東京大学 工学系研究科 機械工学専攻 機械工学第1部 2016年8月実施 問題 2 Author 祭音Myyura (Based on SN's answer refined with GPT 6 Astra)
Description 水槽の液面から深さ H H H L L L d d d d ≪ H , L d\ll H,L d ≪ H , L ρ \rho ρ g g g M M M
(1) 流出を止めるのに必要な最小押付け力 F 0 F_0 F 0
次に物体を出口から離す。直径 d d d V J V_J V J
(2) V J V_J V J F 1 F_1 F 1 P max P_{\max} P m a x F 0 F_0 F 0
管入口損失係数を ζ \zeta ζ λ \lambda λ Δ P / ρ = λ ( L / d ) u ˉ 2 / 2 \Delta P/\rho=\lambda(L/d)\bar u^2/2 Δ P / ρ = λ ( L / d ) u ˉ 2 /2
(6) 層流速度分布 u ( r ) = α [ ( d / 2 ) 2 − r 2 ] u(r)=\alpha[(d/2)^2-r^2] u ( r ) = α [( d /2 ) 2 − r 2 ] u ˉ \bar u u ˉ λ 0 \lambda_0 λ 0 μ \mu μ
(8) ζ = 0.5 , λ = 0.02 , L = 2.5 m , d = 0.05 m \zeta=0.5,\lambda=0.02,L=2.5\,\mathrm m,d=0.05\,\mathrm m ζ = 0.5 , λ = 0.02 , L = 2.5 m , d = 0.05 m F 2 F_2 F 2 F 1 F_1 F 1
以後は出口速度 V J V_J V J t = 0 t=0 t = 0 v ( t ) v(t) v ( t )
(10) 物体を曲面の羽根に変更し、相対噴流が速さを保ったまま + x +x + x θ \theta θ F 3 F_3 F 3
题目描述
恒定水头 H H H d d d L L L M M M θ \theta θ
Kai A = π d 2 / 4 A=\pi d^2/4 A = π d 2 /4
(1)–(4)
静水圧から F 0 = ρ g H A \boxed{F_0=\rho gHA} F 0 = ρ g H A V J = 2 g H \boxed{V_J=\sqrt{2gH}} V J = 2 g H
静止物体で噴流の x x x
F 1 = ρ A V J 2 = 2 ρ g H A , P max = 1 2 ρ V J 2 = ρ g H . \boxed{F_1=\rho AV_J^2=2\rho gHA},\qquad
\boxed{P_{\max}=\frac12\rho V_J^2=\rho gH}. F 1 = ρ A V J 2 = 2 ρ g H A , P m a x = 2 1 ρ V J 2 = ρ g H . (5)
物体速度を v v v ρ A ( V J − v ) \rho A(V_J-v) ρ A ( V J − v ) V J − v V_J-v V J − v
ρ A ( V J − v ) 2 = F 0 , v = ( 2 − 1 ) g H . \rho A(V_J-v)^2=F_0,\qquad
\boxed{v=(\sqrt2-1)\sqrt{gH}}. ρ A ( V J − v ) 2 = F 0 , v = ( 2 − 1 ) g H . (6)
u ˉ = 2 π A ∫ 0 d / 2 α ( d 2 4 − r 2 ) r d r = α d 2 8 . \boxed{\bar u=\frac{2\pi}{A}\int_0^{d/2}\alpha\left(\frac{d^2}{4}-r^2\right)r\,dr
=\frac{\alpha d^2}{8}}. u ˉ = A 2 π ∫ 0 d /2 α ( 4 d 2 − r 2 ) r d r = 8 α d 2 . (7)
τ w = μ α d \tau_w=\mu\alpha d τ w = μα d Δ P A = τ w π d L \Delta P\,A=\tau_w\pi dL Δ P A = τ w π d L Δ P = 4 μ α L \Delta P=4\mu\alpha L Δ P = 4 μαL
λ 0 = 512 μ ρ α d 3 = 64 R e d , R e d = ρ u ˉ d μ . \boxed{\lambda_0=\frac{512\mu}{\rho\alpha d^3}
=\frac{64}{\mathrm{Re}_d}},\qquad \mathrm{Re}_d=\frac{\rho\bar u d}{\mu}. λ 0 = ρ α d 3 512 μ = Re d 64 , Re d = μ ρ u ˉ d . (8)
損失を含むエネルギー式は g H = ( 1 + ζ + λ L / d ) V J 2 / 2 gH=(1+\zeta+\lambda L/d)V_J^2/2 g H = ( 1 + ζ + λ L / d ) V J 2 /2
F 2 F 1 = 1 1 + ζ + λ L / d = 1 2.5 = 0.4 . \boxed{\frac{F_2}{F_1}=\frac1{1+\zeta+\lambda L/d}=\frac1{2.5}=0.4}. F 1 F 2 = 1 + ζ + λ L / d 1 = 2.5 1 = 0.4 . (9)
M d v d t = ρ A ( V J − v ) 2 , v ( 0 ) = 0. M\frac{dv}{dt}=\rho A(V_J-v)^2,\qquad v(0)=0. M d t d v = ρ A ( V J − v ) 2 , v ( 0 ) = 0. 変数分離して積分すると、
v ( t ) = V J ( 1 − 1 1 + ρ A V J t / M ) . \boxed{v(t)=V_J\left(1-\frac1{1+\rho AV_Jt/M}\right)}. v ( t ) = V J ( 1 − 1 + ρ A V J t / M 1 ) . (10)
相対流出速度の x x x ( V J − v ) cos θ (V_J-v)\cos\theta ( V J − v ) cos θ
F 3 = ρ A ( V J − v ) 2 ( 1 − cos θ ) , F_3=\rho A(V_J-v)^2(1-\cos\theta), F 3 = ρ A ( V J − v ) 2 ( 1 − cos θ ) , v = V J − F 3 ρ A ( 1 − cos θ ) = V J − 2 d F 3 π ρ ( 1 − cos θ ) . \boxed{v=V_J-\sqrt{\frac{F_3}{\rho A(1-\cos\theta)}}
=V_J-\frac2d\sqrt{\frac{F_3}{\pi\rho(1-\cos\theta)}}}. v = V J − ρ A ( 1 − cos θ ) F 3 = V J − d 2 π ρ ( 1 − cos θ ) F 3 . Reference 