This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: W
W What is the Reynolds number of a flow of oil at a flow rate of 0.6 m3ls in a 15 cm. pipe
if the dynamic viscosity of the oil is p. = 0.999 stlm2 and its Speciﬁc gravity is 0.89?
Is the ﬂow laminar or turbulent? (T = 20°C.) TI '
AW
W on @ 290 K from Table A—lO; p = 890 kg/m2
[,1 = 99.9 x 102 Nslm2
ALY '
, 2 _ 2
v = Q(volume flow rate) where AP = 1: d = I: (0.15) = 17.67 x loam:
_—_AP — 4 4
v =——M— = 3.395 x 101sz
17.67 x 103
. . 1
Re = v d p =(3.395 x 10 “0.151(890) = 4.53., x 103
it 99.9 x 102 =‘=‘=‘=‘=‘= Since Re >>2300 the flow is turbulent. ¢=¢=¢=¢=<=<= BLE  M The Reynolds number of an incompressible ﬂuid ﬂow in a 20 cm. diameter pipe is 1900. IND ' What is the Reynolds number in the 12 cm. diameter pipe that is connected to the larger
diameter pipe by a reducing ﬁtting? What flow regime. laminar or turbulent. exists in the two pipes?
W
A PTI : Assume inCOmpressible, p1 = p2, and SSSF, m1 = mg.
W
W PiAiVl = PzAzVz : PzVz = Pl’vl (Al/A2) = PiVl (du'dz)2 Therefore p2V2 d2 = plV] d] (dlldz) (1) .v .
Rel = “1—13 = 1900; plVrdl = 1900p
11 (2) Substituting (2) into (1):
pzoV2 d2 = l900tt(d1/d2) = l900tt(20/12) = 31671.1
Pz'Vzdz = 316711 = 3167 V'I'herefore R62 = l1 1
FOI’ R6] = 1900 < 2300; flow is laminar where d] = 20 cm =¢=¢=¢=¢=¢= For Re: = 3167 > 2300; flow is turbulent where d; = 12 cm 7 ¢=¢=¢=<=c=c= ERQBLEM fZS MN; Water ﬂows from reservoir A to reservoir B through 280 m of straight tubing. Both
reservoirs are open to the atmosphere and the water temperature is 20°C. A flow rate of 0.009 m3/s is required through the 75 mm. diameter drawn tubing. mm Neglecting minor losses at the exit and entrance to the reservoirs, calculate the required
difference in reservoir water level to maintain the ﬂow rate. SQHEMATIC;
I N '
ERQEEBIIESJ Water @ 20°C from Table A—9; p = 998.3 kg/m3
0 = 1.003 x 103 Ns/m2
W Ap = nd2/4 = m(75i100)2l4 = 4.413 x 103 m2; V = (Vol. ﬂow rate)/Ap= 0.009] (4.418 x 10'3 ) = 2.037 this Re = Vdp = (2.037H75/1000H9933) = 1511 x 103 11 1.003 x 103
Since Re > 2300. the ﬂow is turbulent: using eq. 71; EA. EA: 2 :23. EB: ' = = =5
I.‘,'g+2g+ A p.g+2'g+ZB+hL , butPA PBandVA VB 0 zAzﬁhL wherehL=%%% eqn.713.wim2K=0 From Table 71 for drawn tubing h1 = 0.0015 mm: hrld = 0.0015175 = 20 x 10'5 From Figure 74 for Re = 1.521 x 105; h,Jd = 20 x 10'5; f= 0.0169 ZA  ZA = 111. = 'E = (00169)I(2—280 —‘E‘(2'037 = 13.35 m 2og (75/1000) 2(9.807) ¢=¢=<=<=<=¢= W [SEQ E N; A pump delivers 0.01 m3 of water per second through a 10 cm. pipeline of new
commercial steel pipe, Fig. 1777, with regular screwed elbows. The water temperature is 20°C.
E1311; If the pump discharge (point A) pressure is 690 kPa absolute, what is the pressure at point B? E A '
regular 45' screwed
Figure P77. Pipeline.
AWE:
ERTI ' Watet @ 20"C from Table A9; p = 998.3 kg/m3
[1 = 1.003 x 10'3 Ns/m2 W Total Pipe Length = 190 m; d = 10 cm; PA = 690 kPa; (Vol. flow rate) = 0.01m3/s
AP = nd214 = TE'(10/100)2/4 = 7.854 x 10’3 m2; V = (Vol. flow rate)/Ap = 0.01! (7.854 x 10'3 ) = 1.273 mls R6 = Vdp =(1.273}(101100059933) z 1265, x 103 11 1.003 x 10‘3
Since Re 2» 2300, the flow is turbulent: using eq. 71 between A and B. P v 2 P V 2
AMA—+2 =—B—+—B—+z +h ; butP =P andV =v =0
93 23 A m 2'8 B L 1" B A B V A = VB (constant AP, incompressible flow) ZA = 0.0; 23 = 30 sin 45° = 21.21 m hL E {f'L + 2x] (eqn. 713) = 2‘g T
For commercial pipe from Table 7 1; hr = 0.046 mm; h.ld = 0.0461100 = 0.00046
Figure 74 for Re = 126.7 x 103; f: 0.0198 Table 7—2 for 45" Regular Screwed Elbows; K = 0.29 therefore 2 elbows 2K = 0.58 hL JEEP. {09993 190 + 0.53] = 3.156 219.807) 0.10 M + 0.0 :1 ——PL_
998(9.307) 9939307) P]; = 690 x 103 — (24.366)998(9.807) = 451.5 x 103 Pa + 21.21 + 3.156 ¢=¢==¢=<== PR B E 71
WED; What is the magnitude of a steady force required on a 1 cm. diameter syringe to create a
flow rate of 0.5 cm3ls through the 0.25 mm. diameter needle which is 5.0 cm. long. The ﬂuid a = 0.0025 kgIms and p = 950 kg/m3. SQHEMATIQ; ASSIJMPTIQNS; (1) The flow in the needle is steadystate and fully developed. ’
(2) The only losses in the syringelneedle system are those due to friction in the needle.
