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Unformatted text preview: Homework Assignment #1 Read Chapter 1 in Wilcox (required). Read Chapter 1 Sections 11, 12, 13, 16 and Chapter 2 Sections 2—1, 24, 25, 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 Ns/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 liquidair 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 WWWé vay
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‘ﬂbmo lb nowcsﬁw 4:9 mow; {Ac p/avéc. a4 a «ma(J a1 //~.1/5 Q/wocj» éa£ [a 2/8x/55Ns/MZ, Us “1/;
——————> >K‘_‘—_IM_—"'”"i O lb '2
J h M A": 2
F=Aﬁivu m: zmwégNé/ML 10 CHAPTER 1. INTRODUCTION 1.18 Chapter 1, Problem 18 For a perfect gas, we can compute the perfectgas constant as follows. p=pRT => 3:1”— pT The given conditions are p = 400 kPa = 4105 Pa, T = 15°C = 288.16 K, and p = 3
kg/m3. Thus,
41 5 N 2 J 0 /m = 463 —— R = _________.
(3 kg/m3) (288.16 K) kg  K where we note that a Joule is 1 Nm. 1.19 Chapter 1, Problem 19 For a perfect gas, we can compute the perfectgas constant as follows. =pRT => 12:1 pT The given conditions are p = 50 psi = 7200 lb/ft2, T = 50° F = 50967" R, and p = 5.110‘3
slug/ft3. Thus, 1 2 ftlb
7200 b/ft = 2770 (5.1 10—3 slug/ft3) (509.670 R) slug ° R R: 1.20 Chapter 1, Problem 20 The perfectgas constant for air is R = 1716 ft'lb/slug0R. 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  107
527.67 + 198.6 9 ft  sec Hence, to determine the coefﬁcient C in the powerlaw Viscosity, we have Slug = C = 4.708  10‘9 Slug C 527.67°R 07 = 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 Sutherlandlaw and powerlaw Viscosities, with u in slug/ftsec and
T in degrees Fahrenheit. 107”
12 Q500 O 500 10001500 2000 T Figure 1.3: Comparison of Sutherlandlaw (—) and powerlaw (  ) viscosities. ...
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 Fall '09
 BRUNGART,TIMOTHY

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