由假设得到公式 * B1 a' \/ X- ]6 J f8 z% r1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）/ \4 E x" U% x( X% `) M
# f: _; G u4 K0 h: c, o
公式4 _: v9 \) J* ]: L9 L
% B/ U _/ j5 v+ g0 n, I
Where % P& f# H" J9 t8 R' l5 u" ~* E0 k4 F8 B7 V) t. E
符号解释7 ~+ }+ f( ?; \* ^7 C
! k# E0 P+ G' K$ N, e; P3 o( NAccording to the assumptions, at every junction we have （由于假设）: K7 J3 d3 U# I+ e
5 t% y% g p" J# N( g, _
公式, k8 }; ?" x3 q; T( [
$ L9 K1 } U' v2 |# l
由原因得到公式 6 x8 Y5 i( Z0 E2．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）； * W6 j3 s- g+ A8 q% t/ f4 P& e - Q" C9 Y! I/ d9 ]公式( r3 Y- {! V' ]1 I% v% ^' s
; Q3 H0 D* [: Q7 D$ ]+ Y8 I9 K2 U1 zSince the fluid is incompressible(由于液体是不可压缩的), we have 7 z1 z( i8 S+ F: a% J( R 3 \( D0 D9 }# g% d+ _公式 & b- M! g3 R8 Z- R X( y. R2 Q5 H1 H
Where! \ k; z3 x9 h- r: ^
# k$ Z4 V& s% D" ~ b公式 8 y' Y: D' ~& k0 l5 ]/ J/ ] 1 q/ F W) q4 r& J, E用原来的公式推出公式. a! d0 Z. F0 U# z7 P5 K
3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）$ ^0 C; P: _9 L0 s
1 a5 i, X+ v p1 Q( a/ b! p- J
公式' L4 X7 h; Y; U4 G0 Z0 c& F
: r5 A; n' \/ i- v! Y11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields： . | k' T- V; [9 } / n7 P7 p' i1 q公式 ' n- x( G" a) P0 T6 ^ + K5 G! Q3 |* b+ y3 ]12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）+ R4 m9 C, K+ _1 k. ^
c0 b/ b7 l5 o6 S1 ?3 X1 G+ S公式 & i) e9 ~: x9 ~4 l* K {+ @. c n% C7 y
Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have $ O3 Q' h8 d( }4 ~6 ]6 s0 T, A- Z$ v% f# X- c3 I0 \
公式6 \/ E& Q* t/ h8 ^/ ^
: g, D) b! |3 b
Putting these into (1) ,we get（把这些公式代入1中）: Z6 E. A/ E) L: _3 a7 C
7 o9 O7 j+ u# L公式7 [$ Z, O$ L( @* |0 V
6 b1 u% c s* ^) {+ U9 T9 z5 oWhich means that the / t1 d# t* h1 f3 L1 U ' W y a8 k8 l2 wCommonly, h is about' S7 u; u7 g n* ~+ {+ S
# J- }, C. K; J) v7 ^/ s6 VFrom these equations, （从这个公式中我们知道）we know that ………0 c2 s) D! y4 D: F7 n
* Y, ^2 U( v0 L 2 q1 Z$ ~" d% Y4 ?+ L2 d! y( g & g+ { g" ?8 {6 n$ f7 V; b$ A引出约束条件, ] C: H+ i0 h& _, O
4．Using pressure and discharge data from Rain Bird 结果," B( u6 C5 J% T/ z6 ?
6 g! v; X" b9 E3 z9 ^ R
We find the attenuation factor （得到衰减因子，常数，系数） to be L* `7 G5 m, f) \% F$ m7 M- H0 L- t: j9 }* f Y
公式 4 s* X- e+ C) ~5 P7 C! L9 L* N+ h! I$ m+ ^/ K
计算结果* F' d6 E8 X; ~# y$ x' k) c4 }0 v k
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） ; H3 @/ K4 J F# Z2 S- }" o+ Y6 j2 B
公式 + P4 \3 D; E1 r; y. h* y) {4 S% C/ w8 t. ~+ L
Where * M' v# }5 n* X4 r+ A 3 g0 G8 J- u O6 W8 x0 e- }() is ;;7 \ k7 } | C+ n/ _3 @
