A Heat Transfer Textbook
Fifth Edition
Solutions Manual for Chapter 1
by
John H. Lienhard IV
and
John H. Lienhard V
Phlogiston
Press
Cambridge
Massachusetts
Professor John H. Lienhard IV
Department of Mechanical Engineering
University of Houston
4800 Calhoun Road
Houston TX 77204-4792 U.S.A.
Professor John H. Lienhard V
Department of Mechanical Engineering
Massachusetts Institute of Technology
77 Massachusetts Avenue
Cambridge MA 02139-4307 U.S.A.
Copyright ©2020 by John H. Lienhard IV and John H. Lienhard V
All rights reserved
Please note that this material is copyrighted under U.S. Copyright Law. The
authors grant you the right to download and print it for your personal use or
for non-profit instructional use. Any other use, including copying, distributing
or modifying the work for commercial purposes, is subject to the restrictions
of U.S. Copyright Law. International copyright is subject to the Berne
International Copyright Convention.
The authors have used their best efforts to ensure the accuracy of the methods,
equations, and data described in this book, but they do not guarantee them for
any particular purpose. The authors and publisher offer no warranties or
representations, nor do they accept any liabilities with respect to the use of
this information. Please report any errata to the authors.
Names: Lienhard, John H., IV, 1930– | Lienhard, John H., V, 1961–.
Title: A Heat Transfer Textbook: Solutions Manual for Chapter 1 /
by John H. Lienhard, IV, and John H. Lienhard, V.
Description: Fifth edition | Cambridge, Massachusetts : Phlogiston
Press, 2020 | Includes bibliographical references and index.
Subjects: Heat—Transmission | Mass Transfer.
Published by Phlogiston Press
Cambridge, Massachusetts, U.S.A.
For updates and information, visit:
http://ahtt.mit.edu
This copy is:
Version 1.0 dated 7 August 2020
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Problem 1.18 A small instrument package is released from a space vehicle. We can
approximate it as a solid aluminum sphere, 4 cm in diameter. The sphere is initially at 303 K
and it contains a pressurized hydrogen component that will condense and malfunction at
30 K. If we approximate outer space to be at 0 K, how long will the instrumentation package
function properly? Is it legitimate to use the lumped-capacity method?
Solution
13
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
1
5
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Problem 1.39 At what minimum temperature does a black radiator have its
maximum monochromatic emissive power in the visible wavelength range? Look
at Fig. 10.2; then describe the difference between what you might see looking at
this object in comparison to looking at the sun.
Solution In accordance with eqn. (1.28), and using λmax.visible = 0.00008 cm,
0.00008 T = 0.2898 cm K. Therefore T = 3623 K
The sun radiates at about 5777 K (see Fig. 10.2). This is a substantially higher
temperature. It also delivers its maximum eλ at a much smaller wavelength – one
at the lower end of the visible range.
Problem 1.40 The local heat transfer coefficient during the laminar flow of fluid
over a flat plate of length L is equal to F/x1/2, where F is some function of fluid
properties and the flow velocity. How does the average heat transfer coefficient
compare with h(x=L) if x is the distance from the leading edge of the plate?
Solution: We use the definition of the average to get:
𝐿
𝐹 1/2
ℎ̅ = ∫0 ℎ 𝑑𝑥 = 2 𝑥 |
𝐿
𝐿
1
𝐿
=2
0
𝐹
√𝐿
= 2 ℎ(𝑥 = 𝐿)
Therefore, the average heat transfer coefficient = ℎ̅ = 2 ℎ(𝑥 = 𝐿)
Problem 1.41 An object is initially at a temperature higher than its surroundings.
We have seen that many kinds of convective processes will bring the object into
equilibrium with its surroundings. Describe the characteristics of a process that
will do so with the smallest net increase in the entropy of the universe.
Solution Entropy is not a path function. Any process connecting the initial to the
final states will yield the same increase of entropy.
26
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Problem 1.44: Room temperature objects at 300 K and the sun at 5777 K each radiate thermal
energy; but Planck’s law, eqn. (1.30), shows that the wavelengths of importance are quite different.
a) Find 𝜆max in micrometers for each of these temperatures from Wien’s Law, eqn. (1.29).
b) Using a spreadsheet or other software, plot eqn. (1.30) for 𝑇 = 300 K as a function of
wavelength from 0 to 50 µm and for 𝑇 = 5777 K for wavelengths from 0 to 5 µm.
c) By numerical integration, find the total area under each of these curves and compare the
value to the Stefan-Boltzmann law, eqn. (1.28). Explain any differences.
d) Show that about 1/4 of the area under each curve is to the left of 𝜆 max (in other words, 3/4
of the energy radiated is on wavelengths greater than 𝜆 max ).
e) What fraction of the energy radiated by the 300 K surface is carried on wavelengths less
than 4 µm? What fraction of the energy radiated by the 5777 K surface is on wavelengths
greater than 4 µm?
Solution.
a)
(
9.9592 µm at 300 K
2987.77 µm · K
=
𝜆 max =
𝑇K
0.5172 µm at 5777 K
b) The plotting and integration can be done in various ways depending upon what software
is used. The results in Fig. 1 are from an Excel spreadsheet with a step size of 0.2 µm at
300 K and of 0.02 µm at 5777 K.
Blackbody Emissive Power [W/m²-μm]
Blackbody Emissive Power [W/m²-μm]
35.0
30.0
25.0
20.0
15.0
10.0
5.0
0.0
9.0E+7
8.0E+7
7.0E+7
6.0E+7
5.0E+7
4.0E+7
3.0E+7
2.0E+7
1.0E+7
0.0E+0
0
10
20
30
40
50
0
Wavelength [μm]
1
2
3
4
5
Wavelength [μm]
(a) 300 K
(b) 5777 K
Figure 1. Plots of Planck’s law at two temperatures
c) Using the values from Excel, a trapezoidal rule integration gives the area under the curve:
(
445.2 W/m2
at 300 K
Integrated area =
7
2
6.283 × 10 W/m at 5777 K
The Stefan-Boltzmann law yields (with 𝜎 = 5.670374 × 10−8 W/m2 K4 )
(
459.3 W/m2
at 300 K
𝜎𝑇 4 =
7
2
6.316 × 10 W/m at 5777 K
28
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
At 5777 K, the integrated value is 99.48% of the Stefan-Boltzmann law, and at 300 K, it
is 96.93%. The principal reason that these values are low is that energy is also radiated at
wavelengths higher than the range of integration.
d) By integrating up to 𝜆 max from part a),
(
∫ 𝜆max
1
28.4% at 300 K
𝑒𝜆,𝑏 (𝑇) 𝑑𝜆 =
4
29.6% at 5777 K
𝜎𝑇 0
These values are bit more than 1/4 of the total energy, but are often stated as “about 1/4”.
e) Similar integrations show that a 300 K surface radiates only 0.33% on wavelengths below
4 µm and that a 5777 K surface radiates 99.0% on wavelengths less than 4 µm (or 1% on
wavelengths above 4 µm). This fact enables the design of materials that selectively absorb
or reflect solar energy (see Section 10.6).
29
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
Problem 1.46: Integration of Planck’s law, eqn. (1.30) over all wavelengths leads to the StefanBoltzmann law, eqn. (1.28). Perform this integration and determine the Stefan-Boltzmann constant
in terms of other fundamental physical constants. Hint: The integral can be written in terms of
Riemann’s zeta function, 𝜁 (𝑠), by using this beautiful relationship between the zeta and gamma
functions
∫ ∞ 𝑠−1
𝑡
𝜁 (𝑠) Γ(𝑠) =
𝑑𝑡
𝑡
0 𝑒 −1
for 𝑠 > 1. When 𝑠 a positive integer, Γ(𝑠) = (𝑠 − 1)! is just a factorial. Further, several values of
𝜁 (𝑠) are known in terms of powers of 𝜋 and can be looked up.
Solution
(1)
∫∞
𝑒 𝑏 (𝑇) =
𝑒𝜆,𝑏 𝑑𝜆
0
∫∞
(2)
(3)
(4)
2𝜋ℎ𝑐2𝑜
=
𝑑𝜆
5
0 𝜆 [exp(ℎ𝑐 𝑜 /𝑘 𝐵𝑇𝜆) − 1]
∫∞
2𝜋ℎ𝜈 3
𝑑𝜈
=
2
0 𝑐 𝑜 [exp(ℎ𝜈/𝑘 𝐵𝑇) − 1]
∫
2𝜋𝑘 4𝐵𝑇 4 ∞ 𝑥 3
=
𝑑𝑥
ℎ3 𝑐2𝑜 0 𝑒 𝑥 − 1
We are given
∫∞
𝜁 (𝑠) Γ(𝑠) =
0
𝑡 𝑠−1
𝑑𝑡
𝑒𝑡 − 1
For our case, 𝑠 = 4 and Γ(4) = 3! = 6. Hence:
(5)
𝑒 𝑏 (𝑇) =
(6)
=
2𝜋𝑘 4𝐵𝑇 4
ℎ3 𝑐2𝑜
12𝜋𝑘 4𝐵
6ℎ3 𝑐2𝑜
𝜁 (4) 3!
𝜁 (4) 𝑇 4
Zeta is a famous function, and the value at 4 has been established to be:
𝜋4
𝜁 (4) =
90
Hence:
!
2𝜋 5 𝑘 4𝐵
(7)
𝑒 𝑏 (𝑇) =
𝑇4
15ℎ3 𝑐2𝑜
(8)
= 𝜎 𝑇4
where we have also found the Stefan-Boltzmann constant in terms of fundamental physical constants.
31
Copyright 2020 John H. Lienhard, IV and John H. Lienhard, V.
A Heat Transfer Textbook
Fifth Edition
Solutions Manual for Chapter 2
by
John H. Lienhard IV
and
John H. Lienhard V
Phlogiston
Press
Cambridge
Massachusetts
Professor John H. Lienhard IV
Department of Mechanical Engineering
University of Houston
4800 Calhoun Road
Houston TX 77204-4792 U.S.A.
Professor John H. Lienhard V
Department of Mechanical Engineering
Massachusetts Institute of Technology
77 Massachusetts Avenue
Cambridge MA 02139-4307 U.S.A.
Copyright ©2020 by John H. Lienhard IV and John H. Lienhard V
All rights reserved
Please note that this material is copyrighted under U.S. Copyright Law. The
authors grant you the right to download and print it for your personal use or
for non-profit instructional use. Any other use, including copying, distributing
or modifying the work for commercial purposes, is subject to the restrictions
of U.S. Copyright Law. International copyright is subject to the Berne
International Copyright Convention.
The authors have used their best efforts to ensure the accuracy of the methods,
equations, and data described in this book, but they do not guarantee them for
any particular purpose. The authors and publisher offer no warranties or
representations, nor do they accept any liabilities with respect to the use of
this information. Please report any errata to the authors.
Names: Lienhard, John H., IV, 1930– | Lienhard, John H., V, 1961–.
Title: A Heat Transfer Textbook: Solutions Manual for Chapter 2 /
by John H. Lienhard, IV, and John H. Lienhard, V.
Description: Fifth edition | Cambridge, Massachusetts : Phlogiston
Press, 2020 | Includes bibliographical references and index.
Subjects: Heat—Transmission | Mass Transfer.
Published by Phlogiston Press
Cambridge, Massachusetts, U.S.A.
For updates and information, visit:
http://ahtt.mit.edu
This copy is:
Version 1.0 dated 7 August 2020
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
2.7 Simplified explanation of the “critical radius” for a nontechnical person.
Suppose you have a hot pipe and want to keep the heat inside. You can
wrap insulating material around it, the same way you wrap clothes around
yourself on a cold day. If the insulating material doesn’t block the heat
very well, or if there’s not much wind to blow the heat away from the
outside surface, then something bad can happen:
The insulation can make the outside of the pipe a lot bigger so the wind
blows more heat away. This means that, in some cases, insulation can help
heat flow. (Of course, that would only be true up to a certain thickness.
Beyond that, insulation will once again keep reducing the heat flow.)
5
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Finally, the percentage of the heat flowing through the brick will be proportional to its relative
resistance. So:
Percentage of heat through brick = R2/(R2 + R3) = 0.197/(0.197 + 0.217)100 = 47.6 %
7
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
2.10
Compute Q and u for the wall shown in Fig. 2.17 if L = 0.3 m,
2
2
= 0.05 m /m into the paper,
Apine
.
A d. = 0.08 m /m, h 1 =
·
s.
2
2
= 10 ° c. Let
10 W/m - 0 c, hr = 18 W/m - 0 c, T
30 ° C and T
i
oor
there be 5 layers each of pine and sawdust.
00
From the example we get the formula for U.
Using the numbers
2
above and A = (0.05 + 0.08) = 0.13 m /m we get:
u =
1 + 1
li3
10
Q = UAL'IT =
2 .11
2.18
1
1
0.14 0.05
0.06 0.08
0.3 0.13 + o":°"3 0.13
m
w
2_ o
2
K
where:
1
2 o
= L _ 0.00 2 m° = 0:000118 m _ c
R
S.S.
k - 17W/rn- C
w
2 o
= k = 0.003 = 0.00000775 m _ c
R
cu
w
387
so
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
8
w
rn
0) ° C =
[5(0.13 �
m )] (30-1
Find U for the wall in Example 1. 2.
u =
= 2.18
w
u = 4103 ___
2o
m
c
2_ o
K
28.3
w
m
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
No
t
i
c
et
ha
tv
e
r
yi
na
c
c
ur
a
t
er
e
s
i
s
t
a
nc
e
sdo
mi
na
t
eb
o
t
ho
ft
he
s
eo
v
e
r
a
l
l
he
a
tt
r
a
ns
f
e
rc
o
e
f
f
i
c
i
e
nt
s
.Wec
o
ul
dno
tr
e
a
s
o
na
b
l
yb
a
s
ea
nyp
r
e
c
i
s
e
c
a
l
c
ul
a
t
i
o
nsup
o
nt
he
m.
10
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.21 Derive an expression for the thermal resistance of a spherical shell
of inner radius ri and outer radius ro.
We give the solution to the heat conduction equation for this case in Problem 2.18.
The resulting heat transfer (considering it to flow from the inside to the outside) is:
Q =4π rorik ΔT/(ro – ri)
The thermal resistance is then
Rt = ΔT/Q = [(ro – ri)/4πrorik] K/W
—————————
17
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.26 We must illuminate a Space Station experiment in a large tank of
water at 20 oC. What is the maximum wattage of a submerged 3 cm diameter
spherical light bulb that will illuminate the tank without boiling the surrounding
water. The bulb is an LED that coverts 70% of the power to light. Bear in mind
that this will occur in zero gravity.
Solution The problem of heat conduction from a spherical cavity in an infinite
medium is solved completely in the solution to Problem 2.5. The result is:
Q = 4πkR(Tcavity wall – Tꝏ) = 4π(0.653)(0.015)(100 – 20) = 9.85 W
where we use k = 0.653 W/m-K at an average water temperature of 60oC. The
power of the bulb is therefore 9.85W/0.30 = 32.8 W
__________________________________________________________________
Problem 2.27 A cylindrical shell is made of two layers: The inner one has an
inner radius ri and an outer radius rc. The outer shell has inner and outer radii of rc
and ro. There is a contact resistance, hc, between the layers. The layers have
different conductivities. T(r = ri) = Ti and T(r = ro) = To. What is the inner
temperature of the outer shell in terms of Ti and To?
Solution
22
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.30 A contact resistance experiment involves pressing two slabs of
different materials together, imposing a given heat flux through them, and
measuring the outside temperatures of each slab. Write the general expression for
hc in terms of known quantities. Then determine hc if: The slabs are 2 cm thick
copper and 1.5 cm thick aluminum, q is 30,000 W/m2, and the two temperatures
are 15oC and 22.1oC.
Problem 2.31 A student working heat transfer problems late at night needs a cup
of hot cocoa to stay awake. She puts milk in a pan on an electric stove. To heat
the milk as fast as possible without burning it, she turns the stove on high and stirs
the milk continuously. Use an analogous electric circuit to explain how this works.
Is it possible to bring the entire bulk of the milk up to the burn temperature without
burning part of it?
_
The student wants the resistance 1/hA to decrease so the temperature drop between
the pan and the milk will stay smaller than Tmilk-burn –Tmilk-bulk. She accomplishes
this by stirring to increase the heat transfer coefficient. But she can do so, only up
to a point. There will always have to be some temperature drop between the pan
wall and the liquid bulk.
Therefore, it will never be possible to bring the milk temperature all the way up to
the burning temperature, without first burning the milk at the bottom of the pan.
24A
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.34 Use data from Fig. 2.3 to create an empirical equation for k(T) in ammonia
vapor. (Be aware that, while the data form a nearly straight line, the coordinates are semilogarithmic. The curve-fit must thus take an exponential form.) Then imagine a hot
horizontal surface parallel to a cold surface a distance H below with ammonia vapor between
them. Derive equations for T(x) and q, with x = 0 at the cold surface and x = H at the hot
surface. Compute q if Thot = 150oC, Tcold = -5oC, and H = 0.15m.
Solution We first seek an equation of the form k = Ae+BT to fit the almost-straight-line lnk
vs. ToC data in the Fig. 2.3 coordinates. Picking two points on the graph, and solving for A
and B, we get k = 0.0213exp(+0.00392T). Then:
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.41: You are in charge of energy conservation at your plant. A 300 m run of 6 in.
