Non-Steady-State Diffusion What is the value of erf(0.527) ? Value is between two data points Erf(z) 0.5633 University of Strathclyde Engineering Z erf(z) z erf(z) ? 0 0 0.55 0.5633 0.025 0.0282 0.60 0.6039 0.5205 0.05 0.0564 0.65 0.6420 0.10 0.1125 0.70 0.6778 0.15 0.1680 0.75 0.7112 0.50 0.55 0.20 0.2227 0.80 0.7421 0.527 Z 0.25 0.2763 0.85 0.7707 0.30 0.3286 0.90 0.7970 (0.527 0.50 0.35 0.3794 0.95 0.8209 erf(z) = 0.5205+ (0.5633 -0.5205) == 0.55 -0.50 0.40 0.4284 1.0 0.8427 0.45 0.4755 1.1 0.8802 0.50 0.5205 1.2 0.9103 erf(0.527) 0.5436 = Non-Steady-State Diffusion University of Strathclyde Engineering Cx-Co x 1-erf Cs-Co 2√Dt Cx-0 = 1 erf 0.1-0 (25 0.001 Cx 0.1 2√2.5 x 10-11 x 36000, - = 1 − erf(0.527) What is the value of erf(0.527) ? Co position, x Co zero (pure iron) C₁ = 0.1 x 1mm = 0.001m D = 2.5 x 10-11 m²/s t = 10 hours = 36000s
Non-Steady-State Diffusion What is the value of erf(0.527) ? Value is between two data points Erf(z) 0.5633 University of Strathclyde Engineering Z erf(z) z erf(z) ? 0 0 0.55 0.5633 0.025 0.0282 0.60 0.6039 0.5205 0.05 0.0564 0.65 0.6420 0.10 0.1125 0.70 0.6778 0.15 0.1680 0.75 0.7112 0.50 0.55 0.20 0.2227 0.80 0.7421 0.527 Z 0.25 0.2763 0.85 0.7707 0.30 0.3286 0.90 0.7970 (0.527 0.50 0.35 0.3794 0.95 0.8209 erf(z) = 0.5205+ (0.5633 -0.5205) == 0.55 -0.50 0.40 0.4284 1.0 0.8427 0.45 0.4755 1.1 0.8802 0.50 0.5205 1.2 0.9103 erf(0.527) 0.5436 = Non-Steady-State Diffusion University of Strathclyde Engineering Cx-Co x 1-erf Cs-Co 2√Dt Cx-0 = 1 erf 0.1-0 (25 0.001 Cx 0.1 2√2.5 x 10-11 x 36000, - = 1 − erf(0.527) What is the value of erf(0.527) ? Co position, x Co zero (pure iron) C₁ = 0.1 x 1mm = 0.001m D = 2.5 x 10-11 m²/s t = 10 hours = 36000s
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Hi, Can you help me understand why the result of erf(0.527) is 0.5436? I understand because the value of erf(Z) is in between two values on the table (the ones highlighted in purple), but I'm trying to make sense of the formula used to find the value erf(z) which then gives the value of .05436. Is it a standard formula that can be applied to different values in a different problem? I hope you can help.
Could you explain it to me? I have an exam in the morning.
![Non-Steady-State Diffusion
What is the value of erf(0.527) ?
Value is between two data points
Erf(z)
0.5633
University of
Strathclyde
Engineering
Z
erf(z)
z
erf(z)
?
0
0
0.55
0.5633
0.025
0.0282
0.60
0.6039
0.5205
0.05
0.0564
0.65
0.6420
0.10
0.1125
0.70
0.6778
0.15
0.1680
0.75
0.7112
0.50
0.55
0.20
0.2227
0.80
0.7421
0.527
Z
0.25
0.2763
0.85
0.7707
0.30
0.3286
0.90
0.7970
(0.527 0.50
0.35
0.3794
0.95
0.8209
erf(z) = 0.5205+ (0.5633 -0.5205)
==
0.55 -0.50
0.40
0.4284
1.0
0.8427
0.45
0.4755
1.1
0.8802
0.50
0.5205
1.2
0.9103
erf(0.527) 0.5436
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4efc2a5-039b-466f-b774-04edf61d54e8%2F15db50ef-5c15-4504-8450-85ab3e00947f%2Facw2iz_processed.png&w=3840&q=75)
Transcribed Image Text:Non-Steady-State Diffusion
What is the value of erf(0.527) ?
Value is between two data points
Erf(z)
0.5633
University of
Strathclyde
Engineering
Z
erf(z)
z
erf(z)
?
0
0
0.55
0.5633
0.025
0.0282
0.60
0.6039
0.5205
0.05
0.0564
0.65
0.6420
0.10
0.1125
0.70
0.6778
0.15
0.1680
0.75
0.7112
0.50
0.55
0.20
0.2227
0.80
0.7421
0.527
Z
0.25
0.2763
0.85
0.7707
0.30
0.3286
0.90
0.7970
(0.527 0.50
0.35
0.3794
0.95
0.8209
erf(z) = 0.5205+ (0.5633 -0.5205)
==
0.55 -0.50
0.40
0.4284
1.0
0.8427
0.45
0.4755
1.1
0.8802
0.50
0.5205
1.2
0.9103
erf(0.527) 0.5436
=
![Non-Steady-State Diffusion
University of
Strathclyde
Engineering
Cx-Co
x
1-erf
Cs-Co
2√Dt
Cx-0
= 1 erf
0.1-0
(25
0.001
Cx
0.1
2√2.5 x 10-11 x 36000,
- = 1 − erf(0.527)
What is the value of erf(0.527) ?
Co
position, x
Co zero (pure iron)
C₁ = 0.1
x 1mm = 0.001m
D = 2.5 x 10-11 m²/s
t = 10 hours = 36000s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4efc2a5-039b-466f-b774-04edf61d54e8%2F15db50ef-5c15-4504-8450-85ab3e00947f%2F51tc3k_processed.png&w=3840&q=75)
Transcribed Image Text:Non-Steady-State Diffusion
University of
Strathclyde
Engineering
Cx-Co
x
1-erf
Cs-Co
2√Dt
Cx-0
= 1 erf
0.1-0
(25
0.001
Cx
0.1
2√2.5 x 10-11 x 36000,
- = 1 − erf(0.527)
What is the value of erf(0.527) ?
Co
position, x
Co zero (pure iron)
C₁ = 0.1
x 1mm = 0.001m
D = 2.5 x 10-11 m²/s
t = 10 hours = 36000s
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