∆y (1) n ∆z ∆x 1.pdf · rorrer g. & foster d.g. (2015) fundamentals of momentum, heat,...
TRANSCRIPT
Y
Z X
(1)
V
Ө
∆y (1) (2) (left & right)
n ∆z
∆x
(X,Y,Z)
How to compute CS
pe V n dA
uv v------ (A)
cosV n v n uv v
at plane (1): cos 1 1xV n V V uv v
------- (B)
n Ө
V
At plane (2), cos 1 1xV n v V uv v
------- (C)
V
Ө
N
(2)
Y
Z X
Y
Z X
2
2
:2
V:
2:
Vmean kinectic energy
e gy U gy potential energy
U energy
internal
------ (D)
Applying (A),(B),(C),(D) at planes (1) and (2), one obtains:
zyp
UgyVp
UgyV xxxxx
2
V
2
V 22
(4)
y ( bottom & top) (3) z
x
(X,Y,Z)
Similarly, applying (A), (B),(C) & (D) at planes (3) and (4), one obtains:
2 2V V
2 2y y y y y
p pV gy U V gy U x z
∆y (5) (6) (front & back)
∆z
∆x
(X,Y,Z)
Applying (A), (B),(C) & (D) at planes (5) and (6), one obtains:
2 2V V
2 2z z z z z
p pV gy U V gy U x y
Dt
DTCTk vq
where
z
T
y
T
x
TT ,, , ------gradient of a scalar is vector
, ,
, ,
T T Tk T k k k
x y z
T T Tk T k k k
x y z
zyx,,
? A where A= (a1,a2,a3)
z
a
y
a
x
a
321 ------ divergence of a vector is scalar
vectorscalar
gradient
divergence
*
)()()(z
Tk
zy
Tk
yx
Tk
xTk
:Dt
DT Substantial derivative
z
Tu
y
Tu
x
Tu
t
Tzyx
This term vanishes if there is no convective flow contribution
q : Heat generation per unit volume unit time
Ф: dissipation function
222222
2z
V
x
V
y
V
z
V
x
V
y
Vμ
z
V
y
V
x
Vμ xzzyyxzyx
Dear CN2125 Students:
Attached please find the three questions for the CN2125 assignment #1. The numbers cited
refer to the following main textbook listed in the module website. (WRF) WELTY J.R.,
RORRER G. & FOSTER D.G. (2015) Fundamentals of Momentum, Heat, and Mass
Transfer, International Student Version, 6th Edition. Publisher John Wiley & Son, New
York, and (ID) INCROPERA F.P.; DEWITT D.P.; BERGMAN T.L.; LAVINE
A.S.(2013) Fundamentals of Heat and Mass Transfer, 7th Edition. Publisher John Wiley &
Son, New York.
(1) Problem 1: WRF Chapter 14 Question (15)
Liquid nitrogen at 77K is stored in a cylindrical container having an inside diameter of
25cm. The cylinder is made of stainless steel (thermal conductivity = 17.3W/m.K) and
has a wall thickness of 1.2cm. Insulation is to be added to the outside surface of the
cylinder to reduce the nitrogen boil-off rate to 25% of its value without insulation. The
insulation to be used has a thermal conductivity of 0.13W/m.K. Energy loss through the
top and bottom ends of the cylinder may be presumed negligible. Neglecting radiation
effects, determine the thickness of insulation when the inner surface of the cylinder is at
77K, the convective heat-transfer coefficient at the insulation surface has a value of 12
W/m2.K, and the surrounding air is at 25oC.
Hint:
(2) Problem 2: ID 3.58.
An electrical current of 700A flows through a stainless steel cable having a diameter of
5mm and an electrical resistance of 6 x 10-4 /m (i.e. per meter of cable length). The cable
is in an environment having a temperature of 30oC, and the total coefficient associated with
convection and radiation between the cable and the environment is approximately 25
W/m2-K.
(a) If the cable is bare, what is the surface temperature?
