: D3 N* N: u, a/ J% _5 J) E符号解释4 @: V7 j, G6 _2 q. h) v
0 Z$ p" w9 h& [0 e7 k4 x
According to the assumptions, at every junction we have (由于假设)6 c }) Z' M/ S8 f* w$ J( |4 l N `
* u0 v. A7 |: M3 p2 L' r公式" V) E. _; t+ P [. X" R+ l
+ V2 s' I1 t$ @9 i2 ~由原因得到公式2 I' m2 ]* Q: l$ X
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); 6 ]' d$ S! g+ ~4 m# ^, P ! Z& w, O( S6 C0 v) {* y1 A公式 + S4 Z+ i F4 l 0 n8 H3 h) q3 Q6 KSince the fluid is incompressible(由于液体是不可压缩的), we have8 v0 j5 _ e5 y& E6 h, t: W8 _6 A8 }
6 A" d( ~" t' k% l/ I( |
公式6 t& G9 r' [/ l9 Q% c' R7 w) ^7 u. ]
1 g# G! I2 q! H: t8 k$ H, y
Where, t. `# V) E \
# M6 w' R K' C# ]公式" H( [$ c. b6 o0 x4 ^
& c/ ]9 i5 h* @! d' \6 Z
用原来的公式推出公式( n+ R7 G2 v. K: }% J
3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到) 4 c4 Q( l: J, e5 u* h% H9 o* J1 J: z& V2 g6 g8 E
公式 : }3 C! C, U6 _+ k& U ; V+ T: z9 _# `11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields: 9 x, Y- ^5 C, X* H7 H2 n( T % k3 {6 R+ y) J2 P0 r3 v: D公式 " K, x+ f+ b$ y; y* Y) q% p0 q$ _ I u) a* l
12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得) , H4 k) g( _" \ : y2 D3 M% C% Y) `# I公式; W9 y; \8 S" t
: Q4 w! g$ z9 g' fPutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have % `( [4 B4 ]* T$ \# o& ^( c! b# U" O! j9 ]3 h
公式! s' U9 b' M* C' q; S
" \$ t% R' l! {7 c0 rPutting these into (1) ,we get(把这些公式代入1中) & M/ |/ T% |- q5 u& q. j3 ~7 ]7 Z6 j$ W( j! y" X
公式 ' K |6 w( O/ \, I2 ^+ U8 K @6 v' Z! i, g) D% \
Which means that the* d9 E- M, O6 I7 O; \& A
- J, w/ Q# C( Y, H, B9 R- e
Commonly, h is about9 ^3 h, h/ w( ?: ^2 V$ ]- j e7 b
( q4 Q' p: _# I: d2 V7 G9 z
From these equations, (从这个公式中我们知道)we know that ………% \- D" g' \+ M! U
$ d0 M" z0 K& k% |' ]引出约束条件- V3 X# R" ^/ }4 ~
4.Using pressure and discharge data from Rain Bird 结果, . j- I7 F4 {' S& r+ { c2 ?; c, `. G4 J( D
We find the attenuation factor (得到衰减因子,常数,系数) to be # h# ~' M. E. _7 P. _3 `0 K8 \6 [, W& ^' o- R( Y3 |5 E
公式 8 V& Z/ C( k! L- `$ m 4 |" \4 S9 J/ G1 V& R; W4 f计算结果5 I) [2 r, V* ]
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)+ Z u0 X* `' C
, ^$ o; f- r7 T, L+ e Z公式 ' C$ I3 h, x/ |% a9 @ 2 B! v @$ }7 z& t1 z; a, K4 V: j }2 m6 ZWhere # g2 S0 ^ S* t! W" B: H/ e2 l7 w; h1 N8 t8 E$ V; L& y* A/ P$ `3 B
() is ;; ! Z7 `# j3 X; O% J: Z7 b X; z W) U7 N5 u, K- I; `8 k7.Solving for VN we obtain (公式的解)6 t( i/ j( e, G9 x7 `! r
" A# \6 D# Y# ~公式 8 j5 P( G8 }! U( I9 s " `$ r! t, @" L* T/ `Where n is the ….. 9 B U3 w; [- |% A# M6 N2 ?- a2 C " ^& X( | F/ j# M7 z! ~/ O : ?: V' B, M7 \' X 7 |4 X" Q; z. U9 I# w1 ?8.We have the following differential equations for speeds in the x- and y- directions:( o; W1 V' s* p, h( | y
) v5 s+ o& E' D" h公式' r4 I c* i, G0 l: ^2 \. R
1 }( f2 `& w9 pWhose solutions are (解)1 Z" {( d1 _# l$ @+ T
5 b" Z0 ^) x: ?7 K
公式 0 L1 q+ {% R8 N3 o1 g# z& B0 z ) F1 j$ N% Y& u2 K, j9.We use the following initial conditions ( 使用初值 ) to determine the drag constant:, w5 j9 s7 g3 C, t5 W' H/ D
6 ]2 w+ A! w" v8 E3 e0 g Z% D
公式 B, p# \" @0 }; \
2 S1 m- b; ]6 `& n2 x8 C: V( ~1 |
根据原有公式 0 E D/ L V/ @10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is7 W% I, x( p( e; q, _
