由假设得到公式+ J+ v9 v) W( K. h8 k0 b& h
1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）2 m: a- d& q4 F/ ^7 Q) {
3 B' ~( b8 R" h+ ~# C X/ Z7 r
公式; X! m2 R8 X% d8 \2 o9 y k
' P; ^/ T1 D5 i, {8 T5 s
Where( a# d A; f7 I
6 t- W. q" \. U+ `& H9 G% u8 m5 z4 T
符号解释 / q, X8 d- `9 s0 M/ S$ U5 F! S8 K* c# d+ U& F
According to the assumptions, at every junction we have （由于假设） . w$ V) d) O! e) D " B: p. r% @' y C4 w) ]2 D! v+ g公式; U+ Q. u! r! K( J
, ^/ r B1 }& }" x) h4 [
由原因得到公式! d/ p; `" W: Z0 F8 y* ]) r& n
2．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）；5 Z# _, z7 i3 I: P ], ~& Y, O
& j% `* ~( `& n' V* ?- ?
公式' n/ o1 \( P. D$ b. m+ {, K
6 T$ J) B. C' J
Since the fluid is incompressible(由于液体是不可压缩的), we have( J' a& l6 \8 r* W2 {% r# J2 U, d
2 e$ R$ M6 d& P1 }
公式4 G, f2 s4 Z0 q; _' u
8 x+ [3 S% k0 c7 U& }; nWhere 3 j) ]' b. K7 n1 r F* o! F- @5 z/ a: v0 s
公式 ! w! d3 j9 R M7 U: V3 ] 2 H9 ?( n2 a( A) b用原来的公式推出公式 % A8 h' ]) ?& A4 s, B3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）8 v$ j+ w9 @7 A, w4 M# S
q3 a7 ~; u! ]% M9 q; X' V公式 1 X0 X5 w: o! S% D6 T9 F: Y% D+ \7 C8 S3 [ g t! `
11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：& W+ T* t$ W2 H- d; u4 g
8 ]' S; g7 L% |
公式 % |/ R( i3 e9 }% p( B4 ^' e1 S A+ x5 O" p$ D
12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）/ W" N* \2 \0 J( w# k
3 n2 a- R1 m- s8 U3 r7 }) G0 {; ~7 J
公式 & c0 ?6 I! Y9 V( S; M: `0 |! D7 a6 ]" F7 C1 o+ k& B
Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have . \3 C( N: \4 k2 D: E' ^ - N% }0 M5 X4 ?公式% K8 s2 D2 h$ v k6 g
9 `: ]7 [0 s$ U( W# fPutting these into (1) ,we get（把这些公式代入1中） ) I9 k6 y4 x/ C( b' x, P& |" P7 b- a5 J! N m/ l
公式( r% e, a- f& z5 n R4 r
" a; F& K& m4 ?2 m. p2 ]Which means that the ( l, ~# E- {: ~' f/ f6 x 6 T0 K7 D% l& S J) |) l( mCommonly, h is about " n1 A% X$ X/ Q' S( R' B, b6 x4 v- }; G
From these equations, （从这个公式中我们知道）we know that ………, ^+ M+ q- X) j* A# X# K
4 S% S+ w$ x% q. _& D" F* l
3 j& i3 _$ v5 ^- ]1 U t1 _
: u% [0 R3 T6 d" \引出约束条件6 E7 ~/ i! A1 R4 {' v( c, \+ a
4．Using pressure and discharge data from Rain Bird 结果, & o/ H' Z( [* b4 x9 r8 N: J$ k) F1 E# h5 M
We find the attenuation factor （得到衰减因子，常数，系数） to be* q/ m6 o4 ^3 a& {6 i
" ?* o5 n$ R) h1 G
公式& D6 X3 w7 D6 |( p! {/ ^7 ?