(b) If a very thin coating of electrical resistance is applied to the cable, with a constant
resistance of 0.02 m2-K/W, what are the insulation and cable surface temperatures?
(c) There is some concern about the ability of the insulation to withstand elevated
temperatures. What thickness of this insulation (k = 0.5 W/m-K) will yield the lowest
value of the maximum insulation temperature? What is the value of the maximum
temperature when the thickness is used?
(3) Problem 3: WRF Chapter 17 Question (17)
How long will a 1-ft concrete wall subject to a surface temperature of 1500oF on one side
maintain the other side below 130oF? The wall is initially at 70oF. Hint: You may solve
the problem by the solution for semi-infinite wall with negligible surface resistance. In
general, the accuracy is greater than 99.5% when the following relationship is true:
L/2(t)^0.5 > 2. Otherwise, the solution accuracy is compromised and the solution is only
an approximate solution.
Due 12:00noon, Feb. 12, 2020@ CN2125 mailbox in the ChBE main office (E5-02-09)
or by e-mail submission to [email protected]
The graded assignments will be returned to you by e-mail roughly 2 weeks after your
submission.
Sincerely yours,
Chi-Hwa Wang, CN2125 Co-instructor
Grader’s Remark Q1
Grader: LI He Email: [email protected]
Comment:
Most students did well in this question. Students can get partial marks if they can show correct
equations about the relationship of thermal resistances. Students can obtain full marks if the final
result is correct and accurate.
Common Mistakes:
1. Many students failed to use the correct number in the equations. For example: Wrong
radius or diameter, wrong thermal conductivity.
2. For students who use trial and error method to solve the equation, they may get an
inaccurate answer as the deviation was not small enough in their final trial. Their result of
the radius of the cylinder is around 175cm or 180cm.
3. Several students did not show a correct equation to calculate the total resistances
with/without the insulation layer. The correct ones are shown below.
Solution:
Liquid nitrogen at 77K is stored in a cylindrical container having an inside diameter of 25cm.
The cylinder is made of stainless steel (thermal conductivity = 17.3W/m.K) and has a wall
thickness of 1.2cm. Insulation is to be added to the outside surface of the cylinder to reduce the
nitrogen boil-off rate to 25% of its value without insulation. The insulation to be used has a thermal
conductivity of 0.13W/m.K. Energy loss through the top and bottom ends of the cylinder may be
presumed negligible. Neglecting radiation effects, determine the thickness of insulation when the
inner surface of the cylinder is at 77K, the convective heat-transfer coefficient at the insulation
surface has a value of 12 W/m2.K, and the surrounding air is at 25oC.
Answer:
R𝑖𝑛𝑠𝑖𝑑𝑒 = 12.5 𝑐𝑚 Routside = 13.7 𝑐𝑚
Without Insulation:
RA = 𝑅𝑐𝑜𝑛𝑣𝑒𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑡𝑒𝑒𝑙 + 𝑅𝑠𝑡𝑒𝑒𝑙−𝑎𝑖𝑟
=1
2πL [ln (
13.712.5
)
17.3+
1
12 ∙ 0.137]
With Insulation:
RB = 𝑅𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑡𝑒𝑒𝑙 + 𝑅𝑠𝑡𝑒𝑒𝑙−𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 + 𝑅𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛−𝑎𝑖𝑟
=1
2πL [ln (
13.712.5
)
17.3+
1
12 ∙ 𝑟0+
ln (𝑟0
13.7)
0.13]
Because
4qw = 𝑞𝑤𝑜 ∆T = constant
4𝑅𝐴 = RB
ln (13.712.5
)
17.3+
1
12 ∙ 𝑟0+
ln (𝑟0
13.7)
0.13= 2.454
𝐾
𝑊
By trial and error,
r0 = 17.7 𝑐𝑚
𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 < 0.008𝐾
𝑊
Question 2.
Grader: Teoh Jia Heng. Email: [email protected]
In this question, students were required to determine the surface temperature of the cable and
insulation under different configurations, where the thermal resistance in each case was varied.
