由假设得到公式 t: r; j# z |) r6 I; f+ |; }1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式) " k0 }6 n+ k; i/ T3 R 7 ]+ j+ I7 L, x# S% Z5 T公式! r; R0 s8 W$ r9 F. A
$ h9 ^6 X# F. A3 {4 [
Where2 f: t, |) F, ~3 r3 d: @9 _$ C
% P4 S! |2 Z/ `+ n) ?6 G符号解释 ; @: D X) k9 i3 a7 c& ]7 R 6 Z3 G7 ]6 f, Q nAccording to the assumptions, at every junction we have (由于假设) ) s/ I! ]; W* w. i' W: }8 G, E% @. s2 O1 z
公式 # a1 w/ X! }) P# A6 f9 r# C ( a6 o# H" `& V; J+ Q6 k由原因得到公式: K; D6 d2 k" w
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); " C) i. B5 p( \/ l( s( Y9 z 6 K' Q: s$ _- w8 f公式 & X& }* W" q e 5 O. U% ], S8 v. H$ j% CSince the fluid is incompressible(由于液体是不可压缩的), we have % Z' Z* R. N' V( _/ \/ r! |7 ~ P2 a1 T# d( H0 z8 {5 e7 c/ E
公式, ?% b: V6 ~) }
8 l" m: X, D- g1 e+ ]
Where$ x9 o/ r7 d0 D; x5 t
8 S0 a3 b$ K; L' H! x( f4 X1 s1 J" f$ R
公式 3 f& z) p, [3 e* W0 T2 S4 X; j1 D" ^/ `1 a# [
用原来的公式推出公式 " g+ n* J: S, [- O, \3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到); j# B5 `3 y) W0 m% ^
4 P: ?1 @% p- x1 b; J, \
公式 # C+ R4 c( Y. U/ X' N$ P" }% F9 o; o: g% e% L5 }1 @; \
11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields: & t; Y9 o, f+ p. R9 ~9 x6 ? - U2 ^2 B! S% X公式4 t! s" v+ v7 N7 q9 p% A7 A! T
- _ z7 D+ |, N! b12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)6 K* R3 z: o" v0 M
% T' U9 W8 z, I; ^ t* M公式 $ h) ?& u+ \" o- c " u: j9 ^6 s& }4 S7 M2 mPutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have 6 D' t. I @. H3 k) P7 V$ j4 m p/ V) a
公式1 {# b# X6 u% O O0 S
8 c7 p5 ^6 |5 b9 T
Putting these into (1) ,we get(把这些公式代入1中): z. ^. _( w8 h) _/ j
6 h1 s; u" [" c9 D8 j公式 + V/ I2 ?( c, j: I( D' P: z. s 6 x! b) X! D7 w$ P3 _6 k# NWhich means that the ( |( y" j, |0 q9 [" n2 B, z & v. o7 G& Y( P2 z: S0 ? eCommonly, h is about, [; y8 V7 ?: {- t
+ k9 k! [; e1 n; a$ ~% u
From these equations, (从这个公式中我们知道)we know that ……… . C9 s1 C- S8 a1 e" ]. B p W " m+ A1 U7 t, j 7 d, `. |' g2 ^
* E2 D& [6 c( d' f+ j$ e1 I
引出约束条件6 q y+ a: D) n4 T; z1 }
4.Using pressure and discharge data from Rain Bird 结果,# V* b; O% D) N4 Y
/ `2 R. s: X- F" H6 s$ eWe find the attenuation factor (得到衰减因子,常数,系数) to be1 T8 y6 k4 H! V
* f$ f8 @' M" [$ `
公式 , \( f. j4 s# F; k) O- a, g% |6 y1 K& D5 d
计算结果. w% p4 y' f: r% C0 T
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)( c0 ?7 `! k! e% d$ L+ ^% g
: v; c$ l3 M& S! |+ Q+ D8 f, n; k
