Given Values Heater core surface temp Heater core diameter Thermal conductivity of sleeve Sleeve width at fin base External air temp Convection coefficient Calculated Values 2D conduction shape factor 2D conduction resistance Case Description Baseline Case Double number of fins Baseline Case Double fin thickness Baseline Case Double fin length T₁ D k₁ W₂ T❤ h S' R',20 [-] 16 32 16 16 [°C] [m] [W/m.K] 16 16 [m] [°C] [W/m².K] Number Thickness N t [m] 0.004 0.004 [-] [m-K/W] Fin Parameters 0.004 0.008 0.004 0.004 300 0.02 240 0.04 50 500 Length L [m] 0.02 0.02 0.02 0.02 0.02 0.04 The elemental unit of an air heater consists of a long circular rod of diameter D = 20 mm. The rod is encapsulated by a finned aluminum sleeve and it generates thermal energy by Ohmic heating. The N=16 fins each have thickness t = 4 mm and length L = 20 mm and they are integrally fabricated with the square sleeve of width w, = 40 mm. T, -300°C Too = 50 °C h = 500 W/m².K A'f [m] A'b [m] Airflow Th A't [m] Heater (q. k) Draw a thermal circuit for the system and evaluate the heat rate per unit length from the rod for the following operating conditions: kg - 240 W/m-K w E'S ni Sleeve, k m [m³¹] jiyi Fin Array and Thermal Circuit Calculations Lc [m] -T₂ nf [-] no R' fins [-] [m-K/W] q' [W/m] д 2х [-]
Given Values Heater core surface temp Heater core diameter Thermal conductivity of sleeve Sleeve width at fin base External air temp Convection coefficient Calculated Values 2D conduction shape factor 2D conduction resistance Case Description Baseline Case Double number of fins Baseline Case Double fin thickness Baseline Case Double fin length T₁ D k₁ W₂ T❤ h S' R',20 [-] 16 32 16 16 [°C] [m] [W/m.K] 16 16 [m] [°C] [W/m².K] Number Thickness N t [m] 0.004 0.004 [-] [m-K/W] Fin Parameters 0.004 0.008 0.004 0.004 300 0.02 240 0.04 50 500 Length L [m] 0.02 0.02 0.02 0.02 0.02 0.04 The elemental unit of an air heater consists of a long circular rod of diameter D = 20 mm. The rod is encapsulated by a finned aluminum sleeve and it generates thermal energy by Ohmic heating. The N=16 fins each have thickness t = 4 mm and length L = 20 mm and they are integrally fabricated with the square sleeve of width w, = 40 mm. T, -300°C Too = 50 °C h = 500 W/m².K A'f [m] A'b [m] Airflow Th A't [m] Heater (q. k) Draw a thermal circuit for the system and evaluate the heat rate per unit length from the rod for the following operating conditions: kg - 240 W/m-K w E'S ni Sleeve, k m [m³¹] jiyi Fin Array and Thermal Circuit Calculations Lc [m] -T₂ nf [-] no R' fins [-] [m-K/W] q' [W/m] д 2х [-]
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Please answer this question and solve write out each step at least for the baseline case. If you reference any tables please site them or include. Thanks.
![Given Values
Heater core surface temp
Heater core diameter
Thermal conductivity of sleeve
Sleeve width at fin base
External air temp
Convection coefficient
Calculated Values
2D conduction shape factor
2D conduction resistance
Case Description
Baseline Case
Double number of fins
Baseline Case
Double fin thickness
Baseline Case
Double fin length
Ts
D
ks
W s
T..
h
S'
R' c,2D
[-]
16
32
16
16
[°C]
[m]
[W/m.K]
Fin Parameters
Number Thickness
N
t
16
16
[m]
[°C]
[W/m².K]
[-]
[m.K/W]
[m]
0.004
0.004
0.004
0.008
0.004
0.004
300
0.02
240
0.04
50
500
Length
L
[m]
0.02
0.02
0.02
0.02
0.02
0.04
The elemental unit of an air heater
consists of a long circular rod of
diameter D = 20 mm. The rod is
encapsulated by a finned aluminum
sleeve and it generates thermal energy
by Ohmic heating. The N = 16 fins
each have thickness t = 4 mm and
length L = 20 mm and they are
integrally fabricated with the square
sleeve of width W, = 40 mm.
T, = 300°C
Too = 50 °C
h = 500 W/m².K
A'f
[m]
A'b
[m]
A't
[m]
Airflow
To h
ks
Heater
(q, kn)
Draw a thermal circuit for the system and evaluate the heat rate per unit length from the
rod for the following operating conditions:
Sleeve, k
= 240 W/m.K
m
[m³¹]
ER
Tin
M
Lc
[m]
-Fins, N
Fin Array and Thermal Circuit Calculations
nf
[-]
T₁
L
no
[-]
R' fins
q'
[m.K/W] [W/m]
Δ 2x
[-]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F40d4b8b9-78fc-4c41-88fb-42b8f2c3a636%2F4fded071-744c-4ccd-aea3-635494bc18e0%2Fsavh2k_processed.png&w=3840&q=75)
Transcribed Image Text:Given Values
Heater core surface temp
Heater core diameter
Thermal conductivity of sleeve
Sleeve width at fin base
External air temp
Convection coefficient
Calculated Values
2D conduction shape factor
2D conduction resistance
Case Description
Baseline Case
Double number of fins
Baseline Case
Double fin thickness
Baseline Case
Double fin length
Ts
D
ks
W s
T..
h
S'
R' c,2D
[-]
16
32
16
16
[°C]
[m]
[W/m.K]
Fin Parameters
Number Thickness
N
t
16
16
[m]
[°C]
[W/m².K]
[-]
[m.K/W]
[m]
0.004
0.004
0.004
0.008
0.004
0.004
300
0.02
240
0.04
50
500
Length
L
[m]
0.02
0.02
0.02
0.02
0.02
0.04
The elemental unit of an air heater
consists of a long circular rod of
diameter D = 20 mm. The rod is
encapsulated by a finned aluminum
sleeve and it generates thermal energy
by Ohmic heating. The N = 16 fins
each have thickness t = 4 mm and
length L = 20 mm and they are
integrally fabricated with the square
sleeve of width W, = 40 mm.
T, = 300°C
Too = 50 °C
h = 500 W/m².K
A'f
[m]
A'b
[m]
A't
[m]
Airflow
To h
ks
Heater
(q, kn)
Draw a thermal circuit for the system and evaluate the heat rate per unit length from the
rod for the following operating conditions:
Sleeve, k
= 240 W/m.K
m
[m³¹]
ER
Tin
M
Lc
[m]
-Fins, N
Fin Array and Thermal Circuit Calculations
nf
[-]
T₁
L
no
[-]
R' fins
q'
[m.K/W] [W/m]
Δ 2x
[-]
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