O.D. iron pipe carries steam at 125 psig. The pipe hangs in a room at 25°C, with a natural
convection heat transfer coefficient ℎ = 6 W/m2 K. The pipe has an emittance of 𝜀 = 0.65. The
thermal resistances are such that the surface of the pipe will stay close to the saturation temperature
of the steam. (a) Find the effective heat transfer coefficient between the pipe surface and the room,
and the rate of heat loss from this pipe, in kWh/y. (b) It is proposed to add a 2 in. layer of glass
fiber insulation with 𝑘 = 0.05 W/m·K. The outside surface of the insulation has of 𝜀 = 0.7. What
is the rate of heat loss with insulation? (c) If the installed insulation cost is $50/m including labor
and the cost of thermal energy is $0.03/kWh, what is the payback time for adding insulation?
Solution.
a) The pipe loses heat by natural convection and thermal radiation. The saturation temperature
of steam at 125 psig = 140 psia = 0.963 MPa may be found from a steam table: 178.3 °C.
The radiation heat transfer coefficient, with 𝑇𝑚 = (178.3 + 25.0)/2 = 101.6 °C, is
ℎrad = 4𝜀𝜎𝑇𝑚3 = 4(0.65) (5.67 × 10−8 )(101.6 + 273.15) 3 = 7.76 W/m2 K
The effective heat transfer coefficient is ℎeff = ℎconv + ℎrad = 6 + 7.76 = 13.8 W/m2 K.
The annual heat loss is
𝑄 ann = ℎeff Δ𝑇 𝐴(365.25 × 24 h/y)
= (13.8)(178.3 − 25)𝜋(6)(0.0254)(300)(365.25)(24) = 2.66 × 109 Wh/y
= 2.66 × 106 kWh/y
b) The surface temperature of the insulation is not known yet, but it will be much lower than
the bare pipe. If we guess 40 °C, then
ℎrad = 4(0.7)(5.67 × 10−8 )(40 + 273.15) 3 = 4.87 W/m2 K
and ℎeff = 6 + 4.87 = 10.87 W/m2 K. The heat loss is for two thermal resistances in series,
with 𝑟 𝑜 is the radius of the insulated pipe:
1
𝑟𝑜
1
+
ln
ℎeff (2𝜋𝑟 𝑜 𝑙) 2𝜋𝑙 𝑘 𝑟 pipe
1
1
3+2
+
ln
=
(10.87)2𝜋(3 + 2)(0.0254)(300) 2𝜋(300)(0.05)
3
𝑅𝑡total =
= 3.84 × 10−4 + 5.42 × 10−3 = 5.80 × 10−3 K/W
Then
𝑄 ann =
Δ𝑇
178.3 − 25
(365.25)(24) = 232 kWh/y
(365.25)(24) =
𝑅𝑡total
5.80 × 10−3
We must check whether our estimate of the surface temperature for ℎrad is acceptable.
Because the heat flow through the outside resistance equals that through the total resistance,
the fraction of the temperature drop outside the insulation is
𝑇surface − 25 𝑅𝑡outside 3.84 × 10−4
=
=
= 0.0662
178.3 − 25
𝑅𝑡total
5.80 × 10−3
31
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Solving, the surface temperature is 𝑇surface = 35.1 °C. Recalculating with this value gives
ℎeff = 10.65 W/m2 K, which represents a 2% reduction of the outside resistance which itself
amounts to only 6.6% of 𝑅𝑡total . There is no need to repeat the calculation with this slightly
lower resistance.
We note that the increase in outside diameter after adding insulation would lower the
natural convection resistance very slightly. In Chapter 8, we’ll see that ℎconv ∼ 𝐷 −1/4 . For
the present dimensions, ℎconv would decrease by about 12% if insulation were added, making
ℎeff = 9.93 W/m2 K, a 9% reduction of the outside resistance, but still a very small net
decrease (0.66%) in 𝑅𝑡total .
c) The energy savings is nearly 100%: 2.66 × 106 kWh/y.
Value of energy saved = (2.66 × 106 kWh/y)(0.03 $/kWh) = $79,800/y
Cost of insulation = (300 m)(50 $/m) = $15,000/y
Payback time = (15, 000)/(79, 800) yr = 2.26 months
Adding insulation is an excellent investment. NB: We have not included the cost of capital
because the payback time is very short (the interest on $15,000 over two months will not
increase the cost significantly).
32
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.42 A large tank made of thin steel plate contains pork fat at 400oF, which is
being rendered into oil. We consider applying a 3-inch layer of 85% magnesia insulation
to the surface of the tank. The average heat transfer coefficient is 1.5 Btu/hr-ft2-oF for
natural convection on the outside. It is far larger on the inside. The outside temperature
is 70oF. By what percentage would adding the insulation reduce the heat loss?
Solution: We sketch a section of the tank below, with the dimensions converted to SI
units for convenience (See Appendix B for conversion factors). Thus Tinside = 204.4oC,
Toutside = 21.1oC, the outer heat transfer coefficient is 8.518 W/m2K, and the wall
thickness is 0.0762 m. We get the thermal conductivity of 85% magnesia as 0.80 W/m-K
directly from Table A.2.
We assume that we can neglect the resistance of the thin steel tank. We are also
confident that the thermal resistance offered by the inner heat transfer coefficient is
negligible. This leaves us with only two significant thermal resistances. They are the
insulation, if it is present, and the outer heat transfer coefficient:
Therefore, adding insulation would reduce the heat flux by (1561 -171.3)(100)/1558
or 89 percent
33
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.43: The thermal resistance of a cylinder is 𝑅𝑡cyl = (1/2𝜋𝑘𝑙) ln(𝑟𝑜 /𝑟𝑖 ). If 𝑟𝑜 = 𝑟𝑖 + 𝛿,
show that the thermal resistance of a thin-walled cylinder (𝛿 𝑟𝑖 ) can be approximated by that for
a slab of thickness 𝛿. Thus, 𝑅𝑡thin = 𝛿/(𝑘 𝐴𝑖 ), where 𝐴𝑖 = 2𝜋𝑟𝑖 𝑙 is the inside surface area. How
much error is introduced by this approximation if 𝛿/𝑟𝑖 = 0.2? Plot 𝑅𝑡thin /𝑅𝑡cyl as a function of 𝛿/𝑟𝑖 .
Hint: Use a Taylor series.
Solution.
1
1
𝛿
𝑟𝑜
𝑅𝑡cyl =
ln
=
ln 1 +
2𝜋𝑘𝑙
𝑟𝑖
2𝜋𝑘𝑙
𝑟𝑖
The Taylor expansion of ln(1 + 𝑥) around 𝑥 = 0 is
∞
Õ
𝑥𝑛
1
1
(−1) 𝑛−1 = 𝑥 − 𝑥 2 + 𝑥 3 − · · ·
ln(1 + 𝑥) =
𝑛
2
3
𝑛=0
so that
"
#
2
3
1
1 𝛿
1 𝛿
𝛿 1 𝛿
𝑅𝑡cyl =
−
+
−··· ≈
2𝜋𝑘𝑙 𝑟𝑖 2 𝑟𝑖
3 𝑟𝑖
2𝜋𝑘𝑙 𝑟𝑖
for 𝛿 𝑟𝑖
Letting 𝐴𝑖 = 2𝜋𝑟𝑖 𝑙, we find that, for 𝛿 𝑟𝑖 ,
𝑅𝑡cyl ≈ 𝑅𝑡thin ≡
1 𝛿
𝛿
=
2𝜋𝑘𝑙 𝑟𝑖 𝑘 𝐴𝑖
For 𝛿/𝑟𝑖 = 0.2,
𝛿
2𝜋𝑘𝑙 𝑅𝑡cyl = ln 1 +
= 0.1832 · · ·
𝑟𝑖
and
𝛿
= 0.2000
𝑟𝑖
The thin wall approximation is high by 9.7% when 𝛿/𝑟𝑖 = 0.2.
The plot is below. To avoid numerical problems as 𝛿/𝑟𝑖 → 0: a) use a few terms of the Taylor
expansion of ln(1 + 𝛿/𝑟𝑖 ) in the denominator for 𝛿/𝑟𝑖 < 0.15; and b) plot only for 𝛿/𝑟𝑖 ≥ 0.001.
2𝜋𝑘𝑙 𝑅𝑡thin =
𝑅𝑡thin 𝑅𝑡cyl
1.6
1.4
1.2
1
0
0.2
0.4
0.6
𝛿/𝑟𝑖
34
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
0.8
1
Problem 2.44: A Gardon gage measures radiation heat flux by detecting a temperature difference. The gage consists of a circular constantan membrane of radius 𝑅, thickness 𝑡, and thermal
conductivity 𝑘 ct which is joined to a heavy copper heat sink at its edges. When a radiant heat
flux 𝑞 rad is absorbed by the membrane, heat flows from the interior of the membrane to the copper
heat sink at the edge, creating a radial temperature gradient. Copper leads are welded to the center
of the membrane and to the copper heat sink, making two copper-constantan thermocouple junctions. These junctions measure the temperature difference Δ𝑇 between the center of the membrane,
𝑇 (𝑟 = 0), and the edge of the membrane, 𝑇 (𝑟 = 𝑅).
The following approximations can be made:
• The membrane surface has been blackened so that it absorbs all radiation that falls on it.
• The radiant heat flux is much larger than the heat lost from the membrane by convection or
re-radiation. Thus, all absorbed radiation is conducted to the heat sink, and other loses can
be neglected.
• The gage operates in steady state.
• The membrane is thin enough (𝑡 𝑅) that the temperature in it varies only with 𝑟, i.e.,
𝑇 = 𝑇 (𝑟) only.
Solve the following problems.
a) For a fixed heat sink temperature, 𝑇 (𝑅), qualitatively sketch the shape of the temperature
distribution in the membrane, 𝑇 (𝑟), for two heat radiant fluxes 𝑞 rad 1 and 𝑞 rad 2 , where
𝑞 rad 1 > 𝑞 rad 2 .
b) Derive the relationship between the radiant heat flux, 𝑞 rad , and the temperature difference
obtained from the thermocouples, Δ𝑇. Hint: Treat the absorbed radiant heat flux as if it were
a volumetric heat source of magnitude 𝑞 rad /𝑡 W/m3 .
Solution.
Membrane temp.
a) Since heat flows from the center to the edges, the highest temperature will be at 𝑟 = 0. The
slope of the temperature profile must be zero at 𝑟 = 0 by symmetry. The temperatures will
be higher when the radiant flux is greater, except at 𝑟 = 𝑅 where the temperature is fixed.
𝑇 (𝑅)
𝑞 rad 1
𝑞 rad 2
0
𝑟/𝑅
1
b) With the approximations given, the situation can be modeled as one-dimensional, steady heat
conduction in cylindrical coordinates, with the absorbed radiation acting like a volumetric
heat release. The heat release per unit volume of membrane is 𝑞 rad /𝑡. With eqns. (2.11) and
35
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
(2.13), the heat conduction equation is:
0
7
𝑞
¤
1
𝜕𝑇
∇ 2𝑇 + =
𝑘 𝛼𝜕𝑡
0
0
2
2
7
7
𝑞¤
1 𝜕 𝜕𝑇
1 𝜕 𝑇 𝜕 𝑇
𝑞 rad
𝑟
+ 2 2 + 2 = − = −
𝑟 𝜕𝑟 𝜕𝑟
𝑘
𝑘𝑡
𝑟 𝜕𝜃
𝜕𝑧
1 𝑑 𝑑𝑇
𝑞 rad
𝑟
=−
𝑟 𝑑𝑟 𝑑𝑟
𝑘𝑡
Integrating this o.d.e. twice gives
𝑑𝑇
𝑞 rad𝑟 𝐶1
=−
+
𝑑𝑟
2𝑘𝑡
𝑟
2
0
𝑞 rad𝑟
>
𝑇 (𝑟) = −
+
𝐶
1 ln 𝑟 + 𝐶2
4𝑘𝑡
By symmetry, we require that the temperature gradient 𝑑𝑇/𝑑𝑟 = 0 at 𝑟 = 0 (or equivalently,
that the temperature be finite at 𝑟 = 0), so 𝐶1 = 0.
The temperature difference, 𝑇 (0) − 𝑇 (𝑅), follows by subtraction without finding 𝐶2 :
Δ𝑇 =
𝑞 rad 𝑅 2
4𝑘𝑡
36
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.45: You have a 12 oz. (375 mL) can of soda at room temperature (70 °F) that you
would like to cool to 45 °F before drinking. You rest the can on its side on the plastic rods of the
refrigerator shelf. The can is 2.5 inches in diameter and 5 inches long. The can’s emittance is
𝜀 = 0.4 and the natural convection heat transfer coefficient around it is a function of the temperature
difference between the can and the air: ℎ = 2 Δ𝑇 1/4 for Δ𝑇 in kelvin.
Assume that thermal interactions with the refrigerator shelf are negligible and that buoyancy
currents inside the can will keep the soda well mixed.
a) Estimate how long it will take to cool the can in the refrigerator compartment, which is at
40 °F.
b) Estimate how long it will take to cool the can in the freezer compartment, which is at 5 °F.
c) Are your answers for parts a) and b) the same? If not, what is the main reason that they are
different?
Solution. Use a lumped-capacity solution because the liquid in the can is able to circulate,
minimizing internal temperature gradients. Treat the soda as having the properties of water (App. A,
Table A.3).
Radiation and natural convection act in parallel, just as in Example 2.7, and the effective heat
transfer coefficient is the sum of ℎrad and ℎconv . Both depend on the temperature difference between
the can and the surroundings. While a strict solution would numerically integrate eqn. (1.20)
to account for changes in the heat transfer coefficients as the can temperature drops, we will get
sufficient accuracy by evaluating ℎrad and ℎconv at a single, intermediate temperature difference and
applying eqn. (1.22).
The time constant is
𝜌𝑐𝑉
T =
𝐴(ℎrad + ℎconv )
The volume of the can itself, if calculated, would turn out to be 7% greater than liquid volume.
That’s because an empty "ullage" space is left to accommodate expansion. For this calculation, we
use the true volume of the liquid, since it represents almost all of the mass to be cooled. We’ll use
the entire surface of the can as heat loss area, however, without trying to account for the ullage.
𝐴 = 𝜋𝐷 𝐿 + 2(𝜋𝐷 2 /4) = 49.09 in2 = 0.03167 m2
𝜌𝑐𝑉 = (999)(4190)(375 × 10−6 ) = 1570 J/kg
a) In this case we begin at 70 °F = 21.11 °C and end at 45 °F = 7.22 °C, with 𝑇∞ = 40 °F =
4.44 °C. If we choose an intermediate can temperature of (21.11 + 7.22)/2 = 14.17 °C, we
estimate the heat transfer coefficients as
ℎconv = 2 (14.17 − 4.44) 1/4 = 3.53 W/m2 K
and, with 𝑇𝑚 = (14.17 + 4.44)/2 = 9.31 °C,
ℎrad = 4𝜀𝜎𝑇𝑚3 = 4(0.4)(5.67 × 10−8 )(9.31 + 273.15) 3 = 2.04 W/m2 K
The time constant is
T =
1570
= 8900 s
(0.03167)(3.53 + 2.04)
37
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Then, finally,
𝑇 − 𝑇∞
= 𝑒 −𝑡/T
𝑇0 − 𝑇∞
𝑇 − 𝑇∞
7.22 − 4.44
𝑡cool = −T ln
= −(8900 s) ln
𝑇0 − 𝑇∞
21.11 − 4.44
so
= 1.594 × 104 s = 4 hr 26 min
b) Here 𝑇∞ = 5 °F = −15.00 °C, so
ℎconv = 2 (14.17 + 15.00) 1/4 = 4.65 W/m2 K
and, with 𝑇𝑚 = (14.17 − 15.00)/2 = −0.42 °C,
ℎrad = 4𝜀𝜎𝑇𝑚3 = 4(0.4)(5.67 × 10−8 )(−0.42 + 273.15) 3 = 1.84 W/m2 K
1570
= 7638 s
T =
(0.03167)(4.65 + 1.84)
𝑇 − 𝑇∞
7.22 + 15.00
𝑡cool = −T ln
= −(7638 s) ln
𝑇0 − 𝑇∞
21.11 + 15.00
= 3709 s = 1 hr 2 min
c) The time to cool in the freezer is less than 1/4 the time to cool in the refrigerator. The reason
is that the driving temperature difference for heat transfer is substantially larger throughout
the process when cooling in the freezer. The change in the heat transfer coefficients, on the
other hand, lowers the time constant by only about 15%. Of course, one must not forget to
remove the can from the freezer before the liquid solidifies!
Numerical solutions. Runga-Kutta integrations of eqn. (1.20) are shown in Fig. 1. In
the refrigerator, ℎconv becomes smaller as the can approaches the refrigerator temperature. The
numerical solution thus takes about 1000 s longer than the lumped capacity solution (the lumped
answer is low by 6%). In the freezer, the heat transfer coefficients do not vary as much, and the
lumped result is in good agreement with the numerical solution.
1
1
Numerical integration
Lumped capacity model
𝖳(𝗍) − 𝖳∞
𝖳𝗂 = 𝖳∞
0.8
0.6
0.4
Θ=
Θ=
𝖳(𝗍) − 𝖳∞
𝖳𝗂 = 𝖳∞
0.8
desired temperature
0.2
N
desired temperature
N
0.6
L
0.4
Numerical integration
Lumped capacity model
0.2
L
0
0
5,000
10,000
15,000
20,000
0
0
1,000
2,000
3,000
4,000
Time, 𝘵 [sec]
Time, 𝘵 [sec]
(a) Refrigerator
(b) Freezer
Figure 1. Numerical integration with temperature dependent heat transfer coefficients
38
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
5,000
Problem 2.46: An exterior wall of a wood-frame house is typically composed, from outside to
inside, of a layer of wooden siding, a layer glass fiber insulation, and a layer of gypsum wall board.