6 F! J6 S7 C0 Q* e
计算结果( @6 U! c# Q5 p3 ^# ~- r( P
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）. r% A+ L2 V* \7 X
! ~7 h& p' R6 M公式* r+ S- m* N! H5 Q. j1 H+ m* F
7 k" c% d1 h- R- \8 b
Where* F/ X; ?; u8 Q* J7 k& { z
0 k- A$ d/ \2 Q9 i# Z; g, p) s4 w
() is ;; + w0 I! v: l* l( X; ]* b: M9 A+ Z# [/ r8 q0 U
7．Solving for VN we obtain (公式的解)% \& V o5 e" d$ p! K; f
8 I" l. {0 |4 {9 x公式 " `6 ^# B( ~6 Y- l, D; D# _( \6 @0 A8 B! y3 I- o0 w
Where n is the …..; y1 q2 K/ E: C+ v. X; h e
$ D( @* X) ]* y. _ 3 \; Y2 Q: p& m4 K' X) p. R. u' p8 h
- _) v) F+ h8 m2 c9 P3 d, y" B8．We have the following differential equations for speeds in the x- and y- directions:2 p8 H$ z7 ^! n# G i
" F& K+ S6 D7 i6 S3 n1 r8 r" n7 Y5 Z
公式/ j* M" `' @0 x, e' [
; U& a- {7 e8 m4 E
Whose solutions are （解） `( K1 a) v1 R) o! w1 K ' L4 E! P. [: x! I公式 9 ?4 Y6 N1 Y& [$ E4 V3 m3 @2 P5 m# [4 f S# Y
9．We use the following initial conditions ( 使用初值 ) to determine the drag constant: / R- Y4 W4 n! s/ h7 O+ n7 A# ?9 }. L {2 H2 o' J) l; p- J
公式 $ d9 j- k9 v0 s& d5 l8 K9 d5 S& V# O0 p; D# A
根据原有公式 * k, D4 v$ r; W5 |10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is ( y5 l( l8 F/ x3 K/ O- w" _ 9 M8 ]- ?; a' o1 f e. r4 w+ Y0 Z公式 % q5 z4 \% \5 h7 r" {" e( W1 o. m9 h
The decrease in potential energy is （势能的减少） & `9 h. A2 s2 B1 s / Z. d) [ q3 y# @. O) N. M, s" ~公式 , d5 t' M9 ~4 A1 a3 ^ ' Y0 a$ k/ A1 [The increase in kinetic energy is (动能的增加)% X r" }4 `: T1 Y3 N/ U, u9 n
" w3 L3 }5 Y. m- ?3 b/ ~
公式! @3 {( j; t4 B
1 V+ p) R }1 D* C4 v
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)3 [, e! b2 x. I2 {& a
& B4 h7 M" E( }% `8 P6 j F+ zWhere a is the acceleration vector and m is mass / Q b% H. ]- k; N$ e - `1 U4 l( [# o4 G0 N, s 0 h2 d- ?: [5 x5 j+ ~1 N
+ ~) E: Y: z4 R" ZUsing the Newton's Second Law, we have that F/m=a and: j5 O. r: Q' I& W" v Z) C( m; g l
) k7 u$ j! J9 D) b+ S8 r! k公式 % S. `7 V8 R2 ~# l' h0 y7 l* _9 n# O, X6 W$ w. W
So that, N7 j+ |# W8 s# _$ S7 R
" A9 b, g( ?7 M
公式3 P2 j% [ z6 |# T# w' O) ]
* _* W: P/ o2 E" f Z% ?& o# q+ y! U
Setting the two expressions for t1/t2 equal and cross-multiplying gives " D$ Y! ~/ H2 h7 r: ~6 `# G+ ?' h0 g
公式 * M/ N" E) \' h5 N; B; T % e( B* v! Z' A7 j22．We approximate the binomial distribution of contenders with a normal distribution:; \, ^; Q' G0 x# B
5 a* c5 Q" z ]+ G( ^7 f公式 ! ~3 f" j: I& n: v- ?0 f" v8 b2 S9 I. h. Z) W0 `- h4 ^9 {
Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives + E/ |4 o# ~( h0 H/ Q " z, B! N3 q q公式7 V7 v1 ~* Z6 Q4 s
