== = ⋅


Hence,
()()
1
22
11
U 1000 1140 W / m K 532W / m K.
−
−−

=+ ⋅= ⋅


As shown below, the temperature of
the pharmaceuticals increases with time due to heat transfer from the hot fluid, approaching the inlet
temperature of the hot fluid (and its maximum possible temperature of 200
°
C) at t = 3600s.
Continued …
0 400 800 12001600 2000 2400 280032003600
Tim e(s)
20
40
60
80
100
120
140
160
180
200
Temperature(C)
Pharmaceutical, Tc
H ot fluid , Th
PROBLEM 8.28 (Cont.)
With increasing T
c
, the rate of heat transfer from the hot fluid decreases (from 4.49
×
10

5
W at t = 0
to 6760 W at 3600s), in which case T
h,o
increases (from 125.2
°
C at t = 0 to 198.9
°
C at 3600s). The
time required for the pharmaceuticals to reach a temperature of T
c
= 160
°
C is
c
t 1266s
=
<
With increasing
h
m,
the overall heat transfer coefficient increases due to increasing h
i
and the hot
fluid maintains a higher temperature as it flows through the tube. Both effects enhance heat transfer
to the pharmaceutical, thereby reducing the time to reach 160
°
C from 2178s for
h
m1kg/s

=
to 906s
at 5 kg/s.
For
2
hDi
1 m 5kg /s, 12,700 Re 63, 700 and 565 h 2050W / m K.
≤≤ ≤ ≤ ≤≤ ⋅
COMMENTS:
Although design changes involving the length and diameter of the coil can be used to
alter the heating rate, process control parameters are limited to
h,i h
Tandm.
1 2 3 4 5
Mass flowrate, m doth (kg/s)
800
1000
1200
1400
1600
1800
2000
2200
Tim e, tc(s )
PROBLEM 8.29
KNOWN:
Tubing with glycerin welded to transformer lateral surface to remove dissipated power.
Maximum allowable temperature rise of coolant is 6
°
C.

FIND:
(a) Required coolant rate
m
, tube length L and lateral spacing S between turns, and (b) Effect of
flowrate on outlet temperature and maximum power.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady-state conditions, (2) All heat dissipated by transformer transferred to
glycerin, (3) Fully developed flow (part a), (4) Negligible kinetic and potential energy changes, (5)
Negligible tube wall thermal resistance.
PROPERTIES:
Table A.5, Glycerin (
m
T

≈
300 K):
ρ
= 1259.9 kg/m
3
, c
p
= 2427 J/kg
⋅
K,
µ
= 79.9
×
10
-

2
N
⋅
s/m
2
, k = 286
×
10
-3
W/m
⋅
K, Pr = 6780.
ANALYSIS:
(a) From an overall energy balance assuming the maximum temperature rise of the
glycerin coolant is 6
°
C, find the flow rate as
()
pm,o m,i
qmcT T
=−

()
()
2
pm,o m,i
m q c T T 1000W 2427J kg K 6K 6.87 10 kg s
−
=−= ⋅=×
<

From Eq. 8.43, the length of tubing can be determined,
()
sm,o
p
sm,i
TT
exp PLh mc
TT
−
=−
−

where P =
π
D. For the tube flow, find
2
D
22
4m 4 6.87 10 kg s
Re 5.47
D
0.020m 79.9 10 N s m
πµ
π
−
−
××
== =
×××⋅


which implies laminar flow, and if fully developed,
D
hD
Nu 3.66
k
==

3
2
3.66 286 10 W m K
h 52.3W m K
0.020m
−
×× ⋅
==⋅
()
()
()
(
)
(
)
22
47 30 C
exp 0.020m 52.3W m K L 6.87 10 kg s 2427J kg K
47 24 C
π
−
−
=− × ⋅× × × ⋅

−



L = 15.3 m. <
The number of turns of the tubing, N, is N = L/(
π
D) = (15.3 m)/
π
(0.3 m) = 16.2 and hence the spacing S
will be
S = H/N = 500 mm/16.2 = 30.8 mm. <
Continued
PROBLEM 8.29 (Cont.)
(b) Parametric calculations were performed using the IHT Correlations Toolpad based on Eq. 8.56 (a
thermal entry length condition), and the following results were obtained.
0.05 0.09 0.13 0.17 0.21 0.25
Mass flowrate, mdot(kg/s)
1000
1400
1800
2200
2600
3000
Heat rate, q(W)

0.05 0.09 0.13 0.17 0.21 0.25
Mass flowrate, mdot(kg/s)
25
27

29
31
33
35
Outlet temperature, Tmo(C)
With T
s
maintained at 47
°
C, the maximum allowable transformer power (heat rate) and glycerin outlet
temperature increase and decrease, respectively, with increasing
m
. The increase in q is due to an
increase in
D
Nu
(and hence
h
) with increasing Re
D
. The value of
D
Nu
increased from 5.3 to 9.4 with
increasing
m
from 0.05 to 0.25 kg/s.
COMMENTS:
Since
()

1
D
D
Gz L D Re Pr
−
=
= (15.3 m/0.02 m)/(5.47
×
6780) = 0.0206 < 0.05,
entrance length effects are significant, and Eq. 8.56 should be used to determine
D
Nu
.
PROBLEM 8.30
KNOWN:
Diameter and length of copper tubing. Temperature of collector plate to which tubing is
soldered. Water inlet temperature and flow rate.
FIND:
(a) Water outlet temperature and heat rate, (b) Variation of outlet temperature and heat rate with
flow rate. Variation of water temperature along tube for the smallest and largest flowrates.
SCHEMATIC:
ASSUMPTIONS:
(1) Straight tube with smooth surface, (2) Negligible kinetic/potential energy and
flow work changes, (3) Negligible thermal resistance between plate and tube inner surface, (4) Re
D,c
=
2300.
PROPERTIES:
Table A.6, water (assume
m