Standard glass fiber insulation has a thickness of 3.5 inch and a conductivity of 0.038 W/m·K.
Gypsum wall board is normally 0.50 inch thick with a conductivity of 0.17 W/m·K, and the siding
can be assumed to be 1.0 inch thick with a conductivity of 0.10 W/m·K.
a) Find the overall thermal resistance of such a wall (in K/W) if it has an area of 400 ft2 .
b) The effective heat transfer coefficient (accounting for both convection and radiation) on the
outside of the wall is ℎ 𝑜 = 20 W/m2 K and that on the inside is ℎ𝑖 = 10 W/m2 K. Determine
the total thermal resistance for heat loss from the indoor air to the outdoor air. Also obtain
an overall heat transfer coefficient, 𝑈, in W/m2 K.
c) If the interior temperature is 20°C and the outdoor temperature is −5°C, find the heat loss
through the wall in watts and the heat flux in W/m2 .
d) Which of the five thermal resistances is dominant?
e) The wall is held together with vertical wooden studs between the siding and the gypsum.
The studs are spruce, 3.5 in. by 1.5 in. on a 16 in. center-to-center spacing. If the wall is 8 ft
high, by how much do the studs increase 𝑈?
Solution.
a) The wall consists of three thermal resistances in series, each of which is a slab of resistance
𝐿/𝑘 𝐴. The area is
𝐴 = 400 ft2 = (400)(0.3048) 2 m2 = 37.16 m2
Summing the resistances and converting inches to meters gives
𝑅𝑡equiv = 𝑅siding + 𝑅insul + 𝑅gypsum
(1) (0.0254)
(3.5)(0.0254)
(0.5)(0.0254)
=
+
+
(0.1)(37.16) (0.038)(37.16) (0.17)(37.16)
= 6.835 × 10−3 + 6.296 × 10−2 + 2.010 × 10−3
= 0.0718 K/W
b) Now we must add the two convection resistances, 1 ℎ𝐴, in series to 𝑅𝑡equiv :
𝑅𝑡total = 𝑅conv, out + 𝑅equiv + 𝑅conv, in
1
1
=
+ 0.0718 +
(20)(37.16)
(10)(37.16)
= 0.0758 K/W
Next, since 𝑄 = Δ𝑇/𝑅𝑡total = 𝑈 𝐴 Δ𝑇, we have 𝑈 𝐴 = 1/𝑅𝑡total . So
𝑈=
1
1
=
= 0.355 W/m2 K
𝐴 𝑅𝑡total (37.16)(0.0758)
c)
Δ𝑇
20 − (−5)
=
= 330 W
𝑅𝑡total
0.0758
𝑄
330
𝑞=
=
= 8.88 W/m2
𝐴 37.16
𝑄=
39
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
d) The insulation is dominant, contributing to 0.0630 K/W or 83% of the total resistance.
e) An 8 ft high wall of 400 ft2 area has a length of 50 ft or 600 inches. This allows for 600/16 =
37.5 studs, but the spacing is likely reduced at one end of the wall. Take 38 studs. Each stud
has an area of (1.5/12)(8) = 1 ft2 , so the total stud area is 𝐴studs = 38 ft2 with the remaining
insulated area being 𝐴ins = 400 − 38 = 362 ft2 .
We will approximate the configuration as one dimensional heat conduction through parallel
resistances, one for the studs and one for the insulation. The total resistance of the insulated
portion of the wall is a proportion of that calculated in part b):
400
𝑅𝑡ins =
(0.0758) = 0.0838 K/W
362
The total resistance of the stud portion is, with 𝑘 spruce = 0.11 W/m·K from Appendix A
(Table A.2),
1
1
1 1(0.0254) 3.5(0.0254) (0.5)(0.0254)
𝑅𝑡studs =
+
+
+
+
𝐴studs 20
(0.1)
(0.11)
(0.17)
10
1
(0.050 + 0.254 + 0.808 + 0.0747 + 0.100)
=
38(0.3048) 2
= 0.365 K/W
The parallel resistance is found as in Example 2.7:
1
1
=
= 0.0682 K/W
𝑅𝑡total = −1
−1
𝑅𝑡ins + 𝑅𝑡studs 1/0.0838 + 1/0.365
The studs reduced the total thermal resistance. The overall heat transfer coefficient rises to
1
1
𝑈studs =
= 0.395 W/m2 K
=
𝐴𝑅𝑡total (0.0682)(37.16)
an increase of 11%.
40
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.47: The heat conduction equation in Sect. 2.1 includes a volumetric heat release
rate, 𝑞¤. We normally describe heat as a transfer of energy and entropy across a system boundary,
so the notion of volumetric heat release needs some thought. Consider an electrical resistor
carrying a current 𝐼 with a voltage difference of Δ𝑉 in steady state. Electrical work is done on the
resistor at the rate Δ𝑉 · 𝐼.
a) Use eqn. (1.1) to find the rate of heat and entropy flow out of the resistor. Assume that the
resistor’s surface temperature, 𝑇, is uniform. What is the rate of entropy generation, 𝑆¤gen ?
b) Suppose that the resistor dissipates electrical work uniformly within its volume, V, and that
its thermal conductivity is high enough to provide a nearly uniform internal temperature.
What is the volumetric entropy generation rate, 𝑠¤gen ?
c) By considering the net heat leaving a differential volume 𝑑V, use 𝑠¤gen to define the
volumetric heat release rate, 𝑞.
¤
d) If the resistor has a nonuniform internal temperature but a uniform rate of work dissipation,
does the total entropy generation change? Why or why not?
e) If the resistor is insulated, so that no heat flows out, what is the entropy generation rate?
Assume the resistor’s temperature is nearly uniform, starting at 𝑇0 at time 𝑡 = 0 .
Solution.
a) The heat flow 𝑄 into the resistor is related to the work Wk done by the resistor
0
7
𝑑𝑈
𝑄 = Wk + = − Δ𝑉 · 𝐼
𝑑𝑡
So, the heat flow out of the resistor is Δ𝑉 · 𝐼. The entropy leaving the resistor is simply
𝑄 Δ𝑉 · 𝐼
𝑆¤out = − =
𝑇
𝑇
Therefore, the rate of entropy generation in the resistor, by dissipation of electrical work, is
Δ𝑉 · 𝐼
(1)
𝑆¤gen = 𝑆¤out =
𝑇
b) Dividing eqn. (1) by V, the entropy generated per unit volume is
Δ𝑉 · 𝐼
𝑠¤gen =
V𝑇
c) The entropy generated in a differential volume 𝑑V must be the net entropy transfer out of
that volume. The volume’s surface has the local temperature 𝑇, so the net heat flow out is
𝑑𝑄 out = 𝑇 𝑑 𝑆¤out = 𝑇 𝑠¤gen 𝑑V
The apparent rate of “volumetric heat release” is therefore
Δ𝑉 · 𝐼
𝑞¤ ≡ 𝑇 𝑠¤gen =
V
d) The heat flow out of the surface is unchanged, so the rate of entropy flow out of the resistor
is unchanged. Thus, the total rate of entropy generation is unchanged. However, heat
transfer from hotter parts of the resistor is associated with lower entropy transfer (since 𝑇 is
greater), but heat conduction through a temperature difference generates additional entropy,
cf. eqn. (1.7). The net result is the same overall entropy generation rate.
41
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
e) If the resistor is adiabatic, the system is unsteady. With eqn. (1.1),
0
𝑑𝑈
7 = Wk +
𝑄
𝑑𝑡
𝑑𝑈
= Δ𝑉 · 𝐼
𝑑𝑡
Assuming an incompressible resistor with a uniform temperature, eqn. (1.4) gives
𝑑𝑆
1 𝑑𝑈 Δ𝑉 · 𝐼
=
=
𝑑𝑡 𝑇 (𝑡) 𝑑𝑡
𝑇 (𝑡)
The temperature as a function of time can be found with eqn. (1.3)
𝑑𝑈
𝑑𝑇
= 𝑚𝑐
𝑑𝑡
𝑑𝑡
and an easy calculation leads to
Δ𝑉 · 𝐼
𝑇 (𝑡) = 𝑇0 +
𝑡
𝑚𝑐
Because no entropy was transferred in the heating process, all of the entropy change is by
entropy generation, and 𝑆¤gen has the same form as in the steady case.
42
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.48: If an overall temperature difference of Δ𝑇 is imposed on 𝑁 thermal resistances
in series, show that the temperature difference across the 𝑖 th thermal resistance is
𝑅𝑖
Δ𝑇𝑖 = Í𝑁
Δ𝑇
𝑖=1 𝑅𝑖
Solution. The heat flow through the series of resistors is (cf. Fig. 2.18):
Δ𝑇
Δ𝑇
= Í𝑁
𝑄=
𝑅𝑡equiv
𝑖=1 𝑅𝑖
The same heat flows through the 𝑖 th resistance:
Δ𝑇𝑖
𝑅𝑖
Equating these expressions and rearranging leads to the stated result:
Δ𝑇
Δ𝑇𝑖
= Í𝑁
𝑅𝑖
𝑖=1 𝑅𝑖
𝑄=
𝑅𝑖
Δ𝑇𝑖 = Í𝑁
Δ𝑇
𝑖=1 𝑅𝑖
In electric circuit theory, this is called the voltage divider relationship.
43
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 2.49: An electrical resistor is a 1 mm thick annulus of Inconel (Fig. 1). It dissipates
9.4 kW/m. The resistor is insulated on both sides by a 3 mm layer of epoxy (𝑘 𝑒 = 0.5 W/m·K). A
316 stainless steel pipe inside the resistor is cooled internally by flowing water. The pipe is 5 cm
I.D. and 6 cm O.D. A larger pipe forms an annular passage outside the resistor, through which
water also flows; ℎinside = ℎoutside = 1400 W/m2 K. The outer pipe has 8.7 cm I.D. and a 0.5 cm
wall thickness and is wrapped with 2 cm thick glass-fiber pipe insulation, surrounded outside by
ambient air. If the water temperature inside is 47 °C and that outside is 53 °C, find the resistor’s
temperature.
steel
epoxy
Inconel
epoxy
water
steel
water
insulation
Figure 1. Cross-section of resistor with water cooling
Solution. This problem can be solved with two effective resistances, one from the resistor to
the water inside and one from the resistor to the water outside. (The insulation and outer pipe can
be ignored because the outside water temperature is known.) Further, the epoxy thickness is much
smaller than the radius, so it may be treated as a slab (Prob. 2.43).
The internal resistances, in series, may be summed for a 1 m length
𝑅inside = 𝑅epoxy + 𝑅pipe + 𝑅conv
ln(𝑟 𝑜 /𝑟𝑖 )
1
𝑡
=
+
+
[2𝜋(𝑟 𝑜 + 𝛿/2)𝑙] 𝑘 𝑒
2𝜋𝑘 𝑠𝑠 𝑙
(2𝜋𝑟𝑖 𝑙)ℎinside
0.003
ln(0.03/0.025)
1
=
+
+
2𝜋(0.0315)(0.5)
2𝜋(14)
2𝜋(0.025)(1400)
= 3.03 × 10−2 + 2.07 × 10−3 + 4.55 × 10−3 = 3.69 × 10−2 K/W
The epoxy is clearly the dominant resistance. The exterior resistance is
𝑅outside = 𝑅epoxy + 𝑅conv
0.003
1
=
+
2𝜋(0.0355)(0.5) 2𝜋(0.037)(1400)
= 2.69 × 10−2 + 3.07 × 10−3 = 3.00 × 10−2 K/W
44
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Heat leaving the resistor goes into both effective resistances. With 𝑄 = 9.4 kW,
𝑇resistor − 𝑇water, inside 𝑇resistor − 𝑇water, outside
+
𝑄=
𝑅inside
𝑅outside
𝑇water, inside 𝑇water, outside
1
1
+
=
+
+𝑄
𝑇resistor
𝑅inside 𝑅outside
𝑅
𝑅outside
inside
1
1
47
53
𝑇resistor
=
+9400
+
+
3.69 × 10−2 3.00 × 10−2
3.69 × 10−2 3.00 × 10−2
|
{z
} |
{z
}
Solving, 𝑇resistor = 194 °C
3040
64.28
45
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
A Heat Transfer Textbook
Fifth Edition
Solutions Manual for Chapter 3
by
John H. Lienhard IV
and
John H. Lienhard V
Phlogiston
Press
Cambridge
Massachusetts
Professor John H. Lienhard IV
Department of Mechanical Engineering
University of Houston
4800 Calhoun Road
Houston TX 77204-4792 U.S.A.
Professor John H. Lienhard V
Department of Mechanical Engineering
Massachusetts Institute of Technology
77 Massachusetts Avenue
Cambridge MA 02139-4307 U.S.A.
Copyright ©2020 by John H. Lienhard IV and John H. Lienhard V
All rights reserved
Please note that this material is copyrighted under U.S. Copyright Law. The
authors grant you the right to download and print it for your personal use or
for non-profit instructional use. Any other use, including copying, distributing
or modifying the work for commercial purposes, is subject to the restrictions
of U.S. Copyright Law. International copyright is subject to the Berne
International Copyright Convention.
The authors have used their best efforts to ensure the accuracy of the methods,
equations, and data described in this book, but they do not guarantee them for
any particular purpose. The authors and publisher offer no warranties or
representations, nor do they accept any liabilities with respect to the use of
this information. Please report any errata to the authors.
Names: Lienhard, John H., IV, 1930– | Lienhard, John H., V, 1961–.
Title: A Heat Transfer Textbook: Solutions Manual for Chapter 3 /
by John H. Lienhard, IV, and John H. Lienhard, V.
Description: Fifth edition | Cambridge, Massachusetts : Phlogiston
Press, 2020 | Includes bibliographical references and index.
Subjects: Heat—Transmission | Mass Transfer.
Published by Phlogiston Press
Cambridge, Massachusetts, U.S.A.
For updates and information, visit:
http://ahtt.mit.edu
This copy is:
Version 1.0 dated 7 August 2020
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
7
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
8
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
7
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
10
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
(151
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 3.24: We want to cool air from 150oC to 60oC, but we can’t afford a custombuilt heat exchanger. Instead, we find a used cross-flow exchanger in storage. For this
one, both fluids are unmixed. It was previously used to cool 136 kg/min of NH3 from
200oC to 10oC using 320 kg/min of water at 7oC and its U was 480 W/m2K.
How much air can we cool with this unit, using the same water supply, if U is about the
same? (We would actually want to modify U using the methods of Chapters 6 and 7 once
we had a new flow rate of air; but that’s beyond our scope at the moment.)
Solution: We must first evaluate the area, based on the exchanger’s previous service:
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 3.30: Plot 𝑇oil and 𝑇H2 O as a function of position in a very long counterflow heat
exchanger where water enters at 0°C, with 𝐶H2 O = 460 W/K, and oil enters at 90°C, with 𝐶oil = 920
W/K, 𝑈 = 742 W/m2 K, and 𝐴 = 10 m2 . Criticize the design.
Solution. The capacity-rate ratio is
𝐶min 𝐶H2 O 460 1
=
=
=
𝐶max
𝐶oil
920 2
Substituting into eqn. (3.21), we have
1 − exp −(1 − 𝐶min /𝐶max )NTU
1 − exp −NTU/2
=
𝜀=
1 − (𝐶min /𝐶max ) exp −(1 − 𝐶min /𝐶max )NTU
1 − 21 exp −NTU/2
(1)
We find the water temperature from 𝜀 with eqn. (3.16), noting that water is the cold stream and oil
the hot stream:
𝑇H O,out − 𝑇H2 O,in
𝜀= 2
𝑇oil,in − 𝑇H2 O,in
To make the plot, we can imagine that the area and NTU of the exchanger are increasing as we
move from the water inlet to the final outlet. Let’s call the position 𝑥, so that 𝐴(𝑥) increases from
zero to 10 m2 . We then write NTU in terms of 𝐴(𝑥):
𝑈 𝐴(𝑥) 742 𝐴(𝑥)
NTU𝑥 =
=
= 1.613 𝐴(𝑥)
for 0 ≤ 𝐴(𝑥) ≤ 10 m2
𝐶min
460
The effectiveness accumulated up to that position is
1 − exp −0.8065𝐴(𝑥)
1 − exp −NTU𝑥 /2
𝜀𝑥 =
=
1 − 12 exp −NTU𝑥 /2
1 − 12 exp −0.8065𝐴(𝑥)
The water temperature at each location is 𝑇H2 O, 𝑥 . Putting 𝑇H2 O, 𝑥 for 𝑇H2 O,out , 𝜀𝑥 for 𝜀, and
rearranging eqn. (1)
𝑇H2 O, 𝑥 = 𝑇H2 O,in + 𝜀𝑥 𝑇oil,in − 𝑇H2 O,in = 90𝜀𝑥 °C
The local temperature difference is given by eqn. (3.5b), from which the hot stream temperature is:
𝑇H O, 𝑥 90𝜀 𝑥=10
𝐶𝑐
𝐶𝑐
𝑇oil, 𝑥 = 𝑇H2 O,x + 𝑇oil,in − 1 −
𝑇H2 O,x − 𝑇H2 O,out = 90 + 2
−
𝐶ℎ
𝐶ℎ
2
2
The exchanger is more than twice the size needed—right side adds nothing to the performance.
