Development of Design Method of Centrifugal Pump using Experimental F actor 2001 2
: ( ) : ( ) : ( ) 2000 12 21
Ab str act 1 N om enclature 2 1 5 2 7 2.1 7 2.1.1 7 2.1.2 12 2.1.3 14 2.1.4 17 2.1.5 19 2.1.6 22 2.2 25 2.2.1 1 25 2.2.2 2 27 2.2.3 29 2.2.4 31 2.2.5 33 2.3 39
2.3.1 41 2.3.2 44 2.3.3 45 3 46 4 68 69 70
Development of D esign Method of Centrifugal Pump using Experimental Factor Hyo-Nam IM Department of Mechanical Engineering, Graduate School, Korea M aritime University Abstract This study is focused on the performance prediction and design of a centrifugal pump with optimum shape. Design and analysis of centrifugal pumps rely on experience of designer due to many fluid mechanical and geometrical variables. In this study, a design method was developed with experimental factors and analysed by comparison with 2nd-order vortex panel method. Impeller is the most important component affecting the performance of the centrifugal pump. The predicted total head for three cases, of which designs were determined by this method, agrees well with a particular commercial pump. This study shows that satisfactory performance of an optimal pump shape can be obtained through the automatic design routine. - 1 -
N om e n c lature av : volute angle b : w idth of impeller b 3 : w idth of volute c : ab solut e velocity d : diam eter of im peller d3 : diam eter of base circle E : Young ' s m odulu s g : acceler ation of gr avity H : tot al pum p head [m ] H t h : theoretical pum p head H : Euler head h : height of pump K m : capacity const ant K u : speed const ant K v : volute const ant n : rot ational speed [rpm ] n s : specific speed [m 3 / m in, rpm, m ] Q : flow r ate [m 3 / m in] su : thickn es s of blade u : peripher al velocity of im peller V : volum e v : m eridian velocity of im peller - 2 -
w : r elative velocity z : num ber of blade greek letter s : angle of blade lim it : guide v alue of diam eter r atio : tot al efficiency h : hydr aulic efficiency m : m ech anical efficiency v : volum etric efficiency : angle of im peller : slip factor : acceler ating r atio : coefficient of viscosity : boss r atio : density subscript 1 : in let of im peller 2 : outlet of impeller 3 : b ase circle of volute b : boss d : delivery side of pum p i : in side of im peller m : m eridian component o : out side of impeller - 3 -
s : suction side of pum p u : peripher al component - 4 -
1...,.. ( 1 ) Bezier CAD. ( 2 ), (3 ) 2 Vortex panel m ethod 2. Stepanoff, Preiderer 2 Vortex panel m ethod. 1 2 Archim edes. 90 600 /. - 5 -
.. - 6 -
2 2.1 2.1.1 (ab solute velocity ) (relative velocity ). (peripher al velocity ). 2 c w u..... (Slip factor, V ane efficiency )., Fig. 2.1., (2.1). 2 H th H (2.2). (2.3) (4 ). - 7 -
Fig. 2.1 Ve lo c ity triang le of c e ntrifug al pump - 8 -
H th L V = 1 g ( c u2 u 2 - c u 1 u 1 ) (2.1a) H = 1 g c u2u 2 = 1 g u 2 ( u 2 - c m2 t an 2 ) (2.1b ) H th H (2.2) h = H H th (2.3).. z ( r,, z ) c z r c=c (r, z)., cm cu. cm, cm. cu (r, z). (2.4). (2.4b ) cm 2 (2.5a) (Head coefficient ) (2.5b ) (Capacity coefficient ). - 9 -