T
= (T
m,i
+ T
s
)/2 = 47.5
°
C = 320.5 K):
ρ
= 986 kg/m
3
, c
p
=
4180 J/kg
⋅
K,
µ
= 577
×
10
-6
N
⋅
s/m
2
, k = 0.640 W/m
⋅
K, Pr = 3.77. Table A.6, water (T
s

= 343 K):
µ
s
=
400
×
10
-6
N
⋅
s/m
2
.
ANALYSIS:
(a) For
m
= 0.01 kg/s, Re
D
= 4
mD
π
µ

= 4(0.01 kg/s)/
π
(0.01 m)577
×
10
-6
N

⋅
s/m
2
=
2200, in which case the flow may be assumed to be laminar. With
fd,t
xD

≈
0.05Re
D
Pr =
0.05(2200)(3.77) = 415 and L/D = 800, the flow is fully developed over approximately 50% of the tube
length. With
()()
1/3
0.14
Ds
Re Pr L D
µµ


= 2.30, Eq. 8.57 may therefore be used to compute the
average convection coefficient
0.14
1/3
D
D
s
Re Pr

Nu 1.86 4.27
LD
µ
µ


==




() ()
2
D
h k D Nu 4.27 0.640W m K 0.01m 273W m K
== ⋅=⋅
From Eq. 8.42b,
2
sm,o
s m,i p
TT
DL 0.01m 8m 273W m K
exp h exp
T T mc 0.01kg s 4180J kg K
ππ


−
××× ⋅


=− =−



−×⋅



()
m,o s s m,i
T T 0.194 T T 70 C 8.7 C 61.3 C
=− − = − =
<
Hence,
()
()()
pm,o m,i
q mc T T 0.01kg s 4186J kg K 36.3K 1519W
=−= ⋅ =

<
(b) The IHT Correlations, Rate Equations and Properties Tool Pads were used to determine the
parametric variations. The effect of
m

was considered in two steps, the first corresponding to
m

<
0.011 kg/s (Re

D
< 2300) and the second for
m

> 0.011 kg/s (Re
D
> 2300). In the first case, Eq. 8.57 was
used to determine
h
, while in the second Eq. 8.60 was used. The effects of
m

are as follows.
Continued
PROBLEM 8.30 (Cont.)
0.005 0.006 0.007 0.008 0.009 0.01 0.011
Mass flowrate, mdot(kg/s)
60
61
62
63
64
65
66
67
Outlet temperature, Tmo(C)
Laminar flow (ReD < 2300)

0.01 0.02 0.03 0.04 0.05
Mass flowrate, mdot(kg/s)

69
69.2
69.4
69.6
69.8
70
Outlet temperature, Tmo(C)
Turbulent flow (ReD>2300)
0.005 0.006 0.007 0.008 0.009 0.01 0.011
Mass flowrate, mdot(kg/s)
800
900
1000
1100
1200
1300
1400
1500
1600
1700
Heat rate, q(W)
Laminar flow (ReD < 2300)

0.01 0.02 0.03 0.04 0.05
Mass flowrate, mdot(kg/s)
1500
3500
5500
7500
9500

Heat rate, q(W)
Turbulent flow (ReD>2300)
The outlet temperature decreases with increasing
m

, although the effect is more pronounced for laminar
flow. If q were independent of
m
, (T
m,o
- T
m,i
) would decrease inversely with increasing
m
. In turbulent
flow, however, the convection coefficient, and hence the heat rate, increases approximately as
0.8
m

,
thereby attenuating the foregoing effect. In laminar flow, q ~
0.5
m

and this attenuation is not as
pronounced.
The temperature distributions were computed from Eq. 8.43, with
h
assumed to be independent of x.
For laminar flow (

m

= 0.005 kg/s),
h
was based on the entire tube length (L = 8 m) and computed from
Eq. 8.57, while for turbulent flow (
m

= 0.05 kg/s) it was assumed to correspond to the value for fully
developed flow and computed from Eq. 8.60. The corresponding temperature distributions are as
follows.
Continued
PROBLEM 8.30 (Cont.)
0 2 4 6 8
Axial location, x(m)
20
30
40
50
60
70
Mean temperature, Tm(C)
mdot = 0.005 kg/s

0 2 4 6 8
Axial location, x(m)
20
30
40
50

60
70
Mean temperature, Tm(C)
mdot = 0.05 kg/s
The more pronounced increase for turbulent flow is due to the much larger value of
h
(4300 W/m
2
⋅
K for
m
= 0.05 kg/s relative to 217 W/m
2
⋅
K for
m

= 0.05 kg/s).
PROBLEM 8.31
KNOWN:
Diameter and surface temperature of ten tubes in an ice bath. Inlet temperature and flowrate
per tube. Volume (
∀
) of container and initial volume fraction, f
v,i
, of ice.
FIND:
(a) Tube length required to achieve a prescribed air outlet temperature T
m,o
and time to

completely melt the ice, (b) Effect of mass flowrate on T
m,o
and suitable design and operating conditions.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady-state, (2) Negligible kinetic/potential energy and flow work changes, (3)
Constant properties, (4) Fully developed flow throughout each tube, (5) Negligible tube wall thermal
resistance.
PROPERTIES:
Table A.4, air (assume
m
T
= 292 K): c
p
= 1007 J/kg
⋅
K,
µ
= 180.6
×
10
-7
N
⋅
s/m
2
, k =
0.0257 W/m
⋅
K, Pr = 0.709; Ice:

ρ
= 920 kg/m
3
, h
sf
= 3.34
×
10
5
J/kg.
ANALYSIS:
(a) With Re
D
= 4
m

/
π
D
µ
= 4(0.01 kg/s)/
π
(0.05 m)180.6
×
10
-7
N
⋅
s/m
2

= 14,100 for
m

=
0.01 kg/s, the flow is turbulent, and from Eq. 8.60,
()()
0.8 0.3
0.8 0.3
D
D
D
Nu Nu 0.023Re Pr 0.023 14,100 0.709 43.3
== = =
() ()
2
D
h Nu k D 43.3 0.0257W m K 0.05m 22.2W m K
== ⋅=⋅
With T
m,o
= 14
°
C, the tube length may be obtained from Eq. 8.42b,
()
(
)
()
2
sm,o
s m,i p

0.05m 22.2W m K L
TT
14 DLh
exp exp
T T 24 mc 0.01kg s 1007J kg K
π
π

⋅

−
−

==− =−



−− ⋅




L = 1.56 m <
The time required to completely melt the ice may be obtained from an energy balance of the form,
() ( )
v,i sf
qt f h
ρ
−=∀
where

()
() ()
p m,i m,o
q Nmc T T 10 0.01kg s 1007J kg K 10K 1007W
=−= ⋅=

. Hence,
(
)
(
)
335
6
0.8 10m 920kg m 3.34 10 J kg
t 2.44 10 s 28.3days
1007W
×
==×=
<
(b) Using the appropriate IHT Correlations and Properties Tool Pads, the following results were
obtained.
Continued
PROPERTIES 8.31 (Cont.)
0 0.01 0.02 0.03 0.04 0.05
Mass flowrate per tube, mdot(kg/s)
12
13
14
15
16

17
Outlet temperature, Tmo(C)
Although heat extraction from the air passing through each tube increases with increasing flowrate, the
increase is not in proportion to the change in
m

and the temperature difference (T
m,i
- T
m,o
) decreases. If
0.05 kg/s of air is routed through a single tube, the outlet temperature of T
m,o
= 16.2
°
C slightly exceeds
the desired value of 16
°
C. The prescribed value could be achieved by slightly increasing the tube length.
However, in the interest of reducing pressure drop requirements, it would be better to operate at a lower
flowrate per tube. If, for example, air is routed through four of the tubes at 0.01 kg/s per tube and the
discharge is mixed with 0.01 kg/s of the available air at 24
°
C, the desired result would be achieved.
COMMENTS:
Since the flow is turbulent and L/D = 31, the assumption of fully developed flow
throughout a tube is marginal and the foregoing analysis overestimates the discharge temperature.
PROBLEM 8.32
KNOWN:

Thermal conductivity and inner and outer diameters of plastic pipe. Volumetric flow rate and inlet
and outlet temperatures of air flow through pipe. Convection coefficient and temperature of water.
FIND:
Pipe length and fan power requirement.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady-state, (2) Negligible heat transfer from air in vertical legs of pipe, (3)
Negligible flow work and potential and kinetic energy changes for air flow through pipe, (4) Smooth
interior surface, (5) Constant properties.
PROPERTIES:
Table A-4, Air (T
m,i
= 29
°
C):
3
i
1.155kg / m .
ρ
=
Air
()
m
T25C:
=°
c
p
= 1007
J/kg
⋅

K,
µ
= 183.6
×
10
-7
N
⋅
s/m
2
, k
a
= 0.0261 W/m
⋅
K, Pr = 0.707.
ANALYSIS:
From Eq. (8.46a)
m,o
s
m,i p
TT
UA
exp
TT mc
∞
∞
−
=−
−





where, from Eq. (3.32),
()
()
1
oi
stot
ii oo
ln D / D
1l
UA R
hDL 2Lk h DL
πππ
−
== + +
With
ii
m 0.0289kg / s
ρ
=∀=
and
Di
Re 4m / D 13,350,
π
µ
==
flow in the pipe is turbulent. Assuming
fully developed flow throughout the pipe, and from Eq. (8.60),

()()
4/5 0.3 24/5 0.3
a
iD
i
k 0.0261W / m K 0.023
h 0.023Re Pr 13,350 0.707 7.20W / m K
D 0.15m
⋅×
== =⋅
()
()
1
s
22
l1 ln0.17/0.15 1
UA
L 2 0.15 W / m K
7.21W / m K 0.15m 1500 W / m K 0.17m
π
ππ
−
=++
×⋅
⋅×× ⋅××




()

s
L
UA 2.335L W / K
0.294 0.133 0.001
==
++
()
m,o
m,i
TT
17 21 2.335L
0.333 exp exp 0.0802
T T 17 29 0.0289kg /s 1007 J / kg K
∞
∞
−
−
===− =−
−− ×⋅



()
ln 0.333
L13.7m
0.0802
=− =
<
From Eqs. (8.22a) and (8.22b) and with
()

2
m,i i i
u/D/41.415m/s,
π
=∀ =
the fan power is
()
()
()
2
3
2
im,i
3
i
i
u
1.155kg / m 1.415m /s
P p f L 0.0294 13.7m 0.025m /s 0.078W
2D 2 0.15m
ρ
=∆ ∀≈ ∀= × =
<
where
1/4
D
f 0.316Re 0.0294
−
==
from Eq. (8.20a).

