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Unformatted text preview: Homework Assignment #5 Read Chapters 5 and 6 (except 6.1.3, 6.2.3 and 6.4) in Wilcox (Required) Read Chapters and Sections 41 through 44 and 61 through 64 in Fox and
McDonald (Optional) Do the following problems: Problems 5.16, 5.17 \/ 5.17 For inviscid ﬂow past an airfoil, the streamlines are as shown. If the freestream velocity, U0, is
160 ﬂ/sec, what is the pressure difference, [)2 — 121, at points where the velocities are U1 = 300 ft/sec
and U2 = 250 ft/sec? Assume that the ﬂuid density, p, is .002 slug/ft3 and that Bernoulli’s equation applies. Express your answer in psi. \/ 5.32 The average velocity of water in a nozzle increases from u; = 10 ﬂ/sec to 11.2 = 60 ft/sec. Assuming
the average velocity varies linearly with distance along the nozzle, 2, and that the length of the nozzle
is t = 1 ft, estimate the pressure gradient dp/dz at a point midway through the nozzle. The density of
water is p = 1.94 slug/its. You may assume the ﬂow can be approximated as one dimensional. 5.46 A Pitotstatic tube is placed in a ﬂow of helium with p = 3.2  10‘4 slug/113. The staticpressure
tap reads 12.0 psi and the ﬂow velocity is 300 ft/sec. What is the stagnation pressure? If the stagnation
pressure changes to 12.2 psi and the static pressure is unchanged, what is the corresponding velocity? 5.50 A Venturi meter is a device used to measure ﬂuid velocities and ﬂow rates for incompressible,
steady flow. As shown, the pressure is measured at two sections of a pipe with diﬁ‘erent crosssectional
areas. A straightforward derivation shows that the volumeﬂow rate, Q, is _ 2(P1P2)
9“ where p i: pressure, p is density and A is crosssectional area. Consider a Venturi meter that has
A1 = l m and A2 = 0.7 m2. The attached pressure gage can accurately measure pressures no smaller than 0.1 Pa. If air of density p = l.20 kym3 is ﬂowing, what is the minimum ﬂow rate that can be
accurately measured? 0 Problems 5.48, 5.49, 5.50 292 CHAPTER 5. CONSERVATION 0F IMASS AND MOMENTUM 5.16 Chapter 5, Problem 16 Figure 5.1: Airfoil in an inviscid ﬂow. As stated, we can assume Bernoulli’s equation applies to this ﬂow. Thus, the pressure
difference is given by 1 1 1 2 2
p1+§pU12=p2+ 5W5 = m —p1 = 59 (U1  U2) Hence, for the given values, we have 2 2 k
p2 p1= 3 (1.1 155) (952 i — 752 3—) = 1.87103 g = 1.87 kPa 2 m3 sec2 sec2 m u sec2 5.17 Chapter 5, Problem 17 Figure 5.2: Airfoil in an inviscid flow. As stated, we can assume Bernoulli’s equation applies to this ﬂow. Thus, the pressure
difference is given by 1 1 l
P1+§PU12=P2+§PU22 => P2—P1=§P(U12—U22) Hence, for the given values, we have 1 slug 2 ft2 2 ft2 Slug 
_ =_ , __ __ — =27.5 =o.19 51
[)2 pl 2 (002 ft, ) (300 sec, 250 m2 p 5.32. CHAPTER 5, PROBLEM 32 5.32 Chapter 5, Problem 32 From the onedimensional Euler equation, du_ dp p”; — —d:c The velocity varies linearly from U1 to 1L2 as :1: increases from 0 to 6. Thus, x du u  u
U($)=u1+(u2—u1)Z m 3;: 2e 1
Halfway through the nozzle, we thus have
1 du 11,2 — U1
u=§(u1+u2) and 8;: e Therefore, the pressure gradient is dp_ (u1+u2)(u2—u1)__ ug—ug
(133— p 2 e — p For the given values, dp _ 1 94 slug (60 ft/sec)2  (10 ft/sec)2 __
_( ' ‘fP‘ 2(1 ft) ft3 E;_ 
l
00
co
co
01
l 307 5.45. CIL4PTER 5, PROBLEM 45 327 5.45 Chapter 5, Problem 45 Since we have a Pitotstatic tube, we know that the velocity and stagnation pressure are 2 s a —— 1
U: (ptg p), pstag=p+_pU2
p 2
For the given values
U _ 2(102500 — 101000) N/m2 _ 50 3:
1.2 kg/m3 sec If the velocity changes to 80 m/sec, the stagnation pressure changes to N 1 k 2 N
1231119: 101000 —— + — 1.2 ——g (80 3) = 104840 —— = 104.8 kPa
m2 2 m3 sec m2 5.46 Chapter 5, Problem 46 Since we have a Pitot—static tube, we know that the velocity and stagnation pressure are 2 50— 1
U: W» pstag=p+§pU2 For the given values
. 1 _4 slug ft 2 1 ft2 .
= _ _ . _ _ __ — = 12.1
pstag 12 psr+ 2 <3 2 10 f9 ) (300 sec 144 mg p51
If the stagnation pressure changes to 12.2 psi, the velocity changes to
2(12.2 — 12.0) psi (144 1112/11?) ft 2424 ~— U = 3
3.2 10—4 slug/ft sec 330 CHAPTER 5. CONSERVATION 0F AMSS AND MOMENT UM 5.49 Chapter 5, Problem 49 Since the volumeﬂow rate in a Venturi meter is 2(P1P2)
2A _____.
Q QpU—Aﬂﬁ) the pressure diﬂ‘erence divided by p is Pi—P2___£ 1_r_4_§ P 2 A? A2 Thus, for the given values of Q, A1 and A2, we arrive at p1 — pg 1 100 ft2 ft 2 ft2
= — 1  10 — = 15.28 —
p 2 < 144 ft2 sec sec2 Hence, for air, water and mercury, we have the following. EmulpoMy> m—mawn Air .00234 0.036
296
402 5.50 Chapter 5, Problem 50 As noted in the statement of the problem, the volumeﬂow rate in a Venturi meter is 2(101  P2) Q=A2puA@ﬁ) Then, with (p1 — p2)mm = 0.1 Pa = 0.1 kg/m/secz, we have 2 (0.1 kg/m/sec2) _ 0 40 £3— , = 2 .._..
Qmm (07 m) (1.2 kg/m3) [1  (0.7 m2)2 / (1.0 m2)2] sec ...
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This note was uploaded on 10/28/2009 for the course AERSP 311 at Pennsylvania State University, University Park.
 '09
 BRUNGART,TIMOTHY

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