Microsoft PowerPoint - 粉体特論2018 [互換モード]
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1 8
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6 4 u[m/s] Q ( /)[m] / [s] / s] u[m/s] A[m ] [m Q π N q N q Q c / s] [m N qc N qc Q c Q + Q i t c V Q c
7 44 τa ma Re 4 / ) ( u u t u V ma ) ( ) ( u u A 4 Re m ma Fr V m ) ( Fr 4
8 45
9 Fr u / L u L Fr 46
10 47 [ W ] Rf [N] u[m/s] R f u [W] [J] [s] [N] [m] [s] [W] [N] [m/s] R f CD A u Au ( ) 4 R f u ( ) N p N p 5 N p
11 48 [s] [m] [N] [J] ] [ t s F W W [J/s] [W] T F r r F ) ( ) ( T T ω [ra/s] T [N m] ω r F F r ω
12 t / 49 t t [s] s t [s] t t t / ω π [ra/s] [ra] t [s] / ω π ω θ/t r ω F ω r F kw=.4 H =.6 S H=Horse ower S=fere Starke
13 [N] 50 R C A u C( br) ( r) r=α b=β (0 α, β /) R C( 4 ) () r r b [N m] D T rr () () () T 5 / 0 C T 5 / T C 0 N N 5 5
14 5 K K, α, β, γ, δ, ε [W]=[J s ]=[(N m) s ]=[{(k m s ) m} s ]=[k m s ] K α β ρ γ μ δ ε =[ML T ] [ ] [s ] α =[T ] α [m] β =[L] β [k m ] γ =[ML ] γ [a s] δ =[(N m ) s] δ =[{(k m s ) m } s] δ =[k m s ] δ =[ML T ] δ [m s ] ε =[LT ] ε M L T
15 M 5 δ ε γ α β 5
16 5 K 5 ) )( )( ( 5 K K 5 D Fr E KRe N D, E
17 54 τa u Au ( )( ) R f u f ( ) p R u N Rf CDA u Au ( ) 4
18 55 A B T 5 A, B T 5 A B N A Re B N N A Re B
19 56 N N Fr a lo Re b
20 φ 57 0
21 N 58 p 0.66 {0.5 ( b / D)} A 0.Re H. B (si ) 0.66 Re 0.Re D A B p 4 ( b / D)[670{( 0 [. 4{( b / D) 0.5} / D).4( / D)] 0.6} 85]. 4( b / D).5{( / D) 0.5} 7( b / D) 4 p ( p b) p = b ={( p b)/} N N
22 59 Zwitteri 958
23 60 Ottes&De Jo 97 [#/(m s)]
24 6 Gah&Mersma 995 Rittier
25 6 T. N. Zwieteri, Suspei of soli particles i liqui by aitators, Chemical Eieeri Sciece, 8, 44-5 (958)
26 6 To ascertai whether the solis were completely suspee, the bottom of the vessel was observe; this coul be oe easily by illumiati the cotets a looki throuh a iclie mirror which was place uereath the vessel. At the borer betwee icomplete a complete suspesio there are particles which settle temporarily at the bottom a remai for a short time i a fixe positio relative to each other. Whe such a small pile remaie at rest loer tha or secos before bei broke up the suspesio was jue icomplete. Whe o eposits remaie o the bottom for more tha sec the suspesio was cosiere complete.
