Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Question 2 Based on question (1), sketch the dimensionless temperature profile 8(x) versus aL = 2, € [0, 1] aL = 3, Є [0, 1] L Ꮎ 1.2 1.0 0.8 0.6 0.4 0.2 Ꮎ 1.2- 1.0 0.8 0.6 0.4 0.2 for the two following cases. Ꮎ 1.2ㅏ 1.2 a L=3 1.0 1.0 0.8 0.8 a L=2 0.6 0.6 0.4 a L=2 0.4 a L=3 0.2 a L = 3 0.2 α L=2 Χ x Χ 0.2 0.4 0.6 0.8 1.0 1.2 L 0.2 0.4 0.6 0.8 1.0 1.2 L 0.2 0.4 0.6 0.8 1.0 1.2 L a L=2 a L-3 x 0.2 0.4 0.6 0.8 1.0 1.2 L Question 1 From the following equation, assume that the cross-sectional area A is a constant. d dx dT ка dx KAT) = hC(T - T fluid) Also assume that all other physical parameters k, h, A, C, L, Tw wall' and Tfluid are well known and assumed to be positive constants. (a) Based on these assumptions, the above ODE can be simplified to which of the follow? hc(T-Tfluid) T ка dx2 dA dk dT + A. dx dx dx hc(T-T fluid) dT Ока = (hC(T Tfluid)) dx dx 이은 dA dk d²T + A. = dx dx dx² hc(T-T fluid) dx -(KA) = (hC(T-Tfluid)) dT (b) Solve the ODE based on the following two boundary conditions. For simplicity, let a² Ch for when formulating your solution. (Use T, to represent T fluid and Tw to represent T ка wall in your response.) At x = 0, T = T¸ wall' dT At x = L, = 0. dx T T T(x) @ αχ 2aL 1+e e + (1-T)eal -αx e + 1+e 2aL - If - (c) Define a "dimensionless" variable (x) T(x) T₂ T wall-fluid fluid Based on the T(x) solution, rewrite the solution in terms of (x) and the hyperbolic cosine function, where cosh(y) 8(x) = cosh(x) + 1-e2aL ox 1+e 2aL 1+cosh(x) ey + e˜y = Again, let a² Ch α = 2 for when formulating your solution. ка

Question 2 Based on question (1), sketch the dimensionless temperature profile 8(x) versus aL = 2, € [0, 1] aL = 3, Є [0, 1] L Ꮎ 1.2 1.0 0.8 0.6 0.4 0.2 Ꮎ 1.2- 1.0 0.8 0.6 0.4 0.2 for the two following cases. Ꮎ 1.2ㅏ 1.2 a L=3 1.0 1.0 0.8 0.8 a L=2 0.6 0.6 0.4 a L=2 0.4 a L=3 0.2 a L = 3 0.2 α L=2 Χ x Χ 0.2 0.4 0.6 0.8 1.0 1.2 L 0.2 0.4 0.6 0.8 1.0 1.2 L 0.2 0.4 0.6 0.8 1.0 1.2 L a L=2 a L-3 x 0.2 0.4 0.6 0.8 1.0 1.2 L Question 1 From the following equation, assume that the cross-sectional area A is a constant. d dx dT ка dx KAT) = hC(T - T fluid) Also assume that all other physical parameters k, h, A, C, L, Tw wall' and Tfluid are well known and assumed to be positive constants. (a) Based on these assumptions, the above ODE can be simplified to which of the follow? hc(T-Tfluid) T ка dx2 dA dk dT + A. dx dx dx hc(T-T fluid) dT Ока = (hC(T Tfluid)) dx dx 이은 dA dk d²T + A. = dx dx dx² hc(T-T fluid) dx -(KA) = (hC(T-Tfluid)) dT (b) Solve the ODE based on the following two boundary conditions. For simplicity, let a² Ch for when formulating your solution. (Use T, to represent T fluid and Tw to represent T ка wall in your response.) At x = 0, T = T¸ wall' dT At x = L, = 0. dx T T T(x) @ αχ 2aL 1+e e + (1-T)eal -αx e + 1+e 2aL - If - (c) Define a "dimensionless" variable (x) T(x) T₂ T wall-fluid fluid Based on the T(x) solution, rewrite the solution in terms of (x) and the hyperbolic cosine function, where cosh(y) 8(x) = cosh(x) + 1-e2aL ox 1+e 2aL 1+cosh(x) ey + e˜y = Again, let a² Ch α = 2 for when formulating your solution. ка

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question

I need help with Question 2 described below and an explanation for the solution. (Differential Equations)

