ME235Lecture%206 - Thermodynamics Vm235 Lecture 6 L t 1st...

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1 Thermodynamics Vm235 1 st Law analysis for a CV Chap-6 Lecture 6 5/15/2008 37 Both Mass & Energy can cross the boundaries of a control volume – the mass carries with it energy. 5/15/2008 38
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2 Conservation of Mass: ² ³ ´ ± ² ³ ´ inlets all from C.V. the into rate flow mass Total C.V. the in mass of increase of Rate . . = i V C m dt dm F ext P T ii v e Fl ow Control surface ² ³ ´ ± exits all from C.V. the of out rate flow mass Total e m dm C. V. dt ee e e Fl ow How to calculate m CV , m i , m e depends on the problem: [] 1 : . . kg dV v dV m write may we general In CV CV V C = = ρ 5/15/2008 39 () sec 1 : 2 3 3 3 kg dA V v dA V m Likewise m s m m kg CS local CS local m m kg µ µ² µ µ³ ´ G G ± ´ × × × = = PE KE U E E m elevation of virtue by usually velocity of virtue by usually e temperatur of virtue by usually + + = = + + G µ µ ´ µ µ ´ µ µ ´ state) amic (thermodun energy potential energy kinetic energy Internal : possesses ' ' mass A Conservation of Energy (1 st Law): boundary the at done is work Thus P pressure inlet the against push to has it inlet the through enters fluid the as However e Z g u e mass unit per or + + = 2 2 1 , '. ' , '. ' energy it the with carries it C.S. he through t C.V. the into flows mass this As : V Fl ow 5/15/2008 40 Pv mass unit per work flow or v m P V P dA P dA P W Rate Work Flow V C the to input energy total the get to e to added be must work flow This V C the of exit entrance exit entrance = = = × = = = ± ± G G ± / / . . ' ' . . V V
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3 G ± ² ³ 2 2 1 / . . / , + + + + + = + = PE d KE d dU e m m d dE Z g h Pv u Pv e mass unit per exit inlet the through V C the leaving entering energy total Thus V Conservation of Energy (1 st Law): ( ) () G ´ µ µ¶ µ µ· ¸ G ´ ´ ´ 2 2 1 2 2 1 . . . . . . . . : , & ) ( ) ( ... : + + + + + = + + = = e e e i i i V C V C V C B B A A V C have we hence zero or small are PE KE usually Z g h m Z g h m W Q dt dt dt dt e dt Thus total h V V 5/15/2008 41 µ µ ¸ ´ µ µ ¸ ´ µ µ µ µ ¸ ´ ´ C.V. the of out flow enthalpy C.V. the into flow enthalpy mass fixed or system closed a for Law First . . . . . . + = e e total e i i total i V C V C V C h m h m W Q dt dE = = V C V C V C dm dt dE and dt dm Thus time with changing stop conditions when state Steady . . . . 0 ; 0 , . : Steady State: = = = e i e i m m or m m dt ´ ´ · ¸ ´ · ¸ ´ · ¸ exits all from C.V. the of out rate flow mass Total inlets all from C.V. the into rate flow mass Total C.V. the in mass of increase of Rate . . , 0 5/15/2008 42 + = + = e e total e V C i i total i V C V C h m W h m Q dt dE and ´ ´ ´ ´ . . . . . . 0
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4 Steady State Example: A boiler provides saturated steam at 100 kPa with inlet liquid water at 20 o C flowing in at 1 m/s. What is the velocity of the saturated vapor exiting at 100 kPa if the pipe size is the same? Can the flow then be at constant P?
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This note was uploaded on 09/15/2010 for the course ME 235 taught by Professor Borgnakke during the Fall '07 term at University of Michigan.

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ME235Lecture%206 - Thermodynamics Vm235 Lecture 6 L t 1st...

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