HW3_solution473

HW3_solution473 - ECE 473/TAM 413 Homework Assignment#3...

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Homework Assignment #3 Solutions - 1 ECE 473/TAM 413 Homework Assignment #3 Solutions 1. Equation 5.2.1’s P = ! rT k in Kinsler et al. provides one form of the perfect gas law. Another form is PV = nRT k where n is the number of moles and R is the universal gas constant. Obtain a relationship between r, R and the gas’ molecular weight M. PV = nRT k and P = ! rT k (see Eq. 5.2.1 and Appendix Eq A9.17) V = nRT k P = nR P P ! r " # \$ % & = nR ! r ( r = nR ! V m = ! V (massof gas) " r = nR m and M = m n (molecular weight) Therefore, r = R M or R = rM 2. Problem 5.2.1 in Kinsler et al. (a) P P o = ! ! o " # \$ % & ( = 1 + s ( ) ( ) 1 + ( s + ... = 1 + ( s for s<<1, that is, P = P o 1 + ! s ( ) = P o + P o ! s From (5.2.5) P ! P o = B " ! " o " o # \$ % & ( , P = P o + B ! " ! o ! o # \$ % & ( = P o + Bs , B = ! P o Also note from (5.2.6) p = Bs where B = ! o " P "! # \$ % & ( ! o , B = ! o " "! P o ! ! o # \$ % & ( ) * + , - , . / , 0 , # \$ % % & ( ( ! o = ) P o ! ! o # \$ % & ( ) 1 1 # \$ % & ( ! o = ) P o (b) B = ! P o = ! " rT K ( ) , so B ! T K 3. The planet Jupiter has an atmosphere of methane at a temperature of -130˚C. Estimate the speed of sound there. Use c = ! rT k (from notes) ! = 1.3 for methane M = 16.04 for methane r = R o M = 8314 J /(kg ! K) 16.04 = 518.3 J /(kg ! T k = 273 ! 130 = 143K c = 1.3 518.3 J /(kg ! ( ) = 310 m /s

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Homework Assignment #3 Solutions - 2 4. Problem 5.4.1 in Kinsler et al. (Hint: think total derivative d ! dt ) Take the total derivative of density with respect to time, where density is a function of time and space. d ! dt = "! " t + "! " x " x " t + "!
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HW3_solution473 - ECE 473/TAM 413 Homework Assignment#3...

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