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p0162

# p0162 - π r σ = ⇒ ρ g ∆ h = 2 σ r p a − p b...

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1.62. CHAPTER 1, PROBLEM 62 63 1.62 Chapter 1, Problem 62 To solve, we must balance all of the forces acting on the fluid in the capillary tube shown in Figure 1.7. There are four forces acting, viz., the pressure above the column of fluid, the surface-tension force, the weight of the column of fluid and the pressure from below. Figure 1.7: Sealed cylindrical capillary tube of radius r . 1.62(a): Since the capillary tube is cylindrical with radius r ,wehave p b π r 2 ± 1 Pressure above +2 π r σ cos φ ± 1 Surface tension ρ g h π r 2 ± 1 Weight + p a π r 2 ± 1 Pressure below =0 Thus, rearranging terms and noting that cos φ =1 when φ =0 o ,the refo l lows ( ρ g h + p b p a ) π r

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Unformatted text preview: π r σ = ⇒ ρ g ∆ h = 2 σ r + p a − p b Therefore, the height, ∆ h , is ∆ h = 2 σ ρ gr + p a − p b ρ g 1.62(b): If the atmospheric pressure is such that σ = 2 p a r , then ∆ h is given by ∆ h = 4 p a ρ g + p a − p b ρ g = 5 p a − p b ρ g This equation tells us that before we pressurize the fluid in the tube, i.e., when p b = p a , the height is ∆ h o = 4 p a ρ g = ⇒ ∆ h ∆ h o = 1 4 X 5 − p b p a ~ 64 CHAPTER 1. INTRODUCTION So, to halve the height, we must have 1 2 = 1 4 X 5 − p b p a ~ Solving for p b /p a yields p b p a = 3...
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p0162 - π r σ = ⇒ ρ g ∆ h = 2 σ r p a − p b...

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