Water flows through a horizontal lobe as shown in figure , If the difference of heights of water column in the vertical lobes is $2 c m$ and the areas of cross-section at A and B are $4 c m^{2}$ and $2 c m^{2}$ respectively. Find the rate of flow of water across any section.

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Solution

$\frac{1}{2} ρ v_{B}^{2} - \frac{1}{2} ρ v_{A}^{2} = ρ g (h_{A} - h_{B})$
or $v_{B}^{2} - v_{A}^{2} = 2 g (h_{A} - h_{B})$
$= 2 \times 10 \times 0.02$
or $v_{B}^{2} - v_{A}^{2} = 0.4$ m $^{2}$ /s $^{2}$ (i)
$v_{A} A_{A} = v_{B} A_{B}$
or $4 v_{A} = 4 v_{B}$
$∴$ $v_{B} = 2 v_{A}$
Solving Eqs. (i) and (ii), we get
$v_{A} = 0.363$ m/s
Volume flow rate $= v_{A} A_{A}$
$= (0.365) (4 \times 10^{- 4})$
$= 1.46 \times 10^{- 4}$ m $^{3}$ /s
$= 146$ cm $^{3}$ /s

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Water flows through a horizontal lobe as shown in figure , If the difference of heights of water column in the vertical lobes is 2cm and the areas of cross-section at A and B are 4cm2 and 2cm2 respectively. Find the rate of flow of water across any section.

12ρv2B−12ρv2A=ρg(hA−hB)or v2B−v2A=2g(hA−hB) =2×10×0.02or v2B−v2A=0.4 m2/s2 (i)vAAA=vBABor 4vA=4vB∴ vB=2vASolving Eqs. (i) and (ii), we getvA=0.363 m/sVolume flow rate=vAAA =(0.365)(4×10−4) =1.46×10−4 m3/s =146 cm3/s

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