H = h u 2 c u2 g (2.4a) ( g H u 2 2 ) = h { 1 - ( c m2 / u 2) t an 2 } (2.4b ) ( gh u 2 2 ) (2.5a) ( c m 2 u 2 ) (2.5b ).,., g,, (2.6) Buckingham (2.7) 3 (2.8) (5 ). f ( n, Q, gh, D,, ) = 0 (2.6) 1 Q 1, 2 = (g H ) 2 D 2 n Q (gh ) 1 2 3 4, 3 = Q D (2.7) F 1 ( 1, 2, 3) = 0 (2.8). (2.9) (2.10). 1 Q, (gh ) 1/ 2 D 2 2 nd (g H ) 1/ 2 (2.9) - 10 -
F 2 ( 1, 2 ) = 0 (2.10) n s = n M / Q M = 1 H M = 1 D u = D M / Q M = 1 (2.11), (2.12). H M = 1 n s = D u = n Q 1/ 2 H 3/ 4 (2.11) DH 1/4 Q 1/ 2 (2.12) (2.11) (2.12) (Specific speed), (Specific diam eter ).. - 11 -
2.1.2,. F ig. 2.2., Stepanoff,,, (6 ). - 12 -
Fig. 2.2 Calc ulatio n pro c edure of des ig n s oftware - 13 -
2.1.3. Fig. 2.3 Stepanoff. K u 2 o, K u 2 m u 2 o u 2 m (2.13). d2 o d2 m (2.14). (2.15). cm 2 b 2 (2.16). z (2.17) Stepanoff Pfeiderer. F ig. 2.4 flow chart. u 2o = K u2 o 2g H (2.13a) u 2m = K u2 m 2g H (2.13b ) d 2o = 60 u 2 o n (2.14a) d 2m = 60 u 2 m n (2.14b ) c m2 = K m2 2gH (2.15) b 2 = V ( d 2 m - z s u2 ) c m 2 (2.16) z 2( ) 3 (2.17a) d z k { 2 + d 1 d 2 - d 1 } s in ( 1 + 2 ) 2 k 6.0 6.5 (2.17b ) - 14 -
Fig. 2.3 Co ns tant of impe lle r - 15 -
Fig. 2.4 De c is io n of o utlet s tandard - 16 -
2.1.4 (2.18).. ( d 1o / d 2o ), ( d 1o / d 2 m ). Stepanoff ( d 1o / d 2o ) (2.19). c 1m (2.20). (2.21) db d 1 o. (2.22). F ig. 2.5 flow chart. K m 1 = c m 1 / 2g H (2.18a) K m2 = c m2 / 2g H (2.18b ) d 1o = ( d 1o / d 2o ) d 2o (2.19) c m 1 = V d 1m b 1 (2.20) d B d 1o (2.21a) ( d 1m d 1o )( b 1 ) = 1-2 d 1o 4 (2.21b ) b 1 = 1-2 4 d 1 (2.22) - 17 -
Fig. 2.5 De c is io n of inle t s tandard - 18 -
2.1.5.. cu 2 (2.23) H (2.24). c u2 = u 2m - c m2 t an 2m (2.23) H = u 2m c u2 g (2.24) Stodola W iesner. Stodola. cu 2 cu 2 (2.25). W iesner (2.26)., (d 1/ d2 )< lim (2.26c) (d 1/ d2 )> lim (2.26d). (2.26f). (2.27) (2.28). (2.29). h, h =0.80 0.90. Stodola W eisner. F ig. 2.6 flow chart. c u = u 2 m s in 2 z (2.25a) - 19 -
= H th H = c u2 c u2 = 1 - u 2m s in 2 c u2 z (2.25b ) c u = u 2 m ( 1 - ) (2.26a) lim = 1 8. 16 s in ex p { 2 z } (2.26b ) = 1 - s in 2 z 0.7 (2.26c) = 1 - s in 2 z 0.7 { 1 - ( d 1 d 2 - lim 1 - lim ) 3} (2.26d) c u2 = c u2 - u 2m ( 1 - ) (2.26e) = c u2 c u2 = 1 - u 2m c u2 ( 1 - ) = 1-1 ( 1 - ) (2.26f) = H / ( u 2m 2 / g ) (2.27) H th = c u2 u 2m g = {c u2 - u 2m ( 1 - ) }u 2m g (2.28) H = hh th (2.29) - 20 -
Fig. 2.6 He ad c he c k - 21 -
2.1.6 (2.30) (2.32). F ig. 2.7. A ( ) d3. F ig. 2.8. d3 Stepanoff. Archim edes. K v = c v / 2gH (2.30) A ( ) = 360 ( Q ) (2.31) c v = d 3 - d 2 d 2 (2.32) - 22 -
Fig. 2.7 Co ns tant of v o lute - 23 -
(a) Starting point of tongue (b) Width of volute Fig. 2.8 S tandard of v o lute - 24 -
2.2 2.2.1 1 d 1, 1, d2, 2 (2.33). R = d 2 2 - d 1 2 4 ( d 2 cos 2 - d 1 cos 1) (2.33) 1... 1 Fig. 2.9 (9 ). (1) B 2 P. (2) OB ( 1 + 2) C. (3) B C C A. (4) P (2.36) B A. - 25 -
