PROBLEM 5.102 KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with no internal generation; suddenly a uniform generation, q=108W/m³, occurs when the element is inserted into the core while the surfaces experience convection (Too,h). FIND: Temperature distribution 1.5s after element is inserted into the core. SCHEMATIC: Fuel element x=5x10 m³/s k-30W/m-K Coolant Too 250°C h=1100W/m² K xحا T(x,0)=250°C 9-10°W/m³ at t>0 2.3.4.5.h k+ax=2mm Coolant Too.h L-10mm PROBLEM 5.102 (Cont.) To be well within the stability limit, select At = 0.3s, which corresponds to At 5×10 m²/sx0.3s Fo= Ax² t=pAt=0.3p(s). (0.002m) = 0.375 Substituting numerical values with q =108W/m³, the nodal equations become T=0.375[21" +10°W/m²³ (0.002m)²/30W/m-K]+(1-2×0.375) Tf TP+1=0.375[2TP =0.375 21+13.33]+0.25 T TP+1 = 0.375[T+TE+13.33] +0.25 TP тр = T-0.375 1+1+13.33]+0.25 T ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) q=0, initially; at t > 0, q is uniform. ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used. Using the nodal network of Example 5.8, the same finite-difference equations may be used. Interior nodes, m = 1, 2, 3, 4 TP+1 Fo TP m-1 +TP ġ(Ax)² + m+1 2 +(1-2 Fo)T (1) T=0.375 1+1+13.33] +0.25 T T=0.375 1+1+1333] +0.25 T T+1=2×0.375 +0.0733×250+ +(1-2×0.375-2×0.0733×0.375)T 13.33 2 TP+1=0.750 [T+24.99] +0.195 Tg. Midplane node, m = 0 Same as Eq. (1), but with TP TP Surface node, m = 5 T}+1=2 Fo| T} +BiTot m-1 = m+1' 9(Ax)2 2k +(1-2F0-2Bi-Fo)T. The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the following parameters: Bi Fo≤ hAx 1100W/m² Kx(0.002m). k 1/2 (1+Bi) 30W/m-K = 0.466 Fo(Ax)² (0.002m)² ΔΙΣ α =0.466- = 0.0733 = 0.373s. 5x10 6m²/s 2 Continued..... The initial temperature distribution is T₁ = 250°C at all nodes. The marching solution, following the procedure of Example 5.8, is represented in the table below. (2) P t(s) Το Τι T₂ T3 T4 T5(°C) 0 0 250 250 250 250 250 1 0.3 255.00 255.00 2 0.6 3 0.9 260.00 260.00 265.00 265.00 255.00 255.00 255.00 260.00 260.00 260.00 259.72 250 254.99 4 1.2 270.00 270.00 265.00 265.00 264.89 264.39 270.00 269.96 269.74 268.97 5 1.5 275.00 275.00 274.98 274.89 274.53 273.50 The desired temperature distribution T(x, 1.5s), corresponds to p = 5. COMMENTS: Note that the nodes near the midplane (0,1) do not feel any effect of the coolant during the first 1.5s time period. (3) (4) (5) (6) (7) (8)
PROBLEM 5.102 KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with no internal generation; suddenly a uniform generation, q=108W/m³, occurs when the element is inserted into the core while the surfaces experience convection (Too,h). FIND: Temperature distribution 1.5s after element is inserted into the core. SCHEMATIC: Fuel element x=5x10 m³/s k-30W/m-K Coolant Too 250°C h=1100W/m² K xحا T(x,0)=250°C 9-10°W/m³ at t>0 2.3.4.5.h k+ax=2mm Coolant Too.h L-10mm PROBLEM 5.102 (Cont.) To be well within the stability limit, select At = 0.3s, which corresponds to At 5×10 m²/sx0.3s Fo= Ax² t=pAt=0.3p(s). (0.002m) = 0.375 Substituting numerical values with q =108W/m³, the nodal equations become T=0.375[21" +10°W/m²³ (0.002m)²/30W/m-K]+(1-2×0.375) Tf TP+1=0.375[2TP =0.375 21+13.33]+0.25 T TP+1 = 0.375[T+TE+13.33] +0.25 TP тр = T-0.375 1+1+13.33]+0.25 T ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) q=0, initially; at t > 0, q is uniform. ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used. Using the nodal network of Example 5.8, the same finite-difference equations may be used. Interior nodes, m = 1, 2, 3, 4 TP+1 Fo TP m-1 +TP ġ(Ax)² + m+1 2 +(1-2 Fo)T (1) T=0.375 1+1+13.33] +0.25 T T=0.375 1+1+1333] +0.25 T T+1=2×0.375 +0.0733×250+ +(1-2×0.375-2×0.0733×0.375)T 13.33 2 TP+1=0.750 [T+24.99] +0.195 Tg. Midplane node, m = 0 Same as Eq. (1), but with TP TP Surface node, m = 5 T}+1=2 Fo| T} +BiTot m-1 = m+1' 9(Ax)2 2k +(1-2F0-2Bi-Fo)T. The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the following parameters: Bi Fo≤ hAx 1100W/m² Kx(0.002m). k 1/2 (1+Bi) 30W/m-K = 0.466 Fo(Ax)² (0.002m)² ΔΙΣ α =0.466- = 0.0733 = 0.373s. 5x10 6m²/s 2 Continued..... The initial temperature distribution is T₁ = 250°C at all nodes. The marching solution, following the procedure of Example 5.8, is represented in the table below. (2) P t(s) Το Τι T₂ T3 T4 T5(°C) 0 0 250 250 250 250 250 1 0.3 255.00 255.00 2 0.6 3 0.9 260.00 260.00 265.00 265.00 255.00 255.00 255.00 260.00 260.00 260.00 259.72 250 254.99 4 1.2 270.00 270.00 265.00 265.00 264.89 264.39 270.00 269.96 269.74 268.97 5 1.5 275.00 275.00 274.98 274.89 274.53 273.50 The desired temperature distribution T(x, 1.5s), corresponds to p = 5. COMMENTS: Note that the nodes near the midplane (0,1) do not feel any effect of the coolant during the first 1.5s time period. (3) (4) (5) (6) (7) (8)
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
100%
### Given the resolution of the problem 5.102 below, develop a python program (without using numpy) that shows the found temperature distribution
![PROBLEM 5.102
KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with
no internal generation; suddenly a uniform generation, q=108W/m³, occurs when the
element is inserted into the core while the surfaces experience convection (Too,h).
FIND: Temperature distribution 1.5s after element is inserted into the core.
SCHEMATIC:
Fuel element
x=5x10 m³/s
k-30W/m-K
Coolant
Too 250°C
h=1100W/m² K
xحا
T(x,0)=250°C
9-10°W/m³ at t>0
2.3.4.5.h
k+ax=2mm
Coolant
Too.h
L-10mm
PROBLEM 5.102 (Cont.)
To be well within the stability limit, select At = 0.3s, which corresponds to
At 5×10 m²/sx0.3s
Fo=
Ax²
t=pAt=0.3p(s).
(0.002m)
= 0.375
Substituting numerical values with q =108W/m³, the nodal equations become
T=0.375[21" +10°W/m²³ (0.002m)²/30W/m-K]+(1-2×0.375) Tf
TP+1=0.375[2TP
=0.375 21+13.33]+0.25 T
TP+1 = 0.375[T+TE+13.33] +0.25 TP
тр
=
T-0.375 1+1+13.33]+0.25 T
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3)
q=0, initially; at t > 0, q is uniform.
ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used.
Using the nodal network of Example 5.8, the same finite-difference equations may be used.
Interior nodes, m = 1, 2, 3, 4
TP+1
Fo TP
m-1
+TP
ġ(Ax)²
+
m+1
2
+(1-2 Fo)T
(1)
T=0.375 1+1+13.33] +0.25 T
T=0.375 1+1+1333] +0.25 T
T+1=2×0.375 +0.0733×250+ +(1-2×0.375-2×0.0733×0.375)T
13.33
2
TP+1=0.750 [T+24.99] +0.195 Tg.
Midplane node, m = 0
Same as Eq. (1), but with TP TP
Surface node, m = 5
T}+1=2 Fo| T} +BiTot
m-1
=
m+1'
9(Ax)2
2k
+(1-2F0-2Bi-Fo)T.
The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the
following parameters:
Bi
Fo≤
hAx 1100W/m² Kx(0.002m).
k
1/2
(1+Bi)
30W/m-K
= 0.466
Fo(Ax)² (0.002m)²
ΔΙΣ
α
=0.466-
= 0.0733
= 0.373s.
5x10
6m²/s
2
Continued.....
The initial temperature distribution is T₁ = 250°C at all nodes. The marching solution,
following the procedure of Example 5.8, is represented in the table below.
(2)
P
t(s)
Το
Τι
T₂
T3
T4
T5(°C)
0
0
250
250
250
250
250
1
0.3
255.00 255.00
2
0.6
3
0.9
260.00 260.00
265.00 265.00
255.00 255.00 255.00
260.00 260.00 260.00 259.72
250
254.99
4
1.2
270.00 270.00
265.00 265.00 264.89 264.39
270.00 269.96 269.74 268.97
5
1.5
275.00 275.00
274.98 274.89 274.53 273.50
The desired temperature distribution T(x, 1.5s), corresponds to p = 5.
COMMENTS: Note that the nodes near the midplane (0,1) do not feel any effect of the
coolant during the first 1.5s time period.
