由假设得到公式 2 {7 H, F h% I! J1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）# G V6 D- w6 J$ L
2 `; A) V; ^* X4 V7 c) W9 m( T2 ~' _公式 + X; e. Q& Y! t# e1 G2 K1 E% }/ {3 @
Where" ` u5 S0 a* A% D# T1 S0 n) b8 D' p
9 _/ M# X/ x* B
符号解释* V& V4 Y% l9 \' Q3 V8 q
7 \. u' [2 [ @* ^' W% j
According to the assumptions, at every junction we have （由于假设） 3 ]8 u& D/ m3 u ~! Z7 }% q. }( _# K2 M3 A$ r3 [ b( h5 X3 t K
公式 ) R6 k J! x/ ^% U. I. m0 M3 d9 l% t* ?# y6 t1 B+ n
由原因得到公式 6 g- ?' R0 A, q; ^: h n, a/ G2．Because our field is flat, we have公式, so the height of our source relative to our sprinklers does not affect the exit speed v2 （由原因得到的公式）；0 f! Z/ d: [: M4 g
- B# s( A/ `6 m公式; M( I h6 K; d
8 I; C5 T0 v! s! C. Y7 \3 a' gSince the fluid is incompressible(由于液体是不可压缩的), we have & @5 i! i6 F9 [% y3 _' M Q9 I. ?" i, h% O6 f公式 8 l1 h4 b% n* d8 H B ; D- M1 e0 z8 r" pWhere ' {/ X: h- {7 o! t( W, ~+ s/ ~8 X# |" x7 C
公式3 j6 ]" T1 j7 t+ K) s
( J* y+ G# }1 P7 \9 x
用原来的公式推出公式# S! i \( X$ W* I! n
3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）) d- O- ~4 w, r+ X
5 z: o2 N/ z* z4 _% b' O# t
公式 , d) ~- B& Y! P; K6 Y/ E0 B 2 J. I- Z7 K8 ]11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields： * [* q) g6 a5 b: ]1 e- @9 r , {$ j7 T8 l1 z公式" | L1 I6 d! C+ x
3 Y7 W* |& q2 t6 O+ v12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得） & q d* t" R8 M) J9 k1 C' W, h9 R6 a7 Z; g# I
公式: @3 G: x- r0 C
' W3 e% @; Y; o. y/ u/ }8 S+ vPutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have' u( e o0 K, A. T+ M. E
( X8 `0 h& U. V
公式; a6 n* Q, N1 g) c5 s' X" H
# E# a! [& \- R+ N; E7 S
Putting these into (1) ,we get（把这些公式代入1中） * P9 T7 K6 n. ?( |6 P0 ]' ]6 ^- \/ l2 ^ v9 |, E
公式 ( @+ u& g$ a& o+ f' u O* ? ) L! u* J; w: w. q" KWhich means that the; X7 ]/ ]0 n L, ]2 ~$ l3 j/ ?
