126_pdfsam_math 54 differential equation solutions odd

126_pdfsam_math 54 differential equation solutions odd - H...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 3 Solving this separable equation yields T ( t )=19 . 17 + Ce 3 t . When t =0weareg iven T (0) = 55. Using this information to solve for C gives C =35 . 83. Hence, the van temperature is given by T ( t )=19 . 17 + 35 . 83 e 3 t . To Fnd out when the temperature in the van will reach 27 C, we let T ( t ) = 27 and solve for t . Thus, we see that 27 = 19 . 17 + 35 . 83 e 3 t e 3 t = 7 . 83 35 . 83 0 . 2185 t ln(0 . 2185) 3 0 . 5070 (hr) or 30 . 4m in . 13. Since the time constant is 64, we have K =1 / 64. The temperature in the tank increases at therateo f2
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: H ( t ) = 2 2 = 4 per hr . We are given that T (0) = 110, and that the temperature M ( t ) outside the tank is a constant 80 . Hence the temperature in the tank is governed by dT dt = 1 64 [80 T ( t )] + 4 = 1 64 T ( t ) + 5 . 25 , T (0) = 110 . Solving this separable equation gives T ( t ) = 336 + Ce t/ 64 . To Fnd C , we use the initial condition to see that T (0) = 110 = 336 + C C = 226 . This yields the equation T ( t ) = 336 226 e t/ 64 . 122...
View Full Document

This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

Ask a homework question - tutors are online