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| g ¢o) 2.2 A closed, 5-m-tall tank is filled with water to a depth of 4 m. The top portion of the tank is filled with air which, as indicated by @ pressure gage at the top of the tank, is at a pressure of 20 kPa, Determine the pressure that the water exerts on the bottom of the tank. 3M 2Ox/0 = = fh aS i 2 20x10 SR2% 3N +9 80x10 53 (4m) OM, = 92 kPa " a S\= f= ppt th ~ me 2.3 A closed tank is partially filled with glycerin. If the air pressure in the tank is 6 Ib/in.? and the depth of glycerin is 10 ft, what is the pressure in Ib/ft? at the bottom of the tank?’ prehe B= (724 4, Vo ft) + (6, \(0%8 fa* ) f2* = & = (650 7 2.4 2.4 Blood pressure is usually given as a ratio of the maximum pressure (systolic pressure) to the minimum pressure (diastolic pressure), As shown in Videu ¥2.3 such pressures are commonly measured with a mercury mano- meter. A typical value for this ratio for a human would be 120/70, where the pressures are in mm Hg. (a) What would these pressures be in pascals? (b) If your car tire was inflated to 120 mm Hg, would it be sufficient for normal driving? . p= xh (A) For 120 ram Hg? p= (133 x 107% lo, 120.m)* b0#RR For To mm Hai p= (133 xX Vo. 070m)= TE ; 7 IN ~# ibfin® (6) For 120 mmie: ps (14.0 x10" )[igsox10 a) > 4,32 pse Since a typical +ire pressure ss Fo-95 PSE, 120mm Mh Js pot subbicient for normal drag. 2.6 2.6 — The water level in an'open standpipe is #0 ft above the ground. What is the static pressure at a'fire hydrant that is con- nected to the standpipe and located at ground level? Express your answer in psi. posh +p Since the shindpipe 1s. open B2°, ana therehvre P= b24 Bs leony t ) = 34.7 psc Mth > 2-4 27 2.7 How high a column of SAE 30 oil would be required to give the same pressure as 700 mm Hg? 2.9 2.4 Bathyscaphes are capable of submerging to great depths in the ocean. What isthe pressure at a depth of 5 km, assum- ing that seawater has a constant specific weight of 10.1 KN/m?? Express your answer in pascals and psi. et At the surface f =0 40 That pez (10.1 4107 & NE 2103m) = 50.5 x 10° Also, p= (505 x10 2 Mise xis” ins Pas wv 2 = 50.5 MP, 7320 pse 2-7 en) 2.10 For the great depths that may be en- countered in the ocean the compressibility of sea- water may become an important consideration. (a) Assume that the bulk modulus for seawater is constant and derive a relationship between pressure and depth which takes into account the change in fluid density with depth. (b) Make use Ca) . d -y - Gn 2-3 Thus, dp se dz : ir of part (a) to determine the pressure at a depth of 6 km assuming seawater has a bulk modulus’ " of 2.3 x 10? Pa, and a density of 1030 kg/m? at the surface. Compare this result with that ob- tained by assuming a constant density of 1030 kg/m’. : (£¢, 2.4%) (1) Tf p 3 a tunchon of P, we must determne P=F(p) behre infegrating gu), Since, E Le (Eg, 413) then - €/P d cp = Ey 4 @ (e so Thet ° L pe E, bn % #& Thus, Pf? fh ete Where lal at p=o ° Pzo at surfuce From £901) t 4 o 4 Zz tp + # . ~g dz zz 4 ), Sol ® J [ or ° 2? % 2 & ap -” a dz z 4 1 so That >= — ~ 442 ) Where 4 Bd, the p - , in (1 £y depth below ‘sarlece. (con?) a-§ 2 tt ee 2.11 Sometimes when tiding an elevator or driving up or down a hilly road a person’s ears “pop” as the pressure difference between the inside and outside of the ear is equalized.. Determine the pressure difference (in psi) associated with this phenomenon if it occurs during a 150 ft elevation change. Af = =%'sh = 0.0768 2, (Sot) ui _ LHF = NeS FE Ges in® ) =O, raps Bul2 2.12 Develop an expression for the pressure variation in a liquid in which the specific weight increases with depth, h, as y = Kh + yo, where Kis a constant and y is the specific weight at the free surface. d ap 78 ( Ee. 24) let 4>2,-2 so that dh =-dz Thus dp= ¥dh Ghel ne ae 4. Z —--y ae CFF CL LE LEE 2-10 R213 ( Con t ) 723 In acertein liquid at rest, measurements of the spe- ” 60 107 cific weight at various depths show the following variation: 70 110 Q Id h(t) v (bit) 100 115 0 70 10 o The depth h = 0 Corresponds to a free surface at atmo- 20 01 spheric pressure. Determine, through numerical integration x 97 of Eq. 2.4, the corresponding variation in pressure and show 4 the results on a plot of Pressure (in psf) versus depth (in 50 