$ N- V) d* O0 z0 z6 _9 l9 a
公式 6 E; W# C7 e0 m7 @ : p; \( r! w/ ^* k9 k: {6 AThe decrease in potential energy is (势能的减少)& r h3 H. Q' Q3 j
) R/ D& y4 u, a! z I公式 : _" ^3 H9 `. O3 s* F1 r. L2 d- ^3 S9 i* d5 r) i Q7 O
The increase in kinetic energy is (动能的增加)& Q" V/ F M4 |5 R' I3 v. Z
6 Q, u. J* j+ v7 o. u0 g @公式3 U7 c" p" ^; w4 R- @& y0 E
2 O9 u3 ^( A/ `2 z& }4 @
Drug acts directly against velocity, so the acceleration vector from drag can be found Newton's law F=ma as : (牛顿第二定律) / y. [% h$ \" V7 h1 @2 c $ c# z+ w0 m$ L# G$ F) Y* tWhere a is the acceleration vector and m is mass 1 m; {& s' g S0 u$ X C , u, x, {( W+ }/ _0 z) S* l3 j ) y' e# \6 T" U5 t$ D( `
. D1 {) t3 s. q' E& p2 XUsing the Newton's Second Law, we have that F/m=a and. E: R' m5 ^+ i$ t9 F% u
! A7 t: H [8 w8 x- `& D5 ^公式 1 m0 X$ V7 y" C2 v# Z) s ' l2 x, }6 r: g- ~( }0 x5 JSo that$ u# b- y) k4 ~. V2 _; ?
; I# ~# m2 e4 I6 Z$ K E- v. S# y
公式 " F/ u5 d# f9 @" k; j. r- l( ]0 n1 H
Setting the two expressions for t1/t2 equal and cross-multiplying gives7 x8 t: G& ?5 T' p* U7 ]
5 N7 \) P+ k: r$ E# m# N公式& R- \! `1 k* ^' X
3 Q5 q+ P" f" p0 X22.We approximate the binomial distribution of contenders with a normal distribution: % n$ o: S7 U0 [6 L' H: B% }/ K: d! L: |- n
公式& T3 k1 T3 ]) J1 ?/ L+ H2 ~
# y" B) M+ a) B- u) rWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives' d c/ a$ n r
* j$ d. ~; o! D. a5 {公式 + Z+ }% `* C6 E& z" B; ^8 k# J9 s i+ T& a9 }
As an analytic approximation to . for k=1, we get B=c$ Z) R, g% [/ f* J5 T+ @
d# ?$ d9 r* }8 u1 g
$ S( s- i) y! U" R* n( R1 y! }. ~3 x% B7 K
26.Integrating, (使结合)we get PVT=constant, where ) X m* X9 Z* g7 D$ c 9 B2 S: D2 T1 i, n# E公式% S4 d8 Z7 y9 t& |- Y* }: V/ z
- ?3 ^% ^; m6 M+ x
The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so: U8 P4 z; l2 h. Q) y0 I9 z2 }
8 c: F; m; v# v! K 7 i# t4 h6 T% f i) Q # H1 V3 e/ {8 \) F9 ~23.According to First Law of Thermodynamics, we get# I7 F" O Z$ L. S- {4 A0 G2 l
* w2 _ L2 @% }" {
公式3 E I! w6 z' Z. N) h
2 g4 }* u9 k8 B3 D9 IWhere ( ) . we also then have4 | Q5 v! P; O5 _
6 Y& k) t: V9 q9 {) V) l- w公式. a/ X' x* U. l9 g4 O1 J6 l
0 }+ h# U4 v9 eWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:1 N7 s1 h, W7 V
8 m1 G8 h; m9 A- P
公式 0 a% P) {% J" h9 M4 H3 A& a : p% a6 |! w1 h" R( nWhere k/ N. z; o' B& F( H
1 m/ H% r3 J! |8 C0 m 5 V. `! C- u v3 L2 D1 w
# T+ ], i+ G. r' W$ d' f3 U6 @
对公式变形6 L7 F/ q( e |; w
13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到)+ i& b5 X' z; @- Y/ ]! B; U
1 t% z& @" ]5 E, _' V# }: w公式0 N) o; _2 T$ l
7 U8 j4 W5 `9 A: T8 o
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 - k0 t0 `" S- P/ o7 j; ~: a + z+ Y9 K2 y. `( b公式7 e; \/ b8 m9 h+ _% ^5 O
1 {: C4 e" ]1 V3 s2 G5 k9 y1 A6 u( j使服从约束条件2 h9 O; {0 @- H( ]4 _ G- E
14.Subject to the constraint (使服从约束条件) , F0 l f+ o& \0 z: v) A. X 7 ^# F7 t5 s0 O5 u公式/ x7 t. Q! [; X" C
, x. o Z4 ]+ T- Z% ^
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)$ f7 G9 R4 i0 b0 N
# p* o9 {. ]8 R( p- p公式9 {$ F2 T/ E) w) G* @& ^
( P$ r: F0 N- \8 h, q3 w