Fig. 2.9 Co ns truc tio n of o ne arc blade - 26 -
2.2.2 2 2 2 1. (2.34) (2.35). (2.34) (2.35). F ig. 2.10 (2.36) 2. (2.36) R 1, r 1, 1, 1 r F, F. 2, 1. R 2 = 1 2 r 2 2 - r 2 2 r 2 cos 2 - r 1 cos F (2.34) R 1 d 1 s in 1 2 F = 360 / z ( = 2 / z ) (2.35) (2.36a) = ( F + 1) - F (2.36b ) 2 R 1 2 ( 1 - cos ) = r F 2 + r 1 2-2 r F r 1 cos F (2.36c) r F 2 2R 1 r F cos + R 1 2 = r 1 2 + R 1 2-2 r 1 R 1 cos 1 (2.36d) - 27 -
Fig. 2.10 Co ns truc tio n of tw o arc blade - 28 -
2.2.3. F ig. 2.11 3. 3 (2.37). r1 = b1 3 (2.37a) r2 = ( sy2 - y o1) 2 + (x s2 - x 1) 2 (2.37b ) r3 = sx3 2 + ( sy3 - db/ 2) 2 (2.37c) - 29 -
Fig. 2.11 Co ns tructio n of s hro ud - 30 -
2.2.4 b 3 = ( 1.5 2.0) b 2 b 3 = 2.0 b 2 b 3 = 1.75 b 2 450 b 3 = 1.6 b 2. a a 14 r/ h 0.2 (6 ) (8 ) (9 ). Fig. 2.12. A ( ) = 360 ( Q ) c v Archim edes. - 31 -
Fig. 2.12 Co ns truc tio n of v o lute - 32 -
2.2.5. (Don ath ). F ig. 2.13 h. r. z 0. r (2.38a). (2.38b ). (2.39) (2.40) (2.39). d dr ( r r) - + g 2 r 2 = 0 (2.38a) r d dr ( b - b r ) - ( 1 + )( b r - b ) = 0 (2.38b ) r = C 1 + C 2 r 2-3 + 8 g 2 r 2 (2.39a) = C 1 - C 2 r 2-1 + 3 8 g 2 r 2 (2.39b ) b r = C 3 + C 4 r 2 (2.39c) b = C 3 - C 4 r 2 (2.39d) - 33 -
C 1 = r + 2 + 1 + 4g 2 r 2 (2.40a) C 2 = r 2 2 ( r - ) + 1-8g 2 r 4 (2.40b ) C 3 = b r + 2 b (2.40c) C 4 = r 2 2 ( b r - b ) (2.40c) - 34 -
(a) div is io n of s hro ud (b) ce ntrifug al fo rc e and s tres s Fig. 2.13 S c he matic of s hro ud - 35 -
F ig. 2.14 n. n M P (2.41). Fig. 2.14 +. r1 = - P 2 r 1 h 1 (2.41a) r1 b = 6M 1 2 r 1 h 1 2 (2.41b ) rn = P n + 1 2 r n + 1 h n (2.41c) b rn = 6M n + 1 2 r n + 1 h 2 n (2.41d) (2.41). (1) F ig. 2.14 n. (2) 1 1, b 1 C 1 C 4. (3) 1. (4) 2. (5) (3), (4) n. (2) 1, b 1. 1, b 1 (2.42) b r, r, r (2.43) 1, 1. - 36 -
1 = 1, b1 = 0, M 1 = 0, P 1 = 0, = 0 (2.42a) 1 = 0, b1 = 1, M 1 = 0, P 1 = 0, = 0 (2.42b ) 1 = 0, b1 = 0, M 1 = M 1, P 1 = P 1, = (2.42c) r1 1 + r b 1 = P n + 1 2 r n + 1 h n - r (2.43a) b r1 1 + b r b 1 = - 6P n + 1 2 r n + 1 h 2 n - b r (2.43b ) - 37 -
Fig. 2.14 n div ide d d is k - 38 -
2.3, (Vortex panel m ethod). 2 Fig. 2.15 m.,. Qb (Source),. NB. m, 2 ( )., km j k i (x k i, y k i) (2.44). - = Q b 2 N B k m = 1 ln ( x 2 k i + y 2 k i) 1/ 2 m j = 1 1 2 ( s kmj ) t an - 1 ( y ki - y kmj x ki - x kmj )ds k mj (0, S kmj ) (2.44) (2.44) ( s k mj ) (2.45). ( s k mj ) = kmj + ( k m (j + 1) - k mj ) s k mj S kmj (2.45) - 39 -
Fig. 2.15 Re plac e me nt of pane l o n a blade - 40 -
2.3.1 0 Kutt a. (2.46). C = =W + r (2.46a) W = - r (2.46b ) (2.46) (2.47). n W =n - n r (2.47) i 0 (2.48). ( n k i ) - ki s in ( ki ) = 0 (2.48) (2.44) (2.48), r 2 ( ) ( ' ). (2.49). - N B km = 1 m j = 1 0 S km j { CN 1( k, i, km,j ) ' kmj - 41 -
- CN2 ( k, i, km,j ) ' k m (j + 1) } = A R H S k i : ' k mj = k mj / 2 r 2 km (j + 1) = km (j + 1) / 2 r 2 (2.49) [Kutt a ] Kutt a. (1), (2), (3).., Kutt a (2.50). ' km 1 + ' km ( m + 1) = 0 (2.50) (2.49) m NB ' Kutt a (2.50) (2.51). N B ( m + 1) j 1 = 0 A N ( i 1, j 1 ) ' (j 1 ) = A R H S ( i 1 ) (2.51) i 1( k = 1 N B, i = 1 m + 1) j 1 ( km = 1 N B, j = 1 m + 1) (2.51) A N ( i 1, j 1 ) ( k, i) - 42 -