Overall, most students were able to solve each of the parts in this question without much difficulty.
Considering that the maximum temperature mentioned in part c) is ambiguous, marks are awarded
if students can determine either the maximum cable surface temperature or the maximum
insulation surface temperature.
Common mistakes include the following:
1) When calculating the maximum cable surface temperature in c) using the surrounding
temperature, some students did not include the resistance of the thin coating that was
introduced in b). Ignoring the resistance of the thin coating is only acceptable if the
maximum insulation surface temperature is being calculated using the surrounding
temperature instead.
2) Some students have mistaken the maximum cable surface temperature as the maximum
insulation surface temperature and vice versa in part b).
3) Some students have mistaken the critical radius as the thickness of the insulation when
instead, the critical radius should be the summation of both the radius of the cable and the
radius of the insulation.
An electrical current of 700A flows through a stainless steel cable having a diameter of 5mm and
an electrical resistance of 6 x 10-4 W/m (i.e. per meter of cable length). The cable is in an
environment having a temperature of 30oC, and the total coefficient associated with convection
and radiation between the cable and the environment is approximately 25 W/m2-K.
(a) If the cable is bare, what is the surface temperature? (3 marks)
Assumptions:
1) System is at steady state.
2) Heat conduction is only occurring in one dimension (radial dimension).
3) All properties are constant throughout the cable.
From the energy balance
𝐻𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑏𝑙𝑒 = 𝐻𝑒𝑎𝑡 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
𝑞𝐿 = 𝐼2𝑅𝑒𝐿 = ℎ𝐴(𝑇𝑠 − 𝑇∞) (𝟏𝒎)
𝐼2𝑅𝑒𝐿 = ℎ(𝜋𝐷𝐿)(𝑇𝑠 − 𝑇∞)
𝐼2𝑅𝑒 = 𝜋ℎ𝐷(𝑇𝑠 − 𝑇∞)
𝑇𝑠 =𝐼2𝑅𝑒
𝜋ℎ𝐷+ 𝑇∞ (𝟏𝒎)
𝑇𝑠 =(700 𝐴)2 (6 × 10−4 𝑊
𝑚 )
𝜋 ( 25𝑊
𝑚2𝐾) (5 × 10−3 𝑚)
+ (303.15 𝐾) = 1051.8 𝐾 (𝟏𝒎)
If temperature is expressed in terms of °𝐶
𝑇𝑠 =(700 𝐴)2 (6 × 10−4 𝑊
𝑚 )
𝜋 ( 25𝑊
𝑚2𝐾) (5 × 10−3 𝑚)
+ (30°𝐶) = 778.7 °𝐶
(b) If a very thin coating of electrical resistance is applied to the cable, with a constant
resistance of 0.02 m2-K/W, what are the insulation and cable surface temperatures? (4
