311hw1_soln

# 311hw1_soln - Homework Assignment#1 Read Chapter 1 in...

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Unformatted text preview: Homework Assignment #1 Read Chapter 1 in Wilcox (required). Read Chapter 1 Sections 1-1, 1-2, 1-3, 1-6 and Chapter 2 Sections 2—1, 2-4, 2-5, 2- 6 in Fox and McDonald (optional). Do the following problems: Problem 1A: An air table is a large ﬂat surface used to support heavy objects above a thin layer of air which allows the objects to be moved with low friction. A1r is blown through close y spaced holes in the table surface. Consider a heavy plate 1 meter square suspended 1 tram above the table on a cushion of air. What force is necessary to move the plate at a spee of 1 m/s along the table? Assume p. = 2.18 x 10'5 N-s/mz. 1.18 Assuming water vapor can be treated as a perfect gas, what is its gas constant, R, if, for a pressure of 400 kPa and a temperature of 15° C, the density is 3 kg/m3? 1.31 A spherical soap bubble has an inside radius, R, ﬁlm thickness, t < R, and surface tension 0. Find the difference between internal and ambient pressure. Note that a bubble has two liquid-air interfaces. 1.45 The viscosity of air can be approximated by a power law according to y % CTOJ where T is absolute temperature and C is a constant. (a) Determine C by insisting that the value of n at T = 68° F matches the value given by Sutherland 's equation. (b) Make a graph of [1. according to the power law developed in Part (a) and Sutherland's equation for —250°F g T 5 2000° F. HON awor)‘ Rubafgﬁmmw'..-_m___Jﬂ__-_m__,..______L______._mmw_m.........__n.m,_.~ . ( no M 4323):... ,3 ,4 My, 454 swim, mg 140 WWW-é vay oEJcoé above a 'tiAm /ay¢r' oxam \lecb snows {+2. OIDJGQE 4!: be Novas] whélw /OLJ ‘ﬁhc—éﬁon. ,4"- A> b/uwp ~Hvruu31~z Ls‘lcgwnfy “upgde helm [m £115, «5545b duréimcw. Cowlclam 52‘ hmwy p/a-éc ﬂaccid ‘ “Zahara 6U¢P<3fﬁéjcwj lmm abowa «(é/w, “LAW/23¢ <3»; an cab/9104 017.2": Wbé‘é ‘ﬂbmo lb nowcsﬁw 4:9 mow; {Ac p/av-éc. a4 a «ma-(J a1 //~.1/5 Q/wocj» éa£ [a 2/8x/55N-s/MZ, Us “1/; —————-—> >K‘_‘—_IM_—"'”"i O lb '2 J h M A": 2 F=Aﬁivu m: zmwégN-é/ML 10 CHAPTER 1. INTRODUCTION 1.18 Chapter 1, Problem 18 For a perfect gas, we can compute the perfect-gas constant as follows. p=pRT => 3:1”— pT The given conditions are p = 400 kPa = 4-105 Pa, T = 15°C = 288.16 K, and p = 3 kg/m3. Thus, 4-1 5 N 2 J 0 /m = 463 —-— R = _________. (3 kg/m3) (288.16 K) kg - K where we note that a Joule is 1 N-m. 1.19 Chapter 1, Problem 19 For a perfect gas, we can compute the perfect-gas constant as follows. =pRT => 12:1 pT The given conditions are p = 50 psi = 7200 lb/ft2, T = 50° F = 50967" R, and p = 5.1-10‘3 slug/ft3. Thus, 1 2 ft-lb 7200 b/ft = 2770 (5.1- 10—3 slug/ft3) (509.670 R) slug -° R R: 1.20 Chapter 1, Problem 20 The perfect-gas constant for air is R = 1716 ft'lb/slug-0R. Thus, the effective molecular weight is ft - lb 497 00 slug -° R ft - lb 1716 slug -° R AA = 7?, — = 28.96 R 18 CHAPTER I. INTRODUCTION 1.31 Chapter 1, Problem 31 Make an imaginary cut through the center of the soap bubble as shown in Figure 1.1. The forces acting are the atmospheric pressure, p0, the internal pressure, [)0 + A p, and the surface tension on the inner and outer surfaces of the bubble, a. Figure 1.1: Force diagram for a soap bubble. The net contribution from the atmosphere is 100 times the projection of the hemisphere onto a vertical plane. Since the soap bubble thickness t < R, the radii of both inner and outer surfaces are m R. Noting that surface tension acts on both inner and outer surfaces, we have 2F = (Po + 1319)sz — pow}?2 — 2 (27rR0) = 0 Simplifying, we ﬁnd Apﬂ'R2 = 47chr Therefore, the pressure difference is 26 CHAPTER I. INTRODUCTION 1.45 Chapter 1, Problem 45 1.45(a): From Sutherland’s law, the viscosity at T = 680 F = 527.670 R is 2.27 ~ 10‘8(527.67)3/2 M = ———————— slug = 3.78 - 10-7 527.67 + 198.6 9 ft - sec Hence, to determine the coefﬁcient C in the power-law Viscosity, we have Slug = C = 4.708 - 10‘9 Slug C 527.67°R 0-7 = 3.789 - 10‘7 ____ ( ) ft - sec ft - sec(°R)0‘7 1.45(b): First, note that all computations must be done in terms of absolute temperature. which means that —250°F S T S 2000°F corresponds to 209.67°R g T S 2459.67° R. Figure 1.3 compares Sutherland-law and power-law Viscosities, with u in slug/ft-sec and T in degrees Fahrenheit. 107” 12 Q500 O 500 10001500 2000 T Figure 1.3: Comparison of Sutherland-law (—) and power-law (- - -) viscosities. ...
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311hw1_soln - Homework Assignment#1 Read Chapter 1 in...

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