* m, j0 e5 Z, Q& B9 F( p7 j+ ~Commonly, h is about 6 G& f. M3 r4 k" y: x8 W2 F( _ + O0 G& @. l& k1 u' A9 ^$ HFrom these equations, （从这个公式中我们知道）we know that ……… ; p$ U( ?- n3 s) }+ w( R b3 L, U
# [/ r8 f2 H1 v# ^1 z+ V3 w4 |4 T
, z; d$ _9 J k6 o- X引出约束条件 ' R l. D* U6 l0 z7 W* g2 k- {4．Using pressure and discharge data from Rain Bird 结果, 1 r, k' T" u m: K ) a- Y- D; m _( C/ vWe find the attenuation factor （得到衰减因子，常数，系数） to be 2 w) V+ r9 ~7 h$ g 9 d8 E1 K b9 P公式 0 k9 X2 C' K/ k 8 B0 o& R4 B# A3 d' Q } A计算结果7 M' e1 E6 c* V' t1 P" ~% c
6．To find the new pressure ,we use the ( 0 0),which states that the volume of water flowing in equals the volume of water flowing out : （为了找到新值，我们用什么方程）' R3 |5 Y E3 @1 ?4 P7 K" y
% j& m; i) V5 g' l公式$ ~% l/ v; D z$ ]6 u2 V) K2 u
2 t% z: N/ W$ A+ s4 q$ n& h2 ~
Where5 g+ [2 M& X- N' T/ S
- }" ]* p6 a' k9 P; F1 Q" y" |
() is ;;8 {. _5 K2 L1 z
3 i# E1 p. A7 p# L; e$ m- z
7．Solving for VN we obtain (公式的解) % i# @$ Q9 m' P( W. l: Z4 m 0 Z3 H, u' _- \ B0 D公式# [9 S7 }1 I$ k# M% A' e5 [8 `, j
) |( n3 h5 c j, p9 FWhere n is the …..) E1 N7 L8 O4 G1 g: }
' B* c, I7 S- }% O Z # m. O/ x9 g( R" v! K & J, z) S) A9 m- l" f: q8．We have the following differential equations for speeds in the x- and y- directions:: A% R, |( v" p- g" \
# b, M& Y) U {5 j公式 ) Y1 e; t( ?9 E. Y% [0 c3 r% T: M( R. i
Whose solutions are （解）) Q0 T$ X) d! Y( Z! I
0 d. O+ I, Q& U, N# @公式 3 \' N% |. m' W* @5 u + @5 J; D/ ^ `& h9 T( l9．We use the following initial conditions ( 使用初值 ) to determine the drag constant: 9 K2 {# C# ]$ c+ A9 v) }8 r+ H9 J! @& ^7 F2 T" g, l2 B
公式" T6 P6 r# | T/ s D
+ T' v3 ~+ l# Y) m7 | _" s根据原有公式3 {# c' {2 L' H. V3 ?; C
10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is, c$ ]. w7 q- l, J2 f" z" j
" l2 j% }+ I% c% l2 N公式 ; u. p; l3 l1 ^$ Z. @" ?3 x6 X2 K3 @5 T
The decrease in potential energy is （势能的减少） + B H' k' H' c! y) A1 K) Z) W8 @0 C% [5 q. ^3 f+ B
公式 ! x* ]- x1 {1 ?/ L" r: Z5 e& I 3 U j! D, b( J% wThe increase in kinetic energy is (动能的增加) 7 G$ \: J) g4 y6 K. m2 |1 O. O$ ^2 {% {
公式1 \! J7 X5 A$ X0 L/ X
) A/ H, k: k6 N- |! w
Drug acts directly against velocity, so the acceleration vector from drag can be found Newton's law F=ma as : (牛顿第二定律)9 K3 J( \. ~. K4 z3 T9 c" R' ~% H
3 _) [5 s$ W h0 b/ ~4 R; AWhere a is the acceleration vector and m is mass 7 {. [/ k9 ] i7 Y9 y3 X# p* B& q! }3 q7 }) b" C7 j+ k
) o: R5 x9 Z" C1 r9 x l2 |( C) ^8 f5 K& L" g) P* J" r4 F( \
Using the Newton's Second Law, we have that F/m=a and" X0 |1 c _2 P) ^6 Y+ y
' r) x) i$ R; J8 e, `: |( R公式 ! k/ g: p$ q: A1 J# F ' V( ?7 ]4 R5 C$ f c3 ]4 fSo that' F" }& k- G, {& E- k3 x
, g+ h6 a5 Z. k% f% j3 g
公式; y5 @: A# R. A( x7 [
# i7 B# y+ p4 E: i) g6 R/ a9 S" A- vSetting the two expressions for t1/t2 equal and cross-multiplying gives/ F$ A' d8 S3 J& w4 e9 F4 T5 |
; E0 c# Q9 k4 p2 _. c公式7 }; L- F4 d9 g
+ ]8 z5 g; G$ @' Q
22．We approximate the binomial distribution of contenders with a normal distribution: : A4 V# U( O1 ^$ j6 M3 N7 [# B: N% H0 F! ]( b. Q+ C. v
公式' a% h: V4 y! h3 U4 }% ?
! u& w6 d: \5 XWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives1 w7 h6 E6 {2 U' M
* T; K& ?. ]- b. y2 u公式 6 g6 \7 n* l; b# _& N" |9 W' [ 3 l, x& j5 K$ M4 ?& r7 J+ PAs an analytic approximation to . for k=1, we get B=c2 n7 D4 P2 ]3 q: {
: L4 u. }! V3 ^- i4 s0 [ 2 l, N( T( d6 G% E+ z5 m
4 u8 ?9 @4 A' ?6 G! l; A26．Integrating, （使结合）we get PVT=constant, where + e: u- z4 @* q' k 8 R+ \% b+ q/ W( [- p公式6 ?% D" \0 ~" p' N# h8 X; c: j- _1 s
; P. @% W0 J P' X9 KThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so $ I( j6 A" t" m. l! Z# G* y8 n/ |+ d- `! f0 B8 \ \! I
( V0 P: |4 D2 S3 P- g) m 6 ]' w. ^& y/ S5 m0 c23．According to First Law of Thermodynamics, we get4 G; p! L6 ~" c' C
) q9 Z9 t4 T. ?: q- m5 B
公式 5 F3 }1 u, |! O: a1 n" |+ ?$ ^( x, m( }0 q) q
Where ( ) . we also then have* b7 L) l, l" k: j
1 {. S- z/ p! w* b
公式. y, @0 L9 [# S7 Q1 o
0 t- Y, A* W0 _# r' r- y
Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula: 5 \. B0 _! {" Z7 \, t3 Y- s# x0 [# r! D, V: D, r; u4 g
公式' X; L* k6 |: v" ~% E+ R" q
+ O* o" K9 I) i0 B; C" v
Where7 Y" Z2 H4 e: [
1 A- ?5 p8 l- _/ ?3 lWe maximize E for each layer, subject to the constraint (2). The calculations are easier if we minimize 1/E.（为了得到最大值，求他倒数的最小值） Neglecting constant factors （忽略常数）, we minimize 4 n, V, [8 [1 N2 f' p( E$ x+ X6 O8 c! j) x8 C c
公式: k* S5 t1 N( @; g
. H% C& R8 f. K9 G
使服从约束条件 , U4 B( G' P; i! u7 I3 G) }/ o8 t14．Subject to the constraint （使服从约束条件）& j9 b7 o) t! t
% G* `( N* [9 ` L4 v4 X* G公式2 ?; l5 C8 F1 W: s/ y1 i7 m& n
9 z: n3 B3 f; w! N& [( z7 WWhere B is constant defined in (2). However, as long as we are obeying this constraint, we can write （根据约束条件我们得到） # O7 \% Y/ T- }$ _7 p+ k1 ^, I! E! Q! ?% X
公式) [# D/ T5 D8 p7 T
: r5 w0 N/ O: o& i# m
And thus f depends only on h , the function f is minimized at （求最小值）- H6 ?0 b& j1 o% f9 k, s6 E
/ U% @7 e! `% t3 f( j' b0 j
公式 7 j$ S" v3 n5 R1 `) f: v% Q* Z7 ^8 `$ c7 U1 @9 O4 m" o
At this value of h, the constraint reduces to5 E8 X6 X3 m4 a( d8 x4 U+ |
1 o& C6 E1 i' Q6 n$ N* f* [
公式5 v2 F& s8 e9 K
3 I. u4 v$ b0 _( x. \
结果说明 1 K5 j$ ?1 j8 S15．This implies（暗示） that the harmonic mean of l and w should be ( v8 ?) K9 K8 n 7 q0 h* [2 o, d$ A7 B: n. ?! Z公式3 \+ }% g! t q( V
! s# E! @4 k9 s* o. m, `* l
So , in the optimal situation. ………0 M% n8 u9 G) a, O- h+ C9 |
% W N5 Z6 |5 l7 o5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is ; u! a' H6 N4 u& u & L; ?! P1 R* u! X# x8 ^$ h公式 , C$ A2 Z5 q+ z$ D7 n$ V [/ f5 c. C) n
