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Unformatted text preview: Homework Assignment #2 Read Chapter 3 Sections 3.1, 3.2, 3.3 in Wilcox (required). Read Chapter 3 Sections 31, 32, 33 in Fox and McDonald (optional). Do the following problems: Problem 3A An aircraft ﬂying at 30,000 ft. has a lift force of L. Determine what will be the lift force at
36,000 ft. (in terms of L), assuming that the velocity field is the same and the lift is directly proportional to the density (i.e., L = pUF). Here U is the velocity and I‘ is the
circulation  both are held constant between altitudes. 3.4 Consider the tank shown with three layers of unmixed ﬂuids. The temperature is 20°C and the
pressure at the bottom of the tank is 259 kPa. Also, the tank is open to the atmosphere at the top. What
is the density of the unknown ﬂuid, pa, if h = 1 meter? Problem 3.4 3.26 An inclined manometer is a large spherical container with an inclined tube of small diameter
attached. The tube is inclined at an angle a to the horizontal and has a scale with markings separated by
1 cm. If the liquid in the manometer is carbon tetrachloride, what must the angle a be if each marking
corresponds to a change in pressure, Apa, of 100 Pa? Assume the temperature is 20" C. Problems 3.26, 3.27 r' it.
\AMPAD 5C: SHEETS
00 :‘inET «33‘: 22142 3
22—3434 203 SHEE7§ 22' l ”W' ‘D M An aircraLL 431mg 3+ 30,0901; 5494; 5% m 45m; (3471.. Damme.
Wm; WM! be, «Hm lmﬁk 4tym 9% 39,000 # (in ‘eﬁr‘r‘dé 0% L), aeéuMI/xj
in»; #76, Wicca» ﬁeld K, 4,474. we», Qno’ ch. MIL ,t, awed/I»,
prwpamiamal ~40 Jive, (:46?wa [L67 L= Pu l"), .Hcr‘c, u )5 +111; “ vc)0¢.}«IY god F 1%; 411:, cured/04300 — 50qu are; he's/C] wnb‘w’ﬂl' 13(31me QH’iJUAab , 3b 00044 399(2)th L_ = P! an F. seioooﬁ L
gamma—Hr. P=PRT 2; F: g; 22 5'00 N/Mz LLJ W = 0'35““3’” 2873/05.“)
F] = .IOLOOON/Mzg \— ELEM—1 2§QWJ$25 == 04505‘11m"
30.000434. 3 (2.83 K —@,_50 HMXQ ’44 14.13) . 287%5'IA L] =0.7‘1L 3&0va 130 CHAPTER 3. EFFECTS OF GRAVITY 0N PRESSURE 3.4 Chapter 3, Problem 4 rviiéiw Figure 3.1: Threelayer tank of ﬂuid. The pressure at the bottom of the tank is the atmospheric pressure plus the weight of the
ﬂuid above, i.e., P 2 Po '1' (PHzo + Pu + pHg) 9h
Thus, the density of the unknown ﬂuid is P _po
Pu gh — pHQO “ PHg 2 _ k W — 998 k—g3  13550 __g3
(9.807 m/sec ) (1 m) m m 16111 kg; — 14548 3%
m m Therefore, the density of the unknown ﬂuid is kg
pu = 1563 E 3.26. CHAPTER 3, PROBLEM 26 151 3.26 Chapter 3, Problem 26 Figure 3.16: Inclined manometer:
From the hydrostatic relation, we know that
p + pgz = constant => pa = pa + pgz Thus, since atmospheric pressure is a constant, changes in pa are related to changes in 2
according to Apa = pgAz But, the distance measured in the direction tangent to the inclined tube, 8, is related to .2
by
A2 = A3 sina => pgAs sinoz = Apa a = sin’1 AP”
pgAs Now, for carbon tetrachloride, p = 1590 kg/m3. Also, we are given A3 = 1 em = 0.01 m
and Apa = 100 Pa. Thus, Therefore, the angle is _ _1 [ 100 N/m2
(1 = $111 (1590 kg/m3) (9.807 m/sec2) (0.01 m)] = 3990 ...
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 Fall '09
 BRUNGART,TIMOTHY

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