COMMENTS:
(1) With L/Di = 91, the assumption of fully developed flow throughout the pipe is
justified. (2) The fan power requirement is small, and the process is economical. (3) The resistance
to heat transfer associated with convection at the outer surface is negligible.
PROBLEM 8.33
KNOWN:
Flow rate, inlet temperature and desired outlet temperature of water passing through a tube of
prescribed diameter and surface temperature.
FIND:
(a) Required tube length, L, for prescribed conditions, (b) Required length using tube diameters
over the range 30
≤
D
≤
50 mm with flow rates
m

= 1, 2 and 3 kg/s; represent this design information
graphically, and (c) Pressure gradient as a function of tube diameter for the three flow rates assuming the
tube wall is smooth.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady-state conditions, (2) Negligible potential energy, kinetic energy and flow
work changes, (3) Constant properties.
PROPERTIES:
Table A.6, Water (
m
T
= 323 K): c
p

= 4181 J/kg
⋅
K,
µ
= 547
×
10
-6
N
⋅
s/m
2
, k = 0.643
W/m
⋅
K, Pr = 3.56.
ANALYSIS:
(a) From Eq. 8.6, the Reynolds number is
()
5
D
62
4m 4 2kg s
Re 1.16 10
D
0.04m 547 10 N s m
πµ
π
−
×

== =×
×⋅

.(1)
Hence the flow is turbulent, and assuming fully developed conditions throughout the tube, it follows
from the Dittus-Boelter correlation, Eq. 8.60,
()
()
4/5
4/5 0.4 5 2
0.4
D
k0.643WmK
h 0.023Re Pr 0.023 1.16 10 3.56 6919 W m K
D0.04m
⋅
== ×=⋅
(2)
From Eq. 8.42a, we then obtain
()
()
()
()
poi
2
2kg s 4181J kg K n 25 C 75 C
mc n T T
L 10.6m
Dh
0.04m 6919W m K

π
π
⋅
−∆∆
==− =
⋅
.
<
(b) Using the IHT Correlations Tool, Internal
Flow, for fully developed Turbulent Flow, along
with appropriate energy balance and rate
equations, the required length L as a function of
flow rate is computed and plotted on the right.
30 35 40 45 50
Tube diameter, D (mm)
5
10
15
Tube length, L (m)
Flow rate, mdot = 1 kg/s
mdot = 2 kg/s
mdot = 3 kg/s
Continued
PROBLEM 8.33 (Cont.)
(c) From Eq. 8.22a the pressure drop is
2
m
pu
f
x2D

ρ
∆
=
∆
(4)
The friction factor, f, for the smooth surface condition, Eq. 8.21 with 3000
≤
Re
D

≤
5
×
10
6
, is
()
()
2
D
f 0.790 n Re 1.64
−
=−

(5)
Using IHT with these equations and Eq. (1), the
pressure gradient as a function of diameter for
the selected flow rates is computed and plotted
on the right.
30 40 50

Tube diameter, D (mm)
0
1000
2000
3000
4000
5000
dP/dx (Pa/m)
Flow rate, mdot = 1 kg/s
mdot = 2 kg/s
mdot = 3 kg/s
COMMENTS:
(1) Since L/D = (10.6/0.040) = 265, the assumption of fully developed conditions
throughout is justified.
(2) The IHT Workspace used to generate the graphical results are shown below.
// Rate Equation Tool - Tube Flow with Constant Surface Temperature:
/* For flow through a tube with a uniform wall temperature, Fig 8.7b, the
overall energy balance and heat rate equations are */
q = mdot*cp*(Tmo - Tmi) // Heat rate, W; Eq 8.37
(Ts - Tmo) / (Ts - Tmi) = exp ( - P * L * hDbar / (mdot * cp)) // Eq 8.42b
// where the fluid and constant tube wall temperatures are
Ts = 100 + 273 // Tube wall temperature, K
Tmi = 25 + 273 // Inlet mean fluid temperature, K
Tmo = 75 + 273 // Outlet mean fluid temperature, K
// The tube parameters are
P = pi * D // Perimeter, m
Ac = pi * (D^2) / 4 // Cross sectional area, m^2
D = 0.040 // Tube diameter, m
D_mm = D * 1000
// The tube mass flow rate and fluid thermophysical properties are

mdot = rho * um * Ac
mdot = 1 // Mass flow rate, kg/s
// Correlation Tool - Internal Flow, Fully Developed Turbulent Flow (Assumed):
NuDbar = NuD_bar_IF_T_FD(ReD,Pr,n) // Eq 8.60
n = 0.4 // n = 0.4 or 0.3 for Ts>Tm or Ts<Tm
NuDbar = hDbar * D / k
ReD = um * D / nu
/* Evaluate properties at the fluid average mean temperature, Tmbar. */
Tmbar = Tfluid_avg (Tmi,Tmo)
// Properties Tool - Water:
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
x = 0 // Quality (0=sat liquid or 1=sat vapor)
rho = rho_Tx("Water",Tmbar,x) // Density, kg/m^3
cp = cp_Tx("Water",Tmbar,x) // Specific heat, J/kg·K
nu = nu_Tx("Water",Tmbar,x) // Kinematic viscosity, m^2/s
k = k_Tx("Water",Tmbar,x) // Thermal conductivity, W/m·K
Pr = Pr_Tx("Water",Tmbar,x) // Prandtl number
// Pressure Gradient, Equations 8.21, 8.22a:
dPdx = f * rho * um^2 / ( 2 * D )
f = ( 0.790 * ln (ReD) - 1.64 ) ^ -2
PROBLEM 8.34
KNOWN:
Flow rate and inlet temperature of water passing through a tube of prescribed length,
diameter and surface temperature.
FIND:
(a) Outlet water temperature and rate of heat transfer to water for prescribed conditions, and (b)
Compute and plot the required tube length L to achieve T
m,o
found in part (a) as a function of the surface

temperature for the range 85
≤
T
s

≤
95
°
C.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy,
potential energy and flow work effects, (4) Fully developed flow conditions.
PROPERTIES:
Table A.6, Water (
m
T