𝑇 (𝑥) [°C ]
80
60
𝑇oil
𝑇water
40
20
0
0
1
2
3
7
4
5
6
Position, 𝐴(𝑥) [m2 ]
19
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
8
9
10
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
22
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
3
Copyright 2020, John H. Lienhard, IV and John H. Lienhard,2
V
Problem 3.36: Both C’s in a parallel-flow heat exchanger are equal to 156 W/K,
U = 327 W/m2 K, and A = 2 m2. The hot fluid enters at 140oC. If we cut both C’s
in half, what will the exit temperature of the hot fluid be?
Solution: NTU = 327(2)/156 = 4.19 and Cmin/Cmax is 1.00. Cutting the C’s in half
will make the NTU still larger. We see, in Fig. 3.16, that ɛ is constant in this NTU
range. Thus, the exiting hot water temperature is unchanged:
Thot-out = 90 oC
_________________________________________________________________
Problem 3.37: A 1.68 ft2 crossflow heat exchanger with one fluid mixed
condenses steam at atmospheric pressure (h = 2000 Btu/ft2-hr-oF) and boils
methanol (Tsat = 170 oF and h = 1500 Btu/ft2-hr-oF) on the other side. Evaluate U
(neglecting the metal’s resistance), F, LMTD, & Q. Can we evaluate NTU and ɛ?
U = [1/2000 + 1/1500]-1 = 857 Btu/ft2-hr-oF
LMTD (per Example 3.2) = Tsteam – Tmethanol = 212 -170 = 42 oF
From Fig. 3.14d, F for P = 0 and any R is equal to 1.0
So, using eqn. (3.14), Q = UAΔTF = (857)(1.68)(42)(1) = 60,470 Btu/hr
NTU and ɛ are not meaningful, since neither Cmin nor Cmax is known or
relevant. Flow rates have no bearing on Q in this case. This configuration is a
simple case of conduction through a wall with two significant resistances.
__________________________________________________________________
24
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 3.38 We can calculate the effectiveness of a crossflow heat exchanger
with neither fluid mixed using the approximate formula:
ɛ ≈ 1 – exp{[exp(-NTU0.78R) – 1][NTU0.22/R]}
where R ≡ Cmin/Cmax. How closely does this correspond to exact results for known
limiting cases? Present results graphically.
Solution As R goes to 0, ɛ therefore approaches 1 – exp(-NTU)
[This is exactly the single stream result, eqn. (3.22)]
We evaluate the equation numerically for R = 1, and compare it with Fig. 3.17a in
the following graph. It shows this approximation to be very good at this value of
R. And it will approach being exactly the same, as R is lowered.
25
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 3.41 To make lead shot, molten droplets of lead are showered into the top of a
tall tower. The droplets fall through air and solidify before they reach the bottom of the tower,
where they are collected. Cool air is introduced at the bottom of the tower and warm air flows
out the top. For a particular tower, 5,000 kg/hr of 2.8 mm diam. droplets are released at their
melting temperature of 600 K. The latent heat of solidification is 23.1 kJ/kg. The dropping pan
size produces 6,700 droplets/m3 in the tower. Air enters the bottom at 20 °C with a mass flow rate
of 2,400 kg/hr. The tower has an internal diameter of 0.6 m with adiabatic walls.
a) Sketch, qualitatively, the temperature distributions of the shot and the air along the height of
the tower.
b) If it is desired to remove the shot at a temperature of 60 °C, what will be the temperature of
the air leaving the top of the tower?
c) Determine the air temperature at the point where the lead has just finished solidifying.
d) Determine the height that the tower must have in order to function as desired. The heat
transfer coefficient between the air and the droplets is ℎ = 170 W/m2 K.
Solution (a) This is a counterflow heat exchanger. The lead shot is the hot stream and the
air is the cold stream. However, while the lead is solidifying, it remains at its freezing temperature.
The small metal beads will be isothermal (a calculation shows that the Biot number is 1).
Temperature [K]
700
600
Lss
Lts
600
Molten
500
Solid lead
400
Air
333
293
Bottom
300
Top
Position
(b) The energy given up by the lead goes to the air.
( 𝑚𝑐
¤ 𝑝 )air (𝑇air,𝑜 − 𝑇air,𝑖 ) = 𝑚¤ lead ℎ 𝑠𝑓 + 𝑐𝑝,lead (𝑇lead,𝑖 − 𝑇lead,𝑜 )
(2400)(1008)(𝑇air,𝑜 − 20) = (5000) [23100 + (148)(600 − 333)]
Solving, 𝑇air,𝑜 = 149.4 °C.
(c) We may consider the air between the freezing point and the outlet:
( 𝑚𝑐
¤ 𝑝 )air (𝑇air,𝑜 − 𝑇air,solid ) = 𝑚¤ lead ℎ 𝑠𝑓
(2400)(1008)(149.4 − 𝑇air, solid ) = (5000)(23100)
Solving, 𝑇air,solid = 101.7 °C.
(d) The height must be determined in two pieces. One is a single stream exchanger with lead at
its freezing temperature. The other is a counterflow exchanger for the section in which the solid
27
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
lead cools. For each section, we need to find the NTU that gives that section the effectiveness
needed to attain the indicated temperatures. The NTU determines the contact area and therefore
the height of each section.
For both sections the overall heat transfer coefficient is simply the air-side heat transfer coefficient,
since the lead beads have negligible thermal resistance: 𝑈 = 170 W/m2 K. The surface area per
meter height is
𝜋
𝐴 = 6700 𝜋(0.0028) 2 (0.6) 2 = 0.0467 m2 /m
4
For the single-stream side, with 𝐶min = 𝐶air , the effectiveness is
𝑇air,𝑜 − 𝑇air,solid 149.4 − 101.7
=
= 0.1554
𝜀=
𝑇lead,𝑖 − 𝑇air,𝑖
600 − 293
We can solve eqn. (3.22) for NTU = 0.1688. If the single-stream side has length 𝐿 ss
𝑈𝐴
(170)(0.0467)𝐿 ss
NTUss =
=
= 0.1688
𝐶min (2400/3600)(1008)
so that 𝐿 ss = 14.3 m.
For the two-stream section, 𝐶min = 𝐶lead , and
𝑇lead,𝑖 − 𝑇lead,𝑜 600 − 333
𝜀=
= 0.870
=
𝑇lead,𝑖 − 𝑇air,𝑖
600 − 293
With 𝐶min /𝐶max = (5000)(148)/(2400)(1008) = 0.306, we may read from Fig. 3.16: NTU = 2.5.
(Substitution into eqn. (3.21) confirms this result.) Solving
𝑈𝐴
(170)(0.0467)𝐿 ts
NTUts =
=
= 2.5
𝐶min (5000/3600)(148)
so that 𝐿 ts = 64.7 m.
The total height of the tower, 𝐿, is
𝐿 = 𝐿 ss + 𝐿 ts = 14.3 + 64.7 = 79 m
28
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 3.42 The entropy change per unit mass of a fluid taken from temperature 𝑇𝑖 to
temperature 𝑇𝑜 at constant pressure is 𝑠 𝑜 − 𝑠𝑖 = 𝑐 𝑝 ln(𝑇𝑜 /𝑇𝑖 ) in J/K·kg. (a) Apply the Second Law of
Thermodynamics to a control volume surrounding a counterflow heat exchanger to determine the
rate of entropy generation, 𝑆¤gen , in W/K. (b) Write 𝑆¤gen /𝐶min as a function of 𝜀, the heat capacity
rate ratio, and 𝑇ℎ,𝑖 /𝑇𝑐,𝑖 . (c) Show (e.g., by plotting) that 𝑆¤gen /𝐶min is minimized when 𝐶min = 𝐶max
(balanced counterflow) for fixed values of 𝜀 and 𝑇ℎ,𝑖 /𝑇𝑐,𝑖 .
Solution (a) The entropy generation is just the difference between the entropy carried out by
the flows and the entropy carried in by the flows, so
𝑇
𝑇
𝑇
𝑇
𝑐,𝑜
ℎ,𝑜
𝑐,𝑜
ℎ,𝑜
+ ( 𝑚𝑐
¤ 𝑝 )𝑐 ln
= 𝐶ℎ ln
+ 𝐶𝑐 ln
𝑆¤gen = 𝑚¤ ℎ Δ𝑠 ℎ + 𝑚¤ 𝑐 Δ𝑠𝑐 = ( 𝑚𝑐
¤ 𝑝 ) ℎ ln
(1)
𝑇ℎ,𝑖
𝑇𝑐,𝑖
𝑇ℎ,𝑖
𝑇𝑐,𝑖
Note that the entropy of the hot stream decreases.
(b) We’ll need to distinguish between the cases when 𝐶ℎ > 𝐶𝑐 = 𝐶min and 𝐶𝑐 > 𝐶ℎ = 𝐶min when
using eqn. (3.16). For 𝐶ℎ > 𝐶𝑐 :
𝐶𝑐
𝑇ℎ,𝑜 = 𝑇ℎ,𝑖 − 𝜀 (𝑇ℎ,𝑖 − 𝑇𝑐,𝑖 )
𝐶ℎ
and
𝑇𝑐,𝑜 = 𝑇𝑐,𝑖 + 𝜀(𝑇ℎ,𝑖 − 𝑇𝑐,𝑖 )
Substituting into eqn. (1) gives
𝑆¤gen 𝐶ℎ
𝑇𝑐,𝑖
𝑇ℎ,𝑖
𝐶𝑐
=
ln 1 − 𝜀
1−
+ ln 1 + 𝜀
−1
(2)
𝐶min 𝐶𝑐
𝐶ℎ
𝑇ℎ,𝑖
𝑇𝑐,𝑖
For the case 𝐶ℎ < 𝐶𝑐 :
𝑇ℎ,𝑜 = 𝑇ℎ,𝑖 − 𝜀(𝑇ℎ,𝑖 − 𝑇𝑐,𝑖 )
and
𝑇𝑐,𝑜 = 𝑇𝑐,𝑖 + 𝜀
𝐶ℎ
(𝑇ℎ,𝑖 − 𝑇𝑐,𝑖 )
𝐶𝑐
Substituting into eqn. (1) as before gives
𝑆¤gen
𝑇𝑐,𝑖
𝐶𝑐
𝐶ℎ 𝑇ℎ,𝑖
= ln 1 − 𝜀 1 −
+
ln 1 + 𝜀
−1
𝐶min
𝑇ℎ,𝑖
𝐶ℎ
𝐶𝑐 𝑇𝑐,𝑖
Both eqn. (2) and (3) have the form:
𝑆¤gen
𝐶ℎ 𝑇ℎ,𝑖
= fn 𝜀, ,
𝐶min
𝐶𝑐 𝑇𝑐,𝑖
(c) For 𝑅 = 𝐶min /𝐶max . Then 0 < 𝑅 ⩽ 1. For 𝐶ℎ > 𝐶𝑐 :
𝑆¤gen
𝑇𝑐,𝑖
𝑇ℎ,𝑖
1
= ln 1 − 𝑅 𝜀 1 −
−1
+ ln 1 + 𝜀
𝐶min 𝑅
𝑇ℎ,𝑖
𝑇𝑐,𝑖
| {z } |
{z
}
≡𝑎, constant>0
1
= ln(1 − 𝑎𝑅) + 𝑏
𝑅
29
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
≡𝑏, constant>0
(3)
The easiest way to think about this function is to plot the first term for a few values of 𝑎, noting that
0 ⩽ 𝑎 ⩽ 1. From the following plot, it’s clear that the lowest values of 𝑆¤gen /𝐶min will be at 𝑅 = 1.
1
𝑎=0
ln(1 − 𝑎𝑅)/𝑅
0
−1
𝑎=1
−2
−3
−4
0
0.1
0.2
0.3
0.4
0.5
𝑅
0.6
0.7
0.8
0.9
1
For 𝐶ℎ < 𝐶𝑐 :
𝑆¤gen
𝑇𝑐,𝑖
𝑇ℎ,𝑖
1
= ln 1 − 𝜀 1 −
+ ln 1 + 𝑅 𝜀
−1
𝐶min
𝑇ℎ,𝑖
𝑅
𝑇𝑐,𝑖
|
{z
}
| {z }
≡𝑐, constant<0
≡𝑑, constant>0
1
ln(1 + 𝑅𝑑)
𝑅
Plotting the relevant part of this expression, we see that it also has a minimum at 𝑅 = 1.
=𝑐+
ln(1 + 𝑅𝑑)/𝑅
3
2
𝑑 = 2.25
1
0
𝑑=0
−1
0
0.1
0.2
0.3
0.4
30
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
0.5
𝑅
0.6
0.7
0.8
0.9
1
Another way to approach this is to plot 𝑆¤gen /𝐶min for fixed 𝑇ℎ,𝑖 /𝑇𝑐,𝑖 and several values of 𝜀.
0.10
1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
𝑆¤gen /𝐶min
𝜀=1
0.05
0.00
𝜀=0
𝑇ℎ,𝑖 /𝑇𝑐,𝑖 = 1.5, 𝐶𝑐 < 𝐶ℎ
−0.05
0
0.1
0.2
0.3
0.4
0.5
𝑅
0.6
0.7
0.8
0.9
0.10
𝑆¤gen /𝐶min
𝜀=1
0.05
0.00
𝜀=0
𝑇ℎ,𝑖 /𝑇𝑐,𝑖 = 1.5, 𝐶𝑐 > 𝐶ℎ
−0.05
0
0.1
0.2
0.3
0.4
0.5
𝑅
0.6
0.7
0.8
0.9
We can see clearly that balancing the exchanger, so that 𝐶max = 𝐶min , minimizes the entropy
generation rate for a given effectiveness and given inlet temperatures. (Note that fixing the effectiveness does not fix the size of the heat exchanger: for any operating point, the exchanger would
need to be sized so as to provide the desired effectiveness.)
Reference: G.P. Narayan, J.H. Lienhard V, and S.M. Zubair, “Entropy Generation Minimization
of Combined Heat and Mass Transfer Devices,” Int. J. Thermal Sciences, 49(10):2057-2066, Oct.
2010.
31
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 3.43 Entropy generation in a power cycle lowers efficiency relative to the Carnot
efficiency. Heat exchangers contribute to this loss. As seen in Problem 3.42, balanced counterflow
heat exchangers can help to limit entropy generation. Let’s look at the entropy generation of a
balanced exchanger.
a) Let Δ𝑇 = 𝑇ℎ − 𝑇𝑐 𝑇c, in (in kelvin). Show that the entropy generation rate in a small area
𝑑𝐴 = 𝑃𝑑𝑥 (with 𝑃 the perimeter) of the exchanger is
1
1
𝑈𝑃Δ𝑇 2
00
¤
𝑑 𝑆gen = 𝑑𝑄
−
≃
𝑑𝑥
𝑇𝑐 𝑇ℎ
𝑇𝑐2
b) Show that the total entropy generation rate is
Δ𝑇
𝑆¤gen ≃ 𝑄
𝑇h,in 𝑇c,in
c) If a fixed heat load, 𝑄, needs to be transferred, how can entropy generation be reduced?
Discuss how cost and other considerations affect your answer.
Solution (a) Equation (1.7) gives the rate entropy generation when heat flow from one
temperature to another. For a heat transfer per unit area of 𝑑𝑄 (W/m2 ) going from 𝑇ℎ to 𝑇𝑐 ,
eqn. (1.7) becomes
1
1
00
¤
−
𝑑 𝑆gen = 𝑑𝑄
𝑇𝑐 𝑇ℎ
If Δ𝑇 = 𝑇ℎ − 𝑇𝑐 and Δ𝑇 𝑇c, in , then
1
1
1
Δ𝑇
1
1
1
−
−
− (1 − Δ𝑇/𝑇𝑐 + · · · ) ≃ 2
=
=
𝑇𝑐 𝑇ℎ
𝑇𝑐 𝑇𝑐 + Δ𝑇
𝑇𝑐 𝑇𝑐
𝑇𝑐
With 𝑑𝑄 = 𝑈Δ𝑇 𝑑𝐴 from eqn. (3.2) and 𝑑𝐴 = 𝑃𝑑𝑥 , we get
Δ𝑇 𝑈𝑃Δ𝑇 2
=
𝑑𝑥
(1)
𝑇𝑐2
𝑇𝑐2
(b) For a balanced counterflow exchanger, the temperature difference between hot and cold
streams is constant through the whole length of the exchanger (because 𝐶𝑐 = 𝐶ℎ , the streams have
the same temperature change in response to heat transfer between them). Therefore, the temperature
varies as a straight line from inlet (𝑥 = 0) to outlet (𝑥 = 𝐿) with a slope 𝑎 = (𝑇𝑐,out − 𝑇𝑐,in )/𝐿:
𝑑 𝑆¤00gen ≃ 𝑑𝑄
𝑇𝑐 (𝑥) = 𝑇𝑐,in + 𝑎𝑥/𝐿
Putting this into eqn. 1 and integrating from 0 to 𝐿
∫𝐿
2
2
𝑈𝑃Δ𝑇
𝑈𝑃Δ𝑇
1
1
𝑄Δ𝑇
𝑄Δ𝑇
𝑆¤gen ≃
𝑑𝑥 =
−
=
≃
(2)
2
𝑎
𝑇𝑐,in 𝑇𝑐,out
𝑇𝑐,in𝑇𝑐,out
𝑇𝑐,in𝑇ℎ,in
0 (𝑇𝑐,in + 𝑎𝑥/𝐿)
where the last step follows from 𝑇ℎ,in = 𝑇𝑐,out + Δ𝑇 ≃ 𝑇𝑐,out . (Recall from Example 3.2 that a
balanced exchanger has LMTD = Δ𝑇 so that 𝑄 = 𝑈 𝐴 Δ𝑇.)
