Estimate the discharge of kerosene (sp gravity=0.8) through the given venturimeter shown in Fig. 5. specific gravity of mercury(Hg) is 13.55.
Solution:
Applying Bernoulli’s equation to section 1 and 2
(p1/γ) + (v1 2 /2g) = (p2/γ) + (v2 2 /2g) + ∆z
or (p1/γ) – ((p2/γ) + ∆z) = (v2 2 /2g) - (v1 2 /2g)
Hence ((p1/γ)- (p2/γ) + ∆z) = 15 v1 2 /2g
Equating the pressures at section AA in the two limbs of the manometer
(p1/γ) + (x + 0.3) = ((p2/γ) + 0.4) + x + (0.3 *(13.85/0.8))
(p1/γ) - (p2/γ) + 0.4) = 5.19 – 0.30 = 4.78 m
15 v1 2 /2g = 5.29 or v1 = 2.63 m/s
Hence Q = 0.785*0.01*2.5
= 0.0196 m3 /s = 19.6 l/s
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