公式. \, ^: u: j) V4 s \
5 u$ K) ^! U8 [8 ZWhere " C( _! a" D% U: V" a( S( M" ]( e, U0 R f2 q- ^
() is ;; 3 t! t/ F' f' a# C, _+ L! p' U" B
7.Solving for VN we obtain (公式的解) ! M( q( v( o8 o* z1 D + g# C, @* F8 B+ y& E; a公式 " e2 q' O8 q+ b- c/ A) \8 t; F" P, g( N. w" z
Where n is the ….. + f; W4 t' X0 k% w/ | 8 ?9 x4 N6 b) F! F* t+ i 9 K) j& Q4 p$ r* Q6 M8 Q( g3 W; F
4 J7 ]* \8 ?8 `, o: C- g$ R8.We have the following differential equations for speeds in the x- and y- directions: # ^# K3 _7 q/ w - w7 v$ o F w公式 ! V; @" B$ X8 r( H" K; v! U& ^* V: J5 ?7 W9 s0 I; i/ t5 ^
Whose solutions are (解)2 V( L" m2 k' l) ~ H% Z
1 s' X" U) S# v& m# t- t- H
公式 9 P; K2 Y6 }' t9 e% P9 u+ }2 T1 c7 v" @! R4 q3 m; A8 A0 V
9.We use the following initial conditions ( 使用初值 ) to determine the drag constant: ' _+ w# D+ B/ s6 J ( d9 m# C$ `3 z5 I7 A9 [# F公式 ! @9 r8 G5 q1 l- n% } 9 }% k' R7 L! }1 [8 V6 r7 {根据原有公式. ] Q- g2 r: f
10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is ; m0 J% ^1 y4 m$ t( f+ [ d; ?
公式 ' E- H) e. ?4 ]5 ^ " T5 h- T% r: t3 LThe decrease in potential energy is (势能的减少)( Z1 V; `6 [% w5 I r
; @9 Z: J0 Z/ ^& [2 F" o. B: W$ x公式 7 k' {' m8 Z2 M4 h# Q3 v . Q' U! Y& w$ t; nThe increase in kinetic energy is (动能的增加) 5 R- \4 u. p& _5 ]! y3 x! R4 f7 \' g2 ~1 a9 E
公式2 h( b: Z! L- S4 t
) F0 Z; n& |* E3 `3 n& x
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 a. G. M9 U4 ]' x. o/ x( a' k* m; E1 H% p8 N+ Y# M% |
Where a is the acceleration vector and m is mass ! { j2 q: e K. B# @( Q$ }4 v ~% w+ S0 U: N- F, e( X) [. @2 ^% O , c+ W6 l7 j8 C* c0 E* ^6 c# S
; ?' ^4 ~- E2 a( VUsing the Newton's Second Law, we have that F/m=a and : k% l c4 V. N- H4 Z+ ^3 d9 D 0 b' e' ^9 x, p+ n公式 , K. t$ F% g! e+ R- v/ C" Y. `! _$ \& M j( x
So that p4 N% \& Q0 [% M( g9 Y6 Q + f6 |- Z7 s' \4 H5 D公式: h& ]' U) e' V0 B4 g
/ t: J& R7 m, J" a7 k8 y4 JSetting the two expressions for t1/t2 equal and cross-multiplying gives4 f" ~7 c" z7 R3 N8 q0 H
2 Q) N) F8 ?' `& b
公式/ n0 k. w4 w# q! w; f, B
( v2 }6 `" `' b' P5 L
22.We approximate the binomial distribution of contenders with a normal distribution: . z, N; X: H7 `4 p, @" Z( O! r) @ % S5 B9 ]; r0 @( G! a公式 7 j4 {( M' x# ? 9 h& \& Q, w2 U3 x: L, S7 u0 G, _Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives & O6 E! v. `- j4 Q2 b- X # J2 B. t/ N0 M, Z6 }公式) ?% s2 J) e9 m! ~8 Z/ W