27 cm m μm μm μm
28 N JS N JS K T a D b c p X ( T / C cost.) T D C p X N JS, T, C, D, p D T C ( Fr) ( Re) JS JS JS N z N z N z z [m] T, D, p ν [m /s] μ/ρ N JS z ( Fr Re ) N JS z ( Fr Re ) NJS z
29 N JS z 66 i) α β k i α β 0 (α, β) (, ), (4, ), k i, 6, 9, p p (α, β) (p, p) p p ( ) Fr Fr Re p Re Fr Re p z p JS JS JS p N N N Fr Re N JS
30 ii) α β 0 α β k i (α, β) (, ), (, 4), 67 k i, 6, 9, q q (α, β) ( q, q) ( Fr Re ) q Re q Re q Fr Fr q q JS q N z z z Re Fr z
31 68 z T, C, D, p T C D p N JS c b a X D T K N p JS ) ( p JS e b a X D D T K N e c e c c e a Zwieteri c=0.0, =0., e= e= ) ( (0.0) e c e c
32 69 N JS 0.0 S L L 0.45 N JS K K K 0.0 p X 0. N 0.0 JS D { ( 0.85 S L ) L } 0.45 K T D a ( S) N JS S p ( Δ D 0.85 L ) 0.45 X 0. Zwieteri
33 0.0 p X 0. N 0.0 JS D { ( 0.85 S L ) L } 0.45 K T D a ( S) S ( T / D) a a=0.8 (propeller),. (isc turbie) T. N. Zwieteri, Suspei of soli particles i liqui by aitators, Chem. E. Sci., 8, 44-5 (958)
34 a S ( T / D) a=0.8 (propeller),. (isc turbie) 7 N JS SD 0.85 T a D (0.85 a) T N JS D (0.85 a ) ( T cost.) JS N JS D 5 D (0.85 a) D 5 D.45 a JS D 0.0 (propeller) JS D.45 (isc turbie)
35 7 N JS SD JS NpNJS D Np SD ) ( D N S D 5 p JS /( JS ) propeller NpS D /( NpS D ) propeller Table Relative JS at optimum eometry Impeller S N p D/T C /T N p S D.45 (N p S D.45 )/(N p S D.45 )propeller propeller / / vae isc / / isc turbie / / pale( blaes, /D). 5.9 / / S N p
36 7 / b /b D /D H /H Re Re Re (ρ =ρ, μ =μ ) 5 5 ) ( ) (
37 Fr 74 Fr Fr / ( ( ) ) / / 7 / 7 / 7 / 7 / (ρ =ρ ) ρ, μ Re Fr /V ( ) V ( ) ( / 4) D H ( / 4) D V H (/V) / ={π(d /) H }/{π(d /) H }=(D /D ) (H /H )
38 75 D /D = / H /H = / N p Re N 5 5 N / =(ρ 5 )/(ρ 5 ) / =( / ) (ρ =ρ ) ( / ) =( 5 )/( 5 ) ( / ) =( / ) /
39 76 u u π π u ) ( ) ( (ρ =ρ )
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Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e
7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z
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(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
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1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2
2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6
微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.
微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
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2007 I II III 1, 2, 3, 4, 5, 6, 7 5 10 19 (!) 1938 70 21? 1 1 2 1 2 2 1! 4, 5 1? 50 1 2 1 1 2 2 1?? 2 1 1, 2 1, 2 1, 2, 3,... 3 1, 2 1, 3? 2 1 3 1 2 1 1, 2 2, 3? 2 1 3 2 3 2 k,l m, n k,l m, n kn > ml...?
) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)
4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
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127 3 II 3.1 3.1.1 Φ(t) ϕ em = dφ dt (3.1) B( r) Φ = { B( r) n( r)}ds (3.2) S S n( r) Φ 128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds
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QMII_10.dvi
65 1 1.1 1.1.1 1.1 H H () = E (), (1.1) H ν () = E ν () ν (). (1.) () () = δ, (1.3) μ () ν () = δ(μ ν). (1.4) E E ν () E () H 1.1: H α(t) = c (t) () + dνc ν (t) ν (), (1.5) H () () + dν ν () ν () = 1 (1.6)
t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z
I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)
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2001 3 5 COD 1 1.1 u d2 u + ku =0 (1) dt2 u = a exp(pt) (2) p = ± k (3) k>0k = ω 2 exp(±iωt) (4) k
18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb
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r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)
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II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2
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II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x
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Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν k ω/c k x ωt ω k α c, k k x ωt η αβ k α x β diag + ++ x β ct, x O O x O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k
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6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2