≈
325 K): c
p
= 4182 J/kg
⋅
K,
µ
= 528
×
10
-6
N

⋅
s/m
2
, k = 0.645
W/m
⋅
K, Pr = 3.42.
ANALYSIS:
(a) From Eq. 8.6, the Reynolds number is
()
5
D
62
4m 4 2kg s
Re 1.21 10
D
0.04m 528 10 N s m
πµ
π
−
×
== =×
×⋅

.
Hence the flow is turbulent, and assuming fully developed conditions throughout the tube, it follows
from the Dittus-Boelter correlation, Eq. 8.60,
(
)
()

4/5
0.4
4/5 0.4 5 2
D
k 0.645W m K
h 0.023Re Pr 0.023 1.21 10 3.42 7064W m K
D 0.04m
⋅
== ×=⋅
.
From the energy balance relation, Eq. 8.42b,
()
m,o s s m,i
p
DL
TTTTexp h
mc
π

=− − −




(
)
2
m,o
0.04m 4m
T 90 C 90 C 25 C exp 7064 W m K 47.5 C

2kg s 4182J kg K
π
××
=− − − ⋅=
×⋅



<
From the overall energy balance, Eq. 8.37,
()
()
pm,o m,i
q mc T T 2kg s 4182J kg K 47.5 25 C 188kW
=−=×⋅−=

. <
(b) Using the IHT Correlations Tool, Internal Flow, for fully developed Turbulent Flow, along with the
energy balance and rate equations used above, the required length, L, to achieve T
m,o
= 44.9
°
C (see
comment 1 below) as a function of tube surface temperature is computed and plotted below.
Continued
PROBLEM 8.34 (Cont.)
85 90 95
Tube temperature, Ts (C)
40
42

44
46
48
50
Outlet temperature, Tmo (C)
From the plot, the outlet temperature increases nearly linearly with the surface temperature. The
convection coefficient and heat rate show similar behavior for this range of conditions.
COMMENTS:
(1) The mean temperature T
m
= 325 K was overestimated in part (a). Another iteration
is recommended and the results with
m
T
= 309 K are:
h
= 6091 W/m
2
⋅
K, T
m,o
= 44.9
°
C and q = 167
kW.
(2) The IHT Workspace used to generate the graphical results are shown below.
// Rate Equation Tool - Tube Flow with Constant Surface Temperature:
/* For flow through a tube with a uniform wall temperature, Fig 8.7b, the overall energy balance and heat
rate equations are */
q = mdot*cp*(Tmo - Tmi) // Heat rate, W; Eq 8.37

(Ts - Tmo) / (Ts - Tmi) = exp ( - P * L * hDbar / (mdot * cp)) // Eq 8.42b
// where the fluid and constant tube wall temperatures are
Ts = 90 + 273 // Tube wall temperature, K
Ts_C = Ts - 273
Tmi = 25 + 273 // Inlet mean fluid temperature, K
//Tmo = // Outlet mean fluid temperature, K
Tmo_C = Tmo - 273
// The tube parameters are
P = pi * D // Perimeter, m
Ac = pi * (D^2) / 4 // Cross sectional area, m^2
D = 0.040 // Tube diameter, m
D_mm = D * 1000
L = 4 // Tube length, m; unknown
// The tube mass flow rate and fluid thermophysical properties are
mdot = rho * um * Ac
mdot = 2 // Mass flow rate, kg/s
// Correlation Tool - Internal Flow, Fully Developed Turbulent Flow (Assumed):
NuDbar = NuD_bar_IF_T_FD(ReD,Pr,n) // Eq 8.60
n = 0.4 // n = 0.4 or 0.3 for Ts>Tm or Ts<Tm
NuDbar = hDbar * D / k
ReD = um * D / nu
/* Evaluate properties at the fluid average mean temperature, Tmbar. */
Tmbar = Tfluid_avg (Tmi,Tmo)
// Properties Tool - Water:
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
x = 0 // Quality (0=sat liquid or 1=sat vapor)
rho = rho_Tx("Water",Tmbar,x) // Density, kg/m^3
cp = cp_Tx("Water",Tmbar,x) // Specific heat, J/kg·K
mu = mu_Tx("Water",Tmbar,x) // Viscosity, N·s/m^2

nu = nu_Tx("Water",Tmbar,x) // Kinematic viscosity, m^2/s
k = k_Tx("Water",Tmbar,x) // Thermal conductivity, W/m·K
Pr = Pr_Tx("Water",Tmbar,x) // Prandtl number
PROBLEM 8.35
KNOWN:
Diameters and thermal conductivity of steel pipe. Temperature and velocity of water flow
in pipe. Temperature and velocity of air in cross flow over pipe. Cost of producing hot water.
FIND:
Daily cost of heat loss per unit length of pipe.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady state, (2) Constant properties, (3) Negligible radiation from outer
surface, (4) Fully-developed flow in pipe.
PROPERTIES:
Table A-4, air (p = 1 atm, T
f

≈
300K): k
a
= 0.0263 W/m
⋅
K,
ν
a
= 15.89
×
10
-6
m

2
/s,
Pr
a
= 0.707. Table A-6, water (T
m
= 323 K):
ρ
w
= 988 kg/m
3
,
µ
w
= 548
×
10
-6
N
⋅
s/m
2
, k
w
= 0.643
W/m
⋅
K, Pr
w
= 3.56.

