I need help on solving this question:A steel plate having a thickness of 100 mm is suddenly exposed to a hot gas at 1000 oC ina furnace. One surface of the plate is heated while the other surface of the plate can beapproximated to be adiabatic. The initial temperature of the steel plate is 20 oC. The thermalconductivity and thermal diffusivity of the steel plate are 34.8 W/m K and 0.555 x 10-5 m2/srespectively. The convective heat-transfer coefficient is 174 W/m2 K. (The pictures are the conduction charts given)a) Find the time necessary to raise the surface temperature of the steel plate to 500 oC. b) Find the maximum temperature difference in the cross-section of the steel plate at the time evaluated in part a). c) Determine the heat energy transferred to the steel plate per unit wall surface area by the time evaluated in part a). 1.00.7T(0,1)-T, 0.5040.3T-T,0.20.10.070.050.040.030.021.00.90.8(0) 0.70.60.50.40.30.20.010.00700050.0040.0000.0000.001T(xt)-T, 1.0T(0,1)-7 0.90.80.70.600.1010-50.11Bi=0.40.50.60.5 0.70.4 0.80.30.22x/L=0.200.01SETitHIND0.91.0alk0.13Bi4 8 12 16 20 24 28 40 60 80Bi=alk1.0 101 1007 Bi10-4 10-3 10-2 10-1J110Bi=GaL100 120 140 200 300 400 500 600Foutk/pcl²102103Bi Fo=1² kpcFigure 1 - Dimensionless Transient Temperatures and Heat Transfer for an Infinite Slab1041.00.70.50.4T(0,1)-T, 03T-T, 020.10.070.050.040.030.020.010.0070.0050.0040.0030.0020.0010.50.401.0T(r.)-T 0.9T(0,1)-T, 0.8 0.40.7 0.50.6 0.6.r/R=0.20.30.2 0.9-0.11.0a) 0.9(1)(0) 0.80.70.60.50.40.70.800.0141.0120.1aRBi==kf6512065573 4 8 12 16 201.0Bi=aR100.30.20.1010-5 10-4 10-3 10-2Bi100Bi24 28 401Bi=1060 80 100 120 140 200 300Fo=tktk/pcR²10-1Bi³ Fo=to/kpcFigure 2 - Dimensionless Transient Temperatures and Heat Transfer for a Long CylinderaRallk102 103104 1.00.70.50.4T(0,1)-T, 030.2T-T,0.010.0070.0050.0040.0030.0020.80.70.60.10.070.050.040.030.020.0011.020.9O(0)1.0T(r.1)-T, 0.9T(0,1)-T, 0.8 0.40.7 0.50.60.50.400.5 1.0 1.5r/R=0.20.60.70.3 0.80.2 0.90.10.50.40.30.20.1010-51.000.01Bi =aRk0.110-42.02.5Bi = 0.0010.0020.0050.01-10-31.00.02-3 4 5 6 7 8 9 10BiBi0.05-0.110-210 Bi 10010-1ולי30401070110Bi=Fo-tk102aRk130 170***PCR²210 250103104Bi² Fo=to²/kpcFigure 3 - Dimensionless Transient Temperatures and Heat Transfer for a Sphere

Answered: Image /qna-images/answer/2c74d62d-b01e-4192-a2cd-550df2ef7556.jpg

Answered: A steel plate having a thickness of 100…

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

See similar textbooks

Related questions

Question

I need help on solving this question:

A steel plate having a thickness of 100 mm is suddenly exposed to a hot gas at 1000 ^oC in
a furnace. One surface of the plate is heated while the other surface of the plate can be
approximated to be adiabatic. The initial temperature of the steel plate is 20 ^oC. The thermal
conductivity and thermal diffusivity of the steel plate are 34.8 W/m K and 0.555 x 10^-5 m²/s
respectively. The convective heat-transfer coefficient is 174 W/m² K.

(The pictures are the conduction charts given)

a) Find the time necessary to raise the surface temperature of the steel plate to 500 ^oC.

b) Find the maximum temperature difference in the cross-section of the steel plate at the time evaluated in part a).

c) Determine the heat energy transferred to the steel plate per unit wall surface area by the time evaluated in part a).

1.0
0.7
T(0,1)-T, 0.5
04
0.3
T-T,
0.2
0.1
0.07
0.05
0.04
0.03
0.02
1.0
0.9
0.8
(0) 0.7
0.6
0.5
0.4
0.3
0.2
0.01
0.007
0005
0.004
0.000
0.000
0.001
T(xt)-T, 1.0
T(0,1)-7 0.9
0.8
0.7
0.6
0
0.1
0
10-5
0.1
1
Bi=
0.4
0.5
0.6
0.5 0.7
0.4 0.8
0.3
0.2
2
x/L=0.2
0
0.01
SE
Tit
HIND
0.9
1.0
al
k
0.1
3
Bi
4 8 12 16 20 24 28 40 60 80
Bi=
al
k
1.0 101 100
7 Bi
10-4 10-3 10-2 10-1
J
1
10
Bi=
G
aL
100 120 140 200 300 400 500 600
Foutk/pcl²
102
103
Bi Fo=1² kpc
Figure 1 - Dimensionless Transient Temperatures and Heat Transfer for an Infinite Slab
104
1.0
0.7
0.5
0.4
T(0,1)-T, 03
T-T, 02
0.1
0.07
0.05
0.04
0.03
0.02
0.01
0.007
0.005
0.004
0.003
0.002
0.001
0.5
0.4
0
1.0
T(r.)-T 0.9
T(0,1)-T, 0.8 0.4
0.7 0.5
0.6 0.6.
r/R=0.2
0.3
0.2 0.9-
0.1
1.0
a) 0.9
(1)
(0) 0.8
0.7
0.6
0.5
0.4
0.7
0.8
0
0.01
4
1.0
12
0.1
aR
Bi==
k
f
65
1206557
3 4 8 12 16 20
1.0
Bi=
aR
10
0.3
0.2
0.1
0
10-5 10-4 10-3 10-2
Bi
100
Bi
24 28 40
1
Bi=
10
60 80 100 120 140 200 300
Fo=tk
tk/pcR²
10-1
Bi³ Fo=to/kpc
Figure 2 - Dimensionless Transient Temperatures and Heat Transfer for a Long Cylinder
aR
all
k
102 103
104

1.0
0.7
0.5
0.4
T(0,1)-T, 03
0.2
T-T,
0.01
0.007
0.005
0.004
0.003
0.002
0.8
0.7
0.6
0.1
0.07
0.05
0.04
0.03
0.02
0.001
1.0
20.9
O(0)
1.0
T(r.1)-T, 0.9
T(0,1)-T, 0.8 0.4
0.7 0.5
0.6
0.5
0.4
0
0.5 1.0 1.5
r/R=0.2
0.6
0.7
0.3 0.8
0.2 0.9
0.1
0.5
0.4
0.3
0.2
0.1
0
10-5
1.0
0
0.01
Bi =
aR
k
0.1
10-4
2.0
2.5
Bi = 0.001
0.002
0.005
0.01-
10-3
1.0
0.02-
3 4 5 6 7 8 9 10
Bi
Bi
0.05-
0.1
10-2
10 Bi 100
10-1
ולי
30
40
10
70
110
Bi=
Fo-tk
102
aR
k
130 170
***
PCR²
210 250
103
104
Bi² Fo=to²/kpc
Figure 3 - Dimensionless Transient Temperatures and Heat Transfer for a Sphere

Expert Solution

Step by step

Solved in 4 steps with 7 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions