Thủy lực 1 ( Nxb Nông nghiệp ) - Chương 8 potx

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w#l#vK+E(y#(#l#vh +h95#N#jy#j#l#m"95#E # # ### c##l#K+ (# Eh + l#h + Nj hh9 5 + l## 9x#E# ### Rl# c w l# m"95 9x #l## 9h!h#EN## qD&(#$#5(M7#,;&'#5(_,#nN#nN#PSPa=84QB:B9#5S#/)A,#$#l#m696x##E }95 /QN## Vt.g###O#l#w#$ R lm"95#N#m696x#N# h!h9 #l#jm}m#E j /Q # V>g### Q#l#O B #l#jm}m }}} 9} l#mx9hj#E j /Q# ### Q#l#mx9hj#E j /#QN# [...]... 0 - m ữ (1 + s)R ố ứ 292 (8 -3 2) (8 -3 3) Thay h tính theo (8 - 3 1) và b tính theo (8 -3 2) vào (8 -2 7), ta được: (1 + ) 2 m0R2 w s R2 = m 0 (1 + s)2 w= hay (8 -3 4) (8 -3 5) Từ (8 -3 3) và (8 - 3 1) ta được: b m0 = m h m0 s= +m b= hay (8 -3 6) Từ các công thức (8 -2 9) đến (8 -3 6) ta thấy rằng nếu biết s thì quan hệ giữa các yếu tố của mặt cắt sẽ được xác định 2 Đặc trưng s của mặt cắt có lợi nhất về thủy lực Ta biết... 14 ,43 1, 523 0,45 18 ,03 3, 314 0,60 21, 31 6 ,11 9 0,75 24 ,19 10 , 01 0,90 27,03 15 ,20 1, 05 29,34 21, 55 1, 20 32 ,17 29,93 1, 35 40 ,1 71, 0 1, 50 12 , 28 1, 0 25,4 K0 (m /s) W0 (m/s) 3 H (m) 28, 9 20 ,12 1, 2 32 ,1 30,40 1, 4 35 ,1 43,60 1, 6 38, 0 59,6 1, 8 Bảng 8- 4 Trị số K0 và W0 của ống hình lòng máng n = 0, 013 10 ,82 K0 (m /s) 3 H (m) Bảng 8- 3 Trị số K0 và W0 của hình trứng n = 0, 013 0,25 H = d(m) Bảng 8- 2 Trị số K0 và... 0,00 01 (hình 8- 9) Giải: Ta chia mặt cắt kênh ra ba phần như hình vẽ (hình 8- 9): 0,50 ử ổ Phần 1: w1 = 10 + ữ 0,50 = 5 ,12 5m2, 2 ứ ố c1 = 10 + 0,5 1 + 12 = 10 ,70m; R1 = 5 ,12 5 = 0,48m; 10 ,70 C1 R1 = 23 ,80 ; K1 = w1C1 R1 = 5 ,12 5 23 ,80 = 12 2m3/s; Q1 = K1 i = 12 2 0,00 01 = 1, 22 m3/s 3 1 h1=0,5 h3=0,5 h2=4,5 2 b3=6 b1 =10 b2=5 Hình 8- 9 Phần 2: w2 = (5 + 4) 4 + 0,5 (5 + 2 4) = 42,5 m2; c2 = 5 + 2 4 1 + 12 ... tròn n = 0, 013 40,7 78, 8 2,0 34,43 42,69 1, 50 44,3 10 6,5 1, 75 46,2 12 8, 0 2,4 37,0 51, 43 1, 65 48, 5 15 2,5 2,0 50,9 19 3,0 2 ,8 38, 92 64,39 1, 80 52,4 2 08 2,25 55,7 276,0 3,2 40,42 78, 58 1, 95 56,2 276 2,50 59 ,8 375,0 3,6 42, 68 92 ,19 2 ,10 59,7 354 2,75 63,9 495,0 4,0 44, 81 115 ,8 2,25 63,3 447 3,0 Giải: Với a = 0 ,8, tra biểu đồ (8 - 1) (hình 8 -1 0 ), ta được A = 1, 0 Vậy K0 = K Q 3 = 47,4 m3/s = = A A i 1 0, 004... nhất về thủy lực sẽ có R lớn nhất Vì w, m0 đ cho trước, nên từ (8 -3 5) thấy rằng, để có R lớn nhất s lớn nhất cần có (1 + s) 2 Muốn thế cần có: d ộ s ự 1 2s =0 ờ ỳ= ds ở (1 + s)2 ỷ (1 + s)2 (1 + s)3 Giải ra ta được: sln = 1 (8 -3 7) Vậy điều kiện để có mặt cắt có lợi nhất về thủy lực của mặt cắt hình thang là s = 1 Tuy rằng có quan hệ khác nhau (8 - 2 1) và (8 -3 7) để biểu thị mặt cắt có lợi nhất về thủy lực. .. hình thang (b, h) sao cho mặt cắt đó có lợi nhất về thủy lực Các số liệu lấy ở thí dụ 3: Q = 1, 1 m3/s; m = 1, 5; n = 0,0275 và i = 0,0006 Giải: Tính Rln: f(Rln) = 4m 0 i 8, 424 0,0006 = = 0 , 18 8 Q 1, 1 Tra phụ lục (8 - 1) ra: Rln = 0,495 m Với mặt cắt có lợi nhất về thủy lực thì h = 2 Rln [xem (8 -2 2)] Vậy h = 0,495 2 = 0,99 m, và tra phụ lục (8 - 3) (với s = 1) ta có: b = 1, 21 R ln Vậy: b = 1, 21 0,495 =... 1 + m 2 = b mh + 2h 1 + m 2 = b + (2 1 + m 2 m)h = b + m0h ở đây: Vậy m0 = 2 1 + m 2 m R= (8 -2 8) h bh = mh b + m0h 1+ 0 b Để cho tiện, ta đặt: s= m0h , b (8 -2 9) s là một số không thứ nguyên, biểu thị quan hệ giữa b, h, m, nghĩa là biểu thị hình dạng mặt cắt, nên gọi là đặc trưng mặt cắt Vậy: R= h 1+ (8 -3 0) hay: h = (1 + s) R (8 - 3 1) Từ (8 -2 9) ta có: b = m0h m 0 = (1 + )R ổm ử b = b mh = ỗ 0 -. .. 0, 013 1. 0 h a=_ H 1, 0 n a= _ h 0 ,8 0 .8 A B 0.6 0,6 A 0.4 0,4 B 0,2 0 0.2 0,2 0,4 0,6 0 ,8 1, 0 1, 2 A,B Hình 8 -1 1 0 A,B 0.2 0.4 0.6 0 .8 1. 0 1. 2 Hình 8 -1 2 Thí dụ 12 : Xác định đường kính của ống tròn bằng bê tông cốt thép sao cho a = h Ê 0 ,80 H Biết Q = 3 m3/s; i = 0,004; n = 0, 013 303 304 0,595 12 ,1 K0 (m3/s) W0 (m/s) 19 ,2 3,76 0,50 25,3 11 ,2 0,75 30,5 24,0 1, 0 35,6 43,7 1, 25 0,497 0,30 W0 (m/s) 14 ,43 1, 523... thước mặt cắt kênh hình thang (b, h) sao cho b = b/h = 5 Cho biết: Q = 19 ,6 m3/s; m = 1; i = 0,0007 và n = 0,02 Giải: Tìm Rln: f(Rln) = 4m 0 i 7, 312 0,0007 = = 0,00 988 Q 19 ,6 Từ phụ lục (8 - 1) ta được: Rln = 1, 30 m Tìm s từ (8 -3 6): s= 1, 82 8 m0 = = 0,305 +m 5 +1 Có s tra phụ lục (8 - 3) ta được: b = 5,75, do đó b = 5,75 1, 3 = 7,5 m, R ln h = 1, 152, do đó h = 1, 152 1, 3 = 1, 5 m R ln 297 Thí dụ 9: Xác định... Bảng 1 h (m) w (m 2) c (m) R (m) R (m0, 5) 0,70 1, 58 3,73 0,425 0 ,85 2 ,11 4,26 0 ,83 2,03 0 ,84 2,07 C (m0,5/s) K (m3/s) 0,652 29,5 30,5 0,495 0,704 30 ,8 45,6 4,20 0, 480 0,695 30,5 43 ,1 4,03 0,490 0,700 30,7 44,6 Qua bảng 1 ta thấy với h = 0 ,84 m thì w C R = 44,6 ằ Q i = 44,9 Vậy độ sâu chảy đều là h = 0 ,84 m 289 Thí dụ 4: Xác định kích thước của kênh hình thang (b, h) sao cho mặt cắt đó là lợi nhất về thủy . #l#mm9"#E j /QN# $(7 #EZ5#Q8#'B3#56@ #(# 5D& ;(# 6S#O#l#w$ R #5)a&'#_&'9#5HB#=z,#&>7#KU&'#mm9"#=>#/)A,N# u8#=BC1#5D& ;(# 573&#/)A,#'(B#=`B#p#K-&'#hN# B¶ng 1 # h (m) w (m 2 ) c (m) R (m) R (m 0,5 ) C (m 0,5 /s) K (m 3 /s) 0,70 1, 58 3,73 0,425 0,652 29,5 30,5 0 ,85 2 ,11 4,26 0,495 0,704 30 ,8 45,6 0 ,83 2,03 4,20 0, 480 0,695 30,5 43 ,1 0 ,84 . jy#5S#,2#EZ5# 4 () a&'#56W& ;(# (SB#È&# Q8#vK9#(y9#Pt.#,c&# 4 (- B#,2#EZ5# 4 () a&'#56W& ;(# 5(_ #(SB#&< ;1# 5(& lt;E#EZ5#E8B#q1S&#(C#K9 #(# &fSN# $2#5(r#,2#(SB#56)T&'#(A4#P0# 4 () a&'#56W& ;(# 5(_ #(SB#/2g## hN# $(7 #KBY5#5û#Q8# b #l# ( K N#O(B#/2#5(S.#ELB#56@#Q8#K#567&'# 4 () a&'#56W& ;(# vxwmy#KU&'## K#l#b(9#5S#,2#EZ5# 4 () a&'#56W& ;(9 #EZ5#È&#Q8 #(# P>#56p#P0#56)T&'#(A4#K>B#573&#5WE #(9 #:(B#/e#KBY5# Q9#K9#E9#&9#B#/e#&2B#p#56<&N#$2#5(r#=R.#5û#Q8# b #KU&'# b =& #=z,#/2# 4 () a&'#56W& ;(# 5(_ #(SB#=># 4 () a&'# 56W& ;(# vxw. hy#(S.#K-&'#vxwhy#/r#5WE# b =& N# VHB#E#l#h95#5WE#/)A,# b =& #l# ln h b ÷ ø ö ç è æ l#}96}6N# b =& 9#E9#&#/e#KBY5#&<&#O#,(s# 4(] # 5(1 Z,#P>7#(N# $(7 #EZ5#Q8#'B3#56@ #(# P>#5D& ;(# 6S#ON# OY5#q 1-# 5D& ;(# 573&#/)A,#'(B#=`B#567&'#K-&'# N# B¶ng 2 # h (m) w (m 2 ) R (m) C (m 0,5 /s)