 # CN2125 FAQ a temperature 404K?

```CN2125 FAQ
1. WWWR P.212 Example 3: How to know that the saturated steam (0.276MPa) has
a temperature 404K?
2. How to read the values of thermal conductivity, heat transfer coefficients and
other relevant transport properties from the textbook?
homework assignments, quizzes and final examination, please also give the page
number of your reference for this. It is anticipated that some variation could be
found because most of these values come from experiments.
3. The derivation for differential heat transfer equations are rather complicated
(Chapter 16 WWWR) and (Chapter 2 ID), do we need to know the detailed
derivation?
Answer: No, you should understand just once the derivation for rectangular
coordinate, the derivations for cylindrical and spherical coordinates can be realized
in a similar way and there is no need to worry about the detailed derivations. Instead,
the focus should be on understanding the physical meanings about the respective
terms in these equations, for instance, which terms contribute to conduction in the
radial direction, etc. You should go through the WWWR 16.1, 16.2 and ID 2.2 and
2.3.
‐‐‐‐‐Original Message‐‐‐‐‐ From: Wang Chi‐Hwa Sent: Friday, February 01, 2013 9:38 AM To: Poh Chaur Ann Subject: RE: Queries on straight fin Dear Chaur Ann: Thanks very much for your e‐mail. Please refer to the attached e‐mail to Professor Welty, the first author of the book regarding our view on this matter. Please visit me during the tutorial session next week for verbal explanation on this matter. Sincerely yours, Chi‐Hwa Wang CN2125 ‐‐‐‐‐Original Message‐‐‐‐‐ From: Wang Chi‐Hwa Sent: Wednesday, January 26, 2011 8:46 AM To: '[email protected]' Cc: Lim Wai Kwan; Zhou Liwen; '[email protected]'; '[email protected]'; '[email protected]' Subject: Possible Typo Error in Your Book, Fundamentals of Momentum, Heat, and Mass Transfer, 5th edition" by Welty, Wicks, Wilson and Rorrer Dear Professor Welty: I am teaching a undergraduate course Heat and Mass Transfer at the National University of Singapore. Chemical engineering undergraduate students of mine [Class size = 315 (CN2125) + 72 (Cn2125E) = 387 students] have been using the book "Fundamentals of Momentum, Heat, and Mass Transfer, 5th edition" by Welty, Wicks, Wilson and Rorrer. Some of my students have noticed that there has been a possible error in the book. In Chapter 17 (Steady‐State Conduction), the Example 3 on page 238 is reproduced as below: Water and air are separated by mild‐steel plane wall. It is proposed to increase the heat‐transfer rate between these fluids by adding straight rectangular fins of 1.27‐mm thickness and 2.5‐cm length, spaced 1.27 cm apart. the air‐side and water‐side heat‐transfer coefficients may be assumed constant with values of 11.4 and 256 W/m^2.K respectively. Determine the percent change in total heat transfer when fins are placed on (a) the water side, (b) the air side, and (c) both sides. Two different printings of the 5th Edition give out the half‐thickness of the fin to be 1.27 mm and 0.64 mm respectively in the evaluation of fin efficiency. There is quite obvious difference between these 2 values. After inspection, both of the books which has different values but identical final answer. Both share the same title, edition, authors and publisher. Then, we have tried using both different values of half‐
Sent: Friday, January 21, 2011 8:08 AM
To: Li Xuan
Subject: RE: Question Regarding An Example in ID Book
Dear Xuan:
Thank you for your e-mail. The governing equation does not contain any convection
because for this problem, the governing equation is applied to the solid phase where there
is no convection. Please take note that general differential energy reduces to Heat /
Diffusion Equation when there is no convective terms. Conversely, the boundary
condition is the matching of conduction and convection flux at the interface.
Sincerely yours,
Chi-Hwa Wang
CN2125
From: Li Xuan
Sent: Thursday, January 20, 2011 10:13 AM
To: Wang Chi-Hwa
Subject: Question Regarding An Example in ID Book
Dear Prof Wang,
I have some doubt regarding Example 2.3 on P79 of ID Book (6th edition, section 2.4
Boundary and Initial Conditions). I am wondering why Heat Differential Equation is used
instead of the General Form of The Differential Energy Equation when there’s
convection involved in this question? Would the result be more accurate using the
general form?