% r3 S) R/ j% Q0 j9 n2 K
As an analytic approximation to . for k=1, we get B=c" c$ u# r7 ?& `& `+ U! ^
2 g' R8 d1 i: K3 n! O 0 J. H; |- l O+ u2 X" B # l0 G$ I# k/ U' f$ m Q26．Integrating, （使结合）we get PVT=constant, where- r. @) j; [6 l$ g. ]( a7 f
: N' h+ D* t# }4 m( J3 H
公式 " U) `& }& j* ]) a; A& \* @+ p7 C. T1 X) p/ r- Q, L$ ]8 ^2 _: B) f
The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so5 a+ N' D! w3 ?% A+ U$ y0 ^9 m9 f
' D" g% Y; r' S3 b) f; k 5 g2 {) b& O3 ~. D5 }* N2 w* P8 M, y5 p' |! A0 h+ o
23．According to First Law of Thermodynamics, we get. T m& G: f/ f, U2 {1 ]
2 q3 |3 Z! z* ^4 p6 H0 K9 N
公式 % V2 m3 }- M" f: E3 R0 k5 C }. y w' @5 Q' q6 j4 S9 k# c
Where ( ) . we also then have) G) X! {1 Z' j; t- U F
8 T# E' j# `" ~- C! x" {2 M" I0 [& Y公式' J: b0 I8 x$ Y. n4 M
4 J# g- c- I3 K' Y- RWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula: & S) V: T1 o4 ~' Z 4 j1 H$ H" q3 Q# c. ~& _7 o" Z公式5 H* W- Z" S+ u z/ S6 e
( D& B! P3 \: H" q0 i, P% gWhere" u1 k9 E8 `6 ?' u
0 n: T3 O; e2 r) W1 @( h% n! v3 S7 X , m# p4 m a2 o7 c$ M4 H5 V5 `2 |8 a
. ?6 f7 \4 a) C& y$ i
对公式变形 1 d: p. u3 q2 a* T13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到） ) R: O6 n j- j" ^+ Y4 K+ v7 V ( @2 E) N* ~' {4 {/ Z! A+ H1 Y公式* M) ]! ^! t8 ]
1 ~: X3 C) E! ~- F1 S2 _1 L
We 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( k' G" Z3 c% i
% b1 ]+ Z; ?0 y4 }+ H7 N0 V. ]公式6 d: g- q8 k; B2 y: n" F3 F2 [ o
6 ]+ {+ V; x/ `. B# o) @
使服从约束条件 : ^; J8 P+ m9 R0 ^- {; V8 L14．Subject to the constraint （使服从约束条件） : s k/ Q3 E$ Z: B( n b3 ]- e {2 q, J- Z0 S3 C% ^公式6 B2 F g; G0 k8 o3 ~* R
! m9 ]5 r' c7 w0 E( K* x$ [Where 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） ! l6 ^$ R* @: ? p* l* H+ a- o( ^1 |& n# V- I
公式 2 ~, t! j/ `9 b6 h$ p# {0 @1 H- G$ c0 t4 ~* }( v
And thus f depends only on h , the function f is minimized at （求最小值）2 U/ d, G' f' |3 C8 V8 g
* f) i& P# z! _7 k. [- x公式- Q b/ d% e0 a) j7 e4 s
9 j1 c. Q3 H* U& s9 w* W$ }+ HAt this value of h, the constraint reduces to# S) ]8 J; q( a! b9 [5 y
+ X* ]8 q2 U, P" {
公式 $ j% m) z! R+ k" x8 b' x/ E ' g' Q* i3 w% t, T; ~1 @6 ?结果说明 ) {2 @4 |! W, D7 i9 S15．This implies（暗示） that the harmonic mean of l and w should be 9 V- C0 l ~0 o# E % r* U/ [9 [" A0 F公式 ' E# s1 k: q7 p0 L5 l/ r * ^# x% o3 k# V; a. DSo , in the optimal situation. ……… 0 q+ m0 L7 w, z8 u0 V7 ^7 u( w }. A + z# S0 F4 F9 Z4 t& Q5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is ' I: O; u; y( C3 _* d, L% D! S& h0 `/ z2 m( B; N3 M2 F