(c) From eqn. (2), entropy generation can only be reduced if Δ𝑇 is reduced. Holding 𝑄 = 𝑈 𝐴 Δ𝑇
fixed, a reduction in Δ𝑇 can be achieved by increasing the area or by increasing 𝑈. Increasing the
area means a larger heat exchanger and greater capital cost. We’ll see in Chapter 7 that increasing
𝑈 at fixed flow rate generally means roughening the surface or reducing the size fluid passages; but
those changes can increase pressure drop, susceptibility to fouling, and/or cost.
32
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 3.44 Water at 100 °C flows into a bundle of 30 copper tubes. The tubes are 28.6 mm
O.D. and 3 m long with a wall thickness of 0.9 mm. Air at 20 °C flows into the bundle, perpendicular
to the tubes. The mass flow rate of water is 17 kg/s and that of air is 25 kg/s. (a) Determine the
outlet temperature of the water if ℎwater = 7200 W/m2 K and ℎair = 110 W/m2 K. (b) To improve
the heat removal, aluminum fins are placed on the outside of the tubes (see Fig. 3.6b). The surface
area of the fins and tubes together is now 81 m2 . Explain in words why the fins improve heat
removal. If the conduction resistance of the fins is small and ℎair is unchanged, what is the new
outlet temperature of the water? Hint: See Problem 3.38.
Solution
(a) This cross-flow heat exchanger has the water stream unmixed. First find 𝑈 𝐴:
𝑈 𝐴 = (𝑅water + 𝑅tube + 𝑅air ) −1
1
1
𝑡
! −1
+
+
ℎwater (30𝜋𝐷 𝑖 𝐿) 30𝑘 𝜋𝐷 𝐿
ℎair (30𝜋𝐷 𝑜 𝐿)
−1
0.0009
1
1
+
+
= 30𝜋(3)
7200(0.0268) (396)(0.0277) (110)(0.0286)
−1
= 874.9 W/K
= 282.7 5.182 × 10−3 + 8.205 × 10−5 + 3.179 × 10−1
=
Observe that the air-side resistance is the largest by two orders of magnitude, and that the tube wall
resistance is entirely negligible.
We have 𝐶ℎ = ( 𝑚𝑐
¤ 𝑝 )H2 O = (17)(4210) = 7.16 × 104 W/K and 𝐶𝑐 = ( 𝑚𝑐
¤ 𝑝 )air = (25)(1007) =
4
2.52 × 10 W/K. The NTU is
𝑈𝐴
874.9
NTU =
=
= 0.0347
𝐶𝑐
2.52 × 104
This is very, very low! Figure 3.17b shows that 𝜀 will be tiny. Neither stream will experience much
change in temperature, and the water leaves at about 100 °C.
(b) Fins are added in order to increase the area on the air-side, thereby lowering the air-side
thermal resistance. If the conduction resistance of the fins is negligible, they are isothermal at the
tube surface temperature. (In Section 4.5, we show how to calculate a fin’s conduction resistance.)
−1
1
0.0009
1
𝑈𝐴 =
+
+
(7200)30𝜋(0.0268)(3) (396)30𝜋(0.0277)(3) (110)(81)
−1
= 1.833 × 10−5 + 2.902 × 10−7 + 1.122 × 10−4 +
= 7, 642 W/K
The NTU is
𝑈𝐴
7642
=
= 0.303
𝐶𝑐
2.52 × 104
In this case, neither stream is mixed, and 𝑟 = 𝐶min /𝐶max = 0.352. With Fig. 3.17a, 𝜀 ≈ 0.25. More
precisely, we may use the equation given in Problem 3.38:
𝜀 = 1 − exp exp −(0.303) 0.78 (0.352) − 1 (0.303) 0.22 (0.352) = 0.246
NTU =
Using eqn. (3.16), we find
𝑇water, out = 𝑇ℎin − 𝜀
𝐶𝑐
(𝑇ℎ − 𝑇𝑐in ) = 100 − (0.246)(0.352)(100 − 20) = 93.1 °C
𝐶ℎ in
33
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
A Heat Transfer Textbook
Fifth Edition
Solutions Manual for Chapters 4–11
by
John H. Lienhard IV
and
John H. Lienhard V
Phlogiston
Press
Cambridge
Massachusetts
Professor John H. Lienhard IV
Department of Mechanical Engineering
University of Houston
4800 Calhoun Road
Houston TX 77204-4792 U.S.A.
Professor John H. Lienhard V
Department of Mechanical Engineering
Massachusetts Institute of Technology
77 Massachusetts Avenue
Cambridge MA 02139-4307 U.S.A.
Copyright ©2020 by John H. Lienhard IV and John H. Lienhard V
All rights reserved
Please note that this material is copyrighted under U.S. Copyright Law. The
authors grant you the right to download and print it for your personal use or
for non-profit instructional use. Any other use, including copying, distributing
or modifying the work for commercial purposes, is subject to the restrictions
of U.S. Copyright Law. International copyright is subject to the Berne
International Copyright Convention.
The authors have used their best efforts to ensure the accuracy of the methods,
equations, and data described in this book, but they do not guarantee them for
any particular purpose. The authors and publisher offer no warranties or
representations, nor do they accept any liabilities with respect to the use of
this information. Please report any errata to the authors.
Names: Lienhard, John H., IV, 1930– | Lienhard, John H., V, 1961–.
Title: A Heat Transfer Textbook: Solutions Manual for
Chapters 4–11 / by John H. Lienhard, IV, and John H.
Lienhard, V.
Description: Fifth edition | Cambridge, Massachusetts : Phlogiston
Press, 2020 | Includes bibliographical references and index.
Subjects: Heat—Transmission | Mass Transfer.
Published by Phlogiston Press
Cambridge, Massachusetts, U.S.A.
For updates and information, visit:
http://ahtt.mit.edu
This copy is:
Version 1.01 dated 3 December 2020
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
The fin efficiency, ηf = tanh(mL)/mL = 0.8913/1.428 = 0.624 = 62.4%
The fin effectiveness, є = ηf (fin surface area)/fin cross-sectional area
є = 0.624(2πrL/π r2) = 1.248L/r = 25
18
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
= 0.836m
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 5.33 A lead bullet travels for 0.5 seconds within a shock wave that heats
the air near the bullet to 300oC. Approximate the bullet as a cylinder 0.8 mm in
diameter. What is its surface temperature at impact if h = 600 W/m2K and if the
bullet was initially at 20oC? What is its center temperature?
Solution The Biot number 600(0.004)/35 = 0.0685, so we can first try the lumped
capacity approximation. See eqn. (1.22):
(Tsfc – 300)/(20 – 300) = exp(-t/T), where T = mc/hA
So T = ρc(area)/h(circumf.) = 11,373(130)π(0.004)2/hπ(0.008) = 4.928 seconds
And (Tsfc – 300)/(20 – 300) = exp(−0.5/4.928).
So Tsfc = 300 - 0.903(280) = 47.0oC
In accordance with the lumped capacity assumption,
47.0oC is also the center temperature.
Now let us see what happens when we use the exact graphical solution, Fig. 5.8:
for Fo = αt/ro2 = 2.34(10−5)(0.5)/0.0042 = 0.731 and r/ro = 1, we get:
(Tsfc – 300)/(20 – 300) = 0.90,
And at r/ro = 0, (Tctr – 300)/(20 – 300) = 0.92,
So Tsfc = 48.0oC
& Tctr = 42.4oC
We thus have good agreement within the limitations of graph-reading accuracy. It
also appears that the lumped capacity assumption is accurate within around 6
degrees in this situation.
132b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 5.52 Suppose that 𝑇∞ (𝑡) is the time-dependent temperature of the environment
surrounding a convectively-cooled, lumped object.
a) When 𝑇∞ is not constant, show that eqn. (1.19) leads to
(𝑇 − 𝑇∞ )
𝑑𝑇∞
𝑑
(𝑇 − 𝑇∞ ) +
=−
𝑑𝑡
T
𝑑𝑡
where the time constant T is defined as usual.
b) If the object’s initial temperature is 𝑇𝑖 , use either an integrating factor or Laplace transforms
to show that 𝑇 (𝑡) is
∫𝑡
−𝑡/T
𝑑
−𝑡/T
𝑇 (𝑡) = 𝑇∞ (𝑡) + 𝑇𝑖 − 𝑇∞ (0) 𝑒
−𝑒
𝑒 𝑠/T 𝑇∞ (𝑠) 𝑑𝑠
𝑑𝑠
0
Solution
a) From eqn. (1.19) for constant 𝑐, with 𝑇∞ (𝑡) not constant:
𝑑 𝑑𝑇
𝜌𝑐𝑉 (𝑇 − 𝑇ref ) = 𝑚𝑐
−ℎ𝐴(𝑇 − 𝑇∞ ) =
𝑑𝑡
𝑑𝑡
𝑑 (𝑇 − 𝑇∞ )
𝑑𝑇∞
= 𝑚𝑐
+ 𝑚𝑐
𝑑𝑡
𝑑𝑡
Setting T ≡ 𝑚𝑐 ℎ𝐴 and rearranging, we obtain the desired result:
𝑑
(𝑇 − 𝑇∞ )
𝑑𝑇∞
(𝑇 − 𝑇∞ ) +
=−
𝑑𝑡
T
𝑑𝑡
(1)
b) The integrating factor for this first-order o.d.e. is 𝑒 𝑡/T . Multiplying through and using the
product rule, we have
i
𝑑 h 𝑡/T
𝑑𝑇∞
𝑒 (𝑇 − 𝑇∞ ) = −𝑒 𝑡/T
𝑑𝑡
𝑑𝑡
Next integrate from 𝑡 = 0 to 𝑡:
∫𝑡
𝑑𝑇∞
𝑡/T
𝑒 (𝑇 − 𝑇∞ ) − 𝑇𝑖 − 𝑇∞ (0) = − 𝑒 𝑠/T
𝑑𝑠
𝑑𝑠
0
Multiplying through by 𝑒 −𝑡/T and rearranging gives the stated result:
∫𝑡
−𝑡/T
𝑑𝑇∞
−𝑡/T
𝑒 𝑠/T
𝑑𝑠
𝑇 (𝑡) = 𝑇∞ (𝑡) + 𝑇𝑖 − 𝑇∞ (0) 𝑒
−𝑒
𝑑𝑠
0
Alternate approach: To use Laplace transforms, we first simplify eqn. (1) by defining
𝑦(𝑡) ≡ 𝑇 − 𝑇∞ and 𝑓 (𝑡) ≡ −𝑑𝑇∞ /𝑑𝑡:
𝑑𝑦 𝑦
+ = 𝑓 (𝑡)
𝑑𝑡 T
Next, we apply the Laplace transform ℒ{..}, with ℒ{𝑦(𝑡)} = 𝑌 ( 𝑝) and ℒ{ 𝑓 (𝑡)} = 𝐹 ( 𝑝):
n 𝑑𝑦 o
n𝑦o
ℒ
+ℒ
= ℒ{ 𝑓 (𝑡)}
𝑑𝑡
T
1
𝑝𝑌 ( 𝑝) − 𝑦(0) + 𝑌 ( 𝑝) = 𝐹 ( 𝑝)
T
144
Solving for 𝑌 ( 𝑝):
1
1
𝑦(0) +
𝐹 ( 𝑝)
𝑝 + 1/T
𝑝 + 1/T
Now take the inverse transform, ℒ −1 {..}:
n
o
n
o
1
1
ℒ −1 {𝑌 ( 𝑝)} = ℒ −1
𝑦(0) + ℒ −1
𝐹 ( 𝑝)
𝑝 + 1/T
𝑝 + 1/T
With a table of Laplace transforms, we find
n
o
1
ℒ −1
= 𝑒 −𝑡/T
|{z}
𝑝 + 1/T
| {z }
≡𝑔(𝑡)
𝑌 ( 𝑝) =
≡𝐺 ( 𝑝)
and with 𝐺 ( 𝑝) and 𝑔(𝑡) defined as shown, the last term is just a convolution integral
∫𝑡
o
n
1
−1
−1
𝐹 ( 𝑝) = ℒ {𝐺 ( 𝑝)𝐹 ( 𝑝)} =
𝑔(𝑡 − 𝑠) 𝑓 (𝑡) 𝑑𝑠
ℒ
𝑝 + 1/T
0
Putting all this back into eqn. (2), we find
∫𝑡
−𝑡/T
𝑦(𝑡) = 𝑒
𝑦(0) + 𝑒 −(𝑡−𝑠)/T 𝑓 (𝑡) 𝑑𝑠
0
and putting back the original variables in place of 𝑦 and 𝑓 , we have at length obtained:
∫𝑡
−𝑡/T
𝑑𝑇∞
−𝑡/T
𝑇 (𝑡) = 𝑇∞ (𝑡) + 𝑇𝑖 − 𝑇∞ (0) 𝑒
−𝑒
𝑒 𝑠/T
𝑑𝑠
𝑑𝑠
0
Extra credit. State which approach is more straightforward!
145
(2)
Problem 5.61
1.0000
Theta
Bi = 0.5
Bi = 1
Bi = 2
Bi = 5
Bi = 10
0.1000
0.0100
0.2
0.4
0.6
0.8
1
Fourier number, Fo
160
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
1.2
1.4
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 6.12 (a) Verify that eqn. (6.120) follows from eqn. (6.119). (b) Derive an equation
for liquids that is analogous to eqn. (6.119).
Solution
a) Beginning with
∫𝐿
1
𝑞 𝑤 𝑑𝑥
ℎ=
𝐿Δ𝑇 0
∫ 𝑥𝑙
∫ 𝑥𝑢
∫𝐿
1
=
ℎlaminar 𝑑𝑥 +
ℎtrans 𝑑𝑥 +
ℎturbulent 𝑑𝑥
𝐿 0
𝑥𝑙
𝑥𝑢
(6.119)
we may evaluate each integral separately. For a uniform temperature surface, the Nusselt
numbers are given by these equations:
1/3
Nulam = 0.332 Re1/2
𝑥 Pr
Re𝑥 𝑐
Nutrans = Nulam Re𝑙 , Pr
Re𝑙
0.6
Nuturb = 0.0296 Re0.8
𝑥 Pr
(6.58)
(6.114b)
for gases
(6.112)
The three integrals are thus
r
∫ 𝑥𝑙
∫ r
0.332 𝑘Pr1/3
𝑢∞𝑥𝑙
𝑘
1
0.332 𝑘Pr1/3 𝑥𝑙 𝑢 ∞
𝑑𝑥 =
2
= 0.664 Re1/2
ℎlam 𝑑𝑥 =
𝑙 Pr
𝐿 0
𝐿
𝜈𝑥
𝐿
𝜈
𝐿
0
∫
∫
1 𝑥𝑢
𝑘 Nulam Re𝑙 , Pr 𝑢 ∞ 𝑐 𝑥𝑢 𝑐−1
𝑘 Nulam Re𝑙 , Pr 𝑢 ∞ 𝑐 1 𝑐
𝑥𝑢 − 𝑥 𝑙𝑐
ℎtrans 𝑑𝑥 =
𝑥 𝑑𝑥 =
𝑐
𝑐
𝐿 𝑥𝑙
𝐿
Re𝑙
𝜈
𝐿
Re𝑙
𝜈 𝑐
𝑥𝑙
𝑘 Nulam Re𝑙 , Pr 1
𝑘 1
𝑐
𝑐
Re
=
−
Re
Nu
Re
,
Pr
−
Nu
Re
,
Pr
=
turb
𝑢
lam
𝑙
𝑢
𝑙
𝐿
Re𝑙𝑐
𝑐
𝐿𝑐
where the last step follows because eqn. (6.114b) intersects Nuturb at Re𝑢 , and
∫
∫
0.0296 𝑘Pr0.6 𝑢 ∞ 0.8 𝐿 −0.2
0.0296 𝑘Pr0.6 0.8
1 𝐿
ℎturb 𝑑𝑥 =
𝑥
𝑑𝑥 =
Re 𝐿 − Re𝑢0.8
𝐿 𝑥𝑢
𝐿
𝜈
(0.8)𝐿
𝑥𝑢
𝑘
0.8
= 0.037 Pr0.6 Re0.8
𝐿 − Re𝑢
𝐿
Collecting these terms, we find:
Nu 𝐿 ≡
ℎ𝐿
0.8
1/3
= 0.037 Pr0.6 Re0.8
−
Re
+ 0.664 Re1/2
𝑢
𝐿
𝑙 Pr
𝑘
1
1/2 1/3
0.8 0.6
+ 0.0296 Re𝑢 Pr − 0.332 Re𝑙 Pr
𝑐
|
{z
}
for gases (6.120)
contribution of transition region
b) For a liquid flow, the turbulent correlation should be eqn. (6.113):
0.43
Nuturb = 0.032 Re0.8
𝑥 Pr
for nonmetallic liquids
189a
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
(6.113)
and the integral in the turbulent range changes to
∫
∫
0.032 𝑘Pr0.43 𝑢 ∞ 0.8 𝐿 −0.2
0.032 𝑘Pr0.43 0.8
1 𝐿
0.8
ℎturb 𝑑𝑥 =
𝑥
𝑑𝑥 =
Re 𝐿 − Re𝑢
𝐿 𝑥𝑢
𝐿
𝜈
(0.8)𝐿
𝑥𝑢
𝑘
0.8
= 0.040 Pr0.43 Re0.8
−
Re
𝑢
𝐿
𝐿
Collecting these terms, we find:
Nu 𝐿 ≡
ℎ𝐿
1/3
0.8
= 0.040 Pr0.43 Re0.8
−
Re
+ 0.664 Re1/2
𝑢
𝐿
𝑙 Pr
𝑘
1
1/2 1/3
0.8 0.43
+ 0.032 Re𝑢 Pr
− 0.332 Re𝑙 Pr
𝑐
|
{z
}
contribution of transition region
189b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
for nonmetallic liquids
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 6.16 Air at −10 °C flows over a smooth, sharp-edged, almost-flat, aerodynamic
surface at 240 km/hr. The surface is at 10 °C. Turbulent transition begins at Re𝑙 = 140,000 and
ends at Re𝑢 = 315,000. Find: (a) the 𝑥-coordinates within which laminar-to-turbulent transition
occurs; (b) ℎ for a 2 m long surface; (c) ℎ at the trailing edge for a 2 m surface; and (d) 𝛿 and ℎ
at 𝑥 𝑙 .