Best Regards,
Li Xuan
‐‐‐‐‐Original Message‐‐‐‐‐ From: Wang Chi‐Hwa Sent: Thursday, January 20, 2011 8:41 AM To: Du Wei Subject: RE: Questions about CN2125 Dear Wei: Thanks for your e‐mail. (1) "Incompressible Flow" and "Isobaric flow" represent two typical systems usually encountered in flow process. These two are named and compared here for governing equation. Although same in the formula, they represent different physical meanings. (2) Thank you very much for your correction on this typographical mistake. I have amended it on the CN2125 Website. It should be dT/dx instead. Sincerely yours, Chi‐Hwa Wang CN2125 ‐‐‐‐‐Original Message‐‐‐‐‐ From: Du Wei Sent: Tuesday, January 18, 2011 8:21 PM To: Wang Chi‐Hwa Subject: Questions about CN2125 Dear Prof.Wang, I'd like to ask you some questions regarding heat conduction. 1. For the special forms of differential energy equation, the two special cases you showed us are A. an incompressible fluid without energy sources and with constant k, B. isobaric flow without energy sources and with constant k. Both A and B will lead to the formula: density*Cv*DT/Dt=k*laplacian T. I'm wondering that how does the condition of an incompressible flow or isobaric flow contribute to simply the original energy equation? In other words, I think the conditions of no energy sources and constand k alone will reduce to that form of equation. Hope you could explain to me. 2. On the slide 4 of lecture notes 2, for 1D, steady state transfer by conduction, the simplified expression is d(x^i*partial T partial x)/dx=0, why is it still partial derivative inside the bracket? should it be an ODE? Thank you very much! Best regards, Du Wei From: Wang Chi-Hwa
Subject: RE: clarification
Dear Kennedy:
Thanks for your e-mail. (1) Yes. Ts = T infinity when the value of h is very big meaning there is
negligible surface resistance.
(2) No, the boundary condition is x -> infinity, t-> To.
Sincerely yours,
Chi-Hwa Wang
From: Devid Desfreed Kennedy
Subject: clarification
Hi prof,
I am wondering can we assume that Ts = T infinity when the value of h is very big meaning there
is negligible surface resistance?
One more is that when we are looking at semi-infinite case, are we assuming that the other end
of the object will be always To thus can we simply just assume that it is insulated?
Thanks
cheers,
Devid
From: Wang Chi-Hwa
Subject: RE: some questions from today's lecture
Dear Leslie:
(1) Yes, Appendix F is a universal method for all as long as the three assumptions on WWWR
Page 261 are satisfied.
(2) Yes, you can check the Biot number and then decide your preferred method. There could be
more than one method for the same problem with full credit.
Sincerely yours,
Chi-Hwa Wang
From: Yeo Qi Fu Leslie
Subject: RE: some questions from today's lecture
Dear Prof Wang,
I have some further questions.
1. Appendix F can be used to solve for problems of all sorts of Biot number is it? Even if the
biot number is very small or very big? A universal method??
2. If the above answer is yes, then how is it different from case 1 (small biot number) and
case 2 (big biot number) in terms of solution application. This is because in example 4 of
your notes (I have read the appendix as well), I wouldn’t know which method to use…is
case 1 applicable to multi-dimension heat transfer as well? If yes, I would have used
method 1 cause it simpler…but would I get full credit for this problem?
Thank you sir for your time.
Cheers,
Leslie
From: Wang Chi-Hwa
Subject: RE: some questions from today's lecture
Dear Leslie:
(1) Yes, only for center line of solid. Other position -> Appendix F.
(2) Case 4: There is no definition of Biolt modulus. Case 2 solution should agree with Case 4
when L-> infinity.
(3) It is a 2D problem
(4) In this case, x is measured from the fixed interface, labeled as x = 0.
Sincerely yours,
Chi-Hwa Wang
CN2125 Co-instructor
From: Yeo Qi Fu Leslie
Subject: some questions from today's lecture
Dear Prof Wang,
Hello, i have some queries which i hope you can help me with.
1. For large Biot numbers, you said we can use the central temperature history chart to
solve. May I check is it only applicable to temperatures given at the center line of the
solid? If yes, then how do we solve the time taken for any other position to reach a
certain temperature, for large biot numbe cases? Is it using the temperature time charts
in appendix F?
2. You said that case 2 and case 4 have similar solution when L is long. Case 2’s solution is
applicable when biot number is large…how about in case 4? Is the biot number also large
in case 4 because L here is infinitely long?
3. In your sketch for transport in a rectangular bar with insulated ends. Is it a 2D or 3D
heat transfer problem?
4. for heat transfer into a semi infinite medium, there is no defined length scale right? But
may I know is the ‘x’ term in the error function the position along the infinite axis given in
the question ?