16． We use a similar process to find the position of the droplet, resulting in9 r: P4 N" R+ y, ~
j( B- Y: w: I& `8 }
公式 {/ ?5 T8 ?5 j
5 J; R% }4 }' X6 p0 i2 G; hWith t=0.0001 s, error from the approximation is virtually zero. * y! D4 k0 m' S- f% _' j) U E7 y- o K; ] 7 d1 ^0 K# ~& ]( z+ f' r/ M; `; `2 A8 i0 `* S: L, R6 q
17．We calculated its trajectory(轨道) using6 h6 j1 J1 U7 S* b
8 M8 \3 C8 d: O
公式3 y$ _8 E% ?: M
- q5 Y+ e9 s0 } W/ U4 o18．For that case, using the same expansion for e as above, # h# G5 y! a1 \6 I' s8 J6 x4 _; Q: [ % @. z' v: z( ?# w5 d% R ~公式 ( T6 [" {+ V" @- N }7 Q5 T$ E2 V: d! k$ S" f
19．Solving for t and equating it to the earlier expression for t, we get( S; z0 {% ^* M9 U) B; l
' Z0 M0 v- I6 p( b4 I8 p v
公式( u; q1 z1 H6 h- l/ k
* j. U: J+ B( v: u' M) h5 W5 X
20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is" g+ I* m2 |1 ~3 v% w0 ?
+ J9 W% Y" a3 `( \9 i4 O8 q公式( [- o; U- I4 v6 O
) a! b& S: N/ _+ U ]1 T% e7 v
As v=…, this equation becomes singular (单数的). " O# U5 {- k9 B# A! s5 Z + p8 }4 R- J; f0 {+ m! Y( \7 C% [ 0 O) `3 ?6 H, }, W) u$ L R0 U/ g0 i$ e, v: J
由语句得到公式( ]. V' p8 t3 ?; Q$ B% i
21．The revenue generated by the flight is1 q: C& q. {+ { n1 z! F, S
$ l8 n' f' x- a. W, n b
公式 ' |5 i o+ P% [, g+ J Y/ O2 u* T3 g7 n( h+ }, S
4 p& s, v) D+ c4 z
/ v4 x- T! y( b- S* k
24．Then we have3 I! T* @ l6 _9 E3 O) H9 `! j G
4 o3 ^( u9 j2 C
公式 # V$ J4 y) W* B& y9 T: Y' P1 |
We differentiate the ideal-gas state equation7 ?8 ?0 D( t: K: e! h; T! N
: T7 B$ F. a& r( q& `2 Y0 O
公式 ' m- z: x/ T2 `; w) M3 ] ! V; g+ R! g6 N! oGetting 1 m( q! T4 `, ^7 H: G$ ?1 L * O+ m2 b6 w' \) Q4 W' P0 M9 r0 i公式 4 t' |1 R0 w+ ~/ s' R/ V$ ]. N % `% |7 J9 i) R( }4 Z \25．We eliminate dT from the last two equations to get （排除因素得到）4 f0 {, }' m7 @0 y, `
* j" r, c8 ?" |6 e( y. d7 Z- c
公式: u* g& H) Y \& s6 K) p, B+ J4 N
* x4 _5 i* x& a7 Q4 ^! g# g
0 O5 V' X3 l7 n# L) w
. f. ^. ?. B% s! V* [% z22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations c1 L; I: d F
4 h- Q ]4 l! a7 s5 R# C( y7 Q! S
公式 7 h7 m6 T9 x) k4 z8 V2 Q+ Z3 C7 S/ K a8 w. c4 f
Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）9 ?. P: Z7 w* ]* z; |
2 W- @* O$ m( B# p
公式 " j6 S1 w9 `4 m" ^1 Y9 ^, C———————————————— ; W; [: L1 Y& t6 L3 h$ w u版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。 6 g! c! w- {# y) y8 L原文链接：https://blog.csdn.net/u011692048/article/details/774743863 S) O, B9 T0 w
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