( km, j ) ' k mj. (2.51) ' kmj. - 43 -
2.3.2 (2.46) (2.54). t W = t - t r (2.52) t W = ( t ki ) +rk i cos k i (2.53) (2.52) (2.53), (2.53) (2.49) (2.54). V T ( x k i, y ki ) = ( r k i r 2 ) cos k i + Q b 2 r ki ( 1 r 2 s in ki ) + GA T ( k, i, km,j ) ' kmj GA T : (2.54) (2.54). k i (2.55). CP ki = p k i - p T ( r 2 ) 2 = 1 2 [ ( r ki r 2 ) 2 - ( V T ki ) 2 ] (2.55) - 44 -
2.3.3 (circulation ). = m j = 1 1 2 ( kmj + k mj + 1 ) S kmj (2.56) H th (2.57). H th = Z 2 g (2.57) - 45 -
3. T able 1 case.,. case1. case2 0.035m 3 / m in, 4m, 2600rpm. case3 case2 case1. case 36. T ab le 1 Calculation re s u lt case1 case2 case3 0.035 0.035 0.035 m 3 /min 2600 2600 2600 rpm 29 26.8 26 mm 55 64.9 55 mm 22 25.9 31.0 42 22.5 22.5 6.3 5.4 5.1 mm 4.2 3.2 4.3 mm 8 8 8 3.79 5.78 3.42 m F ig. 3.1 Layout. - 46 -
,,,,,,,,. F ig. 3.2 Fig. 3.4 case2. F ig. 3.5 F ig. 3.7 case2,. 0.035 m 3 / min, 2600rpm, 4m n s 171.9 64.9m m. F ig. 3.8(a). Fig. 3.8(b ) 73.8m m 16 19.6mm 73.8m m.. 5. - 47 -
Fig. 3.1 Layo ut of c e ntrifug al pump de s ig n pro g ram - 48 -
Fig. 3.2 Calc ulatio n of s pec ific s pe ed - 49 -
Fig. 3.3 Calc ulatio n of inle t/o utle t s tandard - 50 -
Fig. 3.4 Calc ulatio n of inle t/o utle t w idth - 51 -
Fig. 3.5 Ve lo c ity triang le of o utlet - 52 -
Fig. 3.6 Draw ing of impe lle r and s hro ud - 53 -
Fig. 3.7 Draw ing of vo lute c as ing - 54 -
(a) Calc ulatio n of s tres s (b) Dis tributio n of r Fig. 3.8 Calc ulatio n of S hro ud S tres s - 55 -
F ig. 3.9 F ig. 3.11 case1 case3. case2 / case1 case1. case1, case2, case3 case1 case2, case3.. case1 case2. - 56 -
Fig. 3.9 Grid g e ne ratio n of c as e 1-57 -
Fig. 3.10 Grid g e ne ratio n of c as e 2-58 -
Fig. 3.11 Grid g e ne ratio n of c as e 3-59 -
F ig 3.12. 1 18 19 36. case1 case2. case2 case1. F ig. 3.13 Fig. 3.15 case1, case2, case3.. F ig. 3.16 Fig. 3.18 case1, case2, case3.. - 60 -
Fig. 3.12 Pres s ure c o effic ie nt of blade - 61 -
Fig. 3.13 Ve lo c ity dis tributo n of c as e 1-62 -
Fig. 3.14 Ve lo c ity dis tributo n of c as e 2-63 -
Fig. 3.15 Ve lo c ity dis tributo n of c as e 3-64 -
Fig. 3.16 Pre s s ure dis tributio n of c as e 1-65 -
Fig. 3.17 Pre s s ure dis tributio n of c as e 2-66 -
Fig. 3.18 Pre s s ure dis tributio n of c as e 3-67 -
4 2..,,. - 68 -
(1),,, (1997). (2),,, 2,, Vol.17 N o.2 (1993), pp.41-51. (3),,, 2,, (1999), pp.264-270. (4),,,, (1996), pp.25-142. (5),,,,, (2000), pp.800-846. (6)A. J. Stepanoff, "Centrifug al and A xial F low Pum ps" 2nd Edition, John W iley & Son s. Inc, (1957), pp.1-137. (7)Chen Cichang, "A Com puter Inter ated M anufacturing Sy stem for Pump", T hird International Conference on Pum ps and F an s, (1998), pp.103-109. (8)Victor L. Streeter, "Handbook of F luid Dyn am ics", M cgr aw - Hill Book Com pany. Inc, (1982), p.19. (9), " ",, (1964), pp.39-113. - 69 -
2.,...,,.. - 70 -