公式% R: J9 U+ v) D. e. g& C2 p, h
6 \; P# R3 [2 |5 U0 u16． We use a similar process to find the position of the droplet, resulting in ( [' z, B& T& `7 r8 l; z 7 z6 V* n( i; q( [( b! h: `& k公式: R' l' S, [, ?- b, R
, k3 p. R* {, y9 J5 v
With t=0.0001 s, error from the approximation is virtually zero. ' x) w7 m" _: `( L! @8 s d& R- R5 s( y; O# E' r - d' M6 W+ ~; B3 O+ V2 a $ z2 [3 R7 Q9 A; V, P17．We calculated its trajectory(轨道) using+ {# g7 h8 @4 d0 b6 H
$ A) N1 T I# H4 ^公式% n" F) I% B# G2 P* M% m1 X
; z4 P4 M v- @/ t18．For that case, using the same expansion for e as above, : |( Q& z) o+ W; b ?3 R. ` + `- o& o0 b. U0 W8 L7 X3 }# W公式 7 w( ?2 a ^# F. V7 T R2 N8 K1 @! r& M2 i3 m
19．Solving for t and equating it to the earlier expression for t, we get % d" a1 m; l0 X, n) `9 X* T% D0 C4 f* Z" y
公式 & V7 S1 S+ L3 S) i$ u4 A# ^+ h% K( j& [0 N" ]; S* F; K' z
20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is : ~9 z" q5 B: V! m+ x# B" k" }& W+ G- \# m! S$ D
公式 6 m; v- A; O; F. m; T: j8 z5 ?% p" l8 D) a J
As v=…, this equation becomes singular (单数的).# @/ f" w9 |2 ?
, n6 s. Y% f9 e- f3 A 9 C2 p2 ~! I: r0 d$ k, } $ q2 {; M: a5 H3 `. ?8 V$ }由语句得到公式/ k4 M4 C0 @) X. h
21．The revenue generated by the flight is , e$ e9 Y. Y: G, a* h: G+ P$ ~4 o; c, }* M) z+ R* f
公式 1 S) c$ K& h% [: ~2 C4 y2 B) P4 `! k3 l: \ V
7 X- ?2 _7 U* P8 C- K' A e/ S" `+ _/ ^$ N# H3 s7 K; K
24．Then we have+ O( H( c1 x' j
/ P5 L. V8 P' `' [& c( e" l$ a公式 ( N1 ~) s( x$ ^ . _$ E. v% i! |( Q* FWe differentiate the ideal-gas state equation ; L: T1 W1 {6 M 7 h* y# E+ B) ~公式5 \/ E! M* C: t# J( a" M
1 q& N1 a( W( X( i9 G, }
Getting $ G3 l& k E2 ~: y# I1 C0 N6 {+ q4 O% J5 g0 Q7 `- n' k
公式* m' k: \$ T# S1 S* t
! ~4 {9 a& {7 [' A% i; B25．We eliminate dT from the last two equations to get （排除因素得到） ) Y* p1 H* g9 ?; c+ z8 f5 f) p. v+ F* k" H, A( G
公式7 D3 D7 g3 R! K: I8 L+ q
/ H4 l, g! ]1 i8 a7 i J: \ 0 T+ |( k4 H( a& {
2 B9 b: ]. j& ~4 r
22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations3 p/ Y3 _! d. _2 F2 J! _
5 S, ^# t& n$ {+ _8 v% r公式 , Q* W. }( `" D- W) `4 @. |0 K ( m; M$ p. V. ~Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值） 0 ~: {1 {! ]0 @- k+ c1 P* {! ^% S: U6 b2 Q
公式 & s o8 N" ^% ]2 n# m" z————————————————3 [& y* {( F b! }
版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。 + C' A/ o1 D6 `/ K `原文链接：https://blog.csdn.net/u011692048/article/details/77474386 8 p# E3 i. S1 ?- ?# T* x8 P* k0 n
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