ANALYSIS:
The heat loss per unit length of pipe is
()
()
()
mm
11
oi
cnv,w cnd cnv,a
wi ao
p
TT TT
q
ln D /D
RRR
hD hD
2k
ππ
π
∞∞
−−
−−
′
==
′′′
++
++
With
362
D,w w m i w

Re u D / 988kg / m 0.5 m /s 0.084m /548 10 N s / m 75,700,
ρµ
−
==×××⋅=
flow is
turbulent, and for fully developed conditions, the Dittus-Boelter correlation yields
()()
w
0.8 0.3
0.8 0.3 2
w
ww
D
i
k
0.643W / m K
h 0.023Re Pr 0.023 75,700 3.56 2060W / m K
D 0.084m
⋅
== =⋅
With
()
62
D,a o a
Re VD / 3m / s 0.1m /15.89 10 m / s 18,880,
ν
−
==× × =
the Churchill-Bernstein
correlation yields

()
4/5
1/2 1/3
5/8
a
D,a D,w
2
a
a
1/4
o2/3
a
0.62Re Pr
Re
k
h h 0.3 1 20.1W / m K
D 282,000
1 0.4/Pr






== + + = ⋅








+




Hence,
()
()
33 3
50 C 5 C
q 342 W / m 0.342kW / m
1.84 10 0.46 10 158.3610 K / W
−− −
°−−°
′
===
×+×+
The daily energy loss is then
Q 0.346kW / m 24h /d 8.22kW h /d m
′
=×=⋅⋅
and the associated cost is
()()
C 8.22kW h /d m $0.05/ kW h $0.411/m d
′
=⋅⋅ ⋅= ⋅
<
COMMENTS:

Because
cnv,a cnv,w
RR,
′′
>>
the convection resistance for the water side of the pipe
could have been neglected, with negligible error. The implication is that the temperature of the pipe’s
inner surface closely approximates that of the water. If
cnv,w
R
′
is neglected, the heat loss is
q 346 W / m.
′
=
PROBLEM 8.36
KNOWN: Inner and outer diameter of a steel pipe insulated on the outside and experiencing
uniform heat generation. Flow rate and inlet temperature of water flowing through the pipe.
FIND: (a) Pipe length required to achieve desired outlet temperature, (b) Location and value of
maximum pipe temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic
energy, potential energy and flow work changes, (4) One-dimensional radial conduction in pipe wall,
(5) Outer surface in adiabatic.
PROPERTIES: Table A-1, Stainless steel 316 (T ≈ 400K): k = 15 W/m⋅K; Table A-6, Water
(
)
m
T303K:
= c

p
= 4178 J/kg⋅K, k = 0.617 W/m⋅K, µ = 803 × 10
-6
N⋅s/m
2
, Pr = 5.45.
ANALYSIS: (a) Performing an energy balance for a control volume about the inner tube, it follows
that
(
)
( )
(
)
22
pm,om,io
i
m cTTqq/4 DDL
π
−==−
&&
(
)
( )
( )
( ) ( )
( ) ( ) ( )
pm,om,i
22
22
63

o
i
m cTT
0.1 kg/s4178J/kgK20C
L
q/4 DD
10 W/m/4 0.04m0.02m
π
π
−
⋅
==

−
−


o
&
&
L8.87m.
=
<
(b) The maximum wall temperature exists at the pipe exit (x = L) and the insulated surface (r = r
o
).
From Eq. 3.50, the radial temperature distribution in the wall is of the form
( )
2
12

q
TrrCn rC.
4k
=−++
&
l
Considering the boundary conditions;
o
2
o1
oo1
o
rr
qrdTqC
rr: 0 r C
dr2kr2k
=

===−+=


&&
Continued …
PROBLEM 8.36 (Cont.)
( )
22
22
oo
iisi22is
ii

q rq r
qq
rr: TrTrn rC Crn rT.
4k2k4k2k
===−++=−+
&&
&&
ll
The temperature distribution and the maximum wall temperature (r = r
o
) are
( )
(
)
2
22
o
s
i
i
q r
qr
Trrrn T
4k2kr
=−−++
&&
l
( )
(
)

2
22
oo
w,maxoos
i
i
q rrq
TTrrrn T
4k2kr
==−−++
&&
l
where T
s
, the inner surface temperature of the wall at the exit, follows from
( )
(
)
(
)
( )
2222
oo
ii
ssm,o
ii
q/4 DDLqDD
qhTT
DL4 D
π

π
−−
′′
===−
&&
where h is the local convection coefficient at the exit. With
( )
D
62
i
4 m40.1 kg/s
Re7928
D
0.02m80310Ns/m
πµ
π
−
×
===
×⋅
&
the flow is turbulent and, with (L/D
i
) = (8.87 m/0.02m) = 444 >> (x
fd
/D) ≈ 10, it is also fully
developed. Hence, from the Dittus-Boelter correlation, Eq. 8.60,
(
)
( )

4/5
4/50.40.42
D
i
k0.617 W/mK
h0.023 RePr0.02379285.451840 W/mK.
D0.02 m
⋅
===⋅
Hence, the inner surface temperature of the wall at the exit is
(
)
( ) ( )
( )
22
63
22
o
i
sm,o
2
i
10 W/m0.04m0.02m
qDD
TT40C48.2C
4 h D
41840 W/mK0.02m

