PS1_sol_S08 - Problem Set 1 (2.006, Spring 08) Part A 1. =...

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Problem Set 1 (2.006, Spring 08) Part A 1. () 2 3 1/ 0.144 / 883 / 4 mk g s ms A D kg m υ ρ π == = ⎛⎞ ⎜⎟ ⎝⎠ & 3 41 / 4 Re 25.31 (Laminar flow) 503 10 (0.1 ) m D kg s A Dm D Pa s m ρυ μμ μ = ×⋅ & & 2 3 2 5 (883 / ) 0.144 / 64 64 1000 2.53 2.53 2.32 10 Re 25.31 2 0.1 2 kg m m s Lm f Pf P a = Δ = = = × 2. 0 hA t cV TT T Te ∞∞ −= ( ) ( ) 2 2 41 3 3 4 33 ( 1 0 / ) 3.489 10 4 8933 / 385 / 0.025 3 hR hA h W m K s cV cR kg m J kgK m cR ρρ ρπ = = × 0 20 10 1 ln ln 1987 30 10 3.489 10 oo TT cV C C ts h A C C s −− ⎛⎞ ⎛⎞ =− = = ⎜⎟ ⎜⎟ × 3. Only the initial and final states are concerned. ( ) ( ) 22 11 5 5 ln ln 330 2 10 1 5231 / ln 1 2078 / ln 942 / 300 10 p TP Sm c m R KP a kg J kg K kg J kg K J K a Δ= × =⋅ −⋅ = 4. ( ) ( )( )( ) 21 1 3153 / 330 300 94.6 E mc T T kg J kg K K K kJ − = = 5. Entropy does not change during the process. 2078 / 6 5231 / 2 5 1 10 ln ln 0 300 748.8 10 p R Jk g K c g K p P P a c m R T T K K P P a = ⇒ = = = 6. For the gates of unit width, 0 3 3 223 o Lg h h M gh L h h L m ∑= + = = = h 1 3 h 1 2 L L P gh = 1 2 ave P = Yellow arrows show resultant forces by the given pressure distribution
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7. () 2 Crossection area 4 ww a FP D g h P π ρ ≈⋅ + (vacuum inside the sphere) ( ) ( ) 2 32 5 0.3 1000 / 9.806 / 100 10 76383 4 mk g m m s m P a N =⋅ + = (Note: If the pressure inside the sphere is atmospheric, then 2 69316 4 w FD g h N = ) 8. c) is correct. For the ice cube, the weight , w ice immersed Wg V = . After melted, the weight of water wm e l t e d V = , ice immersed melted VV = 9. By Bernoulli equation, the velocity of the exiting stream is maintained at 0 υ ( a P P = Q , z~constant) Momentum equation: CV d Fd V dt ρυ ∑= ur r ( ) ( ) rr CS CS dA m υ ρυ +⋅ = = r r r r r & And ( ) ( ) 34 2 0 13280 / 10 10 / 13.28 / in out m A kg m m m s kg s m == = = && In y-direction: ( ) ( ) 00 0 3 sin30 1 sin30 10 / 13.28 / 2 oo in out Wm m m m s k g s υυ =− + = + = & 199.2 WN = 10. p 0 1 P 2 P F W , ice immersed V melted V
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Attach CV on the moving piston. By mass conservation, 0 0 P P A A υ = By Bernoulli equation, () 22 2 2 2 12 02 1 0 0 0 11 1 1 2 2 PP P υρ ρ ρρ += + = < < Q 2 21 0 1 2 P P A A ρυ ⎛⎞ −≈ ⎜⎟ ⎝⎠ By using momentum equation CV d Fd V dt ∑= ur r ( ) r CS dA +⋅ r ru r and 0 0 P A A , LHS: 2 3 2 2 00 P P AA FF PPA F A F ∑=+ − =− RHS: ( ) ( ) ( ) 2 2 0 0 0 0 1 r P P P P P P P P CS A A A A A A A A ρ υ ⋅= + = = rr u r 2 33 3 3 2 2 0 2 2 2 0 0 1 1 2 2 P P P P P P P A A A FA A A A ρυρ −= =+ = + 11. Heat engine and heat transfer to water stream is reversible. CV gen CS dS Q S dt T δ & & 1 ln 0 in in out out T Q ms m c TT +⋅ −⋅ = + = & && & ( ) 1 1 1200 1 1 300 ln 0 0.1 / 4192 / ln 0 0.01 / 320 reserv in out KT T K mc kg s J kgK RT T K W T K + = & 1 944 TK = 12. 700 600 24985 / ln ln 5 1 2 2 64 / oo oi CC Q Wm r r cm cm L kW m K ππ == = & 13. d) only a): no change in internal energy after a complete cycle no matter whether the cycle is reversible or not b): Irreversible engines waste more thermal energy to low temperature reservoir.
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This note was uploaded on 10/06/2011 for the course MECHANICAL 2.006 taught by Professor Blah during the Spring '08 term at MIT.

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PS1_sol_S08 - Problem Set 1 (2.006, Spring 08) Part A 1. =...

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