3 X5 b6 [3 T4 |- u4 J& R1 J7．Solving for VN we obtain (公式的解)$ P6 m+ O3 y, b, I' Q' x; w9 ?
% k8 w0 e) [- ?( z
公式0 ^0 X+ [ _- s& A) L# V/ O0 X7 c n
0 t# z: k% ]1 U5 V+ o
Where n is the ….. 2 v% ]" h) `% |% x2 H ' F( Q0 u8 A5 k( M/ \ ! @7 q" k8 B3 o
' Q0 c }. o i m4 d, d
8．We have the following differential equations for speeds in the x- and y- directions:$ R% C2 t) T6 q8 F7 t# }6 A7 [" @: I
9 q6 o6 ^6 O$ Z( D9．We use the following initial conditions ( 使用初值 ) to determine the drag constant: - b. w) \1 t6 A; D 1 I# K+ t4 d4 N( j4 Y7 Y0 e; o+ D `公式. B6 }5 j9 W7 T6 b/ ], Q
/ s" ^, c8 Q8 b, U根据原有公式' M3 Z* A$ h$ Z9 V
10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is) s0 v% s' y8 r
9 V' O* ] J- F3 Y2 q4 O公式. ]$ f0 L1 ?1 [ l9 k' s
! \, q3 @1 J2 R" W4 Z% A' y* hThe decrease in potential energy is （势能的减少） # N9 m# ]9 M+ G g, e' Z/ e5 j4 a$ \; V9 R% O公式3 Q$ y+ j0 \1 A
, v( x' D9 ?7 iThe increase in kinetic energy is (动能的增加)- m2 b; W! u$ k3 Q
5 _+ N, j8 J8 x. O3 U
公式 , D7 X. [$ N3 P3 ?2 o+ d K l" f& m
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) $ X8 \7 \$ n6 Y- O; h0 w, R+ l1 X$ H& `- A) Z
Where a is the acceleration vector and m is mass6 o x1 N* V* G9 w
% T/ F- X3 I) s3 ]
) @) S; @" a* v/ q- F' f, m% C0 p3 @: M 5 j4 J/ h' V- @' v5 \+ dUsing the Newton's Second Law, we have that F/m=a and8 L: Q u' W4 ]
2 @7 x1 b" `7 S8 ?
公式 . U) O* V2 G* k! V. e3 |, e! [+ p# k* |/ v$ A
So that : h' G/ f& j7 |7 O7 C/ D. @/ G7 D! P! F [ z( u
公式( V; H( t P" u3 p9 P
8 g7 x" k& _0 g% I3 uSetting the two expressions for t1/t2 equal and cross-multiplying gives4 P/ |4 W2 @( h' U% T( m; c6 B- b
2 Y* v, l n1 j0 k5 e6 l3 }) t
公式9 W, Q/ U8 L1 n) U* ^
+ l0 l' Z+ u/ G" v. X$ h7 \0 y( ?
22．We approximate the binomial distribution of contenders with a normal distribution: d2 [! a' k4 T' P' ?! C& X0 j. q& N* S. i$ C% Z
公式& K& r. _# Q1 s0 e; |
: E( Z& V' f' R" Q
Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives2 m% n! ?( l/ U6 s
o% \- }% T, {, e公式 ; d0 J/ e4 l5 k% W* q 9 F7 H: F- I- FAs an analytic approximation to . for k=1, we get B=c1 \- ~! c9 T9 U* U7 }6 S
- h+ e% K5 D! N
* E' j7 l! U+ ]8 e7 t: X: ~5 o3 \% J/ }1 J( M: d+ }
26．Integrating, （使结合）we get PVT=constant, where! Z' V6 f' l8 t% F4 t. |
( P: `, i2 c, Q) I3 y4 E0 d公式2 |: |3 S3 M2 l1 G. _. y
0 j7 _0 e. J. M4 bThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so% h1 F6 k/ y, p! j" C/ m0 i
9 k9 p5 g. I0 p2 P5 T$ h/ n $ N) D3 k; j* ~$ d: D# o5 L9 r: j% r* ~, }0 r W2 _
23．According to First Law of Thermodynamics, we get 3 U( a/ |1 L) d. g G' q/ c `. R6 {" q0 E
公式 7 c7 }* |* V9 e4 [# [$ { 5 g. t/ g* X3 w7 _" b" GWhere ( ) . we also then have, E: `7 c0 G- m) Y! A i0 c
* U; i0 D/ `7 c3 D) ]* DWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula: 3 `! S* t- h ?& {# A$ V6 @# u! ?$ D/ \. V3 n* |1 s! |
公式 0 y2 s9 V# P- a( M ( k8 { l0 _& Y* ~+ ~. ]Where " X; i" s' }) S. P5 L- [- u# d3 h 7 i0 J% F2 C' w- e" {2 r3 P # ^4 z$ U, l$ [/ U# u- [ 9 `, u$ t1 i- V4 n; Z$ w对公式变形- H, _. a, e8 s, c: I. k* ~
13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）, @- W: |# b6 {! j9 d; M; N1 t
% \, V0 J" M& Q4 v' r+ j% u
公式 7 ~- R' I( z6 H d2 v, `: S5 O' d- a$ j- T. A+ d3 W1 X8 f( @
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 minimize4 m, T6 q% N8 O, T; g+ h3 H1 x
