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p0468

# p0468 - stationary so that u cv = u letting ρ denote cup...

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564 CHAPTER 4. KINEMATICS 4.68 Chapter 4, Problem 68 Problem: At a county fair, a vender is selling souvenir coffee cups. The cups are stacked on a table and they are selling at a rate ˙ n sold . Because of lax security, cups are being stolen at a rate ˙ n stolen = α ˙ n sold , where α = 1 / 10 . Finally, because the cups have not been carefully stacked, some are being broken at a rate that is 5% of the rate at which the number of unbroken cups on the table is changing. Letting N cup denote the number of unbroken cups on the table, use the Reynolds Transport Theorem to determine dN cup /dt as a function of ˙ n sold . Solution: To begin, we must define the system and the control volume. Let the system be all unbroken cups and define the control volume to be the unbroken cups remaining on the table. Also, let N denote the number of unbroken cups in the system and N cup the number of unbroken cups in the control volume. Then, since our control volume is
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Unformatted text preview: stationary so that u cv = u , letting ρ denote cup number density, the Reynolds Transport Theorem tells us that dN dt = d dt 888 V ρ dV + 8 s 8 S ρ u · n dS By definition of ρ , necessarily N cup = \$\$\$ ρ dV . Also, the net flux of cups out of the control volume is 8 s 8 S ρ u · n dS = ˙ n sold + ˙ n stolen Now, the rate of change of the number of cups in the system is the rate at which cups are broken so that dN/dt = λ dN cup /dt , where λ = 0 . 05 = 1 / 20 . So, our basic conservation law is λ dN cup dt = dN cup dt + ˙ n sold + ˙ n stolen Since we are given ˙ n stolen = α ˙ n sold , we find (1 − λ ) dN cup dt = − (1 + α ) ˙ n sold Therefore, dN cup dt = − w 1 + α 1 − λ W ˙ n sold Finally, since α = 1 / 10 and λ = 1 / 20 , we conclude that dN cup dt = − X 11 / 10 19 / 20 ~ ˙ n sold = − 22 19 ˙ n sold...
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