(3)
(4)
(5)
(6)
(7)
(8)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F452fd15a-9cba-484f-a9d5-07fdd6d75662%2F451e2e37-7131-4dc9-a5c1-7d1dcfe58f7e%2Fxdjt9ts_processed.png&w=3840&q=75)
Transcribed Image Text:PROBLEM 5.102
KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with
no internal generation; suddenly a uniform generation, q=108W/m³, occurs when the
element is inserted into the core while the surfaces experience convection (Too,h).
FIND: Temperature distribution 1.5s after element is inserted into the core.
SCHEMATIC:
Fuel element
x=5x10 m³/s
k-30W/m-K
Coolant
Too 250°C
h=1100W/m² K
xحا
T(x,0)=250°C
9-10°W/m³ at t>0
2.3.4.5.h
k+ax=2mm
Coolant
Too.h
L-10mm
PROBLEM 5.102 (Cont.)
To be well within the stability limit, select At = 0.3s, which corresponds to
At 5×10 m²/sx0.3s
Fo=
Ax²
t=pAt=0.3p(s).
(0.002m)
= 0.375
Substituting numerical values with q =108W/m³, the nodal equations become
T=0.375[21" +10°W/m²³ (0.002m)²/30W/m-K]+(1-2×0.375) Tf
TP+1=0.375[2TP
=0.375 21+13.33]+0.25 T
TP+1 = 0.375[T+TE+13.33] +0.25 TP
тр
=
T-0.375 1+1+13.33]+0.25 T
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3)
q=0, initially; at t > 0, q is uniform.
ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used.
Using the nodal network of Example 5.8, the same finite-difference equations may be used.
Interior nodes, m = 1, 2, 3, 4
TP+1
Fo TP
m-1
+TP
ġ(Ax)²
+
m+1
2
+(1-2 Fo)T
(1)
T=0.375 1+1+13.33] +0.25 T
T=0.375 1+1+1333] +0.25 T
T+1=2×0.375 +0.0733×250+ +(1-2×0.375-2×0.0733×0.375)T
13.33
2
TP+1=0.750 [T+24.99] +0.195 Tg.
Midplane node, m = 0
Same as Eq. (1), but with TP TP
Surface node, m = 5
T}+1=2 Fo| T} +BiTot
m-1
=
m+1'
9(Ax)2
2k
+(1-2F0-2Bi-Fo)T.
The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the
following parameters:
Bi
Fo≤
hAx 1100W/m² Kx(0.002m).
k
1/2
(1+Bi)
30W/m-K
= 0.466
Fo(Ax)² (0.002m)²
ΔΙΣ
α
=0.466-
= 0.0733
= 0.373s.
5x10
6m²/s
2
Continued.....
The initial temperature distribution is T₁ = 250°C at all nodes. The marching solution,
following the procedure of Example 5.8, is represented in the table below.
(2)
P
t(s)
Το
Τι
T₂
T3
T4
T5(°C)
0
0
250
250
250
250
250
1
0.3
255.00 255.00
2
0.6
3
0.9
260.00 260.00
265.00 265.00
255.00 255.00 255.00
260.00 260.00 260.00 259.72
250
254.99
4
1.2
270.00 270.00
265.00 265.00 264.89 264.39
270.00 269.96 269.74 268.97
5
1.5
275.00 275.00
274.98 274.89 274.53 273.50
The desired temperature distribution T(x, 1.5s), corresponds to p = 5.
COMMENTS: Note that the nodes near the midplane (0,1) do not feel any effect of the
coolant during the first 1.5s time period.
(3)
(4)
(5)
(6)
(7)
(8)
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 1 steps
Recommended textbooks for you
Database System Concepts
Computer Science
ISBN:
9780078022159
Author:
Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:
McGraw-Hill Education
Starting Out with Python (4th Edition)
Computer Science
ISBN:
9780134444321
Author:
Tony Gaddis
Publisher:
PEARSON
Digital Fundamentals (11th Edition)
Computer Science
ISBN:
9780132737968
Author:
Thomas L. Floyd
Publisher:
PEARSON
Database System Concepts
Computer Science
ISBN:
9780078022159
Author:
Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:
McGraw-Hill Education
Starting Out with Python (4th Edition)
Computer Science
ISBN:
9780134444321
Author:
Tony Gaddis
Publisher:
PEARSON
Digital Fundamentals (11th Edition)
Computer Science
ISBN:
9780132737968
Author:
Thomas L. Floyd
Publisher:
PEARSON
C How to Program (8th Edition)
Computer Science
ISBN:
9780133976892
Author:
Paul J. Deitel, Harvey Deitel
Publisher:
PEARSON
Database Systems: Design, Implementation, & Manag…
Computer Science
ISBN:
9781337627900
Author:
Carlos Coronel, Steven Morris
Publisher:
Cengage Learning
Programmable Logic Controllers
Computer Science
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education