102 feet), ap == 2.4 5b x (Eq. 2.4) let 2: £,-4 (see figure) Se = That cda=—ah and There dre £ ~s= dp = ~rd2 = rdh Thus, 2 he a (he fou a ° ¢ a a a aes or , he ri = x dA Cl) 2 Where zp: is the Pressyve at depth Ae Equation C1) Can be sateg rated Numeric the trape Borda! rule, 2; e., 1 z= Z 2+ G5, Nr, -x,) A here 4nt » ee h, and N= humber of data Pornks . ally using (cont) 2-l F248 OE Re SERS ONES IETS Elevation (ft) Temperature (°F) 5000 50.1 (base) *25 Under normal conditions the temperature of the ooo 53% atmosphere decreases with increasing elevation. In some 6400 ae situations, however, a temperature inversion may exist so 7100 67.0 that the air temperature increases with elevation. A series ‘ of temperature probes on a mountain give the elevation $00 O84 ~temperature data shown in Table P2.15. If the barometric 8600 700 pressure’ at the base of the mountain is 12.1 psia, determine 9200 a by means of numerical integration the pressure at the top of . this ftioun tat, 9900 67.1 (top) SR ee SEETHER TABLE P25 From £9, 4.4, z, Ly t+ if da mM 2 = R ls + 1 In the table below the temperature in OR is Jiven and the ate grand Ter) tabula ted, Elevation, to 7,°F TPR 1/ TR) 5000 50.1 509.8 0.001962 5500 55.2 514.9 0.001942 6000 60.3 520.0 0.001923 6400 62.6 522.3 0.001915 7100 67.0 526.7 0.001899 7400 68.4 528.1 0.001894 8200 70.0 529.7 0.001888 8600 69.5 529.2 0.00189 9200 68.0 $27.7 0.001895 9900 67.1 526.8 0.001898 The approximate Value of The witegral In £9. 2.7 Us G, 34 ob tained using tre Trapeze dal rules &.&, I= ¢ ba ac + Iu C4, He) where Yr Vr, x w elevation, ana an tsi bon of data pote / [900 fk (L)de = 9.34 a Soon te , so trad [tne g° = 82.2 tts? and ae 1716 FE tb Jag: R dn Be (32.28 CEE) __p 1953) 1716 Rafat (con t) 2-13 A215 Ceont ) th flows trom £91) with A= /2.) psca Met 0 OUTER. $B: (72.1 psia) eC 10,2 psa (ote ’ Since the tempera ture variation 4s not very sarge it would be expected That The assump De: tion of & “Constonk temperature would give 9a0d results. Lf The temperature ts @ssymed to be tonstant at The base temperature ($0/F), f=10.1 psia , which 1s Only shohtly differest from the result Given above, 2-14 2/8 2.18 What would be the barometric pressure reading, in mm Hg, at an elevation of 4 km in the U.S. standard atmosphere? (Refer to Table C.2 in Appendix C.) . At an elevation of 44m j p= blk xi” C trom Table C2 th Appendix CJ), Since paw y eM 4 ee SCE EE = DHbH mm = Haman 1BBLX ID port cad 2-16 214 2.1@ An absolute pressure of '7 psia.corré- sponds to what gage pressure for standard at- mospheric pressure of 14.7 psia? fp labs) = f) Gage) t platm) P lgage) Plats) - ~ latm) = T pova- TET psc Thus, - 7.7 pse 2.2) 2.21 Pikes Peak near Denver, Colorado has an elevation of 14,110 ft. (a) Determine the pres- sure at this elevation, based on Eq. 2.12. (b) If the air is assumed to have a constant specific weight of 0.07647 lb/ft’, what would the pressure be at this altitude? (c) If the air is assumed to have a constant temperature of 59 °F what would the pressure be at this elevation? For all three | cases assume standard atmospheric conditions at sea level (see Table 2.1). r (a) pea (i- )* (bg, 2.12) - k oR _ fe for f* Alle.aZs , 3 =000357 » pr tele , 7 918.67°R , R= 171 EH , and Stage Re fe 4 321m 2 = $7262 Kp (i ge [os0i57$ ) Hen 5.252 = ft) p= (ane.2 A) _ (2.00387 BM 40 518.67 R = 1/2406 Re Cats ) iT] (6) — ps poy ae.2 £.~ (087047 4 Niiy no 44) L = /oy4o 2 (abs) - Ee 2,10) (2) Prt e (eq. 10 - feet B)(m y10-F) } = tb $bTb oR)s (Atte2 Je Gm Se, sree? R) a 7 = 1270 #, labs) 2,22 | 2.22 Equation 2.12 provides the relationship between pressure and elevation in the atmo- | sphere for those regions in which the temperature . Varies linearly with elevation. Derive this equa- tion and verify the value of the pressure given in | Table C.2 in Appendix C for an elevation of 5 km. A Az : dp . e/a a ei (Ep, 24) 2, 4 let prep for 420, A~p for B22, and TeTp-pe. Thus, P dp P Z Ln (nse « i [ & coos) ~h| be (1~ #) and tuking logarithm of beth sides of Cguation yields p= # (1-@)* (£5, 2.12) For 22 5hm with Btl0) 338A. » la = 288, eK f? $807 4 s%2 « : ez 0.00050 kz ay Lf 4yk 7.507 287 7 YG ooesX) (0. 00654 rent" Rye be MO OO mt = . 2 I a p (on day] 2PS, GK 4 = S40 x /0 pt 4 (From Table Cad si Appendix Cc, pr S405x10'L -) 2720 