Cheers,
Leslie
From: Wang Chi-Hwa
Subject: RE: Enquiry about infinte wall
Dear Yee Ling:
The problem is finite surface resistance if a h value is given.
Chi-Hwa Wang
From: Tan Yee Ling
Dear Prof Wang,
I would like to ask how do we know for a problem about infinite wall problem, we should use the
equation for infinite surface resistance or finite surface resistance ? Thank you for your time.
Regards,
Yee Ling
-----Original Message----From: Wang Chi-Hwa
Subject: RE: CN2125
Dear Kendrick:
Appendix F is still valid for Bi module < 0.1. When Bi is of order 0.1
(say 0.042 for this case), there is very small temperature difference
across the solid object.
If you decide to use the lump-parameter analysis, you might get a final
answer (505K) which is not too far away from the present solution by
referring to Appendix F (533K).
Please note: Lump parameter solution is obtained by ignoring
temperature variation within the object.
Sincerely yours,
Chi-Hwa Wang
-----Original Message----From: Wong You-Ping Kendrick
Subject: RE: CN2125
thanks for the clarification. I have another question to ask and this
pertains to Chapter 18 lecture notes example 4. In this question, the
approach used was 2D method using the graphical charts. However, I
calculated the Bi number and it turns out to be approximately 0.04
(<0.1). Why is the lumped parameter analysis not applied here? Thanks.
regards,
Kendrick
-----Original Message----From: Wang Chi-Hwa
Subject: RE: Question regarding conduction-convection balance at
surface
Dear Barry:
(1) "When conduction and convection balance is stated as one of the
boundary conditions, does that mean that heat transfer at that surface
is steady? I.e, the surface temperature will remain at To all the time?
If not, then what's the physical meaning of this type of boundary
condition?"
Yes, this condition implies that the heat transfer rate through that
surface is quasi-steady as there is no accumulation of energy at that
surface.
(2) "Secondly, the third requirement for temperature-time chart
(Appendix F) is that the temperature of the boundary / adjacent fluid
is changed to a new value T(infinity) for t>=0, but when you look at,
for example figure F.1, n=1 curve (x=x1, boundary) for any m does have
finite amount of slope, which means that Y actually changes with X at
the boundary (i.e temperature changes with time). Why is this so?
Should the third requirement be interpreted differently?"
Yes, the fluid temperature is different from the temperature on the
solid surface. The solid surface temperature is allowed to change with
time.
Sincerely yours,
Chi-Hwa Wang
-----Original Message----From: Barry Tanny
Subject: Question regarding conduction-convection balance at surface
Dear Dr. Wang,
When conduction and convection balance is stated as one of the boundary
conditions, does that mean that heat transfer at that surface is
steady? I.e, the surface temperature will remain at To all the time? If
not, then what's the physical meaning of this type of boundary
condition?
Secondly, the third requirement for temperature-time chart (Appendix F)
is that the temperature of the boundary / adjacent fluid is changed to
a new value T(infinity) for t>=0, but when you look at, for example
figure F.1, n=1 curve (x=x1, boundary) for any m does have finite
amount of slope, which means that Y actually changes with X at the
boundary (i.e temperature changes with time). Why is this so? Should
the third requirement be interpreted differently?
Thank you for you attention, sorry for this long questions.
Regards,
Barry T.
From: Wang Chi-Hwa
Subject: RE: Fin efficiency
Dear Chian Jing:
Thank you for your e-mail. There is no temperature drop along the axis of
fin. However, there is still temperature difference between the solid fin and
surrounding air (fluid). Henceforth, there is constant convective heat transfer rate
"BETWEEN" fin and surrounding air. Since everwhere on the fin, the driving
force for convection heat transfer is the same "T(s)-T(inf).
Chi-Hwa Wang
From: Tee Chian Jing
Subject: Fin efficiency
hi sir,
this is regarding fin efficiency on the lecture u gave last friday. in the definition of fin efficiency, the
Qfin,max is the ideal heat transfer from the fin if the entire fin were at base temperature. i dont understand
here, if the entire fin were at the same temperature, would there be any heat transfer? as from what i know,
heat transfer happens when there is temperature difference. since the temperature is the same throughout
the fin why does heat transfer still happen?
-----Original Message----From: Wang Chi-Hwa
Subject: RE: Heat Transfer from extended surfaces
Dear Edwin:
kA(dT/dX)|x+dx - kA(dT/dX)|x -hS(T-T(inf))=0 (17-27)(Pg 237)
actually refers to
-(q2) + (q1) - (q3) = 0.
q2 = -kA(dT/dX)|x+dx
q1 = -kA(dT/dX)|x
q3 = hS(T-T(inf)).
