index(5)

index(5) - Amount of material going through per unit time w...

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Typical Steady State Control Volume Problem Chart (Not complete material coverage) Infow point i a · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · Normally 4 variables are needed to fully determine it (since there is also an unknown velocity.) · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 2 known intensive variables (May get away with T only in some approximations, eg, when using saturated values as an approximation for compressed liquids.) a a a Tables, eg B.1.1-B.1.4; pv or Tv diagrams; v = v f + x ( v g - v f ) and similar for u and h . Ideal gas (applicable?) pv = RT (4 forms) Tables A5-A8 for u , h . Make do with the for- mulae for diferences in intensive variables listed below under “Device”? a Remaining intensive variables p,T,v, ( x, ) u,h . · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·
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Unformatted text preview: Amount of material going through per unit time w = W/ m q = Q/ m m = V /v = A V /v A = 4 D 2 a a a Device or Control Volume C1: Type of Device? Given that Q = 0? Do the given device characteristics add info about i a or e a ? C2: Mass: m i = m e Adds info about i a or e a ? Energy: Q + m i ( h i + 1 2 V 2 i ? + gZ i ? ) = m e ( h e + 1 2 V 2 e ? + gZ e ? ) + W Adds info about i a or e a ? C3: Device has W = 0? For ideal gasses: h 2-h 1 = i 2 1 C p d T C p ave ( T 2-T 1 ) For compressed liquids, by approximation , best at constant pressure: h 2-h 1 C ( p ) ave ( T 2-T 1 ) Exit point e a Same procedures as entrance point i a...
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