HWS_CH7_EGN3358_Sleiti - W W What is the Reynolds number of...

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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 10-2 N-slm2 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 10-3 . . 1 Re = v d p =(3.395 x 10 “0.151(890) = 4.53., x 103 it 99.9 x 10-2 =‘=‘=‘=‘=‘= 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 Pi-Ai-Vl = Pz-Az-Vz : Pz-Vz = Pl’vl (Al/A2) = Pi-Vl (du'dz)2 Therefore p2-V2 -d2 = pl-V] -d] (dlldz) (1) .v . Rel = “1—13 = 1900; pl-Vrdl = 1900p 11 (2) Substituting (2) into (1): pzoV2 -d2 = l900-tt-(d1/d2) = l900-tt-(20/12) = 31671.1 Pz'Vz-dz = 3167-11 = 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 fZ-S 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 10-3 N-s/m2 W Ap = n-d2/4 = m(75i100)2l4 = 4.413 x 10-3 m2; V = (Vol. flow rate)/Ap= 0.009] (4.418 x 10'3 ) = 2.037 this Re = V-d-p = (2.037H75/1000H9933) = 1511 x 103 11 1.003 x 10-3 Since Re > 2300. the flow is turbulent: using eq. 7-1; EA. EA: 2 :23. EB: ' = = =5 I.‘,'g+2-g+ A p.g+2'g+ZB+hL , butPA PBandVA VB 0 zA-zfihL wherehL=%-%% eqn.7-13.wim2K=0 From Table 7-1 for drawn tubing h1- = 0.0015 mm: hrld = 0.0015175 = 20 x 10'5 From Figure 7-4 for Re = 1.521 x 105; h,Jd = 20 x 10'5; f= 0.0169 ZA - ZA = 111. = 'E = (0-0169)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. 177-7, 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 P7-7. Pipeline. AWE: ERTI ' Watet @ 20"C from Table A-9; p = 998.3 kg/m3 [1 = 1.003 x 10'3 N-s/m2 W Total Pipe Length = 190 m; d = 10 cm; PA = 690 kPa; (Vol. flow rate) = 0.01m3/s AP = n-d214 = 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 = V-d-p =(1.273}(101100059933) z 1265, x 103 11 1.003 x 10‘3 Since Re 2» 2300, the flow is turbulent: using eq. 7-1 between A and B. P v 2 P V 2 AMA—+2 =—B—+—B—+z +h ; butP =P andV =v =0 9-3 2-3 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. 7-13) = 2‘g T For commercial pipe from Table 7- 1; hr = 0.046 mm; h.ld = 0.0461100 = 0.00046 Figure 7-4 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) 993-9307) P]; = 690 x 103 — (24.366)-998-(9.807) = 451.5 x 103 Pa + 21.21 + 3.156 ¢=¢==¢=<== PR B E 7-1 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 kgIm-s and p = 950 kg/m3. SQHEMATIQ; ASSIJMPTIQNS; (1) The flow in the needle is steady-state 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 = V-deJ/p) = l0.l9-(0.00025)/(0.0025l950) = 968.1 (flow is laminar) r = 64/Red (Figure 74) r = 64/968,! = 0.0661 P1! [3-3 + V12/2-g = P2] p-g + V22/2-g + hL (Eq. 74) V1 = V2 by Conservation of mass Pi - P2 = P'B‘hL = P‘s-f-(L/dlwz l2-8) 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 sea-level conditions (EDT and P = 14.6 lbflinz). ELSE; Determine the pressure and the power required if the propeller fan is 70% efficient. SQflEMAfIfIQ; Figure P7-l 1E. Open Circuit Wind Tunnel. W The flow is fully developed and steady-state. W Air @ 60 “F from Table 3-2; 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. 7-2. 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 = V-dh Iv = 120-(l.865)l(1.581 x 104) = 1.416 x 106 f: 0.014 (Fig. 7-4 @ Red = 1.416 x 106 and hrldh = 0.0002) p1 — 92 = p-hL = p-f-(Ud)-(V2 lZ-g) = 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 - le-V-(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 non-circular 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 well-rounded 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 (A-9); p = 999.8 lag/m3 u. = 1.308 x 10'3 N-slm2 _ 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 +Y-L+z +hL P-e 2': 1 9-3 2-3 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 7-6, K = 0.05; The exit loss for deB = 0 from Figure 7-7 (sudden expansion) K = 1.0. 1 1.05-v2 ‘ = . . = . ' =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.o-s.0=2o ' 2-e 2-3 2-3 ...
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HWS_CH7_EGN3358_Sleiti - W W What is the Reynolds number of...

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