(3) Station 1 is the entrance to the needle and station 2. the exit of the needle. W ll = 0.0025 kg/m's
p = 950 kg/m3
g = 9.807 mils2 V = Q/A = 0.5/[n(0.02§)2/4] = 1019 cmls = 10.19 m/s Red = VdeJ/p) = l0.l9(0.00025)/(0.0025l950) = 968.1 (flow is laminar)
r = 64/Red (Figure 74) r = 64/968,! = 0.0661 P1! [33 + V12/2g = P2] pg + V22/2g + hL (Eq. 74) V1 = V2 by Conservation of mass Pi  P2 = P'B‘hL = P‘sf(L/dlwz l28) P1  P2 = 950(9.807)'(0.066l)'(0.50/0.025)(10.19)2 [19.61 P1  P2 = 65,217 Nlmz Since there are no head losses (pressure drop) in the 1 cm syringe portion, the pressure on the
face of the piston equals P1. The pressure P2 equals the atmospheric pressure. as does the
pressure on the outward side of the syringe face. The force on the piston caused by the pressure difference across the piston.
force = (P1  P2)(area of the piston) = (P1  P2)lt~(dpismn)2 I4 <=¢==t=¢=¢= force = 5.122 N I ru '2'" W ESQ EN; A wind tunnel. Fig. P7—l 1E has wooden wall test section (h, = 0.004 in.) which is 3.3
x 1.3 ft. in cross section. The tunnel has the same pressure dro as an 170 ft. long
c0nstant cross sectiOn test section. The average velooity throug the tunnel is 120st of air at standard sealevel conditions (EDT and P = 14.6 lbflinz). ELSE; Determine the pressure and the power required if the propeller fan is 70% efﬁcient.
SQﬂEMAfIfIQ;
Figure P7l 1E. Open Circuit Wind Tunnel. W The ﬂow is fully developed and steadystate.
W Air @ 60 “F from Table 32; V = 0.5692 fu’hr = 1.58 x 104 it2 Is p = 0.0763 lbm/t't3 g = 32.17 {052
W Since the test section is rectangular it will be analyzed as an equivalent hydraulic diameter, eq. 72.
dh = 4(3.3)(l.3)l(3.3+3.3+l.3+l.3) = 1.850 ft; hrldh = (0.004112)ll.865 = 0.0002 Red = Vdh Iv = 120(l.865)l(1.581 x 104) = 1.416 x 106
f: 0.014 (Fig. 74 @ Red = 1.416 x 106 and hrldh = 0.0002)
p1 — 92 = phL = pf(Ud)(V2 lZg) = 0.0763(0.014)(170/l.865)(120)2 [64.34
p1  P2 = 21.80 11311112
Power required = (power delivered to air)/(efﬁciency of the fan propeller)
=i(P1  leV(ttidn)2 10/550110
= [21.80(120)n(1.865)2 l(4(550))]l0.7 Power required = 18.56 horsepower ¢=¢=¢=¢=c=¢= QﬂMMENTS DISQUSSIQE; This problem demonstrates the use of the frictional loss charts for a noncircular duct by deﬁning an
equivalent diameter. The pressure rise (P1  P2) is the static energy added to the ﬂow. The power
required is that delivered to the propeller fan. including its 70% efﬁciency. I’ll lllll l'l ‘ g 7 g ' A large cistern is ﬁlled with water to a depth of 25.0 m. A wellrounded nozzle is located in the side of the cistern at a distance of 5 m above the bottom of the tank The depth of the water is maintained constant by continuously adding water at a temperature
Ol 10°C ' .1 i. ‘ t l
.1?" D' Detennmehevelocttyofhewaterleavinglleostem:. I (a) If the nozzle has an inside throat diameter of 1.3,cm.
(b) When a 10 to. horizontal length of smooth pipe is added to the nozzle of part (a). . W Water @ 10°C from Table (A9); p = 999.8 lag/m3
u. = 1.308 x 10'3 Nslm2
_ AHALXSLS;
‘ (a) If water exits from the nozzle at 2, then P] = P2 (atmospheric). V1 = 0. 21 = 25. 12 = 5
2 2
3L+XL+2 = P +YL+z +hL
Pe 2': 1 93 23 2 2 V 2 .
25.0:32’?g+5.0+hL; hL=¥?g—[%+£K] In this case L = 0: For a well rounded nozzle from Figure 76, K = 0.05;
The exit loss for deB = 0 from Figure 77 (sudden expansion) K = 1.0. 1 1.05v2
‘ = . . = . ' =v . = 2
..2K 005+10 105, h], —1—2.g {0+105] 2.8
2 _ 2 , '2 25.0:1’1—+s.0+1'05V2  105‘“ =2s.os.0=2o ' 2e 23 23 ...
View Full
Document
 Fall '07
 sleiti

Click to edit the document details