Solution
a) We evaluate physical properties at the film temperature, 𝑇 𝑓 = (−10 + 10)/2 = 0 °C: 𝜈 =
1.332 × 10−5 m2 /s, Pr = 0.711, and 𝑘 = 0.244 W/m·K. Also, 𝑢 ∞ = 240(1000)/(3600) =
66.7 m/s. Then:
𝑥𝑙 =
Re𝑙 𝜈 (140000)(1.332 × 10−5 )
=
= 0.0280 m
𝑢∞
(66.7)
Re𝑢 𝜈 (315000)(1.332 × 10−5 )
=
= 0.0629 m
𝑢∞
(66.7)
Observe that the flow is fully turbulent over 1.937/2.00 = 96.9% of its length.
𝑥𝑢 =
b) First, we need Re 𝐿 :
𝑢∞ 𝐿
(66.7)(2)
=
= 1.00 × 107
𝜈
1.332 × 10−5
Then we get 𝑐 from eqn. (6.115):
Re 𝐿 =
𝑐 = 0.9922 log10 (140, 000) − 3.013 = 2.09
Now we may use eqn. (6.120):
Nu 𝐿 = 0.037(0.711) 0.6 (1.00 × 107 ) 0.8 − (3.15 × 105 ) 0.8
+ 0.664 (1.40 × 105 ) 1/2 (0.711) 1/3
i
1 h
5 0.8
0.6
5 1/2
1/3
+
0.0296(3.15 × 10 ) (0.711) − 0.332 (1.40 × 10 ) (0.711)
2.09
= 11248.9 + 221.8 + 236.0 = 1.171 × 104
Thus
ℎ=
𝑘
(0.0244)(1.171 × 104 )
Nu 𝐿 =
= 143 W/m2 K
𝐿
2
c) With eqn. (6.112),
0.6
Nu 𝐿 = 0.0296 Re0.8
= 0.0296 (1.00 × 107 ) 0.8 (0.711) 0.6 = 9603
𝐿 Pr
so
𝑘
(0.0244)(9603)
Nu 𝐿 =
= 117 W/m2 K
𝐿
2
d) The flow is laminar here. From eqn (6.58):
ℎ(𝐿) =
1/3
Nu𝑥𝑙 = 0.332 Re1/2
= 0.332 (1.40 × 105 ) 1/2 (0.711) 1/3 = 110.9
𝑙 Pr
so
ℎ(𝑥 𝑙 ) =
𝑘
(0.0244)(110.9)
Nu𝑥𝑙 =
= 96.6 W/m2 K
𝑥𝑙
0.0280
191
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
With eqn (6.2), we find that the boundary layer here is very thin:
4.92 𝑥 𝑙 4.92(0.0280)
= √
= 0.000368 m = 0.37 mm
𝛿= p
Re𝑥𝑙
1.4 × 105
191b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 6.17 Find ℎ in Example 6.9 using eqn. (6.120) with Re𝑙 = 80, 000. Compare with
the value in the example and discuss the implication of your result. Hint: See Example 6.10.
Solution
Equation (6.120) is
ℎ𝐿
1/3
0.8
= 0.037 Pr0.6 Re0.8
−
Re
+ 0.664 Re1/2
𝑢
𝐿
𝑙 Pr
𝑘
1
1/3
+ 0.0296 Re𝑢0.8 Pr0.6 − 0.332 Re1/2
Pr
(6.120)
𝑙
𝑐
From Example 6.9, we have Re 𝐿 = 1.270 × 106 and Pr = 0.708. We may find 𝑐 from eqn. (6.115):
Nu 𝐿 ≡
𝑐 = 0.9922 log10 (80, 000) − 3.013 = 1.85
We also need Re𝑢 , which we can find following Example 6.10:
0.0296(0.708) 0.6 (80, 000) 1.85
0.332(80, 000) 1/2 (0.708) 1/3
Solving, Re𝑢 = 184, 500. Substituting all this into eqn. (6.120):
Nu 𝐿 = 0.037(0.708) 0.6 (1.270 × 106 ) 0.8 − (1.845 × 105 ) 0.8 + 0.664 (8.0 × 104 ) 1/2 (0.708) 1/3
i
1 h
0.0296(1.845 × 105 ) 0.8 (0.708) 0.6 − 0.332 (8.0 × 104 ) 1/2 (0.708) 1/3
+
1.85
Evaluating, we find the contributions of the turbulent, laminar, and transition regions:
Re𝑢1.85−0.8 =
Nu 𝐿 = 1806.6 + 167.4 + 167.1 = 2, 141
| {z } |{z} |{z}
turb.
lam.
trans.
The transition region contributes 7.8% of the total. The average heat transfer coefficient is
2141(0.0264)
ℎ=
= 28.26 W/m2 K
2.0
and the convective heat loss from the plate is
𝑄 = (2.0)(1.0)(28.26)(310 − 290) = 1130 W
The earlier transition to turbulence increases the heat removal by [(1130+22)/(756+22)−1]×100 =
48%.
192
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 6.46 Two power laws are available for the skin friction coefficient in turbulent
flow: 𝐶 𝑓 (𝑥) = 0.027 Re−1/7
and 𝐶 𝑓 (𝑥) = 0.059 Re−1/5
. The former is due to White and the latter
𝑥
𝑥
to Prandtl [6.4]. Equation (6.102) is more accurate and wide ranging than either. Plot all three
expressions on semi-log coordinates for 105 ⩽ Re𝑥 ⩽ 109 . Over what range are the power laws in
reasonable agreement with eqn. (6.102)? Also plot the laminar equation (6.33) on same graph for
Re𝑥 ⩽ 106 . Comment on all your results.
Solution
expressions:
The figure shows the two power laws and the mentioned turbulent and laminar
𝐶𝑓 = 0.455
ln(0.06 Re𝑥 )
0.664
𝐶𝑓 = √
Re𝑥
(6.102)
2
(6.33)
The 1⁄ 7 power law is within 5% of eqn. (6.102) for 3.5 × 105 ⩽ Re𝑥 ⩽ 109 , while the 1⁄ 5 power law
is within 5% for 105 ⩽ Re𝑥 ⩽ 5 × 107 . We also observe that skin friction in laminar flow is far less
than in turbulent flow.
0.007
Eqn. (6.102), Cf = 0.455/[ ln (0.06 Rex )]2
0.006
1/7
Cf = 0.027/Rex
1/5
Cf = 0.059/Rex
1/2
Nusselt number, Cf
0.005
Eqn. (6.33), Cf = 0.664/Rex
0.004
0.003
turbule
nt
0.002
0.001
0.000
105
lam
ina
r
106
107
Reynolds number, Rex
205b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
108
109
Problem 6.47 Reynolds et al. [6.27] provide the following measurements for air flowing over
a flat plate at 127 ft/s with 𝑇∞ = 86 °F and 𝑇𝑤 = 63 °F. Plot these data on log-log coordinates as
Nu𝑥 vs. Re𝑥 , and fit a power law to them. How does your fit compare to eqn. (6.112)?
Re 𝑥 ×10−6
St×103
Re 𝑥 ×10−6
St×103
Re 𝑥 ×10−6
St×103
0.255
0.423
0.580
0.736
0.889
1.045
1.196
2.73
2.41
2.13
2.11
2.06
2.02
1.97
1.353
1.507
1.661
1.823
1.970
2.13
2.28
2.01
1.85
1.79
1.84
1.78
1.79
1.73
2.44
2.60
2.75
2.90
3.05
3.18
3.36
1.74
1.75
1.72
1.68
1.73
1.67
1.54
Solution The film temperature is 𝑇 𝑓 = (63 + 86)/2 = 74.5 °F = 23.6 °C = 296.8 K. At this
temperature, Table A.6 gives Pr = 0.707. We can convert the given data to Nu𝑥 = St Re𝑥 Pr using a
spreadsheet.
To make a fit, we must recognize that Pr does not vary. We have no basis for fitting a Pr exponent.
So, we can fit to
Nu𝑥 = 𝐴 Re𝑥𝑏
This fit may be done by linear regression if we first take the logarithm:
ln Nu𝑥 = ln 𝐴 + 𝑏 ln Re𝑥
Using a spreadsheet, we can calculate the logarithms and perform the linear regression to find
𝐴 = 0.0187 and 𝑏 = 0.814 (𝑟 2 = 0.9978), or
Nu𝑥 = 0.0187 Re0.814
𝑥
The fit is plotted with the equation, and the agreement is excellent.
With some additional effort, we may use the spreadsheet to find that the standard deviation of
the data with respect to the fit is 𝑠𝑥 = 2.81%, which provides a 95% confidence interval (two-sided
𝑡-statistic for 21 points, ±2.08𝑠𝑥 ) of ±5.8%.
Equation (6.112) for Pr = 0.712,
0.6
Nu𝑥 = 0.0296 Re0.8
= 0.0240 Re0.8
𝑥 Pr
𝑥
(6.112)
is also plotted in the figure, but it is systematically higher than this data set and our fit. (Reynolds
et al. had 7 other data sets and reported an overall 𝑠𝑥 = 4.5% for a ±9% uncertainty at 95%
confidence.)
205c
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
5 × 103
Nusselt number, Nux
Pr = 0.707 (air)
Tw = constant
103
Reynolds et al., Run 1
Nux = 0.0296 Re0x .8 Pr0.6 = 0.0240 Re0x .8
My fit, Nux = 0.0187 Re0x .814
300
105
106
Reynolds number, Rex
205d
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
107
Problem 6.48 Blair and Werle [6.36] reported the b.l. data below. Their experiment had a
uniform wall heat flux with a 4.29 cm unheated starting length, 𝑢 ∞ = 30.2 m/s, and 𝑇∞ = 20.5°C.
a) Plot these data as Nu𝑥 versus Re𝑥 on log-log coordinates. Identify the regions likely to be
laminar, transitional, and turbulent flow.
b) Plot the appropriate theoretical equation for Nu𝑥 in laminar flow on this graph. Does the
equation agree with the data?
c) Plot eqn. (6.112) for Nu𝑥 in turbulent flow on this graph. How well do the data and the
equation agree?
d) At what Re𝑥 does transition begin? Find values of 𝑐 and Re𝑙 that fit eqn. (6.116b) to these
data, and plot the fit on this graph.
e) Plot eqn. (6.117) through the entire range of Re𝑥 .
Re 𝑥 ×10−6
St×103
Re 𝑥 ×10−6
St×103
Re 𝑥 ×10−6
St×103
0.112
0.137
0.162
0.183
0.212
0.237
0.262
0.289
0.312
0.338
2.94
2.23
1.96
1.68
1.56
1.45
1.33
1.23
1.17
1.14
0.362
0.411
0.460
0.505
0.561
0.665
0.767
0.865
0.961
1.06
1.07
1.05
1.01
1.05
1.07
1.34
1.74
1.99
2.15
2.24
1.27
1.46
1.67
2.06
2.32
2.97
3.54
4.23
4.60
4.83
2.09
2.02
1.96
1.84
1.86
1.74
1.66
1.65
1.62
1.62
Solution
a) Calculate the Nusselt number from the values of Stanton number using Nu𝑥 = St Pr Re𝑥 .
This is easily done with software (or by hand if you are patient) using Pr = 0.71. The results
are plotted on the next page. The regions can be identified from the changes in slope and
curvature (part b makes the laminar regime more obvious).
b) The appropriate formula is eqn. (6.116) for a laminar b.l. with an unheated starting length:
1/3
0.4587 Re1/2
𝑥 Pr
Nulam = 1/3
1 − (𝑥 0 /𝑥) 3/4
(6.116)
We have only Re𝑥 , not 𝑥. However,
𝑥0 Re𝑥0
𝑢 ∞ 𝑥 0 (30.2)(0.0429)
=
and Re𝑥0 =
=
= 8.546 × 104
𝑥
Re𝑥
𝜈
1.516 × 10−5
With this, the expression can be plotted. The agreement is pretty good. (Equation (6.71) is
shown for comparison.)
c) The equation,
0.6
Nuturb = 0.0296 Re0.8
(6.112)
𝑥 Pr
is plotted in the figure, with excellent agreement.
d) To use eqn. (6.114b), we can start by visualizing a straight line through the transitional data
on the log-log plot to determine the slope, 𝑐. This slope can be determined iteratively if using
206
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
software, or by drawing the line if working by hand. The slope is well fit by 𝑐 = 2.5. Once
the slope is found, we find the point at which this line intersects the laminar, unheated starting
length curve. That point is well represented by Re𝑙 = 500,000 and Nulam (Re𝑙 , Pr) = 321.
Hence,
2.5
Re𝑥 𝑐
Re𝑥
Nutrans = Nulam Re𝑙 , Pr
= 321
(6.114b)
Re𝑙
500, 000
This equation is plotted in the figure, with very good agreement. Note that slightly different
values of Re𝑙 and Nulam may produce a good fit, if they lie on the same line. The best approach
is to find Re𝑙 and then calculate Nulam from eqn. (6.116).
e) Equation (6.117) uses the laminar, transitional, and turbulent Nusselt numbers from parts
(b), (c), and (d):
−1/2 1/5
5
−10
−10
Nu𝑥 (Re𝑥 , Pr) = Nu𝑥,lam + Nu𝑥,trans + Nu𝑥,turb
(6.117)
This equation is plotted in the figure as well, with very good agreement.
104
Eqn. (6.71), 0.4587 Re1/2 Pr1/3
Eqn. (6.114b), c = 2.5, Rel = 500,000
c=
tu
nsi
tion
al
103
nt
le
rbu
tra
Nusselt number, Nux
Eqn. (6.112), 0.0296 Re0.8 Pr0.6
Eqn. (6.117)
Blair and Werle, u0r /u∞ = 1.0%
2.5
Eqn. (6.116) for x0 = 4.29 cm
laminar
102
Pr = 0.71 (air)
q = constant
Increased h caused by
unheated starting length
105
106
Reynolds number, Rex
206b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
107
Problem 6.49 Figure 6.21 shows a fit to the following air data from Kestin et al. [6.29]
using eqn. (6.117). The plate temperature was 100 °C (over its entire length) and the free-stream
temperature varied between 20 and 30 °C. Follow the steps used in Problem 6.48 to reproduce that
fit and plot it with these data.
Re 𝑥 ×10−3
Nu 𝑥
Re 𝑥 ×10−3
Nu 𝑥
Re 𝑥 ×10−3
Nu 𝑥
60.4
76.6
133.4
187.8
284.5
42.9
66.3
85.3
105.0
134.0
445.3
580.7
105.2
154.2
242.9
208.0
289.0
71.1
95.1
123.0
336.5
403.2
509.4
907.5
153.0
203.0
256.0
522.0
Solution
a) The results are plotted on the next page. The regions can be identified from the changes in
slope.
b) The appropriate formula is eqn. (6.58) for a laminar b.l. on a uniform temperature plate:
1/3
Nulam = 0.332 Re1/2
𝑥 Pr
(6.58)
The film temperature is between 60 and 65 °C, so Pr = 0.703. This equation is plotted on
the figure. Only two data points touch the line, but they are in excellent agreement.
c) The appropriate equation,
0.6
Nuturb = 0.0296 Re0.8
𝑥 Pr
(6.112)
is plotted in the figure, with very good agreement.
d) To use eqn. (6.114b), we can start by visualizing a straight line through the transitional data
on the log-log plot to determine the slope, 𝑐. The slope is well fit by 𝑐 = 1.7. Once the slope
is found, we find the point at which this line intersects the laminar, unheated starting length
curve. That point is well represented by Re𝑙 = 60,000 and Nulam (Re𝑙 , Pr) = 72.3. Hence,
1.7
Re𝑥 𝑐
Re𝑥
Nutrans = Nulam Re𝑙 , Pr
= 72.3
(6.114b)
Re𝑙
60000
This equation is plotted in the figure, with good agreement. Note that the most consistent
approach is to find Re𝑙 and then calculate Nulam from eqn. (6.58).
e) Equation (6.117) uses the laminar, transitional, and turbulent Nusselt numbers from parts
(b), (c), and (d):
−1/2 1/5
5
−10
−10
Nu𝑥 (Re𝑥 , Pr) = Nu𝑥,lam + Nu𝑥,trans + Nu𝑥,turb
(6.117)
This equation is plotted in the figure as well, with very good agreement in the turbulent and
transitional ranges. The laminar fit looks good with one data point, but not the other one.
The data themselves make a sharp leap between Re𝑥 of 66,300 and 85,300. (Kestin et al.
varied the Reynolds number between these data by increasing the air speed, 𝑢 ∞ —these data
are not from spatially sequential points (unlike the data of Blair in Problem 6.48). The onset
of turbulence is an instability, and the change in flow conditions may well have affected the
transition.)