Chi-Hwa Wang
________________________________
From: Toh Wei Quan Edwin
Subject: RE:Heat Transfer from extended surfaces
Dear Prof Wang,
I have a question with regards to the lecture on Friday, 20 Jan. After the lecture and doing
my own review, I am disturbed by the equation found in both the notes and the
text(WWWR). Question as follows,
Consider heat flow rates of fig 17.7(WWWR pg 237),
q(1)=q(2)+q(3) (17-25)
Further developments,
kA(dT/dX)|x+dx - kA(dT/dX)|x -hS(T-T(inf))=0 (17-27)(Pg 237)
Why is the second term evaluated at x and not at x+dx ? since all energy leaving the
control volume is suppose to be negative.
Referring to the above formula, q2 should be evaluated as "-kA(dT/dX)|x+dx". The
negative sign comes from the original fourier rate equation: Heat transfer direction is
Should not the first and second term both interchange? Would this not have any impact
If you see the orders in the first equation.
-(q2) + (q1) - (q3) = 0.
Thank you and regards,
Edwin Toh
_______________________________________________________________________
From: Wang Chi-Hwa
Subject: RE: Question regarding heat flux
Dear Qiu Rong:
Thank you for your e-mail. NO. There is no contradiction. In the planar wall problem,
both Qx (heat transfer rate, in the unit of W) and Qx" (heat flux, in the unit of W/m^2) are
conserved because the heat transfer areas are the same.
For the heat conduction in the radial direction for a cylinder, the conserved quantity is Qr
(heat transfer rate, in the unit of W) rather than Qr" (heat flux, in the unit of W/m^2). In this
case, the heat transfer area varies with the radial direction.
Sincerely yours,
Chi-Hwa Wang
From: Pek Qiu Rong
Subject: Question regarding heat flux
Dear Sir,
I have a few queries regarding heat flux which i hope you could clarify. I refer to slide 36 of
Lecture 1 which states that under steady-state, one dimensional conditions with no energy
generation, the heat flux is a constant in the direction of transfer. However, in example 1 on page
206 of the textbook by WWW, where there is also steady state with no energy generation and
with heat transfer only in the radial direction, the heat flux is different when calculated for the
inner and outer surface areas of the cylinder. Is this a contradiction?
I understand that q is always constant under steady state conditions but why is heat flux a
constant if in the case of a cylinder or a sphere where the surface areas are increasing in the
Regards
From: Wang Chi-Hwa
Subject: RE: FURTHER PROBLEMS ON TUT AND HW 1
Dear Hong Wei:
Attached please find my brief reply. You can approach me for more elaborated explanations.
Sincerely yours,
Chi-Hwa Wang
From: Lee Hong Wei
Subject: FURTHER PROBLEMS ON TUT AND HW 1
Importance: High
Dear Prof Wang,
After going through the tut/hw set for week 1, I still have further problems and I would like to
clarify them further with you.
For WWR 15.4, with the extra celotex layer, can I make the assumption that the surface
temperature of the fire clay wall and the interface temperature between the steel and fire-clay to
be the same? NO, both will be changed. I suppose this is not true after the addition of the
celotex layer. Further, will the convective heat transfer coefficient change with the addition of the
celotex layer? I suppose it will but I got the same solution as in the outline by assuming that it is a
It is assumed that the convective heat transfer coefficient a very weak function of solid material
surface. It remains the same value after the addition of celotex layer.
For WWR 15.11, can we express the area of the cross-section to vary as a linear function of the
position x, instead of expressing each side of the cross-section to be a linear function of the
position x, then taking the square of it? I tried both ways but the former didn’t work.
As mentioned in the lecture today, WWWR 15.11 involves linear expansion in two perpendicular
directions and hence the heat transfer area is a quadratic function for x. You can contrast this to
the non-uniform cross section fin problem mentioned today where there is shrinking in only one
direction and hence area is a linear functio of x. These two cases are different.
For ID 1.11, why is it that the change in T is expressed in deg C, instead of K, though the
parameters involved in the evaluation of the change in T involve K?
In this case 1oC = 1K. There is no difference between these two expressions.
For WWR 15.25, we would require the value of k for the air in between the 2 glass panes. It is
given in the solution outline as 0.0245. How do we determine this, since k is a function of T (nonlinear relationship for gases)? Possible to take the average of the 2 temperatures of the inner
surfaces of the glass panes? Or we should use interpolation from the data set of k values for air
from the Appendix, which requires us to know the temperature of air in the space, which is the
last part to the question? To find the average temperature of the air within the glass panes, can I
just take the average of the surface temperatures of the inner surfaces of the glass
panes? Finally, is there any convection between the 0.8 cm air space and the inner surfaces of
the glass panels? I suppose I can assume there is no convection since no information on the
convective coefficients between the region and the inner surfaces were mentioned. The provided
convective coefficients refer to those of the inside and outside temperature of the surroundings,
which I suppose.