Question 2 Based on question (1), sketch the dimensionless temperature profile 8(x) versus aL = 2, € [0, 1] aL = 3, Є [0, 1] L Ꮎ 1.2 1.0 0.8 0.6 0.4 0.2 Ꮎ 1.2- 1.0 0.8 0.6 0.4 0.2 for the two following cases. Ꮎ 1.2ㅏ 1.2 a L=3 1.0 1.0 0.8 0.8 a L=2 0.6 0.6 0.4 a L=2 0.4 a L=3 0.2 a L = 3 0.2 α L=2 Χ x Χ 0.2 0.4 0.6 0.8 1.0 1.2 L 0.2 0.4 0.6 0.8 1.0 1.2 L 0.2 0.4 0.6 0.8 1.0 1.2 L a L=2 a L-3 x 0.2 0.4 0.6 0.8 1.0 1.2 L
Transcribed Image Text:Question 2 Based on question (1), sketch the dimensionless temperature profile 8(x) versus aL = 2, € [0, 1] aL = 3, Є [0, 1] L Ꮎ 1.2 1.0 0.8 0.6 0.4 0.2 Ꮎ 1.2- 1.0 0.8 0.6 0.4 0.2 for the two following cases. Ꮎ 1.2ㅏ 1.2 a L=3 1.0 1.0 0.8 0.8 a L=2 0.6 0.6 0.4 a L=2 0.4 a L=3 0.2 a L = 3 0.2 α L=2 Χ x Χ 0.2 0.4 0.6 0.8 1.0 1.2 L 0.2 0.4 0.6 0.8 1.0 1.2 L 0.2 0.4 0.6 0.8 1.0 1.2 L a L=2 a L-3 x 0.2 0.4 0.6 0.8 1.0 1.2 L
Question 1 From the following equation, assume that the cross-sectional area A is a constant. d dx dT ка dx KAT) = hC(T - T fluid) Also assume that all other physical parameters k, h, A, C, L, Tw wall' and Tfluid are well known and assumed to be positive constants. (a) Based on these assumptions, the above ODE can be simplified to which of the follow? hc(T-Tfluid) T ка dx2 dA dk dT + A. dx dx dx hc(T-T fluid) dT Ока = (hC(T Tfluid)) dx dx 이은 dA dk d²T + A. = dx dx dx² hc(T-T fluid) dx -(KA) = (hC(T-Tfluid)) dT (b) Solve the ODE based on the following two boundary conditions. For simplicity, let a² Ch for when formulating your solution. (Use T, to represent T fluid and Tw to represent T ка wall in your response.) At x = 0, T = T¸ wall' dT At x = L, = 0. dx T T T(x) @ αχ 2aL 1+e e + (1-T)eal -αx e + 1+e 2aL - If - (c) Define a "dimensionless" variable (x) T(x) T₂ T wall-fluid fluid Based on the T(x) solution, rewrite the solution in terms of (x) and the hyperbolic cosine function, where cosh(y) 8(x) = cosh(x) + 1-e2aL ox 1+e 2aL 1+cosh(x) ey + e˜y = Again, let a² Ch α = 2 for when formulating your solution. ка
Transcribed Image Text:Question 1 From the following equation, assume that the cross-sectional area A is a constant. d dx dT ка dx KAT) = hC(T - T fluid) Also assume that all other physical parameters k, h, A, C, L, Tw wall' and Tfluid are well known and assumed to be positive constants. (a) Based on these assumptions, the above ODE can be simplified to which of the follow? hc(T-Tfluid) T ка dx2 dA dk dT + A. dx dx dx hc(T-T fluid) dT Ока = (hC(T Tfluid)) dx dx 이은 dA dk d²T + A. = dx dx dx² hc(T-T fluid) dx -(KA) = (hC(T-Tfluid)) dT (b) Solve the ODE based on the following two boundary conditions. For simplicity, let a² Ch for when formulating your solution. (Use T, to represent T fluid and Tw to represent T ка wall in your response.) At x = 0, T = T¸ wall' dT At x = L, = 0. dx T T T(x) @ αχ 2aL 1+e e + (1-T)eal -αx e + 1+e 2aL - If - (c) Define a "dimensionless" variable (x) T(x) T₂ T wall-fluid fluid Based on the T(x) solution, rewrite the solution in terms of (x) and the hyperbolic cosine function, where cosh(y) 8(x) = cosh(x) + 1-e2aL ox 1+e 2aL 1+cosh(x) ey + e˜y = Again, let a² Ch α = 2 for when formulating your solution. ка
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS