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Chapter 2

Unit Operation & Transfer Phenomena. Chapter 2. Fluid Statics. U O T P. P y+ Δ y. P z+ Δ z. P x. P x+ Δ x. Δ z. Δ y. Δ x. P y. P z. Pressure in a Static Fluid. U O T P. P+dP. S. P. Pressure in a Static Fluid. PS = (P+dP)S +ρSdyg /g c dP+ρdyg /gc = 0. dy. 單 元 操 作. 靜壓.

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Chapter 2

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  1. Unit Operation & TransferPhenomena Chapter 2 Fluid Statics

  2. UOTP Py+Δy Pz+Δz Px Px+Δx Δz Δy Δx Py Pz Pressure in a Static Fluid

  3. UOTP P+dP S P Pressure in a Static Fluid PS = (P+dP)S +ρSdyg /gc dP+ρdyg /gc = 0 dy

  4. 單元操作 靜壓 • 靜止流體 : 靜壓四面八方相等 • 流動流體 : 靜壓與流動方向垂直 • 旋轉流體 : 靜壓 = 離心壓力

  5. 單元操作 靜壓 • 靜止流體 : 靜壓四面八方相等 • 流動流體 : 靜壓與流動方向垂直 • 旋轉流體 : 靜壓 = 離心壓力

  6. Examp-1 • Pb=Pc • Pa-Pb= • Patm-Pc= • Pa-Pc= Patm Pa d2 d1 5cm Pc Pb p

  7. 例題一 • 空氣流過閥，產生壓降ΔP，水密度ρ=1000kg/m3，空氣密度可忽略， 以U形管量出水柱差5cm，求ΔP=Pa-Pb=? N/m2 • P=Pa+Za ·ρ·g/gc =Pb+ Zb·ρ·g/gc • Pa-Pb = (Zb-Za)·ρ·g/gc • ΔP=Pa-Pb = (Zb-Za)·ρ·g/gc =0.05m×1000kg/ m3 × 9.8m / s2 / Z Pb Pa 5cm Zb Za p

  8. 例題二 • 水流過閥，以 四氯化炭-U形管 測其壓降ΔP，水密度=1000kg/m3，CCl4密度=1600 kg/m3，量出CCl4液柱差5cm，求ΔP=Pa-Pb=? N/m2 • Pa-Pb = (Zb-Za)·(ρ CCl4–ρH2O)·g/gc =0.05m×(1600-1000)kg/ m3× 9.8m / s2 / Z Pb Pa 5cm Zb Za p

  9. 例題三 • 水流過閥，以 四氯化炭-斜管 測其壓降ΔP，水密度=1000kg/m3，CCl4密度=1600 kg/m3，量出CCl4斜管傾斜30˚液柱差10cm，求ΔP=Pa-Pb=? N/m2 • Pa-Pb = Rmxsinαx(ρ CCl4–ρH2O)xg/gc =0.1m×0.5x(1600-1000)kg/ m3× 9.8m / s2 / Z Pb Pa 10cm α=30˚ p

  10. UOTP Examp-2 Pressure in Static gas

  11. Examp-3 • 1. at rest • 2. pumping Patm g-a Fuel g a d -a g To fuel pump B

  12. Examp-4 α PG Centroid of area

  13. Examp-4 1.5ft 1ft

  14. Examp-5 水 1 m 泥（含水） y

  15. Examp-6 浮力＆重力 截下 y P2 F x dA z (xz)最大橫截面 P1

  16. Prob-2.8 Pa=? Patm Pa oil 10’ 15’ water 1’ Pc Pb Hg

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