marks)
Determining the cable surface temperature (𝑇𝑐𝑠):
𝑞𝐿 = 𝐼2𝑅𝑒𝐿 = (1
1ℎ
+ 𝑅) 𝐴(𝑇𝑐𝑠 − 𝑇∞)
𝐼2𝑅𝑒 = (1
1ℎ𝜋𝐷
+𝑅
𝜋𝐷
) (𝑇𝑐𝑠 − 𝑇∞)
𝑇𝑐𝑠 = 𝐼2𝑅𝑒 (1
ℎ𝜋𝐷+
𝑅
𝜋𝐷) + 𝑇∞ (𝟏𝒎)
= (700 𝐴)2 (6 × 10−4𝑊
𝑚 ) (
1
( 25𝑊
𝑚2𝐾) 𝜋(5 × 10−3 𝑚)
+0.02
𝑚2𝐾𝑊
𝜋(5 × 10−3 𝑚)) + 303.15 𝐾
= 1426.1 𝐾 (𝟏𝒎)
If temperature is expressed in terms of °𝐶
= (700 𝐴)2 (6 × 10−4𝑊
𝑚 ) (
1
( 25𝑊
𝑚2𝐾) 𝜋(5 × 10−3 𝑚)
+0.02
𝑚2𝐾𝑊
𝜋(5 × 10−3 𝑚)) + 30°𝐶
= 1153 °𝐶
For the insulation surface temperature (𝑇𝑖𝑠):
𝑞𝐿 = 𝐼2𝑅𝑒𝐿 = (1
𝑅) 𝐴(𝑇𝑐𝑠 − 𝑇𝑖𝑠)
𝐼2𝑅𝑒 = (𝜋𝐷
𝑅) (𝑇𝑐𝑠 − 𝑇𝑖𝑠)
𝑇𝑖𝑠 = 𝑇𝑐𝑠 −𝐼2𝑅𝑒𝑅
𝜋𝐷 (𝟏𝒎)
= 1426.1𝐾 −(700 𝐴)2 (6 × 10−4 𝑊
𝑚 ) (0.02𝑚2𝐾
𝑊)
𝜋(5 × 10−3𝑚)= 1051.8 𝐾 (𝟏𝒎)
If temperature is expressed in terms of °𝐶
= 1153°𝐶 −(700 𝐴)2 (6 × 10−4 𝑊
𝑚 ) (0.02𝑚2𝐾
𝑊)
𝜋(5 × 10−3𝑚)= 778.7°𝐶
(c) There is some concern about the ability of the insulation to withstand elevated
temperatures. What thickness of this insulation (k = 0.5 W/m-K) will yield the lowest
value of the maximum insulation temperature? What is the value of the maximum
temperature when the thickness is used? (3 marks)
The maximum insulation temperature could be reduced by reducing the resistance to hear transfer
from the outer surface of the insulation. Such a reduction is possible if the thickness of the cable
is less than the critical thickness i.e. 𝐷𝑖 < 𝐷𝑐𝑟
𝑟𝑐𝑟 =𝑘
ℎ=
0.5𝑊
𝑚𝐾
25𝑊
𝑚2𝐾
= 0.02 𝑚
∴ 𝐷𝑐𝑟 = 2𝑟𝑐𝑟 = 0.04 𝑚 > 𝐷𝑖 = 5 × 10−3 𝑚 (𝟏𝒎)
To minimize the maximum temperature which exists at the inner surface of the insulation, the
thickness of the insulation,
𝑡 =(0.04 𝑚 − 0.005 𝑚)
2= 0.0175 𝑚 (𝟏𝒎)
Determining the cable surface temperature (𝑇𝑐𝑠):
𝑞 = 𝐼2𝑅𝑒 = [1
1ℎ𝜋𝐷𝑐𝑟
+𝑅
𝜋𝐷𝑖+
12𝜋𝑘
ln (𝐷𝑐𝑟
𝐷𝑖)
] (𝑇𝑐𝑠 − 𝑇∞)
𝑇𝑐𝑠 =[1
ℎ𝜋𝐷𝑐𝑟+
𝑅
𝜋𝐷𝑖+
1
2𝜋𝑘ln (
𝐷𝑐𝑟
𝐷𝑖) ] 𝐼2𝑅𝑒 + 𝑇∞
= [1
𝜋 (25𝑊
𝑚2𝐾) (0.04 𝑚)
+0.02
𝑚2𝐾𝑊
𝜋(0.005 𝑚)
+1
2𝜋 (0.5𝑊
𝑚𝐾)ln (
0.04 𝑚
0.005 𝑚)] (700 𝐴)2 (6 × 10−4
𝑊
𝑚) + 303.15 𝐾
= 965.7 𝐾
If temperature is expressed in terms of °𝐶
= [1
𝜋 (25𝑊
𝑚2𝐾) (0.04 𝑚)
+0.02
𝑚2𝐾𝑊
𝜋(0.005 𝑚)
+1
2𝜋 (0.5𝑊
𝑚𝐾)ln (
0.04 𝑚
0.005 𝑚)] (700 𝐴)2 (6 × 10−4
𝑊
𝑚) + 30°𝐶
= 692.5°𝐶
For the insulation surface temperature (𝑇𝑖𝑠):
𝑞𝐿 = 𝐼2𝑅𝑒𝐿 = (1
𝑅𝑡,𝑐) 𝐴(𝑇𝑐𝑠 − 𝑇𝑖𝑠)
𝐼2𝑅𝑒 = (𝜋𝐷
𝑅𝑡,𝑐) (𝑇𝑐𝑠 − 𝑇𝑖𝑠)
𝑇𝑖𝑠 = 𝑇𝑐𝑠 −𝐼2𝑅𝑒𝑅𝑡,𝑐
𝜋𝐷
= 965.7 𝐾 −(700 𝐴)2 (6 × 10−4 𝑊
𝑚 ) (0.02𝑚2𝐾
𝑊)
𝜋(5 × 10−3𝑚)= 591.4 𝐾 (𝟏𝒎)
If temperature is expressed in terms of °𝐶
= 692.5°𝐶 −(700 𝐴)2 (6 × 10−4 𝑊
