Question

怎么算？求解答过程

Answer 1

用伯努力方程
h₁+p₁/(ρg)+u₁²/(2g)= h₂+p₂/(ρg)+u₂²/(2g)+∑hf
(1) 高位槽水位不变，有
h₁=8, p₁=p0, u₁=0
h₂=2, p₂=p0, u₂=u, ∑hf=6.5u²
∴8=2+u²/(2g)+6.5u²
解得u=√[12g/(1+13g)]
=0.916(m/s)
流量Q=u*πd²/4
=0.916*3.14*0.1²/4
=0.0072(m³/s)
≈26(m³/h)
(2) 高位槽水位下降，有
h₁=h, p₁=p0, u₁=u₁
h₂=2, p₂=p0, u₂=u,
∑hf=6.5u²
∴h+u₁²/(2g)=2+u²/(2g)+6.5u²
由u₁S₁=u₂S₂得
u₁=S₂/S₁*u₂=0.01u
u₁²=0.0001u²≈0
解得u=√[2g(h-2)/(1+13g)]
由流量守恒有
-S₁dh=uS₂dt=S₂√[2g(h-2)/(1+13g)]*dt
积分可得
-S₁/S₂*√[(1+13g)/(2g)]*∫dh/√(h-2)=∫dt
-S₁/S₂*√[(1+13g)/g]*√(h-2)=t+C
t=0时，h=8
∴C=-S₁/S₂*√[(1+13g)/g]*√6
∴t=S₁/S₂*√[(1+13g)/g]*[√6-√(h-2)]
液面下降3m时，h=5
∴t=S₁/S₂*√[(1+13g)/g]*[√6-√(5-2)]
=1²/0.1²*√[(1+13*9.8)/9.8]/*[√6-√3]
=100*3.62*1.732*0.414
≈260(s)
=4.3(min)