& n. O# a2 S0 H( h6 M% AAs an analytic approximation to . for k=1, we get B=c ; M1 @+ L) w6 [8 X , ^0 g0 S( o8 D" p; Q6 C0 u / G) h0 O5 s/ P1 {* A+ ^8 [( v
) |1 ^; [. [5 [8 _- X
26.Integrating, (使结合)we get PVT=constant, where 7 M' p0 I( y$ _3 A. i: T4 B3 J7 X) j& o# v
公式 4 |6 \# D4 e0 T0 n2 k' V( W9 `1 Q$ P $ X& P; p# k- V- U3 YThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so* P7 Y: @, F, O* G- Z! h" j3 U
# w/ P2 n. d6 t. S. g( l
! Q9 I7 g! X) \4 v4 U' F& B& Q 9 ]5 z4 ?& k8 ^, m9 X/ {7 e l23.According to First Law of Thermodynamics, we get! \1 G7 z1 a8 B7 B1 A
4 R& f& e! C6 ^7 S公式, H# f7 X5 N% v5 l4 z
4 W R G) Z4 a3 n* s9 z+ LWhere ( ) . we also then have; x) e" ]) |/ f! ]5 Q! x7 ? o
8 n5 b. J0 l* X" f, [. x0 q公式( O% G( p ~* H/ c+ j2 j& | g
& C- u' B( U2 _/ U
Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:6 q2 n' A* n/ q: X
* l* b: G2 E4 D' Y公式1 s" v( q" L5 F0 L1 V: a
5 {9 s1 i3 a, l8 {% t2 |7 y( d kWhere + A/ O$ z: t: I$ Z7 s, Q1 u) x. I$ b9 u4 h5 W% o. v3 x8 s! |
5 f2 h/ _, [7 R" G# O2 u6 \( ~" O5 T: y5 }: M1 B
对公式变形 8 b7 r/ t& }4 U* e: S& p13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到)' z4 B$ O! S' ]( n# u& }# z
! t% e# V) y# H+ _
公式 ! a/ I( y0 h. u8 I1 G) ~8 N$ S8 V" v) [+ h! W' s s; `0 R
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+ W+ J1 H2 w$ J7 ]# \% a
4 p( ]# R: P2 h公式+ {1 u: n6 R" v$ j
% c/ `# }/ J" L
使服从约束条件 f' j$ `0 E9 ~5 f
14.Subject to the constraint (使服从约束条件) % b! k5 B+ z, B" t6 q. J5 x* r" Y5 q5 [1 M, g! g8 v
公式2 d; R' E% w* T B
0 n% K$ h- W) Y9 E2 RWhere 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)2 {! W% O& l/ k
' X1 r% J4 u' Y( K" o$ a+ f m% a
公式 6 x9 v6 ]3 G7 ]' ^6 u4 Q( I $ u. ~3 C3 f: b* O" R- O; SAnd thus f depends only on h , the function f is minimized at (求最小值)/ r2 a# [' c& U, ~% w
; P6 d K- `" j
公式 ! u& F, Y: C, K: p! l ; W; f- T/ |2 G u' S4 NAt this value of h, the constraint reduces to - S! B l4 s, J/ T, w. [$ U) C 9 a6 y6 c. Z7 y& Y5 Z( e F公式5 e, |' g6 D' U' s, c3 b$ ?: x
$ u0 R* T. W% o2 v, m' F结果说明6 E9 d) E( U6 h
15.This implies(暗示) that the harmonic mean of l and w should be; \/ Y; k/ |- Z M: U9 Y
) `4 u. t; u4 p% l& D
公式& s" Z" b: g+ H; ]6 }, V
4 M0 o) w/ g @% b0 H8 T% x0 V" F
So , in the optimal situation. ………/ x- z/ S4 r Q
4 ~/ V9 [9 N P6 M8 N
5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is ) o j6 j( `' C4 e# w - }, k) J9 r# F" E$ R公式 , }2 @: L' I, }4 Y C1 O" v7 ^; U9 N1 f" f \; `9 f