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1 6 6.1 (??) (P = ρ rad /3) ρ rad T 4 d(ρv ) + PdV = 0 (6.1) dρ rad ρ rad + 4 da a = 0 (6.2) dt T + da a = 0 T 1 a (6.3) ( ) n ρ m = n (m + 12 ) m v2 = n (m + 32 ) T, P = nt (6.4) (6.1) d [(nm + 32 ] )a
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8- My + Cy + Ky = f () t 8. C f () t ( t) = Ψq( t) () t = Ψq () t () t = Ψq () t = ( q q ) ; = [ ] y y y q Ψ φ φ φ = ( ϕ, ϕ, ϕ,3 ) 8. ψ Ψ MΨq + Ψ CΨq + Ψ KΨq = Ψ f ( t) Ψ MΨ = I; Ψ CΨ = C; Ψ KΨ = Λ; q
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x E E E e i ω t + ikx k λ λ π k π/λ k ω/v v n v c/n k nω c c ω/π λ k πn/λ π/(λ/n) κ n n κ N n iκ k Nω c iωt + inωx c iωt + i( n+ iκ ) ωx c κω x c iω ( t nx c) E E e E e E e e κ e ωκx/c e iω(t nx/c) I I
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.1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {
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1 2 (a 1, a 2, a n ) (b 1, b 2, b n ) A (1.1) A = a 1 b 1 + a 2 b 2 + + a n b n (1.1) n A = a i b i (1.2) i=1 n i 1 n i=1 a i b i n i=1 A = a i b i (1.3) (1.3) (1.3) (1.1) (ummation convention) a 11 x
O x y z O ( O ) O (O ) 3 x y z O O x v t = t = 0 ( 1 ) O t = 0 c t r = ct P (x, y, z) r 2 = x 2 + y 2 + z 2 (t, x, y, z) (ct) 2 x 2 y 2 z 2 = 0
9 O y O ( O ) O (O ) 3 y O O v t = t = 0 ( ) O t = 0 t r = t P (, y, ) r = + y + (t,, y, ) (t) y = 0 () ( )O O t (t ) y = 0 () (t) y = (t ) y = 0 (3) O O v O O v O O O y y O O v P(, y,, t) t (, y,, t )
2 1 1 α = a + bi(a, b R) α (conjugate) α = a bi α (absolute value) α = a 2 + b 2 α (norm) N(α) = a 2 + b 2 = αα = α 2 α (spure) (trace) 1 1. a R aα =
1 1 α = a + bi(a, b R) α (conjugate) α = a bi α (absolute value) α = a + b α (norm) N(α) = a + b = αα = α α (spure) (trace) 1 1. a R aα = aα. α = α 3. α + β = α + β 4. αβ = αβ 5. β 0 6. α = α ( ) α = α
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- k k k = y. k = ky. y du dx = ε ux ( ) ux ( ) = ax+ b x u() = ; u( ) = AE u() = b= u () = a= ; a= d x du ε x = = = dx dx N = σ da = E ε da = EA ε A x A x x - σ x σ x = Eε x N = EAε x = EA = N = EA k =
[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s
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chap9.dvi
9 AR (i) (ii) MA (iii) (iv) (v) 9.1 2 1 AR 1 9.1.1 S S y j = (α i + β i j) D ij + η j, η j = ρ S η j S + ε j (j =1,,T) (1) i=1 {ε j } i.i.d(,σ 2 ) η j (j ) D ij j i S 1 S =1 D ij =1 S>1 S =4 (1) y j =
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18 1 12 2016 キトラ古墳天文図の観測年代と観測地の推定 2015 5 15 2015 10 7 Estimating the Year and Place of Observations for the Celestial Map in the Kitora Tumulus Mitsuru SÔMA Abstract Kitora Tumulus, located in Asuka, Nara
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338 7 7.3 LCR 2.4.3 e ix LC AM 7.3.1 7.3.1.1 m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x k > 0 k 5.3.1.1 x = xt 7.3 339 m 2 x t 2 = k x 2 x t 2 = ω 2 0 x ω0 = k m ω 0 1.4.4.3 2 +α 14.9.3.1 5.3.2.1 2 x
D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j
![D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j](/thumbs/103/158001152.jpg "D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j")
6 6.. [, b] [, d] ij P ij ξ ij, η ij f Sf,, {P ij } Sf,, {P ij } k m i j m fξ ij, η ij i i j j i j i m i j k i i j j m i i j j k i i j j kb d {P ij } lim Sf,, {P ij} kb d f, k [, b] [, d] f, d kb d 6..
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1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
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1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2
1. ( ) 1.1 t + t [m]{ü(t + t)} + [c]{ u(t + t)} + [k]{u(t + t)} = {f(t + t)} (1) m ü f c u k u 1.2 Newmark β (1) (2) ( [m] + t ) 2 [c] + β( t)2
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基礎数学I
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Microsoft Word - 11問題表紙（選択）.docx
A B A.70g/cm 3 B.74g/cm 3 B C 70at% %A C B at% 80at% %B 350 C γ δ y=00 x-y ρ l S ρ C p k C p ρ C p T ρ l t l S S ξ S t = ( k T ) ξ ( ) S = ( k T) ( ) t y ξ S ξ / t S v T T / t = v T / y 00 x v S dy dx
タイトル
Flud Flow Smulton wth Cellulr Automt 00N2100008J 2002 225 1. cellulr utomton n t+ 1 t t = f( r, L, + r (1 t t+ 1 f t r t +1 Prllel Vrtul Mchne Messge-Pssng Interfce 1 2. 2. 1 t t 0 t = 1 = 1 = t = 2 2