−
−



=+=+=
×⋅
oo
&
and
( ) ( )
63
22
w,max
10 W/m
T0.02m0.01m
415 W/mK

=−−


×⋅
( )
2
63
10 W/m0.02m
0.02
n48.2C52.4C.
215 W/mK0.01
++=
×⋅
oo
l <

COMMENTS: The physical situation corresponds to a uniform surface heat flux, and T
m
increases linearly with x. In the fully developed region, T
s
also increases linearly with x.
PROBLEM 8.37
KNOWN:
Dimensions and thermal conductivity of concrete duct. Convection conditions of ambient
air. Flow rate and inlet temperature of water flow through duct.
FIND:
(a) Outlet temperature, (b) Pressure drop and pump power requirement, (c) Effect of flow rate
and pipe diameter on outlet temperature.
SCHEMATIC:
ASSUMPTIONS:
(1) Steady-state, (2) Fully developed flow throughout duct, (3) Negligible pipe
wall conduction resistance, (4) Negligible potential energy, kinetic energy and flow work changes for
water, (5) Constant properties.
PROPERTIES:
Table A-6, water
()
m
T360K:
≈

3
967 kg / m ,
ρ
=

p

c 4203J / kg K,
=⋅

6
324 10
µ
−
=×
2
Ns/m,
⋅

w
k 0.674 W / m K, Pr 2.02.
=⋅=
ANALYSIS:
(a) The outlet temperature is given by
()
()
m,o m,i p
TTTTexpUA/mc
∞∞
=+ − −

where
()
()
1
1
tot cnv,w cnd cnv,a

UA R R R R
−
−
==++
()()
()
4
cnd
ln 1.08w / D ln 1.08 0.30m /0.15m
R 8.75 10 K/W
2 kL 2 1.4W / m K 100m
ππ
−
×
== =×
⋅
()
(
)
1
1
24
cnv,a
R 4w Lh 4 0.3m 100m 25W /m K 3.33 10 K / W
−
−
−
==×××⋅=×
With
()

(
)
62
D
Re 4m/ D 4 2kg/s / 0.15m 324 10 N s/m 52,400,
πµ π
−
==××××⋅=

()()
4/5 0.3 2
4/5 0.3
w
wfd D
k 0.674 W / m K 0.023
h h 0.023Re Pr 52,400 2.02 761W / m K
D0.15m
⋅×
≈= = = ⋅
()
(
)
1
1
25
cnv,w w
R DLh 0.15m 100m 761W /m K 2.79 10 K/W
ππ
−
−

−
==×××⋅=×
(
)
1
544
UA 2.79 10 8.75 10 3.33 10 K / W 809W / K
−
−−−

=×+×+× =


m,o
809W / K
T 0 C 90 C exp 81.7 C
2kg /s 4203J /kg K

=°+ ° − = °

×⋅

<
Continued …
PROBLEM 8.37 (Cont.)
(b) With f = 0.0206 from Fig. 8.3 and
2
m
u m / D / 4 0.117 m/ s,
ρ

π
==
()
2
3
2
24
m
967kg / m 0.117m /s
u
p f L 0.0206 100m 91N / m 8.98 10 bars
2D 2 0.15m
ρ
−
∆= = = = ×
×
<
With
33
m / 2.07 10 m / s,
ρ
−
∀= = ×
the pump power requirement is
(
)
233
P p 91N /m 2.07 10 m /s 0.19W
−
=∆ ∀= × =


<
(c) The effects of varying the flowrate and duct diameter were assessed using the IHT software, and
results are shown below.
Although
cnv,w
R,
and hence
tot
R,
decreases with increasing
m
, thereby increasing UA, the effect is
significantly less than that of
m
to the first power, causing the exponential term,
()
p
exp UA / mc ,
−
to
approach unity and
m,o
T
to approach
m,i
T.
The effect can alternatively be attributed to a reduction
in the residence time of the water in the pipe (u
m

increases with increasing
m
for fixed D). With
increasing D for fixed
m
and w,
m,o
T
decreases due to an increase in the residence time, as well as a
reduction in the conduction resistance,
cnd
R.
COMMENTS:
(1) Use of
m
T 360K
=
to evaluate properties of the water for Parts (a) and (b) is
reasonable, and iteration is not necessary. (2) The pressure drop and pump power requirement are
small.
0 1 2 3 4 5
Mass flow rate, mdot(kg/s)
60
65
70
75
80
85
90
Outlet temperature, Tmo(C)

0.05 0.1 0.15 0.2 0.25
Duct diameter, D(m)
75
78
81
84
87
90
Outlet temperature, Tmo(C)
PROBLEM 8.38
KNOWN:
Water flow through a thick-walled tube immersed in a well stirred, hot reaction tank
maintained at 85
°
C; conduction thermal resistance of the tube wall based upon the inner surface area
is
2
cd
R 0.002 m K / W.
′′
=⋅
FIND:
(a) The outlet temperature of the process fluid, T
m,o
; assume, and then justify, fully
developed flow and thermal conditions within the tube; and (b) Do you expect T
m,o
to increase or
decrease if the combined thermal entry condition exists within the tube? Estimate the outlet
temperature of the process fluid for this condition.