There is no convection in the tight air gap. Based on my answer to commonly asked
questions (Appendix 3, see quotation below), you can select a reference temoperature in
the domain of interests and find the corresponding k value. This should be listed down as your
assumption. This simplification is justified because the temperature on the two sides of air gap is
not known yet.
1. How to read the values of thermal conductivity, heat transfer coefficients and
other relevant transport properties from the textbook?
homework assignments, quizzes and final examination, please also give the page
number of your reference for this. It is anticipated that some variation could be
found because most of these values come from experiments.
Best regards,
Hong Wei
-----Original Message----From: Wang Chi-Hwa
Subject: RE: CN2125 Questions....
Dear Pei Li:
(1) X1 value remains the same because both refer to the same "virtual block" as
mentioned in my lecture. In this virtual block, the center of wall is located at x/x1 = 0.5
and the insulated surface is at x/x1=0. For both cases, the "x" is counted by the
distance from the center of the symmetry center.
(2) The table can be constructed by referring to the initial condition and subsequently,
equation 18.29. The special cases that involve boundary points are shown in eq.
18.30. Just start from t = 0, you calculate the values for t = 1, etc till t = 23. You
proceed row by row until the central plane temperature has reached 425K.
Sincerely yours,
Chi-Hwa Wang
-----Original Message----From: Ong Pei Li
Subject: RE: CN2125 Questions....
Hi,
1. For example 3 in the lecture notes, the question asks us under two different scenario, one at
the centre and the other one at the insulated surface. Why is the Xi value for both cases the same?
Since we have imagine the box to be twice the size in the second case. Is it because of the way
we define the X coordinate, where X=0 at the insulated surface? (with reference to the appendix
F page 707, the small diagram at the top right hand corner, the way Xi is defined is from the
centre, where X is defined to be zero?)
2. Also, is it always true that we define X=0 at the centre of every shape and geometry? We can
define where it will be for our convenience sake right?
3. For example 5 regarding the numerical analysis, last page of the lecture notes, the table......
How do we go about evaluating the value T1,t at different succeeding times? Basically, how do
we calculate the values in the table, through equation (18-30) and ????
Regards,
Ong Pei Li
-----Original Message----From: Wang Chi-Hwa
Subject: RE: some qns.
Dear Win Win:
(a) Steady state means q = q (x) but q does not change with t.
(b) Yes. Central temperature only. Bi > > 0.1 when internal (conduction) resistance is important.
(c) We refer to Appendix F for example 18.3 because figure 18.3 gives only symmetric condition
for central plane only. Example 18.3 asks about the temperature at the center of wall (not on the
symmetric center) and insulated plane. Appendix F gives close fit to the given problem. If both
top and bottom plates are insulated, you only have heat transfer in the other two directions.
(d) Example 2 has two cases for radial transfer (1-D), and both radial and axial (2-D)
transfer. The 1D/2D transfer is included in Figure 18.3 for the two different cylinders chosen. One
is infinitely long (1D) while the other is finite (2D, including heat transfer from the caps).
Sincerely yours,
Chi-Hwa Wang
-----Original Message----From: Ng Win Win
Subject: RE: some qns.
Dear Dr. Wang,
Thank you for your reply. But I have a few more questions.
a)So does steady state means constant q?
b) For figure 18.3 in lecture notes (Y against relative time), is this chart only applicable to
find T in the centre point? and also what's the requirement/limit for Bi in order for this
chart to be applicable(eg. 2)?
c) Hence, for eg.3, do we have to use appendix F (WWW), not figure 18.3 to find the
time? Just to clarify, the reason we use the insulated surface as the midpoint is because of
symmetry purposes. What if both top and bottom are insulated? How do we create the
imaginary cuboid?
d) I'm confused that both eg. 2 and eg. 4 ask to find the time required for the center of the
cylinder to reach a certain temperature. Why is it that for eg.2, we only consider one case
whereas for eg.4 we consider Ycylinder and Ya? Is it because that the two examples are
using different chart , the former figure 18.3 while the latter from appendix F?
Sorry for asking so many questions. If you find it difficult to explain to me via email. I
hope that i can come to consult you in your office.
Thanks.
Win Win
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