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Unformatted text preview: 1.11 In Fig. P111. sensor Areacls 1.3 kPa
(gage). All ﬂuids are at 213°C. Determine
the elevations E in meters of the liquid le'I.'els in the open piezonieter tubes E
andC. Solution: [13} Let piezonietei' tube B be
an arbitrary distance H above the gasoline
gljtcerin interface. The speciﬁc weights are
m s 12.0 H1113. 133301111 = sst0 mm , and
gums: _= 12350 100113. than apps; the Fig. P2.“
hFdI'DSlETJﬂ fonnula from point A to point B: 1500 N"in: + {12.0 1011113 3:10 111} + 00:01:10 — H) — 550123 —H —1_t)j= p3 = 0 {gage} Solve for EB = 1'33 In [33 cm ahos‘e the gasoline—air inteiface} Ans. Solution EC]: Let piezonietei' tube C be an arbitraijr distance Y above the bottom. Then
1300 + 110(10} + 66TD{1.3]+12360[1.U — Yj—1236D{EC — Y] = [10 = [l (gage) Solve for EC = 1.93 In {93 cm above the go rsoli.ne—gl'_ll.rcei‘i.n inteiface} Ans. [cj 2.13 In Fig. P213 the ETC water and gasoline are open to the atmosphere and are at
the same elevation. What is the height h in the third liquid? Solution: Take water = $090 H.301} and gasoline = ﬁﬁTD Ntmg. The bottom pressure
must he the same whether we move down through the water or through the gasoline into
the third ﬂuid: Fig. 02.13 pme = {QT‘90 N.'m3){l.5 m) + 1.ﬁﬂ{9?90){1.{}j = 1.50{9T9D}h+ ﬁﬁ?ﬂ{2.5 — h)
Solve for h = 1.52 In Ans. 2.1?r All ﬂuids in Fig. P21? are at NT. 11‘ p = 19130 psf at point A, determine the
pressures at E, C. and D in psf. Solution: Using a speciﬁc weight of 52.4 lhftFt'J’ for water: we ﬁrst compute pB and
FD: Fig. Flt? p3 = pit —ywm[zE —z_,L) asap—514m: ft} = 1333 lbﬂftl Ans. {pt B)
pB = pA + ym (2A — Zn) =19m + s2.4{s it e) = :21: there‘ Ans. (pt. D) Finally: moving up ﬁmn D to C: we can neglect the air speciﬁc weight tcr grind accuracy:
C = p3. — ,rwmug — 23.) = 2212 — 52.430 ft] = 208? lhfjfrl Airs. (pt. C}
The air near C has y: [LGH lhftﬁ‘J’ times 5 ft yields less than t}. 5 psf correction at C. P126 For gases over large changes in height, the linear approximation Eq. [2.14). is
inaccurate. Expand the troposphere powerlaw, Eq. (2.213): into a power series and show
that the linear approximation p 3 pa moi1 g: is adequate when 23“ where h‘ — i
ﬁ—l B 1 RB
{ ]' Solurion: The powerlaw term in Eq. {2.20) can be expanded into a series: :3: :3: (PET? = l — HE + HEN_1}{£)2 — .... ._ 1where it: i
r; a a! 1; RB 84 solutions Manual  Fluid l‘u'Ieehanies, Fifth Edition Multiply h}; Pa. as in Eq. {2.29}: and note that pmﬁ’fu = ﬁns"REE: = on g: Then the
series may be rewritten as follows: n—lB: = _ '_ l — ——+ .....
P as pssi 2 TD )1 For the linear law to be accurate. the 2“ term in parentheses must be nntch less than unit}: If
the starting point is not at : = I]. then replace : by EC: j
3:" «I, or: 5: on: I” 0 Arts. 11—1
2 2 .31 In Fig. P231 determine np between points A. and B. All ﬂuids are at 210°C. Kerosene it
A El cm
Mia'snry —
.43.?—
ll cm
Water
Fig. P231 Solution: Take the specific weights to he
Benzene: ssio N."n13 Mercury: issioo N..'m3 Kerosene: sass mud Water: sass mini
and fair will he small, prohalol'j,r around 12 3131113. Work your way around ﬂoor A. to E: 1:}; + [SoWERE n1}— [133 l'U'U'JEDﬂS] — [TEEﬂIﬁﬂij +{9TQD){C.25:I — [1 31:13.99)
= p13. or. after cleaning up. plEl —pB 2 SHIN] Pa Ans. 2.3? The inchnetl manometer in Fig. P23? ﬂ'l
contains Meriani red oil, 3G = (132?.
Assume the reservoir is very large. If the inclined arm has gtaduations 1 inch apart.
what should ﬁhe if each graduation repre— sents l psf of the pressure p.51? R: il‘l'Iﬂll' Fig. Past Chapter 2 I Pressure Distribution in a Fluid 3'1 Solution: The speciﬁc 1weight of the oil is [D.Sl?}{ol4} = 51.15 lbf"ft3. If the reservoir
level does not change and El = 1 inch is the scale marking, then lhf . 'x, lhf][ l J .
=1 1 = nz=' nL 5': 1.15— —ft 5'.
PA'EEEEJ fr fat Km. 5111 L3 ﬁg 12 5111
or: sin 5 = {12325 or: H: 13.45” Am. 2.41 Thc system in Fig. P2.41 is at EDDIE.
Detcrmjne thc prcssute at point A. in
pounds per squarc foot. (Jul. 5“  ".5" Solution: Take thc speciﬁc weights of
water and mercury from Table 2.1. Write the hydrostatic formula ﬂﬂﬂl point A thI the ............ . .. water stuface: rimu;
Fig. P2 .41 _ ~ s (Int 5 as: 152.41bﬁ‘ft’ [—ﬂ)— 34s — 152.4 [—]= = 14.? 144 ps+t it )12 I: ltuj+t )12 Pam { )t }
Solve for p}; = ETTIthftftl Ans. 1131f
ﬂl PEAS The system in Fig. P149 is open to l atmon the right side.
{a} IfL = 120 cm, what is the air pressure in container A? {33] Conversely. ifpg = 135 era.
what is the length L‘?‘ Solution: (:1) The vertical elevation of the water surface in the slanted tube is
{l.2n1}{sin55°j = 0.93311 Thenthe pressure at the 18—an level of the water, point D. is N m = pm + gmaz = 101350Pa+{91‘90 3){0.933—0.18m) = 109200Po it: Going up ﬁoin D to E‘ in air is negligible, less than 2 Pa. Thus pg 3 P3. = 109200 Pa.
Going down from point C to the level of point B increases the pressure in mercury: r
p” = pt. +ymn,a:r._ﬂ =iosaoo+nanoo%){asa4.15m)=1s1snura Annie]
' m This is the answer. since again it is negligible to go up to point A in lotus—densityr air.
{33] Givenpa = 135 kPa, go down how point A. to point B with negligible air—pressure
change. then jtunp across the mercury U—tube and go up to point C with a decrease: pg = 3:3 — ymmaeH = 135000 —{133100jlj0.31—0.15) = 1124mm Once again p; 2 pm 3 112400 Pa, jump across the water and then go up to the surface: PM = PH _ Fuhrer = _ ngﬂ'I—Tmiﬁca
Solve for ' e 1.30ﬁm ‘I nuﬁlrc Then the slanted distance L = l.306m."sin55 = LitHm Astrid} — 0.13m) = 1013 50H: ...
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