206c
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
103
nt
le
bu
tur
tra
ns
itio
na
l
Nusselt number, Nux
Pr = 0.703 (air)
Tw = constant
102
r
ina
lam
2
Eqn. (6.58), 0.332 Re1/2 Pr1/3
Eqn. (6.114b), c = 1.7, Rel = 60,000
c=
Eqn. (6.112), 0.0296 Re0.8 Pr0.6
Eqn. (6.117), c = 1.7, Rel = 60,000
Kestin et al. (1961)
105
106
Reynolds number, Rex
206d
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 6.50 A study of the kinetic theory of gases shows that the mean free path of a
molecule in air at one atmosphere and 20 °C is 67 nm and that its mean speed is 467 m/s. Use
eqns. (6.45) obtain 𝐶1 and 𝐶2 from the known physical properties of air. We have asserted that
these constants should be on the order of 1. Are they?
Solution
We had found that
𝜇 = 𝐶1 𝜌𝐶ℓ
(6.45c)
𝑘 = 𝐶2 𝜌𝑐 𝑣 𝐶ℓ
(6.45d)
and
We may interpolate the physical properties of air from Table A.6: 𝜇 = 1.82 × 10−5 kg/m·s,
𝑘 = 0.0259 W/m·K, 𝜌 = 1.21 kg/m3 , and 𝑐 𝑝 = 1006 J/kg·K. In addition, the specific heat capacity
ratio for air is 𝛾 = 𝑐 𝑝 /𝑐 𝑣 = 1.4.
Rearranging:
1.82 × 10−5
𝐶1 =
=
= 0.481
𝜌𝐶ℓ (1.21)(467)(67 × 10−9 )
𝜇
and
𝐶2 =
𝑘𝛾
𝜌𝑐 𝑝 𝐶ℓ
=
(0.0259)(1.4)
= 0.952
(1.21)(1006)(467)(67 × 10−9 )
The constants are indeed 𝒪(1).
206e
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 7.5 Compare the h value computed in Example 7.3 with values predicted
by the Dittus-Boelter, Colburn, McAdams, and Sieder-Tate equations. Comment
on this comparison.
Solution: Taking values of components from Example 7.3, we get:
hDB = (k/D)(0.0243)(Pr)0.4(ReD)0.8
= (0.661/0.12)(0.0243)(3.61)0.4(412,300)0.8 = 6747 W/m2-K
hColburn = (k/D)(0.023)(Pr)1/3(ReD)0.8
= (0.661/0.12)(0.023)(3.61)1/3(412,300)0.8 = 6193 W/m2-K
hMcdams = (k/D)(0.0225)(Pr)0.4(ReD)0.8 = (0.0225/0.0243)hDB
= 6247 W/m2-K
hST = hColburn(μb/μw)0.14 = 6193(1.75)0.14 = 6193(1.081)
= 6698 W/m2-K
The more accurate Gnielinski equation gives h = 8400 W/m2-K. Therefore, these
old equations are low by roughly 20%, 26%, 26%, and 25%, respectively.
Why such consistently large deviations? It is because the old correlations
represent much more limited data sets than Gnielinski’s correlation. In this case,
ReD = 412,000 was a good deal higher than the ReD values used to build the old
correlations.
208a
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 7.17 Air at 1.38 MPa (200 psia) flows at 12 m/s in an 11 cm I.D. duct. At one
location, the bulk temperature is 40 °C and the pipe wall is at 268 °C. Evaluate ℎ if 𝜀/𝐷 = 0.002.
Solution We evaluate the bulk properties at 40°C = 313.15 K. Since the pressure is elevated,
we must use the ideal gas law to find the density of air with the universal gas constant, 𝑅 ◦ , and the
molar mass of air, 𝑀:
𝑝𝑀
(1.38 × 106 )(28.97)
=
= 15.36 kg/m3
𝑅 ◦𝑇
(8314.5)(313.15)
The dynamic viscosity, conductivity, and Prandtl number of a gas depend primarily upon temperature. At 313 K, 𝜇 = 1.917 × 10−5 kg/m·s, 𝑘 = 0.0274 W/m·K, and Pr = 0.706. Hence,
𝜌𝑢 av 𝐷 (15.36) (12) (0.11)
= 1.058 × 106
=
Re𝐷 =
𝜇
1.917 × 10−5
The friction factor may be calculated with Haaland’s equation, (7.50):
(
"
1.11 # ) −2
6.9
0.002
𝑓 = 1.8 log10
= 0.02362
+
3.7
1.058 × 106
𝜌=
We can see from Fig. 7.6 that this condition lies in the fully rough regime, as confirmed by
eqns. (7.48):
r
r
𝑢∗ 𝜀
𝜀
𝑓
0.02362
6
Re𝜀 ≡
= Re𝐷
= (1.058 × 10 )(0.002)
= 114.9 > 70
𝜈
𝐷 8
8
Next, we may compute the Nusselt number from eqn. (7.49):
𝑓 /8 Re𝐷 Pr
Nu𝐷 =
p
0.5
1 + 𝑓 /8 4.5 Re0.2
Pr
−
8.48
𝜀
0.02362/8 (1.058 × 106 )(0.706)
=
p
1 + 0.02362/8 4.5(114.9) 0.2 (0.706) 0.5 − 8.48
= 2061
The temperature difference is quite large, so we should correct for variable properties using
eqn. (7.45):
0.47
0.47
𝑇𝑏
313.15
Nu𝐷 = Nu𝐷
= (2061)
= 1594
𝑇𝑏 𝑇𝑤
541.15
Finally,
𝑘
0.0274
ℎ = Nu𝐷 =
(1594) = 397 W/m2 K
𝐷
0.11
215b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 7.51 Consider the water-cooled annular resistor of Problem 2.49 (Fig. 2.24). The
resistor is 1 m long and dissipates 9.4 kW. Water enters the inner pipe at 47 °C with a mass flow rate
of 0.39 kg/s. The water passes through the inner pipe, then reverses direction and flows through
the outer annular passage, counter to the inside stream.
a) Determine the bulk temperature of water leaving the outer passage.
b) Solve Problem 2.49 if you have not already done so. Compare the thermal resistances
between the resistor and each water stream, 𝑅𝑖 and 𝑅𝑜 .
c) Use the thermal resistances to form differential equations for the streamwise (𝑥-direction)
variation of the inside and outside bulk temperatures (𝑇𝑏,𝑜 and 𝑇𝑏,𝑖 ) and an equation the local
resistor temperature. Use your equations to obtain an equation for 𝑇𝑏,𝑜 − 𝑇𝑏,𝑖 as a function
of 𝑥.
d) Sketch qualitatively the distributions of bulk temperature for both passages and for the resistor.
Discuss the size of: the difference between the resistor and the bulk temperatures; and overall
temperature rise of each stream. Does the resistor temperature change much from one end
to the other?
e) Your boss suggests roughening the inside surface of the pipe to an equivalent sand-grain
roughness of 500 µm. Would this change lower the resistor temperature significantly?
f) If the outlet water pressure is 1 bar, will the water boil? Hint: See Problem 2.48.
g) Solve your equations from part (c) to find 𝑇𝑏,𝑖 (𝑥) and 𝑇𝑟 (𝑥). Arrange your results in terms
of NTU𝑜 ≡ 1/( 𝑚𝑐
¤ 𝑝 𝑅𝑜 ) and NTU𝑖 ≡ 1/( 𝑚𝑐
¤ 𝑝 𝑅𝑖 ). Considering the size of these parameters,
assess the approximation that 𝑇𝑟 is constant in 𝑥.
Solution
a) The answer follows directly from the 1st Law, 𝑄 = 𝑚𝑐
¤ 𝑝 𝑇𝑏,out − 𝑇𝑏,in ):
Δ𝑇𝑏 = 𝑄/( 𝑚𝑐
¤ 𝑝 ) = 9400/(0.39 · 4180) = 5.77 °C
so 𝑇𝑏,out = 47 + 5.77 = 52.8 °C.
b) The inside thermal resistance, 𝑅𝑖 = 3.69 × 10−2 K/W, is 23% greater than the outside
resistance, 𝑅𝑜 = 3.00 × 10−2 K/W.
c) With eqn. (7.10), putting (𝑞 𝑤 𝑃)inside = (𝑇𝑟 − 𝑇𝑏,𝑖 )/𝑅𝑖 𝐿 and (𝑞 𝑤 𝑃)outside = (𝑇𝑟 − 𝑇𝑏,𝑜 )/𝑅𝑜 𝐿
where the tube length is 𝐿 = 1 m:
𝑑𝑇𝑏,𝑖 𝑇𝑟 − 𝑇𝑏,𝑖
=
(1)
𝑚𝑐
¤ 𝑝
𝑑𝑥
𝑅𝑖 𝐿
𝑑𝑇𝑏,𝑜 𝑇𝑟 − 𝑇𝑏,𝑜
−𝑚𝑐
¤ 𝑝
=
(2)
𝑑𝑥
𝑅𝑜 𝐿
Recalling the solution of Problem 4.29, we can divide the resistance equation by 𝐿 to obtain
a local result (assuming that ℎ is equal to ℎ along the entire passage):
𝑇𝑟 − 𝑇𝑏,𝑖 𝑇𝑟 − 𝑇𝑏,𝑜 𝑄
+
=
= constant
(3)
𝑅𝑖 𝐿
𝑅𝑜 𝐿
𝐿
Each of 𝑇𝑏,𝑖 , 𝑇𝑏,𝑜 , and 𝑇𝑟 are functions of 𝑥.
By adding eqn. (1) to eqn. (2), and then using eqn. (3),
𝑑 (𝑇𝑏,𝑜 − 𝑇𝑏,𝑖 ) 𝑄
−𝑚𝑐
¤ 𝑝
=
𝑑𝑥
𝐿
232
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
and integrating (with 𝑇𝑏,𝑜 = 𝑇𝑏,𝑖 at 𝑥 = 𝐿), we find
𝑇𝑏,𝑜 − 𝑇𝑏,𝑖 =
𝑄
(1 − 𝑥/𝐿)
𝑚𝑐
¤ 𝑝
(4)
d) From working part (a) and Problem 2.49, we already know that the resistor will be much
hotter than the water on either side (194 °C at the end where the water enters and exits). At
any point, 𝑇𝑟 − 𝑇𝑏 𝑇𝑏,𝑜 − 𝑇𝑏,𝑖 , so that 𝑇𝑟 − 𝑇𝑏,𝑖 ≃ 𝑇𝑟 − 𝑇𝑏,𝑜 ≃ constant, along the entire
passage. From eqns. (1) and (2), then, the bulk temperature of each stream has a nearly
straight line variation in 𝑥, but the outer passage temperature rises a bit faster because the
thermal resistance on that side is lower. Similarly, eqn. (3) shows that the resistor temperature
varies by no more than do the bulk temperatures.
e) Your solution to Problem 2.49 shows that the epoxy layers provide the dominant thermal
resistance on each side. Roughness will make the convection resistance smaller, but convection resistance is only about 10% of the overall resistance. Your boss’s idea will add
cost and pressure drop, but it won’t lower the resistor temperature much. (Suggestion: Find
a diplomatic way to tell him that.)
f) The water will not boil if the highest temperature of the epoxy is below 𝑇sat . The hottest
point for the epoxy is in the outlet stream at the exit (where the bulk temperature is greatest).
From the solution to Problem 2.49, using the voltage divider relation from Problem 2.48,
𝑇epoxy − 𝑇𝑏,outlet = (𝑇𝑟 − 𝑇𝑏,outlet )
𝑅conv
0.00307
= (194 − 52.8)
= 14.4 K
𝑅outside
0.0300
The water will not boil.
g) Rearranging eqn. (3) with eqn. (4):
𝑅𝑖
𝑅𝑖
= 𝑄𝑅𝑖 − (𝑇𝑏,𝑜 − 𝑇𝑏,𝑖 )
𝑇𝑟 − 𝑇𝑏,𝑖 + (𝑇𝑟 − 𝑇𝑏,𝑖 )
𝑅
𝑅𝑜
𝑜
𝑅𝑖
𝑄𝑅𝑖
(1 − 𝑥/𝐿)
(𝑇𝑟 − 𝑇𝑏,𝑖 ) 1 +
= 𝑄𝑅𝑖 −
𝑅𝑜
𝑚𝑐
¤ 𝑝 𝑅𝑜
𝑅𝑜
1
𝑇𝑟 − 𝑇𝑏,𝑖 = (𝑄𝑅𝑖 )
1−
(1 − 𝑥/𝐿)
𝑅𝑜 + 𝑅𝑖
𝑚𝑐
¤ 𝑝 𝑅𝑜
(5)
From eqn. (3), we may estimate that 𝑄𝑅𝑖 ≈ (𝑇𝑟 − 𝑇𝑏,𝑖 )/2; thus, we can see that the second
term on the right is very small and could be neglected entirely.
Upon substituting eqn. (5) into eqn. (1) we have:
𝑑𝑇𝑏,𝑖 𝑄
𝑅𝑜
1
=
(1 − 𝑥/𝐿)
𝑚𝑐
¤ 𝑝
1−
𝑑𝑥
𝐿 𝑅𝑜 + 𝑅𝑖
𝑚𝑐
¤ 𝑝 𝑅𝑜
Integration gives:
𝑄
𝑅𝑜
𝑥
1
𝑥
𝑥2
𝑇𝑏,𝑖 (𝑥) − 𝑇𝑏,in =
−
−
𝑚𝑐
¤ 𝑝 𝑅𝑜 + 𝑅𝑖 𝐿 𝑚𝑐
¤ 𝑝 𝑅𝑜 𝐿 2𝐿 2
Because the second term in the square brackets is small, we see that the bulk temperature has
an essentially straight line variation.
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
232b
More precisely, we may think of this arrangement as a heat exchanger, where 𝑈 𝐴 = 1/𝑅𝑜
so that
1
1
𝑈𝐴
=
=
= 0.020 1
NTU𝑜 =
−2
𝑚𝑐
¤ 𝑝 𝑚𝑐
¤ 𝑝 𝑅𝑜 (3.00 × 10 )(0.39) (4180)
From Chapter 3, we recall that a heat exchanger with very low NTU causes very little change
in the temperature of the streams, as is the case here. Putting our result in terms of the outside
and inside NTUs:
𝑅𝑜
𝑥2
𝑥
𝑥
𝑇𝑏,𝑖 (𝑥) − 𝑇𝑏,in = (𝑄𝑅𝑖 )NTU𝑖
(6)
− NTU𝑜
−
𝑅𝑜 + 𝑅𝑖 𝐿
𝐿 2𝐿 2
Substituting eqn. (6) into eqn. (5):
𝑥
𝑥2
𝑅𝑜
𝑥
𝑥
𝑇𝑟 − 𝑇𝑏,in = (𝑄𝑅𝑖 )
1 − NTU𝑜 1 −
− NTU𝑖
− NTU𝑜
−
𝑅𝑜 + 𝑅𝑖
𝐿
𝐿
𝐿 2𝐿 2
Since NTU𝑖 has a similar value to NTU𝑜 , the resistor temperature is indeed nearly constant,
with variations on the order of NTU0 = 0.02.
232c
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 8.13 The side wall of a house is 10 m in height. The overall heat transfer coefficient
between the interior air and the exterior surface is 2.5 W/m2 K. On a cold, still winter night
𝑇outside = −30 °C and 𝑇inside air = 25 °C. What is ℎconv on the exterior wall of the house if 𝜀 = 0.9?
Is external convection laminar or turbulent?
Solution The exterior wall is cooled by both natural convection and thermal radiation.
Both heat transfer coefficients depend on the wall temperature, which is unknown. We may solve
iteratively, starting with a guess for 𝑇𝑤 . We might assume (arbitrarily) that 2⁄3 of the temperature
difference occurs across the wall and interior, with 1⁄3 outside, so that 𝑇𝑤 ≈ (25 + 30)/3 − 30 =
−11.7 °C = 261.45 K. We may take properties of air at 𝑇𝑓 ≈ 250 K, to avoid interpolating Table A.6:
Properties of air at 250 K
thermal conductivity
thermal diffusivity
kinematic viscosity
Prandtl number
𝑘
𝛼
𝜈
Pr
0.0226
W/m·K
1.59 × 10−5 m2 /s
1.135 × 10−5 m2 /s
0.715
The next step is to find the Rayleigh number so that we may determine whether to use a correlation
for laminar or turbulent flow. With 𝛽 = 1/𝑇𝑓 = 1/(250) K−1 :
Ra𝐿 =
𝑔𝛽(𝑇𝑤 − 𝑇outside )𝐿 3
(9.806) (−11.7 + 30) (103 )
=
= 3.98 × 1012
𝜈𝛼
(250)(1.59)(1.135)(10−10 )
Since, Ra𝐿 > 109 , we use eqn. (8.13b) to find Nu𝐿 :
(
)2
0.387 Ra1/6
𝐿
Nu𝐿 = 0.825 + 8/27
1 + (0.492/Pr) 9/16
(
)2
0.387(3.98 × 1012 ) 1/6
= 0.825 + = 1738
8/27
1 + (0.492/0.715) 9/16
Hence
0.0226
= 3.927 W/m2 K
10
The radiation heat transfer coefficient, for 𝑇𝑚 = (261.45 + 243.15)/2 = 252.30 K, is
ℎconv = (1738)
ℎrad = 4𝜀𝜎𝑇𝑚3 = 4(0.9)(5.6704 × 10−8 )(252.30) 3 = 3.278 W/m2 K
The revised estimate of the wall temperature is found by equating the heat loss through the wall
to the heat loss by convection and radiation outside:
(2.5)(25 − 𝑇𝑤 ) = (3.927 + 3.278)(𝑇𝑤 + 30)
so that 𝑇𝑤 = −15.8 °C, which is somewhat lower than our estimate. We may repeat the calculations
with this new value (without changing the property data) finding Ra𝐿 = 3.09 × 1012 , Nu𝐿 = 1799,
ℎconv = 4.065 W/m2 K, 𝑇𝑚 = 250.3 K, and ℎrad = 3.201 W/m2 K. Then
(2.5)(25 − 𝑇𝑤 ) = (4.065 + 3.201)(𝑇𝑤 + 30)
so that 𝑇𝑤 = −15.9 °C. Further iteration is not needed. Since the film temperature is very close to
250 K, we do not need to update the property data.
To summarize the final answer, ℎconv = 4.07 W/m2 K and most of the boundary layer is turbulent.
