用伯努力定理
测压管液位差为
∆h=r₂-r₁+h₂-h₁
A,B截面压头损失为
∆z=70mmH₂O=0.07mH₂O
对A截面和B截面用伯努力方程,可得
p₁+1/2*ρv₁²=p₂+1/2*ρv₂²+ρg∆z ①
∵p₁=ρg(r₁+h₁),
p₂=ρg(r₂+h₂)
∴p₂-p₁=ρg∆h ②
又Q₁=πd₁²/4*v₁
Q₂=π₂d₂²/4*v₂
流量守恒Q₁=Q₂
∴v₂=d₁²/d₂²*v₁ ③
将②③代入①可得
ρg∆h=1/2*ρ(v₁²-v₂²)-ρg∆z
∴∆h=1/(2g)*(v₁²-v₂²)-∆z
=1/(2g)*v₁²*[1-(d₁/d₂)⁴]-∆z
=1/(2*9.8)*2.5²*[1-(33/47)⁴]-0.07
=0.241-0 07
=0.171(m)
=171(mm)
即B管液面比A管高,相差171mm