SCHEMATIC:
ASSUMPTIONS:
(1) Flow is fully developed, part (a), (2) Constant properties, (3) Negligible
kinetic and potential energy changes and flow work, and (4) Constant wall temperature heating.
PROPERTIES:
Table A-6, Water (T
m
= (T
m,o
+ T
m,i
)/2 = 337 K): c
p
= 4187 J/kg
⋅
K,
µ
= 4.415
×
10
-4
N
⋅
s/m
2
, k = 0.6574 W/m
⋅
K, Pr = 2.80; (T
s
= 358 K):

µ
s
= 3.316
×
10
-4
N
⋅
s/m
2
.
ANALYSIS:
(a) The outlet temperature is determined from the rate equation, Eq. 8.46a, written as
sm,o
s
s m,i p
TT
UA
exp
TT mc

−
=−


−


(1)
where the overall coefficient, based upon the inner surface area of the tube is expressed in terms of

the convection and conduction thermal resistances,
cd,i
11
R
Uh
′′
=+
(2)
To estimate
h,
begin by characterizing the flow
D
Re 4m/ D
πµ
=

(3)
()
42
D
Re 4 33/3600 kg/s / 0.012 m 4.415 10 N s/m 2210
π
−
=×××⋅=
Consider the flow as laminar, and assuming fully developed conditions, estimate
h
with the
correlation of Eq. 8.55,
D
Nu hD/ k 3.66

==
(4)
2
h 3.66 0.6574W / m K /0.012 m 201 W /m K
=× ⋅ = ⋅
From Eq. (2),
1
22 2
U 1/ 201 W /m K 0.002 m K / W 143.1 W /m K
−

=⋅+⋅=⋅


and from Eq. (1), with A
s
=
π
DL, calculate T
m,o
.
Continued …
PROBLEM 8.38 (Cont.)
2
m,o
85 T
143.1 W / m K 0.012 m 8 m
exp
85 20 33/3600 kg /s 4187 J/ kg K
π


−
⋅×× ×

=−

−×⋅

m,o
T64C
=° <
Fully developed flow and thermal conditions are justified if the tube length is much greater than the
fully developed length x
fd,t
. From Eq. 8.23,
fd,t
D
x
0.05 Re Pr
D
=
fd,t
x 0.012 m 0.05 221.0 2.41 3.20 m
=×××=
That is, the length is only twice that required to reach fully developed conditions.
(b) Considering combined entry length conditions, estimate the convection coefficient using the
Sieder-Tate correlation, Eq. 8.56,
()
()
D

D
2/3
D
0.0668 D/ L Re Pr
Nu 3.66
1 0.04 D /L Re Pr
=+

+

(5)
substituting numerical values, find
D
2
Nu 4.05 h 222 W / m K
==⋅
which is a 10% increase over the fully developed analysis result. Using the foregoing relations, find
2
m,o
U 154 W / m K T 65.5 C
=⋅ =° <
COMMENTS:
(1) The thermophysical properties for the fully developed correlation are evaluated at
the mean fluid temperature T
m
= (T
m,o
+ T
m,i
)/2. The values are shown above in the properties

section.
(2) For the Sieder-Tate correlation, the properties are also evaluated at T
m
, except for
µ
s
, which is
evaluated at T
s
.
(3) For this case where the tube length is about twice x
fd,t
, the average heat transfer 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
∆
p for the range of diameters, 0.1
≤
D
≤
0.2 m, maintaining the total surface area, A
s
=
π

DL, 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 (
m
T

≈
310 K, 1 atm):
ρ
= 1.128 kg/m
3
, c
p
= 1007 J/kg
⋅
K,
µ
= 189
×
10
-
7
N
⋅
s/m

2
, k = 0.027 W/m
⋅
K, Pr = 0.706.
ANALYSIS:
(a) With
()
D
72
4m 4 0.04kg s
Re 17,965
D
0.15m 189 10 N s m
πµ
π
−
×
== =
×⋅

the flow is turbulent. Assuming fully developed conditions throughout the tube, it follows from the
Dittus-Boelter correlation, Eq. 8.60, that
()()
4/5 0.4
4/5 0.4 2
D
k 0.027 W m K
h 0.023Re Pr 0.023 17,965 0.706 9.44W m K
D 0.15m
⋅

== =⋅
.
Hence, from the energy balance relation, Eq. 8.42b,
()
m,o s s m,i
p
DL
TTTTexp h
mc
π

=− − −




()
(
)
()
2
m,o
0.15m 10m 9.44W m K
T 15 C 45 Cexp 29.9 C
0.04kg s 1007J kg K
π

⋅

=+ − =


⋅


<
From the overall energy balance, Eq. 8.37, it follows that
()
()
pm,o m,i
q mc T T 0.04kg s 1007J kg K 29.9 60 C 1212W
=−=×⋅−=−

. <
From Eq. 8.22a, the pressure drop is
2
m
u
pf L
2D
ρ
∆=
Continued
PROBLEM 8.39 (Cont.)
and for the smooth surface conditions, Eq. 8.21 can be used to evaluate the friction factor,
()
()
()
()
22
D

f 0.790ln Re 1.64 0.790ln 17,965 1.64 0.0269
−−
=−= −=
Hence, the pressure drop is
()
2
3
2
1.128kg m 2.0m s
p 0.0269 10m 4.03N m
2 0.15m
∆= × =
×
<
where u
m
=
c
mA
ρ

=
(
)
322
0.04kg s 1.128kg m 0.15 m 4 2.0m s
π
×=
.
(b) For the prescribed conditions of part (a), A

s
=
π
DL =
π
(0.15 m)
×
10 m = 4.712 m
2
, 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
∆
p are calculated and
plotted below.
100 125 150 175 200
Tube diameter, D (mm)
5
10
15
Tube length, L (m)

100 125 150 175 200
Tube diameter, D (mm)
0
10
20
30
40
50

deltaP (Pa)
From above, as D increases, L decreases so that
A
s
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
Heat rate, q (kW)
COMMENTS:
(1) To check the calculations, compute q from Eq. 8.44, where
m
T
∆
is given by Eq.
8.45. It follows that
m
T
∆
= -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
m
T
=
318 K:
h
= 9.42 W/m
2
⋅
K, T
m,o
= 303 K = 30
°
C, q = -1211 W, f = 0.0271 and
∆
p = 4.20 N/m
2
.