241
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 8.15 In eqn. (8.7), we linearized the temperature dependence of the density difference. Suppose that a wall at temperature 𝑇𝑤 sits in water at 𝑇∞ = 7 °C. Use the data in Table A.3
to plot |𝜌𝑤 − 𝜌∞ | and |−𝜌𝑓 𝛽 𝑓 (𝑇𝑤 − 𝑇∞ )| for 7 °C ⩽ 𝑇𝑤 ⩽ 100 °C, where (..)𝑓 is a value at the film
temperature. How well does the linearization work?
Solution With values from Table A.3, we may perform the indicated calculations and make
the plot. The linearization is accurate to within 10% for temperature differences up to 40 °C, and
within 13% over the entire range.
Properties of water from Table A.3
𝑇 °C
𝜌 kg/m3
Density difference, ρ ∞ − ρ w [kg/m3 ]
7
12
17
22
27
32
37
47
67
87
100
999.9
999.5
998.8
997.8
996.5
995.0
993.3
989.3
979.5
967.4
958.3
𝛽 K−1
(𝜌𝑤 − 𝜌∞ ) −𝜌𝑓 𝛽 𝑓 (𝑇𝑤 − 𝑇∞ )
0.0000436
0.000112
0.000172
0.000226
0.000275
0.000319
0.000361
0.000436
0.000565
0.000679
0.000751
0.0
−0.4
−1.1
−2.1
−3.4
−4.9
−6.6
−10.6
−20.4
−32.5
−41.6
0.000
−0.389
−1.08
−2.02
−3.18
−4.52
−6.05
−9.54
−18.1
−28.4
−36.2
40
ρ f β f (Tw − T∞ )
|ρ w − ρ ∞ |
30
20
10
0
0
10
20
30
40
50
60
70
Temperature difference, Tw − T∞ [K]
C
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
80
90
100
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 8.53 An inclined plate in a piece of process equipment is tilted 30◦ above horizontal
and is 20 cm long in the inclined plane and 25 cm wide in the horizontal plane. The plate is held
at 280 K by a stream of liquid flowing past its bottom side; the liquid is cooled by a refrigeration
system capable of removing 12 W. If the heat transfer from the plate to the stream exceeds 12 W,
the temperature of both the liquid and the plate will begin to rise. The upper surface of the plate
is in contact with ammonia vapor at 300 K and a varying pressure. An engineer suggests that any
rise in the bulk temperature of the liquid will signal that the pressure has exceeded a level of about
𝑝 crit = 551 kPa.
a) Explain why the gas’s pressure will affect the heat transfer to the coolant.
b) Suppose that the pressure is 255.3 kPa. What is the heat transfer (in watts) from gas to the
plate, if the plate temperature is 𝑇𝑤 = 280 K? Will the coolant temperature rise?
c) Suppose that the pressure rises to 1062 kPa. What is the heat transfer to the plate if the plate
is still at 𝑇𝑤 = 280 K? Will the coolant temperature rise?
Solution
a) Sufficiently high pressures can cause condensation of the NH3 vapor on the plate. In addition,
before condensation occurs, pressure changes may cause significant properties variations in
the NH3 vapor.
b) At 255.3 kPa, the saturation temperature is 𝑇sat = 260 K < 280 K; condensation will not
occur. Replacing 𝑔 with an effective gravity 𝑔 cos 60◦ , the Rayleigh number is
𝑔 cos 60◦ 𝛽Δ𝑇 𝐿 3 9.81 × (1/2) × 0.00345 × 20 × 0.23
Ra𝐿 =
=
≃ 9.07 × 107
−6
−6
𝜈𝛼
(5.242 × 10 )(5.690 × 10 )
The Nusselt number is
"
Nu𝐿 = 0.68 + 0.67Ra𝐿1/4
0.492
1+
Pr
9/16 # −4/9
"
= 0.68 + 0.67 × (9.07 × 107 ) 1/4
0.492
1+
0.92
9/16 # −4/9
≃ 52.3
Then,
ℎ = Nu𝐿
𝑘
0.0244
= 52.3 ×
= 6.38 W/m2 K
𝐿
0.2
and the heat transfer is
𝑄 = ℎ𝐴(𝑇∞ − 𝑇𝑤 ) = 6.38 × 0.2 × 0.25 × (300 − 280) ≃ 6.38 W < 12 W
and the plate and liquid temperatures will not rise.
c) At a pressure of 1062 kPa, the saturation temperature is 𝑇sat = 300 K > 280 K; condensation
occurs. The Nusselt number is
"
# 1/4
𝜌𝑓 (𝜌𝑓 − 𝜌𝑔 )𝑔 cos 60◦ ℎ0𝑓 𝑔 𝐿 3
Nu𝐿 = 0.9428
= 1814
𝜇𝑘 (𝑇sat − 𝑇𝑤 )
266h
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
The heat transfer coefficient is
ℎ = Nu𝐿
𝑘
= 4353 W/m2 K
𝐿
The heat transfer rate is
𝑄 = ℎ𝐴(𝑇sat − 𝑇𝑤 ) = 4353 W 12 W
and the plate and liquid temperatures will rise.
266i
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 8.54 A characteristic length scale for a falling liquid film is ℓ = (𝜈 2 /𝑔) 1/3 . If the
Nusselt number for a laminar film condensing on plane wall is written as Nuℓ ≡ ℎℓ/𝑘, derive an
−1/3
expression for Nuℓ in terms of Re𝑐 . Show that, when 𝜌𝑓 𝜌𝑔 , Nuℓ = 3Re𝑐
.
Solution
Starting with eqns. (8.58) and (8.72), we have
ℎ𝑥 𝑥
=
Nu𝑥 =
𝑘
𝛿
and
𝜌𝑓 𝜌𝑓 − 𝜌𝑔 𝑔𝛿3 𝜌𝑓 Δ𝜌 𝑔𝛿3
Re𝑐 =
=
3𝜇2
3𝜇2
Then, by replacing 𝑥 by ℓ
ℎℓ ℓ
Nuℓ =
=
𝑘
𝛿
and, by rearranging Re𝑐 ,
1/3
3𝜇𝜈
Re𝑐
𝛿=
𝑔Δ𝜌
So
2 1/3 1/3
1/3
𝜈
𝑔Δ𝜌
Δ𝜌
−1/3
Nuℓ =
Re−1/3
Re𝑐
=
𝑐
𝑔
3𝜇𝜈
3𝜌𝑓
and when 𝜌𝑓 𝜌𝑔 , Δ𝜌 ≃ 𝜌𝑓 so
Nuℓ ≃ (3Re𝑐 ) −1/3
266g
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
for 𝜌𝑓 𝜌𝑔
(8.58)
(8.72)
Problem 8.59 Using data from Tables A.4 and A.5, plot 𝛽 for saturated ammonia vapor for
200 K ⩽ 𝑇 ⩽ 380 K, together with the ideal gas expression 𝛽IG = 1/𝑇. Also calculate 𝑍 = 𝑃/𝜌𝑅𝑇.
Is ammonia vapor more like an ideal gas near the triple point or critical point temperature?
Solution
0.020
0.018
0.016
Data from Table A.4
Ideal gas, β IG = 1/T
0.014
β [K−1 ]
0.012
0.010
0.008
0.006
0.004
0.002
0.000
200
220
240
260
280
300
320
340
360
380
Temperature [K]
With 𝑝 and 𝜌 from Table A.5, and using 𝑅 = 𝑅 ◦ /𝑀NH3 = 8314.5/17.031 = 488.2 J/kg-K, we
find 𝑍 as below. For an ideal gas, 𝑍 = 1.
𝑇 [°C]
𝑍
𝑇 [°C]
𝑍
200
220
240
260
280
0.9944
0.9864
0.9722
0.9505
0.9198
300
320
340
360
380
0.8788
0.8263
0.7606
0.6784
0.5716
Saturated ammonia vapor only behaves like an ideal gas for temperatures close the triple point
temperature (195.5 K) and is highly non-ideal in the vicinity of the critical point temperature
(405.4 K). This behavior underscores the importance of using data for 𝛽 when dealing with vapors
near saturation conditions.
286h
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 9.37
P_vap (Pa)
1000
10000
100000
t (C)
6.97
Delta T (K)
1
2
3
4
5
6
7
8
T_sat (K)
280.12
318.96
372.76
T_sat (C)
6.97
45.81
99.61
q (kW/m^2) h (kW/m^2K)
25.1
25.1
52.9
26.5
83.7
27.9
117.2
29.3
153.6
30.7
192.9
32.1
234.9
33.6
279.8
35.0
45.81
1
2
3
4
5
6
7
8
113.0
238.8
377.3
528.7
692.9
869.8
1059.6
1262.1
113.0
119.4
125.8
132.2
138.6
145.0
151.4
157.8
99.61
1
2
3
4
5
6
7
8
210.3
444.5
702.5
984.2
1289.8
1619.2
1972.4
2349.4
210.3
222.2
234.2
246.1
258.0
269.9
281.8
293.7
286a
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Heat Flux vs Temp Diff (99.61 C)
Heat Flux [kW/m^2]
2500.0
2000.0
1500.0
1000.0
500.0
0.0
0
1
2
3
4
5
6
Temperature Difference [K]
286b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
7
8
9
Problem 9.38
Surface at 100 C
Delta T
0
1
2
3
4
5
6
7
8
9
10
T_0
p_0
R
coef
sigma
factor1
hfg
P_0
101420.0
105090.0
108870.0
112770.0
116780.0
120900.0
125150.0
129520.0
134010.0
138630.0
143380.0
Delta P
0
3670
7450
11350
15360
19480
23730
28100
32590
37210
41960
factor
mdot (kg/m^2s) q (MW/m^2)
0.000345552
0.0
0
0.000345231
1.3
3
0.000344907
2.6
6
0.000344564
3.9
9
0.000344241
5.3
12
0.000343941
6.7
15
0.000343615
8.2
18
0.000343299
9.6
22
0.000343
11.2
25
0.000342697
12.8
29
0.000342402
14.4
32
rho_0
0.051242
0.053871
0.056614
0.059474
0.062457
0.065565
0.068803
0.072176
0.075688
0.079343
0.083147
factor
mdot (kg/m^2s) q (MW/m^2)
0.000372416
0.0
0.0
0.00037186
0.1
0.3
0.000371306
0.3
0.7
0.000370756
0.5
1.1
0.000370213
0.6
1.5
0.00036967
0.8
1.9
0.000369146
1.0
2.3
0.000368579
1.2
2.8
0.000368067
1.4
3.2
0.000367534
1.6
3.7
0.000366984
1.8
4.2
373.15
101420
461.404
1.6678
0.31
3.0329914
2246000 treat as constant
Surface at 40 C
Delta T
0
1
2
3
4
5
6
7
8
9
10
P_0
7384.9
7787.8
8209.6
8650.8
9112.4
9595
10099
10627
11177
11752
12353
Delta P
40 C
T_0
p_0
R
coef
sigma
factor1
hfg
rho_0
0.59817
0.61841
0.6392
0.66056
0.6825
0.70503
0.72816
0.7519
0.77627
0.80127
0.82693
313.15
7384.9
461.403996
1.6678
0.31
3.0329914
2306000 treat as constant
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
0
402.9
824.7
1265.9
1727.5
2210.1
2714.1
3242.1
3792.1
4367.1
4968.1
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
291b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 10.52: The fraction of blackbody radiation between wavelengths of 0 and
𝜆 is
𝜆
1
∫
𝑓=
𝑒𝜆,𝑏 𝑑𝜆
(11)
𝜎𝑇4 0
a) Work Problem 10.51.
b) Show that
15 ∞
𝑡3
𝑓(𝜆𝑇) = 4 ∫
𝑑𝑡
(12)
𝜋 𝑐2 /𝜆𝑇 𝑒𝑡 − 1
where 𝑐2 is the second radiation constant, ℎ𝑐/𝑘𝐵 , equal to 1438.8 µm⋅K.
c) Use the software of your choice to plot 𝑓(𝜆𝑇) and check that your results match
Table 10.7.
Solution. Following the solution to Problem 10.51:
𝜆
1
∫
𝑒𝜆,𝑏 𝑑𝜆
𝜎𝑇4 0
𝜆
1
2𝜋ℎ𝑐𝑜2
∫
=
𝑑𝜆
𝜎𝑇4 0 𝜆5 [exp(ℎ𝑐𝑜 /𝑘𝐵 𝑇𝜆) − 1]
𝑓=
(13)
(14)
=
∞
1
2𝜋ℎ𝜈3
∫
𝑑𝜈
𝜎𝑇4 𝑐𝑜 /𝜆 𝑐𝑜2 [exp(ℎ𝜈/𝑘𝐵 𝑇) − 1]
(15)
=
𝑡3
1 2𝜋𝑘4𝐵 𝑇4 ∞
∫
𝑑𝑡
𝑡
𝜎𝑇4 ℎ3 𝑐𝑜2
𝑐2 /𝜆𝑇 𝑒 − 1
(16)
=
15 ∞
𝑥3
∫
𝑑𝑥
𝜋4 𝑐2 /𝜆𝑇 𝑒𝑥 − 1
(17)
15 ∞ 𝑥3
15 𝑐2 /𝜆𝑇 𝑥3
∫
∫
𝑑𝑥
−
𝑑𝑥
𝜋4 0 𝑒𝑥 − 1
𝜋4 0
𝑒𝑥 − 1
15 𝑐2 /𝜆𝑇 𝑥3
=1− 4 ∫
𝑑𝑥
𝜋 0
𝑒𝑥 − 1
=
(18)
(19)
The numerical integration can be done in various ways, depending on the software available. (On a sophisticated level, the last integral can be written in terms of the Debye
function which is available in the Gnu Scientific Library.) This equation is plotted in
Fig. 1.
Problem 10.53: Read Problem 10.52. Then find the central range of wavelengths that
includes 80% of the energy emitted by blackbodies at room temperature (300 K) and at
the solar temperature (5777 K).
Solution. From Table 10.7, 𝑓 = 0.10 at 𝜆𝑇 = 2195 µm⋅K and 𝑓 = 0.90 at 𝜆𝑇 = 9376
µm⋅K. Dividing by the absolute temperatures gives:
𝑇 [K] 𝜆0.1 [µm] 𝜆0.9 [µm]
300
5777
7.317
0.380
311b
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
31.25
1.62
1.0
0.8
0.6
f(λT)
0.4
0.2
0.0
0.2
0.4
0.6
0.8
1.0
λT [cm⋅K]
1.5
2.0
Figure 1. The radiation fractional function
Problem 10.54: Read Problem 10.52. A crystalline silicon solar cell can convert photons to conducting electrons if the photons have a wavelength less than 𝜆band = 1.11
µm, the bandgap wavelength. Longer wavelengths do not produce an electric current,
but simply get absorbed and heat the silicon. For a solar cell at 320 K, make a rough
estimate of the fraction of solar radiation on wavelengths below the bandgap? Why is
this important?
Solution. The relevant temperature is that of the sun, 5777 K, not that of the solar
cell. We approximate the sun as a blackbody at 5777 K, ignoring atmospheric absorption
bands.
𝜆band 𝑇 = (1.11)(5777) µm ⋅ K = 6412 µm ⋅ K
Referring to Table 10.7, a bit less than 80% of solar energy is on these shorter wavelengths
(with a more exact table, 77%). This is significant because the solar cell can convert less
than 80% of the solar energy to electricity; additional considerations lower the theoretical
efficiency still further, to less than 50%.
311c
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Problem 10.55 Two stainless steel blocks have surface roughness of about 10 µm and 𝜀 ≈ 0.5.
They are brought into contact, and their interface is near 300 K. Ignore the points of direct contact
and make a rough estimate of the conductance across the air-filled gaps, approximating them as two
flat plates. How important is thermal radiation? Compare your result with Table 2.1 and comment
on the relative importance of the direct contact that we ignored.
Solution The gaps are very thin, so little circulation will occur in the air. Heat transfer
through the air will be by conduction. Radiation and conduction act in parallel across the gap.
The temperature difference across the gap will likely be small, so we may use a radiation thermal
resistance. The conductance is the reciprocal of the thermal resistance, per unit area, so ℎgap =
ℎcond + ℎrad .
Letting the gap width be 𝛿 = 10 µm and taking 𝑘 air = 0.0264 W/m·K, we can estimate
𝑘
0.0264
ℎcond ≈ =
= 2, 640 W/m2 K
𝛿 10 × 10−6
With eqns. (2.29) and (10.25):
−1 −1
1
2
1
1
F1–2 =
=
=
+
−1
−1
𝜀1 𝜀2
0.5
3
ℎrad = 4𝜎𝑇𝑚3 F1–2 = 4(5.67 × 10−8 )(300) 3 (0.3333) = 2.041 W/m2 K
Then
ℎgap = ℎcond + ℎrad = 2640 + 2.041 = 2, 642 W/m2 K
This conductance is on the lower end of the range of given in Table 2.1. Conduction through contacting points will add significantly to the heat transfer, although it will be highly multidimensional
and not easily calculated. Thermal radiation, however, is negligible.
311d
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V
Copyright 2020, John H. Lienhard, IV and John H. Lienhard, V