16. We use a similar process to find the position of the droplet, resulting in 9 r5 p7 A3 F7 v Y& w: _( @+ E, m: v4 b; K+ y
公式 ! p, \. {, O/ F; Q. A& o5 U5 b5 x4 o) w) s( l2 q8 T! X/ A$ p
With t=0.0001 s, error from the approximation is virtually zero.9 ]: f8 S# ~/ n9 H" `, Z
2 l/ w. G+ T) w+ `9 |; D18.For that case, using the same expansion for e as above, 4 M& G/ S( M; V& w+ q2 e7 d) \* H+ X& h& Q! G' a5 e
公式 s+ r; P/ O$ ?1 g6 c/ {2 |$ S o" A! ^" A- \- k) L4 w
19.Solving for t and equating it to the earlier expression for t, we get2 c: L: D# n" H9 [. b3 ^
' a" }( w* J1 U/ @* }公式 % W' \4 q3 F8 Q 8 E+ x' v& t0 s) C* j20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is 3 |& B) n& D0 D" \8 r 4 V- k7 N0 `$ b+ O* p公式 " v, D, e+ u) j) b8 c$ z y8 Z/ i1 J
As v=…, this equation becomes singular (单数的). p5 m2 W8 J6 ~0 G
3 m# z9 O, \ m }( p7 f! H * o% L. J" R# T; I: q1 s
( o+ m/ _& d: v" [, \
由语句得到公式9 }3 o* ?7 _+ E/ B# K1 v
21.The revenue generated by the flight is5 R1 _6 ]) N4 z$ }2 ^( a1 t
?2 O' l" d/ W: z! a公式 6 |4 J6 \; A4 s/ G2 J4 W; S' C0 A, W- p! W @# N) I' L# q
8 P) B1 r& Z6 z( y+ L9 k3 t
4 E. D$ Q/ s$ h2 g" W9 b U N
24.Then we have W* Q7 D. i6 y; m- C0 _5 u& V5 U5 }: t/ D+ D6 s$ P- f
公式! m# I9 M' ]) ^6 p) i2 s$ m
! `6 m- t8 R2 B3 Y; j6 H7 k, ?% `We differentiate the ideal-gas state equation * j# F: |3 y% {; T: M/ f5 \' {3 r' m7 E/ d& I
公式% h6 Z/ F c9 L5 c. f
8 G/ H9 t/ y( B# z' L( T( h+ V9 vGetting0 N# A/ j9 x( _, W; C. ]1 `* b) P
# t% e9 R# Y l# c- Y# Y( W% r公式 / H9 Y4 W# x" D+ A$ D$ g, ?2 V) M" Q
25.We eliminate dT from the last two equations to get (排除因素得到) . j# A5 J# t2 |6 n1 }; _$ f % D* d" W9 F9 W; o" {* R9 Y公式$ j* u$ X* ^8 a% @2 n4 b
1 y5 |0 s# p* g; i, r$ K3 c- ]; ?
3 @$ |1 _* \3 }
$ _8 o9 j8 }. u! e
22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations g1 I- O I1 U' W* K* e
, c7 s& v) G' X! S: z; R
公式 , c* |# h5 g" [8 L $ D" z1 ]. c: {$ _Where P is the relative pressure. We must first find the speed v1 of water at our source: (找初值) p: y6 k6 Q3 h$ h# o+ @9 ?6 ]: O
" Z. w! u- W! R* i4 S$ {
公式8 ^6 {3 b$ l5 j8 s2 U( \4 e
————————————————( f1 W4 B; p4 T- r
版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。 7 r3 C" r0 l) \9 Y) v原文链接:https://blog.csdn.net/u011692048/article/details/77474386 1 v' i) z1 C6 I" U$ E6 c