𝑚 ) (0.02𝑚2𝐾
𝑊)
𝜋(5 × 10−3𝑚)= 318.2°𝐶
Comments: Use of the critical insulation thickness in lieu of a thin coating has the effect of
reducing the maximum insulation temperature from 778.7°𝐶 to 318.2°𝐶 . Use of the critical
insulation thickness also reduces the cable surface temperature from 778.7°𝐶 or 1153°𝐶 with a
thin coating to 692.5°𝐶.
Grader’s Remark Q3
Grader: LI He Email: [email protected]
Comment:
A large number of students did well in this problem. Students can obtain partial marks if they
are able to solve this problem by using the solution for semi-infinite wall. The rest of the marks
will be given if the student can calculate the time correctly and validate that the accuracy of this
result is lower than 99.5%.
Common Mistakes:
1. Some students did not achieve the physical properties of concrete from the Appendix of
WRF. Most of them just use a thermal diffusivity which is 0.75x10-6 m2/s. However, the
correct thermal diffusivity is around 0.6x10-6m2/s or 0.0231 ft2/h.
2. A large number of students did not estimate the accuracy of result.
3. A few students did not get the accurate result when they were solving erf(x)=0.9580.
Most of them got x=1.4 or 1.5 only which made their answer wrong.
4. A few students were confused with less-than sign and greater-than sign when they were
validating the accuracy. Correct result should be t<2.7hr but they got t>2.7hr or t<27hr.
Solution:
How long will a 1-ft concrete wall subject to a surface temperature of 1500oF on one side
maintain the other side below 130oF? The wall is initially at 70oF. Hint: You may solve the
problem by the solution for semi-infinite wall with negligible surface resistance. In general, the
accuracy is greater than 99.5% when the following relationship is true: L/2(αt)^0.5 > 2.
Otherwise, the solution accuracy is compromised, and the solution is only an approximate
solution.
All of the repaired physical properties (@ 68 oF) of concrete can be found in WRF Appendix H.
𝜌 = 144𝑙𝑏𝑚
𝑓𝑡3 𝑐𝑝 = 0.21
𝐵𝑇𝑈
𝑙𝑏𝑚 °𝐹 𝑘 = 0.70
𝐵𝑇𝑈
ℎ 𝑓𝑡 °𝐹
𝛼 =𝑘
𝜌 𝑐𝑝= 0.0231
𝑓𝑡2
ℎ
Semi-infinite wall solution:
erf (1 𝑓𝑡
2√𝛼𝑡) =
𝑇𝑠 − 𝑇
𝑇𝑠 − 𝑇0
1500 − 130
1500 − 70= erf (
3.29
𝑡12
) = 0.958
3.29
𝑡12
≈ 1.44
𝑡 ≅ 5.2 ℎ
Validate:
1𝑓𝑡
2√0.231𝑡≈ 0.46
The result is compromised and an approximate solution.