And thus f depends only on h , the function f is minimized at (求最小值)7 \/ ~; ?) @% K1 z" F& \
. n- `- T1 m S* t/ g
公式 6 m. C( V& K! S8 h& K+ b/ B! @/ h7 {" i9 O8 S/ P; p9 Z" T
At this value of h, the constraint reduces to0 m) ]( V! C X/ Q
( B+ a {' q4 u2 [+ P公式% }5 J) @# E, S: n- M2 n
0 V: _3 Y$ z# [1 c% [结果说明 & \0 \" M/ G* g. e; Q15.This implies(暗示) that the harmonic mean of l and w should be& { q$ O- ]3 Q2 ?' f: {6 l4 n, [
2 E2 p3 h s' K' W. y1 K
公式! X' E" ?3 \; P) C/ [2 o m
3 V. _ G5 U- l" vSo , in the optimal situation. ………8 D" j" t& ~2 ~( B0 N
$ C0 G$ N, Y' W5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is : o! m" q" P$ _* I3 E9 k7 w 3 x. O( H: B: M3 G. D+ \* g6 w公式5 A2 j/ L5 l& F
U$ \' l1 d: h0 Q) k' L% f- k: H
16. We use a similar process to find the position of the droplet, resulting in ' F% s4 a* A7 w$ j/ H* e' l4 _' y$ b4 f3 v
公式 % o m; U v( Y1 W7 M ( |0 c+ N6 [; d2 hWith t=0.0001 s, error from the approximation is virtually zero.7 Y; D8 n( E2 L Q# ], F6 A! c+ L* W
8 C8 `+ }3 H$ d & k' g4 j% r4 n/ V+ w0 D8 x( E0 a8 h( E5 P; l& M% Y
17.We calculated its trajectory(轨道) using& k, o# b0 G& x" m$ H
; ]; g9 h Q" b5 b公式0 o2 }# t/ S' K1 S2 H( Z/ R
1 _6 F6 B$ ?, \; D& V18.For that case, using the same expansion for e as above, , Y; y H! i5 c, m0 R$ m1 l % Z. {! y" z1 s公式+ I! D: x/ E+ r/ m4 Q0 ~9 e
5 N- g6 t- O9 {- E# f6 d+ N19.Solving for t and equating it to the earlier expression for t, we get : Z3 K, x4 C7 R! n- g& |& `. R3 W9 l; _1 |* {$ y$ ^) P2 F2 I1 W
公式2 G8 @& E+ z, }5 |! l
* w D0 `' T0 j: h) v
20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is * C: S8 P5 k: X# N- h7 |) K# A) x6 c9 w% v/ ^
公式 # q9 T) u! p) a0 s: [8 r2 X' r
As v=…, this equation becomes singular (单数的).( Y$ B, e7 a Q2 j6 Q& L8 i
3 d$ K0 p' {0 R! M9 A% C
9 n9 ]% O: p' G Q' b
9 J+ J' \! d% `* R2 h4 j1 N, l& O由语句得到公式 2 t' ]# Y; g7 K21.The revenue generated by the flight is 0 H# X" b* _7 R + q4 e# h; F+ b% }2 v% L. J公式 ! v, q4 X9 q' m: k3 M+ H2 l5 C8 H" \
6 J4 A6 i. U5 f; K& f' R. w4 W) Y % S0 t7 G3 z! q: y8 p N24.Then we have - y. l9 b6 d# e6 V+ X; c 1 R( C" a' v& P公式 ; P) @% ?' J" N6 z% H4 K, ~1 C# N& z# G# _# ?
We differentiate the ideal-gas state equation & \4 e( S5 z0 C" v - V# ?3 v0 f% Z- s, b公式( m) X3 k% j# K& Q$ z. P6 k
0 A8 ~/ m; o0 w' D9 b# P4 [
Getting4 ^5 J1 T! i, n# m3 X
9 F. m# e* v$ n' z- |8 o0 _公式" |4 @) x1 G3 o" E8 u0 V- H
% M* k; w! l: {: M* V1 c
25.We eliminate dT from the last two equations to get (排除因素得到)0 ?: s! ?% ~9 x0 @# Q$ _) p
- ^+ H3 [- J9 o/ _ c7 B# z公式 7 g2 t$ F' a% D1 ^$ ] * D O/ n! ~/ k . @6 }0 X/ t; v) ~: P5 i1 N4 T
22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations 2 U0 E! p7 k/ K0 S4 n$ F- S+ a7 Q2 q2 P
公式* N: {2 \2 ?0 }0 t) Z
, ^8 |: F, k! I+ D t v2 L9 q9 oWhere P is the relative pressure. We must first find the speed v1 of water at our source: (找初值) ) L$ v8 M8 s3 X2 g6 L/ I3 A, ^4 f; ]; m' G/ _9 r
公式 r. R7 B" M0 O$ Y———————————————— 7 Z. U; r2 f7 r/ c% e版权声明:本文为CSDN博主「闪闪亮亮」的原创文章。 8 ]7 Z- O1 g. S# W; L% M9 V0 J: E原文链接:https://blog.csdn.net/u011692048/article/details/77474386 8 t' d. E+ j0 u( Y