9 q# R5 _% O5 v% X) u
公式 * ~9 w6 @9 j1 t/ q; \' E" E& T) q/ r
使服从约束条件 $ d2 R& j0 z$ x# N8 r14．Subject to the constraint （使服从约束条件） 4 r1 z& g7 Q3 P5 l$ u0 \5 Q) v5 t9 t' I1 N- x
公式: w, ^/ f0 s1 ]; \: W
- s7 s) w7 l7 u% [& v, n5 i. |7 C0 D1 tWhere 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）$ h0 i2 D9 N: `! m/ v
+ t( |/ W3 C% `3 s! ?7 k
公式, i9 R9 c5 \' n$ [9 K6 O4 Z
/ t$ z( k$ K9 F: e; a2 M# U
And thus f depends only on h , the function f is minimized at （求最小值）2 G% I: d. p5 [0 j* x
/ x1 m: w+ a N5 Z0 n! W$ V公式 0 j5 ^, E* H3 u% ] & t& ]- `* Q/ DAt this value of h, the constraint reduces to / _# H; ?# U( i% ~' |: r$ C; y/ J! U% V3 g
公式) w/ q: n1 @# a5 f7 s
' J& U3 Q0 X4 ~8 k- t结果说明 ! a+ }' K, n( b& R15．This implies（暗示） that the harmonic mean of l and w should be ' B3 v. B& f5 T* m- q t0 k: `, x$ v2 @) `: V* r
公式4 O, g# s/ c* V( @
% g' J5 K1 d0 s. t7 F$ CSo , in the optimal situation. ………% L1 p% E/ h4 {. S
3 R3 N% f$ S" i) c7 J& {
5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is/ `+ ^: A1 V+ J8 B
r; o1 v! c7 U4 _5 {
公式* w8 v) A! V+ P
5 F8 ]" q/ A/ X5 [' x+ k; g K16． We use a similar process to find the position of the droplet, resulting in0 Q; y b9 q7 a9 n
/ `( p8 c# F3 R8 k: V0 g6 K: m/ N公式& }7 g$ U& f3 g
) @, }) y( |- |) q% I
With t=0.0001 s, error from the approximation is virtually zero.6 Y ^2 n/ `+ [. R/ I
- Z# z8 T# l" L$ u! L3 G/ T$ R 2 c/ R7 y5 f' C5 R Y2 M" Z
( `( C; U1 @" l2 {19．Solving for t and equating it to the earlier expression for t, we get2 [: {) B \9 W$ M9 \
! l+ ~# r( c- N% ~9 p& P
公式 3 t; ~' @# W: f$ ]- T `; o0 A( D; {' r8 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. @4 Y4 p4 t) J( x* n3 m
" h- D) [- m8 W: A$ Y. I- J" H; x公式 4 H6 a% F8 V( u2 E2 I, c" g. u- i; [8 C; Z$ y' O# R
As v=…, this equation becomes singular (单数的).. f+ {( N' O* i7 c
6 ~# n0 @0 a9 K" u( x 5 W/ N1 D# K v" T }4 c% N- z( D% Z * N! T& B6 d, O, U4 |3 F1 N由语句得到公式* A+ _$ a) j% g
21．The revenue generated by the flight is . e6 r# Q. s$ }. J/ Z# ?* j3 M" c, ?; I, n; l. m
公式 - `; S) p, `/ h/ d! t' w! Y" `9 z( v6 g, h
: ^5 A9 e; z! g/ A: s+ R0 o( }# t& a/ A+ b N
24．Then we have+ h4 e. g! I5 u' G4 a# m- V
6 c+ K0 u N4 h) P0 h' e- C公式( O: |, l! O+ ~9 ?! q& P% e
: T! E* v* ` X, Q! Z( \We differentiate the ideal-gas state equation 8 u; @/ T0 C$ o8 c& T- j i% r' T6 Z) [0 L# z! x& A) o1 x
公式 % g2 J$ T9 J+ S 1 q- ]8 E% I; g8 d! x+ ?; _* A% nGetting * n8 t7 z6 G2 h2 @4 B# m! e3 Q) d# ?9 r6 e5 U" G, }
公式 % L( x. M) b8 e: {7 R: i" Q0 m. x6 c5 V8 ~ V
25．We eliminate dT from the last two equations to get （排除因素得到） % f4 ]: r) \! h! I' }- Y ^3 p8 O: c+ n! U3 D$ {
公式 ' S% Q' g( Y' o$ w1 ~8 B# m; M; D& p6 {
2 X$ N M3 z3 B/ ^! D6 i& \& ?. ]5 ?. U* [. P
22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations. ^: K: t" ?( T+ C4 d7 z4 C
% M! W6 [: f6 K+ B. p
公式7 Z l7 v* q# P/ Z; V9 W$ @$ v3 T
+ h5 m0 B3 p% s! ]1 N% [Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）( Y5 g7 j- E& h! Z
) [% j- l0 A, ?" Q# H
公式0 j* v- q$ u9 a+ i+ p2 y9 k* T4 e
————————————————) k, u! J( Z$ P+ w/ a
版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。4 O3 c* ^6 {0 }* r3 G
原文链接：https://blog.csdn.net/u011692048/article/details/77474386( Q& @* ?. Y9 q p3 W( f! s
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